Astronomia
BRITANNICA
:
The
second
Book
.
CHAP.
1.
Of
the
year
Civil
and
Astronomical
.
THe
Altitudes
of
the
Planets
being
given
to
find
their
places
in
the
Zodiack
,
hath
been
already
shewed
in
the
Doctrine
of
the
Sphere
,
&
thence
their
annual
or
periodical
revolutions
,
together
with
their
middle
motions
must
be
sought
,
but
how
to
state
them
so
exactly
as
that
we
may
thereby
find
their
true
or
apparent
places
for
any
time
required
,
is
that
which
many
have
endevoured
,
but
none
have
as
yet
found
out
,
at
least
not
so
,
as
that
their
places
computed
by
their
rules
,
shall
exactly
agree
with
observation
,
nor
was
Astronomy
brought
to
that
perfection
in
which
it
now
stands
but
by
degrees
,
and
although
there
hath
been
very
much
done
of
late
towards
the
perfecting
thereof
,
yet
shall
it
not
perhaps
come
to
its
full
Acme
in
this
our
age
.
That
which
we
intend
,
is
not
to
shew
you
from
what
small
beginnings
it
hath
been
increased
,
or
by
whose
labours
,
it
hath
from
time
to
time
been
still
corrected
and
amended
,
but
to
shew
you
how
to
compute
the
places
of
heavenly
Bodies
,
by
the
plainest
,
speediest
,
and
exactest
ways
that
are
as
yet
made
publike
.
And
in
order
hereunto
we
will
shew
you
first
the
usual
way
of
finding
out
the
time
in
which
the
Planets
make
their
Annual
or
Periodical
revolutions
,
and
how
from
thence
to
compute
their
middle
Motions
,
that
their
annual
revolutions
may
be
known
,
the
time
of
their
entrance
into
one
and
the
same
point
of
the
Zodiack
,
taken
in
divers
years
by
observation
must
be
given
,
with
a
considerable
interval
of
time
between
these
Observations
.
And
because
the
Observations
taken
in
any
one
Meridian
(
that
are
as
yet
published
)
are
not
sufficient
for
our
present
purpose
we
must
of
necessity
,
use
the
observations
made
in
diverse
places
,
but
the
intervall
of
time
between
those
observations
cannot
be
had
,
unlesse
we
can
reduce
the
time
of
an
observation
made
in
one
account
to
the
like
time
in
another
;
Although
the
periodicall
revolutions
therefore
of
the
planets
are
the
onely
proper
yeares
and
first
in
nature
;
yet
since
the
civil
yeare
in
every
nation
is
somewhat
different
from
them
,
we
will
first
shew
the
quantity
of
that
in
most
nations
,
and
how
to
reduce
the
day
of
the
moneth
given
in
one
accont
,
to
the
correspondent
time
in
another
.
The
Civil
yeare
then
,
though
it
doth
not
exactly
agree
,
yet
hath
it
some
proportion
with
the
motions
of
the
Sunne
or
Moone
in
every
nation
;
Twelve
Moones
or
Moneths
is
the
common
measure
of
the
year
in
Turkey
,
in
every
moneth
they
have
29
or
30
dayes
,
in
the
whole
yeare
354
,
and
in
every
third
yeare
355
dayes
.
The
Persians
and
Egyptians
do
also
account
12
Months
to
their
yeare
,
but
their
months
are
proportioned
to
the
time
of
the
Suns
continuance
in
every
of
the
12
signes
;
in
their
year
therefore
which
is
Solar
,
there
are
alwayes
365
dayes
.
And
the
Julian
yeare
,
which
is
the
account
of
all
Christendome
doth
differ
from
the
other
onely
in
this
,
(
that
by
reason
of
the
Suns
excesse
in
motion
above
365
dayes
,
which
is
about
5
hours
49
minutes
)
it
hath
a
day
intercalated
once
in
4
yeares
,
and
by
reason
of
this
intercalation
it
is
more
agreeable
with
the
motion
of
the
Sun
then
the
former
,
and
yet
here
is
a
considerable
difference
between
them
,
which
hath
occasioned
the
Church
of
Rome
to
make
some
further
amendment
of
the
Solar
year
,
but
hath
not
brought
it
to
that
exactness
which
is
desired
,
nor
will
(
as
is
to
be
feared
)
be
over-hastily
brought
to
that
exactnes
which
it
might
;
taking
these
accounts
therefore
as
they
now
stand
,
if
we
will
reconcile
that
discrepancy
that
is
between
them
,
there
must
be
some
beginning
appointed
to
every
of
these
accounts
,
and
that
beginning
must
be
referred
to
some
one
,
as
to
the
common
measure
of
the
rest
.
The
most
natural
beginning
of
all
accounts
,
is
the
time
of
the
Worlds
Creation
,
but
they
who
could
not
attain
the
Worlds
beginning
,
have
reckoned
from
their
own
,
as
the
Romanes
,
ab
urbe
condita
,
or
from
some
great
name
or
notable
event
;
so
the
Greeks
account
from
their
Olympicks
,
and
the
Assyrians
from
Nabonasser
,
and
all
Christians
from
the
birth
of
Christ
,
the
beginning
of
which
and
all
other
the
most
notable
Epochaes
,
as
others
formerly
,
so
we
now
have
also
ascertained
to
their
correspondent
times
in
the
Julian
Period
,
which
Scaliger
contrived
by
the
continual
multiplication
of
three
circles
all
in
former
times
of
good
use
,
&
two
of
them
do
yet
remain
;
the
Circles
yet
in
use
are
those
of
the
Sun
and
Moon
,
the
one
to
wit
the
Sun
is
a
Circle
of
28
years
,
in
which
time
the
Sunday
Letter
makes
all
the
varieties
that
it
can
have
by
reason
of
the
Bisextile
or
Leap-year
,
and
the
Circle
of
the
Moon
is
the
revolution
of
19
years
,
in
which
time
,
though
not
precisely
,
the
Lunations
do
recur
;
it
is
called
the
Golden
Number
,
and
was
made
Christian
by
the
Fathers
of
the
Nicene
Council
,
as
being
altogether
necessaay
to
the
finding
out
of
the
Neomenia
Paschalis
,
upon
which
the
Feast
of
Easter
,
and
the
rest
of
the
moveable
Feasts
depend
.
The
third
Circle
which
now
serves
for
no
other
use
,
then
the
constituting
of
the
Julian
Period
,
is
the
Roman
Indiction
,
or
a
Circle
of
15
years
,
for
if
you
multiply
28
the
Cycle
of
the
Sun
,
by
19
the
Cycle
of
the
Moon
,
the
product
will
be
532
,
&
this
by
15
,
the
product
will
be
7980
the
number
of
years
in
the
Julian
Period
;
whose
admirable
condition
is
to
distinguish
every
year
within
the
whole
Circle
,
by
a
several
certain
character
,
the
year
of
the
Sun
,
Moon
,
and
Indiction
,
being
never
the
same
again
until
the
revolution
of
7980
years
be
gon
about
:
this
Period
,
the
Authour
fixed
in
the
Tohu
,
or
eternal
Chaos
of
the
World
,
764
Julian
years
before
the
most
reputed
time
of
Creation
;
which
being
,
premised
,
we
will
now
by
example
shew
you
how
to
reduce
the
years
of
Forreigners
to
our
Julian
years
,
and
the
contrary
.
1
Example
.
I
desire
to
know
at
what
time
in
the
Turkish
account
,
the
5
of
June
1649
,
falls
.
The
work
is
this
The
years
compleat
are
1648
,
and
are
thus
turned
into
Dayes
,
by
the
table
of
Dayes
,
and
Decimals
of
Dayes
in
Julian
Years
.
1000
Julian
yeares
give
dayes
-
365250
600
years
give
-
219150
40
years
give
-
14610
8
years
give
-
292●
May
Compleat
-
151
Dayes
-
5
The
Summe
-
602088
Now
because
the
Turkish
account
began
July
the
16.
Anno
Christi
622
,
convert
these
yeares
into
dayes
also
thus
600
Julian
years
give
-
219150
20
years
give
-
7305
1
year
gives
-
365
June
Compleat
-
181
Dayes
-
15
The
Summe
subtract
-
227016
From
-
602088
There
rests
-
375072
900
Turkish
years
gives
-
318930
There
rests
-
56142
150
years
gives
-
53155
There
rests
-
2987
8
years
give
-
2835
There
rests
-
152
Giumadi
I.
gives
-
148
There
rests
-
4
Therefore
the
5th
.
of
June
1649
,
in
our
English
accompt
,
falls
in
the
Turkish
accompt
,
in
the
year
of
Mahomet
,
or
their
Hegira
,
1058
,
the
4th
.
day
of
Giumadi
II.
2
Example
.
I
desire
to
know
upon
what
day
of
our
Julian
year
,
the
23
day
of
the
moneth
Pharmuthi
in
the
1912
year
currant
of
the
Aegyptian
accompt
from
the
death
of
Alexanders
fall
.
The
beginning
of
this
Epoch●
is
from
the
beginning
of
the
Julian
Period
in
compleat
dayes
.
-
1603397
1000
Egyptian
years
give
-
365000
900
yeares
give
-
328500
10
years
give
-
3650
1
yeare
gives
-
365
Phamenoth
compleat
-
2●0
Dayes
-
23
The
summe
-
2301145
6000
Julian
yeares
-
2191500
There
rests
-
109645
300
yeares
give
-
109575
There
rests
-
70
April
compleat
-
59
There
rests
-
11
It
therefore
fell
out
in
the
yeare
of
the
Iulian
period
6300
the
11
of
March
,
that
is
subtracting
from
that
period
,
4712
in
the
yeare
of
Christ
1588.
He
that
understands
this
may
by
like
method
convert
the
yeare
,
of
other
Epochaes
into
our
Julian
yeares
and
the
contrary
.
Next
to
the
tables
which
concern
the
reduction
of
years
in
general
,
we
annexed
tables
for
the
perpetual
finding
of
the
Sunday
letter
,
Golden
number
and
Epact
in
both
the
Old
Julian
,
and
New
Gregorian
accompt
,
with
the
fixed
and
moveable
Feasts
,
and
a
Catalogue
of
some
famous
places
with
their
latitude
and
distance
in
longitude
from
the
meridian
of
London
,
whose
use
is
so
obvious
that
it
needs
but
little
explanation
;
yet
to
take
away
all
difficulty
we
have
added
these
directions
.
The
Cycle
of
Sun
,
Sunday
Letter
,
Golden
Number
and
the
Epact
in
both
accounts
are
set
against
the
yeare
of
our
Lord
,
and
when
those
years
are
out
,
they
may
be
renewed
againe
as
oft
as
you
please
,
thus
for
the
yeare
1656
the
Cycle
of
the
Sun
1513
,
the
Sunday
letters
in
the
English
account
are
F
E
,
in
the
Gregorian
B
A
the
prime
or
Golden
number
in
both
is
4
,
the
Epact
in
the
English
accompt
is
14
in
the
Gregorian
4.
And
now
to
find
the
movable
Feast
,
seek
the
English
Epact
,
in
the
first
Columne
of
that
Table
towards
the
left
hand
,
and
the
first
F
that
follows
▪
will
shew
you
that
the
3d.
of
February
is
L
X
X
Sunday
,
the
17
of
February
▪
L
Sunday
,
&
the
20th
of
February
Ashwednesday
,
&
the
first
E
that
follows
will
shew
that
Easter
day
is
the
6th
.
of
April
,
Ascension
day
the
15th
.
day
of
May
,
Whitsunday
the
25
of
May
,
Corpus
Christi
the
fifth
of
June
,
Advent
Sunday
,
November
the
30th
.
But
in
the
Gregorian
,
the
Epact
and
Sunday
Letters
must
be
sought
in
the
first
Columne
towards
the
right
hand
,
so
shall
the
Sunday
Letters
B
A
shew
the
Feast
of
Easter
to
be
on
the
9th
of
their
April
,
and
the
rest
as
in
that
line
they
are
set
down
.
The
fixed
Feasts
,
together
with
the
Week-day
Letters
,
are
set
against
their
proper
dayes
in
every
moneth
of
the
Julian
year
,
knowing
therefore
the
Sunday
Letter
,
you
may
easily
know
upon
what
day
of
the
Week
any
Feast
or
day
of
the
moneth
shall
be
.
The
Catalogue
of
places
doth
serve
to
shew
the
height
of
the
Pole
in
those
places
,
and
the
Difference
of
the
Meridians
of
any
place
in
the
Catalogue
from
that
of
London
.
The
Letter
S
notes
that
the
distance
is
Westward
,
A
that
it
is
Eastward
,
the
figures
under
the
title
of
Time
are
Hours
and
Decimal
parts
of
an
houre
,
the
Earth
or
any
Starre
comes
sooner
or
later
to
the
Meridian
of
that
place
then
that
of
London
.
If
the
time
of
a
Lunar
Eclipse
then
or
other
appearance
be
given
at
London
,
afternoon
8
hours
,
23
parts
,
and
the
time
when
this
happens
at
Uraniburge
be
inquired
,
there
is
found
in
the
Catalogue
for
Uraniburge
0
hour
83
parts
A
,
if
therefore
according
to
the
letter
A
,
83
parts
be
added
to
the
time
given
it
makes
9
houres
06
parts
for
the
time
at
Uraniburge
.
But
if
the
time
of
another
place
be
to
be
reduced
to
the
time
at
London
,
the
difference
is
to
be
applied
with
the
contrary
title
.
And
that
these
Reductions
whether
in
time
or
motion
may
be
the
better
compared
with
those
bookes
that
are
written
in
the
old
Sexagenary
forme
,
we
have
added
tables
for
the
ready
converting
of
Sexagenary
parts
into
Decimall
and
the
contrary
,
the
first
of
these
tables
is
for
the
converting
of
the
Minutes
and
Seconds
,
&c.
of
a
Degree
in
motion
;
and
the
other
of
the
parts
of
a
day
in
time
,
an
example
in
each
will
be
a
sufficient
explanation
.
Let
it
be
required
to
find
the
Decimall
answering
to
37′
25″
16‴
5
'
'
'
'
29
'
'
'
'
'
in
motion
.
In
the
first
page
of
the
table
I
find
37′
12″
which
is
the
nearest
lesse
,
and
62
answering
thereunto
,
and
in
the
third
columne
of
the
second
page
in
the
top
of
the
page
I
find
12″
,
in
which
columne
I
find
25
seconds
,
and
in
the
sixt
and
last
columne
of
that
page
right
against
25″
,
I
find
this
number
36111111
,
which
being
annexed
to
62.
The
Decimall
of
37
minutes
25
seconds
is
-
6236111111
And
the
Decimall
of
16
thuds
-
0000740741
The
Decimall
of
5
fourths
-
0000003858
The
Decimall
of
29
fifths
-
0000000373
Their
summe
-
6236856083
is
the
Decimall
sought
2.
Example
Again
,
if
it
be
required
to
find
the
Decimall
of
8
hours
,
17
minutes
,
8
seconds
,
5
thirds
,
12
fourths
,
9
fifts
.
In
the
first
columne
of
the
table
entituled
,
A
Table
to
convert
the
hours
and
minutes
of
a
day
into
Decimalls
,
I
find
7
hours
12
minutes
,
and
in
the
second
columne
the
figure
3
,
then
looking
the
12
minutes
in
the
top
of
the
pages
,
I
cast
mine
eye
downward
in
that
column
till
I
come
to
8
hours
17
minutes
,
and
in
the
last
columne
of
the
page
against
8
houres
17
minutes
,
I
find
this
number
451388889
and
therefore
,
The
Decimall
of
8
hours
17
minutes
is
-
3451388889
The
Decimall
of
8
seconds
-
925926
The
Decimall
of
5
thirds
-
009645
The
Decimall
of
12
fourths
-
0387
The
Decimal
of
9
fifths
-
0005
Their
aggregate
-
3452324852
Is
the
decimall
sought
.
To
find
the
parts
of
a
degree
in
motion
,
or
of
a
day
in
time
answering
to
any
Decimal
given
,
is
but
the
contrary
worke
to
the
former
;
Example
.
As
if
it
were
required
to
find
the
parts
of
a
degree
answering
to
6236856083
,
the
2
first
figures
of
this
Decimall
are
62
which
being
sought
in
the
first
page
of
the
table
give
me
37′
.
