Institutio mathematica, or, A mathematical institution shewing the construction and use of the naturall and artificiall sines, tangents, and secants in decimal numbers, and also of the table of logarithms in the general solution of any triangle, whether plain or spherical, with their more particular application in astronomie, dialling, and navigation / by John Newton. Newton, John, 1622-1678. This text is an enriched version of the TCP digital transcription A52264 of text R20441 in the English Short Title Catalog (Wing N1061). Textual changes and metadata enrichments aim at making the text more computationally tractable, easier to read, and suitable for network-based collaborative curation by amateur and professional end users from many walks of life. The text has been tokenized and linguistically annotated with MorphAdorner. The annotation includes standard spellings that support the display of a text in a standardized format that preserves archaic forms ('loveth', 'seekest'). Textual changes aim at restoring the text the author or stationer meant to publish. This text has not been fully proofread Approx. 558 KB of XML-encoded text transcribed from 240 1-bit group-IV TIFF page images. EarlyPrint Project Evanston,IL, Notre Dame, IN, St. Louis, MO 2018 A52264 Wing N1061 ESTC R20441 12355076 ocm 12355076 60119 2240936251

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Early English books online. (EEBO-TCP ; phase 1, no. A52264) Transcribed from: (Early English Books Online ; image set 60119) Images scanned from microfilm: (Early English books, 1641-1700 ; 217:14) Institutio mathematica, or, A mathematical institution shewing the construction and use of the naturall and artificiall sines, tangents, and secants in decimal numbers, and also of the table of logarithms in the general solution of any triangle, whether plain or spherical, with their more particular application in astronomie, dialling, and navigation / by John Newton. Newton, John, 1622-1678. [9], 420 p., [4] leaves of plates (2 folded) : ill. Printed by R. & W. Leybourn, for George Hurlock ... and Robert Boydel ..., London : 1654. First ed. Cf. NUC pre-1956. "The second part" has special t.p. Errata on p. [9]. Reproduction of original in British Library. eng Geometry -- Early works to 1800. Trigonometry -- Early works to 1800. Logarithms. Mathematics -- Problems, exercises, etc. eebo Institutio mathematica. Or, A mathematical institution. Shewing the construction and use of the naturall and artificiall sines, tangents, and secants, Newton, John, 1654 417 79130 157 0 18 0 0 0 22.12 C The rate of 22.12 defects per 10,000 words puts this text in the C category of texts with between 10 and 35 defects per 10,000 words. 2005-05 Assigned for keying and markup 2005-05 Keyed and coded from ProQuest page images 2005-06 Sampled and proofread 2005-06 Text and markup reviewed and edited 2005-10 Batch review (QC) and XML conversion

Institutio Mathematica . OR , A MATHEMATICAL Institution . Shewing the Construction and Use of the Naturall and Artificiall Sines , Tangents , and Secants , in Decimal Numbers , and also of the Table of Logarithms . In the general solution of any Triangle whether Plain or Spherical . WITH Their more particular application in ASTRONOMIE , DIALLING , and NAVIGATION . By JOHN NEWTON .

LONDON , Printed by R. & W. Leybourn , for George Hurlock , at Magnus Church corner , and Robert Boydel in the Bulwark neer the Tower : MDCLIV .

TO THE COVRTEOVS Reader .

ALthough Mathematicall studies have for these many years been much neglected , if not contemned , yet have there been so many rare inventions found , even by men of our own Nation , that nothing now seems almost possible to be added more : as in other studies so may we say in these , nil dictum quod non dictum prius . We at the least must needs acknowledge that in this we have presented thee with nothing new , nothing that is our own . Ex integrâ Graecâ integram Comoediam , hodie sum acturus , Heautonti morum enon , saith Terence , that famous Comoedian : translation was his apologie , transcription and collection ours : this only we have endeavoured , that the first principles and foundations of these studies ( which until now were not to be known , but by being acquainted with many Books ) might in a due method and a perspicuous manner , be as it were at once , presented to thy view , and serve as a perfect INSTITUTION MATHEMATICAL , unto such as have as yet learned nothing but Arithmetick .

To that purpose we have first laid down such propositions Geometricall , out of Euclide , Pitiscus , and others , as must be known to such as would understand the nature and mensuration of all Triangles . Next we have proceeded to the affections of Triangles in the generall , and thence to the composition of the Sines , Tangents , and Secants Naturall , in which we have for the most part followed the Rules prescribed by Pitiscus , in some things we have taken the direction of Snellius , and in the trisection and quinquisection of an angle we have proceeded Algebraically , with those two famous Mathematicians of our age and Nation , Brigges , and Oughtred ; and because the Algebraicall work is of it selfe abstruse and intricate , to those that are not acquainted with it , we have insisted the more upon it , and by our explanation we have endeavoured to make it plain and easie ; and that nothing may be wanting , which either former ages or our own ( by Gods blessing and their industry ) have afforded to us , we have to the composition of the Natural Canon , added out of Briggs and Wintage the construction of the Logarithms of any numbers , and consequently how to make the Logarithms of the Naturall Sines , Tangents , and Secants . This done , the proportions in the usuall Cases of all Triangles both Plain and Sphericall , we have first cleared by Demonstration out of Pitiscus , Gelibrand , Norwood , and others ; and then explained the manner of the work in Natural and Artificial Numbers both , and so conclude the first Part of our Institution .

And in the second Part we have made our application of all the former unto Astronomie first , and then to Dialling and Navigation . In our application to Astronomie , we have furnished you with a Table of the Suns Motion , whereby to calculate his place in the Zodiac in Decimal numbers , and without which most of the other Problemes would be found ( if not useless ) yet very intricate and obscure , that being , for the most part , one of the three terms supposed to be given in Astronomical computations .

In the Chapter of Dialling , you have the Sphears projection , according to the directions of Wels , in his Art of Shadows , and how to draw the houre-lines of all the severall Dials which he hath contrived from thence , we have briefly shewed ; and in the finding of all the arches in these Cases necessary , we have kept our selves to our own CANON , which doth exhibit the degrees of the Quadrant in Centecimal parts or minutes .

In the Chapter of Navigation , you have first the division of the Sea-mans Compasse , next the description and making of the Sea-Chart , as Edward Wright our worthy Countrey-man hath given us the Demonstration thereof in his Book entituled , The correction of Errours in Navigation : to these we have added such other Problemes as are now amongst our Sea-men of most frequent use ; annexing thereunto a Table of meridionall parts , and other Tables usefull as well in Dialling as in Navigation , and all these in Decimall numbers , it being indeed our aim ( as much as in us lieth ) not only to promote these studies by this our Compendium of the first rudiments of Mathematical learning , as in relation to the matter therein to be considered , but by such expeditious and advantagious wayes of working also , as have been lately found , or former ages have commended to us ; amongst which there is none more excellent then that which is performed by Decimal numbers ; fully to explicate the manner and worth whereof were matter enough for a whole Treatise , and therefore not to be expected in a short Epistle : It would indeed be very impertinent to intermeddle any further with it here , then in our Institution it selfe is already explained , in which thou maist perceive Addition and Subtraction of Degrees , Minutes , and Seconds , to be performed as in Vulgar numbers , without any Reduction to their severall Denominations , Multiplication is performed by the addition of Cyphers , and Division by the cutting off of Figures . Others that have either spent more time , or made a farther progresse in these ravishing Studies , might ( if they would have taken the Pains ) have haply presented thee with more , and in a lesser room : The most of this was at the first collected for our private use , and now published for the good of others .

John Newton .
ERRATA .

Page Line 6 25 Read , or termes . 17 For B. 9 A , r. B. 9 C. 16 20 read , 3 times 6 is 18. 22 12 r. one third 22 25 r. one third 30 9 r. triangles in the following figure . 32 21 r. by the 18 th . 34 21 r. are equiangled . 38 24 r. the arch CGK . 44 13 r. 21 of the first . 47 5 r. of 120 degrees . 49 4 r. by the 18. 55 10 r. 18 of the first . 60 9 r. by 18. 69 20 r. 77 16 r. 147. 77 21 r. 3913. 77 22 r. 206087. 91 3 r. 24986. 108 1 r. KP . 112 26 r. BD 156 6 r. BD. 7 r. DF. 8 r. BF

A MATHEMATICAL Institution .
CHAP. I. Geometricall Definitions .

OF things Mathematicall there are two principall kindes , Number and Magnitude ; and each of these hath his proper Science . The Science of Number is Arithmetick , and the Science of Magnitude is commonly called Geometry , but may more properly be termed Megethelogia , as comprehending all Magnitudes whatsoever , whereas Geometry , by the very Etymologie of the word , doth seeme to confine this Science to Land-measuring only .

Of this Megethelogia Geometry , or Science of Magnitudes , we will set down such grounds and principles as are necessary to be known , for the better understanding of that which follows , presuming that the reader hereof hath already gotten some competent knowledge in Arithmetick .

Concerning then this Science of Magnitude , two things are to be considered : First , the severall heads to which all Magnitudes may be referred : And then secondly , the terms and limits of those Magnitudes .

All Magnitudes are either Lines , Plains , or Solids , and do participate of Length , Breadth , or Thicknesse .

1. A Line is a supposed length , or a thing extending it self in length , without breadth or thickness , whether it be a right line or a crooked ; and may be divided into parts in respect of his length , but admitteth no other division : as the line AB .

2. The ends or limits of a line are points , as having his beginning from a point , and ending in a point , and therefore a Point hath neither part nor quantity , it is only the term or end of quantity , as the points A and B are the ends of the aforesaid line AB , and no parts thereof .

3. A Plain or Superficies is the second kind of magnitude , to which belongeth two dimensions , length , and breadth , but not thickness .

4. As the ends , limits or bounds of a line are points confining the line , so are lines the limits , bounds and ends inclosing a Superficies ; as in the figure you may see the plain or Superficies here inclosed with four lines , which are the extreams or limits thereof .

5. A Body or Solid is the third kinde of magnitude , and hath three dimensions belonging to it , length , breadth , and thickness . And as a point is the limit or term of a line , and a line the limit or term of a Superficies , so likewise a Superficies is the end or limit of a Body or Solid , and representeth to the eye the shape or figure thereof .

6. A Figure is that which is contained under one or many limits , Under one bound or limit is comprehended a Circle , and all other figures under many .

7. A Circle is a plain figure contained under one round line , which is called a circumference , as in the Figure following , the Ring CBDE is called the circumference of that Circle .

8. The Center of a Circle is that point which is in the midst thereof , from which point , all right lines drawn to the circumference are equal the one to the other ; as in the following figure , the lines AB , AC , AD , and AE , are equal .

9. The Diameter of a Circle , is a right line drawn through the center thereof , and ending at the circumference on either side , dividing the Circle into two equal parts , as the lines CAD and BAE , are either of them the diameter of the Circle BCDE , because that either of them doth passe through the center A , and divideth the whole Circle into two equal parts .

10. The Semidiameter of a Circle is half the Diameter , and is contained betwixt the center and one side of the Circle ; as the lines AB , AC , AD , and AE , are either of them the Semidiameters of the Circle CBDE .

11. A Semicircle is the one halfe of a Circle drawn upon his Diameter , and is contained by the half circumference and the Diameter ; as the Semicircle CBD is halfe the Circle CBDE , and contained above the Diameter CAD .

12. A Quadrant is the fourth part of a Circle , and is contained betwixt the Semidiameter of the Circle , and a line drawn perpendicular unto the Diameter of the same Circle , from the center thereof , dividing the Semicircle into two equal parts , of the which parts the one is the Quadrant or fourth part of the same Circle . Thus , the Diameter of the Circle BDEC is the line CAD , dividing the Circle into two equal parts , then from the center A raise the perpendicular AB , dividing the Semicircle likewise into two equal parts , so is ABD , or ABC , the Quadrant or fourth part of the Circle .

13. A Segment or portion of a Circle is a figure contained under a right line and a part of the circumference of a Circle , either greater or lesser than the Semicircle ; as in the former figure , FBGH is a segment or part of the Circle CBDE , contained under the right line FHG lesse than the Diameter CAD .

14. By the application of several lines ●terms of a Superficies one to another , are made Parallels , Angles , and many sided Figures .

15. A Parallel line is a line drawn by the side of another line , in such sort that they may be equidistant in all places , and of such there are two sorts , the right lined parallel , and the circular parallel .

Right lined Parallels are two right lines equidistant in all places one from the other , which being drawn to an infinite length would never meet or concur ; as may be seen by these two lines , AB and CD .

A Circular Parallel is a Circle drawn within or without another Circle , upon the same center , as you may plainly see by the two Circles BCDE , and FGHI , these Circles are both of them drawn upon the same center A , and therefore are parallel one to the other .

16. An Angle is the meeting of two lines in any sort , so as they both make not one line ; as the two lines AB and AC incline the one to the other , and touch one another in the point A , in which point is made the angle BAC . And if the lines which contain the angle be right lines , then it is called a right lined angle ; as the angle BAC . A crooked lined angle is that which is contained of crooked lines ; as the angle DEF : and a mixt angle is that which is contained both of a right and crooked line ; as the angle GHI : where note that an angle is ( for the most part ) described by three letters , of which the second or middle letter representeth the angular point ; as in the angle BAC , A representeth the angular point .

17. All Angles are either Right , Acute , or Obtuse .

18. When a right line standeth upon a right line , making the angles on either side equal , either of those angles is a right angle , and the right line which standeth erected , is a perpendicular line to that upon which it standeth . As the line AB ( in the following figure ) falling upon the line CBD perpendicularly , doth make the angles on both sides equal , that is , the angle ABC is equal to the angle ABD , and either of those angles is therefore a right angle .

19. An acute angle is that which is lesse than a right angle ; as the angle ABE is an acute angle , because it is less than the right angle ABD , in the former figure .

20. An Obtuse Angle is that which is greater than a right angle ; CBE in the former figure is greater than the angle ABC by the angle ABE , and therefore it is an obtuse angle .

21. The measure of every angle is the arch of a Circle described on the angular point , as in the following figure , the arch CD is the measure of the right angle CED . The arch BC is the measure of the acute angle BEC . And the arch BCD is the measure of the obtuse angle BED . But of their measure there can be no certain knowledge , unlesse the quantity of those arches be exprest in numbers .

22. Every Circle therefore is supposed to be divided into 360 equall parts , called Degrees , and every Degree into 60 Minutes , every Minute into 60 Seconds , and so forward . This division of the Circle into 360 parts we shall retain , but every Degree we will suppose to be divided into 100 parts or Minutes , & every Minute into 100 Seconds : and thus all Calculations will be much easier , and no lesse certain .

23. A Semicircle is the halfe of a whole circle containing 180 degrees . A Quadrant or fourth part of a circle is 90 degrees . And thus the measure of the right angle CED is the arch CD 90 degrees . The measure of the acute angle BEC is the arch BC 30 degrees . And the measure of the obtuse angle BED is the arch BD 120 degrees .

24. The complement of an angle to a Quadrant is so much as the angle wanteth of 90 degrees , as the complement of the angle AEB 60 degrees is the angle BEC 30 degrees ; for 30 and 60 do make a Quadrant or 90 degrees .

25. The complement of an angle to a Semicircle is so much as the said angle wanteth of 180 degrees , as the complement of the angle BED 120 degrees , is the angle AEB , 60 degrees ; for 60 and 120 do make 180 degrees .

26. Many sided figures are such as are made of three , four , or more lines , though for distinction sake , those only are so called which are contained under five lines or terms at the least .

27. Four sided figures are such as are contained under four lines or terms , and are of divers sorts . 1. There is the Quadrat or Square whose sides are equall and his angles right . 2. The Long Square whose angles are right , but the sides unequal . 3. The Rhombus or Diamond , having equall sides but not equal angles . 4. The Rhomboides , having neither equal sides nor equal angles , and yet the opposite sides and angles equal .

All other figures of four sides are called Trapezia or Tables . The dimension whereof as also of all figures whatsoever , dependeth upon the knowledge of three sided figures , or Triangles , of which in the Chapter following .

CHAP. II. Of the nature and quality of Triangles .

1. A Triangle is a figure consisting of three sides and three angles .

2. Every of the two sides of any Triangle are the sides of the angle comprehended by them , the third side is the Base , as in the figure following , the sides AC and BC and sides of the angle BCA , and AB is the Base of the said angle .

3. Every side is said to subtend the angle that is opposite to that side ; as the side AB subtendeth the angle ACB , the side AC subtendeth the angle ABC , and the side BC subtendeth the angle BAC : the greater sides subtend the greater angles , the lesser sides lesser angles , and equal sides equal angles .

4. Of Triangles there are diverse sorts ; as ,

1. There are Equilateral Triangles , having three equal sides .

2. There is an Isoscheles , which is a Triangle that hath two equal sides .

3. Scalenum , which is a Triangle whose sides are all unequal .

4. An Orthigonium , or a right angled Triangle , having one right angle .

5. An Ambligonium , or an obtuse angled Triangle , having in it one obtuse angle .

6. An Oxigonium , or an acute angled Triangle , having all his angles acute .

7. All these Triangles are either Plain or Spherical .

8. The sides of Plain Triangles in Trigonometria are right lines only , concerning which we have added these Theorems following .

9. Theorem . If one right line cut through two parallel right lines , then are the angles opposite one against another equal .

In the following Scheme the two lines WX and YZ are parallel , and therefore the angles XIC , and ICY are equal .

Demonstration .

The two angles XIC and WIC are equal to two right angles , as also ICY and ICZ , because on the parallel lines at the points I and C there may be drawn two Semicircles , each of which are the measures of two right angles . If then the angle XIC be lesse than ICY , the angle WIC must as much exceed the angle ICZ , and the angles XIC and ZCI would be lesse than two right angles , and consequently the lines WX and YZ may be extended on the side X and Z til at length they shall concur together , and then the lines WX and YZ are not parallels , as is supposed here , and therefore the angles XIC and ICY are equal .

10. Theor , If four right lines be proportionall , the right angled figure made of the two means , is equal to the right angled figure made of the two extreams .

Let the four proportional lines be AB two foot , EF three foot , FG six foot , and BC nine foot : I say then that the right angled figure made of the two means EF and FG , that is , the right angled figure EF GH , is equal to the right angled figure made of the extreams AB and BC , that is , to the right angled figure ABCD ; for as twice 9 is 18 , so likewise three times is 18.

11. Theor. If three right lines be proportionall , the Square made of the mean is equal to the rightangled figure made of the extreams .

Demonstration .

The Demonstration of this proposition is all one in effect with the former , the difference is , that here is spoken of three lines , there of four , and therefore if we take the mean twice , of which the square is made , the work will be the same with that in the former proposition . As if the length of the first line were two foot , the second four , and the third eight ; it is evident , that as four times four is 16 , so two times eight is 16 , and therefore what hath been said of four proportionals , is to be understood of three proportionals also .

12. Theor. If a right line being divided into two equal parts , shall be continued at pleasure , then is the right angled figure made of the line continued , and the line of continuation , with the square of one of the bi-segments , equal to a square made of one of the bi-segments and the line of continuation .

The line PQ is divided into two equal parts , the midst is C , to the same is added a right line , as QN ; and of the whole line PQ , and the added line QN is made PN as one line , and of this line PN , and the added line QN is inclosed the right angled figure ●M , and upon the halfe line CQ and the line of continuation QN is made the square CF. Now if you draw the line QG parallel to NF , and equal to the same , then is the right angled figure ●M with the square of CQ that is , the square IG , equal to the square of CN , that is , the square CF.

Demonstration .

Forasmuch as CQ is equal unto MF , the which is also equal unto IO or I L , it followeth that IG is a Square , which with the right angled figure PM is equal to the square CF , because the right angled figure GM is equal to CO , which is also equal to PI.

13. Theor. To divide a right line in two parts , so that the right , angled figure made of the whole line and one part shall be equal to the square of the other part .

The right line given is AB , upon the same line AB , make a square , as ABCD ; and divide the side AD in two equal parts , the midst is M , from M draw a line to B , and produce AD to H , so that MH be equal to MB ; and upon AH make a square , as AHGF. Then extend GF to E , and then is the right angled figure FC , being made of the whole line FE ( which is equal to AB ) and the part BF , equall to the square of the other part AF , that is , to the square AHGF.

Forasmuch as by the last aforegoing , the right angled figure comprehended of HD and HA , or the right angled figure of HD and HG , as the figure GHED , with the square of AM , are together equal to the square of HM , being equall to BM : it followeth , that if we take away the square ●f AM , common to both , that the square ●f AB , that is , the square ABCD is equal ●o the right angled figure HGED , and the ●ommon right angled figure AE being taken from them both , there shall remain the right angled figure FC , equal to the square ●HFG , which was to be proved .

14 Theor. To divide a right line given by extream and mean proportion .

A right line is said to be divided by an extream and mean proportion , when the whole is to the greater part , as the greater is to the lesse . And thus a right line being divided , as the right line AB is divided in the preceding Diagram in the point F , it is divided in extream and mean proportion ; that is , As AB , is to AF : so is AF , to BF .

Demonstration .

Forasmuch as the right lined figure included with AB and FB , as the figure FBCE is equal to the square of AF , that is , to the square AFGH ; it followeth , by the eleventh Theorem of this Chapter , that the line AB is divided in extream and mean proportion ; that is , As AB , is to AF : So is AF , to FB .

15 Theor. In all plain Triangles , a line drawn parallel to any of the sides , cutteth the other two sides proportionally .

As in the plain Triangle ABC , KL being parallel to the base BC , it cutteth off from the side AC one fourth , and also it cutteth off from the side AB one third part : the reason is , because the right line EH cutteth off one third part from the whole space DGFB , & therefore it cutteth off one third part from all the lines that are drawn quite through that space .

And hereupon parallel lines bounded with parallels are equal ; as the parallels ED and GH being bounded with the parallels DG and HE are equal , for since the whole lines DB and GF are equall , DE and GH being one fourth part thereof , must needs be equal also .

16 Theor. Equiangled Triangles have their sides about the equall angles proportionall , and contrarily .

Let ABC and ADE be two plain equiangled Triangles , so as the angles at B and D , at A and A , and also at C and E be equal one to the other ; I say , their sides about the equal angles are proportionall ; that is , 1 As AB , is to BC : So is AD , to ED. 2 As AB , is to AC : So is AD , to AE. 3 As AC , is to CB : So is AE , to ED.

Demonstration .

Because the angles BAC and DAE are equal by the Proposition ; therefore if A + B be applied to AD , AC shall fall in AE ; and by such application is this figure made . In which , because that AB and AD do meet together , and also that the angles at B and D are equall , by the Proposition ; therefore the other sides BC and DE are parallel ; and , by the last aforegoing , BC cutteth the sides AD and AE proportionally : and therefore ,

As AB , to AD : So is AC , to AE.

Moreover , by the point B , let there be drawn the right line BF parallel to the base AE , and it shall cut the other two sides proportionally in the points B and F , and therefore ,

1. As AB to AD : so is EF to ED ,

Or thus .

As AB to AD : so is CB to ED : because that FE and BC are equal , by the last aforegoing .

1. Theor. In all right angled plain Triangles , the sides including the right angle are equal to the the third side .

In the right angled plain triangle ABC , right angled at B , the sides AB and BC are equal in power to the third side AC ; that is the squares of the sides AB and BC , to wit , the squares ALMB and BEDC added together , are equal to the square of the side AC , that is to the square ACKI .

Demonstration .

18. Theor. The three angles of a right lined Triangle are equal to two right angles .

As in the following plain Triangle ABC the three angles ABC , ACB , and CAB are equal to two right angles . Let the side AB be extended to D , and let there be a semicircle drawn upon the point B , and let there be also dawn a line parallel unto AC , from B unto G.

Demonstration .

I say that the angle GBD is equal to the angle BAC , by the 9 th hereof , and the angle CBG is equal to the angle ACB by the same reason , and the angles CBG and GBD , are together equal to the angle CBD , which is also equal to the angle ABC , by the 18 th . of the first : and therefore ; the three angles of a right lined Triangle are equal to two right angles , which was to be proved .

19. Theor. If a plain Triangle be inscribed in a Circle , the angles opposite to the circumference are halfe as much as that part of the Circumference which is opposite to the angles .

As if in the circle ABC the circumference BC be 120 degrees , then the angle BAC which is opposite to that circumference shall be 60 degrees . The reason is , because the whole circle ABC is 360 degrees , and the three angles of a plain triangle cannot exceed 180 degrees , or two right angles , by the last aforegoing , therefore , as every arch is the one third of 360 , so every angle opposite to that arch is the one third of 180 , that is 60 degrees .

Or thus , From the angle ABC , let there be drawn the diameter BED , and from the center E to the circumference , let there be drawn the two Radii or semidiameters AE and AC , I say then that the divided angles ABD and DBC are the one halfe of the angles AED and DEC: for the angles ABE and BAE are equall , because their Radii AE and EB are equall , and also the angle AED is equal to the angles ABE and BAE added together , for if you draw the line EF parallel to AB , the angle FED shall be equal to the angle ABE by the 9 th . hereof ; and by the like reason the angle AEF is also equal to the angle BAE . and therefore the angle AED is equal to the angles ABE , and BAE : or , which is all one , the angle AED is double to the angle ABD .

In like manner , the angles EBC and ECB are equal , and the angle DEC is equal to them both : therefore the angle DEC is double to the angle DBC . Then because the parts of the angle AEC are double to the parts of the angle ABC ; therefore also the whole angle AEC is double to the whose angle ABC ; and thereupon the angle ABC is half the angle AEC ; and consequently , half the arch ADC ; is the measure of the angle ABC , as was to be proved . Hence it followeth ,

1. If the side of a plain Triangele inscribed in a circle be the diameter , the angle opposite to that side is a right angle , that is , 90 degrees ; for that it is opposite to a semicircle , which is 180 degrees .

2. If divers right lined triangles be inscribed in the same segment of a circle upon one base ; the angles , in the circumference are equal . As the two triangles ABD and ACD being inscribed in the same segmeut of the circle ABCD , upon the same base AD are equiangled in the points B and C , falling in the circumference . For the same arch AD is opposite to both those angles ; that is , to the angle ACD , and also to the angle ABD .

20 Theor. If two plaine Triangles inscribed in the same segment of a circle , upon the same base , be so joyned together in the top , ( or in the angles falling in the circumference ) that thereof is made a four-sided figure , intersected with Diagonals , the right angled figure made of the Diagonals , is equall to the right angled figures made of the opposite sides added together .

Let ABD and ACD be two triangles , inscribed in the same segment of the circle ABCD upon the same base AD so joyned in the top by the right line BC , that thereupon is made the four sided figure ABCD. I say , that the right angled figures made of the opposit sides AB and DC , and also of the sides BC and AD added together are equall to the right angled figure made of the Diagonals AC and BD.

Demonstration .

If at the point B you make the angle ABE equall to the angle DBC , and so cut the Diagonall AC into two parts by the right line EB at the point E , then shall the angles ABD and EBC be equall , because the angles ABE and DBC are equal by the proposition , and the angle EBD common to both , and the angles ADB & ECB are equal , because the arch AB is the double measure of them both by the last aforegoing , and therefore the triangles ABD & EBC are equiangled and there sides proportional by the 18 th and 16 th Theoremes of this chapter , that is , as BD to DA , so is BC to CE , and therefore also the rectangles of BD in CE is equal to the rectangle of DA in BC by the 10 th hereof .

And because the angles DBC and ABE are equal by the proposition , and the angles BDC and EAB equall because the arch BC is the double measure of them both by the last aforegoing , the triangles BDC , & EAB are equiangled and there sides proportional by the 18 th and 16 Theoremes of this chapter , that is as BD to DC so is AB to AE ; and therefore also the rectangle of BD in AE is equal to the rectangle of DC in AB .

And because the rectangled figure 〈…〉 of AD and DF is equall to the two rectangled figures of AD in DC and BC in CF , therefore also the rectangled figure of BD in AC is equal to the rectangled figures of BD in AE and BD in EC . From hen●● and the two former proportions the proposition is thus Demonstrated .

1. BD in CE is equal to DA in BC 2. BD in AE DC in AB

And by Composition .

BD in CE more by BD in AE is equal to DA in BC more by DC in AB , now then because BD in AC is equal to BD in 〈◊〉 EC more by BD in EA , therefore also BD in AC is equall to DA in BC more by DC in AB which was to be demonstrated .

21. Theor. If two right lines inscribed in a circle cut each other within the circle , the rectangle under the segments of the one , is equall to the rectangle under the segment of the other .

Let the two lines be FD and BC , intersecting each other in the point A ; I say , the triangles ABF and ADC are like , because of their equal angles AFB and ACD , which are equal , because the arch BD is the double measure of them both , and because of their equal angles BAF and DAC , which are equal by the ninth hereof , and where two are equal , the third is eqaul by the 18 aforegoing ; therefore AD in AF is equal to AC in AB , which was to be proved .

Theor. 22. In a plain right angled Triangle , a perpendicular let fall from the right angle upon the Hypothenuse , divides the triangle into two triangles , both like to the whole , and to one another .

The triangle ABC is right angled at B , the hypotenuse or side subtending the right angle is AC , upon which from the point B is drawn the perpendicular BD which divideth the triangle ABC into two triangles , ADB and BDC , each of them like to the whole triangle ABC , and each like to one another also , that is equiangled one to another .

Demonstration .

In the triangle ABD , the angles ABD and ADB are equal to the angles ACB and ABC , because of their common angle at A , and their right angles at B and D , and in the triangle CDB , the angle . CBD and BDC are equal to the angles ABC and CAB , because of their common angle at C , and their right angles at B and D ; these triangles are therefore each of them like to the whole triangle ABC , and by consequence like to one another .

23. Theor. If two sides of one triangle be equal to two sides of another , & the angle comprehended by the equal sides equal , the third side or base of the one , shall be equal to the base of the other , and the remaining angles of the one equal to the remaining angles of the other .

Of these two triangles CBH and DEF , the sides CB & BH in the one are equal , to DE & EF in the other and the angle CBH equal to the angle DEF , therefore CH in one is equal to DF in the other , for if the base CH be greater then the base CF from CH let be taken .

CG equal to DF and let there be drawne the right line BG , now if BC and BG be equal to DE and EF , yet the angle CBG cannot be equal to the angle DEF by the angle GBH which is contrary to the Proposition , and therefore CH must be equal to DF , and consequently the angle BCH equal to EDF , and CHB equal to DFE which was to be proved .

24. Theor. An Isocles or triangle of two equal sides , hath his angles at the base equal the one to the other , and contrarily .

25. Theor. If the Radius of a circle be divided , in extreame and mean proportion , the greater segment shall be the side of a Decangle , in the same circle .

These foundations being laid we will proceed to the making of the tables , whereby any triangle may be measured .

Chap. III of Trigonometria , or the measuring of all Triangles .

THe dimension of triangles , is performed by the Golden Rule of Arithmetick , which teacheth of four numbers proportional one to another , any three of them being given , to finde out a fourth .

Therefore for the measuring of all triangles there must be certain proportions of all the parts of a triangle one to another , and these proportions must be explained in numbers .

2. And the proportions of all the parts of a triangle one to another cannot be certain unlesse the arches of circles ( by which the angles of all triangles , and of Spherical triangles , also the sides are measured ) be first reduced into right lines , because the proportions of arches one to another , or of an arch to a right line , is not as yet found out .

3. The arches of every circle are after a sort reduced to right lines , by defining the quantity , which the right lines to them applied have , in respect of Radius , or the Semidiameter of the circle .

4. The arches of a circle thus reduced to right lines are either Chords , Sines , Tangents , or Secants ,

5. A Chord or Subtense is a right line inscribed in a circle , dividing the whole circle into two segments , and in like manner subtending both the segments .

6. A chord or subtense is either the greatest or not the greatest .

7. The greatest Subtense is that which divideth the whole circle into two equal segments , as the right line GD , and is also commonly called a diameter .

8. A subtense not the greatest , is that which divideth the whole circle into two unequal segments : and so on the one side subtendeth an arch less then a semicircle ; and on the other side subtendeth an arch more then a semicircle , as the right line CK on the one side subtendeth the arch CDK , lesse then a semicircle ; and on the other side subtendeth the arch CGK more then a semicircle .

9. A sine is either right or versed .

10. A right sine is the one half of the subtense of the double arch , as the right sine of the arches CD and CG is the right line AC , be●●● half the chord or subtense of the double arches of CD and CG , that is , half of the right line CAK , which subtendeth the arches CDK and CGK ; whence it is manifest , that the right sine of an arch lesse then a Quadrant , is also the right sine of an arch greater th●n a Quadrant . For as the arch CD is lesse then a Quadrant by the arch CE ; so the arch CG doth as much exceed a Quadrant , the right line AC being the right sine unto them both . And hence instead of the obtuse angle GBC , which exceeds 90 degrees , we take the acute angle CBA , the complement thereof to 180 : and so our Canon of sines doth never exceed a quadrant or 90 d.

11. Againe , a right 〈◊〉 is either Sinus totus , that is the Radius or whole sine , as in the triangle ABC , AC is the Radius , semidiameter , or whole sine , Or else the right sine is the Sinus simpliciter , that is , the first sine , as CA or BA , the one whereof is alwayes the complement of the other to 90 degrees ; we usualy call them sine and co-sine .

12. The versed sine of an arch is that part of the diameter , which lieth between the right sine of that arch and the circumference . Thus AD is the versed sine of the arch CD , and AG the versed Sine of the arch CEG ; therefore of versed Sines some are greater , and some are lesse .

13. A greater versed Sine is the versed Sine of an arch greater then a Quadrant , as AG is the versed Sine of the arch CEG greater then a Quadrant .

14. A lesser versed sine is the versed Sine of an arch lesse then a Quadrant , as AD is the versed sine of the arch CD less then a Quadrant .

15. A tangent of an arch or angle is a right line drawn perpendicular to the Radius or semidiameter of the circle of the triangle , so as that it toucheth the outside of the circumference , And thus the right line FD is the tangent of the arch DC .

16. A secant is a right line proceeding from the center of the circle , and extended through the circumference to the end of the tangent ; and thus BF is the Secant of the arch DC .

17. The definition of the quantity which right lines applyed to a circle have , is the making of the Tables of Sines , tangents and secants ; that is to say , of right Sines and not of versed ; for the versed Sines are found by the right without any labour .

18. The lesser versed sine with the sine of the complement is equal to the Rad●●● as the lesser versed sine AD with the right sine of the complement AB is equal to the Radius BD ; therefore if you substract the right sine of the complement AB from the Radius BD , the remainder is the versed sine AD.

19. The greater versed sine is equal to the Radius added to the right sine of the excesse of an arch more then a Quadrant , as the greater versed sine AG is equal to the Radius BG with the sine of the excess AC : therefore if you adde the right sine of the excess AB to the Radins BG , you shal have the versed sine of the arch CEG , & so there is no need of the table of versed sines , the right sines may thus be made .

20. The Tables of Sines , Tangents , and Secants may be made to minutes , but may , by the like reason , be made to seconds , thirds , fourths , or more , if any please to take that paines : for the making whereof the Radius must first be taken of a certain number of parts , and of what-parts soever the Radius be taken , the Sines , Tangents , and Secants are for the most part irrational●● i● , that is , they are inexplicable in any true whole numbers or fractions precisely , because there are but few proportional parts to any Radius , 〈…〉 , whose square root multiplied in it self will produce the number from whence it was taken , without some fraction still remaining to it , and therefore the Tables of Sines , Tangents , and Secants cannot be exactly made by any meanes ; and yet such may and ought to be made , wherein no number is different from the truth by an integer of those parts , whereof the Radius is taken , as if the Radius be taken of ten Millions , no number of these Tables ought to be different from the truth by one of ten Millions .

That you may attain to this exactnesse , either you must use the fractions , or else take the Radius for the making of the Tables much greater then the true Radius , but to work with whole numbers and fractions is in the calculation very tedious ; besides here no fractions almost are exquisitely true : therefore the Radius for the making of rhese Tables is to be taken so much the more , that there may be no errour , in so many of the figures towards the left hand as you would have placed in the Tables ; and as for the numbers superfluous , they are to be cut off from the right hand towards the left after the ending of the supputation , Thus , to finde the numbers answering to each degree and minute of the Quadrant to the Radius of 10000000 or ten millions , I adde eight ciphers more , and then my Radius doth consist of sixteen places .

This done , you must next finde out the right Sines of all the arches lesse then a Quadant , in the same parts as the Radius is taken of , whatsoever bignesse it be , and from those right Sines the Tangents and secants must be found out .

21. The right Sines in making of the Tables are either primary or secondary . The primarie Sines are those , by which the rest are found , And thus the Radius or whole Sine is the first primary Sine , the which how great or little soever is equall to the side of a six-angled figure inscribed in a circle , that is , to the subtense of 60 degrees , the which is thus demonstrated .

Out of the Radius or subtense of 60 degrees the sine of 30 degrees is easily found , the halfe of the subtense being the measure of an angle at the circumference opposite thereunto by the 19 of the second ; if therefore your Radius consists of 16 places being 1000.0000.0000.0000 . The sine of 30 degrees will be the one half thereof , to wit , 500.0000.0000.0000 .

22. The other primary sines are the sines of 60 , 45 , 36 , and of 18 degrees , being the halfe of the subtenses of 120 , 90 , 72 , and of 36 degrees .

23. The subtense of 120 degrees is the side of an equilateral triangle inscribed in a circle , and may thus be found .

The Rule .

Substract the Square of the subtense of 60 degrees , from the Square of the diameter , the Square root of what remaineth is the side of an equilateral triangle inscribed in a circle● or the subtense of 120 degrees .

The reason of the Rule .

The subtense of an arch with the subtense of the complement thereof to 180 with the diameter , make in the meeting of the two subtenses a right angled triangle . As the subtense AB 60 degrees , with the subtense AC 120 degrees , and the diameter CB , make the right angled triangle ABC , right angled at A , by the 19 of the second . And therfore the sides including the right angle are equal in power to the third side , by the 〈◊〉 of the second . Therefore the square of AB being taken from the square of CB , there remaineth the square of AC , whose squar root is the subtense of 〈◊〉 degrees or the side of an equilateral triangle inscribed in a circle ,

Example .

Let the diameter CB be 2000.0000 . 0000.0000 . the square thereof is 400000. 00000.00000.00000.00000.00000 . The subtense of AB is 100000.00000.00000 . The square thereof is 100000.00000.00000 . 00000.00000.00000 , which being substracted from the square of CB , the remainder is 300000.00000.00000.00000.00000.00000 , whose square root 173205.08075.68877 . the subtense of 120 degrees .

CONSECTARY .

Hence it followeth , that the subtense of an arch lesse then a Semicircle being given , the subtense of the complement of that arch to a Semicirc●e is also given .

24. The Subtense of 90 degrees is the side of a square inscribed in a circle , and may thus be found .

The Rule .

Multiply the diameter in it self , and the square root of half the product is the subtense of 90 degrees , or the side of a square inscribed in a circle .

The reason of this Rule .

The diagonal lines of a square inscribed in a circle are two diameters , and the right angled figure made of the diagonals is equal to the right angled figures made of the opposite sides , by the 20 th . of the second , now because the diagonal lines AB and CD are equal , it is all one , whether I multiply AC by it self , or by the other diagonal CD , the p●oduct will be still the same , then because the sides AB , AC , and BC do make a right angled triangle , right angled at C , by the 〈◊〉 of the second , & that the 〈◊〉 AC and ●B are equal by the work , the half of the square of AB must needs be the square of AC or CB , by the 17 th . of the second , whose square rootes the subtense of CB , the side of a square or 90 degree .

Example .

Let the diameter AB be 200000.00000 . 00000 , the square thereof is 400000.00000 . 00000.00000.00000.00000 , the half whereof is 200000.00000.00000.00000.00000 . 00000. whose square root 14142● . 356●3 . 73095. is the subtense of 90 degrees , or the side of a square inscribed in a Circle .

25. The subtense of 36 degrees is the side of a decangle , and may thus be found .

The Rule .

Divide the Radius by two , then multiply the Radius by it self , and the half thereof by it self , and from the square root of the summe of these two products substract the half of Radius , what remaineth is the side of a decangle , or the subtense of 36 degrees .

The reason of the rule .

For example .

Let the Radius EB be 100000.00000.00000 . then is BH , or the half thereof 500000. 00000.00000 . the square of EB is 100000 00000.00000.00000.00000.00000 . and the square of BH 250000.00000.00000.00000 . 00000.00000.00000 . The summe of these two squares , viz 125000.00000.00000 . 00000 , 00000. 00000 , is the square of HE or HK , whose square root is 1118033● 887●9895 , from which deduct the halfe Radius BH 500000000000000 , and there remaineth 618033988749895 , the right line KB , which is the side of a decangle , or the subtense of 36 degrees .

26 The subtense of 72 degrees is the side of a Pentagon inscribed in a circle , and may thus be sound .

The Rule .

Substract the side of a decangle from the diameter , the remainer multiplied by the Radius , shall be the square of one side of a Pentagon , whose square root shall be the side it self , or subtense of 72 degrees .

The Reason of the Rule .

In the following Diagram let AC be the side of a decangle , equal to CX in the diameter , and let the rest of the semicircle be bisected in the point E , then shall either of the right lines AE or EB represent the side of an equilateral pentagon , for AC the side of a decangle subtends an arch of 36 degrees the tenth part of a circle , and therefore AEB the remaining arch of a semicircle is 144 degrees , the half whereof AE or EB is 72 degrees , the fift part of a circle , or side of an equilateral pentagon , the square whereof is equal to the oblong made of DB and BX .

Demonstration .

Draw the right lines EX , ED , and EC , then will the sides of the angles ACE and ECX be equal , because CX is made equal to AC , and EC common to both ; and the angles themselves are equal , because they are in equal segments of the same circle by the 19 of the second ; and their bases AE and EX are equal by the 23 of the second ; and because EX is equal to AE , it is also equal to EB , and so the triangle EXB is equicrural , and so is the triangle EDB , because the sides ED and DB are Radii , and the angles at their bases X and B , E and B , by the 24th . of the second , and because the angles at B is common to both , therefore the two triangles , EXB and EDB are equiangled , and their sides proportional , by the 18 th . and 16 th . Theoremes of the second Chapter , that is as DB to EB ; so is EB to BX , and the rectangle of DB in BX is equal to the square of EB , whose square root is the side EB , or subtense of 72 degrees ,

Example .