12
and
62
being
subtract
▪
from
6236856083
,
the
remainder
will
be
36856083
which
being
sought
in
the
last
columne
,
my
nearest
number
is
36111111
,
and
right
against
that
number
under
12
in
the
top
of
the
page
I
find
25
,
therefore
37′
25″
are
the
parts
of
a
degree
answering
to
the
Decimall
given
,
but
if
you
would
find
the
thirds
,
fourths
and
fifths
,
from
36856083
Subtract
-
36111111
The
remainder
is
-
749972
Which
being
sought
amongst
the
Decimals
of
the
thirds
,
gives
me
16
thirds
,
and
this
number
to
be
subtracted
from
it
740741
;
and
the
remainder
004231
being
sought
amongst
the
Decimals
of
the
fourths
gives
me
5
fourths
,
and
this
number
to
be
subtracted
from
it
3858
,
and
the
remainder
373
sought
amongst
the
Decimals
of
the
fifths
gives
me
29
fifths
,
and
so
the
parts
of
a
degree
answering
to
the
Decimall
given
are
37
minutes
,
25
seconds
,
16
thirds
,
11
fourths
,
and
29
fifths
.
Thus
may
you
also
find
the
parts
of
a
day
in
time
answering
to
any
Decimall
given
.
The
next
thing
to
be
done
towards
the
finding
of
the
annuall
revolutions
of
the
planets
is
to
find
their
entrance
into
any
point
of
the
Zodiack
desired
,
and
that
may
be
done
thus
.
Having
the
place
of
the
planet
taken
by
observation
before
and
after
its
entrance
into
the
point
desired
,
subtract
the
observed
place
next
before
from
the
observed
place
next
after
,
and
the
remainder
shall
shew
you
the
apparent
motion
answerable
to
the
time
between
those
observation
,
subtract
also
the
former
place
,
from
the
place
in
the
point
desired
,
and
note
their
difference
:
for
as
the
former
remainder
,
that
is
the
apparent
motion
between
the
observations
,
is
to
the
time
between
those
observations
:
so
is
this
difference
,
to
the
time
between
the
first
observation
,
and
the
planets
entrance
into
the
point
desired
:
thus
we
are
to
deal
with
those
observations
that
we
our selves
shall
make
,
but
one
mans
age
not
being
distance
enough
between
the
observations
from
whence
the
middle
motions
may
be
rightly
stated
,
we
must
take
some
observations
upon
trust
;
and
find
the
middle
motions
by
comparing
the
observations
made
in
former
ages
with
those
of
our
owne
,
of
the
Sun
or
Earth
,
take
this
Example
following
.
The
vernal
Equinox
observed
by
Hypparchus
in
the
year
from
the
death
of
Alexander
178
,
was
Mechir
the
26
day
,
and
95833333
,
that
is
at
London
86746111.
And
the
vernal
Equinox
observed
at
Uraniburge
by
Tycho
1588
was
March
the
9th
.
86458333
,
that
is
at
London
82986111.
And
that
the
intervall
of
time
between
these
two
vernall
Equinoctialls
may
be
known
,
the
9
of
March
1588
must
be
reduced
to
the
correspondent
time
in
Egyptian
yeares
from
the
death
of
Alexander
,
which
according
to
the
former
directions
is
thus
.
The
Christian
Aera
began
in
the
4713
complete
yeare
of
Julian
period
,
to
which
1587
being
added
,
it
makes
6●00
from
the
beginning
of
the
Julian
period
therefore
to
the
11
of
March
1588
,
there
are
dayes
as
followeth
.
6000
Julia●
yeares
give
-
2191500
300
years
give
-
109575
February
-
59
Dayes
-
08
The
Summe
-
2301142
The
Aera
Alexandri
began
in
the
12
of
November
in
4390
yeare
of
the
Julian
period
in
which
there
are
dayes
,
4000
-
1461000
300
-
109575
80
-
29226
9
-
3287
October
-
304
Dayes
-
11
Which
being
subtracted
-
1603397
From
-
2301142
There
rests
-
697745
1000
Egyptian
yeares
give
-
365000
There
rests
-
332745
900
yeares
give
-
328500
There
rests
-
4246
10
years
give
-
3650
There
rests
-
596
1
yeare
gives
-
365
There
rests
-
231
Phamenoth
compleate
-
210
There
rests
-
21
Therefore
the
11
of
March
2588
in
our
English
account
,
falls
in
the
1912
yeare
of
the
Aera
Alexandri
the
●1
day
of
Pharmuthi
.
In
which
space
of
time
There
are
dayes
-
697746
And
from
the
death
of
Alexander
to
the
26
of
Mechir
178
,
there
are
-
64781
There
rests
-
632965
From
days
-
697746.
829●6111
Subtract
-
64781.
86746111
There
rests
-
632964.
96240000
And
in
this
time
the
Earth
or
Sun
hath
gone
1733
circles
,
〈◊〉
623880
degrees
.
Hence
to
find
the
mean
motion
for
a
year
or
365
days
I
say
▪
If
632964.
9624
d
▪
Give
623880
degrees
;
How
many
degrees
shall
365
dayes
give
?
And
the
answer
is
359
deg.
7611456036.
That
is
in
Sexagenary
numbers
359
deg.
45
minutes
,
41
seconds
,
1
third
,
27
fourths
.
Again
,
to
find
the
mean
motion
for
a
day
I
say
,
If
365
dayes
gives
359
degrees
,
7611456036
,
what
shall
one
day
give
?
And
the
answer
is
0.
9856469743.
That
is
in
Sexagenary
numbers
0
deg
,
59
minutes
,
8
seconds
,
19
thirds
,
44
fourths
.
And
what
is
here
done
for
the
middle
motion
of
the
Earth
or
Sun
,
may
be
done
for
the
other
planets
.
CHAP.
2.
Of
the
figure
which
the
planets
describe
in
their
Motion
.
HAving
shewed
in
the
former
Chapter
,
by
what
means
the
Annuall
or
Periodicall
revolutions
of
the
Planets
may
be
knowne
,
with
their
mean
or
equal
motion
,
for
any
part
of
those
revolutions
,
we
should
now
shew
you
,
how
by
those
equal
motions
to
find
their
true
or
apparent
places
.
But
we
can
never
hope
to
find
the
true
and
exact
Phenomenon
of
the
planets
,
unlesse
we
first
know
the
figure
in
which
they
move
;
And
this
must
be
collected
from
such
affections
,
as
are
by
the
constant
observations
of
all
ages
found
to
be
proper
and
naturall
to
them
,
or
may
be
rationally
collected
from
them
.
-
1
That
the
planets
have
one
onely
motion
,
in
one
onely
line
,
and
that
those
motions
are
equal
,
constant
and
perpetual
,
hath
been
confirmed
by
the
observation
of
all
ages
,
-
2
And
therefore
they
must
needs
be
regular
,
their
motions
must
be
in
a
circle
or
some
other
line
returning
into
it selfe
,
or
else
their
motions
could
not
be
perpetuall
.
-
3
Their
equall
motions
must
have
some
place
assigned
(
which
the
planets
naturally
behold
)
to
be
the
beginning
of
this
equall
motion
.
-
4
And
because
the
apparent
place
of
a
planet
taken
by
observation
,
is
generally
different
from
the
place
reckoned
in
its
middle
motion
,
the
inequality
of
this
middle
and
apparent
motion
must
be
referred
to
the
center
of
the
Zodiacke
,
●s
to
the
point
of
that
inequ●lity
.
-
5
And
because
the
center
of
the
Zodiacke
and
of
the
world
is
to
out
appearance
the
same
,
the
point
of
this
inequality
must
be
referred
to
the
center
of
the
world
.
-
6
And
because
of
this
difference
between
the
middle
and
apparent
motion
,
the
center
of
the
world
cannot
be
the
true
and
exact
center
of
the
planets
,
but
the
center
of
that
figure
which
the
planets
describe
in
their
motion
,
must
be
some
other
point
then
the
center
of
the
Zodiacke
.
-
7
And
though
the
planets
to
our
appearance
are
observed
to
be
sometimes
swifter
in
motion
,
then
at
other
some
,
yet
the
cause
of
this
inequality
of
motion
must
not
be
such
as
shall
alter
the
natural
and
equal
motion
of
the
planet
,
it
must
be
such
as
shall
make
the
planet
to
be
slower
in
its
furthest
distance
from
the
center
of
the
world
,
and
swifter
at
his
nearest
,
without
transposing
the
equal
motion
into
any
other
then
the
first
place
assigned
,
whether
superficies
or
circle
.
-
8
And
further
the
apparent
motions
of
the
planets
in
their
nearest
and
furthest
distances
from
the
center
of
the
world
being
the
same
with
their
middle
,
the
way
of
the
planets
must
be
such
,
that
when
they
have
gone
90
degrees
from
their
farthest
distance
in
their
middle
motions
,
their
apparent
motions
must
be
lesse
then
90
by
the
quantity
of
that
whole
inequality
between
the
middle
and
apparent
motion
;
And
when
the
planets
have
gone
a
quadrant
in
their
apparent
motions
,
their
difference
between
their
motions
shall
be
that
whole
inequality
also
,
and
therefore
the
center
of
that
figure
which
the
planets
describe
in
their
motions
must
be
in
the
middle
between
the
points
of
their
equal
and
apparent
motions
.
-
9
And
because
the
mean
motion
from
the
point
of
a
planets
farthest
distance
from
the
center
of
the
world
,
to
the
first
quadrant
is
greater
then
the
apparent
,
therefore
the
apparent
motion
must
be
greater
then
the
mean
,
from
the
first
quadrant
to
the
point
of
the
planets
nearest
distance
,
and
consequently
a
greater
portion
of
the
line
in
which
the
planets
move
,
must
be
allowed
to
the
apparent
from
the
first
quadrant
to
the
point
of
nearest
distance
,
then
from
the
point
of
farthest
distance
to
the
first
quadrant
.
-
10
And
because
the
equal
motion
must
not
change
and
that
the
apparent
motion
doth
increase
from
the
point
of
the
planets
farthest
distance
from
the
center
of
the
world
,
the
angles
of
the
middle
motion
must
be
reckoned
,
in
the
arches
of
many
parallel
circles
,
which
shall
also
increase
from
the
points
of
farthest
to
the
point
of
their
nearest
distance
to
the
center
of
the
world
,
and
the
line
of
the
apparent
motion
,
must
containe
those
circles
in
one
and
the
same
superficies
,
and
therefore
that
line
must
be
excentricall
from
those
circles
of
apparent
motion
,
and
so
placed
that
all
the
parts
of
apparent
motion
may
proportionably
answer
to
all
the
parts
of
equal
,
yet
so
as
that
the
least
circles
of
equal
motion
,
shall
agree
with
the
point
of
the
planets
farthest
distance
,
and
the
greatest
circles
with
the
point
of
the
planets
nearest
distance
from
the
center
of
the
world
.
Seeing
now
that
these
circles
of
middle
motion
must
be
parallel
,
succeeding
one
another
in
a
continued
series
,
and
are
not
one
within
another
,
and
that
the
apparent
motion
must
in
the
farthest
distance
answer
to
the
least
circles
,
&
in
the
nearest
distance
from
the
center
of
the
world
to
the
greatest
,
there
is
none
but
a
solid
Superficies
that
can
be
capable
of
those
greater
and
lesser
circles
:
And
that
an
unequal
sided
Cone
may
be
so
cut
,
as
that
the
figure
upon
the
plaine
of
that
Section
shall
truly
represent
these
affections
of
the
planets
,
the
learned
Bullialldus
doth
Demonstrate
,
and
for
a
preparation
thereunto
he
sheweth
first
,
How
two
equal
right
lines
may
be
so
drawn
in
an
unequal
sided
Triangle
,
as
that
the
one
shall
bisect
the
other
.
An
unequal
sided
Cone
being
cut
through
the
Axis
by
a
plaine
perpendicular
to
the
plain
of
the
base
,
shall
make
an
unequal
sided
Triangle
,
and
let
A
B
C
be
such
a
Triangle
,
whose
base
B
C
let
be
bisected
in
I
,
and
parallel
therunto
draw
the
line
P
S
,
which
being
within
the
triangle
shall
be
also
bisected
in
the
point
R
,
and
from
a
point
taken
in
this
line
at
pleasure
,
suppose
at
H
,
to
the
Axis
of
the
Cone
A
I
,
draw
the
line
H
M
so
,
as
that
the
angles
H
M
R
and
M
R
H
may
be
equal
;
then
shall
H
M
and
H
R
be
equal
also
,
and
let
the
line
H
M
being
continued
to
the
sides
of
the
Triangle
A
B
and
A
C
be
bisected
in
the
point
X
,
and
by
the
point
of
bisection
at
X
,
draw
the
line
V
X
T
parallel
to
the
base
B
C
,
then
are
the
right
lines
E
K
and
V
T
equal
,
and
E
K
is
bisected
by
V
T
in
the
point
X.
And
now
if
this
unequal
sided
Cone
be
cut
through
the
bisected
line
E
K
,
the
figure
made
on
that
plane
by
such
section
shall
be
an
Ellipsis
one
of
whose
umbilique
points
shall
fall
in
the
Axis
of
the
Cone
.
For
V
T
being
equal
to
E
K
,
and
E
K
being
bisected
in
X
,
the
right
line
X
O
in
the
circle
V
O
T
shall
be
the
conjugate
diameter
in
the
Ellipsis
E
K
,
because
it
is
perpendicular
to
the
line
E
K
in
the
point
X
,
and
reacheth
to
the
extremitie
of
the
plain
V
O
T
,
and
it
is
also
a
mean
proportionall
between
V
X
and
X
T
,
because
it
cuts
the
line
V
T
at
right
angles
in
the
point
X.
And
therefore
the
square
of
X
O
is
equal
to
the
rectangle
V
X
T
more
by
the
square
of
Z
X
but
the
rectangle
V
X
T
more
by
the
square
of
Z
X
shall
be
equal
to
the
square
of
Z
V
,
and
the
square
of
Z
V
shall
be
equal
to
the
square
of
E
X
,
because
Z
V
and
E
X
are
equal
,
and
therefore
the
squares
of
X
O
and
Z
X
,
shall
be
equal
to
the
square
of
E
X
,
but
E
X
is
the
greatest
semidiameter
,
X
O
the
lesser
,
and
the
square
of
Z
X
or
M
X
the
difference
between
the
squares
of
E
X
and
X
O
,
and
therefore
M
X
shall
be
the
distance
of
the
umbilique
point
from
the
center
X
,
and
M
the
umbilique
which
by
construction
is
placed
in
the
Axis
of
the
Cone
A
B
C
,
as
was
to
be
proved
.
And
because
the
center
of
the
Ellipsis
at
X
is
distant
from
the
Axis
A
I
,
by
the
quantity
of
X
Z
equal
to
M
X
,
the
plane
which
bisects
the
Ellipsis
by
the
conjugate
Diameter
O
X
N
shall
not
bisect
the
circles
,
but
shall
cut
off
V
X
a
greater
part
towards
E
,
and
X
T
a
lesser
part
towards
K.
Let
the
equal
motion
of
the
planet
therefore
be
about
the
Axis
of
the
Cone
A
I
,
and
through
all
the
circles
which
are
intercepted
between
E
D
and
P
K
,
and
let
the
centers
of
those
circles
be
in
the
Axis
of
the
Cone
,
and
upon
those
centers
let
the
planets
be
conceived
to
make
equal
angles
in
equal
portions
of
time
,
but
the
terme
of
apparent
motion
to
which
it
is
referred
,
let
be
the
other
umbilique
at
H
,
the
place
of
the
Sun
(
if
we
suppose
him
according
to
our
new
Astronomy
to
be
the
center
of
the
World
.
)
.
Then
is
the
Aphelion
or
part
remotest
from
the
Sun
at
E
,
the
Perihelion
or
nearest
part
at
K.
And
now
while
the
planets
describing
the
Ellipsis
shall
be
equally
moved
about
the
Axis
of
the
Cone
,
in
their
equal
or
middle
motions
,
they
shall
have
gone
a
quadrant
about
the
Cone
,
when
yet
they
shall
want
of
a
quadrant
in
the
Ellipsis
by
the
quantity
of
the
right
line
M
X
,
and
shall
be
in
the
point
Y
,
which
point
is
in
the
circle
F
M
G
,
and
in
the
Ellipsis
,
and
the
right
line
M
Y
which
is
drawn
from
the
center
of
the
circle
F
G
,
is
set
at
right
angles
,
upon
the
Diameters
both
of
its
owne
circle
F
G
and
of
the
Ellipsis
E
K
,
and
consequently
the
angle
F
M
Y
is
a
right
angle
,
and
therefore
the
planets
shall
move
90
deg.
of
middle
motion
about
the
axis
of
the
Cone
when
they
come
to
Y
,
but
in
the
Ellipsis
they
shall
not
move
so
much
by
the
arch
Y
O
,
or
the
right
line
M
X.