Let AC , the side of a decangle or the subtense of 36 degrees , be as before : 618033988749895 , which being substracted from the diameter BC 200000.00000 , 00000. the remainer is XB , 1381966011151105 , which being multiplied by the Radius DB , the product 1381966011251105 00000.00000.0000 , shall be the square of EB whose square root 1175570504584946 is the right line EB , the side of a Pentagon or subtense of 72 degrees .

CONSECTARY .

Hence it followes , that the subtense of an arch lesse then a semicircle being given , the subtense of half the complement to a semicircle is given also ,

Thus much of the primarie Sines , the secondary Sines or all the Sines remaining may be found by these and the Propositions following

27. The subtenses of any two arches together lesse then a semicircle being given , to finde the subtense of both those arches .

The Rule .

Finde the subtense of their complements to a semicircle , by the 23 hereof ; then multiply each subtense given by the subtense of the complement of the other subtense given , the sum of both the products being divided by the diameter , shall be the subtense of both the arches given .

The reason of the Rule .

Example .

Let AI , the side of a square or subtense of 90 degrees be 141421.35623.73059 . And EO , the side of a triangle , or subtense of 120 degrees , 173205.08075.68877 , the product of these two will be 2449489742783 ▪ 77659465844164315. Let AE , the side of a sixangled figure , or the subtense of 60 degrees be 100000 00000.00000 . And IO , the side of a square , or subtense of 90 degrees 141421.35623.73059 the product of these two will be 141421.35623.73059 . 00000.00000.00000 . the summe of these two products 3863703305156272659465844164315 . And this summe divided by the diameter AO , 200000.00000.00000 . leaveth in the quotient for the side EI , or subtense of 150 degrees , 1931851652578136. the half whereof 965925826289068 , is the Sine of 75 degrees .

28 The subtenses of any two arches lesse then a Semicircle being given , to finde the subtense of the difference of those arches ,

The Rule .

Finde the subtenses of their complements to a semicircle , by the 23 hereof , as before , then multiply each subtense given , by the subtense of the complement of the other subtense given ; the lesser product being substracted from the greater , and their difference divided by the diameter , shall be the subtense of the difference of the arches given .

The Reason of the Rule .

Let the subtenses of the given arches be AE and EI , and let the subtense sought be the right line EI ; then because the right angled figure made of the diagonals . AI and EO is equal to the right angled figures made of their opposite sides , by the 20 of the second ; therefore if I subtract the right angled figure made of AE and IO , from the right angled figure made of AI and EO the remainer will be the right angled figure of AO and EI , which being divided by the diameter AO , leaveth in the quotient EI.

Example ,

29. The sine of an arch lesse then a Quadrant being given , together with the sine of half his complement , to finde the sine of an arch equal to the commplement of the arch given , and the half complement added together .

The Rule .

Multiply the double of the sine given , by the sine of half his complement , the product divided by the Radius , will leave in the quotient , a number , which being added to the sine of the half complement shall be the sine of the arch sought .

The reason of the Rule .

That is , as AI , is to AM : so is CP , to PM ; and so is PS , to PN , and then by composition , as AI , AM : so is CS , to MN . Now then let ES be the arch given , and SI the complement thereof to a Quadrant , then is CG or IB , being equal to EY , the half of the said complement SI , and AM is the Sine thereof , and the Sine of ES is the right line HS , and the double CS , MN is the difference between AM , the Sine of CG or IB , and AN the Sine of SB , and AI is the Radius , and it is already proved , that AI is in proportion to AM , as CS , is to MN , therefore if you multiply AM by SC , and divide the product by AI , the quotient will be NM , which being added to AM , doth make AN , the Sine or the arch sought .

Example .

Let ES , the arch given , be 84 degrees , and the Sine thereof 9945219 , which doubled is 19890438 , the Sine of 3 degrees , the halfe complement is 523360 , by which the double Sine of 84 degrees being multiplied , the product will be 104098●9 . 631680 , which divided by the Radius , the quotient will be 10409859 , from which also cutting off the last figure , because the Sine of 3 degrees was at first taken too little , and adding the remainer to the Sine of 3 degrees , the aggregate 1564345 is the Sine of 6 degrees , the complement of 84 , and of 3 degrees , the halfe complement added together , that is , it is the sine of 9 degrees .

30. The subtense of an arch being given , to find the subtense of the triple arch .

The Rule .

Multiply the subtense given by thrice Radius square , and from the product substract the cube of the subtense given , what remaineth shall be the subtense of the triple arch .

The reason of the Rule .

Now then we have already proved , that the square of AO divided by Radius , is equal to OX , and also that OX is equal to SA , and therefore SN is less then twice Radius by the right line AS ; or thus , NS is twice Radius less by AO square divided by Radius : and NS multiplied by SA is the same with twice Radius lesse by AO square divided by Radius , multiplied into AO square divided by Radius , and NS multiplied by SA is equal to SC multiplied by OS ; and therefore twice Radius less AO square divided by Rad. multiplied by AO square divided by Radius , is equal to SC , multiplied by SO : or thus , 2 Radius less AO square divided by Radius , multiplied into AO square divided by Radius , and divided by AO or SO is equal to SC. All the parts of the first side of this Equation are fractions , except AO and the two Radii , as will plainly appear , by setting it down according to the form of Symbolical or specious Arithmetick ; thus . . Which being reduced into an improper fraction , by multiplying 2 Radius by Radius , the Equation will run thus :

And then these two fractions having one common denominator , they may be reduced into one after the manner of vulgar fractions , that is , by multiplying the numerators , the product will be a new numerator , and by multiplying the denominators the product will be a new denominator ; thus multiplying the numerators , 2 Rad. aa − AO aa by the numerator AO aa , the product is 2 Rad. square into AO square , less AO square square , as doth appear by the operation ;

And then the denominators being multiplied by the other , that is , Radius being multiplied by Radius , the product will be Radius aa for a new denominator ; and then the Equation will run thus ; : but before this fraction can be divided by AO , AO being a whole number , must be reduced into an improper fraction , by subscribing an Unite , and then the Equation will be ; . Now as in vulgar fractions , if you multiply the numerator of the dividend by the denominators of the divisor , the product shall be a new numerator ; again , if you multiply the denominator of the dividend by the numerator of the divisor , their product shall be a new denominator , and this new fraction is the Quotient sought in this example , the numerator will be still the same , and the denominator will be Radius square multiplied in AO , and the fraction will be . And in its least termes it is . In words thu● : Twice Radius square multiplied in AO , lesse by the cube of AO divided by Radius square is equal to SC. And by adding AO to both sides of the Equation , it will be , twice Radius square in AO , lesse AO cube divided by Radius square , more AO , is equal to SC more AO , that is , to OC . Here again AO , the last part of the first side of this Equation is a whole number , and must be reduced into an improper fraction , by being multiplied by Radius square , the denominator of the fraction ; and then it will be Radius square in AO divided by Radius square , which being added to twice Radius square in AO , divided by Radius , the summe will be 3 Radius square in AO divided by Radius square , and the whole Equation , the subtense of the triple arch .

For Example .

Let AO or AB , 17431. 14854. 95316. the subtense of 10 degrees be the subtense given , and let the subtense of 30 degrees be required ; the Radius of this subtense given consists of 16 places , that is , of a unite and 15 ciphers , and therefore thrice Rad. square is 3 , and 30 ciphers thereunto annexed , by which if you multiply the subtense given , the product will be 52293. 44564. 85948. 00000. 00000. 00000. 00000. 00000. 00000. the square of this subtense given is 3038449397 . 55837.60253.85793.9856 , and the cube 529.63662.80907.48519.77452.00270 . 23994. 54977. 14496 , which being substracted from the former product , there will remain 51763.80902.05040.51480.22547.99729 . 76005.45022.85504 . this remainer divided by the square of Radius , will leave in the quotient , 51763.80902.05040 . for the subtense of 30 degrees .

31. The subtense of an arch being given , to finde the subtense of the third part of the arch given .

The Rule .

Multiply the subtense given by Radius square , and divide the product by thrice Radius square , substracting in every operation the cube of the figure placed in the quotient from the triple thereof ; so shall the quotient in this division be the subtense of the third part of the arch given .

The reason of the Rule .

The reason of the rule is the same with the triple arch , but the manner of working is more troublesome , the which I shall endeovour to explain by example .

Let there be given the subtense of 30 degrees , 517638090205040 , and let the subtense of 10 degrees be required : First , I multiply the subtense given by the square of Radius , that is , I adde 30 ciphers thereunto , and for the better proceeding in the work , I distinguish the subtense given thus inlarged by multiplication into little cubes , setting a point between every third figure or cipher , beginning with the last first , and then the subtense given will stand thus : 517.638.090.205.040.000.000.000.000.000 000.000.000.000.000 . And so many points as in this manner are interposed , of so many places the quotient wil consist , the which in this example is 15 , and because here are too many figures to be placed in so narrow a page , we will take so many of them onely as will be necessary for our present purpose ; as namely , the 15 first figures , which being ordered , according to the rules of decimal Arithmetick , may be divided into little cubes , beginning with the first figure , but then you must consider whether the number given to be thus divided be a whole number or a fraction , if it be a whole number , you must set your point after or over the head of the first figure , if it be a fraction , place as many ciphers before the fraction given , as will make it consist of equal places with the denominator of the Fraction given ; thus the subtense given being a fraction , part of the supposed Radius of a circle , the which , as hath been said doth consist of 16 , and the subtense given but of fifteen , I set a cipher before it , and distinguish that cipher from the subtense given by a point or line , and every third figure after , so will the subtense given be distinguished into little cubes , as before . This done , I place my divisor thrice Radius square , that is , 3 with ciphers ( or at least supposing ciphers to be thereunto annexed ) as in common division under the first figure of the subtense given , that is , as we have now ordered it under the cipher , and ask how often 3 in nought , which being not once , I put a cipher in the margine , and move my divisor a place forwarder , setting it under 5 , and ask how often 3 in 5 , which being but once , I place one in the quotient , and the triple thereof being 3 , I place under 3 my divisor , and the cube of the figure placed in the quotient , which in this case is the same with the quotient it self , I set under the last figure of the first cube , and supposing ciphers to be annexed to the triple root , I substract this cube from it , and there doth remain 299 , which is my divisor corrected ; with this therefore I see whether I have rightly wrought or not , by asking , how often 299 is contained in the first cube of the subtense given , 517 , which being but once , as before , the former work must stand , & this divisor corrected must be subtracted from the first cube in the subtense given , and there will rest 218 , and so have I wrote once . To this remainer of the first cube 218 , I draw down 638 , the figures of the next cube & moving my divisor a place forwarder , I ask , how often 3 in 21 , which being 7 times , I put 7 in the quotient , and under the first figure of this second cube , that is , under 6 I set the triple square of the first figure in the quotient , that is , 3 , for the quotient being but one , the square is no more , and the triple thereof is 3 ; under the second figure of this second cube I set the triple quotient , the which in this example is likewise 3 , and both these added together , do make 33 , which being substracted from my divisor 3000 , there will remain 2967 , for the divisor corrected , and by this also I finde the quotient to be 7 , and yet I know not whether my work be right or not , I must therefore proceed , and set the triple of the figure last placed in the Quotient under the first figure of the remainer of the first cube , that is , I must set 21 , the triple of 7 under 2 , the first figure of 218 , and now having two figures in the quotient , for distinction sake I call the first a , and the second e , that so the method of the work may the better be seen in the margine , and I set 3 aae , that is , 3 , the square of the first figure noted with the letter a , viz. 1. multiplied by the second figure , noted with the letter (e) to wit , 7 , under the first figure of the next cube , now the square of (a) that is , of one is one , and the triple of this square is 3 , and 3 times 7 is 21 , which is ( 3 aae ) or thrice (a) square in e , the last figure whereof , to wit , one , I place under 6 , the first figure of the next cube 638 : next I set ( 3 aee ) that is , three times one multiplied by the square of 7 , that is , 3 multiplied by 49 , which is 1●● under the 2 figure of the cube 638 : and lastly , I set ( eee , that is ) the cube of e , that is , the cube of 7 , viz. 343 , under the last figure of the cube 638 , and these 3 sums added together do make 3●●3 , which being substracted from the triple root , that is , from 21 , supposing ciphers to be thereunto annexed , as before , there will remain 〈…〉 , and because this may be subtracted from the 2d . cube , & the remainer of the first , I finde that 7 is the true figure to be placed in the quotient , and such a subtraction being made , the remainer will be 12511 , and so have I wrote twice . The work following must be done in all things , as this second , save onely in this particular , that both the figures in the quotient are reckoned but as one , which for distinction sake I called a , and the figure to be found by division I called e , and therefore in this third work 3 aa , or thrice a square is the square of 17 , that is 289 , 3 a or thrice a is 3 times 17 , that is , 51 , and so of the rest , in the fourth work the three first figures must be called a , in the fifth work the four first figures found , and so forward , till you have finished your division , and therefore this second manner of working being well observed , there can be no difficulty in that which followes .

32. The subtense of an arch being given to finde the subtense of the arch quintuple , or of an arch five times as much .

The Rule .

From the product of the subtense given , multiplied by 5 times Radius square square , subtract the cube of the subtense given multiplied by 5 times Radius square , the squared cube of the subtense given being first added thereunto , the remainer divided by Radius square square , shall leave in the quotient the subtense of the arch quintuple , or the arch 5 times as much .

The reason of the Rule .

Multiply the numerator of the fractions in the second place by the numerator of the fraction in the third , and their product will be a new numerator , the numerator of the fraction in the second term is 2 Rad. − AO aa And in the third , 3 Rad. aa × AO − AO aaa

That one of these termes may be the better multiplied by the other , the first of the second term , 2 Rad. must be reduced into an improper fraction , by the multiplication thereof by Radius , the denominator of that fraction , and then the 2d . term will be 2 Rad. aa − AO aa , and because this second term is the lesse , we will multiply the third thereby , the work stands thus :

Thrice Radius square in AO multiplied by twice Radius square , doth make 6 Rad. square squares , and AO cube multiplied by 2 Radius square is 2 Rad. square in AO cube , and because it hath the signe lesse , therefore the first product is 6 Rad. aaaa × AO − 2 Rad. aa × AO aaa . Again , 3 R. aa in AO , multiplied by AO aa , doth make 3 R. aa in AO aaa , & AO aa multiplied by AO aaa , doth make AO aaaaa , & because it hath the sign less , therefore the 2. product is 3 R. square × AO aaa + AO aaaaa , and so both the products will be 6 Rad. square of squares multiplied by AO lesse by 5 Rad. square in AO cube more by AO square cube . And if you multiply Rad. square , the denominator of the third term by Rad. the denominator of the second , the product will be Rad. cube , and the whole product will stand thus ,

To divide this product by twice Radius , twice Radius being a whole number must be first reduced into an improper fraction , by subscribing an unite thus , . then if you multiply the numerator of the product by one , the denominator of this fraction , the product will be still the same , and if you multiply the denominator of the product Rad. aaa by 2 Radius , the numerator of this improper fraction , the product will be 2 Rad. square square for a new denominator , and the Quotient will be the quantity of the right line OT , the double whereof is which is the quantity of the right line OE more by CB , and therefore CB or AO being deducted , the remainer will be the right line OE , which is the quintuplation of an angle , and to this end AO must be reduced into an improper fraction of the same denomination , that is , by multiplying thereof by 2 Rad. aaaa , and then the fraction will be and this being deducted from the remainer will be . And this reduced into its least terms , will be , which was to be proved .

For example .

Let AO or AB 349048 , the subtense of 2 degrees be given , and let the subtense of 10 degrees be demanded , 5 times Radius square square is 50000000.000000.00000 00.0000000 . by which if you multiply the subtense given , the product will be 1745240 0000000.0000000.0000000.0000000 . The Cube of the subtense given multiplied by 5 times Radius square is 212630453781992960 . 0000000.0000000 . the squared cube of the subtense given is 5184639242824921385360723968 , the which being added to the product of 5 Rad. aa in AO , that is , to 212630453781992960.0000000.0000000 the summe will be 21268230017442120921385360723968 . And this being subtracted from the product of the subtense given multiplied by 5 times Radius square square , the remainer will be 17431141769982557879078614639276032 , and this remainer divided by Radius square square , that is , cutting off 28 figures , their quotient will be 1743114 , the subtense of 10 degrees .

33. The subtense of an arch being given to finde the subtense of the fift part of the arch given .

The Rule .

Divide the subtense given by five roots , lesse 5 cubes , more one Quadrato cube , the quotient shall be the fift part of the arch given .

The reason of the rule depends upon the foregoing Probleme , in which we have proved , that the subtense of five equall arches is equall to 5 roots , lesse 5 cubes , more by one quadrato cube , of which 5 roots one of them is the subtense of the fift part of the arch given . And consequently , if I shall divide the subtense of five equall arches by 5 roots , lesse 5 cubes , more one quadrato cube , the quotient shall be the subtense of the fift part of the arch .

The manner of the work is thus : First , consider whether the subtense given to be divided doth consist of equal , or of fewer places then the Radius thereof , if it consist of equal places , set a point over the head of the first figure of the subtense given , if of fewer places , make it equal , by prefixing as many ciphers before the subtense given as it wanteth of the number of places of the Radius thereof .

For example .

Let the subtense of 10 degrees be given , viz. 0.17431.14854.95316.34711 . This is lesse then the Rad. by one place , and therefore I have set one cipher before , and have distinguished it from the subtense given by a point set between , the which is all one , as if it had been put over the head thereof : next you must distinguish the subtense given into little cubes , & into quadratocubes , which may be conveniently done thus ; having found the place of the first point , which is alwayes the place of the Radius , the subtense given must be distinguished into little cubes , by putting a point under every third figure , as in the trisection of an angle : thus in this example the first cubick point will fall under the figure 4 , and the subtense given must be distinguished into quadrato cubes , by setting a point over the head , or else between every fift figure from the place of the Radius : thus in this example the first quadrato cubick point must be set over the head , or after the figure of 1 , the second after 4 , as here you see .

After this preparation made , you must place your two divisors , 5 roots and 5 cubes in this manner , the first as in ordinary division under the first figure of the subtense given , the other 5 under the first cubick point , and they will stand as in the work you see ; then ask how often 5 in one , which being not once , I put a cipher in the quotient , and remove my first divisor a place forwarder , as in ordinary division , but the other 5 I remove to the next cubick point , then , as before , I ask how often 5 in 17 , which being 3 times , I set 3 in the quotient , and of this quotient I seek the quadrato cube , and finde it to be 243 , the last figure whereof , namely , 3 , I set under the last figure of the second quadrato cubick point ( because there are but 3 figures between my divisor 5 and the first cubick point , whereas there must be alwayes four at the least ) then I multiply the figure 3 placed in the quotient by my divisor 5 , and the product thereof is 15 , the first figure whereof I place under my said divisor 5 , to which having annexed ciphers , or at least supposing them to be annexed , ( as to the triple root in the trisection ) I draw the quadrato cube of the figure in the quotient , and these 5 roots or 5 quotients into one summe , the which is 1500000243 , under this summe I draw a line , so have we five roots more one quadrato cube , from which I must subtract 5 cubes , I therefore seek the cube of 3 , the figure placed in the quotient , and finde it to be 27 , which multiplied by 5 , the product will be 135 , the last figure of these five cubes , viz. 5 , I set under my second 5 or cubick divisor , and substracting these 5 cubes from the 5 roots more one quadrato cube , the remainer will be 〈…〉 , which remainer being also substracted from the figures of the subtense given standing over the head thereof , the remainer of the subtense given will be 244464611 , and so have I wrought once .

To this remainer of the two first quadrato cubes , I draw down 95316 , the figures of the next quadrato cube , and setting my first divisor a place forwarder , I ask how often 5 in 24 , which being four times , I set 4 in the quotient , not knowing yet whether this be the true quotient or not , but with this I proceed to correct my divisor , and first I seek the quadrato quadrat of 3 , the first quotient , and finde it to be 81 , this multiplied by 5 , will make 405 , this product I set under my divisor , and 5 the last figure thereof I set under 9 , the first figure of the 3 quadrato quadrate ; next I seek the cube of 3 , & finde it to be 27 , which being multiplied by 10 , the product will be 270 , and this I set a place forwarder under the former product : thirdly , I seek the square of 3 which is 9 , and this multiplied by 10 is 90 , which I set a place forwarder under the second product 270. Lastly , I multiply 3 , the figure in the quotient by 5 my divisor , this product which is 15 , I set a place forwarder under 90 , the third product , and now these 4 products together with my divisor and ciphers thereunto annexed , being gathered into one summe , will be 500000432915 , under which I draw a line . And thrice the square of 3 , multiplied by 5 , which is 135 , I set under this summe , the last figure thereof 5 , under the first figure of the third cubick point , that is , under 4 , and the triple of 3 multiplied by 5 , which is 45 , I set under the former summe 135 , a place forwarder , and my cubick divisor 5 under the last summe a place forwarder , that is , under the third cubick point , these drawn into one summe will be 13955 , and being substracted from the former summe 500000432915 , the remainer 498.60493 . + 2915 is my divisor corrected , and yet I know not whether I have a true quotient or not ; under this remainer therefore I draw a line , and work with 4 , which I suppose to be the true quotient in manner following ; and that the manner of the work may be the more perspicuous , ( as in the trisection of an angle , so here ) 3 the first figure found I call (a) and 4 the second figure I call (e) the square of three I note with aa , the cube with aaa , the quadrate quadrat with aaaa , the quadrato cube with aaaaa , so likewise the square of 4 the second figure I note with ee , the cube with eee , the quadrato quadrate with eeee , the quadrato cube with eeeee ; my first divisor I note with ffff , because this Equation is quadrato quadratick , and 5 my second divisor , I note with cc , because the divisor it self is cubick : these things premised , I proceed thus : First , I multiply 405 , which is 5 aaaa or 5 times the quadrato quadrate of 3 by e , that is , by 4 , and the product thereof 1620 , I set under my divisor corrected , so as the last figure thereof may stand under the first figure of the third quadrato cubick number , and against this number I put in the margine 5 aaaae , that is , five times the quadrato quadrate of 3 multiplied by 4 : next 270 , ten times the cube of 3 , by 16 the square of 4 , and this product 4320 , I set under the former a place forwarder , and 90 , which is 10 times the square of 3 , I multiply by 64 , the cube of 4 , & this product 5760 I set under the last a place forwarder then that , and 15 , which is 5 times 3 , I multiply by 256 , the quadrato quadrate of 4 , & the product thereof 3840 , I set under the third product a place forwarder , and 1024 , the quadrato cube of four under that : lastly , I multiply four , the last figure placed in the quotient by 5 my divisor , and the last figure of this product I set under 5 my divisor , and supposing ciphers to be thereunto annexed , I collect these several products into one summe , and their aggreagate 20000021135424 , is five roots more one quadrato quadrate , under which I draw a line , and seek the five cubes to be substracted , thus .

First , I multiply 135 ( which is thrice the square of three multiplied by five my cubick divisor ) by four , the figure last placed in the quotient , and the product thereof 540 I set under the last summe , so as the last figure thereof may be under the first figure of the third cube ; next I multiply 45 that is , five times the triple of three , by 16 the square of four , and this product 720 I set under the former a place forwarder , and under that 320 , which is five times the cube of 4 , a place forwarder too , these products drawn into one summe do make 61520 , the five cubes to the substracted from the five roots more one quadrato quadrate before found , which being done , the remainer will be 19938501135424 , and this remainer being substracted from the figures of the subtense given over the head thereof , the remainer will be 450.79600 . 59892 , and because such a substraction may be conveniently made , I conclude , that I have found the true quotient , and so have I wrought twice .

34 The Sines of two arches equally distant on both sides from 60 degrees , being given , to finde the Sine of the distance .

The Rule .

Take the difference of the Sines given , and that difference shall be the Sine of the arch sought .

The reason of the Rule .

Let CN and PN be the two arches given , and equally distant from 60 deg. MN , that is equally distant on both sides from the point M. And let the right lines CK and PL be the Sines of those arches , being drawn perpendicular to the right line AN , and thereupon parallel to one another .

Moreover , let the right line PT be drawn perpendicular upon the right line CK , and so parallel to the right line KL , then this right line TP cutteth from the right line CK another line TK , equal unto PL , by the 15 of the second , and leaveth the right line TC for the difference of the Sines CK and PL. Lastly , the Sines of the distance of either of them from 60 degrees let be the right line CD or DP , I say , that the right line TC is equal to the right line CD or DP .

Demonstration .

Example .

Let the arches CN be 70 degrees , PN 50 , CM or PM 10 degrees ; for so many degrees are the arches of 70 degrees ; and 50 degrees distant from the arch of 60 degrees on both sides . And let first the Sines of 70 degrees and 10 degrees be given , and let the Sine of 50 degrees be demanded .

From the Sine of 70d. CK 9396926 Subtract the Sine of 10d. CD or CT , 1736482 The Remainer will be the Sine of 50d. TK or PL , 7660444

Then let the Sine of 70 degrees and 50 degrees be given , and let the Sine of ten degrees be demanded .

From the Sine of 70 degrees CK , 9396926 Substract the Sine of 50d. TK or PL , 7660444 Remainer is the Sine of 10d . CD , 1736482

Lastly , let the Sines of 50 degrees and 10 degrees be given , and let the Sine of 70 degrees be demanded .

To the Sine of 50d . PL or TK , 7660444 Adde the Sine of 10d . DP or TC , 1736482 Their sum will be the Sine of 70d . 9396926

And thus far of the making of the Tables of right Sines , the Tables of versed Sines are not necessary , as hath been said

CHAP. IV. By the Tables of Sines to make the Tables of Tangents and Secants .

1. AS the Sine of the complement , Is to the Sine of an arch : so is the Radius , to the tangent of that arch .

2. As the Sine of the complement , is to the Radius ; so is the Radius , to the secant of that arch . For , by the 16 th . of the second : 1. As the Sine of the complement AB , is to the Sine CA: so is the Radius BD or BC , to DF the tangent . 2. As the sine of the complement AB , is to the Radius BD or BC : so is the Radius BC , to the secant BF .

Example .

Let the tangent and secant of the arch CD 30 degrees be sought for . The sine AC 30 degrees is 5000000 , the sine of the complement AB 60 degrees is 8660254. Now then if you multiply the sine AC 5000000 , by the Radius CB 10000000 , the product wil be 50000000000000 , which divided by the sine of the complement AB 8660254 : the quotient will be 5773503 , the right line FD or the tangent of the arch of 30 degrees .

2. As the sine of the complement AB 8660254 , Is to the Radius DB 10000000 : so is the Radius BC 10000000 , to FB , the secant of the arch of 30 degrees : and so for any other : but with more ease by the help of these Theorems following .

Theorem 1. The difference of the Tangents of any two arches making a Quadrant , is double to the tangent of the difference of those arches

The Declaration .

Let the two arches making a Quadrant be CD and BD , whose tangents are CG and BP , and let BS be an arch made equall to CD ; and then SD will be the arch of the difference of the two given arches CD or BS , and BD. And also let the tangent BT be equal to the tangent CG , and then the right line TP will be the difference of the tangents given CG or BT , and BP. Lastly , let the arches BL and BO ( whose tangents are BK and BM ) be made equal to the arch SD ; I say , the right line TP being the difference of the two given tangents , CG and BP is double to the right line BK , being the tangent of the difference of the two given arches ; or which is all one , I say , that the right line TP is equal to the right line MK .

Demonstration .

Then that the right line MT is equall to the right line KA is thus proved ; the right line MA is equall to the right line KA , by the work , but the right line MT is equal to the right line MA , and therefore it is also equall to the right line KA .

That the right line MT is equal to the right line MA doth thus appear : for that the angles MAT and MTA are equall ; and therefore the sides opposite unto them are equal , for equall sides subtend equall angles : and the angles MTA and MAT are equal , because the angle MTA is equal to the angle TAC , by the like reason , that the angle KPA is equal to the angle DAC ; and the angle MAT is equall to the angle TAC , by the proposition : for the arches CS and SO are put to be equal : therefore it followes , that they are also equal one to another . Generally therefore , the difference of the tangents of two arches , making a Quadrant , is double to the tangent of the difference of those arches , which was to be demonstrated . And by consequence , the tangents of two arches being given , making a Quadrant , the tangent of the difference of those arches is also given . And contrarily , the tangent of the difference of those two arches being given , together with the tangent of one of the arches ; the tangent of the other arch is also given .

Example .

Let there be given the Tang. of 72 de . 94 m. And the Tang. of its complement , that is , of 17 6 Halfe the difference of these two arches is 27 94

Tangent of 72 de. 94 m. is 32586438 Tangent of 17 6 306●761 Their difference is 29517677 The halfe whereof is 14758838
The Tangent of 55 de. 88 min.

Or let the tangent of the greater arch 72 d. 94 m. be given , with the Tangent of the difference 55 de. 88 m. and let the lesser arch 17 de. 6 m. be demanded . Tangent of 72 de. 94 m. is 32586438 Tang. of 55 de. 88 m. doubled is 29517676 Their difference is 03068762
The Tangent of 17 de. 6 m.

Or lastly , let the lesser arch be given , with the Tangent of the difference , and let the greater arch be demanded . Tang. of 55 de. 88 m. the diff . is 14758838 Which doubled is 29517676 To which the tang . of 17 d. 6 m. ad . 3068761 Their aggregate is 32586437
the tangent of 72 degrees , 94 minutes .

Theor. 2. The tangent of the difference of two arches making a Quadrant , with the tangent of the lesser arch maketh the secant of the difference .

The Reason is

Because the tangent of the difference BL or BO , that is , the right line BK or BM with the tangent of the lesser arch BS , that is , with the right line BT , maketh the right line MT , which is equall to the Secant AK , by the demonstration of the first Theorem . Therefore , the tangent of the difference of two arches making a Quadrant , and the tangent of the lesser arch being given , the secant of the difference is also given . And contrarily .

For example .

Let the tangent of the former difference 55 degrees , 88 minutes , and the tangent of the lesser arch 17 degrees , ●● minutes , be given ; I say , the secant of this difference is also given . Tang. of the diff. 55 de. 88 m. is 14758838 The tangent of 17 06 is 3068762 Their sum is the secant of 55 88 , 17827600

Theor. 3. The tangent of the difference of two arches making a Quadrant , with the secant of their difference , is equal to the tangent of the greater arch .

Because the tangent of the arch BL , being the difference of the two arches BC and DC , making a Quadrant with the secant of the same arch BL , that is , the right line BK with the right line AK , is equal to the right line BP , by the demonstration of the first Theorem : therefore the tangent of the difference of two arches making a Quadrant being given , with the secant of their difference , the tangent of the greater arch is also given .

For example .

Let the tangent of the difference be the tang . of the arch of 55 de. 88 m. viz. 14758838 The secant of this difference is 17827600 Their sum is the tang . of 72 94 , 32586438
the greater of the two former given arches .

And now by the like reason these Rules may be added by way of Appendix .

Rule I.

The double tangent of an arch , with the tangent of half the complement , is equall to the tangent of the arch , composed of the arch given and half the complement thereof .

For if the arch BL be put for the arch given , the double tangent thereof shall be TP , by the demonstration of the first Theorem . And the complement of the arch BL , shall be the arch LC , whose half is the arch LD or DC , whose tangent is the right line GC or BT , but TP added to BT maketh BP , being the tangent of the arch BD , composed of the given arch BL , and half the complement LD , therefore the double tangent , &c.

Rule II.

The tangent of an arch with the tangent of half the complement is equal to the secant of that arch . For if you have the arch BL or BO for the arch given , the tangent of the arch given shall be BM , the tangent of half the complement shall be BT , which two tangents added together , make the right line MT , but the right line MT is equal to the right line AK , by the demonstration of the first Theorem ; which right line AK is the secant of the arch given BL , by the proposition : Therefore the tangent of an arch , &c.

Rule III.

The tangent of an arch with the secant thereof is equal to the tangent of an arch composed of the arch given , and half the complement . For if you have the arch BL for the arch given , BK shall be the tangent , and AK the secant of that arch . But the right line AK and KP are equal , by the demonstration of the first Theorem : therefore the tangent of the arch given BL , that is , the right line BK , with the secant of the same arch , that is , AK is equall to the right line BP , which is the tangent of the arch BD , being composed of the given arch , BL and LD being half the complement .

These rules are sufficient for the making of the Tables of natural Sines , Tangents , & Secants . The use whereof in the resolution of plain & spherical triangles should now folow ; but because the Right Honourable John Lord Nepoir , Baron of Marchiston , hath taught us how by borrowed numbers , called Logarithmes : to perform the same after a more easie and compendious way : we will first speak something of the nature and construction of those numbers , called Logarithmes ; by which is made the Table of the artificial Sines and Tangents , and then shew the use of both .

CHAP. V. Of the nature and construction of Logarithmes .

LOgarithmes are borrowed numbers , which differ amongst themselves by Arithmetical proportion , as the numbers that borrow them differ by Geometrical proportion : So in the first column of the ensuing Table the numbers Geometrically proportional being 1 , 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , &c. you may assigne unto them for bo●rowed numbers or Logarithmes , the numbers subscribed under the letters A , B , C , D , or any other at pleasure ; provided , that the Logarithmes so assigned still differ amongst themselves by Arithmetical proportion , as the numbers of the first column differ by Geometrical proportion : For example . In the column C , if you will appoint 5 to be the Logarithme of one , 8 the Logarithme of 2 , and 11 the Logarithme of 4 , 14 must needs be the Logarithme of 8 , the next proportional , because the numbers 5 , 8 , 11 , and 14 differ amongst themselves by Arithmetical proportion , as 1 , 2 , 4 , and 8 ( the proportional numbers unto which they are respectively assigned ) differ by Geometrical proportion , that is , as the numbers 5 , 8 , 11 , and 14 have equal differences : so the numbers 1 , 2 , 4 , and 8 have their differences of the same kinde : for as the difference between 5 and 8 , 8 and 11 , 11 and 14 , is 3 : so in the other numbers , as 1 is half 2 , so 2 is half 4 , 4 half 8 , &c. The same observation may be made of the Logarithmes placed in the columns , A , B , and D , or of any other numbers which you shall assigne as Logarithmes unto any rank of numbers , which are Geometrically proportional , and these Logarithmes or borrowed numbers you may propound to increase , and to be continued upwards , as those of the columnes A , B , C , or otherwise to decrease , and to be continued downwards , as those of the column D.

A B C D 1 1 5 5 35 2 2 6 8 32 4 3 7 11 29 8 4 8 14 26 16 5 9 17 23 32 6 10 20 20 64 7 11 23 17 128 8 12 26 14 256 9 13 29 11 512 10 14 32 8 1024 11 15 35 5 Log Log Log Log

The numbers continually proportional , which Mr. Briggs ( after a conference had with the Lord Nepeir ) hath proposed to himself in the Calculation of his C●ili●des , are 1 , 10 , 100 , 1000 , &c. to which numbers he hath assigned for Logarithmes 000 , &c. 1000 , and 2000 , and 3000 , that is to say , to 1 , the Logarithme 0.000 , and to 10 , the Logarithme 1,000 , and to 100 the Logarithme 2.000 , as in the table following you may perceive . In the column marked by the letter A , there is a rank of numbers continually proportional from 1 , and over against each number his respective Logarithme in the other column , signed by the letter B.

A B 1 0.00000 10 1.00000 100 2.00000 1000 3.00000 10000 4.00000

Having thus assigned the Logarithme to the proportional numbers of 1 , 10 , 100 , 1000 , &c. in the next place , it is requisite to finde the Logarithmes of the mean numbers situate amongst those proportionals of the same table , viz. of 2 , 3 , 4 , &c. which are numbers scituate betwixt 1 and 10 , of 11 , 12 , 13 , &c. which are placed betwixt 10 and 100 ; and so consequently of the rest : wherefore how this also may be done we intend to explain by that which followeth .

1. § . Make choice of one of the propotional numbers in the Table AB , and by a continued extraction of the square root create a rank of continuall meanes betwixt that number and 1 , in such sort , that the continuall mean which cometh nearest 1 may be a mixt number , lesse then 2 , and so near 1 , that it may have as many ciphers before the significant figures of the numerator , as you intend that the Logarithmes of your Table shall consist of places .

Example .

In the premised Table AB , I take 10 , the second proportional of that Table , then annexing unto it a compent company of ciphers , as twenty and four , thirty and six , fourty and eight , or any other number at pleasure ; onely observe , that the more ciphers you annex unto the number given , the more just and exact the operation will prove ; to make the Logarithmes of a Table to seven places 28 ciphers will be sufficient , they being therefore added to 10 , I extract the square root thereof , and finde it to be 3.16227766016837 ; again , annexing unto this root thus found 14 ciphers more , and working by that entire number so ordered , as if it were a whole number , I extract the root thereof , which I finde to be 1.77827941003892 : and so proceeding successively by a continued extraction , I produce 27 square roots , or continual means betwixt 10 and 1 , and write them down in the first column of the Table hereunto annexed , in which you may observe , that the three last numbers marked by the letters G , H , and L , viz. 1.00000006862238 1.00000003431119 1.00000001715559 are each of them mixt numbers lesse then 2 , and greater then 1 , and likewise to have seven ciphers placed bef●re the significant figures of their numerators , according to the true meaning and intention of this present rule .

2. § . Having thus produced a great company of continual meanes , annex unto them their proper Logarithmes , by halfing first the Logarithme of the number taken , and then successively the Logarithme of the rest .

For example .

1.000000000000000 being assigned the Logarithme of 10 , the number taken 0.500000 , &c. marked by the letter D , in the second column of the following Table , which is the half of 1.0000 , &c. is the Logarithme of the number A , the square root of 10 : in like manner 0.25000 , &c. being half 0.5000 , and is the Logarithme of the number B , and 0.125000 , &c. is the Logarithme of the number C , and so of the rest in their order . So that at last , as you have in the first column of the following Table 27 continuall meanes , betwixt 10 and 1 , as aforesaid : So in the other column you have to each of those continuall meanes , his respective Logarihme .

3. § . When a number which being lesse then 2 , and greater then 1 , comes so neer to 1 , that it hath seven ciphers placed before the significant figures of the numerator , the first seven significant figures of the numerator of such a number , and the first seven significant figures of the numerator of his square root lessen themselves like their Logarithmes , that is , by halfes .

This is proved by the Table following ; for there in the second column thereof , the number N being the Logarithme of the number G , I say , as the Logarithme K is half the Logarithme N , so 3431119 , the first seven figures of the numerator of the number H , are half 6862238 , the first seven significant figures of the numerator of the number G. Any two numbers of this kinde therefore being given , their Logarithmes and the significant figures of their numerators are proportional .

Example . The numerators G and H being given , I say , as 6862238 , the significant figures of the numerator of the number G ; are to 3431119 , the significant figures of the numerator of the number H ; so is 29802322 , the Logarithme of the number G , to 140901161 , the Logarithme of the number H. In like manner , G and L being given , as 6862238 , is to 1715559 , so is 29802322 , the Logarithme of the number G , to 7450580 , the Logarithme of the number L. This holdeth also true in any other number of this kinde , though it be not one of the continual means betwixt 10 and 1 , for the significant figures of the numerator of any such number bear the same proportion to his proper Logarithme , that the significant figures of any of the numbers marked by the letters G , H , or L bear to his .

10.0000 , &c. 1.000000000000000 A 3.16227766016837 0.500000000000000 D B 1.77827941003892 0.250000000000000 C 1.33352143216332 0.125000000000000 1.15478198468945 0.062500000000000 1.07460782832131 0.031250000000000 1.03663292843769 0.015625000000000 1.01815172171818 0.007812500000000 1.00903504484144 0.003906250000000 1.00450736425446 0.001953125000000 1.00225114829291 0.000976562500000 1.00112494139987 0.000488281250000 1.00056231260220 0.000244140625000 1.00028111678778 0.000122070312500 1.00014054851694 0.000061035156250 1.00007027178941 0.000030517578125 1.00003513527746 0.000015258789062 1.00001756748442 0.000007629394531 1.00000878270363 0.000003814697265 1.00000439184217 0.000001907348632 1.00000219591867 0.000000953674316 1.00000109795873 0.000000476837158 1.00000054897921 0.000000238418579 1.00000027448957 0.000000119209289 1.00000013724477 0.000000059604644 G 1.00000006862238 0.000000029802322 N H 1.00000003431119 0.000000014901161 K L 1.00000001715559 0.000000007450580 M

4. § . These things being thus cleared , it is manifest , that a number of this kinde being given , the Logarithme thereof may be found by the Rule of three direct . For as the significant figures of the numerator of any one of the numbers ( signed in the first column of the last Table by the letters G , H , or L ) are to his respective Logarithme : so are the significant figure of the numerator of the number given , to the Logarithme of the same number .

Example . The number 1.00000001021301 being given , I demand the Logarithme thereof : I say then ,

As 6862238 , the significant figures of the numerator of the number G , are to 29802322 , the logarithme of the same number G : so are 1021301 , the significant figures of the numerator of the number given , to 4357281 , the Logarithme sought ; before which if you prefix 9 ciphers , to the intent it may have as many places as the Logarithme in the last premised Table , ( viz. 16 ) the true and entire Logarithme of 1 . 00000001021301 , the number given is 0 . 000000004357281 , as before . And to every Logarithme thus found , you must prefix as many ciphers as will make the said Logarithme to have as many places as the other Logarithmes in the same table : for though you make your Table of Logarithmes to consist of as many places as you please , yet when you are once resolved of how many places the Logarithmes of your Table shall consist , you must not alter your first resolution , as to make the Logarithme of 2 to consist of six places , and the Logarithme of 16 to have seven , but if the significant figures of the numerator of the Logarithme of 2 have not so many places as the significant figures of the Logarithme of 16 , you must prefix a cipher or ciphers to make them equal ; because ( as hath been said , the Logarithmes of this kinde ought all to consist of equal places in the same Table .

5. § . Now then to finde the Logarithme of any number whatsoever , you are first to search out so many continual means betwixt the same number and 1 , till the continual mean that cometh neerest 1 hath as many ciphers placed before the significant figures of his numerator , as you intend the Logarithmes of your Table shall consist of places ; Again , this being done , you are to finde the Logarithme of that continual mean : And lastly , by often doubling and redoubling of that Logarithme so found ( according to the number of the continual meanes produced ) in conclusion you shall fall upon the Logarithme of the number given .