That
this
is
the
true
and
naturall
Hypothesis
may
thus
appeare
,
because
-
1
The
planets
thus
describe
one
onely
line
about
the
axis
of
the
Cone
,
in
their
equal
constant
and
perpetual
motions
.
-
2
Their
motions
thus
are
regular
,
though
not
in
a
perfect
circle
,
yet
in
a
line
returning
into
it selfe
.
-
3
Their
equal
motions
have
their
beginning
alwayes
in
one
place
,
●●at
is
,
in
the
Axis
of
the
Cone
.
-
4
,
5
The
apparent
motion
of
the
planets
is
referred
to
the
Sun
at
H
,
as
to
the
center
of
the
Zodiac
.
6
And
the
whole
inequality
between
the
middle
and
apprent
motion
,
is
between
the
umbilique
points
M
and
H.
-
7
The
motion
of
the
Planets
are
thus
made
to
be
flower
in
the
Aphelion
,
then
in
the
Perihelion
,
and
yet
the
equal
motion
is
not
reckoned
any
where
but
in
the
first
place
assigned
the
axis
of
the
Cone
.
-
8
The
Planets
in
their
middle
motions
will
thus
goe
90
degrees
about
the
axis
,
being
come
to
Y
,
when
yet
they
want
of
a
quadrant
in
the
Ellipsis
,
the
arch
Y
O
,
or
M
X
,
and
so
the
center
of
the
Ellipsis
is
in
X
,
the
middle
of
the
whole
Inequality
.
-
9
There
equal
motions
from
the
Aphelion
at
E
to
the
first
quadrant
are
greater
then
the
apparent
,
but
from
the
first
quadrant
to
the
Perihelion
at
K
,
the
apparent
motions
are
greater
then
the
mean
,
and
therefore
a
smaller
portion
of
the
line
which
the
Planets
describe
is
allowed
to
the
apparent
motion
from
the
Aphelion
to
the
first
quadrant
,
to
wit
,
F
Y
,
and
a
greater
part
from
the
first
quadrant
to
the
Perihelion
,
to
wit
,
Y
O
R.
-
10
And
lastly
,
because
the
circles
of
middle
motion
,
F
G
,
V
T
,
&c.
do
increase
from
the
Aphelion
to
the
Perihelion
,
&
that
the
Planets
notwithstanding
make
equal
angles
in
equal
portions
of
time
about
the
axis
of
the
cone
,
their
motions
in
the
Ellipsis
do
increase
also
from
the
Aphelion
to
the
Periheliō
,
because
these
greater
angles
are
subtēded
by
greater
lines
in
greater
circles
,
by
lesser
lines
in
lesser
circles
,
&
because
the
lesser
circles
are
placed
towards
the
top
of
the
Cone
at
A
,
&
Aphelion
at
E
,
the
greater
towards
the
base
and
Perihelion
at
K
,
the
motion
in
the
Ellipsis
is
slower
about
the
Aphelion
and
swister
towards
the
Perihelion
;
And
thus
the
middle
motions
are
not
reckoned
in
one
onely
circle
,
but
in
many
parallel
circles
comprehended
between
E
D
and
P
K
,
these
circles
are
contained
in
one
plain
Superficies
,
and
by
these
circles
the
planets
describing
an
Ellipsis
doe
continually
passe
,
and
yet
they
are
all
of
them
excentricall
in
respect
of
the
figure
which
the
planets
describe
,
as
was
before
required
.
Thus
then
there
is
an
admirable
Harmony
between
the
motions
of
the
planets
in
this
figure
,
and
their
motions
in
the
heavens
found
by
observation
;
probably
therefore
we
may
conclude
that
the
figure
which
the
planets
describe
in
their
motions
is
an
Ellipsis
.
CHAP.
3.
Of
the
Lines
and
Method
to
be
used
for
the
finding
of
a
Planets
true
longitude
from
the
Aphelion
in
this
figure
.
HAving
resolved
upon
the
figure
which
the
Planets
describe
in
their
motions
,
we
come
now
to
shew
you
what
lines
must
be
drawn
,
and
method
used
for
the
finding
a
planets
true
longitude
from
the
Aphelion
in
this
figure
;
and
in
order
thereunto
,
we
will
shew
you
first
the
order
of
the
spheares
in
which
the
planets
move
,
and
how
mechanically
to
draw
this
Ellipticall
figure
of
their
motions
upon
a
plane
.
As
to
the
Spheares
,
1
We
suppose
that
the
Sun
is
placed
in
the
middle
of
the
world
in
or
about
the
center
of
the
Spheare
of
the
fixed
Starres
,
and
hath
no
circular
motion
but
centrall
onely
.
2
That
the
Earth
is
one
of
the
planets
,
and
with
her
annual
motion
about
the
Sun
describeth
her
Orbe
between
the
Orbs
of
Mars
and
Venus
.
3
That
the
Moon
is
moved
about
the
Earth
,
as
her
center
,
and
so
in
her
annuall
motion
hath
respect
both
to
the
Earth
,
and
to
the
center
of
the
Earths
orbe
the
Sun
.
4
That
the
Orbe
of
Venus
is
next
under
the
Orbe
of
the
earth
,
and
the
Orbe
of
Mercury
between
the
Sun
&
the
Orbe
of
Venus
.
Next
above
the
Orbe
of
the
earth
we
suppose
the
orbe
of
Mars
,
the
Orbe
of
Jupiter
next
above
Mars
,
and
the
Orbe
of
Saturn
next
to
the
Orbe
of
the
fixed
Stars
.
According
to
these
supposed
principles
,
we
would
have
immediately
shewed
the
method
of
calculation
,
but
that
the
Mechanicall
way
of
drawing
an
Ellpsis
,
doth
if
not
demostrate
,
yet
at
least
illustrate
that
method
.
An
Ellipsis
by
the
helpe
of
a
thread
may
be
mechanically
made
thus
,
first
draw
a
right
line
to
that
length
which
you
would
have
the
greatest
Diameter
to
be
,
which
let
be
A
P
,
and
from
the
middle
of
this
line
at
X
,
set
off
with
your
Compasses
the
equal
distances
X
M
and
X
H.
Then
take
a
piece
of
thrid
of
the
same
length
with
the
diameter
AP
,
&
fasten
one
end
of
the
thrid
in
the
point
M
,
and
the
other
at
H
,
&
with
your
pen
extending
the
thread
thus
fastened
to
A
,
&
from
thence
towards
P
,
keeping
the
thrid
stiffe
upon
your
pen
,
draw
a
line
from
P
by
B
to
A
,
the
line
so
drawne
shall
be
an
Ellipsis
,
in
which
because
the
whole
thread
is
equal
to
the
Diameter
A
P
therefore
the
two
lines
made
by
the
thread
in
drawing
of
the
Ellipsis
must
in
every
point
of
the
Ellipsis
be
also
equal
to
the
fame
diameter
A
P
,
they
that
desire
a
demonstration
thereof
Geometrically
may
consult
with
Apollonius
Pergaeus
,
Claudius
Mydorgius
,
o●
others
,
in
their
treatises
of
Conicall
sections
;
for
our
present
purpose
this
is
sufficient
,
and
from
the
equality
of
those
two
lines
,
with
the
Diameter
,
a
brief
Method
of
Calculation
,
is
thus
demonstrated
by
Dr.
Warde
.
Let
the
line
M
E
be
equal
to
A
P
,
and
draw
the
lines
H
B
and
H
E
,
then
in
the
plaine
triangle
M
H
E
,
having
the
sides
M
E
equal
to
the
Diameter
,
and
M
H
the
distance
of
the
umbilique
points
,
with
the
angle
H
M
E
,
the
angles
M
E
H
and
M
H
E
shall
be
given
also
,
but
the
angles
B
E
H
and
B
H
E
are
equal
,
because
the
sides
B
H
and
B
E
are
equal
by
construction
,
and
therefore
if
you
subtract
the
angle
B
E
H
from
the
angle
M
H
E
,
there
will
remaine
the
angle
at
the
Sun
M
H
B
,
which
is
a
planets
true
longitude
from
the
Aphelion
or
the
equated
Anomaly
.
And
of
these
three
things
propounded
to
be
given
,
the
side
M
E
is
by
construction
made
equal
to
the
Diameter
A
P
,
how
the
angle
H
M
E
and
the
side
M
H
must
be
had
shall
plainely
appeare
by
that
which
followes
.
CHAP.
4.
Of
the
proportion
by
which
the
motion
of
the
Planets
doe
increase
from
the
Aphelion
to
the
Perihelion
.
THough
the
equal
motions
of
the
planets
are
to
be
reckoned
(
as
hath
been
said
)
in
diverse
parallel
circles
about
the
Axis
of
the
Cone
,
whose
diameters
must
still
increase
from
the
top
of
the
Cone
at
A
,
to
the
base
B
C
,
that
the
motions
of
the
planets
in
the
Ellipsis
may
encrease
also
;
yet
in
the
calculation
we
cannot
conveniently
reckon
the
middle
motion
in
any
more
circles
then
one
,
and
therefore
it
must
be
proved
,
that
the
angle
comprehended
between
the
lines
drawn
from
E
to
that
point
in
the
diameter
of
the
Ellipsis
which
is
made
the
common
Center
of
the
Circles
of
middle
motion
,
and
fr●●
that
Center
to
the
planet
in
the
Ellipsis
,
is
alwayes
equal
to
the
a●gle
of
middle
motion
,
comprehended
between
the
s●midiamet●●
of
the
planets
proper
circle
of
middle
motion
and
the
line
d●●wn
from
the
center
of
that
circle
to
the
planet
in
the
Ellipsis
,
this
Bull●aldus
takes
for
granted
,
and
Dr.
Ward
doth
thus
demonstrate
.
F
M
G
is
a
circle
of
middle
motion
whose
center
is
in
the
Axis
of
the
Cone
,
and
in
the
umbilique
point
of
the
Ellipsis
M
,
but
a
planet
being
in
the
Ellipsis
at
R
the
proper
circle
of
its
middle
motion
is
L
β
Q
and
the
angle
comprehended
by
the
Radius
of
that
circle
Lβ
,
and
the
line
drawn
from
β
the
center
thereof
to
R
the
place
of
the
planet
is
equal
to
the
angle
comprehended
between
the
line
E
M
and
the
line
drawn
from
M
to
R
in
the
Ellipsis
,
for
thus
there
are
made
two
right
angled
triangles
M
H
R
and
β
H
R
,
in
which
the
side
H
R
is
common
to
both
,
for
it
is
set
at
right
angles
in
the
Ellipsis
and
in
the
circle
equant
,
and
the
sides
M
H
and
β
H
are
equal
by
what
hath
beene
already
said
in
the
2
Chapter
,
therefore
the
angle
R
M
H
is
equal
to
the
angle
R
β
H
,
and
by
consequence
R
β
L
and
R
M
E
are
equal
also
,
but
the
angle
R
β
L
is
the
angle
of
the
middle
motion
of
a
planet
from
the
Aphelion
,
or
the
angle
of
the
simple
excentricke
anomaly
,
and
therefore
the
angle
R
M
E
is
the
mean
anomaly
also
,
whose
complement
to
a
semicircle
is
the
angle
M
E
H
in
the
Diagram
of
the
preceding
Chapter
.
When
a
Planet
therefore
descends
from
E
to
K
,
the
angles
at
the
axis
of
the
Cone
,
or
at
M
the
umbilique
point
of
the
Ellipsis
do
alwayes
increase
,
and
therefore
the
meane
anomalie
is
increased
,
for
the
angles
of
its
circle
Equant
do
answer
to
more
degrees
at
the
axis
,
and
those
angles
also
,
are
alwayes
degrees
or
parts
of
greater
circles
,
and
therefore
the
planets
increase
in
the
swiftnesse
of
their
motion
in
such
proportion
as
the
circles
,
Diameters
,
or
Radii
of
those
circles
equant
do
increase
.
For
Demonstration
whereof
let
the
line
E
ψ
be
made
parallel
to
the
axis
of
the
Cone
,
then
shall
the
parts
of
the
semidiameters
of
the
circles
equant
comprehended
between
those
parallels
E
ψ
and
A
I
be
equal
to
the
semidiameter
of
the
least
circle
E
O
,
and
the
parts
intercepted
between
that
parallel
E
ψ
and
the
side
of
the
triangle
A
B
,
shall
be
the
proportionall
excesse
above
the
least
motion
in
E
O.
And
the
mean
acceleration
is
in
that
circle
equant
whose
plain
parallel
to
the
base
of
the
cone
doth
pass
by
the
center
of
the
Ellipsis
,
that
is
in
the
circle
V
D
T
,
for
seeing
that
M
X
&
Z
X
are
equal
,
as
also
V
D
&
Z
X
,
V
D
must
needs
be
the
difference
between
the
semidiameter
of
the
least
circle
E
L
&
the
middle
circle
V
T
,
&
the
excess
of
that
semidiameter
,
above
the
semidiameter
of
the
least
circle
,
must
be
equal
to
M
X
or
the
excentricity
,
but
the
distance
of
this
circle
from
the
Aphelion
is
90
degrees
,
and
may
be
called
the
Diacentrick
circle
,
and
is
the
Radius
of
the
circle
of
the
equated
anomalie
E
X
And
the
difference
between
the
semidiamater
of
V
T
,
and
the
parallel
circle
ω
K
is
also
equal
to
M
X
,
for
as
E
V
to
E
ω
so
is
E
X
to
E
K
,
and
again
,
as
E
V
to
E
ω
so
is
V
D
to
ω
ψ
,
and
because
E
X
is
the
halfe
of
E
K
,
therefore
V
D
shall
be
equal
to
ω
Y
or
to
the
halfe
of
ω
ψ
,
and
therefore
ω
Y
is
also
equal
to
M
X
,
when
a
planet
therefore
is
in
the
circle
V
T
,
it
is
swift
in
motion
,
and
is
in
the
middle
between
the
swiftest
and
the
slowest
motion
,
and
because
that
middle
acceleration
of
the
planets
,
is
the
acceleration
in
a
quadrant
,
therefore
as
E
X
to
V
D
,
so
is
E
M
to
G
π.
That
is
,
as
Radius
to
the
excentricity
,
so
is
the
versed
sine
of
the
distance
from
the
Aphelion
,
to
the
part
proportionall
of
the
planets
acceleration
;
and
therefore
also
(
faith
Bullialdus
)
it
is
as
the
whole
sine
,
to
the
whole
difference
,
so
is
the
sine
of
the
distance
of
the
middle
motion
from
the
Aphelion
,
to
the
part
of
the
difference
answering
to
that
distance
.
From
hence
and
the
two
following
problemes
of
Vieta
,
he
propounds
a
method
for
the
finding
of
the
Aphelion
and
distance
of
the
umbilique
points
.
Probleme
1.
Three
points
in
the
circumference
of
a
circle
being
given
,
to
find
a
diameter
▪
upon
which
there
being
perpendiculars
let
fall
from
the
points
given
;
the
segments
of
the
diameter
intercepted
by
these
perpendiculars
shall
be
proportional
to
another
proportion
given
.
Let
the
line
C
B
be
cut
in
E
,
so
as
C
B
may
be
to
E
B
,
as
S
to
R
,
and
let
D
E
,
cut
the
Diameter
F
G
at
right
angles
in
H.
Then
is
F
G
the
Diameter
sought
,
upon
which
seeing
the
lines
B
I
,
C
K
,
D
H
doe
fall
perpendicularly
,
K
I
shall
be
to
H
I
as
S
to
R.
For
the
right
lines
C
B
are
parallel
,
or
not
parallel
,
if
they
be
parallel
,
C
B
shall
be
equal
to
K
I
,
and
E
B
to
H
I
,
and
then
by
construction
K
I
shall
be
H
I
,
that
is
C
B
to
E
B
,
in
the
given
proportion
as
S
to
R.
But
if
they
can
meet
,
let
the
point
of
their
meeting
be
at
L.
Then
it
shall
be
as
L
C
to
L
K
,
so
L
E
to
L
H
,
and
so
is
L
B
to
L
I.
And
then
dividing
and
changing
the
termes
,
it
shall
be
as
K
I
to
H
I
,
so
is
C
B
to
E
B
,
or
so
is
S
to
R
as
was
required
.
And
thus
likewise
K
H
and
H
I
,
with
the
arches
B
C
,
C
D
,
and
B
D
,
being
given
,
we
may
find
the
arch
B
G
,
and
the
Diameter
F
G
in
the
same
parts
with
K
H
and
H
I
,
for
the
arches
C
D
and
D
B
being
given
,
the
subtenses
of
those
arches
and
angles
opposite
to
them
in
the
triangle
B
C
D
shall
be
given
also
in
the
parts
of
the
Diameter
F
G
,
and
therefore
the
sides
E
B
and
D
B
with
the
angle
E
B
D
being
given
,
the
angle
E
D
B
or
D
B
M
shall
be
given
also
,
which
being
deducted
from
the
arch
B
M
D
shall
leave
B
M
or
the
double
of
B
G.
Probleme
2.
Two
points
in
the
circumference
of
a
given
circle
being
given
,
to
find
the
Diameter
,
upon
which
perpendiculars
being
let
fall
from
the
points
given
,
the
segment
of
the
Diameter
intercepted
by
those
perpendiculars
shall
be
equal
to
a
line
given
.
In
the
circle
B
C
whose
center
is
A
,
let
the
given
points
be
B
C
,
let
the
given
line
be
Z.
Subttend
the
periphery
B
C
,
and
let
the
right
line
B
C
be
made
the
Diameter
of
a
circle
,
in
which
draw
the
line
C
D
equal
to
Z.
But
by
the
center
let
there
be
drawn
E
F
the
Diameter
of
the
circle
B
L
C
parallel
to
D
C
,
and
then
the
perpendicular
B
D
being
let
fall
upon
M
C
,
the
angle
B
D
C
shall
be
a
right
angle
.
And
because
E
F
and
D
C
are
parallel
,
B
D
shall
cut
E
F
at
right
angles
in
the
point
K
,
and
B
K
shall
be
parallel
to
C
I
,
and
therefore
D
C
and
K
I
shall
be
equal
,
therefore
in
the
circle
B
C
,
the
points
B
and
C
being
given
,
there
is
found
the
Diameter
E
F
,
upon
which
the
perpendiculars
B
K
and
C
I
being
let
fall
,
the
segment
K
I
intercepted
by
those
perpendiculars
is
equal
to
C
D
that
is
to
Z
the
line
given
.
Thus
likewise
the
summe
of
the
arches
,
and
the
summe
of
the
Sines
of
these
arches
being
given
,
we
may
distinguish
the
arches
and
the
Sines
if
the
center
A
be
between
K
and
I.
Or
the
difference
of
the
arches
and
the
difference
of
their
sines
being
given
,
we
may
distinguish
the
arches
and
the
sines
,
if
the
point
K
be
betweene
the
center
A
and
the
point
I.
When
three
places
therefore
in
the
apparent
motion
of
the
Sun
orany
other
planet
,
with
the
intervall
of
time
are
given
,
the
middle
motion
of
the
planet
shall
be
given
also
,
with
the
difference
between
the
middle
and
apparent
in
those
intervalls
,
and
the
rest
from
these
.
For
if
the
points
B
C
D
in
the
Diagrams
of
the
first
probleme
be
given
,
to
wit
,
three
places
in
the
apparent
motion
,
&
the
middle
motion
in
the
intervalls
of
time
B
C
,
C
D
,
the
difference
between
the
middle
and
apparent
motion
shall
be
given
also
;
let
K
I
the
part
of
the
diameter
F
G
,
be
the
difference
of
the
middle
from
the
apparent
in
the
intervall
B
C
,
and
let
H
I
be
the
difference
in
the
intervall
B
D.
Then
in
the
triangle
B
C
D
,
the
arches
B
C
and
C
D
are
known
,
and
therefore
the
angles
at
D
and
B
are
known
also
,
and
by
consequence
the
third
angle
at
C
,
and
from
thence
the
subtense
C
B
,
now
that
we
may
find
the
angle
E
D
B
,
the
side
C
B
must
be
cut
in
the
same
proportion
as
the
line
K
I
is
cut
in
H.
That
so
K
I
may
be
to
H
I
as
C
B
to
E
B.
And
so
we
shall
have
E
B
in
the
like
parts
with
B
D
;
and
the
sides
E
B
,
B
D
,
with
the
angle
at
B
,
in
the
triangle
B
E
D
being
given
,
the
rest
shall
be
given
also
,
and
the
angle
at
D
being
given
,
we
shall
have
the
arches
M
B
and
M
D.
And
lastly
,
F
G
shall
be
given
in
the
like
parts
in
which
K
H
,
and
H
I
are
given
.
For
as
the
summe
of
the
Sines
complement
B
I
and
C
K
,
to
K
I
,
so
is
the
Diameter
F
G
,
to
the
same
diameter
in
the
parts
of
K
E
and
so
the
first
inequality
shall
be
given
,
and
the
diameter
from
whence
it
begins
.
CHAP.
5.
Of
the
inequality
of
the
Earths
annual
motion
,
and
of
the
Diameter
in
which
the
Aphelion
and
Perihelion
are
placed
.
THe
inequality
of
the
Earths
annual
motion
,
if
we
suppose
the
motion
to
be
exactly
circular
,
may
from
the
Observations
of
Copernicus
made
at
Fruenburg
in
Prussia
,
1525
,
or
rather
as
they
were
corrected
by
Tycho
,
1584
be
they
found
;
Between
the
Autumne
and
the
Vernal
Equinox
,
according
to
Tycho
there
were
dayes
178.
43333.
33333.
And
between
the
Autumne
Equinox
,
and
the
middle
of
Mars
there
were
dayes
45.
15416.
66667.
Upon
the
Center
F
describe
the
Suns
Orbe
A
B
C
,
let
A
be
the
Vernal
Equinox
the
the
Autumne
,
D
the
middle
of
Mars
,
and
let
D
E
intersect
A
C
in
the
point
G
,
and
draw
the
line
A
D
:
Then
is
the
ark
of
middle
motion
C
D
44
degrees
,
5061111111
,
and
therefore
the
angle
C
A
D
,
22.
25305.
55556.
and
the
angle
of
apparent
motion
C
G
D
45
degrees
,
which
being
deducted
from
180
degrees
,
give
the
obtuse
angle
A
G
D
,
135
degrees
.
And
therefore
the
angle
C
D
G
-
22.
74694.
44444.
And
the
arch
A
E
-
45.
49388.
88889.
Again
,
the
ark
C
D
A
,
175
deg.
87500.
00000.
from
which
deducting
C
D
44.
50611.
11111.
the
remainder
is
D
A
,
1●1
.
36888.
88889.
The
subtense
A
D
18225828.
to
which
A
E
being
added
,
the
whole
is
D
E
,
176.
86277.
77778.
and
the
subtense
of
D
E
,
19992506.
And
because
that
neither
of
these
Segments
do
make
a
Semicircle
,
we
must
find
the
Aphelion
in
the
other
part
of
the
circle
E
B
C.
Let
the
line
E
B
represent
the
Aphelion
and
Perihelion
,
G
F
the
excentricity
,
F
H
then
being
drawne
perpendicularly
shall
bisect
D
E
and
make
right
angles
in
the
point
H.
And
in
the
triangle
A
G
D
,
the
angles
being
given
with
the
side
A
D
,
there
is
also
given
G
D
9763585
,
whch
being
deducted
from
the
halfe
of
D
E
there
shall
remaine
G
H
232668
,
and
the
perpendicular
F
H
may
be
thus
found
,
D
A
E
wants
deg.
3.
13722.
22222
of
a
semicircle
,
the
halfe
whereof
is
1.
56861.
11111
,
whose
sine
273740
is
the
side
F
H
,
now
then
H
G
and
F
H
being
given
,
we
may
find
the
angle
F
G
H
49.
63666.
66667
,
which
with
the
arch
A
E
45.
49388.
88889
makes
95.
13055.
55556
,
therefore
the
Aphelion
is
♋
5.
13055.
55556
,
or
in
sexagenary
numbers
,
in
5
degrees
of
Cancer
7
min.
50
seconds
.
And
the
excentricity
F
G
359261.
But
the
method
for
the
finding
of
the
Aphelion
and
excentricity
which
we
propounded
in
the
last
chapter
,
is
more
sutable
to
the
Elliptical
motion
of
the
planets
,
and
according
to
that
method
the
earths
Aphelion
and
Excentricity
,
or
semidifference
of
the
umbilique
points
,
from
the
accurate
observations
of
Tycho
in
the
year
1588
may
be
thus
found
.
From
the
middle
of
Taurus
to
the
middle
of
Leo
there
were
dayes
in
the
apparent
motion
,
94.
24662.
03703.
From
the
middle
of
Leo
to
the
Autumne
Aequinox
there
were
dayes
46.
40277.
77778.
From
the
middle
of
Taurus
to
the
Autumne
Aequinox
there
were
therefore
dayes
140.
64939.
81481.
In
the
annexed
Diagram
let
C
represent
Taurus
15
deg.
D
15
deg.
of
Leo.
Then
shall
the
arch
of
apparent
motion
C
D
be
90
deg.
and
D
B
45
,
but
in
the
middle
motion
the
arch
C
D
shall
be
92.
89277
,
and
D
B
45.
73777
,
and
CB
138.
63055.
First
,
we
will
worke
with
the
arches
of
apparent
motion
.
Arch.
|
Deg.
|
Angles
|
Deg.
|
Sides
|
C
D
|
90.
00000
|
C
B
D
|
45.
00
|
C
D
|
70711
|
D
B
|
45.
|
D
C
B
|
22.
50
|
D
B
|
38268
|
C
B
|
135.
|
C
D
B
|
67.
50
|
C
B
|
92388
|
The
halfe
differences
of
the
middle
from
the
true
.
1
Deg.
1.
44638
|
K
H
|
2524
|
|
2
Deg.
0.
36888
|
H
O
|
644
|
|
Deg.
1.
81526
|
K
O
|
3168
|
|
As
K
O
3168
co.
arith
.
|
|
|
6.
4992149
|
To
C
B
92388
|
|
|
4.
9656155
|
So
is
H
O
644
|
|
|
2.
8088858
|
To
E
B
18780
|
|
|
4.
●7●7152
|
And
now
in
the
triangle
E
D
B
we
have
the
sides
D
B
38268
,
●
B
18780
,
and
the
angle
E
B
D
45
deg.
to
find
the
angle
E
D
B.
As
the
summe
of
D
B
and
E
B
|
57048
co.
ar
.
|
5.
2437596
|
To
their
difference
|
19488
|
4.
2897673
|
So
the
tangent
of
½
D
and
E
|
67.
50
|
10.
3827756
|
To
the
tang
.
½
the
difference
|
39.
51283
|
9.
9163025
|
The
angle
E
D
B
27.
98717
and
the
double
thereof
is
the
arch
B
M
N
55.
97434
from
whence
taking
M
N
●q●al
to
D
B
45
there
rests
B
M
10.
97434
,
and
the
halfe
of
that
is
B
G
5.
48717
,
and
the
summe
of
D
B
and
B
G
50.
48717
,
now
D
represents
the
15
deg.
of
Leo
and
the
complement
of
D
G
to
a
quadrant
is
the
arch
D
L
39.
51283
from
the
vernall
Equinox
to
the
15
deg.
of
Leo
is
135
deg.
from
which
deducting
D
L
39.
51283
,
there
rests
for
the
point
of
the
Aphelion
at
L
95.
48717
,
that
is
in
Cancer
deg.
5.
48717.
And
now
to
find
the
Semidistance
of
the
umbiliques
,
As
the
summe
of
the
Sin●s
of
the
complements
of
B
O
and
C
K
,
is
to
K
O
,
So
is
the
Radius
A
G
to
the
same
Radius
A
G.
Now
if
you
deduct
15
deg
of
Taurus
from
Cancer
5.
48717
the
arch
C
L
will
be
50.
48717
and
B
G
5.
48717
being
d●ducted
from
L
G
90
,
there
rests
L
B
,
84.
51283
the
sine
of
L
C
is
K
A
77●●8
and
the
sine
of
A
O
99541
and
the
summe
of
K
A
and
A
O
is
176●8●
now
then
,
As
the
summe
of
K
A
and
A
O
|
176689
co.
ar
.
|
4.
7527●94
|
Is
to
the
½
difference
K
O
|
3168
|
3.
5007851
|
So
is
the
Radius
A
G
|
100000
|
|
To
the
same
Radius
A
G
in
the
parts
of
K
O
|
1792
|
8.
2534945
or
degrees
1.
02139
halfe
the
inequality
desired
.
|
But
if
we
take
the
arches
of
Middle
motion
the
calculation
will
be
as
followeth
,
Arch.
|
Deg.
|
Add
|
Ded.
|
Sides
.
|
C
D
|
92.
89277
|
C
B
D
|
46.
446
85
|
C
D.
|
72473
|
D
B
|
45.
73777
|
D
C
B
|
22.
868885
|
D
B.
|
38862
|
C
B
|
138.
63054
|
C
D
B
|
|
C
B.
|
93554
|
Halfe
the
differences
of
the
middle
from
the
true
▪
as
before
.
1
Deg.
1.
44638
|
K
H
|
2524.
14119
|
2
Deg.
0.
36888
|
H
O
|
643.
81259
|
Deg.
1.
81526
|
K
O
|
3167.
95378
|
As
K
O
|
3168
co.
arith
.
|
6.
4992149
|
Is
to
C
B
|
93554
|
4.
9710623
|
So
is
H
O
|
644
|
2.
8088858
|
To
E
B
|
19018
|
4.
2791630
|
And
in
the
triangle
E
D
B
the
angle
at
D
may
be
thus
found
.
The
side
D
B
is
|
38862
|
|
The
sine
E
B
is
|
19018
|
|
The
summe
is
|
57880
|
5.
2374765
|
The
difference
|
19844
|
4.
2976292
|
Tangent
½
D
&
E
|
66.
77681
|
10.
3674618
|
To
the
tangent
½
|
38.
62586
|
9.
9025625
|
28.
15095
the
angle
E
D
B
and
the
double
thereof
is
the
arch
B
M
N
56.
30190
&
from
thence
taking
M
N
equal
to
D
B
45.
73777
There
rests
B
M
10.
56413
and
the
half
of
that
is
B
G
5.
282065.
and
the
summe
of
D
B
and
B
G
51.
01997.
Now
D
represents
the
15
degrees
of
Leo
,
and
the
complement
of
D
G
to
a
quadrant
is
D
L
38.
9800●
.
from
the
Vernal
Equinox
.
to
the
15
deg.
of
Leo
,
is
135
deg.
from
which
deducting
D
L.
38.
98003
,
there
rests
for
the
point
of
the
Aphelion
at
L
,
96.
01997.
that
is
in
Cancer
,
6
deg.
01997
,
from
which
the
difference
between
the
true
and
mean
motion
in
that
interval
being
deducted
73776
,
the
place
of
the
Aphelion
will
be
in
Cancer
,
5.
28221.
And
to
find
the
Semidistance
of
the
umbiliques
deduct
15
deg.
of
Taurus
from
Cancer
5.
28221
,
there
will
remaine
for
the
arch
C
L
50.
28221
,
and
B
G
5.
28206
being
deducted
from
L
G
90
there
rests
for
the
arch
,
L
B
84.
71779
,
the
sine
of
C
L
is
K
A
76921
The
sine
of
L
B
is
A
O
99575
As
their
summe
|
1764964
co.
ar
.
|
4.
7532652
|
Is
to
K
O
|
3168
|
3.
5007851
|
So
is
the
Radius
A
G
|
|
10.
0000000
|
To
the
same
A
G
1795
|
|
8.
2540503
|
This
excentricity
Bullialdus
corrects
by
the
apparent
places
of
the
planets
in
the
center
of
the
Ellipsis
,
and
that
angle
according
to
the
Method
of
our
calculation
may
be
thus
found
.
From
15
degrees
of
Taurus
,
or
from
-
225.
deg.
Deduct
the
Aphelion
-
095.
28221
There
rests
the
angle
M
H
E
-
129.
71779
whose
complement
to
180
deg.
is
-
50.
28221
the
summe
of
M
E
H
and
E
M
H
and
the
halfe
thereof
-
25.
14110
The
side
H
E
|
200000
|
|
The
side
M
H
|
3590
|
|
The
summe
|
203590
|
4.
6912436
|
The
Differ
.
|
196410
|
5.
2931636
|
Tang.
½
summe
|
25.
14110
|
9.
6714590
|
Tang.
½
differ
.
|
24.
35902
|
9.
6558662
|
The
summe
|
49.
50012
E
M
H
|
|
Difference
|
00.
78208
M
E
H
being
doublled
is
the
angle
|
|
M
B
H
|
1.
56416
|
|
|
From
E
M
H
|
49.
50012
|
|
Deduct
M
E
H
or
E
M
B
|
00.
78208
|
|
There
rests
B
M
H
|
48.
71804
|
As
the
sine
of
M
B
H
|
1.
56416
co.
ar
.
|
1.