Example . the number 2 being given , I demand the Logarithme thereof to seven places : Here first in imitation of that which is before taught in the first rule of this Chapter , I produce so many continual meanes between 2 and 1 , till that which cometh nearest 1 hath seven ciphers before the significant figures of the numerator , which after three and twenty continued extractions , I finde to be 1.00000008262958 This continual mean being thus found ( by the direction of the last rule aforegoing ) I finde the Logarithme thereof to be 0.000000035885571 . for , As 6862238 , is to 29802322 : So 8262958 , is to 35885571.

This Logarithme being doubled will produce the Logarithme of the continual mean next above 1.00000008262958 , and so by doubling successively the Logarithme of each continual mean one after another , according to the number of the extractions ( viz. three and twenty times in all ) at last you shall happen upon the Logarithme 0.301029987975168 , which is the Logarithme of 2 the number propounded : The whole frame of the work is plainly set down in the table following ; for in the first column thereof you have 23 continual meanes betwixt 2 and 1 , and in the other column their respective Logarithmes , found by a continual doubling and redoubling of 0.000000035885571 , the Logarithme of the last continual mean in the table .

2.0000 , &c. 0 301029987975168 1.41421356237309 0.150514993987584 1.18920711500272 0.075257496993792 1.19050713266525 0.037628748496896 1.04427378243220 0.018814374248448 1.02189714865645 0 009407187124224 1.01088928605285 0.004703593562112 1.00542990111387 0.002351796781056 1.00271127505073 0.001175898390528 1.00135471989237 0.000587949195264 1.00067713069319 0.000293974597632 1.00033850805274 0.000146987298816 1.00016923970533 0.000073493649408 1.00008461627271 0.000036746824704 1.00004230724140 0.000018373412352 1.00002115339696 0.000009186706176 1.00001057664255 0.000004593353088 1.00000528830729 0.000002296676544 1.00000264415015 0.000001148338272 1.00000132207420 0.000000574169136 1.00000066103688 0.000000287084568 1.00000033051838 0.000000143542284 1.00000016525917 0.000000071771142 1.00000008262958 0.000000035885571

But now because the Logarithme of the number propounded was to consist onely of seven places ; therefore of the Logarithme so found I take onely the first seven figures rejecting the rest as superfluous , and then at the last the proper Logarithme of 2 , the number given will be found to be 0.301029 , and because the eighth figure being 9 , doth almost carry the value of an unit to the same seventh figure , I adde one thereto , and then the precise Logarithme of 2 will be 0.301030 . And thus as the Logarithme of 2 is made , so may you likewise make the Logarithme of any other number whatsoever : Howbeit , the Logarithmes of some few of the prime numbers being thus discovered , the Logarithmes of many other derivative numbers may be found out afterwards without the trouble of so many continued extractions of the square root , as shall appear by that which followes .

6. § . When of four numbers given , the second exceeds the first as much as the fourth exceeds the third ; the summe of the first and fourth is equal to the summe of the second and third ; and contrarily .

As 8 , 5 : 6 , 3. here 8 exceeds 5 , as much as 6 exceeds 3 : therefore the summe of the first and fourth , namely , of 8 and 3 is equall to the summe of the second and third ; namely of 5 and 6 : from whence necessarily followes this Corollary ; When four numbers are proportionall , the summe of the Logarithmes of the mean numbers is equal to the summe of the Logarithmes of the extreams .

Example .

Let the four proportional numbers be those exprest in the first column of the first Table in this Chapter , viz. 4 , 16 , 32 , 128 , in which Table the Logarithme of 4 under the letter A is 3 , the Logarithme of 16 , 5 , the Logarithme of 32 , 6 ; and the Logarithme of 128 is 8. Now as the summe of 5 and 6 , the Logarithmes of the mean numbers do make 11 , so the summe of 3 and 8 , the Logarithmes of the extreames , do make 11 also .

7. § . When four numbers be proportional , the Logarithme of the first substracted from the summe of the Logarithmes of the second and third , leaveth the Logarithme of the fourth .

Example .

Let the proportion be , as 128 , to 32 ; so is 16 , to a fourth number : here adding 5 and 6 , the Logarithmes of the second and third , the sum is 11 , from which substracting 8 , the Logarithme of 128 , the first proportional , the remainer is 3 , the Logarithm of 4 , the fourth proportional .

8. § . If instead of substracting the aforesaid Logarithme of the first , we adde his complement arithmetical to any number , the totall abating that number , is as much as the remainer would have been .

The complement arithmetical of one number to another , ( as here we take it ) is that , which makes that first number equall to the other ; thus the complement arithmetical of 8 to 10 is 2 , because 8 and 2 are 10. Now then whereas in the example of the last Proposition , substracting 8 from 11 , there remained 3 , if instead of substracting 8 , we adde his complement arithmeticall to 10 , which is 2 , the totall is 13 , from which abating 10 , there remains 3 , as before : both the operations stand thus : As 128 , is to 32 : So is 16 , Logar 8 compl . arithmetical 2 6 6 5 5 The aggreg . of 1.2 . 11 Their aggregate is 13 To 4 , 3
from which abate 10 , there remaines 3 , and the like is to be understood of any other .

The reason is manifest , for whereas we should have abated 8 out of 11 , we did not onely not abate it , but added moreover his complement to 10 , which is 2 , wherefore the total is more then if should be by 8 & 2 , that is by 10 ; wherefore abating 10 from it , we have the Logarithme desired ; which rule , although it be generall , yet we shall seldome have occasion to use any other complements , then such as are the complements of the Logarithmes given either to 10,000000 , or to 20,000000 , the ● complement arithmetical of any Logarithme to either of these numbers , is that which makes the Logarithme given equal to either of them . Thus the complement arithmetical of the Logarithme of 2 viz. 0301030 , is 9698970 , because these two numbers added together , do make 10.000000 , and thus the complement thereof to 20 . 000000is 19698970 : if therefore 0301030 be substracted from 10.000000 , the remainer is his complement arithmetical .

But to finde it readily , you may instead of substracting the Logarithme given from 10.000000 , write the complement of every figure thereof unto 9 , beginning with the first figure toward the left hand , and so on , till you come to the last figure towards the right hand , and thereof set down the residue unto 10. Thus for the complement arithmetical of the aforesaid Logarithme , 0301030 ; I write for 0 , 9 : for 3 , 6 : for 0 , 9 : for 1 , 8 : for 0 , 9 : for 3 again I should write 6 : but because the last place of the Logarithme is a cipher , and that I must write the complement thereof to 10 , instead of 6 I write 7 , and for 0 , 0 : and so have I this number , 9698970 , which is the complement arithmetical of 0301030 , as before .

9. § Every Logarithme hath his proper Characteristick , and the Character or Characteristicall root of every Logarithme is the first figure or figures towards the left hand , distinguished from the rest by a point or comma . Thus the Character of the Logarithmes of every number lesse then 10 is 0 , but the Character of the Logarithme of 10 is 1 ; and so of all other numbers to 100 , but the Character of the Logarithme of 100 is 2 ; and so of the rest to 1000 ; and the Character of the Logarithme of 1000 is 3 ; and so of the rest to 10000 : in brief , the Characteristick of any Logarithme must consist of a unite lesse then the given number consisteth of digits or places , And therefore by the Character of a Logarithme you may know of how many places the absolute number answering to that Logarithme doth consist .

10. § . If one number multiply another , the summe of their Logarithme is equal to the Logarithme of the product .

As let the two numbers multiplied together be 2 , and 2 the products is 4 , I say then that the summe of the Logarithmes of 2 and 2 , or the Logarithme of 2 doubled is equal to the Logarithme of 4 , as here you may see . 2. 0.301030 2. 0.301030 4. 0.602060

Again , let the two numbers multiplied together be 2 , and 4 , the product is 8 , I say then that the summe of the Logarithmes of 2 and 4 is equall to the Logarithme of 8 , as here you may also see , 2. 0.301030 4. 0.602060 8. 0.903090
And so for any other .

The reason is , for that ( by the ground of multiplication ) as unit is in proportion to the multiplier : so is the multiplicand , to the product : therefore ( by the sixth of this Chapter ) the sum of the Logarithmes of a unit , and of the product is equall to the summe of the Logarithmes of the multiplier and multiplicand , but the Logarithme of a unit is 0 , therefore the Logarithme of the product alone is equal to the summe of the Logarithmes of the multiplier and multiplicand .

And by the like reason , it three or more numbers be multiplied together , the summe of all their Logarithmes is equall to the Logarithme of the product of them all .

11. § . If one number divide another , the Logarithme of the Divisor being substracted from the Logari●hme of the Dividend , leaveth the Logarithme of the Quotient .

As let 10 be divided by 2 , the quotient is 5. I say then , if the Logarithme of 2 be substracted from the Logarithme of 10 , there will remain the Logarithme of 5 , as here is to be seen . 10. 1.000000 2. 0.301030 5. 0.698970

For seeing that the quotient multiplied by the divisor produceth the dividend , therefore , by the last proposition , the sum of the Logarithmes of the quotient and of the divisor is equal to the Logarithme of the divi●●● if therefore the Logarithme of the divid●●ol , be substracted from the Logarithme of the divi●●● there remaines the Logarithme of the quotient .

12. § . In any continued rank of numbers Geometrically proportionall from 1 , the Logarithme of any one of them being divided by the denomination of the power which it challengeth in the same rank , the quotient will give you the Logarithme of the root . In the rank of the proportional numbers of the Table ABCD , 2 being the root , or first power ; 4 the square or second power , 8 the cube , or third power , 16 the bi-quadrate or fourth , 32 the fifth power , 64 ▪ the sixth power , &c. I say , the Logarithme of 4 , 8 , 16 , 32 , 64 , or of any of the other subsequent proportionals in that rank , being divided by the demonination of the power that the same proportional claimeth in the same rank , you shall finde in the quotient the Logarithme of 2 the root .

For example .

In the same Table the Logarithme of 4. the square or second power , viz. 3. being given , I demand the Logarithme of 2 , the root : here the denomination of the power that the proportional 4 challengeth in that rank ( being the square or second power ) is 2 , wherefore if 3 , the Logarithme of 4 be divided by 2 , the quotient will be 1 , and there will remain 1 for a fraction ; so that you see it cometh very near in the Logarithmes of but one figure , but if you take it to seven places , as in this table is intended , you shall finde it exactly : for then the Logarithme of 4 will be 0.602060 , and this being divided by 2 , the quotient will be 0.301030 , the Logarithme of 2 the root . So likewise 0.903090 , the Logarithme of 8 the third power , being divided by 3 , leaves 0.301030 in the quotient , as before , and so of any other .

13. § . In any rank of numbers Geometrically proportionall from 1 , the Logarithme of the root being multiplied by the denomination of any of the powers , the product is the Logarithme of the same power .

This Rule is the inverse of the last .

For example .

In the rank produced in the last rule 0.301030 , ( the Logarithme of 2 the root ) being doubled , or multiplied by 2 , produceth 0.602060 , the Logarithme of 4 , the square or second power , and the same Logarithme of 0.301030 , being trebled or multiplied by 3 , produceth 0.903090 , the Logarithme of 8 , the cube or third power , and so of the rest .

The truth of these two last rules may thus be proved . In arithmeticall proportion , when the first term is the common difference of the terms , the last term being divided by the number of the terms , the quotient will give you the first term of the rank : again , in this case , the first term multiplied by the number of the termes produceth the last term . So this rank 3 , 6 , 9 , 12 , 15 , 18 , 21 being propounded , wherein three is both the first term and also the common difference of the terms : I say , 21 , the last term being divided by 7 , the number of the termes , the quotient is 3 , the first term . Contrariwise , 3 the first term multiplied by 7 , produceth 21 , the last term ; and by the like reason , 0.301030 being the first term , and also the common difference of the termes , that is , of the Logarithmes of 4 , 8 , 16 , 32 and 64 , the Logarithme of 2 the first term , being multiplied by 6 , the number of the terms , produceth the Logarithme of 64 , the last term , and the garithme of 64 , the last term , being divided by 6 , leaveth in the quotient the Logarithme of 2 the root .

Hence it also followes , that if you adde the Logarithme of 2 , the common difference of the termes , to the Logarithme of any term , their aggregate shall be the Logarithme of the next term . Thus if I adde 0.301030 , the Logarithme of 2 the root or first term , to 0.903090 , the Logarithme of 8 , the third term , their aggregate is 1 . 204120 , the Logarithme of 16 , the fourth term ; and so of the rest .

14. § . Thus having shewed the construction of the Logarithmetical Tables , the converting of the Table of natural Sines , Tangents , and Secants into artificiall cannot be difficult , the artificiall Sines and Tangents being nothing but the Logarithmes of the naturall .

15. § . In the conversion whereof Mr. Briggs in his Trigonometria Britannica , thought fit to make the Radius of his natural Canon to consist of 16 places , and to confine his artificiall to the Radius of eleven , whose Characteristick is 10 , but the Characteristick of the rest of the Sines till you come to the sine of 5 degrees and 73 centesmes is 9 , and from thence to 57 centesmes , the Characteristick is 8 , and from thence 7 , till you come to 5 centensmes , and from thence but 6 , to the beginning of the Canon . The Characteristick still decreasing in the same proportion with the naturall numbers , and the number of the places in the naturall Canon , do therefore exceed the Characteristick in the artificiall , that so the artificiall numbers might be the more exact .

16. § . In the Canon herewith printed , the Characteristick in the artificiall numbers doth exceed the number of places in the naturall , which is not done so much out of necessity as conveniency , for the artificiall numbers in this Canon might in all respects have been made answerable to the natural , and so the Characteristick of the Radius , or whole Sine would have been seven , the Characterick of the first minute 3 , but thus the subduction of the Radius would not have been so ready as now it is , nor yet the Canon it self altogether so exact , and therefore as Master Briggs confined the Radius of his artificiall Canon to eleven places for conveniency sake , though he made the Logarithmes to the Radius of sixteen : so here for conveniency and exactnesse both , the same Characterick is here continued , though the naturall numbers do not require it , if any think this a defect , I answer , that it could not well be avoided here , but may be supplied by Master Briggs his Canon , of which this is an abbreviation : and yet even here there is so small a difference between the Logarithmes of these naturall numbers , and the Logarithmes in the Canon , that any one may well perceive the one to be nothing else but the Logarithme of the other , if they do but change the Characteristick .

And hence we may gather , that the making of this Canon is not so difficult as laborious , and the labour thereof may be much abridged by this Proposition following .

17. § . The Sine of an arch and half the Radius are mean proportionals between the Sine of halfe that arch , and the Sine complement of the same half .

In the annexed Diagram , let DE be the sine of 56 degrees , BC the sine of 28 , AC the sine complement thereof , that is , of 62. DB the subtense of 56. CF perpendicular to the Radius , then are ABC and ACF like triangles , by the 22 of the second , and their sides proportional that is , AB AB BC AO AC DE CF CF

And therefore the oblongs of BC×AC , AO × DE , and AB × CF are equal , and the sides of equal rectangled figures reciprocally proportional , that is , as BC , AO ∷ DE , AC . or as AO , BC ∷ AC , DE.

If therefore you multiply AO , the half Radius , by DE , the sine of the arch given , and divide the product by BC , the sine of half the arch given , the quotient shall be AC , the sine complement of half the given arch .

Or if you multiply BC , the sine of an arch by AC , the sine complement of the same arch , and divide the product by AO , the half Radius , the quotient shall be DE , the sine of the double arch . And therefore the sines of 45 degrees being given , or the Logarithmes of those sines , the rest may be found by the rule of proportion . For illustration sake we will adde an example in naturall and artificiall numbers .

Naturall , As BC 28 , 46947 Is to AO 30 ; 50000 So is DE 56 , 82903 To AC 62 : 88294

Logarith . As BC 28 , 9.671609 Is to AO 30 ; 9.698970 So is DE 56 , 9.918574 To AC 62. 9.945935

18. § . The composition of the naturall Tangents and Secants , by the first and second of the fourth are thus to be made .

1. As the sine of the complement , is to the sine of an arch : So is the Radius , to the tangent of that arch .

2. As the sine of the complement , is to the Radius : so is the Radius , to the Secant of that arch ; and by the same rules may be also made the artificiall ; but with more ease , as by example it will appear .

Let the tangent of 30 degrees be sought . Logarith . As the co-sine of 60 degrees , 9.937531 Is to the sine of 30 ; 9.698970 So is the Radius , 10.000000 To the tangent of 30 : 9.761439

And thus having made the artificiall Tangents of 45 degrees , the other 45 are but the arithmeticall complements of the former , taken as hath been shewed in the eighth rule of the fifth Chapter .

Again , let the secant of 30 degrees be sought . As the co-sine of 60 degrees , 9.937531 Is to the Radius , 10.000000 So is the Radius , 10.000000 20.000000 To the secant of 30 : 10.062469

And thus the Radius being added to the arithmetical complement of the sine of an arch , their aggregate is the secant of the complement of that arch . And this is sufficient for the construction of the naturall and artificiall Canon . How to finde the Sine , Tangent or Secant of any arch given in the Canon herewith printed , shall be shewen in the Preface thereunto : here followeth the use of the naturall and artificiall numbers both ; first , in the resolving any Triangle , and then in Astronomy , Dialling , and Navigation .

CHAP. VI. The use of the Tables of natural and artificial Sines , and Tangents , and the Table of Logarithmes . In the Dimension
I. Of plain right angled Triangles .

THe measuring or resolving of Triangles is the finding out of the unknown sides or angles thereof by three things known , whether angles , or sides , or both ; and this by the help of that precious gemme in Arithmetick , for the excellency thereof called the Golden Rule , ( which teacheth of four numbers proportional one to another , any three of them being given , to finde out a fourth ) and also of these Tables aforesaid .

Of Triangles , as hath been said , there are two sorts ; plain and sphericall . A triangle upon a plain is right lined , upon the Sphere circular . Right lined Triangles are right angled or oblique .

A right angled , right-lined Triangle we speak of first , whose sides then related to a circle are inscribed totally or partially .

Totally , if the side subtending the right angle be made the Radius of a Circle , and then all the sides are called Sines , as in the Triangle ABC .

Partially , if either of the sides adjacent to the right angle be made the Radius of a circle , and then one side of the Triangle is the Radius or whole Sine , the shorter of the other two sides is a Tangent , and the longest a secant . Now according as the right angled Triangle is supposed , whether to be totally or but partially inscribed in a circle ; so is the trouble of finding the parts unknown more or lesse , whether sides or angles ; for if the triangle be supposed to be totally inscribed in a circle , we are in the solution thereof confined to the Table of Sines onely , because all the sides of such a triangle are sines : but if the triangle be supposed to be but partially inscribed in a circle , we are left at liberty to use the Table of Sines , Tangents , or Secants , as we shall finde to be most convenient for the work .

In a right angled plain Triangle , either all the angles with one side are given , and the other two sides are demanded , I say , all the angles , because one of the acute angles being given , the other is given also by con●●quence .

Or else two sides with one angle , that is , the right angle are given and the other two angles with the third side are demanded . In both which cases this Axiome following is well nigh sufficient .

The first AXIOME .

In all plain Triangles , the sides are in portion one to another , as are the sines o● the angles opposite to those sides .

As in the triangle ABC , the side AB is in proportion to the side AC , as the sine of the angle at B is in proportion to the sine of the angle at c and so of the rest .

Demonstration .

The circle ADF being circumscribed about the Triangle ABC , the side AB is made the chord or subtense of the angle ACB , that is , of the arch AB , which is opposite to the angle ACB . The side AC is made the subtense of the angle ABC ; and the side BC is made the subtense of the angle BAC , and are the double measures thereof , by the 19 Theorem of the second Chapter : therefore the side AB is in proportion to the side AC , as the subtense of the angle ACB is in proportion to the subtense of the angle ABC , but half the subtense of the angle ACB is the sine of the angle ACB , and half the subtense of the angle ABC is the sine of the angle ABC ; now as the whole is to the whole ; so is the half , to the half . Therefore in all plain Triangles , &c.

The first Consectary .

The angles of a plain triangle , and one side being given , the reason of the other sides is also given .

The second Consectary .

Two sides of a plain Triangle , with an angle opposite to one of them being given , the reason of the other angles is also given , by this proportion .

If the side of a Triangle be required , put the angle opposite to the given side in the first place .

If an angle be sought , put the side opposite to the given angle in the first place .

For the better understanding whereof we will adde an example , and to distinguish the sides of the Triangle , we call the side subtending the right angle , the Hypothenusall , and of the other two the one is called the perpendicular , and the other the base , at pleasure , but most commonly the shortest is called the perpendicular , and the longer the base . As in the former figure , the side BC is the Hypothenusal , AC the base , and AB the perpendicular .

Now then in the Triangle ABC , let there be given the base AC 768 paces , and the angle CBA 67 degrees , 40 minutes , ( then the angle ACB is also known , it being the complement of the other ) and let there be required the perpendicular : because it is a side that is required , I put the angle opposite to the given side in the first place , and then the proportion is :

As the sine of the angle at the perpendicular , is in proportion to the base : So is the sine of the angle at the base , to the perpendicular .

Now if you work by the natural Sines , you must multiply the second term given , by the third , and divide the product by the first , and then the quotient is the fourth term required , and the whole work will stand thus : As sine the ang. at the perpend . ABC 67 degrees 40 minutes 9232102 Is in proportion to the base AC ; 768 So is sine the angle at the base , ACB 22 degrees 60 minutes 3842953 30743624 23057718 26900671 The product of the 2d . & 3d. 2951387904
Which divided by 9232102 , the first term given , leaveth in the quotient 320 ferè .

But if you work by the artificiall sines , that is , by the Logarithmes of the natural , then you must adde together the Logarithmes of the second and third terms ; given , and from their aggregate substract the Logarithme of the first , and what remaineth will be the Logarithme of the fourth proportional , whether side or angle : the work standeth thus . As sine the angle at the perpendicular B 67 deg. 40 min. 9.9653006 Is in proportion to the base AB 768 ; 2.8853612 So is sine the angle at the base C 22 degree , 60 minutes . 9.5846651 The aggregate of the 2d . & 3d. 12.4700263
From which I substract the first , 9.9653006 , and the remainer which is 2.5047257 , is the Logarithme of the fourth : wherefore looking in the Table for the absolute number answering thereunto , I finde the nearest to be 320 , which is the length of the perpendicular , as before .

The operation it self may yet be performed with more ease , if instead of the Logarithme of the first proportional , we take his complement arithmetical , as hath been shewed in the eighth raile of the fifth Chapter : for then the totall of the arithmaticall complement , and the Logarithme of the second and third proportionals , abating Radius , is the Logarithme of the fourth proportionall , as doth appear in this example . As sine of ABC 67 de. 40m . co. ar . 0.0346994 To the base AC 768 ; 2.8853612 So the sine of ACB 22 de. 60m . 9.5846651 To the perpendic . AB 320 ferè 2.5047257

Thus having sufficiently explained the operation in this first example , we shall be briefer in the rest that follow , understanding the like in them also .

In this manner may all the cases of a plain right angled Triangle be resolved by this proportion , except it be when the base and perpendicular with their contained angle ( that is the right angle ) is given , to finde either an angle or the third side ; in this case therefore we must have recourse to the 17 th . Theorem of the second Chapter , by help whereof the Hypothenusall may be found in this manner : square the sides , and from the aggregate of their squares extract the square root , that square root shall be the length of the Hypothenusal . For example . Let the base be four paces , and the perpendicular 3 , the square of the base is 16 , the square of the perpendicular is 9 , the summe of these two squares is 25 , the square root of this summe is 5 paces , and that is the length of the hypo●enusal ; and this hypothenusal being thus found , the angles also may be ●ound , as before .

Nor are we tied to this way of finding the hypothenusal , unlesse we confine our selves to the Tables of Sines onely ; if we would make use of the Tables of Tangents or Secants , the hypothenusal may not onely be found with more ease , but all the cases of a right angled plain triangle may be also found several wayes , by the help of this Axiome following .

The second AXIOME .

In a plain right angled triangle , any of the three sides may be made the Radius of a circle , and the other sides will be as Sines , Tangents , or Secants . And what proportion the side put as Radius hath unto Radius ; the same proportion hath the other sides unto the Sines , Tangents , or Secants of the opposite angles by them represented .

If you make the hypothenusal Radius , the triangle will be totally inscribed in the circle , and consequently the other two sides shall represent the sines of their opposite angles , that is , the base shall represent the sine of the angle at the perpendicular , and the perpendicular shall represent the sine of the angle at the base , as in the preceding Diagram .

If you make the base Radius , the triangle will be but partially inscribed in the circle , and the other two sides shall be one of them a tangent , and the other a secant . Thus in the first Diagram of this Chapter , the base BD is made the Radius of the circle , the perpendicular D● is the tangent of the angle at the base , and B● is the secant of the same angle .

If you make the perpendicular Radius , the triangle will be but partially inscribed in the circle , as before , and the other two sides will be also the one a tangent and the other a secant . As in this example , the perpendicular AB is made the Radius of the circle , the base AC is the tangent of the angle at the perpendicular , and the hypothenusal BC is the secant of the same angle .

Hence it followes , that if you make AB the Radius , the base and perpendicular being given , the angle at the perpendicular may be found by this proportion .

As the perpendicular , is in proportion to Radius : So is the base , to tang●nt of the angle at the perpendicular ; for the perpendicular being made the Radius of the circle , it must of necessity bear the same proportion unto Radius , as the hypothenusal doth , when that is made the Radius of the circle : and if the perpendicular be the Radius , the base must needs represent the tangent of the angle at the perpendicular .

And the angle at the perpendicular be-being thus found , the hypothenusal may be found by the first Axiome . For ,

As the sine of the angle at the perpendicular , is in proportion to his opposite side the base ; So is Radius , to his opposite side the hypothenusal : and thus you see that the hypothenusal may be found without the trouble of squaring the sides , and thence extracting the square root . And hence also all the cases of a right angled plain triangle may be resolved several wayes : that is to say ,

1. In a plain right angled triangle : the angles and one side being given , every of the other sides is given , by a threefold proportion , that is , as you shall put for the Radius , either the side subtending the right angle , or the greater or lesser side including the right angle .

2. Any of the two sides being given , either of the acute angles is given by a double proportion , that is , as you shall put either this or that side for the Radius : to make this clear , we will first set down the grounds or reasons for varying of the termes of proportion : and then the proportions themselves in every case , according to all the variations .

The reasons for varying of the termes of proportion are chiefly three .

The first reason is , because the Radius of a circle doth bear a threefold proportion to a sine , tangent , or secant ; and contrariwise , a sine , tangent , or secant hath a threefold proportion to Radius , by the second Axiome of this Chapter .

For As sine BC , to Rad. AC in the 1. triangle So Rad. BC , to secant AC in the 3d. tri . So tang . BC , to secant AC in the 2d . tri . & contra

Again , As tang . BC , to Rad. AB in the 2d . triang . So Rad. BC , to tang . AB in the 3d. trian . So sine BC , to sine BA in the first triang . & contra

Lastly , As secant AC , to Rad. BC in the 3d. tri . So Rad. AC , to sine BC in the first trian So secant AC , to tang . BC in the 2d . tri . & contra

Hence then As the sine of an arch or ang. is to Rad. So Rad. to the secant comp. of that arch & so is the tang . of that arch , to his sec. & contr .

Also As the tang . of an arch or ang. is to Rad. So is Rad. to the tangent compl . thereof . And so is the sine thereof , to the sine of its complement . & contra .

Lastly , As the secant of an arch or ang. to Rad. So is Radius , to the sine compl . thereof And so is secant complement to tangent complement thereof . & contra .

Example .

Let there be given the angle at the perpendicular 41 degrees 60 minutes , and the base 768 paces , to finde the perpendicular .

First , by the natural numbers , As the secant of BAC 41 d. 60m . 13372593 Is to Radius , 10000000 So is the base AB 768 To the perpendicular BC ●74 574

By the Artificiall . As the secant of BAC 41.60 . 10.1262157 Is to Radius ; 10.0000000 So is the base 768 , 2.8853612 12.8853612 To the perpendicular 574 : 2.7591455

Secondly , by the naturall numbers . As the Radius , 10000000 To the co-sine of BAC 41.60 . 7477981 So is the base AB 768 To the perpendicular BC 574

By the Artificiall . As the Radius 10.0000000 To the co-sine of BAC 41.60 . 9.8737843 So is the base AB 768 , 2.8853612 To the perpendicular BC 574 : 2.7591455

Thirdly , by the natural numbers . As the co-secant of BAC 41.60 . 15061915 Is to the co-tang . of BAC 41.60 . 11263271 So is the base AB 768 To the perpendicular BC 574

By the artificiall . As the co-secant of BAC 41.60 . 10. 1778802 Is to the co-tang . of BAC 41.60 10.0516645 So is the base AB 768 2.8853612 To the perpendicular BC 574 2.7591455

COROLLARY .

Hence it is evident , that Radius is a mean proportional between the sine of an arch , and the secant complement of the same arch ; also between the tangent of an arch , and the tangent of the complement of the same arch .

The second Reason .

The sines of several arches , and the secants of their complements are reciprocally proportional , that is ,

As the sine of an arch or angle , is to the sine of another arch or angle : So is the secant of the complement of that other , to the co-secant of the former .

For by the foregoing Corollary , Radius is the mean proportional between the sine of any arch and the co-secant of the same arch .

Therefore , whatsoever sine is multiplied by the secant of the complement , is equall to the square of Radius ; so that all rectangles made of the sines of arches and of the secants of their complements are equal one to another ; but equall rectangles have their sides reciprocally pro portional , by the tenth Theorem of the second Chapter . Therefore the sines of several arches , &c.

The third Reason .

The tangents of severall arches , and the tangents of their complements are reciprocally proportional , that is ,

As the tangent of an arch or angle , is to the tangent of another arch or angle , so is the co-tangent of that other , to the co-tangent of the former .

For by the foregoing Corollary , Radius is the mean proportionall between the tangent of every arch and the tangent of his complement .

Therefore the Rectangle made of any tangent , and of the tangent of his complement , is equall to the square of Radius : so that all rectangles made of the tangents of arches , and of the tangents of their complements are equall one to another , but equal rectangles , &c. as before .

To these three reasons a fourth may be added . For in the rule of proportion ; wherein there are alwayes four termes , three given , the fourth demanded : It is all one , whether of the two middle terms is put in the second or third place .

For it is all one , whether I shall say ;

As 2 , to 4 ; so 5 , to 10 : or say , as 2 , to 5 ; so 4 , to 10 : and from hence every example in any triangle may be varied , and thus you see the reasons of varying the termes of proportion , we come now to shew you the various proportions themselves of the severall Cases in right angled plain triangles .

Right angled plain triangles may be distinguished into seven Cases ; whereof those in which a side is required , viz. three , may be found by a triple proportion ; and those in which an angle is required , viz. three , may be found by a double proportion .

CASE 1. The angles and base given , to finde the perpendicular .

First , As sine the angle at the perpendicular , is to the base : so is sine the angle at the base , to the perpendicular .

Or secondly , thus : As Radius , to the base ; so tangent the angle at the base , to the perpendicular .

Or thirdly , thus : As the tangent of the angle at the perpendicular , is to the base : so is Radius , to the perpendicular .

CASE 2. The angles and base given , to finde the hypothenusal .

First , As the sine of the angle at the perpendicular , is to the base ; so is Radius , to the hypothenusal .

Or secondly thus : As Radius , is to the base ; so the secant of the angle at the base , to the hypothenusal .

Or thirdly , thus : As the tangent of the angle at the perpendicular , is to the base : so is the secant of the same angle in proportion to the hypothenusal .

CASE 3. The angles and hypothenusal given , to finde the base .

First , As Radius , to the hypothenusal : so the sine of the angle at the perpendicular , to the base . Or secondly , thus :

As the secant of the angle at the base , to the hypothenusal : so is Radius , to the base .

Or thirdly , thus : As the secant of the angle at the perpendicular , to the hypothenusal : so the tangent of the same angle , to the base .

CASE 4. The base and perpendicular given , to finde an angle .

First , As the base , to Radius : so the perpendicular , to the tangent of the angle at the base . Or secondly , thus :

As the perpendicular , is to Radius : so the base , to the tangent of the angle at the perpendicular .

CASE 5. The base and hypothenusal given , to finde an angle .

1. As the hypothenusal , is to Radius : so is the base , to the sine of the angle at the perpendicular .

Or secondly thus , As the base is to Radius ; so is the hypothenusal , to the secant of the angle at the base .

CASE 6. The base and perpendicular given , to finde the hypothenusal .

First , finde the angle at the perpendicular , by the fourth Case : Then ,

As the sine of the angle at the perpendicular , is to the base : so is Radius , to the hypothenusal .

Otherwise by the Logarithmes of absolute numbers .

From the doubled Logarithme of the greater side , whether base or perpendicular , substract the Logarithme of the lesse , and to the absolute number answering to the difference of the Logarithmes adde the lesse , the half summe of the Logarithmes of the summe , and the lesse side , is the Logarithme of the hypothenusal inquired .

The Illustration Arithmetical .

Let the base be 768 , and the perpendicular 320. The Logarithme of 768 is 2.8853612 This Logarithme doubled is 5.7707224 From which substr . the Log. of 320 , 2.5051500 The remain . is the Log. of 1843 : 3.2655724
To which the lesser side being added 320 , their aggregate is 2163. The Logarithme of 2163 is 3.3350565 The Logarithme of 320 is 2.5051500 The summe is 5.8402065 The half sum is the Log. of 832. 2.9201032
which is the length of the hypothenusal inquired .

CASE 7. The base and hypothenusal given , to finde the perpendicular .

The resolve this Probleme by the Canon , there is required a double operation : First , by the 5 Case , finde an angle . Secondly , by the first Case , finde the perpendicular .

But Mr. Briggs resolves this Case more readily , by the Logarithmes of the absolute numbers , Briggs Arithmetica Logarith . cap. 17.

Take the Logarithmes of the summe and difference of the hypothenusal and side given , half the summe of those two Logarithmes , is the Logarithme of the perpendicular , or side inquired .

As let the hypothen . be 832 The side given 768 Logarith . The summe is 1600 3.2041200 The difference is 64 1.8061800 The summe is , 5.0103000 The half sum is the Logarith . of 320 the side inquired . 2.5051500

The two Axiomes following are true in all plain triangles , but are chiefly intended for the oblique angled ; which now we come to handle .

II. Of plain oblique angled Triangles .

In a plain oblique angled triangle , there are four varieties .

1. All the angles may be given , ( for when two are given , the third is given by consequence ) and one side , and the other two sides demanded .

2. Two sides with an angle opposite to one of them may be given , and the angle opposite to the other , with the third side are demanded . In both which cases the first Axiome is fully sufficient .

3. Two sides with an angle comprehended by them may be given , and the other two angles with the third side demanded . For the solution whereof we will lay down this Axiome following .

The third AXIOME .

As the summe of the two sides , is to their difference : so is the tangent of half the summe of the opposite angles , to the tangent of half the difference .

Let ABC be the oblique angled triangle , in which let the side AB be continued to H , and let the line of continuation BH be made equall to BC , and BK equal to AB ; then is AH the summe of the sides , AB , BC , and KH is their difference , now if you draw the lines BD and KG parallel unto AC , then shall the angle CBH be equal to the two angles of the triangle given ACB and CAB , because the angle CBA common to both is their complement to a Semicircle , and DB being parallel to CA , the angle DBH shall be equall to the angle CAB , and the angle DBC equall to the angle ACB , if therefore you let fall the perpendicular BE , and draw the periphery MEL , the right line CE shall be the tangent of half the summe of the angles ACB and CAB , it being the tangent of half the angle CBH .

Again , if you make E● equal to DE , and draw the right line FB , then shall the angle DBF be the difference between the angles CBD and DBH , or between the angles ACB and CAB , and DE the tangent of half the difference . And because the right sines AC , DB , and KG are parallel , and CD , DG , and FH are equall , and DF equal to GH , and the triangles ACH and KGH are like , and therefore ; As AH is in proportion to HK : so is CH , to HG : or as AH , the summe of the sides , is in proportion to HK , their difference : so is CE the tangent of the half summe of the angles ACB and CAB , to DE , the tangent of half their difference .

Consectary .

Hence it followes , that in a plain oblique angled Triangle ; if two sides and the angle comprehended by them be given , the other two angles and the third side are also given .

As in the triangle ABC , having the sides AC 189 , and AB 156 , whose summe is 345 , and difference 33 , with the angle BAC 22 degrees , 60 minutes , to finde the angle ABC or ACB . The proportion is As the sum of the sides given 345 , 2.5378190 Is to their difference 33 , 1.5185139 So the tangent of half the angles at B & C 78de . 70m . 10.6993616 To the tangent of half their difference 25 degr. 58 minutes 9.6800565

Which being added to the half sum 78 degrees , 70 minutes , the obtuse angle at B , is 104 degrees , 28 minutes ; and substracted from the half summe , it leaveth 53 degrees , 12 minutes for the quantity of the acute angle ACB .

Then to finde the third side BC , the proportion , by the first Axiome , is ,

As the sine of the angle at ● , is in proportion to his opposite side AB ; so is the sine of the angle at A , to his opposite side BC.

4. And lastly , all the three sides may be given , and the angles may be demanded ; for the solution whereof we will lay down this Axiome .

The fourth AXIOME .

As the base , is to the summe of the sides : So is the difference of the sides , to the difference of the segments of the base .

Let BCD be the triangle , CD the base , BD the shortest side ; upon the point B describe the circle ADFH , making BD the Radius thereof , let the side BC be produced to A , then is CA the summe of the sides , because BA and BD are equal , by the work , CH is the difference of the sides , CF the difference of the segments of the base .

Now if you draw the right lines AF and HD , the triangles CHD and CAF shall be equiangled , because of their common angle ACF or HCD , and their equal angles CAF and HDC , which are equall , because the arch HF is the double measure to them both ; and therefore , as CD , to CA ; so is CH , to CF , which was to be proved .

Consectary .

Therefore the three sides of a plain oblique angled triangle being given , the reason of the angles is also given .

For first , the obliquangled triangle may be resolved into two right angled triangles , by this Axiome , and then the right angled triangles may be resolved by the first Axiome .

As in the plain oblique angled triangle , BCD , let the three sides be given , BD 189 paces , BC 156 paces , and DC 75 paces , and let the angle CBA be required .

First , by this Axiome , I resolve it into two right angled triangles ; thus : As the true base BD 189 co. ar . 7.7235382 Is to the sum of BC & DC 231 2.3636120 So the difference of BC & DC 81 1.9084850 To the alternate base BG 99 1.9956352

As the the hypothenusal BC , is to Radius : So is the base AB 144 , to the sine of the angle at the perpendicular , whose complement is the angle at the base inquired .

In like manner may be found the angle at D , and then the angle BCD is found by consequence , being the complement of the other two to two right angles or 180 degrees .

CHAP. VII . Of Sphericall Triangles .

A Sphericall Triangle is a figure described upon a Sphericall or round superficies , consisting of three arches of the greatest circles that can be described upon it , every one being lesse then a Semicircle .

2. The greatest circles of a round or Spherical superficies are those which divide the whole Sphere equally into two Hemispheres , and are every where distant from their own centers by a Quadrant , or fourth part of a great circle .

3. A great circle of the Sphere passing through the poles or centers of another great circle , cut one another at right angles .

4. A spherical angle is measured by the arch of a great circle described from the angular point betwixt the sides of the triangle , those sides being continued to quadrants .

5. The sides of a Spherical triangle may be turned into angles , and the angles into sides , the complements of the greatest side or greatest angle to a Semicircle , being taken in each conversion .

It will be necessary to demonstrate this , which is of so frequent use in Trigonometry . In the annexed Diagram let ABC be a sphericall triangle , obtuse angled at B , let DE be the measure of the angle at A. Let FG be the measure of the acute angle at B , ( which is the complement of the obtuse angle B , being the greatest angle in the given triangle ) and let HI be the measure of the angle at C , KL is equal to the arch DE , because KD and LE are Quadrants , and their common complement is LD . LM is equall to the arch FG , because LG and FM are Quadrants , and their common complement is LF . KM is equal to the arch HI , because KI and MH are Quadrants , and their common complement is KH . Therefore the sides of the triangle KLM are equal to the angles of the triangle ABC , taking for the greatest angle ABC , the complement thereof FBG .

And by the like reason it may be demonstrated , that the sides of the triangle ABC are equal to the angles of the triangle KLM . For the side AC is equall to the arch DI , being the measure of the angle DKI , which is the complement of the obtuse angle MKL. The side AB is equall to the arch OP , being the measure of the angle MLK . And lastly , the side BC is equal to the arch FH , being the measure of the angle LMK , for AD and CI are Quadrants : so are AP and OB , BF and CH. And CD , AO , and CF are the common complements of two of those arches . Therefore the sides of a spherical triangle may be changed into angles , and the angles into sides , which was to be demonstrated .

6. The three sides of any spherical triangle are lesse then two Semicircles .

7. The three angles of a spherical triangle are greater then two right angles , and therefore two angles being known , the third is not known by consequence , as in plain triangles .

8. If a spherical triangle have one or more right angles , it is called a right angled spherical triangle .

9. If a spherical triangle have one or more of his sides quadrants , it is called a quadrantal triangle .

10. If it have neither right angle , nor any side a quadrant , it is called an oblique spherical triangle .

11. Two oblique angles of a spherical triangle are either of them of the same kinde of which their opposite sides are .

12. If any angle of a triangle be neerer to a quadrant then his opposite side : two sides of that triangle shall be of one kinde , and the third lesse then a quadrant .

13. But if any side of a triangle be nearer to a quadrant then his opposite angle , two angles of that triangle shall be of one kinde , and the third greater then a quadrant .

14. If a spherical triangle be both right angled and quadrantal , the sides thereof are equall to the opposite angles .

For if it have three right angles , the three sides are quadrants , if it have two right angles , the two sides subtending them are quadrants ; if it have one right angle , and one side a quadrant , it hath two right angles and two quadrantal sides , as is evident by the third Proposition . But if two sides be quadrants , the third measureth their contained angle , by the fourth proposition . Therefore for the solution of these kindes of triangles , there needs no further rule : But for the solution of right angled , quadrantall , and oblique spherical triangles there are other affections proper to them , which are necessary to be known as well as these general affections common to all spherical triangles . The affections proper to right angled and quadrantal triangles we will speak of first .

CHAP. VIII . Of the affections of right angled Sphericall Triangles .