5638976
|
Is
to
the
side
M
H
|
3590
|
3.
5550944
|
So
is
the
sine
of
B
M
H
|
48.
71804
|
9.
8759126
|
4.
9949046
|
To
the
side
B
H
|
98833
|
|
The
side
X
H
|
1795
|
|
The
Summe
|
100628
co.
ar
.
|
4.
9972811
|
The
Differ
.
|
97038
|
4.
9869418
|
Tang.
½
sum
|
25.
14110
|
9.
6714590
|
9.
6556819
|
Tang.
½
Dif
.
|
24.
34989
|
|
Differ
.
00.
79121
X
B
H.
From
the
15
deg.
of
Taurus
or
from
45
-
45.
Deduct
the
angle
X
B
H
-
00.
79121
There
rests
the
place
required
-
44.
20879
Again
,
from
0
deg.
of
Libra
or
from
-
180.
Subtract
the
Aphelion
-
95.
28221
There
rests
the
angle
M
H
E
-
84.
71779
And
therefore
½
of
M
E
H
and
E
M
H
-
47.
64110
The
side
H
E
|
200000
|
|
The
side
H
M
|
3590
|
|
The
Summe
|
203590
co.
ar
.
|
4.
6912436
|
The
Differ
.
|
196410
|
5.
2931636
|
Tang.
½
summe
|
47.
64110
|
10.
0400951
|
Tang.
½
Differ
.
|
46.
61542
|
10.
0245023
|
Aggregate
-
94.
25652
E
M
H
Difference
-
1.
02568
M
E
H
which
being
doubled
is
the
angle
M
B
H
-
2.
05136
From
E
M
H
-
94.
25652
Deduct
M
E
H
-
1.
02568
There
rests
B
M
H
-
93.
23084
whose
complement
is
A
M
B
-
86.
76916
The
summe
|
101927
co.
ar
.
|
4.
9917108
|
The
differ
.
|
98337
|
4.
9927169
|
Tang.
½
summe
|
47.
64110
|
10.
0400951
|
Tang
½
differ
.
|
46.
61677
|
10.
0245228
|
Differ
.
|
1.
02433
The
angle
X
B
H.
|
|
And
therefore
the
Earths
place
-
181.
02433
Now
then
from
the
Aphelion
-
95.
28221
Subtract
Taurus
,
that
is
-
44.
20879
There
rests
the
arch
C
L
-
51.
07342
And
from
L
G
-
91.
02433
Deduct
B
G
-
5.
28206
There
rests
the
arch
L
B
-
85.
74227
77795
is
the
sine
of
51.
07342
99724
is
the
sine
of
85.
74227
Summe
177519
-
As
the
summe
177519
com
.
ar
.
-
4.
7505576
Is
to
K
O
3168
-
3.
5007851
So
is
A
G
100000
-
5.
0000000
To
the
same
A
G
1784
-
3.
2513427
But
according
to
this
method
,
the
Aphelion
may
be
somewhat
more
exactly
found
,
if
we
take
the
Arithmetical
mean
,
between
the
apparent
and
middle
motion
,
and
so
;
Arch.
|
Deg
|
Angle
|
Deg.
|
Side
|
C
D
|
91.
44638
|
C
B
D
|
45.
723●9
|
C
D
|
71597
|
D
B
|
45.
36889
|
D
C
B
|
22.
68444
|
D
B
|
38565
|
C
B
|
136.
81527
|
C
D
B
|
68.
40763
|
C
B
|
9●983
|
Now
then
to
find
E
B
I
say
,
As
K
O
3160
co.
ar
.
-
6.
4992149
Is
to
C
B
92983
-
4.
9684035
So
is
H
O
644
-
2.
8088858
To
E
B
18902
-
4.
2765042
Then
in
the
triangle
E
D
B
,
we
have
known
the
side
D
B
38565
,
the
side
E
B
18902
,
and
their
contained
angle
E
B
D
45.
72318
,
whose
complement
to
a
semicircle
is
134.
27681.
The
half
summe
67.
13840.
As
the
summe
of
D
B
and
EB
|
57467
co.
ar
.
|
5.
2405815
|
Is
to
their
differ
.
|
19663
|
4.
2936497
|
So
is
Tang.
½
D
and
E
|
67.
13840
|
10.
3750715
|
To
the
Tang
½
differ
.
|
39.
06019
|
|
Differ
.
|
28.
07821
the
angle
E
D
B
|
|
The
angle
E
D
B
-
28.
07821
E
D
B
doubled
is
the
arch
B
M
N
-
56.
17926
From
which
subtract
M
N
or
D
B
-
45.
36889
There
rests
B
M
-
10.
78753
The
halfe
of
B
M
is
the
arch
B
G
-
5.
39376
And
D
B
being
added
to
B
G
,
D
G
is
-
50.
76265
And
the
complement
there
of
is
D
L
-
39.
23735
Which
being
deducted
from
♌
or
-
435.
00000
There
rests
the
Aphelion
at
L
-
95.
76265
And
the
halfe
difference
D
B
-
0.
36889
Being
deducted
there
rests
-
95.
39377
The
Aphelion
then
is
in
Cancer
5
deg.
-
39377
And
to
correct
the
excentricity
from
the
15
degree
of
Taurus
or
adding
a
semicircle
from
-
225.
Deduct
the
Aphelion
-
95.
39●77
There
rests
in
the
Ellipsis
M
H
E
-
129.
60623
whose
complement
to
180
is
the
summe
of
the
opposite
angles
M
E
H
and
EMH
-
50.
39377
In
the
Triangle
therefore
of
the
last
Diagram
M
E
H
,
we
have
,
-
1.
The
side
H
E
200000
-
2.
The
side
H
M
3568
-
3.
The
angle
M
H
E
given
As
the
summe
|
203568
co.
ar
.
|
4.
6912904
|
To
the
difference
|
196432
|
5.
●932122
|
So
tang
.
halfe
summe
|
25.
19688
|
9.
6724222
|
To
tang
.
halfe
differ
.
|
24.
41154
|
9.
6569248
|
Aggregate
|
49.
60842
the
angle
E
M
H
|
|
Difference
|
00.
78534
the
angle
M
E
H
|
|
Differ
.
doubled
is
|
1.
57068
the
angle
M
B
H
|
|
Differ
.
subtracted
|
48.
82308
is
the
angle
B
M
H
▪
|
|
As
the
sine
of
M
B
H
|
1.
57068
co.
ar
.
|
1.
5620900
|
To
the
side
M
H
|
3568
|
3.
5524248
|
So
is
the
sine
of
B
M
H
|
48.
82508
|
9.
8766104
|
To
the
side
H
B
|
97977
|
4.
9911252
|
The
side
X
H
|
1784
|
|
The
summe
|
099761
co.
ar
.
|
5.
0010393
|
The
differ
.
|
96193
|
4.
9831434
|
Tang
½
summe
|
25.
19688
|
9.
6724222
|
Tang.
½
differ
.
|
24.
39566
|
9.
6566049
|
Differ
.
|
00.
80122
the
angle
X
B
H
which
being
subtracted
from
15
degrees
of
Taurus
,
or
from
45
deg.
there
rests
44.
19878
the
Aphelion
|
95.
39377
|
Place
of
the
earth
subtract
|
|
44.
39878
|
There
rests
the
arch
C
L
|
|
51.
19499
|
Againe
from
0
degrees
of
Libra
or
from
180.
Deduct
the
Aphelion
|
|
95.
39377
|
There
rests
in
the
Ellipsis
M
H
E
|
|
84.
60623
|
And
therefore
the
½
summe
of
the
angles
E
M
H
and
M
E
H
|
|
47.
69688
|
As
the
summe
|
203568
co.
ar
.
|
4.
6912904
|
To
the
differ
.
|
196432
|
5.
2932122
|
So
tang
.
½
summe
|
47.
69688
|
10.
0409444
|
To
tang
.
½
differ
.
|
46.
67763
|
10.
0254470
|
Aggregate
|
94.
37451
angle
E
M
H
|
|
Difference
|
01.
01925
angle
M
E
H
|
|
Differ
.
doubled
|
2.
03850
angle
M
B
H
|
|
Differ
.
subtract
|
93.
35526
angle
B
M
H
|
|
As
the
sine
of
M
●
H
|
2.
0385
co.
ar
.
|
1.
4489043
|
To
the
side
M
H
|
3568
|
3.
5524248
|
So
is
the
sine
of
B
M
H
|
86.
64474
|
9.
9992548
|
To
the
side
H
B
|
100134
|
5.
0095839
|
The
side
X
H
|
1784
|
|
The
summe
|
101918
co.
ar
.
|
4.
9917492
|
The
differ
|
98●50
|
4.
9927743
|
The
tang
.
½
summe
|
47.
69688
|
10.
0409444
|
Tang.
½
differ
.
|
46.
67901
|
10.
0254679
|
Difference
|
01.
01787
,
X
B
H
which
being
added
to
♎
the
place
will
be
181.
01787
,
from
which
subtract
B
G
5.
●9376
there
rests
175
or
L
B
85.
62411.
|
|
The
sine
of
C
L
|
77927
|
|
The
sine
of
L
B
|
99707
|
|
As
the
summe
|
177634
co.
ar
.
|
4.
7504740
|
Is
to
K
O
|
3168
|
3.
5007851
|
So
is
A
G
|
10000
|
5.
0000000
|
To
A
G
|
1783
|
3.
2512591
|
which
comes
so
neer
to
the
Excentricity
before
found
that
we
may
without
manifest
error
make
use
of
either
.
CHAP.
6.
Of
Stating
the
Earths
middle
motions
by
sundry
observations
.
TO
find
the
Earths
middle
motion
for
any
time
under
a
yeare
,
the
way
already
prescribed
in
the
first
Chapter
(
as
to
the
use
for
which
it
was
intended
)
is
exact
enough
,
but
to
state
the
true
quantity
of
the
Earths
annual
motion
,
the
apparent
Equinoctials
must
be
reduced
into
the
mean
,
which
cannot
be
done
unless
the
Aphelion
be
first
found
,
having
found
that
therefore
by
the
observations
of
Tycho
,
we
will
now
find
it
by
the
observations
of
Albategnius
,
in
the
year
from
the
death
of
Alexander
,
1206
,
and
the
intervall
of
time
then
between
the
Autumne
and
the
Vernal
Equinox
was
dayes
178.
51250
,
and
the
middle
motion
for
that
time
,
is
deg.
175.
95083.
The
true
motion
is
180.
From
which
subtract
-
175.
95083
Their
difference
is
-
4.
04917
The
half
difference
is
K
L
-
2.
02458
Therefore
as
A
E
|
3568
com
.
ar
.
|
6.
4475752
|
To
A
E
|
100000
|
5.
0000000
|
So
is
K
L
|
3533
|
3.
5481436
|
To
K
L
|
99019
|
4.
9957188
|
Half
the
arch
H
I
L
is
87.
97541
,
whose
sine
99938
is
the
side
H
L
,
and
therefore
,
This
Autumne
Equinox
was
observed
September
the
19th
,
from
the
death
of
Alexander
1206
yeares
,
that
is
in
the
yeare
of
our
Lord
882.
In
the
beginning
therefore
of
the
yeare
of
Christ
883
,
the
Aphelion
was
in
Gemini
24
d.
25176
And
in
June
1588
,
the
Aphelion
was
in
Cancer
-
5.
39377
Their
difference
is
-
11.
14201
And
betweene
both
observations
there
are
706
Egyptian
years
,
now
then
to
find
the
mean
motion
of
the
Aphelion
for
a
yeare
I
say
,
If
706
years
give
11.
14201
,
what
shall
one
yeare
or
365
dayes
give
?
and
the
answer
is
Deg.
0.
0157818838
And
againe
,
if
365
dayes
give
0157818838
one
day
shall
give
o
deg.
0000432380.
In
882
Julian
years
there
are
322150
dayes
,
by
which
if
you
multiply
0000432380
the
product
will
be
deg.
13.
9291217
,
which
being
deducted
from
the
aphelion
before
found
,
Gemini
24.
25176
,
the
aphelion
in
the
beginning
of
the
Christian
Aera
will
be
in
Gemini
10.
3226383
,
that
is
,
19
21
29.
But
from
Hypparchus
,
that
is
from
177
yeare
from
the
death
of
Alexander
to
the
1205
yeare
compleate
in
the
same
account
,
there
are
1028
Egyptian
years
,
and
the
meane
motion
of
the
Aphelion
in
that
time
is
,
Deg.
-
16.
2237765464
Gemini
-
24.
2516600000
Gemini
-
08.
0278834536
which
being
deducted
from
there
rests
for
the
aphelion
at
that
time
.
And
therefore
the
vernall
Equinox
observed
by
Hypparchus
in
the
yeare
from
the
death
of
Alexander
178
Mechir
26.
95833333
,
was
distant
from
the
Aphelion
deg.
68.
027883
,
which
being
deducted
from
a
Semicircle
the
angle
in
the
Ellipsis
of
the
next
Chapter
A
M
E
will
be
found
to
be
111.
972117
,
and
this
angle
is
the
summe
of
the
angles
M
E
H
and
M
H
E
,
and
therefore
the
equation
to
be
subtracted
may
be
thus
found
.
The
side
M
E
|
200000
|
|
The
side
M
H
|
3568
|
Logarithms
|
The
summe
|
203568
co.
ar
.
|
4.
6912905
|
The
Differ
.
|
196432
|
5.
2932122
|
The
tang
.
½
summe
|
55.
98606
|
10.
1707846
|
Tang.
½
differ
|
55.
03186
|
10.
1552873
|
Differ
.
|
00.
95420
angle
M
E
H
|
|
Differ
.
doubled
.
01.
90840
angle
M
B
H
or
the
Equation
sought
:
which
may
be
converted
into
time
thus
,
if
the
parts
of
a
degree
of
equal
motion
,
98564
give
one
day
;
1.
90840
snall
give
1.
93620
,
and
this
being
added
to
the
true
Equinoctial
,
Mechir
26.
95833
the
middle
will
be
Mechir
28.
8945●
,
or
deducting
05625
;
for
the
difference
of
meridians
between
Uraniburge
and
Alexandria
,
it
will
be
at
Uraniburge
;
Mechir
28.
83828.
And
the
vernall
Equinox
observed
by
Tycho
at
Uraniburge
1588
,
was
March
the
9.
86458
,
and
the
Earths
aphelion
then
was
in
Cancer
5.
39377
;
and
therefore
the
arch
answering
to
the
excentricity
3568
,
viz.
deg.
2.
04529
being
converted
into
time
as
before
,
will
be
days
2.
07508
,
which
being
added
to
the
former
time
the
middle
Equinoctial
wil
be
March
the
11.
93966.
And
in
the
Egyptian
account
from
the
death
of
Alexander
it
was
1912
Pharmuthi
23.
93966
,
from
which
if
you
deduct
in
the
same
account
178
,
Mechir
28.
83828
between
both
observations
there
will
be
found
,
1734
Egyptian
years
,
dayes
55.
10138
,
which
being
converted
into
dayes
give
632965.
10138.
Hence
to
find
the
quantity
of
the
Tropicall
yeare
,
I
say
,
if
1733
Zodiacks
give
dayes
632965.
10138
,
that
one
Zodiack
shall
give
dayes
365.
2418357126.
And
to
find
the
earths
middle
motion
for
a
yeare
,
I
convert
1733
Zodiacks
into
degrees
,
and
they
amount
to
623880
degrees
;
then
I
say
,
if
632965.
10138
give
623880
,
that
365
days
shal
give
359.
76106661098
that
is
in
Sexagenary
numbers
359
deg.
45
minutes
,
39
seconds
,
50
thirds
,
24
fourths
.
And
to
find
the
meane
motion
for
a
day
,
I
say
,
if
365
dayes
give
359.
76106661098
,
that
one
day
shall
give
9856467579
,
that
is
in
Sexagenary
numbers
0
degrees
,
59
minutes
,
8
seconds
,
19
thirds
,
41
fourths
,
57
fifths
.
And
the
daily
motion
of
the
Aphelion
is
0000432380
,
which
being
deducted
from
the
diurnall
longitude
gives
the
daily
motion
of
the
Anomaly
985603599
,
these
things
premised
we
will
now
determine
the
Epochaes
of
the
middle
motions
.
The
middle
Equinoctiall
Anno
Christi
1588
,
March
11.
9●966
,
is
from
the
Aera
Nabonassari
2336
Pharmuthi●3
●3
.
93966.
2335
years
being
multiplyed
by
359.
761067
the
product
will
be
840042.
091445
,
and
the
diurnal
motion
985647
,
being
multiplied
by
232
days
,
the
product
will
be
228.
670104
and
the
middle
motion
answering
to
the
parts
of
a
day
,
93966
,
is
926173
,
the
which
being
added
togethea
do
amount
to
840●71
degrees
687722
parts
of
a
degree
,
that
is
,
rejecting
the
whole
circles
31
d.
687722
,
which
being
deducted
from
360
,
the
remainder
328.
312278
or
10
Signes
28
degrees
and
312278
parts
,
is
the
Radix
of
the
earth
or
Suns
mean
longitude
in
the
beginning
of
the
Aera
Nabonassari
.
To
which
if
you
add
deg.
258.
692408
the
middle
motion
for
424
years
,
the
whole
circles
being
rejected
,
the
Radix
of
the
earths
middle
motion
to
the
beginning
of
the
Aera
Alexandri
shall
be
227.
004686
or
7
sines
17
deg.
004686
parts
.
And
adding
to
this
Epocha
,
deg.
51.
944398
,
which
is
the
middle
motion
for
323
years
131
dayes
,
the
whole
circles
being
rejected
,
the
Radix
of
the
earths
middle
motion
in
the
beginning
of
the
Christian
Aera
shall
be
deg.
278.
949084
or
9
signes
,
8
deg.
949084
,
to
which
if
you
add
034223
the
equal
motion
belonging
to
034722
the
difference
between
the
Meridians
of
Uraniburge
and
London
,
the
Radix
of
middle
motion
at
London
will
be
278.
983307
And
the
Aphelion
-
70.
322638
And
the
Mean
Anomaly
-
208.
660669
CHAP.
7.
To
calculate
the
Suns
true
place
and
distance
from
the
Earth
.
HAving
composed
tables
of
the
Suns
middle
motions
,
according
to
the
directions
of
the
last
Chapter
,
his
true
place
in
the
Zodiack
,
and
distance
from
the
earth
may
thus
be
found
.
1
Write
out
the
Epocha
next
before
the
given
time
and
severally
under
that
,
set
the
motions
belonging
to
the
years
,
moneths
,
and
days
compleat
,
and
to
the
houres
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
Unite
)
then
adding
them
altogether
,
the
summe
shall
be
the
Suns
meane
motion
for
the
time
given
.
Example
.
Let
the
time
given
be
May
the
12th
.
houre
11
parts
15
before
noon
at
London
in
the
Bissextile
yeare
1656
,
and
the
Suns
place
to
be
soughts
The
numbers
are
thus
,
|
|
Suns
Longitude
|
Suns
Apogeon
|
|
|
Deg.
|
parts
|
Deg.
|
parts
|
The
Epocha
|
1640
|
291.
|
24777
|
96.
|
22265
|
Years
comp.
|
15
|
359.
|
37294
|
|
23686
|
April
|
118.
|
27760
|
|
519
|
Dayes
|
12
|
011.
|
82776
|
|
52
|
Houres
|
23
|
|
94458
|
|
|
Scruples
|
15
|
|
616
|
|
|
Suns
Mean
Longitude
|
421.
|
87681
|
96.
|
46522
|
2
Subtract
the
Apogaeum
from
the
Mean
Longitude
,
there
rests
the
mean
Anomaly
.
Example
.
The
Suns
mean
longitude
-
421.
67681
Apogaeum
substract
-
96.
46522
Rest
mean
Anomaly
-
325.
21459
Whose
complement
to
a
Circle
-
34.
78541
is
the
angle
A
M
E
in
the
Ellipsis
.
And
the
complement
of
A
M
E
to
a
semicircle
is
the
angle
E
M
H
145.
21459.
The
side
M
E
|
200000
|
|
The
side
M
H
|
3568
|
|
The
summe
|
203568
co.
ar
.
|
4.
6912905
|
Differ
.
|
196432
|
5.
2932122
|
Tang.
½
summe
of
the
opposite
angles
|
17.
39270
|
9.
4958787
|
|
17.
39270
|
|
Tang.
½
Differ
|
16.
81799
|
9.
4803814
|
Differ
|
57471
is
the
angle
M
E
H.
|
|
Difference
doubled
|
1.
14942
is
the
angle
M
B
H
|
|
3
The
mean
Anomaly
being
above
180
deg.
the
Aequation
found
must
be
added
to
the
sunsmeane
longitude
,
so
have
you
the
Suns
true
place
.
Example
.
The
Suns
meane
longitude
-
421.
67681
Aequation
adde
-
1
14942
The
Suns
true
place
-
422.
82623
or
2
Signes
2
degrees
82623
parts
of
a
degree
-
Lastly
,
to
find
his
distance
from
the
earth
,
I
say
,
As
the
sine
of
M
B
H
|
1.
14942
co
▪
ar
.
|
1.
6977118
|
Is
to
the
side
M
H
|
3568
|
3.
5524249
|
So
is
the
sine
of
B
M
H
|
34.
78541
|
9.
7562590
|
To
the
side
B
H
|
|
5.
0063957
|
or
distance
required
|
101483
|
|
Thus
we
have
found
the
Suns
place
by
calculation
,
we
will
now
shew
how
to
reduce
the
Suns
mean
longitude
to
his
true
,
by
the
Table
of
Aequations
of
the
Suns
excentrick
.
The
Suns
Anomaly
in
this
example
is
-
325.
21459
The
Aequation
of
325
is
-
1.
15566
326
-
1.
12648
Difference
is
-
02918
Now
then
I
say
if
one
deg.
co.
ar
.
-
5.
Give
2918
-
3.
4650853
What
shall
21459
-
4.
3316095
The
answer
is
6●6
-
2.
7966948
Aequation
of
325
deg.
-
1.
15566
Part
proportional
subt
.
-
626
Aequation
equated
-
1.
14940
The
Snns
mean
longitude
-
61.
67681
Aequation
adde
-
1.
14940
Suns
true
place
-
62.
82621
And
in
like
manner
the
Logarithme
of
the
Suns
distance
from
the
Earth
will
be
found
to
be
5.
0063633
,
which
being
more
necessary
then
the
distance
it self
,
in
the
calculation
of
the
places
of
the
other
planets
,
we
have
as
most
convenient
placed
in
the
table
.
CHAP.
8.
Of
the
Aequation
of
Civil
Dayes
.
SOme
there
are
of
late
,
which
allow
not
of
any
Aequation
of
Civil
Dayes
,
others
will
have
the
inequality
proceed
from
two
causes
.
First
,
from
the
unequal
motion
of
the
Sun
in
the
Zodiack
,
and
the
other
from
the
Zodiacks
obliquity
;
Tycho
(
whom
we
shall
follow
in
this
particular
,
)
doth
make
the
difference
between
the
Suns
true
longitude
and
his
Right
Ascension
;
to
be
the
absolute
Aequation
of
naturall
dayes
,
the
which
is
also
clearly
demonstrated
,
according
to
the
Copernican
Systeme
by
Thomas
Street
in
his
Ephemeris
for
the
yeare
1655
;
which
being
but
short
is
here
inserted
.
The
Aequation
of
time
demonstrated
.
Let
A
be
the
center
of
the
Sun
,
and
E
of
the
Earth
,
♎
E
the
earths
longitude
from
the
Equinoctiall
point
in
the
ecliptick
,
♎
F
the
like
arch
projected
in
the
Equator
,
♎
B
the
Right
Ascension
of
the
Earths
or
Suns
true
place
,
G
H
is
a
diameter
of
the
Equinoctiall
and
Meridian
of
the
earths
apparent
diurnal
revolution
,
A
B
the
semidiameter
of
the
true
meridian
,
and
equinoctial
supposed
in
the
heavens
;
and
G
H
parallel
to
A
B
(
though
here
they
appeare
as
one
right
line
.
Then
let
C
D
parallel
to
A
F
be
likewise
a
diameter
of
the
Equinoctial
and
Meridian
of
the
meane
or
equal
diurnal
revolution
.
Hence
C
E
G
the
angle
of
the
earths
libration
,
equal
to
B
A
F
the
difference
of
longitude
and
Right
Ascension
is
the
true
Equation
of
time
or
the
difference
between
the
equal
and
apparent
time
.
And
according
to
this
Demonstration
is
our
Table
(
entituled
,
A
perpetual
Table
for
the
Equation
of
time
)
composed
.
In
which
you
must
enter
with
the
signe
and
degree
of
the
Suns
place
either
in
the
uppermost
and
left
hand
columnes
descending
,
or
in
the
lowermost
and
right
hand
ascending
,
and
in
the
common
angle
is
the
Equation
(
according
to
the
titles
)
to
be
added
or
subtracted
to
or
from
the
equal
time
,
that
it
may
be
made
apparent
;
But
to
reduce
the
apparent
to
the
equal
,
take
the
contrary
title
.
CHAP.
9.
Of
the
Theory
and
Motion
of
the
Moon
.
THe
Moon
according
to
our
Hypothesis
is
a
secondary
planet
,
moving
about
the
earth
,
as
the
earth
and
other
planets
doe
about
the
Sun
,
and
so
not
onely
the
earth
,
but
the
whole
Systeme
of
the
Moone
is
also
carryed
about
the
Sun
in
a
yeare
,
And
hence
according
to
Hypparchus
there
ariseth
a
twofold
,
but
according
to
Tycho
a
threefold
inequality
in
the
Moons
motion
.
The
first
is
periodicall
,
and
is
to
be
obtained
,
after
the
same
manner
,
as
was
the
excentrique
Equation
of
the
Sun
or
Earth
;
in
order
whereunto
her
middle
motions
should
be
first
stated
,
the
which
Bullialdus
by
the
rules
delivered
in
the
fourth
and
fifth
Chapters
preceding
hath
for
the
Meridian
of
Uraniburge
determined
to
be
as
followeth
.
From
the
Equinoctiall
to
the
beginning
of
the
Christian
Aera
,
the
Moons
middle
motion
was
|
135d
.
|
16′
|
27″
|
The
Meane
Anomalie
|
355
|
5
|
18
|
And
the
Radix
of
her
latitude
|
366
|
29
|
56
|
These
then
we
will
take
for
granted
,
until
there
be
a
more
exact
,
and
true
Geometrical
way
propounded
to
us
,
;
onely
we
will
convert
them
into
Decimall
numbers
,
and
reduce
them
to
the
Meridian
of
London
.
From
the
Equinoctial
to
the
beginning
of
the
Christian
Aera
,
The
Moons
middle
motion
in
decimal
numbers
at
Uraniburge
was
135.
27417
For
the
Difference
of
Merid.
adde
-
.
45750
The
Moons
mean
longitude
at
London
-
135.
73167
The
Meane
Anomaly
,
At
Uraniburge
-
355.
08833
Differ
.
Merid.
adde
-
.
45361
Mean
Anomaly
at
London
-
355.
54194
The
Radix
of
Latitude
At
Vraniburge
-
366.
49889
Differ
.
Merid.
adde
-
.
45944
Latitude
at
London
-
366.
95833
d.
The
Diurnal
Motion
of
the
Moons
Mean
longitude
-
13.
17639
Anomaly
-
13.
06500
Latitude
-
13.
22944
The
Annual
Motion
.
In
longitude
-
129.
38389
Anomaly
-
88.
71889
Latitude
-
148.
71278
According
to
which
limitations
of
the
Moones
middle
motions
,
we
have
composed
our
Tables
,
by
help
whereof
and
the
Semi-excentricity
of
the
Moons
Orbe
,
which
according
to
Bullialdus
is
4362
the
Moons
excentrique
equation
,
or
place
first
equated
may
be
found
,
as
before
was
shewed
in
the
Sun
.
Save
onely
that
here
the
Moons
Anomaly
is
given
without
subtraction
.
Example
.
Anno
1587
,
August
17
ho.
19.
41667
in
the
apparent
time
,
or
ho.
19.
28973
in
the
middle
time
,
the
Moon
being
in
the
meridian
of
Vraniburge
noble
Tycho
observed
her
in
26
deg.
38333
of
Gemini
with
latitude
5
deg.
23333
S.
from
which
middle
time
if
you
subtract
83333
for
the
difference
of
the
meridians
of
London
and
Vraniburge
,
the
time
in
our
meridian
is
,
ho.
18.
45640.
And
the
Suns
true
place
-
154.
07347.
The
Suns
distance
from
the
Earth
-
100895.
The
Log●rithme
of
that
distance
-
5.
0038707
The
Moons
middle
motions
for
the
same
time
are
as
here
you
see
them
.
Time
given
|
☽
Longitud
|
☽
Anomaly
|
☽
Latitude
|
Years
1500
|
072.
88194
|
313.
06916
|
017.
17805
|
80
|
174.
24805
|
158.
80139
|
281.
61167
|
6
|
069.
48028
|
18●
.
37750
|
185.
50583
|
Iuly
|
27●
.
39555
|
249.
77639
|
284.
62194
|
D●yes
16
|
●10
.
82222
|
209.
0●972
|
211.
66944
|
H●●res
18
|
●9
.
88222
|
9.
79861
|
9.
92222
|
Paris
4564
|
.
250●1
|
.
24848
|
.
25152
|
Mean
Longitude
|
810.
96077
|
1126.
11125
|
990.
76067
|
Ded●ct
|
720.
|
1080.
|
720.
|
There
rests
|
90.
96077
|
46.
11125
|
270.
76067
|
The
Moones
meane
Anomaly
46.
11125
is
the
angle
A
M
E
in
the
preceding
Ellip●is
,
or
the
summe
of
the
angles
M
E
H
and
M
H
E.
Therefore
in
the
triangle
M
E
H
we
have
given
,
1.
The
side
M
E
200000.
2.
The
side
M
H
8724.
3.
The
angle
E
M
H
the
complements
of
the
Moones
Anomaly
,
to
find
M
E
H
,
whose
double
,
is
the
excentrique
Equation
M
B
H.
I
say
then
,
As
the
summe
of
M
E
and
M
H
|
208724
co.
ar
.
|
4.
6804276
|
Is
to
their
difference
|
191276
|
5.
2816605
|
So
is
the
tang
.
half
summe
of
the
opposite
angles
,
|
23.
05562
|
9.
6290228
|
To
the
tang
halfe
diff
.
|
21.
30786
|
9.
5911109
|
Differ
.
|
1.
74776
is
the
angle
M
E
H
|
|
Differ
▪
doubled
3.
49552
is
the
angle
M
B
H
or
the
equation
sought
which
being
subtracted
from
the
Moons
mean
longitude
,
because
the
Anomaly
is
lesse
then
a
semicircle
you
shall
have
the
Moones
place
first
equated
.
Example
.
The
Moones
meane
longitude
-
90.
96077
Equation
subtract
-
3.
49552
The
Moones
place
first
equated
-
87.
46525
And
to
find
her
distance
from
the
Umbilique
point
at
H.
As
the
sine
of
M
B
H
|
3.
49552
co.
ar
.
|
1.
2148808
|
Is
to
the
side
M
H
|
8724
|
3.
9407156
|
So
is
the
Sine
of
B
M
H
|
46.
11125
|
9.
8577468
|
To
the
side
B
H
|
103120
|
5.
0133432
|
But
whilest
the
Moone
is
thus
making
her
owne
periodicall
revolution
,
her
whole
Systeme
is
by
the
motion
of
the
Earth
removed
from
the
proper
seats
thereof
,
whence
there
ariseth
another
revolution
which
is
called
Synodicall
,
the
beginning
whereof
is
the
line
which
passeth
through
the
center
of
the
earth
to
the
Sun
,
and
therefore
the
Moone
in
that
line
is
void
of
this
second
inequality
,
which
is
both
at
the
Conjunction
and
opposition
,
but
being
in
or
about
her
quarters
,
where
she
is
farthest
removed
from
the
said
Zyzigiacal
line
the
angle
of
her
evection
is
2
deg.
50′
as
is
cleerely
proved
by
the
observations
of
Tycho
and
Bullialdus
,
whose
method
we
follow
,
in
our
calculation
of
this
inequality
of
the
Moon
,
making
4362
the
sine
of
the
greatest
evection
to
be
the
Diameter
K
D
in
the
little
circle
K
C
D
H.
The
motion
of
this
libration
of
the
Moone
must
be
measured
by
her
double
distance
from
the
Sun
,
because
she
is
void
of
this
inequality
at
her
conjunctions
and
oppositions
as
was
said
before
.
Now
then
let
the
angle
M
H
B
represent
the
Moones
equated
Anomaly
,
found
by
subtracting
the
former
Equation
from
the
simple
Anomaly
,
which
is
46.
11125
Aequation
subtract
-
3.
49552
Aequated
Anomaly
-
42.
61573
And
if
from
the
place
of
the
moon
first
equated
-
87.