IN all spherical rectangled Triangles , having the same acute angle at the base : The sines of the hypothenusals are proportional to the sines of their perpēdiculars . As in the annexed diagram , let ADB represent a spherical triangle , right angled at B : so that AD is the sine of the hypothenusal , AB the sine of the base , and DB is the perpendicular . Then is DAB the angle at the base , and IH the sine , and LM the tangent thereof : Also DF is the sine of the perpendicular DB , and KB is the tangent thereof : I say then , As AD , is to FD : So is AI , to IH , by the 16 th . Theoreme of the second Chapter .

And because it is all one , whether of the mean proportionals be put in the second place ; therefore I may say : As AD , the sine of the hypothenusal , is in proportion to AI Radius : So is FD , the sine of the perpendicular , to IH the sine of the angle at the base .

2. In all rectangled spherical triangles , having the same acute angle at the base . The sines of the bases , and the tangents of the perpendiculars are proportional .

For as AB , to KB ; so is AM , to ML , by the 16 th . Theorem of the second Chapter : or which is all one ; As AB , the sine of the base , is in proportion to AM Radius : so is BK , the tangent of the perpendicular , to ML , the tangent of the angle at the base .

3. If ● circles of the Sphere be so ordered , that the first intersect the second , the second the third , the third the fourth , the fourth the fift , and the fift the fift at right angles : the right angled triangles made by their intersections do all consist of the same circular parts .

As in this Scheme , let IGAB be the first circle , BLF the second , FEC the third , GAD the fourth , HLEI the fift .

Then do these five circles retain the conditions required . The first intersecting the second in B , the second the third in F , the third the fourth in C , the fourth the fift in H , the fift the angle , we mark or note intersections at B , F , ● to a quadrant . As angles ; therefore I say●nt as the completriangles made by the inte●●or AD we write circles ; namely , ABD , D● write compl . EGI , and GCA do all co●●d AB besame circular parts ; for the circu●●● 〈◊〉 in every of these triangles are , as h●●d by peareth .

In ABD are AB BD c BDA c AD c DA● DHL c HLD c LD c LDH DH HL LFE cō ELF LF FE cō FEL c EL EGI IG cō IGE c GE cō GEI IE GCA c GA c AGC GC CA c CAG

Where you may observe , that the side AB in the first triangle is equal to compl . HLD in the second , or compl . ELF in the third , or IG in the fourth , or com . GA in the fift ; and so of the rest .

To expresse this more plainly , AB in the first triangle is the complement of the angle HLD in the second , or the complement of the angle ELF in the third , or the side IG in the fourth , or the complement of the hypothenusal GA in the fift . And from these premises is deduced this universall proposition .

4. The sine of the middle part and Radius are reciprocally proportional , with the tangents of the extreams conjunct , and with the co-sines of the extreams disjunct .

Namely ; As the Radius , to the tangent of one of the extreames conjoyned : so is tangent of the other extream conjoyned , to the sine of the middle part .

And also ; As the Radius , to the co-sine of one of the extreams dis-joyned : so the co-sine of the other extream dis-joyned , to the sine of the middle part .

Therefore if the middle part be sought , the Radius must be in the first place , if either of the extreams ; the other extream must be in the first place .

For the better Demonstration hereof , it is first to be understood , that a right angled Spherical Triangle hath five parts besides the right angle . As the triangle ABD in the former Diagram , right angled at B , hath first , the side AB : secondly , the angle at A : thirdly , the hypothenusal AD : fourthly , the angle ADB : fifthly , the side DB. Three of these parts which are farthest from the right angle , we mark or no●e by their complements to a quadrant . As the angle BAD we account as the complement to the same angle . For AD we write comp. AD , and for ADB we write compl . ADB . But the two sides DB and AB being next to the right angle , 〈…〉 are not noted by their complements . Of these five parts , two are alwayes given to finde a third , and of these three one is in the middle , and the other two are extreams either adjacent to that middle one , or opposite to it . If the parts given and required are all conjoyned together , the middle is the middle part conjunct , and the extreams the extream parts conjunct . If again any of the parts given or required be dis-joyned , that which stands by it self is the middle part dis-joyned , and the extreames are extream parts dis-joyned . Thus , if there were given in the triangle ABD , the side AB , the angle at A , to finde the hypothenusal AD , there the angle at A is in the middle , and the sides AD and AB are adjacent to it ; and therefore the middle part is called the middle conjunct , and the extreames are the extreames conjunct ; but if there were given the side AB , the hypothenusal AD , to finde the angle at D , here AB is the middle part dis-junct , because it is dis-joyned from the side AD by the angle at A , and from the angle at D by the side DB , for the right angle is not reckoned among the circular parts , and here the extreams are extreams dis-junct .

These things premised , we come now to demonstrate the proposition it self , consisting of two parts : first , we will prove , that the sine of the middle part and Radius are proportional with the tangents of the extreams conjunct .

The middle part is either one of the sides , or one of the oblique angles , or the hypothenusal .

CASE 1.

Let the middle part be a side , as in the right angled spherical triangle ABD of the last diagram , let the perpendicular AB be the middle part , the base DB and comp. A the extreame conjunct , then I say , that the rectangle of the sine of AB and Radius is equal to the rectangle of the tangent of DB , and the tangent of the complement of DAB : for , by the second proposition of this Chapter , As the sine of AB , is in proportion to Radius : so is the tangent of DB , to the tangent of the angle at A. Therefore if you put the third term in the second place , it will be , as the sine of AB , to the tangent of DB : so is the Radius , to the tangent of the angle at A. But Radius is a mean proportional between the tangent of an arch , and the tangent of the complement of the same arch , by the Corollary of the first reason of the second Axiome of plain Triangles : and therefore as Radius , is to the tangent of the angle at A ; so is the tangent complement of the same angle at A unto Radius : Therefore as the sine of AB is in proportion to the tangent of DB ; so is the co-tangent of the angle at A , to Radius : and therefore the rectangle of AB ▪ Radius , is equall to the rectangle of the tangent of DB , and the co-tangent of the angle at A.

CASE 2.

Let the middle part be an angle , as in the triangle DHL of the former Diagram , and let compl . HLD be the middle part , HL and compl . LD the extreames conjunct ; then I say , that the rectangle made of the co-sine of HLD and Radius , is equal to the rectangle of the tangent of HL and the co●tangent of LD . For ▪ by the third proposition of this Chapter , compl . HLD is equal to AB , and compl . LD to DB , and HL to compl . DAB ; and here we have proved before , that the rectangle of the sine of AB and Radius , is equal to the rectangle of the tangent of DB , and the co-tangent of the angle at A ; therefore also the rectangle of the co-sine of HLD and Radius , is equal to the rectangle of the co-tangent of LD , and the 〈◊〉 tangent of HL.

CASE 3.

Let the middle part be the hypothenusal , as in the triangle GCA , let compl . AG be the middle part , compl . AGC , and compl . CAG the extreams conjunct ; then I say , that the rectangle of the co-sine of ▪ AG and Radius , is equal to the rectangle of the co-tangent of AGC , and the co-tangent of CAG : for we have proved before , that the rectangle of the sine of AB and Radius is equal to the rectangle of the tangent of DB and the co-tangent of DAB , but , by the third proposition of this Chapter , compl . AG is equal to AB , compl . AGC to DB , and compl . CAG to compl , DAB ; therefore also the rectangle of the co-sine of AG and Radius , is equal to the rectangle of the co-tangent of AGC and the co-tangent of CAG , which was to be proved .

It is further to be proved , that the sine of the middle part and Radius are proportional with the co-sines of the extreams dis-junct . Here also the middle part is either one of the sides , or the hypothenusal , or one of the oblique angles .

CASE 1.

Let the middle part be a side : as in the triangle ABD , let DB be the middle part , compl . AD and compl . A the opposite extreams : then I say , that the rectangle of the sine of BD and Radius is equal to the rectangle of the sine of AD , and the sine of the angle at A ; for , by the first proposition of this Chapter , as the sine of AD , is to Radius ; so is the sine of DB , to the sine of the angle at A. Therefore , the rectangle of the sine of DB and Radius , is equal to the rectangle of the sine of AD and the sine of the angle at A.

CASE 2.

Let the hypothenusal be the middle part ; as in the triangle DHL , let compl . LD be the middle part , DH and HL the extreams dis-junct . Then I say , that the rectangle of the co-sine of LD and Radius is equal to the rectangle of the co-sine of DH and the co-sine of HL : for compl . LD is equal to DB , and DH is equall to compl . AD , and HL to compl . DAB , by the third proposition of this Chapter : therefore the rectangle of the co-sine of LD and Radius , is equal to the rectangle of the co-sine of DH and the co-sine of HL.

CASE 3.

Let one of the oblique angles be the middle part , as in the triangle IEG , let compl . IGE be the middle part : then I say , that the rectangle of the co-sine of IGE and Radius is equal to the rectangle of the sine of GEI and the co-sine of IE : for compl . IGE is equall to DB , and GEI is equal to AD , and EI to compl . DAB .

5. In any Spherical triangle , the sines of the sides are proportional to the sines of their opposite angles .

Let ABC be a spherical triangle , right angled at C , then let the sides AB , AC , and CB be continued to make the quadrants AE , AF , and CD , and from the pole of the quadrant AF , to wit , from the point D , let be drawn down the other quadrants DF and DH ; so there is made three new triangles BDE , GDE , and the obliquangled triangle BDG . I say , in the right angled triangle ABC , that the sine of the side AB is in proportion to the sine of his opposite angle ACB : as the sine of the side AC , is to his opposite angle ABC ; or as BC , to BAC : likewise in the obliquangled spherical triangle BDG , I say , that as BG , is to BDG : so is BD , to BGD ; or so is DG , to DBG .

For first , in the right angled triangle ABC , the angle ACB and the arch AE are of the same quantity , to wit , quadrants , so likewise the angle BAC and the arch EF are of the same quantity , it being the measure of the said angle . Now then as AB , to AE ; so is BC , to EF , by the first proposition of this Chapter : therefore also as AB , to ACB ; so is BC , to BAC . Then in the obliquangled Triangle BDG , because , by the demonstration of right angled triangles , they are as DB , to DEB ; so is DE , to DBE : and as DG , to DEG ▪ so is DE , to DGE , or to DGB . Therefore changing of the proportional termes , it shall be , as DG , to DB : so is DBE , or DBG , to DGB , which was to be demonstrated .

These foundations being thus laid , the businesse of right angled spherical triangles is easily dispatcht . And the proportions to be used in every case may be discovered either by the first , second and fift propositions ; or by the fourth proposition only . The severall cases in a right angled sphericall triangle are sixteen in number , whereof six may be resolved by the first proposition : seven by the second , and three by the fift ; an example in each will suffice .

In the triangle ABC , let there be given the hypothenusal AB , and the perpendicular BC , to finde the base AC ; then by the first proposition , the Analogie is ,

As the co-sine of the perpendicular , is to Radius : so is the co-sine of the hypothenusal , to the co-sine of the base .

2. Let there be given the base AC , and the angle at the base BAC , to finde the perpendicular BC , by the second proposition , the analogie is :

As Radius , to the sine of the base ; so is the tangent of the angle at the base , to the tangent of the perpendicular .

3. Let there be given the hypothenusal AB , the angle at the base BAC , to finde the perpendicular BC , by the fifth proposition , the analogie is :

As Radius , to the sine of the hypothenusal : so is the sine of the angle at the base , to the sine of the perpendicular : and so of the rest .

By the fourth or universall Proposition , the proportions for right angled sphericall triangles may be found two wayes :

First , by the equality of the Sines and Tangents of the circular parts of a triangle , that is , of the Logarithmes of the natural , thus by the universal proposition in the aforesaid triangle ABC , the hypothenusal AB , and the angles at AM and B being noted by their complements , I say . 1. The sine of AC added to Radius ; is equal to the sine of AB added to the sine of the angle at ● . 2. The cosine of A added to Radius is equal to the co-sine of BC added to the sine of the angle at B. 3. The co-sine of AB added to Radius , is equall to the co-sine of AC added to to the co-sine of BC. 4. The co-sine of AB added to Radius is equal to the co-tangent of A , added to the co-tangent of the angle at B. 5. The cosine of the angle at B added to Radius is equal to the tangent of BC , added to the cotangent of AB . 6. The sine of BC added to Radius is equal to the co-tangent of the angle at B added to the tangent of AC .

And thus he that listeth may set down the equality of the sines and tangents of the other sides and angles , and so there will be ten in all ; but these may here suffice : for to these may the sixteen cases of a right angled spherical triangle be reduced ; namely , three to the first , three to the second , two to the third , two to the fourth , three to the fift , and three to the sixt .

As admit there were given the hypothenusal BA , and the angle at B , to finde the base AC ; then , by the first , seeing that the sine of AB added to the sine of the angle at B , is equal to the sine of AC added to Radius . Therefore , if working by natural numbers I multiply the sine of AB by the sine of B , and divide the product by Radius , the remainer will be the sine of AC : and working by Logarithmes , if from the summe of the sines of AB and B I substract Radius , the rest is the sine of AC .

Secondly , admit there were given AB and AC , to finde B , then seeing that the sine of AC added to Radius is equal to the sines of AB and B. Therefore , if working by naturall numbers I multiply the sine of AC by Radius , and divide the product by AB , the remainer is the sine of B. Or working by Logarithmes , if from the sum of the sines of AC and Radius , I substract the sine of AB , the remainer will be the sine of B.

Or thirdly , if there were given AC and the angle at B , to finde AB : then forasmuch as AC and Radius is equal to the sines of AB and B , therefore if working by natural numbers I multiply AC by the Radius , and divide the product by the sine of B , the remainer is the sine of AB . Or working by Logarithmes , if from the sine of AC and Radius , I substract the sine of B the remainer is the sine of AB : an so of the rest .

Which that you may the better perceive , I have here added in expresse words , the Canons or rules of the proportions of the things given and required in every of the sixteen cases of a right angled sphericall triangle , as they are collected from the Catholick Proposition . And here the side subtending the right angle we call the hypothenusal , the other two containing the right angle we may call the sides ; but for further distinction , we call one of these containing sides ( it matters not which ) the base , and the other the perpendicular .

The base an angle at the base given , to finde 1. The Perpendicular . ] As Radius , to the sine of the base ; so is the tangent of the angle at the base , to the tangent of the perpendicular . 2. Angle at the perpendicular . ] As Radius , to the co-sine of the base ; so the sine of the angle at the base , to the co-sine of the angle at the perpendicular . 3. Hypothenusal . ] As Radius , to the co-sine of the angle at the base : so the co-tangent of the base , to the co-tangent of the hypothenusal .

The perpendicular and angle at the base given , to finde 4. Angle at perpend . ] As the co-sine of the perpendicular , to Radius ; so the co-sine of the angle at the base , to the sine of the angle at the perpendicular . 5. Hypothenusal . ] As the sine of the angle at the base , to Radius ; so the sine of the perpendicular , to the sine of the hypothenusal . 6. The Base . ] As Radius , to the co-tangent of the angle at the base ; so is the tangent of the perpendicular , to the sine of the base .

The hypothenusal and angle at the base given , to finde 7. The base . ] As Radius , to the co-sine of the angle at the base ; so the tangent of the hypothenusal , to the tangent of the base . 8. Perpendicular . ] As Radius , to the sine of the hypothenusal , so the sine of the angle at the base , to the sine of the perpendicular . 9. Angle at perpend . ] As Radius , to the co-sine of the hypothenasal ; so the tangent of the angle at the base , to the co-tangent of the angle at the perpendicular .

The base and perpendicular given , to finde 10. Hypothenusal . ] As Radius , to the co-sine of the perpendicular : so the co-sine of the base , to the co-sine of the hypothenusal . 11. Angle at the base ] As Radius , to the sine of the base : so is the co-tangent of the perpendicular , to the co-tangent of the angle at the base .

The base and hypothenusal given , to finde the 12. Perpendicular . ] As the co-sine of the base , to Radius ; so the co-sine of the hypothenusal , to the co-sine of the perpendicular . 13. Angle at the base . ] As Radius , to the tangent of the base ; so the co-tangent of the hypothenusal , to the co-sine of the angle at the base . 14. Angle at the perpend . ] As the sine of the hypothenusal , to Radius ; so the sine of the base , to the sine of the angle at the perpendicular .

The angles at the base and perpendicular given , to finde 15. The perpendicular . ] As the sine of the angle at the perpendicular , is to Radius : so the co-sine of the angle at the base , to the co-sine of the perpendicular . 16. The hypothenusal . ] As Radius , to co-tangent of the angle at the perpendicular ; so the co-tangent of the angle at the base , to the co-sine of the hypothenusal .

Secondly , the proportions of all the cases of a right angled spherical triangle , may by the aforesaid Catholick Proposition be known thus : If the middle part be sought , put the Radius in the first place ; if either of the extreams , the other extream put in the first place .

And note , that when a complement in the proposition doth chance to concur with a complement in the circular parts , you must take the sine it self , or the tangent it self , because the co-sine of the co-sine is the sine , and the co-tangent of the co-tangent is the tangent .

As in the following triangle ABC , let there be given the base AB , and the angle at C , to finde the hypothenusal BC. Here AB is the middle part , BC and C are the opposite extreams , or the extreams disjunct . Now because the extream BC is sought , therefore I must put the other extream , that is , the angle at C , in the first place ; and because that angle , as also the side sought are noted by their complements , therefore I must not say : As the co-sine of the angle at C , is to Radius : so is the sine of the base AB , to the co-sine of the hypothenusal BC : but thus ;

As the sine of the angle at the perpendicular ACB , is to Radius ; so is the sine of the base AB , to the sine of the hypothenusal BC. The like is to be understood of the rest .

Thus much concerning right angled spherical triangles : as for Quadrantal there needs not much be said , because the circular parts of a quadrantal triangle , are the same with the circular parts of a right angled triangle adjoyning .

As let ABC be a triangle , right angled at A , and let one of the sides thereof ; namely , AC be extended , till it become a quadrant , that is to D ; then draw an arch from D to B ; then is DBC a quadrantal triangle , to which there is a right angled triangle adjoyning , as ABC . I say therefore that the circular parts of the quadrantal triangle BCD are the same with the circular parts of the right angled triangle ABC : for the circular parts of either of them are as here appeareth .

The five circular parts of the triangle . ABC are AC AB co ABC cō B● cō BCA BCD are com CD CDB DBC cō BC cō BCD

Where it is evident , that AD and DB being quadrants , DBA is a right angle , and BA is the measure of the angle at D , so that the side AC in the one is equall to compl . CD in the other : and the side AB in the one is equal to the angle BDC in the other : and compl . ABC in the one is equal to DBC in the other , and compl . BC in the one is the same with BC in the other : and lastly , compl . BCA in the one is the same with compl . DCB in the other ; for the compl . of the acute angle A●● unto a quadrant is also the complement of the obtuse angle BCD , and the circular parts of both triangles being the same , it followes , that that which is here proved touching right angled triangles is also true of quadrantal . And all the sixteen cases thereof may also be resolved by the aforesaid Catholick Proposition .

As let there be given the side DC , and the angle at C , to finde the angle at D , then is the side DC the middle , and the angles at D and C are extreams adjacent ; now because the angle at D , one of the extreams is sought , we must put the other extream , to wit , the angle at C in the first place , and that is noted by its complement : and therefore the Analogie is ▪

As the co-tangent of the angle at C , to Radius ; so the co-sine of DC , to the tangent of the angle at D : and so of the rest ; and what is said of the addition of the artificial numbers is to be understood of the rectangles of the natural .

CHAP. IX . Of Oblique angled Sphericall Triangles .

IN an obliquangled spherical triangle , there are twelve Cases ; two whereof , that is , those wherein the things given and required are opposite , may be resolved by the fift proposition of the last Chapter .

CASE 1. Two angles with a side opposite to one of them being given , to finde the side opposite to the other .

As in the triangle ABC , let there be given the side BC , with his opposite angle at A , and the angle ABC , to finde the side AC . I say then , by the fift proposition of the last Chapter :

As the sine of the angle at A , is to the sine of his opposite side BC : so is the sine of the angle at B , to the sine of his opposite side AC .

CASE 2. Two sides with an angle opposite to one of them being given , to finde an angle opposite to the other .

As the sine of BC , to the sine of his opposite angle at A : so is the sine of AC , to the sine of his opposite angle B.

Other eight cases must be resolved by the aid of two Analogies at the least , and that by reducing the triangle proposed to two right angled triangles , by a perpendicular let fall from one of the angles to his opposite side , which perpendicular falls sometimes within , sometimes without the triangle .

If the perpendicular be let fall from an obtuse angle , it falleth within , but if it fall from an acute angle , it falls without the triangle : however it falleth , it must be alwayes opposite to a known angle .

For your better direction , in letting fall the perpendicular take this generall rule .

From the end of a side given , being adjacent to an angle given , let fall the perpendicular .

As in the triangle , ABC , if there were given the side AB , and the angle at A : by this rule the perpendicular must fall from B upon the side AC ; but if there were given the side AC , and the angle at A ; then AB must be produced to D ; and the perpendicular must fall from C upon the side AD. Thus shall we have two right angled triangles , and the side or angle required may easily be resolved by the Catholick Proposition .

As suppose there were given the side AB , the angles at A and C , and required the side AC ; then the perpendicular must fall from B upon the side AC , as in the first triangle , and divide the oblique triangle ABC into two right angled triangles , to wit , ABF and BFC . And in the triangle ABF we have given the side AB , and the angle at A , to finde the base AF , for which the analogie , by the Catholick Proposition , is ,

As the co-tangent of AB , to Radius : so is the co-sine of the angle at A , to the tangent of AF : that is , by the seventh case of right angled triangles .

Secondly , by the eighth case , finde the perpendicular BF . Lastly , in the triangle BFC , having the perpendicular BF , and the angle at C , by the sixt case of right angled spherical triangles , you may finde the base FC , which being added to AF , is the side AC .

But thus there are three operations required ; whereas it may be done at two : for the obliquangled triangle being reduced into two right angled triangles , by letting fall a perpendicular , as before : the hypothenusal in one of the right angled triangles will be correspondent to the hypothenusal in the other , and the base in the one to the base in the other ; and so the other parts .

Then in one of these right angled triangles ( which for distinction sake we call the first ) there is given the hypothenusal and angle at the base , whereby may be found the base or angle at the perpendicular , as occasion requires ; by the seventh or ninth cases of right angled triangles . And this is the first operation .

For the second , there must ( of the things thus given and required ) two things in one triangle , be compared to two correspondent things in the other triangle , which two in each with the perpendicular make three things in each triangle , either adjacent , that is , lying together , or opposite of which three the perpendicular is alwayes one of the extreams , and the thing required one of the other extreams .

Thus in the triangle ABF , if there were given AF and BF , to finde AB : AB is the middle part , AF and BF are opposite extreams ; and therefore by the Catholick Proposition .

Radius added to the co-sine of AB , is equal to the co-sines of AF and BF .

Then in the triangle BFC , if there were given BF and FC , to finde BC : BC will be the middle part , BF and FC opposite extreames ; and therefore by the Catholik Proposition .

The co-sines of BF and FC are equall to the co-sine of BC and Radius .

But if from equal things we take away equal things , the things remaining must needs be equal ; if therefore we take away the Radius , and co-sine BF in both these proportions , it followes , that the co-sine of AB added to the co-sine of FC is equal to the cosine of BC added to the co-sine AF. And therefore , the middle part AB in the first , and the extream FC in the second , is equall to the middle part BC in the second , and the extream AF in the first : or thus ;

As the middle part in the first triangle , is in proporion to the middle part in the second : so is the extream in the first , to the extream in the second .

Thus by the Catholick Proposition , and the help of this , the eight cases following may be resolved . In the exemplification whereof this sign + signifies addition .

By the Catholick Proposition , it is evident that 1 Rad. + cs AB is equal to cs AF + cs FB cs BF + cs FC cs BC + Rad 2 Rad. + s AF is equal to ct A + t FB t FB + ct C s FC + Rad. 3 Rad. + cs A is equal to s ABF + cs FB cs FB + s FBC cs C + Rad 4 Ra. + cs ABF is equal to ct AB + t FB t FB + ct BC cs FBC + Ra.

Then taking from either side tangent FB and Radius , or co-sine FB and Radius , it followes , by the former proposition , that 1. cs AB + cs FC is equall to cs BC + cs AF. 2. s AF + ct C is equall to s FC + ct A. 3. cs A + s FBC is equall to cs C + s ABF 4. cs ABF + ct BC is equal to cs FBC + ct AB

For seeing that AF and FB are opposite extreams to AB , as CF and FB are to BC : therefore ,

1. As cs AF , to cs FC ; so is cs AB , to cs BC : that is , As co-sine the first base , to co-sine the second ; so co-sine the first hypotheriusal , to co-sine the second . And this serves for the third and seventh cases following .

And seeing that A and FB are adjacent extreams to AF : as C and FB are to FC : therefore ,

2. As s AF , to s FC ; so ct A , to ct C : that is , as the sine of the first base , to the sine of the second ; so co-tangent the first angle at the base , to co-tangent the second , which serves for the fourth and tenth cases .

Again , seeing that ABF and FB are opposite extreams to A , as CBF and FB are to C : therefore ,

3. As s ABF , to s CBF ; so cs A , to ●s C : that is , as the sine of the first angle at the perpendicular , to the sine of the second ; so co-sine the first angle at the base , to co-sine the second : which serves for the fifth and ninth cases .

Lastly , seeing AB and FB are adjacent extreams to ABF , as BC and FB are to CBF : therefore ,

4. As cs ABF , to cs CBF ; so ct AB , to ct BC : that is , as co-sine the first angle at the perpendicular , to co-sine the second ; so co-tangent the first hypothenusal , to co-tangent the second : this serves for the sixth and eighth cases following . And this foundation being thus laid , we come now to the severall Cases thereon depending .

CASE 3. Two sides and their contained angle given , to finde the third side .

First , by the seventh case of right angled triangles , the analogie is :

As Radius , to the co-sine of the angle at the base : so is the tangent of the hypothenusal , to the tangent of the base , or first arch . Which being added to or substracted from the base given , according to the following direction , giveth the second arch .

If the perpendicular fall Within the triangle , subtract AF the base found from AC the base given , the remainer is EC , the second arch . Without , and the contained angle obtuse , adde the arch found to the arch given , and their aggregate is the second arch . Without , and the contained angle acute , substract the arch given from the arch found , the remainer is the second arch .

Then , by the first Consectary aforegoing say : as the co-sine of the first base , to the co-sine of the second ; so the co-sine of the first hypothenusal , to the co-sine of the second : but this we will illustrate by example .

Let there be therefore given in the oblique angled spherical triangle ABC , the side or arch AB 38 degrees 47 minutes , the side AC 74 degrees , 84 minutes , and their contained angle BAC 56 degrees , 44 minutes , to finde the side BC. Now then according to the rules given , I let fall the perpendicular BF , and so have I two right angled triangles , the triangle ABF and the triangle BFC . In the triangle ABF , we have the hypothenusal AB 38 degrees , 47 minutes , and the angle at the base BAF 56 degrees 44 minutes , to finde the base AF. First therefore I say , As the Radius 90 , 10.000000 Is to the co-sine of BAC 56.44 . 9.742576 So is the tangent of AB 38 47. 9.900138 To the tangent of AF 23.72 . 9.642714

Now because the perpendicular falls within the triangle , I substract AF 23 degrees , 7● minutes from AC 74 degrees , 84 min. and there remains FC 51 degrees , 1● minutes , the second arch . Hence to finde BC , I say ; As the co-sine of AF 23. 72. co. ar . 0.038331 Is to the co-sine of FC 51.12 . 9.797746 So is the co-sine of AB 38.47 . 9.893725 To the co-sine of BC 57.53 . 9.729802

2. Example .

In the same triangle , let there be given the side AB 38 degr. 47 min. the side BC 57 degr. 53 min. and their contained angle ABC 107 deg. 60 min. and let the side AC be sought . First , let fall the perpendicular DC , and continue the side AB to D , then in the right angled triangle BDC , there is given the angle DBC 72 deg. 40 min. the complement of the obtuse angle ABC , and the hypothehusal BC 57 degrees 53 minutes : to finde BD , I say first ; As the Radius 90 , 10.000000 Is to the co-sine of DBC 72.40 . 9.480539 So is the tangent of BC 57.53 . 10.196314 To the tangent of BD 25.42 . 9.676853

Now because the perpendicular falls without the triangle , and the contained angle obtuse , I adde BD 25 degrees , 42 minutes to AB 38 deg. 47 min. and their aggregate is AD 63 deg. 89 min. the second arch : hence to finde AC , I say , As the co-sine of BD , 25.42 . 0.044223 Is to the co-sine of 63.89 . 9.643547 So is the co-sine of BC 57.53 . 9.729859 To the co-sine of AC 74.84 . 9.417629

3 Example .

In this triangle , let there be given the side BC 57 deg. 53 min. the side AC 74 deg. 84 min. and their contained angle ACB 37 deg. 92 min. and let the side AB be sought . First , I let fall the perpendicular AE , and the side BC I continue to E , then in the right angled triangle AEC , we have known the angle ACE , and the hypothenus ; al AC , to finde EC , I say then : As the Radius 90 , 10.000000 Is to the co-sine of ACE 37.92 . 9.897005 So is the tangeent of AC 74.84 . 10.567120 To the tangent of EC 71.5 . 10.464125

Now because the perpendicular falls without the tringle , and the contained angle acute , I substract the arch given BC 57 degrees 53 minutes from EC 71 degrees 5 minutes , the arch found , and their difference 13 deg. 52 min. is EB , the second arch . Hence to finde AB , I say : As the co-sine of EC 71.5 . co. ar . 0.488461 Is to the co-sine of EB 13.52 . 9.987795 So is the co-sine of AC 74.84 . 9.417497 To the co-sine of AB 38.47 . 9.893753

CASE 4. Two sides and their contained angle given , to finde one of the other angles .

First , by the seventh case of right angled spherical triangles , I say : As Radius , to the co-sine of the angle at the base ; so is the tangent of the hypothenusal , to the tangent of the base , or first arch : which being added to , or substracted from the base given , according to those directions given in the third case , giveth the second arch ; then by the second Consectary of this Chapter , the proportion is :

As the sine of the first base , to the sine of the second : so is the co-tangent of the first angle at the base , to the co-tangent of the second .

1 Example .

Thus if there were given , as in the first example of the last case , the side AB 38 degrees , 47 minutes , the side AC 74 degrees , 84 minutes , and their conteined angle BAC 56 degrees , 44 min. and ACB , the angle sought , the first operation will in all things be the same , and AF 23 degrees , 72 minutes , the first arch , FC 51 degrees , 12 minutes , the second ; hence to finde the angle ACB , I say : As the sine of AF 23.72 . co. ar . 0.395486 To the sine of FC 51.12 . 9.891237 So is the co-tang . of BAC 56.44 . 9.821771 To the co-tangent of ACB 37.92 . 10.108494

There being no other variation in this case then what hath been shewed in the former , one example will be sufficient .

CASE 5. Two angles , and the side between them given , to finde the third angle .

First , by the ninth case of right angled spherical triangles , the proportion is ; As Radius , to the co-sine of the hypothenusal ; so the tangent of the angle at the base , to the co-tangent of the angle at the perpendicular , which being added to , or substracted from the other given angle , according to the following direction , giveth the second arch .

If the perpendicular fall Within the triangle , substract the angle found from the angle given , the remainer is the second arch . Without , and both the angles given acute , substract the angle given from the angle found , and the remainer is the second arch . Without , and one of the angles given be obtuse , adde the angle found to the angle given , & their aggregate is the second arch .

Then , by the third Consectary of this Chapter , the analogie is ; As the sine of the first at the perpendicular , to the sine of the second angle sound : so is the co-sine of the first angle at the base , to the co-sine of the second .

1 Example .

In the triangle ABC , let there be given the angles BAC 56 degrees 44 minutes , and ABC 107 degrees , 60 minutes , and the side between them AB 38 degrees 47 minutes , to finde the angle ACB . First , let fall the perpendicular BF , and then in the right angled spherical triangle ABF we have known the angle at the base BAF , and the hypothenusal AB , to finde the angle at the perpendicular ABF . First , then I say : As the Radius 90 , 10.000000 To the co-sine of AB 38.47 . 9.893725 So is the tangent of BAF 56.44 . 10.178229 To the co-tangent of ABF 40.28 . 10.071954

Let there be given , as before , the two angles BAC and ABC , with the side between them AB , to finde the angle ACB , and let the perpendicular EA , and let the side BC be continued to E , then in the right angled triangle AEB we have known the hypothenusal AB 38 degrees , 47 minutes , and the angle at the base ABE 72 degrees , 40 minutes , the complement of the obtuse angle ABC , to finde the angle EAB . First then I say : As the Radius 90. 10,000000 To the co-sine of AB 38.47 . 9.893725 So is the tangent of ABE 72.40 . 10.498641 To the co-tangent of EAB 22.6 . 10.392366

And because the perpendicular falls without the triangle , and one of the angles given obtuse , I adde the angle found EAB 22 degrees 6 minutes to the angle given BAC 56 degrees , 44 minutes , and their aggregate 78 degrees 50 minutes is the angle EAC , the second arch ; and hence to finde the angle at C , I say , as before . As the sine of EAB 22.6 . co. ar . 0.425300 To the sine of EAC 78 . 5● 9.991194 So is the co-sine of ABE 72.40 . 9.480538 To the co-sine of ACB 37.92 . 9.897032

3 Example .

Let there be given the angles BAC 56 degrees 44 minutes , and ACB 37 degrees , 92 minutes , with their contained side AC 74 degres , 84 minutes , to finde the angle ABC , let fall the perpendicular CD , and let the side AB be continued to D , then in the right angled triangle ADC , we have known the hypothenusal AC , and the angle at the base DAC , to finde ACD ; first , then I say ; As the Radius 90 10.000000 To the co-sine of AC 74.84 . 9.417497 So is the tangent of DAC 56.44 . 10.178229 To the co-tangent of ACD 68.48 . 9.595726

Now because the perpendicular falls without the triangle , and both the angles given acute , therefore I substract the angle given ACB 37 degrees , 92 minutes from the angle found ACD 68 degrees 48 minutes , and their difference 30 degrees 56 minutes is the angle BCD , the second arch . Hence to finde the angle CBD , I say , as before ; As the sine of ACD 68.48 . co. ar . 0.031382 To the sine of BCD 30.56 . 9.706240 So is the co-sine of DAC 56.44 . 9.742575 To the co-sine of CBD 72.40 . 9.480197

CASE 6. Two angles and the side between them given to finde the other side .

First , by the ninth case of right angled triangles , I say , as before ; As Radius , to the co-sine of the hypothenusal ; so the tangent of the angle at the base , to the co-tangent of the angle at the perpendicular . Which being added to or substracted from the other angle given , according to the direction of the fift case , giveth the second arch .

Then by the fourth Consectary of this Chapter , As the co-sine of the first angle at the perpendicular , to the co-sine of the second ; so is the co-tangent of the first hypothenusal , to the co-tangent of the second .

Example .

If there were given , as in the first example of the last case , the angles BAC 56 degrees 44 minutes , and ABC 107 degrees 60 minutes , with the side AB 38 degrees , 47 minutes , to finde the side BC. The first operation will be in all things the same , and the first arch ABF 40 degrees , 28 minutes ; the second arch FBC 67 degrees , 32 minutes . Hence to finde the side BC , I say : As the co-sine of ABF 40.28 . co. ar . 0.117536 To the co-sine of FBC 67.32 . 9 . 5861●9 So is the cotangent of AB 38.47 . 10 . 09986● To the co-tangent of BC 57.53 9.803517

CASE 7. Two sides with an angle opposite to one of them , to finde the third side .

First , by the seventh case of right angled sphericall triangles , I say ; As Radius , to the co-sine of the angle at the base ; so is the tangent of the hypothenusal , to the tangent of the base , or first arch .

Then , by the first Consectary of this Chapter , the analogie is ,

As the co-sine of the first hypothenusal , to the co-sine of the second ; so the co-sine of the first arch found , to the co-sine of the second . Which being added to or substracted from the first arch found , according to the direction following , their sum or difference is the third side .

If the perpendicular fall Within the triangle , adde the first arch found to the second arch found , and their aggregate is the side required . Without , & the angle given obtuse , substract the first arch found from the second arch found , and what remaineth is the third side . Without , & the given angle acute , substract the second arch found from the first , and what remaineth is the side required .

1 Example .

In the oblique angled triangle ABC , let there be given the sides AB 38 degrees , 47 minutes , and BC 57 degrees , 53 minutes , with the angle BAC 56 degrees , 44 minutes , and let the side AC be required . First , I let fall the perpendicular BF , and then in the right angled triangle ABF , we have given the hypothenusal AB , and the angle at the base BAF , to finde the base AF , for which I say : As the Radius 90 10.000000 To the co-sine of BAF 56.44 . 9.742576 So is the tangent of AB 38.47 . 9.900138 To the tangent of AF 23.72 . 9.642714

Secondly , for FC , I say : As the co-sine of AB 38.47 . co. ar . 0.106275 To the co-sine of BC 57.53 . 9.729859 So is the co-sine of AF 23.72 . 9. ●61669 To the co-sine of FC 51.12 . 9.797803

Now because the perpendicular fell within the triangle , therefore I adde the first arch found AF 23 degrees , 72 minutes to the second arch found FC 51 degrees 12 minutes , and their aggregate 74 degrees , 84 minutes is AC the side required .

2 Example .

In the same triangle ABC , let there be given the sides AB 38 degrees , 47 minutes and AC 74 degrees 84 minutes , and the angle ABC 107 degrees , 60 minutes , and let BC be required . First then , I let fall the perpendicular AE , and continue the side BC to E , and then in the right angled triangle AEB we have given the side AB 38 degrees , 47 minutes , and the angle ABE 72 degrees , 40 minutes , the complement of ABC , to finde EB : for which I say : As the Radius 90 10.000000 To the co-sine of ABE 72.40 . 9.480538 So is the tangent of AB 38.47 . 9.900138 To the tangent of EB 13.51 . 9.380676

Secondly , to finde EC , I say : As the co-sine of AB 38.47 . co. ar . 9.106275 To the co-sine of AC 74.84 . 9. ●17497 So is the co-sine of EB 13.51 . 9.987813 To the co-sine of EC 71.4 . 9.511585

Now because the perpendicular falls without the triangle , and the given angle obtuse , therefore I substract the first arch found EB 13 degrees 51 minutes , from the second arch EC 71 degrees , 4 minutes , and their difference 57 degrees , 53 minutes is BC , the side required .

3 Example .

In the same triangle ABC , let there be given the sides AC 74 degrees , 84 minutes , and BC 57 degrees , 53 minutes , and the angle BAC 56 deg. 44 min. to finde the side AB : I let fall the perpendicular DC , and continue the side AB to D , then in the right angled triangle ADC we have given the hypothenusal AC , and the angle at A , to finde AD.

As the Radius 90 10.000000 To the co-sine of BAC 56.44 . 9.742576 So is the tangent of AC 74.84 . 10.567119 To the tangent of AD 63.89 . 10.309695

Secondly , to finde DB , I say : As the co-sine of AC 74.84 . co. ar . 0.582503 To the co-sine of BC 57.53 . 9.729859 So is the co-sine of AD 63.89 . 9.643547 To the co-sine of DB 25.39 . 9 . 95●909

Now because the perpendicular falls without the triangle , and the angle given acute , therefore I substract the second arch found DB 25 degrees , 39 minutes , from the first arch found AD 63 degrees 89 minutes , and their difference 38 degrees 50 minutes is AB , the side required .

CASE 8. Two sides with an angle opposite to one of them being given , to finde their contained angle .

First , by the ninth case of right angled spherical triangles , I say ; As Radius , to the co-sine of the hypothenusal ; so the tangent of the angle at the base , to the co-tangent of the angle at the perpendicular . Then , by the fourth Consectary of this Chapter , the proportion is :

As the co-tangent of the first hypothenusal , to the co-tangent of the second ; so the co-sine of the first angle at the perpendicular , to the co-sine of the second : which being added to , or substracted from the first arch found , according to the direction of the seventh case , giveth the angle sought .

Example .

If there were given , as in the first example of the last case , the sides AB 38 deg. 47 min. and BC 57 deg. 53 min. with the angle BAC 56 deg. 44 min. to finde the obtuse angle ABC . The perpendicular BF falling within the triangle , then in the right angled triangle ABF , we have knowne the hypothenusal AB , and the angle at A , to finde the angle ABF , I say then , As the Radius 90 , 10.000000 Is to the co-sine of AB 38.47 . 9●93725 So is the tangent of BAF 56.44 . 10.178229 To the co-tang . of ABF 40.28 . 10.071954

Secondly , to finde FBC , I say : As the co-tangent of AB 38.47 . 9.900138 To the co-tangent of BC 57.53 . 9. ●03686 So is the co-sine of ABF 40.28 . 9.882464 To the co-sine of FBC 67.32 . 9.586288

Now because the perpendicular falls within the triangle , I adde the first arch found ABF 40 degrees , 28 minutes , to the second arch found FBC 67 degrees , 32 minutes , and their aggregate is 107 degr. 60 min. the angle ABC required .

CASE 9. Two angles and a side opposite to one of them being given , to finde the third angle .

First , by the ninth case of right angled spherical triangles , I say : As the Radius , to the co-sine of the hypothenusal ; so the tangent of the angle at the base , to the co-tangent of the angle at the perpendicular .

Then by the third Consectary of this Chapter , the proportion is . 〈◊〉 the co-sine of the first angle at the base , to the co-sine of the second ; so is the sine of the first angle at the perpendicular , to the sine of the second : which being added to , or substracted from the first arch found , according to the ●i●ect●on following , their summe or difference is the angle sought .

If the perpendicular fall Within the triangle , adde both arches together . Without , and the angle opposite to the given side acute , substract the first from the second arch . Without , and the angle opposite to the given side obtuse , substract the second from the first .

1. Example .

In the oblique angled Triangle ABC , let there be given the angle BAC 56 deg. 44 min. and ACB 37 deg. 92 min. and the side AB 38 deg. 47 min. to finde the angle ABC . First , let fall the perpendicular FB , then in the right angled triangle AFB we have known , the hypothenusal AB , and the angle at A , to finde the angle ABF , for which I say , As Radius , 90 deg. 10.000000 To co-sine of AB , 38.47 9 . 893●26 So the tangent of BAF , 56. ●4 10.178229 To the co-tangent of ABF , 40.28 . 10.071955

Secondly , to finde FBC , I say , As the co-sine of BAF , 56.44 0.257424 . To the co-sine of ACB , 37.92 9 . 8970●5 So is the sine of ABF , 40.28 9.810584 To the sine of FBC , 67.32 9.965013

Now because the perpendicular fals within the Triangle , I adde the first arch found ABF 40 deg. 28 min. to the second arch found FBC 67 deg. 32 min. and their aggregate is 107 deg. 60 min. the angle ABC required .

2. Example .

In the same Triangle let there be given the angle ACB 37 deg. 92 min. and ABC 107 deg. 60 min. and the side AB 38 deg. 47 min. to finde the angle BAC . First , let fall the perpendicular AE , and let the side BC be continued to E , then in the right angled triangle AEB we have known the Hypothenusal AB , and the angle at B , 72 deg. 40 min. the complement of ABC , to finde EAB , I say then , As the Radius 90 , 10.000000 To the co-sine of AB , 38.47 9.893726 So is the tangent of ABE , 72.40 10.498641 To the co-tangent of EAB , 22. ●6 10,392367

Secondly , to finde EAC , I say , As the co-sine of ABE , 72.40 0.519462 To the co-sine of ACB , 37.92 9.897005 So is the sine of EAB , 22.6 9.574699 To the sine of EAC 78.49 9 . 99●166

Now because the perpendicular falls without the triangle and the angle opposite to the given side acute , I substract the first angle found E. AB 22 deg. 6 min. from the second arch found 78 deg. 49 min. and their difference 56 deg. 43 min. is the angle BAC required .

3 Example .

In the same triangle ABC , let there be given the angles ACB 37 deg. 92 min. and ABC 107 deg. 60 min. and the side AC 74 deg. 84 min. to finde the angle BAC . Let fall the perpendicular AE , and then in the right angled triangle AEC , we have known the hypothenusal AC , and the angle ACB , to finde the angle EAC .

As the Radius 90 , 10.000000 To the co-sine of AC , 74.84 9.417497 So is the tangent of ACE , 37.92 9.891559 To the co-tangent of EAC , 78.49 9 . 3090●6

Secondly , to finde EAB , I say , As the co-sine of ACE , 37.92 0 . 102●95 To the co-sine of ABE , 72.40 9.480538 So is the sine of EAC , 78.49 9.991177 To the sine of EAB , 22.6 9.574790

Now because the perpendicular falls without the Triangle , and the angle opposite to the given side obtuse , therefore I substract the second arch found EAB , 22 deg. 6 min. from the first arch found , EAC 78 deg. 49 min. and their difference 56 deg. 43 min. is the angle BAC required .

CASE 10. Two angles , and a side opposite to one of them being given , to finde the side between them .

First , by the 7th . Case of right angled Sphericall Triangles , I say , As Radius , to the co-sine of the angle at the base ; so is the Tangent of the Hypothenusal , to the Tangent of the Base .

Then by the second Consectary of this Chapter , the proportion is , As the co-tangent of the first angle at the base , to the co-tangent of the second ; so is the sine of the first base , to the sine of the second : which being added to , or substracted from , the first arch found , according to the direction of the 9th . Case , giveth the side required .

Example .

In the oblique angled triangle ABC , let there be given the two angles BAC 56 deg. 44 min. and ACB 37 deg. 92 min. with the side BC 57 deg. 53 min. to finde the side AC . Let fall the perpendicular BF , then in the right angled triangle BCF , we have known the Hypothenusal BC , and the angle FCB , to finde the base FC : say then , As the Radius , 90 10.000000 Is to the co-sine of FCB , 37.92 9.897005 So is the tangent of BC , 57.53 10.196314 To the tangent of FC , 51.11 10.093319

Secondly , to finde AF , I say , As co-tangent FCB , 37.92 , co. ar . 9.891559 To co-tangent of BAC . 56.44 9.821771 So is the sine of FC , 51.11 9.891176 To the sine of AF , 23.72 9.604506

Now because the perpendicular falls within the Triangle , I adde the first arch FC 51 deg. 11 min. to the second arch AF , 23 deg. 72 min. and their aggregate is 74 deg. 83 min. the side AC required .

CASE 11. The three sides given to finde an angle .

The solution of this and the Case following , depends upon the Demonstration of this Proposition .

As the Rectangular figure of the sines of the sides comprehending the angle required ; Is to the square of Radius :

So is the Rectangular figure of the sines of the difference of each containing side taken from the half summe of the three sides given ; To the square of the sine of half the angle required .

Let the sides of the triangle ZPS be known , and let the vertical angle SZP be the angle required , then shall ZS the one be equal ZC . In like manner PS the base of the vertical angle shall be equal to PH or PB , then draw PR the sine of PZ and CK the sine of CZ or ZS . Divide CH into two equal parts in G , draw the Radius AG and let fall the perpendiculars P● and CN which are the sines of the arches PG and CG . The right line EV is the versed sine of a certain arch in a great circle , and SC the versed of the like arch in a less , then if you draw the right line NF parallel to SH bisecting CH in N , it shall also bisect the versed sine SC in F by the 15th . of the second and RM bisecting TP in R , and drawn parallel to TX , shall for the same reason bisect PX in M , and the triangles SCH and FNC shall be like , as also the triangles TPX and RPM are like ; and ZG shall be equal to the half summe of the three sides given , which thus I prove . Of any three unequal quantities given , if the difference of the two lesser be substracted from the greatest , and half the remainer added to the mean quantity , the summe shall be equall to half the summe of the three unequal quantities given .

Example .

Let the quantities given be 9 , 13 , and 16 , the difference between 9 and 13 is 4 , which being substracted from 16 , there remaineth 12 , the half whereof is 6 , which being added to 13 maketh 19 , the half sum of the three unequal quantities . Now then in this Diagram PC is the difference of the two lesser sides , which taken from PH , the remainer is CH , the half whereof is CG , and CG added to CZ , the mean side , giveth GZ the half summe , and if we substract ZP the lesser containing side of the angle required , from ZG the half sum , their difference will be PG , and if we substract ZC the other side , the difference will be CG . Lastly , let the arch IV be the measure of the vertical angle PZS , and the right line OQ bisect the lines EV and IV , and the right line AQ perpendicular to the right line IV , bisecting the same in Q , I say then .

And dividing the two last rectangles by CF , the proportion will be PR × CK And because VO in VA is equal to VQ square ; therefore if you multiply CN by VA , the proportion will be , as PR × CK , to PM × VA ; so is CN × VA , to VO × VA equal to VQ square , which was to be proved . PM × VA CN VO

If then the three sides of an oblique angled spherical triangle be given , and an angle inquired ; do thus : 1. Take the sines of the sides comprehending the angle inquired . Or the Logarithmes of those sines . 2. Take also the quadrat of the Radius , or the Logarithme of the Radius doubled . 3. Substract each side comprehending the angle inquired from the half sum of the three sides given , and take the sines of their differences , or the Logarithmes of those sines . 4. If the rectangle of the first divide the rectangle of the second and third , the side of the quotient is the sine of half the angle inquired .

Or if the sum of the Logarithme of the first be deducted from the sum of the Logarithmes of the second and third , the half difference is the Logarithme of half the angle sought .

Arithmeticall illustration by Naturall Numbers .

In the Oblique angled Triangle SZP , having the Sides PS , 42 deg. 15 min. PZ , 30 00 And SZ , 24 7
To finde the angle PZS .

Sines . The side PZ , 30 deg. 50000 The side SZ , 24 deg. 7 min. 40785 1 The factus of the Sines 2039250000 2 Quadrat of the Radius 10000000000
The summe of the sides 96 deg. 22 min. The half summe , 48 11

Sines . The difference of ZS 24 de. 4 min. 40737 The difference of PZ 18 11 31084 3 Factus of the sines 1266268908

Which being multiplyed by Radids square , 100000.00000 , and divided by 2039250000 , the quotient will be 620●●83●7● , the side whereof is 78802 , the sine of 52 deg. which doubled is 104 , the angle PZS inquired .

Arithmeticall illustration by artificiall numbers .

The side PS , 42.15 . Logar . Sine . The side PZ , 30 9.698970 The side SZ , 24.7 9.610503 Sum of the sides , 96.22 19.309473 The halfe sum , 48.11 Diff. of ZS and the half sum , 24.4 9.609993 Dif . of PZ & the half sum , 18.11 9.492540 The doubled Radius 20.000000 39.102533 From which substract the sum of the Log. of the sides , ●S . PZ 19.309473 There doth remain , 19.793060

The halfe thereof , 9.896530 is the Logarithm of the sine of 52 deg. whose double 104 is the angle PZ Sinquired as before .

Or if instead of the Logarithms of the sines of the sides ●S and PZ , you take their Arithmeticall complements , as was shewed in the 8th . Proposition of the 4th . Chapter , and leave out the doubled Radius , the work may be performed without substraction in this manner . The side PZ , 30 co. ar . 0.301030 The side ZS . 14.7 co. ar . 0.389497 Dif . of ZP and half sum , 18.11 9.492540 Dif . of ZS and half sum , 24.4 9.609993 The summe is 19.793060 The halfe thereof 9.896530
Is the Logarithm of the sine of 52 deg. as before .

CASE 12. The three angles of a Sphericall Triangle given , to finde a side .

This Case is the converse of the former , and to be resolved after the same manner , if so be we convert the angles into sides , according to the fifth of the sixth Chapter . For the two lesser angles are alwayes equal unto two sides of a Triangle comprehended by the arkes of great Circles drawn from their Poles , and the third angle may be greater then a Quadrant , and therefore the complement thereof to a Semicircle must be taken for the third side .

The angle being found , shall be one of the three sides inquired .

As in the Triangle ABC , the poles of those arks L , M , K , which connected do make the Triangle LMK , the sides of the former Triangle being equal to the angles of this latter , taking the complement of the greater angle to a semicircle for one . As AB is equal to the angle at L , or the arke EG . The side BC is equal to the angle at M , or the arch FH . And the side AC is equal to the complement of the angle LKM , or the arch DI. Therefore if the angles of the latter triangle LMK be given , the sides of the former triangle AB , BC , and AC are likewise given . And the angles of the triangle LMK being thus converted into sides , if we resolve the triangle ABC , according to the precepts of the last Case , we may finde any of the angles , which is the side inquired .

Illustration Arithmetical , by the Artificiall Canon .

Let the three angles of the triangle LMK be given . LMK , 104 deg. or the complement of DKI , 76 deg. equal to AC . MLK , or the side AB , 46 deg. 30 min. LMK , or the side BC , 36 deg. 14 min. To finde the side ML , or the angle ABC .

The sides AC 76. AB 46.30 9.859118 BC 36.14 9.770675 Sum of the sides 158.44 19.629893 Halfe sum 79.22 Diff. of AB and the sum 32.92 9.735173 Dif . of BC and half sum 43.08 9.834432 The doubled Radius 20.000000 The summe 39.569605 Sum of the sides substract 19.629893 The difference 19.939712 Halfe difference 9.969856

The Sine of 68 deg. 90 min. which doubled is 137 deg. 80 min. the quantity of the angle ABC , and the complement thereof to a semicircle 42 deg. 20 min. is the angle FBG , or the arch FG , equal to the side ML which was inquired .

Institutio Mathematica : OR , A MATHEMATICALL Institution : The second Part. Containing the application and use of the Naturall and Artificiall SINES and TANGENTS , as also of the LOGARITHMS , IN Astronomie , Dialling , and Navigation . By JOHN NEWTON .

LONDON , Printed Anno Domini , 1654.

A Mathematicall Institution : The second Part.
CHAP. I. Of the Tables of the Suns motion , and of the equation of time for the difference of Meridians .

WHereas it is requisite that the Reader should be acquainted with the Sphere , before he enter upon the practise of Spherical Trigonometri , the which is fully explained in Blundeviles Exercises , or Ch●lmades translation of Hues on the Globes , to whom I refer those that are not yet acquainted therewith : that which I here intend is to shew the use of Trigonometrie in the actuall resolution of so me known Triangles of the Sphere .

And because the Suns place or distance from the next Equinoctial point is usually one of the three terms given in Astronomical Questions , I will first shew how to compute that by Tables calculated in Decimal numbers according to the Hypothesis of Bullialdus , and for the Meridian of London , whose Longitude reckoned from the Canarie or Fortunate Islands is 21 deg. and the Latitude , North , 51 deg. 57 parts ( min. ) or centesms of a degree .

Nor are these Tables so confined to this Meridian , but that they may be reduced to any other : If the place be East of London , adde to the time given , but if it be West make substraction , according to the difference of Longitude , allowing 15 deg. for an houre , and 6 minutes or centesms of an houre to one degree , so will the sum or difference be the time aequated to the Meridian of London , and for the more speedy effecting of the said Reduction , I have added a Catalogue of many of the chiefest Towns and Cities in diverse Regions , with their Latitudes and difference of Meridians from London in time , together with the notes of Addition and Substraction , the use whereof is thus .

Suppose the time of the Suns enterance into Taurus were at London Aprill the 10th . 1654 , at 11 of the clock and 16 centesms before noon , and it be required to reduce the same to the Meridian of Vraniburge , I therefore seeke Vraniburge in the Catalogue of Cities and Places , against which I finde 83 with the letter A annexed , therefore I conclude , that the Sun did that day at Uraniburge enter into Taurus at 11 of the clock and 99 min. or centesms before noon , and so of any other .

Problem 1. To calculate the Suns true place .

THe form of these our Tables of the Suns motion is this , In the first page is had his motion in Julian years compleat , the Epochaes or roots of motions being prefixed , which sheweth the place of the Sun at that time where the Epocha adscribed hath its beginning : the Tables in the following pages serve for Julian Years , Moneths , Dayes , Houres , and Parts , as by their Titles it doth appear . The Years , Moneths , and Dayes , are taken compleat , the Houres and Scruples current . After these Tables followeth another , which contains the Aequations of the Eccentrick to every degree of a Semicircle , by which you may thus compute the Suns place .

First , Write out the Epocha next going before the given time , then severally set under those the motions belonging to the years , moneths , and dayes compleat , and to the hours and scruples current , every one under his like , ( onely remember that in the Bissextile year , after the end of February , the dayes must be increased by an unit ) then adding them all together , the summe shall be the Suns mean motion for the time given .

Example .

Let the given time be 1654 , May 13 , 11 hours , 25 scruples before noon at London , and the Suns place to be sought .

The numbers are thus : Longit. ☉ Aphel . ☉ The Epocha 1640 291.2536 96.2297 Years compl . 13 359.8508 2052 Moneth co. April 118.2775 53 Dayes compl . 12 11.8278 6 Hours 23 9444 Scruples 25 102 Sū or mean motion 782.1643 96.4308

2. Substract the Aphelium from the mean Longitude , there rests the mean Anomalie , if it exceed not 360 degrees , but if it exceed 360 degr. 360 being taken from their difference , as oft as it can , the rest is the mean Anomalie sought .

Example .

The ☉ mean Longitude 782.1643 The Aphelium substracted 96.4308 There rests 685.7335 From whence deduct 360. There rests the mean Anomalie . 325.7335

3. With the mean Anomalie enter the Table of the Suns Eccentrick Equation , with the degree descending on the left side , if the number thereof be lesse then 180 ; and ascending on the right side , if it exceed 180 , and in a straight line you have the Equation answering thereunto , using the part proportional , if need require .

Lastly , according to the title Add or Substract this Equation found to or from the mean longitude ; so have you the Suns true place .

Example .

The Suns mean longitude 782.1643 Or deducting two circles , 720. The Suns mean longitude is 62.1643 The Suns mean Anomalie 325.7335
In this Table the Equation answering to 325 degrees is 1.1525 The Equation answering to 326 degrees is 1.1236
And their difference 289.

Now then if one degree or 10000 Give 289 What shall 7335
Give , the product of the second and third term is 2119815 , and this divided by 10000 the first term given , the quotient or term required will be 212 fere , which being deducted from 1.1525 , the Equation answering to 325 degr. because the Equation decreased , their difference 1.1313 . is the true Equation of this mean Anomalie , which being added to the Suns mean longitude , their aggregate is the Suns place required .

Example .

The Suns mean longitude 62.1643 Equation corrected Add 1.1313 The Suns true place or Longitude 63.2956
That is , 2 Signes , 3 degrees , 29 minutes , 56 parts .

The Suns Equation in this example corrected by Multiplication and Division may more readily be performed by Addition and Substraction with the help of the Table of Logarithmes : for , As one degree , or 10000 , 4.000000 Is to 289 ; 2.460898 So is 7335 , 3.865400 To 212 fere 2.326298

TheSunsmeanMotions. Epochae Longitud ☉ Aphelium ☉ ° ′ ″ ° ′ ″ Per. Jul. 242 99 61 355 85 44 M●●di 248 71 08 007 92 42 Christi 278 98 69 010 31 36 An. Do. 1600 290 95 44 095 58 78 An. Do. 1620 291 10 41 095 90 39 An. Do. 1640 291 25 36 096 21 97 An. Do. 1660 291 40 33 096 53 56 1 356 76 11 0 01 58 2 359 52 22 0 18 17 3 359 28 30 0 04 74 B 4 000 03 00 0 06 30 5 359 79 11 0 07 89 6 359 55 19 0 09 47 7 359 31 30 0 11 05 B 8 000 05 97 0 12 64 9 359 82 08 0 14 22 10 359 58 19 0 15 78 11 359 34 30 0 17 36 B 12 000 08 97 0 18 94 13 359 85 08 0 20 52 14 359 00 19 0 22 11 15 359 37 30 0 23 69 B 16 000 11 97 0 25 25 17 359 88 08 0 26 83 18 359 64 19 0 28 41 19 359 40 28 0 30 00 B 20 000 14 97 0 31 61 40 000 29 91 0 63 19 60 000 44 83 0 94 77 80 000 59 83 1 26 39 100 000 74 80 1 57 97 100 00 74 80 01 57 97 200 01 49 58 03 15 94 300 02 24 39 04 73 94 400 02 99 19 06 31 94 500 03 73 97 07 89 91 600 04 48 77 09 47 92 700 05 23 58 11 05 89 800 05 98 36 12 63 89 900 06 73 17 14 21 86 1000 07 47 97 15 79 86 2000 14 95 92 31 59 69 3000 22 43 89 47 39 55 4000 29 91 82 63 19 41 5000 37 39 80 78 99 25 January 030 55 50 0 00 14 February 058 15 30 0 00 25 March 088 70 83 0 00 39 April 118 27 75 0 00 53 May 148 83 28 0 00 67 June 178 40 19 0 00 80 July 208 95 69 0 00 94 August 239 51 22 0 01 06 September 269 08 17 0 01 19 October 299 63 66 0 01 33 November 329 20 61 0 01 44 December 359 76 11 0 01 58

TheSunsmeanmotionsinDayes. Longit. ☉ Aphel . D ° ′ ″ ′ ″ 1 0 98 55 0 00 2 1 97 14 0 00 3 2 95 69 0 00 4 3 94 25 0 02 5 4 92 83 0 02 6 5 91 39 0 03 7 6 89 94 0 03 8 7 88 52 0 03 9 8 87 08 0 05 10 9 85 63 0 05 11 10 84 22 0 05 12 11 82 78 0 06 13 12 81 33 0 06 14 13 79 91 0 06 15 14 78 47 0 06 16 15 77 03 0 08 17 16 75 61 0 08 18 17 74 16 0 08 19 18 72 72 0 08 20 19 71 30 0 08 21 20 69 86 0 08 22 21 68 41 0 11 23 22 67 00 0 11 24 23 65 56 0 11 25 24 64 11 0 11 26 25 62 94 0 11 27 26 61 25 0 11 28 27 59 80 0 11 29 28 58 36 0 13 30 29 56 94 0 14 31 30 55 50 0 14 32 31 54 05 0 14

Longit. ☉ Long. Long. H ° ′ ″ M ′ ″ M ′ ″ 1 0 04 11 34 1 39 67 2 75 2 0 08 22 35 1 43 68 2 79 3 0 12 31 36 1 47 69 2 83 4 0 16 42 37 1 51 70 2 87 5 0 20 52 38 1 56 71 2 91 6 0 24 63 39 1 60 72 2 96 7 0 28 75 40 1 64 73 3 00 8 0 32 86 41 1 68 74 3 04 9 0 36 97 42 1 72 75 3 08 10 0 41 06 43 1 76 76 3 12 11 0 45 17 44 1 80 77 3 16 12 0 49 27 45 1 84 78 3 20 13 0 53 39 46 1 88 79 3 24 14 0 52 50 47 1 93 80 3 28 15 0 61 61 48 1 97 81 3 32 16 0 65 72 49 2 01 82 3 37 17 0 69 80 50 2 05 83 3 41 18 0 73 91 51 2 09 84 3 45 19 0 78 03 52 2 13 85 3 49 20 0 82 14 53 2 17 86 3 53 21 0 86 25 54 2 21 87 3 57 22 0 90 36 55 2 25 88 3 61 23 0 94 44 56 2 30 89 3 65 24 0 98 55 57 2 34 90 3 69 25 1 02 66 58 2 38 91 3 74 26 1 06 77 59 2 42 92 3 78 27 1 10 88 60 2 46 93 3 82 28 1 14 99 61 2 50 94 3 86 29 1 19 10 62 2 54 95 3 90 30 1 23 21 63 2 58 96 3 94 31 1 27 32 64 2 62 97 3 98 32 1 31 43 65 2 67 98 4 02 33 1 35 54 66 2 71 99 4 06 ′ ′ ″ ‴ ′ ′ ″ 100 4 11 ″ ″ ‴ ' ' ' ' ″ ″ ‴ ″ ″ ‴

TheEquationsoftheSunsEccentrick. Aeq . sub ° ′ ″ 0 0 00 00 360 1 0 03 52 359 2 0 07 03 358 3 0 10 56 357 4 0 14 05 356 5 0 17 53 355 6 0 21 00 354 7 0 24 44 353 8 0 27 89 352 9 0 31 30 351 10 0 34 72 350 11 0 38 17 349 12 0 41 56 348 13 0 44 94 347 14 0 48 30 346 15 0 51 67 345 16 0 55 03 344 17 0 58 36 343 18 0 61 67 342 19 0 64 97 341 20 0 68 24 340 21 0 71 53 339 22 0 74 78 338 23 0 78 03 337 24 0 81 22 336 25 0 84 41 335 26 0 87 56 334 27 0 90 69 333 28 0 94 26 332 29 0 97 30 331 30 1 00 19 330 31 1 03 33 329 32 1 06 41 328 33 1 09 41 327 34 1 12 36 326 35 1 15 25 325 36 1 18 03 324 37 1 20 78 323 38 1 23 50 322 39 1 26 22 321 40 1 28 91 320 41 1 31 58 319 42 1 34 22 318 43 1 36 86 317 44 1 39 50 316 45 1 42 08 315 46 1 44 52 314 47 1 47 05 313 48 1 49 47 312 49 1 51 89 311 50 1 54 16 310 51 1 56 47 309 52 1 58 69 308 53 1 60 86 307 54 1 63 00 306 55 1 65 14 305 56 1 67 25 304 57 1 69 30 303 58 1 71 33 302 59 1 73 28 301 60 1 75 05 300 1 76 92 299 62 1 76 69 298 63 1 80 39 297 64 1 81 97 296 65 1 83 50 295 66 1 85 00 294 67 1 86 44 293 68 1 87 83 292 69 1 89 16 291 70 1 90 44 290 71 1 91 69 289 72 1 92 86 288 73 1 93 96 287 74 1 95 28 286 75 1 96 22 285 76 1 97 14 284 77 1 97 97 283 78 1 98 72 282 79 1 99 61 281 80 2 00 41 280 81 2 01 14 279 82 2 01 72 278 83 2 02 25 277 84 2 02 94 276 85 2 03 14 275 86 2 03 44 274 87 2 03 66 273 88 2 04 05 272 89 2 04 22 271 90 2 04 41 270 91 2 04 47 269 92 2 04 41 268 93 2 04 27 267 94 2 04 11 266 95 2 03 89 265 96 2 03 61 264 97 2 03 33 263 98 2 02 94 262 99 2 02 50 261 100 2 02 03 260 101 2 01 42 259 102 2 00 64 258 103 1 99 83 257 104 1 99 27 256 105 1 98 47 255 106 1 97 64 254 107 1 96 67 253 108 1 95 67 252 109 1 94 55 251 110 1 93 39 250 111 1 92 11 249 112 1 90 89 248 113 1 89 58 247 114 1 88 28 246 115 1 86 89 245 116 1 85 44 244 117 1 83 97 243 118 1 82 39 242 119 1 80 72 241 120 1 79 00 240 121 1 77 19 239 122 1 75 39 238 123 1 73 50 237 124 1 71 50 236 125 1 69 50 235 126 1 67 53 234 127 1 65 39 233 128 1 63 22 232 129 1 61 28 231 130 1 58 77 230 131 1 56 44 229 132 1 54 05 228 133 1 51 64 227 134 1 49 16 226 135 1 46 97 225 136 1 44 16 224 137 1 41 58 223 138 1 38 94 222 139 1 36 31 221 140 1 33 58 220 141 1 30 83 219 142 1 28 08 218 143 1 25 28 217 144 1 22 42 216 145 1 19 55 215 146 1 16 67 214 147 1 13 72 213 148 1 10 61 212 149 1 07 47 211 150 1 04 27 210 151 1 01 00 209 152 0 97 75 208 153 0 94 47 207 154 0 91 19 206 155 0 87 89 205 156 0 84 58 204 157 0 81 28 203 158 0 77 97 202 159 0 74 61 201 160 0 71 25 200 161 0 67 86 199 162 0 64 44 198 163 0 60 97 197 164 0 57 44 196 165 0 53 89 195 166 0 50 33 194 167 0 46 75 193 168 0 43 19 192 169 0 39 64 191 170 0 36 06 190 171 0 32 50 189 172 0 28 91 188 173 0 25 31 187 174 0 21 69 186 175 0 17 08 185 176 0 14 47 184 177 0 10 86 183 178 0 07 25 182 179 0 03 64 181 180 0 00 00 180

A Catalogue of some of the most eminent Cities and Towns in England , Ireland , and other Countreys , wherein is shewed the difference of their Merdians from London , with the height of the Pole Artique .

Names of the Places . Diff. in time Pole ABerden in Scotland S 0 12 58 67 S. Albons S 0 02 51 92 Alexandria in Egypt A 2 18 30 97 Amsterdam in Holland A 0 35 52 42 Athens in Greece A 1 87 37 70 Bethelem A 2 77 31 83 Barwick S 0 10 55 82 Bedford S 0 03 52 30 Calice in France 0 00 50 87 Cambridge A 0 03 52 33 Canterbury A 0 08 51 45 Constantinople A 2 30 43 00 Darby S 0 08 53 10 Dublin in Ireland S 0 43 53 18 Dartmouth S 0 25 50 53 Ely A 0 02 52 33 Grantham S 0 03 52 97 Glocester S 0 15 52 00 Hartford S 0 02 51 83 Hierusalem A 3 08 32 17 Huntington S 0 02 52 32 Leicester S 0 07 52 67 Lincolne S 0 02 57 25 Nottingham S 0 07 53 05 Newark S 0 05 53 03 Newcastle S 0 10 54 97 Northampton S 0 07 52 30 Oxford S 0 08 51 90 Peterborough S 0 03 52 38 Richmond S 0 10 54 43 Rochester A 0 05 51 47 Rochel in France S 0 07 45 82 Rome in Italy A 0 83 42 03 Stafford S 0 13 52 92 Stamford S 0 03 52 68 Sbrewsbury S 0 18 54 80 Tredagh in Ireland S 0 45 53 63 Uppingham S 0 05 52 67 Uraniburge A 0 83 55 90 Warwick S 0 10 52 42 Winchester S 0 08 51 17 Waterford in Ireland S 0 45 52 37 Worcester S 0 15 52 33 Yarmouth A 0 10 52 75 York S 0 07 54 00 LONDON 0 00 51 53

Probl. 2. To finde the Suns greatest declination , and the Poles elevation .

THe Declination of a Planet or other Star is his distance from the Equator , and as he declines from thence either Northward or Southward , so is the Declination thereof counted either North or South .

In the annexed Diagram , GMNB represents the Meridian , LK the Equinoctiall , HP the Zodiac , A the North pole , O the South , MB the Horizon , G the Zenith , N the Nadir , HC a parallel of the Suns diu●nall motion at H , or the Suns greatest declination from the Equator towards the North pole , PQ a parallel of the Suns greatest declination from the Equator towards the South pole . From whence it is apparent , that from M to H is the Suns greatest Meridian altitude , from M to Q his least ; if therefore you deduct MQ , the least Meridian altitude from MH , the greatest , the difference will be HQ , the Suns greatest declination on both sides of the Equator , and because the angles HDL and KDP are equal , by the 9th . of the second , therefore the Suns greatest declination towards the South pole is equall to his greatest declination towards the North ; and consequently , half the distance of the Tropicks , or the arch HQ , that is , the arch HL is the quantity of the Suns greatest declination . And then if you deduct the Suns greatest declination , or the arch HL from the Suns greatest Meridian altitude , or the arch MH , the difference will be ML , or the height of the Equator above the Horizon , the complement whereof to a Quadrant is the arch MO equal to AB , the height of the Pole .

Example .

The Suns greatest meridian altitude at London about the 11 th . of June was found to be 62 00 00 His least December 10. 14 94 00 Their difference is the distance of the Tropicks 47 06 00 Half that the Suns greatest declin . 23 53 00 Whose difference from the greatest Altitude is the height of the Equator 38 47 00 Whose complement is the Poles elevation 51 53 00

Probl. 3. The Suns place and greatest declination given to finde the declination of any point of the Ecliptique .

IN this figure let DFHG denote the Solsticiall Colure , FBAG the Equator , DAH the Ecliptique , I the Pole of the Ecliptique , E the Pole of the Equator , CEB a Meridian line passing from E through the Sun at C , and falling upon the Equator FAG with right angles in the point B. Then is DAF the angle of the Suns greatest declination , AC the Suns distance from Aries the next Equinoctiall point , BC the declination of the point sought .

Now suppose the sun to be in 00 deg. of Gemini , which point is distant from the next Equinoctiall point 60 deg. and his declination be required . In the rectangled spherical triangle we have known , 1 The hypothenusal AC 60 deg. 2 The angle at the base BAC 23 deg. 53 min. Hence to finde the perpendicular BC , by the 8 Case of right angled sphericall triangles , the analogie is , As the Radius , 90 10.000000 To the sine of BAC , 23.53 . 9.601222 So is the sine of AC , 60 9.937531 To the sine of BC , 20.22 9.538753

Probl. 4. The greatest declination of the Sun , and his distance from the next Equinoctial point given , to finde his right ascension .

IN the Triangle ABC of the former diagram , having as before , the angle BAC , and the hypothenusal AC , the Right Ascension of the sun AB may be found by the 7 Case of right angled spherical triangles : for As the Radius , 90 10. ●00000 To the Co-sine of CAB , 23.53 9.962299 So is the tangent of AC , 60 10.238561 To the Tangent of AB , 57.80 10.200860

Only note , that if the Right Ascension of the point sought be in the second Quadrant ( as in ♋ ♌ ♍ ) the complement of the arch found to 180 is the arch sought . If in the third Quadrant ( as in ♎ ♏ ♐ ) adde a semicircle to the arch found ; if in the last Quadrant , substract the arch found from 360 , and their difference shall be the Right Ascension sought .

Probl. 5. The Latitude of the place , and declination of the Sun given , to finde the Ascensionall difference , or time of the Suns rising before or after the houre of six .

THe Ascensionall difference is nothing else but the difference between the Ascension of any point in the Ecliptique in a right Sphere , and the ascension of the same point in an oblique Sphere .

As in the annexed Diagram , AGEV represents the Meridian , EMT the Horizon , GMCV the Equator , A the North Pole , VT the complement of the Poles elevation , BC the Suns declination , DB an arch of the Ecliptique , DC the Right Ascension , MC the Ascensionall difference . Then in the right angled triangle BMC , we have limited , 1 The angle BMC , the complement of the Poles elevation , 38 deg. 47 min. 2 The perpendicular BC , the Suns Declination 20 deg. 22 min.

Hence to finde MC the Ascensional difference , by the 6 Case of right angled Spherical Triangles , the Proportion is , As the Radius , 90 10.000000 To the tangent of BC , 20.22 9.566231 So is co-tangent of BMC , 38.47 10.099861 To the sine of MC , 27.62 9.666092

Probl. 6. The Latitude of the place , and the Suns Declination given , to finde his Amplitude .

THe Suns Amplitude is an arch of the Horizon intercepted between the Equator , and the point of rising , that is , in the preceding Diagram the arch MB , therefore in the right angled Sphericall triangle MBC , having the angle BMC the height of the Equator , 38 deg. 47 min. and BC the Suns declination 20 de. 22 m. given , the hypothenusal MB may be found by the 5 Case of right angled sphericall triangles : for As the sine of BMC , 38.47 9.793863 Is to the Radius , 90 10.000000 So is the sine of BC , 20.22 9.538606 To the sine of MB , 33.75 9.744743

Probl. 7. The Latitude of the place , and the Suns Declination given , to finde the time when he will be East or West .

LEt ABCD in the annexed diagram represent the Meridian , BD the Horizon , FG the Equator , HNK an arch of a Meridian , AC the Azimuth of East and West , or first Verticall , EM , a parallel of declination . Then in the right angled sphericall triangle AHN , we have known , 1 The perpendicular AH , the complement of the Poles elevation , 38 deg. 47 mi. 2 The hypothenusal HN , the complement of the Suns declination , 69 deg. 78 m.

Hence the angle AHN may be found by the 13 Case of right angled sphericall triangles .

As the Radius 90 10,000000 To the tangent of AH 38.47 . 9.900138 So is the co-tangent HN 69.78 . 9.566231 To the co-sine of AHN 72.98 . 9.466369

Whose complement NHZ 17 degr. 2 min. being converted into time , giveth one houre , 13 minutes , or centesmes of an hour , and so much is it after six in the morning when the Sun will be due East , and before six at night , when he will be due West .

Probl. 8. The Latitude of the place and Declination of the Sun given , to finde his Altitude when he cometh to be due East or west .

IN the right angled sphericall triangle NQZ of the last Diagram , we have limited . 1. The perpendicular QN , the Suns declination . 2. The angle at the base NZQ , the Poles elevation 51 degr. 53 min. Hence to finde the hypothenusal NZ , by the fift Case of right angled sphericall Triangles , the proportion is ; As the sine of the ang. NZQ 51.53 . 9.893725 Is to the Radius 90 10.000000 So is the sine of NQ 20.22 . 9.538606 To the sine of NZ 26.20 . 9.644881

Probl. 9. The Latitude of the place , and Declination of the Sun given , to finde the Suns Azimuth at the hour of six .

IN the right angled sphericall triangle AIH of the seventh Probleme , we have known : 1. The base AH , the complement of the Poles elevation 38 degr. 47 min. and the perpendicular IH , the complement of the Suns declination 69 degr. 78 min. Hence to finde the angle at the base HAI the suns Azimuth at the houre of six , by the 11 Case of right angled spherical triangles , the proportion is , As the Radius , 90 10.000000 To the sine of AH , 38.47 9.793863 So the co-tangent of HI , 69.78 9.566231 To the co-tangent of HAI , 77. 9.360094

Probl. 10. The Poles elevation , with the Suns Altitude and Declination given , to finde the Suns Azimuth .

IN the oblique angled Spherical triangle AHS , in the Diagram of the seventh Probleme , we have known , the side AH , the complement of the Poles elevation , 38 deg. 47 min. HS , the complement of the Suns declination , 74 deg. 83 min. And the side SA , the complement of the Suns altitude , 57 deg. 53 min , to finde the angle SAH : Now then , by the 11 Case of Oblique angled Sphericall Triangles , I work as is there directed .

SH , 74.83 HA , 38,47 9.793863 SA , 57.53 9 . 9●6174 Summe of the sides 170.83 19.720037 Halfe summe 85. 41. 50 Dif . of HA & half sum , 46.91.50 9.863737 Dif . of SA & half sum , 27.88.50 9.669990 The doubled Radius 20.000000 Their summe 39.533727 From whence substract 19.720037 There rests 19.813690 The halfe whereof 9.906845

Is the sine of 53 deg. 80 min. which doubled is 107 deg. 60 min. the Suns Azimuth from the north , and 72 deg. 40 min. the complement thereof to a Semicircle is the Suns Azimuth from the South .

CHAP. II. THE ART OF SHADOWS : Commonly called DIALLING . Plainly shewing out of the Sphere , the true ground and reason of making all kinde of Dials that any plain is capable of .
Problem 1. How to divide diverse lines , and make a Chord to any proportion given .

FOrasmuch as there is continuall use both of Scales and Chords in drawing the Scheams and Dials following , it will be necessary first to shew the making of them , that such as cannot have the benefit of the skilful artificers labour , may by their own pains supply that defect .

Draw therefore upon a piece of paper or pastboard a streight line of what length you please , divide this line into 10 equal parts , and each 10 into 10 more , so is your line divided into 100 equal parts , by help where of a line of Chords to any proportion may be thus made .

First , prepare a Table , therein set down the degrees , halves , and quarters , if you please , from one to 90. Unto each degree and part of a degree joyn the Chord proper to it , which is the naturall sine of halfe the arch doubled , by the 19th . of the second of the first part : if you double then the naturall sines of 5. 10. 20. 30. degrees , you shall produce the Chords of 10. 20. 40. 60. degrees : Thus 17364 the sine of 10 de . being doubled , the sum will be 34●28 , the Chord of 20 deg. and so of the rest as in the Table following .

De Chord 1 17 2 35 3 52 4 70 5 87 6 105 7 122 8 139 9 157 10 175 11 192 12 209 13 226 14 244 15 261 16 278 17 296 18 313 19 330 20 347 21 364 22 382 23 398 24 416 25 432 26 450 27 466 28 384 29 501 30 ●18 31 534 32 551 33 568 34 585 35 601 36 618 37 635 38 651 39 668 40 684 41 700 42 717 43 733 44 749 45 765 44 781 47 797 48 813 49 830 50 845 51 861 52 876 53 892 54 908 55 923 56 939 57 954 58 970 59 984 60 1000 61 1015 62 1030 63 1045 64 1060 65 1074 66 1089 67 1104 68 1118 69 1133 70 1147 71 1161 72 1176 73 1190 74 1204 75 1217 76 1231 77 1245 78 1259 79 1273 80 1286 81 1299 82 1312 83 1325 84 1338 85 1351 86 1364 87 1377 88 1389 89 1402 90 1414

This done , proportion the Radius of a circle to what extent you please , make AB equal thereto , in the middle whereof , as in C , erect the perpendicular CD , and draw the lines AD and BD , equal in length to your line of equal parts , so have you made an equiangled Triangle , by help whereof and the Table aforesaid , the Chord of any arch proportionable to this Radius may speedily be obtained .

As for example . Let there be required the Chord of 30 deg. the number in the Table answering to this arke is 518 , or in proportion to this Scale 52 almost , I take therefore 52 from the Scale of equal parts , and set them from D to E and F , and draw the line EF , which is the Chord desired . Thus may you finde the Chord of any other arch agreeable to this Radius . Or if your Radius be either of a greater or lesser extent , if you make the base of your Triangle AB equal thereunto , you may in like manner finde the Chord of any arch agreeable to any Radius given . Only remember that if the Chord of the arch desired exceed 60 deg. the sides of the Triangle AD and DB must be continued from A and B as far as need shall require . In this manner is made the line of Chords adjoyning , answerable to the Radius of the Fundamental Scheme .

And in this manner may you finde the Sine , Tangent or Secant of any arch proportionable to any Radius , by help of the Canon of Naturall Sines , Tangents and Secants , and the aforesaid Scale of equall parts , as by example may more plainly appear .

Let there be required the sine of 44 degrees in the table of natural sines , the number answering to 44 degrees is 694. I take therefore with my compasses 69 from my Scale of equal parts , and set them from D to G and H ; so is the line GH the sine of 44 degrees , where the Radius of the circle is AB .

Again , if there were required the tangent of 44 degrees , the number in the table is 965 ; and therefore 96 set from D to K and L shall give the tangent required ; and so for any other .

Your Scales being thus prepared for the Mechanicall part , we will now shew you how to project the Sphere in plano , and so proceed to the arithmeticall work .

Probl. 2. The explanation and making of the fundamental Diagram .

THis Scheme representeth to the eye the true and natural situation of those circles of the Sphere , whereof we shall have use in the description of such sorts of Dials as any flat or plane is capable of . It is therefore necessary first to explain that , and the making thereof , that the Symetry of the Scheme with the Globe being well understood , the representation of every plane therein may be the better conceived .

Suppose then that the Globe elevated to the height of the Pole be prest flat down into the plane of the Horizon , then will the outward circle or limbe of this Scheme NESW represent that Horizon , and all the circles contained in the upper Hemisphere of the Globe may artificially be contrived , and represented thereon , as Azimuths , Almicanters , Meridians , Parallels , Equator , Tropicks , circles of position , and such like , the which in this Diagram are thus distinguished .

The letter Z represents the Zenith of the place , and the center of the horizontal circle , NZS represents the meridian , P the pole of the world elevated above the North part of the Horizon N here at London , 51 degrees 53 minutes , or centesmes of a degree , the complement whereof PZ 38 degrees , 47 minutes , and the distance between the Pole and the Zenith ; EZW is the prime vertical , DZG and CZV any other intermediate Azimuths , NOS a circle of position , EKW the Equator , the distance whereof from Z is equal to PN , the height of the Pole , or from S equal to PZ , the complement thereof , HBQX the Tropick or parallel of Cancer , LFM , the Tropique of Capricorn , the rest of the circles intersecting each other in the point P , are the meridians or hour-circles , cutting the Horizon and other circles of this Diagram in such manner as they do in the Globe itself .

Amongst these the Azimuths onely in this projection become streight lines , all the rest remain circles , and are greater or lesser , according to their natural situation in the Globe , and may be thus described .

Open your compasses to the extent of the line AB in the former Probleme , ( or to any other extent you please ) with that Radius , or Semidiameter describe the horizontal circle NESW , crosse it at right angles in Z with the lines NZS and EZW .

That done , seek the place of the Pole at P , through which the hour circles must pass , the Equinoctial point at K , the Tropiques at T and F , the reclining circle at O , and the declining reclining at A ; all which may thus be found .