46525
You
subtract
the
Suns
true
place
-
154.
07347
Their
distance
is
-
293.
39178
The
double
distance
-
226.
78356
is
the
arch
K
C
D
H
and
drawing
the
lines
F
H
and
H
K
there
shall
be
made
the
equicrurall
Triangle
H
F
K
,
whose
exteriour
angle
H
F
D
is
known
,
viz.
the
excesse
of
the
Sun
and
Moons
double
distance
above
a
se
micircle
46.
78356.
The
halfe
whereof
is
the
angle
D
K
H
23.
●9178
,
equal
to
the
angle
A
H
G
,
which
being
subtracted
from
the
Equated
Anomaly
A
H
B
42.
61573.
The
angle
G
H
B
or
the
Synodical
Anomaly
will
be
19.
2239●
,
whose
complement
to
a
semicircle
is
the
angle
B
H
K
160.
77605.
The
side
H
B
103120
as
before
,
and
the
side
H
K
may
be
found
in
this
manner
.
The
arch
H
V
K
is
the
complement
of
the
double
distance
of
the
Sun
and
Moon
to
a
whole
circle
13●
.
21644
the
subtense
of
this
arch
is
H
K
,
H
V
the
halfe
arch
is
66.
60822
,
and
H
X
is
the
right
sine
thereof
.
Now
then
,
As
the
Radius
,
|
|
|
Is
to
the
Diameter
K
D
|
4●62
.
|
3.
6396857
|
So
is
the
sine
of
H
V
or
H
X
|
66.
60822
|
9.
9627533
|
To
the
Subtense
H
K
|
4003
|
3.
6024390
|
Therefore
in
the
Triangle
B
H
K
we
have
known
,
1.
The
angle
B
H
K
|
160.
77605.
|
|
2.
The
side
H
B
|
103120
To
find
the
angle
H
B
K.
|
|
3.
The
side
H
K
|
4003
To
find
the
angle
H
B
K.
|
|
As
the
summe
|
107123
co.
ar
.
|
4.
9701173
|
To
the
differ
.
|
99117
|
4.
9961481
|
So
tang
.
½
the
opposite
ang.
|
9.
61197
|
9.
2287638
|
To
the
tang
.
½
differ
.
|
8.
90502
|
9.
1950292
|
Their
difference
0.
70695
is
the
angle
of
the
evection
,
H
B
K
to
be
added
if
the
Synodicall
Anomaly
be
more
then
180
,
and
to
be
subtracted
from
the
place
of
the
Moone
first
equated
when
lesse
,
as
here
it
is
in
our
example
,
and
therefore
to
be
subtracted
from
the
Moons
place
87
46525
,
and
then
her
place
secondly
equated
will
be
86.
7●830
.
And
according
to
this
Analogie
may
be
made
a
table
of
the
Moons
evection
,
when
she
is
in
quadrature
or
90
degrees
distant
from
the
Sun
;
for
then
the
equated
and
Synodicall
Anomalies
are
both
the
same
,
and
therefore
in
the
Triangle
E
M
H
we
have
give
the
angle
E
H
M
,
or
the
equated
Anomaly
which
suppose
25
degrees
,
the
side
M
H
8724
and
the
side
H
E
●00000
,
to
find
the
angle
at
E.
As
the
sum
of
H
E
and
M
H
|
208724
co.
ar
.
|
4.
6804276
|
Is
to
their
difference
|
191276
|
5.
2816605
|
So
is
the
tang
.
half
summe
|
|
10.
6542447
|
To
tang
.
half
differ
.
|
76.
40435
|
10.
6163328
|
Their
differ
.
|
1.
09565
is
the
angle
M
E
H
|
|
Whose
double
is
|
2.
19130
is
the
angle
M
B
H
|
|
Then
as
the
sine
of
M
B
H
|
2.
1913
co.
ar
.
|
1.
4175273
|
To
the
sine
of
B
M
H
|
27.
1913
|
9.
6598808
|
So
is
the
side
M
H
|
.
8724
|
3.
9407156
|
To
the
side
B
H
|
10.
4161
|
5.
0181235
|
or
the
Moons
distance
from
the
umbilique
.
|
|
|
Hence
in
the
first
Diagram
of
this
Chapter
,
in
the
Triangle
B
H
K
we
have
given
A
H
B
the
equated
Anomaly
,
25
degrees
,
the
Moons
distance
H
B
104161
,
with
the
side
H
K
,
or
rather
D
K
,
the
Diameter
of
the
little
circle
4362
,
to
find
the
angle
H
B
K.
As
the
summe
of
H
B
and
H
K
|
108523
co.
ar
.
|
4.
9644782
|
To
their
difference
|
99799
|
4.
9991262
|
So
is
tang
.
halfe
summe
|
12.
50
|
9.
3447552
|
To
tang
.
halfe
differ
.
|
11.
52314
|
9.
3093596
|
whose
difference
.
97686
is
the
evection
sought
and
by
Bullialdus
.
97805
which
is
so
little
differing
from
what
we
have
found
,
that
I
have
taken
his
Table
and
converted
it
into
Decimall
numbers
.
And
for
the
finding
the
part
proportial
between
the
quadrature
and
the
Zyzygia
,
Bullialdus
whom
we
follow
,
hath
annexed
scruples
of
proportion
in
this
manner
.
As
Radius
to
60
minutes
or
one
degree
;
so
is
the
fine
of
halfe
the
degrees
of
the
equated
Anomaly
to
the
scruples
of
proportion
required
.
But
this
proportion
in
the
Sexagenary
Canon
will
not
give
the
scruples
either
so
easily
or
so
exactly
as
the
Decimal
Canon
will
,
because
the
seconds
must
still
be
found
in
that
Canon
by
the
part
proportional
,
which
in
the
beginning
of
the
Canon
cannot
be
true
,
but
working
by
a
Decimal
Canon
the
natural
sines
of
halfe
the
degrees
,
are
the
Decimall
parts
required
,
as
the
naturall
sine
of
4
degrees
;
069756
,
are
the
Decimall
parts
for
8
degrees
of
equated
anomaly
,
and
so
of
the
rest
.
Having
done
with
the
first
and
second
inequalities
of
the
Moon
,
we
come
to
the
third
which
Tycho
calls
the
variation
,
Bullialdus
the
Reflection
for
as
the
Moons
Systeme
is
carryed
about
by
the
earth
,
the
place
of
her
Apogaeon
is
changed
,
or
doth
reflect
contrary
to
the
succession
of
the
Signes
,
by
reason
of
which
reflection
the
angle
of
her
Evection
is
sometimes
more
sometimes
lesse
then
it
will
be
found
by
the
former
directions
,
but
the
quantity
of
this
variation
according
to
Tycho
doth
never
exceed
40′
30″
or
in
Decimal
numbers
67500
,
to
be
added
to
or
subtracted
from
the
place
of
the
Moone
secondly
equated
,
and
the
proportion
by
which
he
finds
it
is
thus
,
As
the
Radius
To
the
sine
of
the
complement
of
the
double
distance
of
the
Sun
and
Moon
if
lesse
then
a
Semicircle
:
To
the
excesse
if
more
:
So
is
the
sine
of
the
greatest
variation
,
or
Reflection
.
To
the
variation
required
,
which
is
to
be
added
to
the
Moons
place
,
secondly
,
equated
if
the
double
distance
be
lesse
then
180
deg.
to
be
subtracted
,
when
it
is
more
.
Therefore
in
our
Example
,
As
the
Radius
|
|
|
To
the
sine
of
H
F
D
|
46.
78356
|
9.
8625917
|
So
is
the
sine
of
|
67500
|
8.
0711591
|
To
the
variation
|
49189
|
7.
9337508
|
The
Moones
place
secondly
equated
-
86.
75830
Variation
subtract
-
49189
The
Moones
place
in
her
Orbe
-
86.
26641
Lastly
,
to
find
the
Moones
latitude
and
place
in
the
Ecliptique
,
take
the
middle
motion
of
her
latitude
for
the
time
propounded
,
the
which
rejecting
the
whole
circles
is
270.
76067
and
in
which
according
to
Tycho
there
is
a
twofold
variation
,
The
first
is
occasioned
by
the
various
intersection
of
the
Moones
orbe
with
the
Zodiack
,
and
the
other
by
the
reciprocal
progression
and
retrogradation
of
the
Nodes
.
In
the
New
and
Full
Moones
the
limits
of
her
greatest
latitude
,
are
4.
97500
,
but
in
her
quarters
5.
29167
,
as
Tycho
hath
experimented
by
many
diligent
and
accurate
observations
,
whose
method
of
calculation
is
as
followeth
.
From
the
meane
motion
of
the
Moones
latitude
-
270.
76067
Subtract
the
Moones
absolute
Equation
-
4.
694
6
The
Equated
latitude
of
the
Moon
-
266.
06631
Then
to
find
the
Equation
of
the
Nodes
,
let
the
line
A
D
or
the
angle
A
T
D
5.
13333
represent
the
meane
inclination
of
the
Moones
Orbe
with
the
Ecliptique
,
let
the
least
inclination
be
represented
by
A
B
4.
97500
,
and
the
greatest
by
A
C
5.
29166.
And
from
the
distance
of
the
Sun
and
Moon
before
found
,
-
293.
39178
Subtract
the
evection
and
variation
-
●●1
.
●9884
True
distance
of
the
Sun
and
Moon
-
292.
19●94
The
double
thereof
is
-
224.
385●8
which
being
numbred
from
B
by
C
to
F
in
the
Triangle
F
D
A
we
have
known
A
D
5.
13333
the
meane
inclination
of
the
Moones
Orbe
.
2.
D
F
15833
the
halfe
difference
of
the
least
and
greatest
inclination
.
3.
With
the
angle
F
D
A
135.
61412
,
the
complement
of
the
double
distance
of
the
Sun
and
Moon
to
a
whole
circle
:
whence
to
find
the
angle
F
A
D
the
Aequation
of
the
Nodes
,
by
the
Doctrine
of
spherical
Triangles
say
,
First
,
As
the
Radius
|
|
|
To
the
cosine
of
F
D
C
|
44.
38588
|
9.
8540905
|
So
is
the
tang
.
of
D
F
|
15833
|
7.
4413175
|
To
the
tang
of
D
C
|
11321
|
7.
2954080
|
Adde
the
arch
A
D
|
5.
13333
|
|
Summe
is
A
C
|
5.
24654
|
|
2
As
sine
D
C
|
11321
c●
.
ar
.
|
2.
7046002
|
To
the
sine
A
C
|
5.
24654
|
8.
9611430
|
So
cotang
.
F
D
C
|
44.
38588
|
10.
0093107
|
To
cotang
.
F
A
D
|
1.
21062
|
11.
6750539
|
From
the
Moones
latitude
equated
|
|
266.
06787
|
Equation
Nodes
subtract
|
|
1.
21062
|
True
motion
of
the
Moones
latitude
.
|
|
264.
85725
|
whose
complement
to
a
whole
circle
|
|
95.
14275
|
And
to
find
A
F
or
the
angle
of
〈◊〉
A
T
●
,
the
〈…〉
of
the
Sun
and
Moone
being
more
than
180
degrees
,
and
lesse
th●●●●●
▪
I
deduct
the
Moones
double
distance
224
d.
3858●
being
numbred
●●
the
little
circle
,
from
B
by
C
to
F
,
from
the
Arch
B
C
F
G
270
,
there
remaines
F
G
45.
61412
,
and
then
the
A●alogie
is
,
As
the
Radius
D
C
|
|
10.
0000000
|
To
the
sine
of
F
G
or
D
F
|
45.
61412
|
9.
8540305
|
So
is
the
sine
of
the
arch
,
D
C
|
.
15833
|
7.
4413159
|
To
the
sine
of
the
arch
D
E
|
.
11319
|
7.
2953464
|
The
aggregate
is
B
E
.
27152
which
being
added
to
the
least
angle
of
inclination
A
T
B
,
or
the
arch
A
B
4.
97500
the
present
inclination
is
A
F
or
the
angle
A
T
F
5.
24652.
Hence
to
find
the
Moones
true
latitude
,
I
say
,
As
Radius
|
|
10.
0000000
|
To
the
sine
of
A
T
F
|
5.
24652
|
8.
9611413
|
So
is
the
sine
of
A
T
|
84.
85725
|
9.
9982481
|
To
the
sine
of
A
F
|
5.
22533
|
8.
9593894
|
or
the
angle
A
S
F.
|
|
|
And
by
these
Analogies
may
be
made
the
Table
of
the
Moons
latitude
wc●
we
have
borrowed
of
Tycho
,
converting
it
onely
into
Decimall
numbers
.
)
For
supposing
the
Moon
to
be
in
her
Syzygial
points
,
the
angle
of
Inclination
is
alwayes
A
T
B
4.
975
,
and
then
her
latitude
for
every
degree
of
her
true
motion
of
latitude
may
be
found
by
the
last
Analogie
;
As
Radius
Example
.
To
the
sine
of
A
T
B
|
4.
975
|
8.
9381242
|
So
is
the
sine
of
A
T
|
45.
|
9.
8494850
|
To
the
sine
of
A
B
|
3.
51564
|
8.
7876092
|
And
her
latitude
when
she
is
in
Quadratu●e
or
90
degrees
distant
from
the
Sun
may
be
found
by
the
same
analogie
,
if
you
make
the
angle
of
Inclination
A
T
C
5.
29166.
Example
.
As
Radius
|
|
|
To
the
sine
of
A
T
C
|
5.
29166
|
8.
9648517
|
So
is
the
sine
of
A
T
|
45.
|
9.
8494850
|
To
the
sine
of
A
F
|
3.
73910
|
8.
8143367
|
Subtract
A
B
3.
51564
there
rests
the
Excesse
to
be
placed
in
the
Table
0.
22346.
The
proportipnal
part
of
which
excesse
to
be
added
to
the
Moones
laitude
must
be
found
by
scruples
of
proportion
,
and
the
Scruples
themselves
for
every
degree
of
the
Moones
distance
from
the
Sunne
may
thus
be
had
.
As
Radius
▪
|
|
|
To
the
Co●ine
of
the
Moones
double
distance
D
H
|
40
|
9.
8842539
|
So
is
the
sine
of
D
B
|
0.
15833
|
7.
4413575
|
To
the
sine
of
D
H
|
0.
12142
|
7.
3256114
|
Their
differ
▪
is
B
H
|
0.
03691
|
|
Then
as
the
Diameter
B
C
|
31666
|
5.
4994068
|
Is
to
the
Diameter
B
C
|
100.
000
|
5.
0000000
|
So
is
B
H
|
0.
03691
|
3.
5671440
|
To
B
H
|
0
,
11656
|
4.
0665508
|
Or
more
readily
thus
D
H
76604
is
the
sine
of
50
or
the
Cosine
of
40
the
Moones
double
distance
from
the
Sun
,
which
being
deducted
from
Radius
,
the
remainder
is
the
versed
sine
B
H
23396
the
halfe
11698
,
are
the
scruples
of
proportion
answering
to
20
deg.
of
the
Moones
single
distance
from
the
Sun
,
From
the
Moones
place
in
her
Orbe
-
86.
26641
Subtract
the
Moones
true
latitude
-
264.
85725
The
Moones
Node
ascending
-
181.
40916
Lastly
,
for
her
Reduction
▪
As
Radius
|
|
|
To
the
Cosine
of
A
T
F
|
5.
24775
|
9.
9981757
|
So
tang
.
of
A
T
|
84.
85725
|
11.
0458587
|
To
tang
.
of
T
F
|
84.
83689
|
11.
0440344
|
Difference
02036
is
the
Reduction
sought
From
the
place
in
her
Orbe
-
86.
26641
Subtract
her
Reduction
-
.
02036
The
Moones
place
in
the
Ecliptique
-
86.
24605
CHAP.
10.
To
calculate
the
true
Motion
of
the
Moon
by
Tables
.
HAving
gathered
the
meane
motions
of
the
Moones
Longitude
,
Anomaly
,
and
argument
of
Latitude
,
as
in
the
last
Chapter
,
by
the
Anomaly
find
the
Moones
Eccentrick
equation
,
and
by
that
her
Eccentrick
place
.
2
Apply
her
eccentrick
Equation
acoording
to
the
title
both
to
the
meane
Anomaly
,
and
to
the
motion
or
argument
of
Latitude
;
So
have
you
the
equated
Anomaly
and
motion
of
Latitude
,
first
equated
.
Example
.
Anno
1587
,
August
17
ho.
18.
45640
the
Moons
meane
Longitude
was
before
found
to
be
90.
96077
Her
meane
Anomaly
|
|
46.