The Zenith in the Globe or Materiall Sphere is the Pole of the Horizon , and Z in the Scheme is the center of the limbe , representing the same , from which point the distance of each circle being given both wayes , as it lyeth in the Sphere , and set upon the Azimuth , or streight line of the Scheme proper thereunto , you may by help of the natural tangents of half their arches give three points to draw each circle by , for if the naturall tangents of both distances from the Zenith be added together , the half thereof shall be the Semidiameters of those circles desired .

The reason why the natural tangent of half the arches are here taken , may be made plain by this Diagram following . Wherein making EZ the Radius , SZN is a tangent line thereunto , upon which if you will project the whole Semicircle SWN , it is manifest , by the work , that every part of the lines ZN or ZS can be no more then the tangent of half the arch desired , because the whole line ZN or ZS is the tang . of no more then half the Quadrant , that is , of 45 degrees , by the 19th . of the second Chapter of the first Part ; and therefore WEA is but half the angle WZA and WEB is but half the angle WZB.

Now then if EZ or Radius of the fundamental Scheme be 1000 , ZP shal be 349 , the natural tangent of 19 degrees , 23 minutes , 50 seconds , the half of 38 degrees , 47 minutes , the distance between the North pole and the Zenith in our Latitude of 51 degrees , 53 minutes , or centesmes of a degree . And the South pole being as much under the Horizon as the North is above it , the distance thereof from the Zenith must be the complement of 38 degrees , 47 minutes to a Semicircle , that is , 141 degrees , 53 minutes ; and as the half of 38 degrees , 47 minutes , viz. 19 degrees , 23 minutes , 50 seconds is the quantity of the angle PEZ , and the tangent thereof the distance from Z to P , so the half of 141 degrees , 53 minutes , viz. 70 degrees , 76 minutes , 50 seconds must be the measure of the angle in the circumference between the Zenith and the South , the tangent whereof 2866 must be the distance also , and the tangents of these two arches added together 3215 , is the whole diameter of that circle , the half whereof 1607 , that is , one Radius , and neer 61 hundred parts of another is the Semidiameter or distance from P to L in the former Scheme , to which extent open the compasses , and set off the distance PL , and therewith draw the circle WPE for the six of the clock hour .

The Semidiameters of the other circles are to be found in the same manner : the distance between the Zenith and the Equinoctiall is alwayes equal to the height of the Pole , which in our Latitude is 51 degr. 53 min. and therefore the half thereof 25 degrees , 76 minutes , 50 seconds is the measure of the angle WEB , and the natural tangent thereof 483 , which being added to the tangent of the complement 2070 , their aggregate 2553 will be the whole diameter of that circle , and 1277 the Radius or Semidiameter by which to draw the Equinoctiall circle EKW .

The Tropique of Cancer is 23 degrees , 53 minutes above the Equator , and 66 degrees 47 minutes distant from the Pole , and the Pole in this Latitude is 38 degrees 47 min. distant from the Zenith , which being substracted from 66 degrees 47 minutes , the distance of the Tropique of Cancer from the Zenith , will be 28 , the half thereof is 14 , whose natural tangent 249 being set from Z to T , giveth the point T in the Meridian , by which that parallel must passe ; the distance thereof from the Zenith on the North side is TN 90 degrees , and substracting 23 degrees , 53 minutes , the height of the Tropique above the Equator , from 38 degrees , 47 minutes , the height of the Equator above the Horizon , their difference is 14 degrees , 94 minutes , the distance of the Tropique from N under the Horizon ; and so the whole distance thereof from Z is 104 degrees , 94 minutes , the half whereof is 52 degrees , 47 minutes , and the natural tangent thereof 1302 added to the former tangent 249 , giveth the whole diameter of that circle 1551 , whose half 776 is the Semidiameter desired , and gives the center to draw that circle by .

The Tropique of Capricorn is 23 degrees , 53 minutes below the Equator , and therefore 113 degrees 53 minutes from the North pole , from which if you deduct , as before , 38 degrees , 47 minutes , the distance of the Pole from the Zenith , the distance of the Tropique of Capricorn from the Zenith will be 75 degrees , 6 minutes , and the half thereof 37 degrees , 53 minutes , whose natural tangent 768 being set from Z to F , giveth the point F in the Meridian , by which that parallel must pass : the distance thereof from the Zenith on the North side is ZN 90 degrees , as before ; and adding 23 degrees , 53 minutes , the distance of the Tropique from the Equator to 38 degrees , 47 minutes , the distance of the Equator from the Horizon , their aggregate is 62 degrees , the distance of the Tropique from the Horizon , which being added to ZN 90 degrees , their aggregate is 152 degrees , and the half thereof 76 degrees , whose natural tangent 4011 being added to the former tangent 768 , giveth the whole diameter of that circle 4.779 , whose half 2.389 is the Semidiameter desired , and gives the center to draw that circle by .

The distance of the reclining circle NOS from Z to O is 40 degrees , the half thereof 20 , whose naturall tangent 3.64 set from Z to O , giveth the point O in the prime vertical EZW , by which that circle must pass ; the distance thereof from the Zenith on the East side is ZE 90 degrees , to which adding 50 degrees , the complement of the former arch , their aggregate 140 degrees is the distance from Z Eastward , and the half thereof 70 degrees , whose natural tangent 2747 being added to the former tangent 364 , their aggregate 3111 is the whole diameter of that circle , and the half thereof 1555 is the Semidiameter desired , and gives the center to draw that circle by .

The distance of the declining reclining circle DAG from the Zenith is ZA 35 deg. the half thereof 17 degrees , 50 minutes , whose natural tangent 315 being set from Z to A , giveth the point by which that circle must passe , and the natural tangent of 7● degr. 50 min. the complement thereof 317● being added thereto is 3486 , the whole diameter of the circle , and the half thereof 1743 , the Semidiameter desired , and giveth the center to draw that circle by .

The streight lines CZA or DZG are put upon the limbe by help of a line of Chords 30 degrees distant from the Cardinal points NESW , and must crosse each other at right angles in Z , representing two Azimuths equidistant from the Meridian and prime verticall .

Last of all , the hour-circles are thus to be drawn ; first , seek the center of the six of clock hour-circle , as formerly directed , making ZE the Radius , and is found at L upon the Meridian line continued from P to L , which cross at right angles in L with the line 8 L 4 , extended far enough to serve the turn , make PL the Radius , then shall 8 L 4 be a tangent line thereunto , and the natural tangents of the Equinoctiall hour arches , that is the tangent of 15 degrees 268 for one hour , of 30 degr. 577 for two , hours , of 45 degrees 1000 for three hours of 60 deg. 1732 for four hours , and 75 deg. 3732 for five hours set upon the line from L both wayes , that is , from L to 5 and 7 , 4 and 8 , and will give the true center of those hour-circles : thus , 5 upon the line 8 L 4 is the center of the hour-circle 5 P 5 , and 7 the center of the hour-circle 7 P 7 ; and so of the rest .

The centers of these hour-circles may be also found upon the line 8 L 4 by the naturall secants of the same Equinoctiall arches , because the hypothenuse in a right angled plain triangle is alwayes the secant of the angle at the base , and the perpendicular the tangent of the same angle : if therefore the tangent set from L doth give the center , the secant set from P shall give that center also . The Scheme with the lines and circles thereof being thus made plain , we come now to the Art of Dialling it self .

Probl. 3. Of the severall plains , and to finde their scituation .

ALL great Circles of the Sphere , projected upon any plain , howsoever situated , do become streight lines , as any one may experiment upon an ordinary bowle thus . If he saw the Bowle in the midst , and joyne the two parts together again , there will remain upon the circumference of the Bowle , some signe of the former partition , in form of a great Circle of the Sphere : now then , if in any part of that Circle the roundnesse of the bowle be taken off with a smoothing plain or otherwise , as the bowle becomes flat , so will the Circle upon the bowle become a streight line ; from whence it follows , that the houre lines of every Diall ( being great Circles of the Sphere ) drawn upon any plain superficies , must also be streight lines .

Now the art of Dialling consisteth in the artificiall finding out of these lines , and their distances each from other , which do continually varie according to the situation of the plain on which they are projected .

Of these plains there are but three sorts . 1. Parallel to the Horizon , as is the Horizontal only . 2. Perpendicular to the Horizon , as are all erect plains , whether they be such as are direct North , South , East or West , or such as decline from these points of North , South , East , or West . 3. Inclining to the Horizon , or rather Reclining from the Zenith , and these are direct plains reclining and inclining North and South , and reclining and inclining East and West , or Declining-reclining and inclining plains .

To contrive the houre lines upon these severall plains , there are certain Spherical arches and angles , in number six , which must of necessity be known , and divers of these are in some Cases given , in others they are sought .

1. The first is an arch of a great Circle perpendicular to the plain , comprehended betwixt the Zenith and the plain , which is the Reclination , as ZT , ZK , and ZF , in the fundamental Diagram . 2. The second is an arch of the Horizon betwixt the Meridian and Azimuth passing by the poles of the plain , as SV or NC in the Scheme . 3. The third is an arch of the plain betwixt the Meridian and the Horizon , prescribing the distance of the 12 a clock houre from the horizontal line , as PB in the Scheme of the 11th . Probl. 4. The fourth is an arch of the plain betwixt the Meridian and the substile , which limits the distance thereof from the 12 a clock houre line , as ZR in the Scheme . 5. The fifth is an arch of a great Circle perpendicular to the plain , comprehended betwixt the Pole of the World ; and the plain , commonly called the height of the stile , as PR in the Scheme . 6. The last is an angle at the Pole betwixt the two Meridians , the one of the place , the other of the plain ( taking the substile in the common sense for the Meridian of the plain ) as the angle ZPR in the fundamental Scheme .

The two first of these arches are alwayes given , or may be found by the rules following .

To finde the Inclination or Reclination of any plain .

If the plain seem to be level with the Horizon , you may try it by laying a ruler thereupon , and applying the side of your Quadrant AB to the upper side of the ruler , so that the center may hang a little over the end of the ruler , and holding up a threed and plummet , so that it may play upon the center , if it shall fall directly upon his level line AC , making no angle therewith , it is an horizontal plain .

If the plain seeme to be verticall , like the wall of an upright building , you may try it by holding the Quadrant so that the threed may fall on the plumb line AC , for then if the side of the Quadrant shall lie close to the plain , it is erect , and a line drawn by that side of the Quadrant shall be a Verticall line , as the line DE in the figure .

If the plain shall be found to incline to the Horizon , you may finde out the quantity of the inclination after this manner . Apply the side of your Quadrant AC to the plain , so shall the threed upon the limbe give you the inclination required .

Suppose the plain to be BGED , and the line FZ to be verticall , to which applying the side of your Quadrant AC , the threed upon the limbe shall make the angle CAH the inclination required , whose complement is the reclination .

To finde the declination of a plain .

To effect this , there are required two observations , the first is of the horizontal distance of the Sun from the pole of the plain , the second is of the Suns altitude , thereby to get the Azimuth : and these two observations must be made at one instant of time as neer as may be , that the parts of the work may the better agree together .

1. For the horizontal distance of the Sun from the pole of the plain , apply one edge of the Quadrant to the plain , so that the other may be perpendicular to it , and the limbe may be towards the Sun , and hold the whole Quadrant horizontal as neer as you can conjecture , then holding a threed and plummet at full liberty , so that the shadow of the threed may passe through the center and limb of the Quadrant , observe then what degrees of the limb the shadow cuts , counting them from that side of the Quadrant which is perpendicular to the horizontal line , those degrees are called the Horizontal distance . 2. At the same instant observe the Suns altitude , by this altitude you may get the Suns Azimuth from the South , by the 10th . Probleme of the first Chapter hereof .

When you make your observation of the Suns horizontal distance , marke whether the shadow of the threed fall between the South , and the perpendicular side of the Quadrant , or not , for , 1. If the shadow fall between them , then the distance and Azimuth added together do make the declination of the plain , and in this case the declination is upon the same coast whereon the Suns Azimuth is . 2. If the shadow fall not between them , then the difference of the distance and Azimuth is the declination of the plain , and if the Azimuth be the greater of the two , then the plain declineth to the same coast whereon the Azimuth is , but if the distance be the greater , then the plain declneth to the contrary coast to that whereon the Suns Azimuth is .

Note here further , that the declination so found , is alwayes accounted from the South , and that all declinations are numbered from North or South , towards East or West , and must not exceed 90 deg.

1. If therefore the number of declination exceed 90 , you must take its complement to 180 , and the same shall be the plains declination from the North. 2. If the declination found exceed 180 deg. then the excesse above 180 , gives the plains declination from the North , towards that Coast which is contrary to the Coast whereon the Sun is .

By this accounting from North & South , you may alwayes make your plains declination not to exceed a Quadrant or 90 de . And as when it declines nothing , it is a full South or North plain , so if it decline just 90 , it is then a full East or West plain .

These precepts are sufficient to finde the declination of any plain howsoever situated , but that there may be no mistake , we will adde an Example .

1 Example .

Now because the line of shadow AG , falleth between P the pole of the plains horizontal line , and S the South point , therefore according to the former direction , I adde the horizontal distance PG 24 deg. to the Suns Azimuth GS 40 deg , and their aggregate is PS 64 deg. the declination sought ; and in this case it is upon the same coast with the sun , that is West , according to the rule given , and as the figure it selfe sheweth , the East and North points being hid from our sight by the plain it selfe ; this therefore is a South plain declining West 64 degrees .

2. Example .

To finde a Meridian line upon an Horizontal plain .

If your plain be levell with your Horizon , draw thereon the Circle BCMP , then holding a threed and plummet , so as the shadow thereof may fall upon the center , and draw in the last diagram the line of shadow HA : then if the Suns Azimuth shall be 50 deg. and the line of shadow taken in the afternoon , set off the 50 deg. from H to S , and the line SN shall be the Meridian line desired .

Probl. 4. To draw the houre lines upon the Horizontal plain .

THis plane in respect of the Poles thereof , which lie in the Vertex and Nadir of the place may be called vertical , in respect of the plane it self , which is parallel to the Horizon , horizontal , howsoever it be termed , the making of the Dial is the same , and there is but one onely arch of the Meridian betwixt the pole of the world and the plane required to the artificiall projecting of the hour-lines thereof , which being the height of the pole above the horizon ( equal to the height of the stile above the plane ) is alwayes given , by the help whereof we may presently proceed to calculate the hour distances in manner following .

This plane is represented in the fundamental Diagram by the outward circle ESWN , in which the diameter SN drawn from the South to the North may go both for the Meridian line , and the Meridian circle , Z for the Zenith , P for the pole of the world , and the circles drawn through P for the hour-circles of 1 , 2 , 3 , 4 , &c. as they are numbred from the Meridian , and limit the distance of each hour line from the Meridian upon the plane , according to the arches of the Horizon , N 11 , N 10 , N 9 , &c. which by the severall Triangles SP 11 , SP 10 , SP 9 , or their verticals NP 11 , NP 10 , NP 9 may thus be found ; because every quarter of the Horizon is alike , you may begin with which you will , and resolve each hours distance , either by the small Triangle NP 11 , or the verticall Triangle KP 11. In the Triangle PN11 , the side PN is alwayes given , and is the height of the pole above the horizon , the which at London is 51 deg. 53 min. and the angle at P is given one hours distance from the Meridian , whose measure in the Equinoctiall is 15 deg. & the angle at N is alwayes right , that is 90 deg. wherefore by the first case of right angled spherical Triangles , the perpendicular N 11 may thus be found . As Radius 90 , 10.000000 To the tangent of NP11 , 15d . 9.428052 So is the sine of PN 51.53 . 9.893725 To the tangent of N11 , 11.85 . 9.321777

Which is the distance of the hours of 1 and 11 , on each side of the Meridian , thus in all respects must you finde the distance of 2 and 10 of clock , by resolving the triangle NP10 , and of 3 and 9 of clock , by resolving the triangle NP9 ; and so of the rest : in which , as the angle at Pincreafeth which for 2 hours is 30 degrees , for 3 hours 45 degr. for 4 hours 60 degr. for 5 hours 75 degr. so will the arches of the Horizon N10 , N9 , N8 , N7 , vary proportionably , and give each hours true distance from the Meridian , which is the thing desired .

Probl. 5. To draw the hour-lines upon a direct South or North plane .

EVery perpendicular plane , whether direct or declining , lieth in some Azimuth or other ; as here the South wall or plane doth lie in the prime vertical or Azimuth of East and West , represented in the fundamental Diagram by the line EZW , and therefore it cutteth the Meridian of the place at right angles in the Zenith , and hath the two poles of the plane seated in the North and South intersection of the Meridian and Horizon ; and because the plane hideth the North pole from our sight , we may therefore conclude , ( it being a general rule that every plane hath that pole depressed , or raised above it , which lieth open unto it ) that the South pole is elevated thereupon , and the stile of this Diall must look downwards thereunto , erected above the plane the height of the Antartick Pole , which being an arch of the Meridian betwixt the South pole and the Nadir , is equall to the opposite part thereof , betwixt the North pole and the Zenith ; and therefore the complement of the North pole above the horizon .

Suppose then that P in the fundamental Scheme , be now the South pole , and N the South part of the Meridian , S the North ; then do all the hour-circles from the pole cut the line EZW , representing the plane unequally , as the hour-lines will do upon the plane it self , and as it doth appear by the figures set at the end of every hour line in the Scheme . Now having already the poles elevation given , as was in the horizontal , there is nothing else to be done , but to calculate the true hour-distances upon the line EZW from the meridian SZN ; and then to proceed , as formerly , and note that because the hours equidistant on both sides the meridian , are equal upon the plane , the one half being found , the other is also had , you may therefore begin with which side you will .

In the triangle ZP11 , right angled at Z , I have ZP given , the complement of the height of the pole 38 deg. 47 min. the which is also the height of the stile to this Diall , and the angle at P15 degrees one hours distance from the meridian upon the Equator to finde the side Z11 , for which by the first case of right angled sphericall triangles , the proportion is , as before . As the Radius 90 , 10.000000 To the sine of PZ 38.47 . 9.793863 So is the tangent of ZP11 , 15d . 9.428052 To the tangent of Z11 , 9.47 . 9.221915

And thus in all respects must you finde the distance of 2 and 10 , of 3 and 9 ; and so forward , as was directed for the houres in the horizontal plane .

The North plane is but the back side of the South , lying in the same Azimuth with it , & represented in the Scheme by the back part of the same streight line EZW , whatsoever therefore is said of the South plane may be applied to the North ; because as the South pole is above the South plane 38 degr. 47 min. so is the North pole under the North plane as much , and each stile must respect his own pole , onely the meridian upon this plane representeth the midnight , and not the noon , and the hours about it 9 , 10 , 11 , and 1 , 2 , 3 , are altogether uselesse , because the Sun in his greatest northern declination hath but 39 degr. 90 min. of amplitude in this our Latitude ; and therefore riseth but 22 min. before 4. in the morning , and setteth so much after 8 at night ; neither can it shine upon this plane longer then 35 min. past 7 in the morning , and returning to it as much before 5 at night , because then the Sun passeth on the North side of the prime vertical , in which this plane lieth , and cometh upon the South .

Now therefore to make this Dial , is but to turn the South Dial upside down , and leave out all the superfluous hours between 5 and 7 , 4 and 8 , and the Diall to the North plane is made to your hand .

The Geometricall projection .

To project these and the Horizontal Dials , do thus : First , draw the perpendicular line CEB , which is the twelve of clock hour , crosse it at right angles with 6C6 , which is the six of clock hour ; then take with your compasses 60 deg. from a line of Chords , and making C the center draw the circle 6E6 , representing the azimuth in which the plane doth lie ; this done , take from the same Chord all the hour distances , and setting one foot of your compasses

in E , with the other mark out those hour distances before found by calculation , both wayes upon the circle 6E6 ; streight lines drawn from the center C to those pricks in the circle are the true hour-lines desired .

Having drawn all the hour-lines , take from the same line of Chords the arch of your poles elevation , or stile above the plane , and place it from E to O , draw the prickt line COA representing the axis or heighth of the stile , from any part of the meridian draw a line parallel to 6C6 , as is BA , & it shall make a triāgle , the fittest form to support the stile at the true height ; let the line 6C6 be horizontal , the triangular stile CBA erected at right angles over the 12 of clock line , and then is the Diall perfected either for the Horizontal , or the direct North and South planes .

Probl. 6. To draw the hour-lines upon the direct East or West planes .

AS the planes of South and North Dials do lie in the Azimuth of East and West , and their poles in the South and North parts of the meridian ; so do the planes of East and West Dials lie in the South and North azimuth , and their poles in the East and West part of the Horizon , from whence these Dials receive their denomination , and because they are parallel to the meridian line in the fundamental Scheme SZN , some call them meridian planes .

And because the meridian , in which this plane lieth , is one of the hour-circles , and no plane that lieth in any of the hour circles can cut the axis of the world , but must be parallel thereunto ; therefore the hour lines of all such planes are also parallel each to other , and in the fundamental Scheme may be represented in this manner .

Let NESW in this case be supposed to be the Eq●inoctiall divided into 24 equall parts , and let the prickt line E 8. 7. parallel to ZS be a tangent line to that circle in E , straight lines drawn from the center Z thorow the equal divisions of the limbe , intersecting the tangent line , shall give points in 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , thorow which parallels being drawn to the prime vertical , or 6 of clock hour line EZW , you have the hour-lines desired , which may for more certainties sake be found by tangents also ; for making ZE of the former Scheme to be the Radius , and E 8. 7. a tangent line , as before ; then shall the naturall tangent of 15 degr. 268 taken from a diagonal scale equal to the Radius , and set both wayes from E upon the tangent line E 8. 7. gives the distance of the houres of 5 and 7 , the tangent of 30 degr. the distance of the hours of 4 and 8 , and the tangent of 45 degr. the distance of the hours of 3 and 9 , &c. from the six of clock hour , as before ; and is a general rule for all Latitudes whatsoever .

The Geometricall projection .

The length of the stile being thus proportioned to the plain , make that the Radius of a Circle , and then the Equator DAE shall be a Tangent line thereunto , and therefore , the naturall Tangent of 15 deg. being set upon the Equinoctiall DAE both wayes from A , shall give the points of 5 and 7 : the Tangent of 30 deg. the points of 8 and 4 , &c. through which streight lines being drawn parallel to the six a clock houre , you have at one work made both the East and West Dials , only remember that because the Sun riseth before 4 in Cancer , and setteth after 8 , you must adde two houres before six in the East Diall , and two houres after six in the West , that so the plain may have as many houres as it is capable of .

The West Dial is the same in all respect with the East , only the arch BD , or the height of the Equator , must be drawn on the right hand of the center A for the West Dial , and on the left for the East , that so the houre lines crossing it at right angles , may respect the Poles of the world to which they are parallel .

Probl. 7. To draw the houre-lines upon a South or North erect plain declining East or West , to any declination given .

EVery erect plain lieth under some Azimuth or other , and those only are said to decline which differ from the Meridian and Prime Vertical . The declination therefore being attained by the rules already given , ( or by what other means you like best ) we come to the calculation of the Diall it selfe , represented in the fundamentall Scheme by the right line GZD , the Poles whereof are C and V , the declination from the South Easterly NC , or North Westerly SV , 25 deg. supposing now S to be North , and N South ; W East , and E the West point , the houre circles proper to this plain are the black lines passing through the Pole P , and crossing upon the plain GZD , wherein note generally that where they run neerest together , thereabouts must the sub-stile stand , and alwayes on the contrary side to the declination , as in this example declining East , the stile must stand on the West side ( supposing P to be the South Pole ) between Z and D , the reason whereof doth manifestly appear ; because the Sun rising East , sendeth the shadow of the Axis West , and alwayes to the opposite part of the Meridian wherein he is , wherefore reason enforceth , that the morning houres be put on the West side of the Meridian , as the evening houres are on the East , and from the same ground that the substile of every plain representing the Meridian thereof , must alwayes stand on the contrary side to the declination of the plain and that the houre-lines must there run neerest together , because the Sun in that position is at right angles with the plain . For the making of this Diall three things must be found . 1. The elevation of the Pole above the plain , represented by PR , which is the height of the stile , and is an arch of the Meridian of the plain , between the Pole of the world and the plain . 2. The distance of the substile from the Meridian , represented by ZR , and is an arch of the plain between the Meridian and the substile . 3. The angle ZPR , which is an arch between the substile PR the meridian of the plain , and the line PZ the meridian of the place , and these are thus found .

Because the substile is the Meridian of the plain , it must be part of a great circle passing through the pole of the world , and crossing the plain at right angles , therefore in the supposed right angled triangle PRZ , ( for yet the place of R is not found ) you have given the base PZ 38 deg. 47 min. and the angle PZR the complement of the declination 65 deg. and the supposed right angle at R , to finde the side PR , which is the height of the stile as aforesaid , but as yet the place unknown : wherefore by the 8 Case of right angled Spherical Triangles the analogie is , As the Radius , 10.00000 To the sine of PZ , 38.47 9.793863 So the sine of PZR , 65 9.957275 To the sine of PR , 34.32 9.751138

Secondly , you may finde ZR the distance of the substile from the meridian , by the 7 case of right angled Spherical Triangles . As the Radius , 90 10.000000 To the Co-sine of PZR , 65 9.629378 So is the tangent of PZ , 38.47 9.900138 To the tangent of ZR , 18.70 9.529516

These things given , the angle at P between the two meridians may be found by the 9 Case of right angled Sphericall Triangles , for the proportion is , As the Radius , 10.000000 To the Co-sine of PZ , 38.47 9.893725 So the Tangent of PZR , 65 10.331327 To the Co-tang . of RPZ , 30.78 10.225052

Having thus found the angle between the Meridians to be 30 deg. 78 min. you may conclude from thence , that the substile shall fall between the 2d . & third houres distance from the Meridian of the place , and therefore between 9 and 10 of the clock in the morning , because the plain declineth East from us , 9 of the clock being 45 deg. from the Meridian , and 10 of the clock 30 deg. distant , now therefore let fall a perpendicular between 9 and 10 , the better to inform the fancie in the rest of the work , and this shall make up the Triangle PRZ before mentioned and supposed , which being found we may calculate all the houre distances by the first case of right angled sphericall Triangles . For , As the Radius , Is to the sine of the base PR ; S● is the Tangent of the angle at the perpendicular , RP 9 , To the tangent of R 9 the perpendicular

The angle at P is alwayes the Equinoctiall distance of the houre line from the substile , and may thus be sound : If the angle between the Meridians be lesse than the houre distance , substract the distance of the substile from the houre distance ; if greater substract the houre distance from that , and their difference shall give you the Equinoctiall distance required .

Thus in our Example , the angle between the Meridians was found to be 30 deg. 78 m. and the distance of 9 of the clock from 1● is three houres , or 45 deg. if therefore I substract 30 deg. 78 min. from 45 deg. the remainder will be 14 deg. 22 min. the distance of 9 of the clock from the substile . Again , the distance of 10 of the clock from the Meridian is 30 deg. and therefore if I substract 30 deg. from 30 deg. 78 min. the distance of 10 of the clock from the substile will be 78 centesms or parts of a degree : the rest of the houres and parts are easily found by a continual addition of 15 deg. for every houre , 7 deg. 50 min. for half an houre , 3 deg. 75 min. for a quarter of an houre , as in the Table following you may perceive , the which consists of three columns , the first containeth the houres , the second their Equinoctiall distances from the substile , the third and last the houre arches , computed by the former proportion in this manner . As the Radius , 90 10.000000 Is to the sine of PR , 34.32 9.751136 So is the tang . of RP 9 , 14.22 9.403824 To the tangent of R 9 , 8.13 9.154960

H Aequ . Arches 4 89 22 88 61 5 74 22 63 38 6 59 22 43 43 7 44 22 28 75 8 29 22 17 50 9 14 22 8 13 merid . substil 10 00 78 00 44 11 15 78 9 05 12 30 78 18 56 1 45 78 30 08 2 60 78 45 23 3 75 78 65 80 4 90 78 88 61

The Geometricall Projection .

Having calculated the hour distances , you may thus make the Diall ; Draw the Horizontall line ACB , then crosse it at right angles in C , with the line CO 12. Take 60 degrees from a Chord , and making C the Center , draw the Semicircle AOB , representing the azimuth GZD in the Scheme , in which the plane lieth ; upon this circle from O to N set off the distance of the substile from the Meridian , which was found before to be 18.70 . upon the West side of the Meridian , because this plane declineth East , then take off the same Chord the severall hour-distances , as they are ready calculated in the table , viz. 8.13 . for 9 , 17.50 . for 8 : and so of the rest ; and set them from the substile upon the circle RNO , as the Table it self directeth ; draw streight lines from the center C to these several points , so have you the true hour lines , which were desired : and lastly , take from the same Chord the heighth of the stile found to be 34. 32. which being set from N to R , and a streight line drawn from C through R representing the axis , the Diall is finished for use .

Nay , thus have you made four Dials in one , viz. a South declining East and West 25 degrees , and a North declining East or West as much ; to make this plainly appear , suppose in the fundamental Scheme if N were again the North part of the horizon , P the North pole , and that GZD were a North declining West 25 degrees , then do all the hour-circles crosse the same plane , as they did the former ; onely DZ which was in the former the East side will now be the West : and consequently the afternoon hours must stand where the forenoon hours did , the stile also , which in the East declining stood between 9 and 10 , must now stand between 2 and 3 of the afternoon hours . And lest there should be yet any doubt conceived , I have drawn to the South declining East 25 , the North declining West as much ; from which to make the South declining West , and North declining East , you need to do no more then prick these hour lines through the paper , and draw them again on the other side , stile and all ; so shall they serve the turn , if you place the morning hours in the one , where the afternoon were in the other .

APPENDIX . To draw the hour lines upon any plane declining far East or West , without respect to the Center .

THe ordinary way is with a Beamcompasse of 16 , 18 , or 20 foot long , to draw the Diall upon a large floor , and then to cut off the hours , stile and all , at 10 , 12 , or 14 foot distance from the center , but this being too mechanical for them that have any Trigonometrical skill , I omit , and rather commend the way following ; by help whereof you may upon half a sheet of paper make a perfect model of your Diall , to what largenesse you please , without any regard at all to the Center .

Suppose the wall or plane DZG , on which you would make a Diall to decline from N to C , that is from the South Easterly 83 degrees , 62 min. set down the Data , and by them seek the Quaesita , according to the former directions .

The Data or things given are two . 1. PS the poles elevation 51 degrees , 53 minutes . 2. SA , the planes declination southeast 83 deg. 62 min.

The Quaesita or things sought are three , 1. PR the height of the stile 3 degrees 97 minutes . 2. ZR , the distance of the substile from the meridian 38 deg. 30 min. 3. ZPR , the angle of the meridian of the plane with the meridian of the place 85 degrees , which being found , according to the former directions , the substile line must fall within five degrees of six of the clock , because 85 degrees wanteth but 5 of 90 , the distance of 6 from 12.

Now therefore make a table , according to this example following , wherein set down the houres from 12 , as they are equidistant from the meridian , and unto them adjoyn their Equinoctial distances , and write Meridian and substile between the hours of 6 and 7 , and write 5 degrees against the hour of 6 , 10 degrees against the hour of 7 , and to the Equinoctial distances of each hour adde the natural tangents of those distances , as here you see . So is the Table prepared for use , by which you may easily frame● the Diall to what greatnesse you will , after this manner . Hours Equ . dist. Tang. 4 8 35 0 700 5 7 20 0 364 6 6 5 0 087 Meridiā Substile 7 5 10 0 176 8 4 25 0 166 9 3 40 0 839 10 2 55 0 1.428 11 1 70 0 2.747 12 12 85 0 11.430

The Geometricall projection .

Proportion the plane BCDE , whereon you will draw the Diall to what scantling you think fit . Let VP represent the horizontal line , upon any part thereof , as at P , make choice of a fit place for the perpendicular stile ( though afterwards you may use another forme ) neer about the upper part of the plane , because the great angle between the two Meridians maketh the substile , which must passe thorow the point P , to fall so near the 6 of clock hour , as that there may be but one hour placed above it , if you desire to have the hour of 11 upon the plane , which is more useful then 4 , let P be the center , and with any Chord ( the greater the better ) make two obscure arches ; one above the horizontal line , the other under it , and with the same Chord set off the arch of 51.70 . which is the angle between the substile and horizon , and is the complement of the angle between the substile and meridian , and set it from V to T both wayes , then draw the streight line TPT , which shall be the substile of this Diall .

Which is 5 inches , and 66 hundred parts for the distance of 11 a clock from the point H , and will be the same with those points set off by the natural tangents in the Table . Having done with this Equinoctiall , you must do the like with another : to finde the place whereof , it will be necessary first to know the length of the whole line from H the Equinoctial to the center of the Diall in parts of the perpendicular stile PO , if you will work by the scale of inches , or else the length in natural tangents , if you will use a diagonall Scale : first therefore , to finde the length thereof in inch-measure , we have given in the right angled plain triangle HOP , the base OP , and the angle at O to finde HP , and in the triangle OP center . We have given the perpendicular OP , and the angle PO center the complement of the former , to finde H center : wherefore , by the first case of right angled plain triangles : As the Radius 90 10.000000 Is to the base OP 206 ; 2.313867 So is the tang . of HOP 3.97 . 8.841364 To the perpendicular PH14 . 1.155231

Again , As the Radius 90 , 10.000000 Is to the perpend . OP 206 , 2.313867 So is the tang . PO center 86.3 . 11.158636 To the base P center 2972 3.472403

Adde the two lines of 014 and 2972 together , and you have the whole line H center 2986 in parts of the Radius PO , viz. 29 inches , and 86 parts ; out of this line abate what parts you will , suppose 343 , that is , 3 inches and 43 parts , and then the remainer will be 2643. Now if you set 343 from H to I , the triangle IO center will be equiangled with the former , and I center being given , to finde LO , the proportion is ; As H center the first base 2986 , co. ar . 6.524911 Is to HO , the first perpend . 206. 2.313867 So is I center the 2d . base 2643 , 3.422097 To IO the 2d . perpend . 182 , 2.260875

Having thus found the length of IO to be one inch , and 82 parts ; make that the Radius , and then NT4 shall be a tangent line thereunto , upon which , according to this new Radius , set off the hour-distances before found , and so have you 2 pricks , by which you may draw the height of the stile OO , and the hour-lines for the Dial.

The length of H center in natural tangents , is thus found , HP 069 is the tangent line of the angle HOP 3 deg. 97 min. and by the same reason P center 14421 is the tangent line of PO center 86.3 . the complement of the other , and therefore these two tangents added together do make 14490 , the length of the substile H center , that is , 14 times the Radius , and 49 parts , out of which substract what number of parts you will , the rest is the distance from the second Equinoctial to the center in natural tangents ; suppose 158 to be substracted , that is , one radius , and 58 parts , which set from H to T , in proportion to the Radius HO , and from the point T draw the line NT4 parallel to the former Equinoctial , and there will remain from T to the center 1291. Now to finde the length of LO , the proportion , by the 16 th . of the second , will be As H center 1449 , co. ar . 6.838932 Is to HO 321 , 2.506505 So is T center 1291 , 3.110926 To TO 286 , 2.456363

Now then if you set 286 from T to O in the same measure , from which you took HO , then may you draw ONO , and the tangents in the Table set upon the line NT in proportion to this new radius TO , you shall have two pricks , by which to draw the hour-lines , as before .

Probl. 8. To draw the hour lines upon any direct plane , reclining or inclining East or West .

HItherto we have only spoken of such planes , as are either parallel or perpendicular to the horizon , all which except the horizontal , lie in the plane of some azimuth or other . The rest that follow are reclining or inclining planes , according to the respect of the upper or nether faces of the planes , in those that recline , the base is a line in the plane , parallel to the Horizon or Meridian , and alwayes scituate in some azimuth or other : thus the base of the East and West reclining planes lie in the Meridian , or South and North azimuth , and the poles thereof in the prime vertical , but the plane it self in some circle of position ( as it is Astrologically taken ) which is a great circle of the Sphere , passing by the North or South intersections of the meridian and horizon , and falling East or West from the Zenith upon the prime vertical , as much as the poles of the plane are elevated and depressed above and under the horizon . And this kinde of plane rightly conceived and represented in the fundamental Scheme by NOS , is no other but an erect declining plane in any Countrey , where the pole is elevated the complement of ours : for if you consider the Sphere , it is apparent , that as all the azimuths , representing the decliners , do crosse the prime vertical in the Zenith , and fall at right angles upon the horizon , so do all the circles of position , representing the reclining and inclining East or West planes crosse the horizon in the North and South points of the Meridian , and fall at right angles upon the prime vertical . From which analogie it commeth to passe , that making a Diall declining 30 degr. from the Meridian , it shall be the same that a reclining 30 degr. from the Zenith , and contrary , onely changing the poles elevation into the complement thereof , because the prime vertical in this case is supposed to be the horizon , above which the pole is alwayes elevated the complement of the height thereof above the horizon .

And therefore the poles elevation and the planes reclination being given , which for the one suppose to be , as before , 51 deg. 53 min. and the other , that is , the reclination 35 degrees towards the West , we must finde ( as in all decliners ) first the height of the pole above the plane , which in the fundamental diagram is PR , part of the meridian of the plane between the Pole of the world and the plane . 2. The distance thereof from the meridian of the place , which is NR part of the plane betwixt the substile and the meridian . 3. The angle betwixt the two meridians NPR , by which you may calculate the hour distances , as in the decliners .

First , therefore in the supposed triangle NPR ( because you know not yet where R shall fall ) you have the right angle at R the side opposite PN 51 degr. 53 min. and the angle at N , whose measure is the reclination ZO 35 degr. to finde the side PR , the height of the stile , or poles elevation above the plane , wherefore , by the eighth case of right angled spherical triangles , the analogie is As the Radius 90 , 10.000000 Is to the sine of PN 51.53 . 9.893725 So is the sine PNR 35. 9.758591 To the sine of the side PR 26.69 . 9.652316

Secondly , you may finde the side NR , which is the distance of the substile from the meridian , by the seventh case of right angled spherical triangles ; for As the Radius 90 , 10.000000 Is to the ●●sine of PNR 35. 9.913364 So is the tangent of PN 51.53 . 10.099861 To the tangent of NR 45.87 . 10.013225

Thirdly , the angle at P between the two meridians m●● be found by the ninth case of right angle● spherical triangles . As the Radius ●● 10.000000 Is to the co-sine ●●N , 51.53 . 9.793864 So is the tangent ●f ●NR 35. 9.845227 To the co-tangent of RPN 66.46 . 9.639091

The angle at P being 66 deg. 46 min. the perpendicular PR must needs fall somewhat neer the middle between 7 and 8 of the clock ; if then you deduct the Equinoctial distance of 8 , which is 60 , from 66 deg. 46 min. the Equinoctial distance of 8 of the clock from the substile will be 6 deg. 46 min. again , if you deduct 66 degr. 47 min. from 75 deg. the distance of 7 from the Meridian , the Equinoctiall distance of 7 from the substile will be 08. deg. 53 min. the rest are found by the continual addition of 15 deg. for an hour : thus , 15 degr. and 6 degr. 47 min. do make 21 deg. 47 min. for 9 of the clock ; and so of the rest . And now the hour distances upon the plane may be found by the first case of right angled spherical triangles : for As the Radius 90 10.000000 Is to the sine of PR 26.69 . 9.652404 So is the tangent of RP . 8 , 6.46 . 9.053956 To the tangent of R 8 , 2.91 . 8.706360

These 2 deg. 91 min. are the true distance of 8 of the clock from the substile . And there is no other difference at all in calculating the rest of the hours , but increasing the angle at ● , acccording to each hours Equinoctial distance from the substile .

The Geometrical Projection .

Having calculated the hour distances , you shall thus make the Diall ; let AD be the base or horizontal line of the plane parallel to NZS , the meridian line of the Scheme . And ADEF the plane reclining 35 degr. from the Zenith , as doth SON of the Scheme ▪ through any part of the plane , but most convenient for the houres , draw a line parallel to the base AD , which shall be GO 12 , the 12 of the clock hour representing NZS of the Scheme ; because the base AD is parallel to the meridian , take 60 degrees from a Chord , and make G the center , and draw the circle PRO , representing the circle of position NOS in the Scheme in which this plane lieth ; from the point O to R Westerly in the East reclining , & Easterly in the West reclining , set off the distance of the substile and meridian formerly found to be 45 degrees , 87 min. and draw the prickt line GR for the substile , agreeable to PR in the Scheme , GO in the Diall representing the arch PN , and OR in the plane the arch NR in the Scheme . From the point R of the substile both wayes set off the hour distances , by help of the Chord , for 8 of the clock 2 degr. 91 min. and so of the rest ; and draw streight lines from the center G through those points , which shall be the true hour lines desired . Last of all , the height of the stile PR 26 degr. 69 min. being set from R to P , draw the streight line

GP for the axis of the stile , which must give the shadow on the diall , Erect GP at the angle RGP perpendicularly over the substile line GR , and let the point P be directed to the North pole , GO12 placed in the Meridian , the center G representing the South , and the plane at EF elevated above the horizon 55 degrees ; so have you finished this diall for use , onely remember , because the Sun riseth but a little before 4 , and setteth a little after 8 , to leave out the hours of 3 and 9 , and put on all the rest .

And thus you have the projection of four Dials in one ; for that which is the West recliner is also the East incliner 〈◊〉 you take the complements of the recliners ●ours unto 12 , and that but from 3 in the afternoon till 8 at night : again , if you draw the same lines on the other side of your ●●per , and change the houres of 8 , 7 , 6 , &c. into 4 , 5 , 6 , &c. you have the East recliner , and the complement of the East recliners hours from 3 to 8 is the West incliner : onely remember , that as the stile in the West recliner beholds the North , and the plane the Zenith ; so in the East incliner , the stile must behold the South , and the plane the Nadir .

Probl. 9. To draw the hour-lines upon any direct South reclining or inclining plane .

AS the base of East and West reclining or inclining planes do alwayes lie in the meridian of the place , or parallel thereunto , and the poles in the prime vertical ; so doth the base of South and North reclining or inclining planes lie in the prime vertical or azimuth of East and West , and their poles consequently in the Meridian . Now if you suppose the circle of position , ( which Astrologically taken is fixed in the intersection of the meridian and horizon ) to move about upon the horizon , till it comes into the plane of the prime vertical , and being fixed in the intersection thereof with the horizon , to be let fall either way from the Zenith upon the meridian , it shall truly represent all the South and North reclining and inclining planes also , of which there are six varieties three of South and three of North reclining ; for either the South plane doth recline just to the pole , and then it becommeth an Equinoctial , because the poles of this plane do then lie in the Equinoctiall ; some call it a polar plane , or else it reclineth more and less then the pole , and consequently the poles of the plane above and under the Equinoctiall , somewhat differing from the former . In like manner , the North plane reclineth just to the Equinoctial , and then becometh a polar plane , because the poles of that plane lie in the poles of the world ; some term it an Equinoctiall plane . Or else it reclineth more or lesse then the Equinoctial , and consequently the poles of the plane above and under the poles of the world , somewhat differing from the former .

Of the Equinoctiall plane .

And because this Diall is no other but the very horizontall of a right Sphere , where the Equinoctial is Zenith , and the Poles of the world in the Horizon ; therefore it is not capable of the six of clock hour ( no more then the East and West are of the 12 a clock hour ) which vanish upon the planes , unto which they are parallel : and the twelve a clock hour is the middle line of this Diall ( because the Meridian cutteth the plane of six a clock at right angles ) which the Sun attaineth not , till he be perpendicular to the plain . And this in my opinion , besides the respect of the poles , is reason enough to call it an Equinoctiall Diall , seeing it is the Diall proper to them that live under the Equinoctiall .

This Diall is to be made in all respects as the East and West were , being indeed the very same with them , onely changing the numbers of the hours : for seeing the six of clock hour in which this plane lieth crosseth the twelve of clock hour at right angles , in which the East and West plane lieth , the rest of the hour-lines will have equall respect unto them both : so that the fifth hour from six of the clock is equal to the fift hour from twelve ; the four to the four ; and so of the ●est . These analogies holding , the hour distances from six are to be set off by the natural tangents in these Dials , as they were from twelve in the East and West Dials .

The Geometricall Projection .

Draw the tangent line DSK , parallel to the line EZW in the Scheme , crosse it at right angles with MSA the Meridian line , make SA the Radius to that tangent line , on which prick down the hours ; and that there may be as many hours upon the plane as it is capable of , you must proportion the stile to the plane ( as in the fifth Problem ) after this manner : let the length of the plane from A be given in known parts , then because the extream hours upon this plane are 5 or 7 , reckoning 15 degrees to every hour from 12 , the arch of the Equator will be 75 degrees : and therefore in the right angled plain triangle SA ♎ , we have given the base A ♎ , the length of the plane from A , and the angle AS ♎ 75 degrees , to finde the perpendicular SA ; for which , as in the fifth Chapter , I say ; As the Radius 90 , 10.000000 Is to the base A ♎ 3.50 . 2.544068 So is the tangent of A ♎ S 15 9,428052 To the perpendicular AS 94 1.972120

At which height a stile being erected over the 12 a clock hour line , and the hours from 12 drawn parallel thereunto through the points made in the tangent line , by setting off the natural tangents thereon , and then the Diall is finished .

Let SA 12 be placed in the meridian , and the whole plane at S raised to the height of the pole 51 degr. 53 min. then will the stile shew the hours truly , and the Diall stand in its due position .

2. Of South reclining lesse then the pole .

This plane is represented by the prickt circle in the fundamental Diagram ECW , and is intersected by the hour circles from the pole P , as by the Scheme appeareth , and therefore the Diall proper to this plane must have a center , above which the South pole is elevated ; and therefore the stile must look downwards , as in South direct planes ; to calculate which Dials there must be given the Poles elevation , and the quantity of reclination , by which to finde the hour distances from the meridian , and thus in the triangle PC 1 , having the poles elevation 51 degr. 53 min. and the reclination 25 degr. PC is given , by substracting 25 degr. from PZ 38 degr. 47 min. the complement of the poles height , the angle CP 1 is 15 degrees , one hours distance , and the angle at C right , we may finde C 1 , by the first case of right angled spherical triangles : for , As the Radius 90 , 10.000000 Is to the sine of PC 13.47 . 9.367237 So is the tangent of CP 1. 15. 9.428052 To the tangent of C 1 3.57 . 8.795289

And this being all the varieties , save onely increasing the angle at P , I need not reiterate the work .

3. Of South reclining more then the pole .

This plane in the fundamental Scheme is represented by the prickt circle EAW , of which in the same latitude let the reclination be 55 degrees , from which if you deduct PZ 38 deg. 47 min. the complement of the poles height , there will remain PA 16 deg. 53 min. the height of the north pole above the plane , and instead of the triangle PC 1 , in the former plane we have the triangle PA 1 , in which there is given as before the angle at P 15 deg. & the height of the pole PA 16 deg. 53 min. and therefore the same proportion holds : for , As the Radius 90 , 10.000000 Is to the sine of PA 16.53 . 9.454108 So is the tangent of A 15. 9.428052 To the tangent of A 1. 4.36 . 8.882160

The rest of the hours , as in the former , are thus computed , varying onely the angle at P.

The Geometricall Projection .

These arches being thus found , to draw the Dials true , consider the Scheme , wherein so oft as the plane falleth between Z and P , the Zenith and the North pole , the South pole is elevated ; in all the rest the North ; the substile is in them all the meridian , as in the direct North and South Dials ; in which the stile and hours are to be placed , as was for them directed : which being done let the plane reclining lesse then the pole , be raised above the horizon to an angle equal to the complement of reclination , which in our example is to 65 degr. and the axis of the plane point downwards ; and let all planes reclining more then the pole have the hour of 12 elevated above the horizon to an angle equal to the complement of the reclination also , that is in our example , to 35 deg. then shall the axis point up to the North pole , and the Diall-fitted to the plane .

Probl. 10. To draw the hour-lines upon any direct North reclining or inclining plane .

THe direct north reclining planes have the same variety that the South had ; for either the plane may recline from the Zenith just to the Equinoctial , and then it is a Polar plane , as I called it before , because the poles of the plane lie in the poles of the world ; or else the plane may recline more or lesse then the Equinoctial , and consequently their poles do fall above or under the poles of the world , and the houre lines do likewise differ from the former .

Of the Polar plain .

This place is well known to be a Circle divided into 24 equall parts , which may be done by drawing a circle with the line of Chords , and then taking the distance of 15 degrees from the same Chord , drawing streight lines from the center through those equall divisions , you have the houre-lines desired . The houre-lines being drawn , erect a streight pin of wier upon the center , of wh●● length you please , and the Diall is finished : yet seeing our Latitude is capable of no more then 16 houres and a halfe , the six houres next the South part of the Meridian , 11 , 10 , 9 , 1 , 2 , and 3 , may be left out as uselesse . Nor can the reclining face serve any longer then during the Suns aboad in the North part of the Zodiac , and the inclining face the rest of the year , because this plain is parallel to the Equinoctial , which the Sun crosseth twice in a year . These things performed to your liking , let the houre of 12 be placed upon the Meridian , and the whole plain raised to an angle equall to the complement of your Latitude , the which in this example is 38 deg. 47 min. so is this Polar plain and Diall rectified to shew the true houre of the day .

2. Of North reclining less then the Equator .

The next sort is of such reclining plains as fall between the Zenith and the Equator , and in the Scheme is represented by the pricked circle EFW , supposed to recline 25 degrees from the Zenith , which being added to PZ 38 deg. 47 min. the complement of the poles elevation , the aggregate is PF , 63 deg. 47 min. the height of the Pole or stile above the plane . And therefore in the triangle PF1 , we have given PF , and the angle at P , to finde F1 , the first houres distance from the Meridian upon the plain , for which the proportion is , As the Radius , 90 , 10.000000 Is to the sine of PF , 63.47 9.951677 So is the tangent of FP1 , 15 9.428052 To the Tangent of F1 , 13.48 9.379729

In computing the other houre distances there is no other variety but increasing the angle at P as before we shewed .

3. Of North reclining more then the Equator .

The last sort is of such reclining plain ; as fall between the Horizon and Equator , represented in the fundamental Scheme by the prickt circle EBW , supposed to recline 70 deg. And because the Equator cutteth the Axis of the world at right angles , all planes that are parallel thereunto have the height of their stiles full 90 deg. above the plane : and by how much any plane reclineth from the Zenith , more then the Equator , by so much less then 90 is the height of the stile proper to it , and therefore if you adde PZ 38 deg. 47 min. the height of the Equator , unto ZB 70 deg. the reclination of the plain , the totall is PB 108 deg. 47 mi. whose complemenc to 180 is the arch BS , 71 deg. 53 min. the height of the pole above the plain . To calculate the houre lines thereof , we must suppose the Meridian PFB and the houre circles P1 , P2 , P3 , &c. to be continued till they meet in the South pole , then will the proportion be the same as before .

As the Radius , 90 , 10.000000 To the sine of PB , 71.53 9.977033 So is the tangent of 1PB , 15 9.428052 To the tangent of B1 , 14.27 9.405085

And so are the other houre distances to be computed , as in all the other planes .

The Geometricall Projection .

The projection of these planes is but little differing from those in the last Probl. for the placing the hours and erecting the stile , they are the same , and must be elevated to an angle above the horizon equall to the complement of their reclinations , which in the North reclining lesse then the Equator is in our example 65 degrees , and in this plane the houres about the meridian , that is , from 10 in the morning till 2 in the afternoon , can never receive any shadow , by reason of the planes small reclination from the Zenith , and therefore needlesse to put them on . In the North reclining more then the Equator , the plane in our example must be elevated 120 degr. above the horizon , and the stiles of both must point to the North pole .

Lastly , as all other planes have two faces respecting the contrary parts of the heavens ; so these recliners have opposite sides , look downwards the Nadir , as those do towards the Zenith , and may be therefore made by the same rules ; or if you will spare that labour , and make the same Dials serve for the opposite sides , turn the centers of the incliners downwards , which were upwards in the recliners ; and those upwards in the incliners which were downwards in the recliners , and after this conversion , let the hours on the right hand of the meridian in the recliner become on the left hand in the incliner , and contrarily ; so have you done what you desired : and this is a general rule for the opposite sides of all planes .

Probl. 11. To draw the hour-lines upon a declining reclining , or declining inclining plane .

DEclining reclining planes have the same varieties that were in the former reclining North and South ; for either the declination may be such , that the reclining plane will fall just upon the pole , and then it is called a declining Equinoctial ; or it may fall above or under the pole , and then it is called a South declining cast and west recliner : on the other side the declination may be such , that the reclining plane shall fall just upon the intersection of the Meridian and Equator ; and then it is called a declining polar ; or it may fall above or under the said intersection , and then it is called a North declining East and West recliner . The three varieties of South recliners are represented by the three circles , AHB falling between the pole of the world and the Zenith : AGB just upon the pole ; and AEB between the pole and the horizon : and the particular pole of each plane is so much elevated above the horizon , ( upon the azimuth ) DZC , crossing the base at right angles ) as the plane it self reclines from the Zenith , noted in the Scheme , with I , K , and L.

1. Of the Equinoctiall declining and reclining plane .

This plane represented by the circle AGB , hath his base AZB declining 30 degrees from the East and West line EZW equal to the declination of the South pole thereof 30 degrees from S the South part of the Meridian Easterly unto D , reclining from the Zenith upon the azimuth CZD the quantity ZG 34 degrees , 53 min. and Passeth through the pole at P. Set off the reclination ZG , from D to K , and K shall represent the pole of the reclining plane so much elevated above the horizon at D , as the circle AGB representing the plane declineth from the Zenith Z , from P the pole of the world , to K the pole of the plane , draw an arch of a great circle PK , thereby the better to informe the fancie in the rest of the work . And if any be desirous , to any declination given , to fit a plane reclining just to the pole : or any reclination being given , to finde the declination proper to it , this Diagram will satisfie them therein : for in the Triangle ZGP , we have limited ,

First , the hypothenusal PZ 38 degrees , 47 min.

Secondly , the angle at the base PZG , the planes declination 30 degrees . Hence to finde the base GZ , by the seventh case of right angled spherical triangles , the proportion is ; As the Radius 90 , 10.000000 To the co-sine of GZP 30 ; 9.937531 So the tangent of PZ 38.47 . 9.900138 To the tangent of GZ 34.53 . 9.837669
the reclination required .

If the declination be required to a reclination given , then by the 13 case of right angled spherical triangles , the proportion is As the Radius 90 , 10.000000 To the tangent of ZG 34.53 . 9.837669 So the co-tangent of PZ 38.47 . 10.099861 o the co-sine of GZP 39. 9.937530

And now to calculate the hour-lines of this Diall , you are to finde two things : first , the arch of the plane , or distance of the meridian and substile from the horizontal line , which in this Scheme is PB , the intersection of the reclining plane with the horizon , being at B. And secondly , the distance of the meridian of the place SZPN , from the meridian of the plane PK , which being had , the Diall is easily made .

Wherefore in the triangle ZGP , right angled at G , you have the angle GZP given 30 degrees , the declination ; and ZP 38 degr. 47 min. the complement of the Pole ; to finde GP : and therefore , by the eighth case of right angled spherical triangles , the proportion is : As the Radius , 90 10.000000 To the sine of ZP , 38.47 9.793863 So is the sine of GZP , 30 9.698970 To the sine of GP , 18.12 9.492833
Whose complement 71 deg. 88 min. is the arch PB desired .

The second thing to be found is the distance of the Meridian of the place , which is the houre of 12 from the substile or meridian of the plane , represented by the angle ZPG , which may be found by the 11 Case of right angled sphericall Triangles , for As the Radius , 90 10.000000 Is to the sine of GP , 18.12 9.492833 So is the co-tang . of GZ , 34.53 10.162379 To the co-tang . of GPZ , 65.68 9.655212
Whose complement is ZPK 24 deg. 32 min. the arch desired .

Now because 24 deg. 32 min. is more then 15 deg. one houres distance from the Meridian , and lesse then 30 deg. two houres distance , I conclude that the stile shall fall between 10 and 11 of the clock on the West side of the Meridian , because the plain declineth East : if then you take 15 deg , from 24 deg. 32 min. there shall remain 9 deg. 32 min. for the Equinoctiall distance of the 11 a clock houre line from the substile , and taking 24 deg. 32 min. out of 30 deg. there shall remain 5 deg. 68 min. for the distance of the houre of 10 from the substile : the rest of the houre distances are easily found by continual addition of 15 deg. Unto these houre distances joyn the naturall tangents as in the East and West Dials , which will give you the true distāces of each houre from the substile , the plane being projected as in the 5 Pro. for the east & west dials , or as in the 8 Prob. for the Equinoctial , according to which rules you may proportion the length of the stile also , which being erected over the substile , and the Diall placed according to the declination 30 deg. easterly , and the whole plain raised to an angle of 55 deg. 47 min. the complement of the reclination , the shadow of the stile shall give the houre of the day desired .

2. To draw the houre lines upon a South reclining plain , declining East or West , which passeth between the Zenith and the Pole .

In these kinde of declining reclining plains , the South pole is elevated above the plane , as is clear by the circle AHB representing the same , which falleth between the Zenith and the North pole , and therefore hideth that pole from the eye , and forceth you to seeke the elevation of the contrary pole above the plain , which notwithstanding maketh the like and equall angles upon the South side objected to it , as the North pole doth upon the North side , ( as was shewed in the 7 Prop. ) so that either you may imagine the Scheme to be turned about , and the North and South points changed , or you may calculate the houres as it standeth , remembring to turn the stile upwards or downwards , and change the numbers of the houres , as the nature of the Diall wil direct you .

In this sort of declining reclining Dials , there are four things to be sought before you can calculate the houres . 1 The distance of the Meridian from the Horizon . 2 The height of the pole above the plain . 3 The distance of the substile from the Meridian . 4 The angle of inclination between the Meridian of the plane , and the meridian of the place .

1 The distance of the Meridian from the Horizon , is represented by the arch OB , to finde which , in the right angled Triangle HOZ , we have HZ the reclination 20 deg. and the angle HZO the declination , to find HO , the complement of OB , for which , by the first case of right angled sphericall triangles , the analogie is , As the Radius , 90 10.000000 o the sine of HZ , 20 9.534051 o is the tangent of HZO , 30 9.761439 o the tangent of HO , 11.17 9.295490
Whose complement 78 deg. 83 min. is OB , the arch desired .

2. To finde the height of the pole above the plane , there is required two operations , the first to finde OP , and the second to finde PR ; OP may be found by the 3 Case of right angled Sphericall Triangles , for , As the Radius , 90 10.000000 Is to the co-sine of HZP. 30 9.937531 So is the co-tang . of HZ , 20. 10.438934 To the co-tangent of ZO , 22.80 10.376465
Which arch being found , and deducted out of , ZP 38 deg. 47 min. there resteth PO 15 deg. 67 min.

Then may you finde PR , by the triangles HZO , and PRO both together , because the sines of the hypothenusals and the sines of the perpendiculars are proportional , by the first of the 7 Chap. of Triangles .

Therefore , As the sine of ZO , 22.80 9.588289 Is to the sine of ZH , 20 9.534052 So is the sine of PO , 15.67 9.431519 To the sine of PR , 13.79 9.377282
The height of the stile desired .

3 The distance of the substile from the Meridian may be found by the 12 Case of right angled sphericall triangles , for As the co-sine of PR , 13 78 9.987298 Is to the Radius , 90 10.000000 So is the co-sine of PO , 15.67 9.983551 To the co-sine of OR , 7.41 9.996253
The arch desired .

4. The angle of inclination between the Meridians , may be found by the 11 Case of right angled Spherical triangles , for , As the Radius , 90 10.000000 Is to the sine of PR , 13.79 . 9.377241 So is the co-tang . of OR 7.51 10.879985 To the co-tang . of OPR , 28.93 10.257226

Now as in all the former works , the angle P between the two Meridians being 28 deg. 93 min. which is more then one houres distance from the Meridian , and lesse then two , you may conclude that the substile must stand between the first & second hours from the Meridian or 12 of the clock Westerly , because the declination is easterly : and 28 deg. 93 min. being deducted out of 30 deg. there resteth 1 deg. 7 min. for the distance of 10 of the clock from the substile ; again , deducting 15 deg. from 28 deg. 93 min. there resteth 13 deg. 73 min. the distance of the 11 a clock houre line from the substile , the rest are found by continuall addition of 15 deg. as before .

And here the true houre distances may be found by the first case of right angled Sphericall triangles , for , As the Radius , 90 10.000000 Is to the sine of PR 13.79 9.377240 So is the tangent of RP , 11.15 9.428052 To the tangent of R 11 , 3.66 8.805292
And so proceed with all the rest .

3. To draw the houre lines upon a South reclining plain , declining East or west , which passeth between the Pole and the Horizon .

In this plain represented by the circle of reclination AFB , the North pole is elevated above the plane , as the South pole was above the other , and the same four things that you found for the former Diall must also be sought for this ; in the finding whereof there being no difference , save only deducting ZP from ZO , because ZO is the greatest arch , as by the Scheam appeareth : to calculate the houres of this plane needeth no further instruction .

Probl. 12. To draw the houre lines upon a polar plain , declining East or west , being the first variety of North declining reclining planes

AS in the South declining recliners , there are three varieties , so are there in the North as many : for either the plane reclining doth passe by the intersection of the Meridian and Equator , and then it is called a declining Polar , which hath the substile alwayes perpendicular to the Meridian ; or else it passeth above or under the intersection of the Meridian and Equator , which somewhat differeth from the former . I will therefore first shew how they lie in the Scheam , and then proceed to the particular making of the Dials proper to them .

1. Of the Polar declining reclining plane .

This plane is in this diagram represented by the circle AGB , ZG is the reclination , ZAE the distance of the Equator from the Zenith , the declination NC , K the pole thereof . Here also as in the last Probl. there may be a reclination found to any declination given , and contrary , by which to fit the plane howsoever declining , to passe through the intersection of the Meridian and Equator , by the 7 and 13 Cases of right angled sphericall triangles .

As the Radius , 90 10.000000 To the co-sine of GZAE , 60 9.698970 So is the tangent of ZAE , 51.53 10.099861 To the tangent of ZG , 32.18 9.798871
The reclination desired .

And , As the Radius , 90 10.000000 To the tangent of GZ , 32.18 9.798831 So is the co-tangent of ZAE , 51.53 9.900138 To the co-sine of GZAE , 60 9.698969
The declination .

And now to calculate the houre lines of this Dial , you must finde , first , the distance of the Meridian from the Horizon , by the 8 Case of right angled Spherical triangles . As the Radius , 90 10.000000 Is to the sine of ZAE , 51.53 9.893725 So is the sine of GZAE , 60 9.937531 To the sine of AEG , 42.69 9.831256
Whose complement 47 deg. 31 min. is AAE the arch desired .

2. You must finde RP , the height of the pole above the plane , by the 2 Case of right angled Sphericall Triangles , for As the Radius , 90 10.000000 Is to the sine of AEZG , 60 9.937531 So is the co-sine of ZG , 32.11 9.927565 To the co-sine of ZAEG , 42.87 9 . 865●9●
Which is the height of the pole above the plane , AER being a Quadrant , PR must needs be the measure of the angle at AE.

3. Because in all decliners ( whose planes passe by the intersection of the Meridian and Equinoctiall ) the substile is perpendicular to the Meridian , therefore you need not seek AER , the distance between the substile and Meridian , which is alwayes 90 deg. and falleth upon the 6 a clock houre .

4. Lastly , the arch AER , which is the distance of the substile from the Meridian : being 90 degrees , the angle at P opposite thereunto must needs be 90 also : from whence it followes , that the houres equidistant from the six of the clock hour in Equinoctial degrees shall also have the like distance of degrees in their arches upon the plane , and so one half of the Diall being calculated , serves for the whole ; these things considered , the true hour-distances may be found , by the first case of right angled spherical triangles : for , As the Radius , 10.000000 Is to the sine of PR 42.87 . 9.832724 So is the tangent of RP 5. 15 d. 9.428052 To the tangent of R 5 ▪ 10.34 . 9.260776

The which 10 degr. 34 min. is the true distance of 5 and 7 from the substile or six of the clock hour , and so of the rest .

The Geometrical projection of this plane needs no direction ; those already given are sufficient , according to which this Diall being made and rectified by the declination and reclination given , it is prepared for use .

2. To draw the hou●● lines upon a North reclining plane , declining East or West , which cutteth the meridian between the Zenith and the Equinoctial .

All North reclining planes howsoever declining , have the North pole elevated above them , and therefore the center of the Diall must be so placed above the plane , that the stile may look upwards to the pole , neither can it be expected that the plane being elevated above the horizon Southward , should at all times of the year be enlightened by the Sun , except it recline so far from the Zenith , as to intersect the Meridian between the horizon and the Tropique of Capricorn ; this plane therefore reclining but 16 degrees from the Zenith , and declining 60 cannot shew many hours , when the Sun is in his greatest Northern declination , partly by reason of the height of the plane above the horizon , and partly by reason of the great declination thereof , hindring the Sun-beams from all the morning houres , which may be therefore left out as useless .

In this second variety , the plane represented by the Circle AMB in the last Diagram , cutteth the Meridian at O between the Zenith and the Equator , ZM being the reclination , 16 deg. ZAE the distance of the Equator from the Zenith , 51 deg. 53 m. and the declination NC 60 as before .

As in the former , so in this Diall , the same four things are again to be found before you can calculate the houre distances thereof . The first is the distance of the Meridian from the Horizon , represented in this plain by the arch A● . The second is PR , the height of the pole above the plane . The third is ●R , the distance of the substile or Meridian of the plane , from the Meridian of the place . The fourth is the angle ●PR between the two Meridians : all which , and the houre distances also , being to be found according to the directions of the last Probl. there needeth no further instruction here .

3 To draw the houre lines upon a North reclining plane , declining East or West , which cutteth the Meridian between the Equator and Horizon .

The last variety of the six declining recliners , represented by the circle ALB , and cutteth the Meridian at H , between the Equator and the Horizon , ZL being the reclination , 54 deg. the declination NC , 60 deg. as before ; and hence the four things mentioned before must be sought ere you can calculate the houre distances .

1 The distance of the Meridian and Horizon , represented by AH . 2 RH the substile . or Meridian of the plane from the Meridian of the place . 3 PR , the height of the pole above the plane . 4 HPR , the angle between the two Meridians .

In finding whereof the proportions are still the same , though the triangles are somewhat altered , for when you have found ZH , it is to be added to ZP to finde PH , both which together do exceed a Quadrant , therefore the sides PN must be continued to X , then is PX the complement of PH to a semicircle , and if RB be continued ●o X also , RX may be found by the 12 Case of right angled spherical triangles as before , whose complement is RH , the distance of the substile from the Meridian ; and hence the angle at P must be found in that triangle also , though the proportion be the same , there being no other variety , I think it needlesse to reiterate the work .

The Geometrical Projection .

There is so little difference between the South & North declining reclining planes , that the manner of making the Dials for both may be shewed at once : Let the example therefore be a Diall for a South plane declining East 30 deg. reclining 20 deg.

And thus have you made four Dials at once , or at least , this Dial thus drawn may be made to serve four sorts of planes , for first , it serves for a South declining East 30 deg. reclining 20 deg. and if you prick the houre lines through the paper , and draw them on the other side stile and all , this Diall will then be fitted for a South plane declining West 30 deg. reclining 20 deg. only remember to change the houres , that is to say , instead of writing 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 1 , 2 , 3 , 4 , from A , the west side of the East declining plane , you must write , 8 , 9 , 10 , 11 , 12 , 1 , 2 , 3 , 4 , 5 , 6 , 7. Again , if you turn the Zenith of your Dial downwards , the South declining East reclining shall in all respects serve for a North declining west inclining as much ; and the South declining West reclining , will likewise serve for a North declining East inclining ; and therefore there needs no further direction either to make the one , or calculate the other .

CHAP. III. Of the Art of NAVIGATION .
Probl. 1. Of the 32 windes , or Seamans Compass .

THe course of a ship upon the Sea dependeth upon the windes : The designation of these depends upon the certain knowledge of one principal ; which considering the situation and condition of the whole Sphere , ought in nature to be North or South , the North to us upon this side of the Line , the South to those in the other hemisphere ; for in making this observation men were to intend themselves towards one fixed part of the heavens or other , and therefore to the one of these . In the South part there is not found any star so notable , and of so neer distance from the Pole , as to make any precise or firm direction of that winde , but in the North we have that of the second magnitude in the tale of the lesser Bear , making so small and incensible a circle about the Pole , that it commeth all to one , as if it were the Pole it self . This pointed out the North winde to the Mariners of old especially , and was therefore called by some the Lead or Lead star ; but this could be only in the night , and not alwayes then . It is now more constantly and surely shewed by the Needle touched with the Magnet , which is therefore called the Load or Lead-stone , for the same reason of the leading and directing their courses to the North and South position of the earth , not in all parts directly , because in following the constitution of the great Magnet of the whole earth , it must needs be here and there led aside towards the East or West by the unequal temper of the Globe ; consisting more of water then of earth in some places , and of earth more or lesse Magnetical in others .

This deviation of the Needle , the Mariners call North-easting , & North-westing , as it falleth out to be , otherwise , and more artificially , the Variation of the Compass , which though it pretend uncertainty , yet proveth to be one of the greatest helps the Seaman hath . And the North and South windes being thus assured by the motion either of direction or variation of the needle , the Mariner supposeth his ship to be ( as it alwayes is ) upon some Horizon or other , the center whereof is the place of the ship .

The line of North and South found ou● by the Needle , a line crossing this at right angles , sheweth the East and West , and so they have the four Cardinall windes , crosse again each of these lines , and they have the eight whole windes , as they call them . Another division of these maketh eight more which they call halfe windes , a third makeeth 16 , which they call the quarter windes , so they are 32 in all . Every one of these Windes is otherwise termed a several point of the Compasse , and the whole line consisting of two windes , as the line of North and South , or that of East and West is called a Rumb . The Windes and Rumbs thus assigned by an equal division of a great Circle into 32 parts , the angle which each Rumb maketh with the Meridian is easily known , for if you divide a quadrant or 90 degrees in eight parts : you have the angles which the eight windes reckoned from North to East or West do make with the meridian ; and those reckoned from South to the East or West are the same , and for your better direction are here exhibited in the Table following .

ATablefortheangleswhicheveryRumb●makethwiththeMeridian. North South D. part South North 02.8125 05.6250 08.4375 N by E S by E 11.2500 S by W N by W 14.0625 16.8750 19.6875 NNE SSE 22.5000 SSW NNW 25.3125 28.1250 30.9375 NE by N SE by S 33.7500 SW by S NW b N 36.5625 39.3750 42.1875 NE SE 45.0000 SW NW 47.8125 50.6250 53.4375 NE by E SE by S 56.2500 SW b W NW b W 59.0625 61.8650 64.6875 ENE ESE 67.5000 WSW WNW 70.3125 73.1250 75.9375 E by N E by S 78.7500 W by S W by N 81.5625 84.3750 87.1875 East East 90.0000 West West

Probl. 2. Of the description and making of the Sea-chart .

THe Sea-mans Chart is a Parallelogram , divided into little rectangled figures , and in the plain Chart are equal Squares , representing the Longitudes and Latitudes of such places , as may be set in the Chart , but the body of the earth being of a Globular form , the degrees of Longitude reckoned in the Equator from the Meridian , are in no place equal to those of the Latitude reckoned in the Meridian from the Equator , save onely in the Equinoctial ; for the degrees of latitude are all equall throughout the whole Globe , and as large as those of the Equinoctial ; but the degrees of Longitude at every parallel of Latitude lessen themselves in such proportion as that parallel is lesse then the Equinoctial : This dis-proportion of longitude and latitude caused for a long time much errour in the practise of Navigation , till at last it was in part reconciled by Mercator , that famous Geographer : and afterwards exactly rectified by our worthy Countreyman Master Edward Wright , in his Book entituled , The Correction of Errours in Navigation : In which he hath demonstrated by what proportion the degrees of Longitude must either increase or decrease in any Latitude , his words are as followeth .

Suppose , saith he , a spherical Superficies , with Meridians , Parallels , Rumbes , and the whole Hydrographial description drawn thereupon , to be inscribed into a concave Cylinder , their axes agreeing in one .

Let this Spherical superficiees swell like a bladder ( whiles it is in blowing ) equally alwayes in every part thereof ( that is as much in Longitude as in Latitude , til it apply and joyn it self ( round about and all along also towards either pole ) unto the concave superficies of the Cylinder : each parallel upon this spherical superficies increasing successively from the Equinoctial towards either pole , until it come to be of equal diameter with the Cylinder , and consequently the Meridians , stil inclining themselves , till they come to be so far distant every where each from other , as they are at the Equinoctial .

Thus it may most easily be understood , how a spherical superficies may by extension be made a Cylindrical , and consequently a plain parallelogram superficies ; because the superficies of a cylinder is nothing else but a plain parallelogram wound about two equal equidistant circles , that have one common axletree perpendicular upon the centers of them both , and the peripheries of each of them equall to the length of the parallelogram , as the distance betwixt those circles , or height of the cylinder is equall to the breadth thereof . So as the Nautical planisphere may be defined to be nothing else but a parallelogram made of the Spherical superficies of an Hydrographical Globe inscribed into a concave cylinder , both their axes concurring in one , and the sphericall superficies swelling in every part equally in longitude and latitude , till every one of the parallels thereupon be inscribed into the cylinder ( each parallel growing as great as the Equinoctial , or till the whole spherical superficies touch and apply it self every where to the concavity of the cylinder .

In this Nautical planisphere thus conceived to be made , all places must needs be situate in the same longitudes , latitudes , and directions or courses , and upon the same meridians , parallels and rumbes , that they were in the Globe , because that at every point between the Equinoctial and the Pole we understand the spherical superficies , whereof this planisphere is conceived to be made , to swell equally as much in longitude as in latitude ( till it joyn it self unto the concavity of the cylinder ; so as hereby no part thereof is any way distorted or displaced out of his true and natural situation upon his meridian , parallel or rumbe , but onely dilated and enlarged , the meridians also , parallels , and rumbes , dilating and enlarging themselves likewise at every point of latitude in the same proportion .

Now then let us diligently consider of the Geometrical lineaments , that is , the meridians , rumbes , and parallels of this imaginary Nautical planisphere , that we may in like manner expresse the same in the Mariners Chart : for so undoubtedly we shall have therein a true Hydrographical description of all places in their longitudes , latitudes , and directions , or respective situations each from other , according to the points of the compasse in all things correspondent to the Globe , without either sensible or explicable errour .

First , therefore in this planisphere , because the parallels are every where equal each to other ( for every one of them is equal to the Equinoctiall or circumference of the circumscribing cylinder ) the meridians also must needs be parallel & streight lines ; and consequently the rumbes , ( making equall angles with every meridian ) must likewise be streight lines .

Secondly , because the spherical superficies whereof this planisphere is conceived to be made , swelleth in every part thereof equally , that is as much in Latitude as in Longitude , till it apply it self round about to the concavity of the cylinder : therefore at every point of Latitude in this planisphere , a part of the Meridian keepeth the same proportion to the like part of the parallel that the like parts of the Meridian , and parallel have each to other in the Globe , without any explicable errour .

And because like parts of wholes keep the same proportion that their wholes have therefore the like parts of any parallel and Meridian of the Globe , have the same proportion , that the same parallel and meridian have .

For example sake , as the meridian is double to the parallel of 60 degrees : so a degree of the meridian is double to a degree of that parallel , or a minute to a minute , and what proportion the parallel hath to the meridian , the same proportion have their diameters and semidiameters each to other .

But the sine of the complement of the parallels latitude , or distance from the Equinoctial , is the semidiameter of the parallel .

As here you see AE , the sine of AH , the complement of AF , the latitude or distance of the parallel ABCD from the Equinoctial , is the semidiameter of the same parallel . And as the semidiameter of the meridian or whole sine , is to the semidiameter of the parallel ; so is the secant or hypothenusa of the parallels latitude , or of the parallels distance from the Equinoctial , to the semidiameter of the meridian or whole sine ; as FK , ( that is AK ) to AE ( that is GK ) so is LK , to KF .

Therefore in this nautical planisphere , the Semidiameter of each parallel being equal to the semidiameter of the Equinoctial , that is , to the whole sine ; the parts of the Meridian at every point of Latitude must needs increase with the same proportion wherewith the secants of the ark , conteined between those points of Latitude and the Equinoctial do increase .

Now then we have an easie way laid open for the making of a Table ( by help of the natural Canon of Triangles ) whereby the meridians of the Mariners Chart may most easily and truly be divided into parts , in due proportion , and from the Equinoctial towards either Pole .

For ( supposing each distance of each point of latitude , or of each parallel from other , to contein so many parts as the secant of the latitude of each point or parallel conteineth ) by perpetual addition of the secants answerable to the latitudes of each point or parallel unto the summe compounded of all the former secants , beginning with the secant of the first parallels latitude , and thereto adding the secant of the second parallels Latitude , and to the summe of both these adjoyning the secant of the third parallels Latitude ; and so forth in all the rest we may make a Table which shall truly shew the sections and points of latitude in the Meridians of the Nautical Planisphere , by which sections the parallels must be drawn .

As in the Table of meridional parts placed at the end of this Discourse , we made the distance of each parallel from other , to be one minute or centesm of a degree : and we supposed the space between any two parallels , next to each other in the Planispere , to contain so many parts as the secant answerable to the distance of the furthest of those two parallels from the Equinoctial ; and so by perpetual addition of the secants of each minute or centesm to the sum compounded of all the former secants , is made the whole Table .

As for example , the secant of one centesm in Master Briggs 's Trigonometrica Britannica is 100000.00152 , which also sheweth the section of one minute or centesm of the meridian from the Equinoctial in the Nautical Planisphere ; whereunto adde the secants of two minutes or centesmes , that is 100000. 00609 , the sum is 200000.00761 . which sheweth the section of the second minute of the meridian from the Equinoctial in the planisphere : to this sum adde the secant of three minutes , which is 100000.01371 , the sum will be 3000●0 . 02132 , which sheweth the section of the third minute of the meridian from the Equinoctial , and so ●orth in all the rest ; but after the Table was thus finished , it being too large for so small a Volume , we have contented our selves with every tenth number , and have also cut off eight places towards the right hand , so that in this Table the section of 10 minutes is 100 , of one degree 1000 , and this is sufficient for the making either of the generall or any particular Chart.

I call that a general Chart , whose line AE in the following figure represents the Equinoctial , ( as here it doth the parallel of 50 degrees ) and so containeth all the parallels successively from the Equinoctial towards either Pole , but they can never be extended very near the Pole , because the distances of the parallels increase as much as secants do . But notwithstanding this , it may be remed general , because a more general Chart cannot be contrived in plano , except a true projection of the Sphere it self .

And I call that a particular Chart which is made properly for one particular Navigation ; as if a man were to sail betwen the Latitude of 50 and 55 degrees , and his difference of Longitude were not to exceed six degrees , then a Chart made , as this figure is for such a Voyage , may be called particular , and is thus to be projected .

Probl. 3. The Latitudes of two places being known , to finde the Meridional difference of the same Latitudes .

IN this Proposition there are three varieties : First , when one of the places is under the Equinoctial , and the other without ; and in this case the degrees and minutes in the Table answering to the latitude of that other place are the meridional difference of those Latitudes .

So if one place propounded were the entrance of the River of the Amazones , which hath no latitude at all , and the other the Lizard , whose latitude is 50 degrees , their difference will be found 57.905 .

2. When both the places have Northerly or Southerly Latitude , in this case if you substract the degrees and minutes in the Table answering to the lesser Latitude , out of those in the same Table answering to the greater Latitude , the remainer will be the Meridional difference required .

Example .

Admit the Latitude of S. Christophers to be 15 deg. 50 parts or minutes , and the Latitude of the Lizard to be 50 degrees . In the Table of Latitudes , the number answering to 15 deg. 50 min. is 15.692 50 deg. is 57.905 Their difference 42.213

3. When one of the places have Southerly and the other Northerly Latitude ; in this case , the sum of the numbers answering to their Latitudes in the Table , is the meridional difference you look for .

So Caput bonae spei , whose latitude is about 36 deg. 50 parts , and Japan in the East Indies , whose latitude is about 30 degrees being propounded , their meridional difference will be found to be 70.724 .

For the meridional parts of 36.50 . 39.252 And the meridional parts of 30 d. 31.472 Their sum is the difference required . 70.724

Probl. 4. Two places differing onely in Latitude , to finde their distance .

IN this proposition there are two varieties .

1. If the two places propounded lie under the same meridian , and both of them on one side of the Equinoctial , you must substract the lesser latitude from the greater , and the remainer converted into leagues , by allowing 20 leagues to a degree , will be the distance required .

2. If one place lie on the North , and the other on the South side of the Equinoctial ( yet both under the same meridian ) you must then adde both the latitudes together , and the sum converted into leagues , will give their distance .

Probl. 5. Two places differing onely in longitude being given , to finde their distance .

IN this proposition there are also two varieties .

1. If the two places propounded lie under the Equinoctial , then the difference of their Longitudes reduced into leagues ( by allowing 20 leagues to a degree ) giveth the distance of the places required .

2. But if the two places propounded differ onely in longitude , and lie not under the Equinoctial , but under some other intermediate parallel between the Eqninoctial and one of the poles : then to finde their distance , the proportion is , As the Radius , Is to the co-sine of the common latitude ; So is the sine of half the difference of longitude , To the sine of half their distance .

Probl. 6. Two places being given , which differ both in Longitude and Latitude , to finde their distance .

IN this Proposition there are three varieties .

1. If one place be under the Equinoctial circle , and the other towards either pole , then the proportion is , As Radius , To the cosine of the difference of longitude ; So is the co-sine of the latitude given , To the co-sine of the distance required .

2. If both the places propounded be without the Equinoctial , and on the Northern or Southern side thereof , then the proportion must be wrought at two operations . 1. Say ; As the Radius , To the cosine of the difference of Longitude So the co-tangent of the lesser latitude , To the tangent of the fourth ark .

Which fourth ark substract out of the complement of the greater latitude , and retaining the remaining ark say , As the co-sine of the ark found , Is to the co-sine of the ark remaining ; So is the sine of the lesser latitude , To the co-sine of the distance required .

3. If the two places propounded differ both in Longitude and Latitude , and be both of them without the Equinoctial , and one of them towards the North pole , and the other towards the South pole , the proportion is , As the Radius , Is to the co-sine of the difference of Longit. So is the co-tangent of one of the Latitudes To the tangent of another ark .

Which being substracted out of the other Latitude , and 90 degrees added thereto , say : As the co-sine of the ark found , Is to the co-sine of the ark remaining ; So is the co-sine of the Latitude first taken , To the co-sine of the distance .

Probl. 7. The Rumbe and distance of two places given , to finde the difference of Latitude .

THe proportion is : As the Radius , Is to the co-sine of the rumb from the meridian : So is the distance , To the difference of Latitude .

Example .

If a ship sail West-north-west , ( that is , upon the sixt rumb from the meridian ) the distance of 90 leagues ; what shall be the difference of Latitude ?

First , I seek in the Table of Angles which every Rumb maketh with the Meridian , for the quantity of the angle of the sixt rumb , which is 67 degr. 50 parts , the complement whereof is 22 degr. 50 parts : therefore , As the Radius , 10.000000 Is to the sine of 22.50 . 9.582839 So is the distance in leagues 90 , 1.954242 To the difference of Latitude 34 , and better 1.537081

And by looking the next neerest Logarithm , the difference of latitude will be 34 leagues , and 44 hundred parts of a league .

And because 5 centesmes of a degree answereth to one league , therefore if you multiply 3444 by 5 , the product will be 17220 , from which cutting off the four last figures , the difference of latitude will be one degree 72 centesmes of a degree , and somewhat more .

Probl. 8. The Rumb and Latitude of two places being given , to finde the difference of Longitude .

THe proportion is : As the Radius ,

Is to the tangent of the rumb from the meridian : So is the proper difference of latitude , To the difference of Longitude .

Example .

If a ship sail West-north-west ( that is , upon the sixt Rumb from the meridian ) so far , that from the latitude of 51 degrees , 53 centesmes , it cometh to the latitude of 49 degrees , 82 centesmes ; what difference of Longitude hath such a course made ?

First , I seek in the Table of Meridional parts what degrees do there answer to each latitude , and to 51 degrees , 53 min. I finde 60. 328 , and to 49 degrees , 82 minutes 57. 629 , which being substracted from 60. 328 their difference is 2. 699 , the proper difference of latitude . Therefore , As the Radius , 10.000000 To the tangent of 67.50 . 10.382775 So is 2.699 . 0.431203 To 6 the difference of Longitude , 0.813978

Or in minuter parts 6. 515 , that is 6 degr. 52 centesmes of a degree fere , which was the thing required .

Here followeth the Table of Meridional parts , mentioned in some of the preceeding Problemes , together with other Tables usefull in the Arts of Dialling and Navigation .

ATableofMeridionalparts. M. Gr. par 0.00 0.000 0.10 0.100 0.20 0.200 0.30 0.300 0.40 0.400 0.50 0.500 0.60 0.600 0.70 0.700 0.80 0.800 0.90 0.900 1.00 1.000 1.10 1.100 1.20 1.200 1.30 1.300 1.40 1.400 1.50 1.500 1.60 1.600 1.70 1.700 1.80 1.800 1.90 1.900 2.00 2.000 2.10 2.100 2.20 2.200 2.30 2.300 2.40 2.400 2.50 2.500 2.61 2.600 2.71 2.700 2.81 2.800 2.91 2.900 3.01 3.000 3.00 3.001 3.10 3.101 3.20 3.201 3.30 3.301 3.40 3.402 3.50 3.502 3.60 3.602 3.70 3.702 3.80 3.803 3.90 3.903 4.00 4.003 4.10 4.103 4.20 4.204 4.30 4.304 4.40 4.404 4.50 4.504 4.60 4.605 4.70 4.705 4.80 4.805 4.90 4.906 5.00 5.006 5.10 5.106 5.20 5.207 5.30 5.307 5.40 5.408 5.50 5.508 5.60 5.609 5.70 5.709 5.80 5.810 5.90 5.910 6.00 6.011 6.10 6.111 6.20 6.212 6.30 6.312 6.40 6.413 6.50 6.514 6.60 6.614 6.70 6.715 6.80 6.816 6.90 6.916 7.00 7.017 7.10 7.118 7.20 7.219 7.30 7.319 7.40 7.420 7.50 7.521 7.60 7.622 7.70 7.723 7.80 7.824 7.90 7.925 8.00 8.026 8.10 8.127 8.20 8.228 8.30 8.329 8.40 8.430 8.50 8.531 8.60 8.632 8.70 8.733 8.80 8.834 8.90 8.936 9.00 9.037 9.10 9.138 9.20 9.239 9.30 9.341 9.40 9.442 9.50 9.543 9.60 9.645 9.70 9.746 9.80 9.848 9.90 9.949 10.00 10.051 10.10 10.152 10.20 10.254 10.30 10.355 10.40 10.457 10.50 10.559 10.60 10.661 10.70 10.762 10.80 10.864 10.90 10.966 11.00 11.068 11.10 11.170 11.20 11.272 11.30 11.374 11.40 11.476 11.50 11.578 11.60 11.680 11.70 11.782 11.80 11.884 11.90 11.986 12.00 12.088 12.10 12.190 12.20 12.293 12.30 12.395 12.40 12.497 12.50 12.600 12.60 12.702 12.70 12.805 12.80 12.907 12.90 13.010 13.00 13.112 13.10 13.215 13.20 13.318 13.30 13.422 13.40 13.523 13.50 13.626 13.60 13.729 13.70 13.832 13.80 13.935 13.90 14.038 14.00 14.141 14.10 14.244 14.20 14.347 14.30 14.450 14.40 14.553 14.50 14.656 14.60 14.760 14.70 14.863 14.80 14.967 14.90 15.070 15.00 15.174 15.10 15.277 15.20 15.381 15.30 15.485 15.40 15.588 15.50 15.692 15.60 15.796 15.70 15.900 15.80 16. ●04 15.90 16.107 16.00 16.211 16.10 16.316 16.20 16 . 42● 16.30 16.524 16.40 16.628 16.50 16.732 16.60 16.836 16.70 16.941 16.80 17.045 16.90 17.150 17.00 17.255 17.10 17.359 17.20 17.464 17.30 17.568 17.40 17.673 17.50 17.778 17.60 17.883 17.70 17.988 17.80 18.093 17.90 18.198 18.00 18.303 18.10 18.408 18.20 18.513 18.30 18.619 18.40 18.724 18.50 18.830 18.60 18.935 18.70 19.041 18.80 19.146 18.90 19.251 19.00 19.356 19.10 19.463 19.20 19.569 19.30 19.675 19.40 19.781 19.50 19.887 19.60 19.993 19.70 20.100 19.80 20.206 19.90 20.312 20.00 20.419 20.10 20.525 20.20 20.632 20.30 20.738 20.40 20.845 20.50 20.952 20.60 21.059 20.70 21.165 20.80 21.272 20.90 21.379 21.00 21.486 21.10 21.593 21.20 21.701 21.30 21.808 21.40 21.915 21.50 21.023 21.60 22.130 21.70 22.238 21.80 22.345 21.90 22.453 22.00 22.561 22.10 22.669 22.20 22.777 22.30 22.885 22.40 22.993 22.50 23.101 22.60 23.210 22.70 23.318 22.80 23.427 22.90 23.535 23.00 23.643 23.10 23.752 23.20 23.861 23.30 23.970 23.40 24.079 23.50 24.188 23.60 24.297 23.70 24.406 23.80 24.515 23.90 24.624 24.00 24.734 24.10 24.844 24.20 24.953 24.30 25.063 24.40 25.173 24.50 25.282 24.60 25.392 24.70 25.502 24.80 25.613 24.90 25.723 25.00 25.833 25.10 25.943 25.20 26.054 25.30 26.164 25.40 26.275 25.50 26.386 25.60 26.497 25.70 26.608 25.80 26.719 25.90 26.830 26.00 26.941 26.10 27.052 26.20 27.164 26.30 27.275 26.40 27.387 26.50 27.499 26.60 27.610 26.70 27.722 26.80 27.834 26.90 27.946 27.00 28.058 27.20 28.283 27.30 28.396 27.40 28.508 27.50 28.621 27.60 28.734 27.70 28.847 27.80 28.959 27.90 29.072 28.00 29.186 28.10 29.299 28.20 29.413 28.30 29.526 28.40 29.640 28.50 29.753 28.60 29.867 28.70 29.981 28.80 30.095 28.90 30.300 29.00 30.324 29.10 30.438 29.20 30.553 29.30 30.667 29.40 30.782 29.50 30.897 29.60 31.012 29.70 31.127 29.80 31.242 29.90 31.357 30.00 31.473 30.10 31.588 30.20 31.704 30.30 31.820 30.40 31.936 30.50 32.052 30.60 32.168 30.70 32.284 30.80 32.409 30.90 32.517 31.00 32.633 31.10 32.750 31.20 32.867 31.30 32.984 31.40 33.101 31.50 33.218 31.60 33.336 31.70 33.453 31.80 33.571 31.90 33.688 32.00 33.806 32.10 33.924 32.20 34.042 32.30 34.161 32.40 34.279 32.50 34.397 32.60 34.516 32.70 34.635 32.80 34.754 32.90 34.873 33.00 34.992 33.10 35.111 33.20 35.231 33.30 35.350 33.40 35.470 33.50 33.590 33.60 35.710 33.70 35.830 33.80 35.950 33.90 36.071 34.00 36.191 34.10 36.312 34.20 36.433 34.30 36.554 34.40 36.675 34.50 36.796 34.60 36.917 34.70 37.039 34.80 37.161 34.90 37.283 35.00 37.405 35.10 37.527 35.20 37.649 35.30 37.771 35.40 37.894 35.50 38.017 35.60 38.140 35.70 38.263 35.80 38.386 35.90 38.509 36.00 38.633 36.10 38.757 36.20 38.880 36.30 39.004 36.40 39.129 36.50 39.253 36.60 39.377 36.70 39.502 36.80 39.627 36.90 39.752 37.00 39.877 37.10 40.002 37.20 40.128 37.30 40.253 37.40 40.379 37.50 40.505 37.60 40 . 63● 37.70 40.757 37.80 40.884 37.90 41.011 38.00 41.137 38.10 41.264 38.20 41.392 38.30 41.519 38.40 41.646 38.50 41.774 38.60 41.902 38.70 42.030 38.80 42.158 38.90 42.287 39.00 42.415 39.10 42.544 39.20 42.673 39.30 42.802 39.40 42.931 39.50 43.061 39.60 43.191 39.70 43.320 39.80 43.451 39.90 43.581 40.00 43.711 40.10 43.842 40.20 43.973 40.30 44.104 40.40 44.235 40.50 44.366 40.60 44.498 40.70 44.630 40.80 44.762 40.90 44.894 41.00 45.026 41.10 45.159 41.20 45.292 41.30 45.425 41.40 45.558 41.50 45.691 41.60 45.825 41.70 45.959 41.80 46.093 41.90 46.227 42.00 46.362 42.10 46.496 42.20 46.631 42.30 46.766 42.40 46.902 42.50 47.037 42.60 47.173 42.70 47.309 42.80 47.445 42.90 47.581 43.00 47.718 43.10 47.855 43.20 47.992 43.30 48.129 43.40 48.267 43.50 48.404 43.60 48.542 43.70 48.681 43.80 48.819 43.90 48.958 44.00 49.097 44.10 49.236 44.20 49.375 44.30 49.515 44.40 49.655 44.50 49.795 44.60 49.935 44.70 50.076 44.80 50.217 44.90 50.358 45.00 50.499 45.10 50 . 64● 45.20 50.783 45.30 50.925 45.40 51.068 45.50 51.210 45.60 51.353 45.70 51.496 45.80 51.639 45.90 51.783 46.00 51.927 46.10 52.071 46.20 52.215 46.30 52.360 46.40 52.505 46.50 52.650 46.60 52.795 46.70 52.941 46.80 53.087 46.90 53.233 47.00 53.380 47.10 53.526 47.20 53.673 47.30 53.821 47.40 53.968 47.50 54.116 47.60 54.264 47.70 54.413 47.80 54.562 47.90 54.711 48.00 54.860 48.10 55.010 48.20 55.160 48.30 55.310 48.40 55.460 48.50 55.611 48.60 55.762 48.70 55.913 48.80 56.065 48.90 56.117 49.00 56.369 49.10 56.522 49.20 56.675 49.30 56.828 49.40 56.981 49.50 57.135 49.60 57.289 49.70 57.444 49.80 57.598 49.90 57.754 50.00 57.909 50.10 58.065 50.20 58.221 50.30 58.377 50.40 58.534 50.50 58.691 50.60 58.848 50.70 59.006 50.80 59.164 50.90 59.322 51.00 59.481 51.10 59.640 51.20 59.800 51.30 59.960 51.40 60.120 51.50 60.280 51.60 60.441 51.70 60.601 51.80 60.763 51.90 60.925 52.00 61.088 52.10 61.250 52.20 61.413 52.30 61.577 52.40 61.740 52.50 61.904 52.60 62.069 52.70 62.234 52.80 62.399 52.90 62.564 53.00 62.730 53.10 62.897 53.20 63.063 53.30 63.231 53.40 63.398 53.50 63.566 53.60 63.734 53.70 63.903 53.80 64.072 53.90 64.242 54.00 64.412 54.10 64.582 54.20 64.753 54.30 64.924 54.40 65.096 54.50 65.268 54.60 65.440 54.70 65.613 54.80 65.786 54.90 65.960 55.00 66.134 55.10 66.308 55.20 66.483 55.30 66.659 55.40 66.835 55.50 67.011 55.60 67.188 55.70 67.365 55.80 67.543 55.90 67.721 56.00 67.900 56.10 68.079 56.20 68 . 25● 56.30 68.438 56.40 68.618 56.50 68.799 56.60 68.981 56.70 69.163 56.80 69.345 56.90 69.528 57.00 69.711 57.10 69.895 57.20 70.080 57.30 70.263 57.40 70.449 57.50 70.635 57.60 70.821 57.70 71.008 57.80 71.195 57.90 71.383 58.00 71.572 58.10 71.761 58.20 71.950 58.30 72.140 55.40 72.331 58.50 72.522 58.60 72.714 58.70 72.906 58.80 73.099 58.90 73.292 59.00 73.486 59.10 73.680 59.20 73.875 59.30 74.071 59.40 74.267 59.50 74.464 59.60 74.661 59.70 74.859 59.80 75.057 59.90 75.256 60.00 75.456 60.10 75.650 60.20 75.857 60.30 76.059 60.40 76.261 60.50 76.464 60.60 76.667 60.70 76.871 60.80 77.076 60.90 77.281 61.00 77.487 61.10 77.694 61.20 77.901 61.30 78.109 61.40 78.317 61.50 78.526 61.60 78.736 61.70 78.947 61.80 79.158 61.90 79.370 62.00 79.583 62.10 79.796 62.20 89.010 62.30 89.225 62.40 89.441 62.50 89.657 62.60 89.874 62.70 81.091 62.80 81.310 62.90 81.529 63.00 81.749 63.10 81.970 63.20 82.191 63.30 82.413 63.40 82.635 63.50 82.860 63.60 83.084 63.70 83.310 63.80 83.536 63.90 83.763 64.00 83.990 64.10 84.219 64.20 84.448 64.30 84.678 64.40 84.909 64.50 85.141 64.60 85.374 64.70 85.607 64.80 85.842 64.90 86.077 65.00 86.313 65.10 86.550 65.20 86.788 65.30 87.027 65.40 87.267 65.50 87.508 65.60 87.749 65.70 87.992 65.80 88.235 65.90 88.480 66.00 88.725 66.10 88.971 66.20 89.219 66.30 89.467 66.40 89.716 66.50 89.967 66.60 90.218 66.70 90.470 66.80 90.723 66.90 90.978 67.00 91.232 67.10 91.489 67.20 91.746 67.30 92.005 67.40 92.264 67.50 92.525 67.60 92.787 67.70 93.050 67.80 93.314 67.90 93.579 68.00 93.846 68.10 94.113 68.20 94.382 68.30 94.652 68.40 94.923 68.50 95.195 68.60 95.468 68.70 95.743 68.80 96.019 68.90 96.296 69.00 96.575 69.10 96.854 69.20 97.135 69.30 97.418 69.40 97.701 69.50 97.986 69.60 98.272 69.70 98.560 69.80 98.849 69.90 99.139 70.00 99.431 70.10 99.724 70.20 100.018 70.30 100.314 70.40 100.612 70.50 100.910 70.60 101.211 70.70 101.513 70.80 101.816 70.90 102.121 71.00 102.427 71.10 102.735 71.20 103.044 71.30 103.356 71.40 103.668 71.50 103.983 71.60 104.299 71.70 104.616 71.80 104.936 71.90 105.257 72.00 105.579 72.10 105 904 72.20 106 230 72.30 106 558 72.40 106 888 72.50 107 220 72.60 107 553 72.70 107 888 72.80 108 226 72.90 108 565 73.00 108 906 73.10 109 249 73.20 109 594 73.30 109 941 73.40 110 290 73.50 110 641 73.60 110 994 73.70 111 349 73.80 111 707 73.90 112 066 74.00 112 428 74.10 112 792 74.20 113 158 74.30 113 526 74.40 113 897 74.50 114 270 74.60 114 645 74.70 115 023 74.80 115 403 74.90 115 786 75.00 116 171 75.10 116 559 75.20 116 949 75.30 117 342 75.40 117 737 75.50 118 135 75.60 118 536 75.70 118 939 75.80 119 345 75.90 119 755 76.00 120 160 76.10 120 581 76 20 121 000 76.30 121 420 76.40 121 843 76.50 12 270 76.60 12 700 76.70 123 133 76.80 123 570 76.90 124 009 77.00 124 452 77.10 124 898 72.20 125 348 77.30 125 801 77.40 126 258 77.50 126 718 77.60 127 182 77.70 127 649 77.80 128 121 77.90 128 596 78.00 129 075 78 10 129 558 78 20 130 045 78 30 130 536 78 40 131 031 78 50 131 530 78 60 132 034 78 70 132 542 78 80 113 055 78 90 113 572 79 00 134 094 79 10 134 620 79 20 135 151 79 30 135 687 79 40 136 228 79 50 136 775 79 60 137 326 79 70 137 883 79 80 138 445 79 90 139 012 80 00 139 585 80 10 140 164 80 20 140 748 80 30 141 339 80 40 141 936 80 50 142 138 80 60 143 147 80 70 143 763 80 80 144 385 80 90 145 014 81 00 145 650 81 10 146 292 81 20 146 942 81 30 147 600 81 40 148 265 81 50 148 937 81 60 149 618 81 70 150 307 81 80 151 003 81 90 151 709 82 00 152 423 82 10 153 147 82 20 153 878 82 30 154 620 82 40 155 372 82 50 156 132 82 60 156 903 82 70 157 685 82 80 158 478 82 90 159 281 83 00 160 096 83 10 160 922 83 20 161 761 83 30 162 612 83 40 163 475 83 50 164 352 83 60 165 242 83 70 166 146 83 80 167 065 83 90 167 999 84 00 168 947 84 10 169 912 84 20 170 893 84 30 171 891 84 40 172 907 84 50 173 941 84 60 174 994 84 70 176 ●67 84 80 177 160 84 90 178 275 85 00 179 411 85 10 180 569 85 20 181 752 85 30 182 960 85 40 184 194 85 50 185 454 85 60 186 743 85 70 188 062 85 80 189 411 85 90 190 793 86 00 192 210 86 10 193 661 86 20 195 151 86 30 196 680 86 40 198 251 86 50 199 867 86 60 201 529 86 70 203 240 86 80 205 005 86 90 206 825 87 00 208 705 87 10 210 649 87 20 212 668 87 30 214 745 87 40 216 909 87 50 219 158 87 60 221 498 87 70 223 938 87 80 226 486 87 90 229 153 88 00 231 95● 88 10 234 891 88 20 237 991 88 30 241 268 88 40 244 744 88 50 248 445 88 60 252 402 88 70 256 652 88 80 261 243 88 90 266 235 89 00 271 705 89 10 277 753 89 20 284 517 89 30 292 191 89 40 301 058 89 50 311 563 89 60 324 455 89 70 341 166 89 80 365 039 89 90 408 011 90 00 Infinite

ATableoftheSunsDeclination,fortheyears1654,1658,1662,1666. Ianu. Febr. Mar Apr. May. June July . Aug. Sep. Octo. Nov Dec. Dayes . south south sout north north north north north nort south south south 1 21 78 13 85 3 48 08 52 18 03 23 18 22 16 15 28 4 50 7 15 17 60 23 13 2 21 62 13 52 3 10 08 88 18 28 23 25 22 ●3 14 98 4 11 7 53 17 86 23 20 3 21 45 13 17 2 70 09 25 18 53 23 30 21 88 14 66 3 73 7 91 18 13 23 26 4 21 27 12 83 2 30 09 60 18 77 23 35 21 73 14 36 3 35 8 28 18 40 23 33 5 21 08 12 50 1 92 09 97 19 00 23 40 21 58 14 05 2 96 8 65 18 66 23 38 6 20 88 12 15 1 52 10 31 19 23 23 43 21 42 13 73 2 56 9 03 18 91 23 43 7 20 68 11 80 1 11 10 67 19 47 23 46 21 25 13 41 2 18 9 40 19 15 23 46 8 20 48 11 43 0 72 11 02 19 68 23 50 21 07 13 08 1 80 9 76 19 40 23 48 9 20 27 11 08 0 33 11 36 19 90 23 51 20 90 12 76 1 40 10 13 19 63 23 50 10 20 05 10 72 0 06 11 70 20 11 23 52 20 71 12 43 1 01 10 48 19 86 23 51 11 19 82 10 37 N 47 12 05 20 31 23 53 20 51 12 10 0 63 10 85 20 08 23 53 12 19 58 09 83 0 85 12 38 20 51 23 52 20 31 11 76 0 23 11 20 20 30 23 51 13 19 35 09 63 1 25 12 72 20 70 23 51 20 11 11 43 0●16 11 57 20 51 23 51 14 19 11 09 25 1 65 13 05 20 88 23 50 19 90 11 08 0 55 11 91 20 71 23 48 15 18 86 08 88 2 03 13 36 21 06 23 46 19 68 10 73 0 95 12 25 20 91 23 46 16 18 61 08 52 2 41 13 68 21 25 23 43 19 47 10 38 1 33 12 60 21 10 23 43 17 18 35 08 13 2 82 14 00 21 41 23 40 19 25 10 03 1 73 12 95 21 28 23 28 18 18 08 07 75 3 20 14 31 21 58 23 35 19 01 09 68 2 11 13 28 21 46 23 33 19 17 81 07 37 3 60 14 63 21 73 23 30 18 78 09 33 2 51 13 61 21 63 23 28 20 17 53 06 98 3 98 14 93 21 88 23 25 18 55 08 96 2 90 13 95 21 80 23 21 21 17 25 06 60 4 37 15 23 22 03 23 18 18 30 08 60 3 30 14 26 21 95 23 13 22 16 96 06 22 4 75 15 53 22 16 23 10 18 05 08 25 3 68 14 60 22 10 23 05 23 16 68 05 83 5 13 15 83 22 30 23 03 17 78 07 88 4 06 14 91 22 25 22 96 24 16 38 05 45 5 51 16 13 22 41 22 95 17 53 07 51 4 46 15 23 22 38 22 86 25 16 08 05 07 5 90 16 41 22 53 22 85 17 26 07 ●5 4 85 15 55 22 51 22 76 26 15 78 04 67 6 28 16 70 22 65 22 75 17 00 06 76 5 23 15 85 22 63 22 65 27 15 46 04 28 6 67 16 97 22 75 22 65 16 71 06 40 5 63 16 15 22 75 22 53 28 15 15 03 88 7 03 17 23 22 85 22 53 16 43 06 01 6 00 16 45 22 85 22 40 29 14 83 7 41 17 50 22 95 22 41 15 86 05 26 6 76 17 03 23 05 22 11 30 14 51 7 78 17 77 23 ●3 22 30 15 86 05 26 6 76 17 03 23 05 22 11 31 14 18 8 15 23 77 15 56 04 88 17 31 21 96

ATableoftheSunsDeclination,fortheyears1655,1659,1663,1667. Ianu. Febr. Mar Apr. May. June July . Aug. Sep. Octo. Nov Dec. Dayes . south south sout north north north north north nort south south south 1 21 81 13 93 3 58 08 43 17 96 23 16 22 20 15 35 4 58 7 06 17 53 23 10 2 21 65 13 60 3 18 08 80 18 21 23 23 22 06 15 05 4 20 7 43 17 80 23 18 3 21 48 13 26 2 80 09 15 18 46 23 30 21 91 14 75 3 81 7 81 18 06 23 25 4 21 30 12 91 2 40 09 51 18 71 23 35 21 76 14 43 3 43 8 20 18 33 23 31 5 21 11 12 58 2 00 09 88 18 95 23 40 21 61 14 13 3 05 8 56 18 60 23 36 6 20 93 12 23 1 61 10 23 19 18 23 43 21 45 13 81 2 66 8 93 18 85 23 41 7 20 73 11 88 1 21 10 58 19 41 23 46 21 28 13 50 2 28 9 30 19 10 23 45 8 20 53 11 53 0 81 10 93 19 63 23 48 21 11 13 16 1 88 9 66 19 33 23 48 9 20 32 11 16 0 43 11 28 19 85 23 50 20 93 12 85 1 50 10 03 19 56 23 50 10 20 10 10 81 ●●03 11 61 20 06 23 51 20 75 12 51 1 10 10 40 19 80 23 51 11 19 88 10 45 0 36 11 96 20 26 23 53 20 56 12 18 0 71 10 76 20 03 23 53 12 19 65 10 08 0 76 12 30 20 46 23 51 20 36 11 85 0 33 11 11 20 25 23 52 13 19 41 09 71 1 15 12 63 20 66 23 51 20 16 11 51 0●06 11 46 20 46 23 51 14 19 16 09 35 1 55 12 96 20 85 23 50 19 95 11 16 0 46 11 81 20 66 23 50 15 18 92 08 96 1 93 13 28 21 03 23 48 19 75 10 81 0 85 12 16 20 86 23 46 16 18 67 08 60 2 33 13 61 21 20 23 45 19 53 10 65 1 25 12 51 21 06 23 43 17 18 41 08 23 2 71 13 93 21 38 23 41 19 30 10 13 1 63 12 86 21 25 23 40 18 18 15 07 85 3 11 14 25 21 55 23 36 19 06 09 76 2 03 13 20 21 41 23 35 19 17 88 07 46 3 50 14 56 21 70 23 31 18 83 09 41 2 41 13 53 21 60 23 28 20 17 60 07 08 3 88 14 86 21 85 23 26 18 60 09 06 2 80 13 86 21 76 23 21 21 17 31 06 70 4 28 15 16 22 00 23 20 18 35 08 70 3 20 14 20 21 91 23 15 22 17 03 06 31 4 66 15 46 22 13 23 13 18 10 08 33 3 60 14 51 22 06 23 06 23 16 75 05 93 5 05 15 76 22 26 23 05 17 85 07 96 3 98 14 83 22 21 22 98 24 16 45 05 53 5 43 16 05 22 40 22 96 17 58 07 60 4 36 15 15 22 35 22 88 25 16 15 05 15 5 81 16 35 22 51 22 88 17 31 07 23 4 76 15 46 22 48 22 78 26 15 85 04 76 6 20 16 63 22 63 22 78 17 05 06 85 5 15 15 76 22 60 22 68 27 15 53 04 36 6 56 16 90 22 73 22 68 16 78 06 48 5 53 16 08 22 71 22 56 28 15 23 03 98 6 95 17 18 22 83 22 56 16 50 06 11 5 91 16 38 22 83 22 43 29 14 91 7 31 17 45 22 93 22 45 16 21 05 73 6 30 16 68 22 93 22 30 30 14 58 7 70 17 71 23 01 22 33 15 93 05 36 6 68 16 96 23 01 22 16 31 14 26 8 06 23 10 15 65 04 98 17 25 22 01

ATableoftheSunsDeclination,fortheyears1656,1660,1664,1668. Ianu. Febr. Mar Apr. May. June July Aug. Sep. Octo Nov Dece . Dayes south south sout north north north north north nort south south south 1 21 85 14 01 3 28 08 70 18 16 23 21 22 08 15 11 4 30 7 35 17 73 23 16 2 21 70 13 68 3 90 09 06 18 41 23 28 21 95 14 81 3 91 7 73 18 00 23 23 3 21 53 13 35 2 50 09 43 18 65 23 33 21 80 14 51 3 53 8 10 18 26 23 30 4 21 35 13 00 2 10 09 78 18 90 23 38 21 65 14 20 3 15 8 48 18 53 23 35 5 21 16 12 66 1 66 10 15 19 13 23 41 21 50 13 88 2 75 8 85 18 78 23 40 6 20 98 12 31 1 33 10 50 19 36 23 45 21 33 13 56 2 36 9 21 19 03 23 45 7 20 78 11 96 0 91 10 85 19 58 23 48 21 16 13 25 1 98 9 58 19 28 23 48 8 20 58 11 61 0 53 11 20 19 80 23 50 20 98 12 91 1 58 9 95 19 51 23 50 9 20 36 11 26 0 13 11 53 20 01 23 51 20 80 12 60 1 20 10 31 19 75 23 51 10 20 15 10 90 N26 11 88 20 21 23 52 20 61 12 26 0 81 10 68 19 98 23 52 11 19 93 10 53 0 66 12 21 20 41 23 53 20 41 11 93 0 41 11 03 20 20 23 53 12 19 70 10 16 1 05 12 55 20 61 23 51 20 21 11 60 0 03 11 38 20 41 23 51 13 ●9 46 09 80 1 45 12 88 20 80 23 50 20 00 11 25 0●36 11 73 20 61 23 50 14 19 23 09 43 1 83 13 20 20 98 23 48 19 80 10 90 0 75 12 08 20 81 23 48 15 18 98 09 06 2 23 13 53 21 16 23 45 19 58 10 56 1 15 12 43 21 01 23 45 16 18 73 08 70 2 63 13 85 21 33 23 41 ●9 35 10 21 1 55 12 78 21 20 23 40 17 18 48 08 31 3 03 14 16 21 50 23 38 ●9 13 9 85 1 93 13 11 21 38 23 35 18 18 21 07 93 3 41 14 48 21 66 23 33 18 90 9 50 2 31 13 45 21 55 23 30 19 17 95 07 55 3 80 14 80 21 81 23 26 18 65 9 15 2 71 13 78 21 7● 23 25 20 17 66 07 16 4 18 15 10 21 96 23 21 18 41 8 78 3 10 14 11 21 88 23 18 21 17 40 06 78 4 56 15 40 22 10 23 15 18 16 8 41 3 50 14 43 22 03 23 10 22 17 11 06 40 4 95 15 70 22 23 23 06 17 91 8 05 3 88 14 76 22 18 23 01 23 16 81 06 01 5 33 15 98 22 36 22 98 17 66 7 68 4 28 15 08 22 31 22 91 24 16 53 05 63 5 71 16 28 22 48 22 90 17 40 7 31 4 66 15 40 22 45 22 81 25 16 23 05 25 6 10 16 56 22 60 22 80 17 11 6 95 5 05 15 70 22 58 22 70 26 15 91 04 86 6 48 16 83 22 71 22 70 16 85 6 56 5 43 16 00 22 70 22 58 27 15 61 04 48 6 85 17 11 22 81 22 60 16 56 6 20 5 81 16 30 22 80 22 46 28 15 30 04 06 7 23 17 38 22 90 22 48 16 28 5 83 6 20 16 60 22 90 22 33 29 14 98 03 68 7 60 17 65 22 98 22 35 16 00 5 45 6 58 16 90 23 00 22 20 30 14 66 7 96 17 90 23 06 22 21 15 71 5 06 6 96 17 ●8 23 08 22 05 31 14 35 8 33 23 15 15 41 4 68 17 46 21 90

ATableoftheSunsDeclination,fortheyears1657,1661,1665,1669. Ianu. Febr. Mar Apr. May. June July Aug. Sep. Octo Nov Dece . Dayes . south south sout north north north north north nort south south south 1 21 73 13 76 3 40 08 60 18 08 23 20 22 13 15 20 4 40 7 25 17 66 23 15 2 21 56 13 43 3 00 08 96 18 33 23 26 22 00 14 90 4 01 7 63 17 93 23 21 3 21 38 13 08 2 61 09 33 18 58 23 31 21 85 14 60 3 63 8 00 18 20 23 28 4 21 21 12 75 2 21 09 70 18 83 23 36 21 70 14 28 3 25 8 36 18 46 23 33 5 21 03 12 41 1 81 10 05 19 06 23 41 21 53 13 96 2 86 8 75 18 71 23 38 6 20 83 12 06 1 41 10 40 19 30 23 45 21 36 13 65 2 48 9 11 18 96 23 43 7 20 63 11 71 1 01 10 75 19 51 23 48 21 20 13 33 2 08 9 48 19 21 23 46 8 20 43 11 35 0 63 11 10 19 73 23 50 21 03 13 01 1 70 9 85 19 45 23 50 9 20 21 11 00 0 23 11 45 11 95 23 51 20 85 12 68 1 31 10 21 19 68 23 51 10 20 00 10 63 N16 11 78 20 16 23 52 20 66 12 35 0 91 10 58 19 91 23 52 11 19 76 10 26 0 55 12 11 20 36 23 53 20 46 12 01 0 53 10 93 20 13 23 53 12 19 53 09 90 0 95 12 46 20 56 23 51 20 26 11 68 0 13 11 30 20 35 23 51 13 19 30 09 53 1 35 12 80 20 75 23 50 20 06 11 35 0●26 11 65 20 56 23 50 14 19 05 09 16 1 73 13 11 20 93 23 48 19 85 11 15 0 65 12 00 20 76 23 48 15 18 80 08 80 2 13 13 45 21 11 23 46 19 63 10 65 1 05 12 35 20 96 23 45 16 18 55 08 41 2 51 13 76 21 28 23 43 19 41 10 30 1 43 12 68 21 15 23 41 17 18 28 08 05 2 90 14 08 21 45 23 38 19 20 9 95 1 83 13 20 21 33 23 36 18 18 03 07 66 3 30 14 40 21 61 23 33 18 96 9 60 2 21 13 36 21 51 23 31 19 17 75 07 28 3 68 14 70 21 76 23 28 18 71 9 25 2 61 13 70 21 68 123 26 20 17 46 06 90 3 08 15 01 21 91 23 23 18 48 8 88 3 00 14 03 21 83 23 20 21 17 18 06 51 4 46 15 31 22 06 23 16 18 23 8 51 3 38 14 35 22 00 23 11 22 16 90 06 13 4 85 15 61 22 20 23 10 17 98 8 15 3 78 14 68 22 15 23 03 23 16 60 05 75 5 23 15 90 22 33 23 01 17 73 7 78 4 16 15 00 22 28 22 95 24 16 30 05 35 5 61 16 20 22 45 22 91 17 46 7 41 4 55 15 31 22 41 22 85 25 16 00 04 96 6 00 16 48 22 56 22 83 17 20 7 05 4 95 15 61 22 55 22 73 26 15 70 04 56 6 36 16 76 22 68 22 73 16 93 6 68 5 33 15 91 22 66 22 61 27 15 38 04 18 6 75 17 03 22 78 22 61 16 65 6 30 5 71 16 21 22 76 22 50 28 15 06 03 78 7 11 17 30 22 88 22 51 16 36 5 93 6 10 16 51 22 86 22 36 29 14 75 7 50 17 56 22 96 22 38 16 08 5 55 6 48 16 81 22 96 22 ●● 30 14 43 7 86 17 83 23 05 22 26 15 80 5 16 6 86 17 10 23 06 22 08 31 14 10 8 23 23 13 15 50 4 78 17 38 21 93

ATableoftheSunsrightAscensioninhoursandminutes. Janu Febr. Mar. Apr. May Iune Jul Aug. Sept. Octo. Nov. Dece . Dayes H. M. H. M. H. M. H M H. M H. M H M H. M H. M H. M H. M H. M 1 19 53 21 68 23 45 1 33 3 21 5 30 7 36 9 40 11 30 13 10 15 10 17 23 2 19 61 21 75 23 51 1 38 3 28 5 36 7 43 9 46 11 35 13 16 15 16 17 31 3 19 68 21 81 23 56 1 45 3 35 5 43 7 50 9 51 11 41 13 23 15 23 17 38 4 19 75 21 88 23 63 1 51 3 40 5 50 7 56 9 58 11 48 13 28 15 30 17 45 5 19 83 21 95 23 70 1 56 3 46 5 56 7 63 9 65 11 53 13 35 15 38 17 53 6 19 90 22 00 23 75 1 63 3 53 5 63 7 70 9 71 11 60 13 41 15 45 17 60 7 19 96 22 06 23 81 1 70 3 60 5 71 7 76 9 76 11 65 13 48 15 51 17 68 8 20 05 22 13 23 86 1 75 3 66 5 78 7 83 9 83 11 71 13 53 15 58 17 75 9 20 11 22 20 23 93 1 81 3 73 5 85 7 90 9 90 11 78 13 60 15 65 17 83 10 20 18 22 26 00 00 1 88 3 80 5 91 7 96 9 96 11 83 13 66 15 71 17 90 11 20 25 22 33 0 05 1 95 3 86 5 98 8 03 10 01 11 90 13 73 15 80 17 98 12 20 33 22 40 0 11 2 00 3 93 6 05 8 10 10 08 11 95 13 80 15 86 18 05 13 20 40 22 45 0 18 2 06 4 0● 6 13 8 16 10 15 12 01 13 85 15 93 18 11 14 20 46 22 51 0 23 2 13 4 06 6 20 8 23 10 20 12 06 13 91 16 00 18 20 15 20 53 22 58 0 30 2 18 4 13 6 26 8 30 10 26 12 13 13 98 16 08 18 26 16 20 60 22 65 0 35 2 25 4 20 6 33 8 36 10 33 12 20 14 05 16 10 18 35 17 20 66 22 70 0 41 2 31 4 26 6 40 8 43 10 38 12 25 14 11 16 21 18 41 18 20 75 20 76 0 48 2 38 4 33 6 40 8 50 10 45 12 31 14 18 16 28 18 50 19 20 81 22 83 0 53 2 45 4 40 6 53 8 56 10 51 12 38 14 23 16 36 18 56 20 20 88 22 90 0 60 2 50 4 46 6 61 8 63 10 56 12 43 14 30 16 43 18 65 21 20 95 22 95 0 66 2 56 4 55 6 68 8 70 10 63 12 50 14 36 16 50 18 71 22 21 01 23 01 0 71 2 63 4 61 6 75 8 76 10 70 12 55 14 43 16 58 18 80 23 21 08 23 08 0 78 2 70 4 68 6 81 8 81 10 75 12 61 14 50 16 65 18 86 24 21 15 23 13 0 83 2 76 4 75 6 88 8 88 10 81 12 68 14 56 16 71 18 93 25 21 21 23 20 0 90 2 81 4 81 6 95 8 95 10 88 12 73 14 63 16 80 19 01 26 21 28 23 26 0 96 2 88 4 88 7 01 9 01 10 93 12 80 14 70 16 86 19 08 27 21 35 23 33 1 01 2 95 4 95 7 10 9 08 11 00 12 86 14 76 16 93 19 16 28 21 41 23 38 1 08 3 01 5 01 7 16 9 15 11 05 12 91 14 83 17 01 19 23 29 21 48 1 15 3 08 5 08 7 23 9 21 11 11 12 98 14 90 17 08 19 30 30 21 55 1 20 3 15 5 16 7 30 9 26 11 18 13 05 14 96 17 16 19 38 31 21 61 1 26 5 23 9 33 11 23 15 03 19 45

DeclinationandRightAscensionoftheStars. Declination Dist. from the pole Right Ascenation The names of the Stars . D. M. D. M. H. M. Brest of Cassiopeia 54 76 N 35 24 0 35 Pole-star 87 48 N 02 52 0 51 Girdle of Andromeda 33 83 N 56 17 0 83 Knees of Cassiopeia 58 41 N 31 59 1 05 Whales belly 12 00 S 78 00 1 58 South , foot by Andr. 40 65 N 19 35 1 70 Rams head 21 81 N 68 19 1 80 Head of Medusa 39 58 N 50 42 2 76 Perseus right side 48 55 N 41 45 2 98 Buls eye 15 75 N 74 25 4 26 The Goat 45 58 N 44 42 4 85 Orions left foot 08 63 S 81 37 4 96 Orions left shoulder 05 98 N 84 02 5 10 First in Orions girdle 00 58 S 89 42 5 25 Second in Orions gird . 01 45 S 88 55 5 31 Third in Orions girdle 02 15 S 87 85 5 38 Wagoners right shold . 44 86 N 45 14 5 70 Orions right shoulder 07 30 N 82 70 5 60 Bright foot of Twins 16 65 N 73 35 6 30 The great Dog 16 21 S 73 79 6 50 Upper head of Twins 32 50 N 57 50 7 20 The lesse Dog 06 10 N 83 90 7 36 Lower head of Twins 28 80 N 61 20 7 40 Brightest in Hydra 07 16 S 82 84 09 16 Lions heart 13 63 N 76 37 09 83 Lions back 22 06 N 65 94 11 50 Lions tail 16 50 N 73 50 11 51 Great Bears rump . 58 72 N 31 28 10 66 First in the great Bears tail next her rump 57 85 N 32 15 12 63 Virgines spike ●9 32 S 80 68 13 11 Middlemost in the great Bears tail 56 75 N 33 25 13 16 In the end of her tail 51 08 N 38 92 13 56 Between Bootes thighs 21 03 N 68 97 14 00 South Ballance 14 55 S 75 45 14 53 North Ballance 08 05 S 81 95 14 96 Scorpions heart 25 58 S 64 42 16 13 Hercules head 14 85 N 75 15 16 98 Serpentaries head 12 86 N 77 14 17 31 Dragons head 51 60 N 38 40 17 80 Brightest in the Harp 38 50 N 51 50 18 41 Eagles heart 08 02 N 81 98 19 56 Swans tail 44 08 N 45 92 20 50 Pegasus mouth 08 32 N 81 68 21 45 Pegasus shoulder 17 38 N 76 62 23 91 The head of Androm . 27 22 N 62 78 23 85

Rules for finding of the Poles elevation by the meridian altitude of the Sun or stars , and by the Table of their Declinations aforegoing .
Case 1.

IF the Sun or star be on the meridian to the southwards , and have south declination . Adde the suns declination to his meridian altitude , and taking that total from 90 degrees , what remaineth is the latitude of the place desired .

As the 7 of February , 1654 , by the aforegoing Table , the suns decl . south . is 11.80 The suns meridian altitude 15.27 The sum or total is 27.07 Which substracted from 90.00 There remains the North latitude 62.93

But when you have added the suns declination to his meridian altitude , if the total exceed 90 : substract 90 degr. from it , and what remaineth is your latitude to the southwards .

As admit the suns declination to be southerly 11.80 And his meridian altitude 87.23 The sum or total is 99.03 From which substracting 90.00 There remains the latitude south . 09.03

Case . 2.

If the sun or star be on the meridian to the southwards , and have north declination .

Substract the suns declination from his meridian altitude , and that which remains substract from 90 , and then the remainer is the poles elevation northerly .

Case 3.

If the sun or star be on the meridian to the northwards , and have north declination ▪

Adde the suns declination to his meridian altitude , the total take from 90 , and what remaineth is the poles elevation southerly .

But when you have added the suns declination to his meridian altitude , if it exceed 90 , substract 90 from it , and what remaineth is your latitude northerly .

Case 4.

If the sun be to the northwards at noon , and declination south .

Substract the suns declination from his meridian altitude , and that which remains substract from 90 , what then remaineth is your latitude southerly .

And what is said of the Sun , is also to be understood of the Stars , being upon the Meridian .

Case 5.

If you observe when the Sun hath no declination .

Substract his meridian altitude from 90 , what remaineth is your latitude .

Case 6.

If you chance to observe when the Sun or star is in the Zenith , that is 90 degrees above the Horizon . Look in the table for the declination of the Sun or of that star , and the same is your latitude .

Case 7.

If the Sun come to the meridian under the Pole .

If you be within the Artick or Antartick circle , and observe the Sun upon the meridian under the Pole ; substract the Suns declination from 90 , the remainer is the Suns distance from the Pole ; which distance added to his meridian altitude , the sum or total is the latitude sought .

And what is here said of the Sun is to be understood of the stars , whose declinations , distances from the pole , and right ascensions we have expressed in the foregoing Table .

FINIS .