11125
|
Her
motion
of
Latitude
|
|
270.
76067
|
Her
Eccentrick
equation
to
|
47
deg.
|
3.
54892
|
|
46
deg.
|
3.
48880
|
Differ
.
|
|
0.
06012
|
Now
then
as
1
deg.
|
|
5.
0000000
|
Is
to
|
06012
|
3.
7790190
|
So
is
|
11125
|
4.
0463000
|
To
|
00668
|
2.
8253190
|
To
the
Equation
of
46
deg.
|
|
3.
48880
|
Adde
the
part
proportionall
|
|
668
|
The
Moones
eccentrick
Equation
|
|
3.
49548
|
which
being
subtracted
from
her
meane
Longitude
,
Anomaly
and
Latitude
.
Her
place
first
equated
is
-
87.
46529
Her
equated
Anomaly
-
42.
61577
Her
Latitude
first
Equated
-
267.
26519
3.
By
her
meane
Anomaly
you
must
also
find
the
Logarithme
o●he●
distance
from
her
umbilique
.
The
Logarithme
to
|
46
Deg.
is
|
5.
0133649
|
|
47
Deg.
is
|
5.
0131474
|
|
Differ
.
|
2175
|
As
if
one
Degree
|
|
5.
0000000
|
Is
to
|
2175
|
3.
3374592
|
So
is
|
11125
|
4.
0463000
|
To
|
241
|
2.
837592
|
Which
being
deducted
from
|
|
5.
0133649
|
The
Moones
Logarithme
is
|
|
5.
0133408
|
4
▪
Subtract
the
true
place
of
the
Sun
,
from
the
Moones
eccentrick
place
,
so
have
you
the
distance
of
the
luminaries
,
with
the
double
whereof
seeke
in
the
Table
the
eccentricity
of
the
Moones
evection
,
and
the
quantity
of
her
variation
or
reflection
,
using
the
part
proportional
if
need
require
.
5
If
the
double
distance
of
the
luminaries
with
which
you
enter
the
Table
be
lesse
then
a
semicircle
,
adde
halfe
the
complement
thereof
unto
the
equated
Anomaly
,
or
if
it
be
more
than
a
semicircle
,
deduct
halfe
the
excesse
above
a
semicircle
from
the
equated
Anomaly
,
then
the
summe
or
difference
if
lesse
then
a
Semicircle
,
is
the
summe
of
the
opposite
angles
,
or
if
it
be
more
,
take
the
complement
,
to
a
whole
circle
.
Example
.
|
☉
Longitude
|
☉
Apogaon
|
Years
1500
|
290.
20076
|
94.
01167
|
80
|
.
59826
|
1.
26342
|
6
|
359.
55205
|
.
09473
|
Iuly
|
208.
95710
|
917
|
Dayes
16
|
15.
77035
|
69
|
Houres
18
|
.
73923
|
|
Parts
4564
|
.
01874
|
|
Suns
Longitude
|
875.
83649
|
95.
37968
|
Apogaeon
subtract
|
095.
37968
|
|
Mean
Anomaly
|
060.
45681
|
|
The
Suns
eccentrick
to
|
61
deg.
|
1.
77254
|
The
Suns
eccentrick
to
|
60
deg.
|
1.
75464
|
Difference
|
|
1790
|
As
one
degree
|
|
5.
0000000
|
Is
to
|
1790
|
3.
2528530
|
So
is
|
45681
|
4.
6597●56
|
To
|
817
|
2.
9125886
|
The
equation
to
60
deg.
|
|
1.
75464
|
Adde
th●
part
proportional
|
|
817
|
The
absolute
Equation
subtract
|
|
1.
76281
|
From
the
Suns
meane
Longitude
|
|
155.
83649
|
The
Suns
true
place
subtract
|
|
154.
07368
|
From
the
Moones
eccentrick
place
|
|
87.
46529
|
The
distance
of
the
Sun
and
Moon
|
|
293.
39161
|
The
double
distance
|
|
226.
78322
|
The
Logarithme
of
the
Eccentrick
to
|
226
deg.
|
3.
6037118
|
The
Logarithme
of
the
Eccentrick
to
|
227
deg.
|
3.
60208●4
|
Difference
|
|
16284
|
As
one
degree
|
|
5.
|
Is
to
|
16284
|
4.
2117610
|
So
is
|
78322
|
4.
8938837
|
To
|
12753
|
4.
1056447
|
which
being
subtracted
from
|
|
3.
6037118
|
The
Logarithme
of
the
Eccentricity
is
|
|
3.
6024365
|
Thus
we
have
found
the
Logarithms
of
the
two
lateral
distances
,
namely
of
the
distance
of
the
Moon
from
her
Umbilique
▪
5.
0133408
and
of
the
Eccentricity
of
her
Evectioon
3.
6024365.
And
because
the
double
distance
of
the
Sun
and
Moone
is
more
then
a
semicircle
226.
78322
Deduct
180
there
rests
-
46.
78322
The
halfe
whereof
-
23
39161
Deduct
from
the
equated
Anomaly
-
42.
61577
There
resteth
the
Synodical
Anomaly
-
●9
.
22416
or
summe
of
the
opposite
angles
.
Hence
to
find
the
Evection
,
Say
,
As
the
greater
Logarithme
co
▪
ar
.
-
4.
9866502
Is
to
the
lesser
-
3.
6024365
So
is
Radius
-
●0
.
000●●●0
To
the
tang
.
of
2
deg.
22494
-
8.
●890●●●
Adde
45
2
Operation
.
As
Radius
|
|
10.
0000000
|
To
the
Cotang
.
|
47.
22494
|
9.
9662367
|
So
tang
.
half
sum
|
9.
61208
|
9.
2287955
|
To
the
tang
.
half
dif
.
|
8.
90555
|
9.
1950324
|
Evection
0.
70653.
Subtract
because
the
Synodical
anomaly
is
lesse
then
180.
Then
for
the
variation
,
I
enter
the
Table
with
the
Moones
double
distance
22678322
;
and
using
the
part
proportional
I
find
it
to
be
0.
49186
subtract
.
The
Evection
to
be
subtracted
is
-
0.
70653
The
Variation
to
be
subtracted
is
-
.
49186
Their
summe
-
1.
19839
Subtract
from
the
Moons
Eccentrick
place
-
87.
46529
There
rests
the
Moones
place
in
her
Orbe
-
86.
26690
Otherwise
thus
.
Enter
the
Table
entit●led
Bullialdus
his
Table
of
Evections
,
with
the
Synodical
Anomaly
19.
224●6
,
and
using
the
part
proportional
you
shall
find
it
to
be
75942
,
subtract
,
then
take
from
thence
with
the
Moones
double
distance
226.
78322
the
scruples
of
proportion
also
,
the
which
observing
the
part
proportional
are
;
91781
,
by
which
the
Evection
found
is
thus
to
be
corrected
.
As
1
degree
is
to
75942
:
so
is
91781
to
the
Evection
sought
69701
subtract
,
and
this
subtracted
according
to
the
title
from
the
Moones
Eccentrick
place
87.
46529
There
rests
her
place
secondly
equated
86.
75828
Then
for
the
variation
enter
this
or
the
former
table
with
the
Moons
double
distance
226.
78322
,
and
using
the
part
proportional
you
shall
find
it
to
be
49186
as
before
,
with
the
title
Subtract
;
and
therefore
the
Moons
place
in
her
Orbe
86.
26742
,
but
little
differing
from
the
former
.
And
from
these
two
Tables
of
Evection
&
Variation
,
we
have
composed
a
third
Table
,
entituled
a
compounded
Table
of
the
Moones
second
and
third
inequalities
,
for
the
yet
more
speedy
finding
of
these
equations
,
whose
construction
is
thus
.
First
,
enter
the
Table
of
the
Moones
Evection
with
her
Synodicall
Anomaly
to
find
the
Evection
,
and
with
her
double
distance
from
the
Sun
to
find
the
Scruples
of
proportion
.
The
Moones
double
distance
may
be
supposed
,
and
the
Synodicall
anomaly
easily
made
from
it
:
as
suppose
the
Moones
distance
from
the
Sun
to
be
3
degrees
thereof
to
a
quadrant
87
being
added
to
the
equated
Anomaly
which
you
may
also
suppose
to
be
what
you
please
,
will
give
the
synodical
Anomaly
,
thus
if
the
equated
Anomaly
be
25
,
adde
87
thereto
,
and
then
the
synodical
Anomaly
will
be
112
and
the
angle
of
evection
answering
thereto
2.
38861
subtract
:
and
the
scruples
of
proportion
answering
to
6
degrees
52335.
Hence
to
finde
the
true
evection
.
I
say
,
As
1
degree
|
1.
00000
|
0.
0000000
|
Is
to
the
evection
in
the
table
|
2.
38861
|
0.
3781452
|
So
are
the
Scruples
of
prop.
|
0.
52335
|
1.
7187912
|
To
the
evection
sought
,
Sub.
|
0.
12500
|
0.
0969374
|
The
variation
to
6
deg.
adde
|
0.
07055
|
|
And
because
the
evection
and
variation
are
of
different
titles
therefore
there
difference
05445
is
the
compounded
equation
sought
,
which
I
place
against
25
degrees
of
equated
Anomaly
in
the
columne
of
3
deg.
of
the
Moons
distance
from
the
Sun
,
with
the
title
subtract
,
because
the
title
of
the
biggest
number
was
subtract
.
And
therefore
the
second
and
the
third
inequalities
of
the
Moone
are
to
be
found
in
this
table
by
entering
it
with
the
Moones
distance
from
the
Sun
in
the
head
or
foot
of
the
table
,
and
with
the
equated
Anomaly
in
the
sides
,
for
so
the
common
angle
using
the
part
proportional
,
if
need
require
,
will
give
the
Equation
sought
to
be
applyed
to
the
Moons
place
according
to
the
title
.
Example
.
Suppose
the
distance
of
the
Sun
and
Moone
were
293.
39161
,
that
is
,
deducting
a
Semicircle
113.
39161
,
and
her
equated
Anomaly
42.
61577
because
the
Moones
distance
from
the
Sun
is
found
in
the
bottome
of
the
table
,
I
number
the
equated
Anomaly
in
the
first
columne
towards
the
right
hand
ascending
,
and
in
the
common
angle
(
by
making
proportion
for
the
intercepted
degrees
)
I
finde
the
Equation
to
be
1.
188
S.
that
is
the
title
subtract
,
and
therefore
this
Equation
being
subtracted
from
the
Moones
Eccentrick
place
,
what
remaineth
shall
be
the
Moones
place
in
her
orbe
.
CHAP.
11.
To
finde
the
Moones
true
Latitude
and
place
in
the
Ecliptick
.
TO
the
motion
of
Latitude
first
Equated
,
according
to
the
title
apply
the
agregate
of
the
Moones
second
and
third
Equations
,
so
have
you
the
motion
of
Latitude
secondly
equated
.
2.
To
the
distance
of
the
Sun
and
Moon
before
found
apply
the
agregate
of
the
Moons
2
d.
and
●
d.
Equations
according
to
the
title
,
so
have
you
the
true
distance
of
the
Sun
and
Moone
.
Example
.
Motion
of
Latitude
first
equated
-
267.
26519
Second
and
third
equations
sub
.
-
1.
19839
Motion
of
Latitude
secondly
equated
-
266.
06680
Distance
of
the
Sun
and
Moone
as
before
-
293.
39161
Second
and
third
Equation
sub
.
-
1.
19839
True
distance
of
the
Sun
and
Moone
-
292.
19322
3.
With
the
true
distance
of
the
Sun
and
Moon
enter
the
table
of
the
Equation
of
the
Nodes
,
and
take
thence
the
Equation
of
the
Nodes
,
which
according
to
the
title
,
apply
to
the
motion
of
Latitude
secondly
Equated
,
and
you
have
the
true
and
absolute
motion
of
Latitude
.
At
the
same
entrance
take
out
also
the
scruples
of
proportion
and
reserve
them
.
Example
.
With
the
true
distance
of
the
Sun
and
Moone
-
292.
19322
I
finde
the
Equation
of
the
Nodes
,
subt
.
-
1.
21103
From
the
Latitude
secondly
Equated
-
266.
06680
So
the
true
motion
of
Latitude
-
264.
85577
And
the
scruples
of
proportion
-
85729
4.
With
the
true
motion
of
Latitude
enter
the
table
of
Latitude
,
and
thence
take
the
Latitude
and
the
excesse
,
then
say
,
as
one
degree
to
the
excesse
in
the
table
:
so
are
the
scruples
of
proportion
before
reserved
,
to
the
excess
sought
,
which
being
added
to
the
Latitude
found
by
the
table
,
the
summe
shall
be
the
true
Latitude
of
the
Moone
,
which
is
North
when
the
true
motion
of
Latitude
is
lesse
then
6
●ignes
,
and
South
when
it
is
more
.
Example
.
The
true
motion
of
Latitude
|
|
264.
85577
|
Gives
The
Latitude
|
|
4.
95490
|
Gives
The
Excess
|
|
31519
|
As
one
degree
|
1.
00000
|
|
To
the
excess
in
the
table
|
31519
|
1.
4985724
|
So
the
scrupls
of
propor
.
|
85729
|
1.
9331277
|
To
the
excess
desired
|
27020
|
1.
4317001
|
Which
being
added
to
the
Latitude
|
|
4.
95490
|
It
gives
the
true
Latitude
▪
South
|
|
5.
22510
|
5.
If
the
true
motion
of
Latitude
be
subtracted
from
the
Moones
true
place
in
her
orbe
,
there
resteth
the
Node
Ascendant
.
As
The
Moones
true
place
-
86.
26690
As
The
Motion
of
Latitude
subt
.
-
264.
85577
The
Node
ascendant
-
181.
41113
6.
With
the
true
motion
of
Latitude
enter
the
table
of
Reduction
,
and
take
out
the
Reduction
,
which
according
to
the
title
apply
to
the
Moones
place
in
her
orbe
,
you
have
her
true
place
in
the
Ecliptick
.
Example
.
The
true
motion
of
Latitude
-
264.
85577
Gives
the
reduction
to
be
subt
.
-
02067
From
the
Moones
place
in
her
Orbe
-
86.
26690
The
Moons
place
in
the
Ecliptick
-
86.
24623
CHAP.
12.
Of
the
motion
of
the
fixed
Starres
.
THe
motions
of
the
fixed
Stars
are
by
the
observations
of
all
ages
found
to
be
equall
,
and
the
quantity
of
that
equal
motion
,
Noble
Tycho
by
comparing
his
owne
observations
with
those
of
the
ancients
hath
determined
to
be
exact
51
seconds
,
Bullialdus
50″
55‴
ferè
,
and
the
place
of
the
first
Star
in
Aries
in
the
yeare
of
our
Lord
1600
compleat
to
be
27
deg.
37
min.
which
being
converted
into
Decimalls
the
Radix
of
the
middle
motions
of
the
fixed
Stars
at
that
time
will
be
27.
61667
and
the
yearely
motion
.
01414
,
that
is
the
decimall
of
50
seconds
55
thirds
.
Hence
to
finde
their
places
at
any
time
assigned
,
we
have
exhibited
a
table
of
the
longitudes
and
latitudes
of
some
of
the
most
notable
fixed
Stars
for
the
yeare
1650
compleat
,
which
by
the
motions
of
the
fixed
stars
in
the
tables
of
the
Suns
meane
motions
,
may
thus
be
done
for
any
other
time
.
Take
the
difference
betweene
the
time
given
and
1650
compleat
,
and
the
motion
agreeing
to
that
difference
,
this
motion
subtract
from
the
place
in
the
table
when
the
time
given
is
before
1650
,
or
else
adde
it
,
and
you
have
the
place
desired
.
The
Latitudes
and
Magnitudes
are
still
the
same
.
Example
.
The
time
given
1683
September
.
Difference
from
1650
compl
.
-
33
yeare
8
Moneths
Motion
Correspondent
-
.
476●8
Place
of
Oculus
♉
1650
compl
.
-
♊
4.
91667
Place
required
-
♊
5.
39305
Latitude
South
-
5.
51667
CHAP.
13.
Of
the
Motion
of
Saturne
.
OUr
Tables
of
Saturns
meane
motions
as
of
the
other
Planets
,
are
he
same
with
those
of
Bullialdus
,
being
onely
reduced
to
the
Meridian
of
London
,
and
converted
into
Decimall
numbers
whose
eccentrick
being
so
easie
to
be
found
,
and
the
investigation
of
his
true
place
,
with
the
places
of
Jupiter
,
Mars
,
Venus
and
Mercury
,
being
out
of
curiosity
,
rather
then
use
▪
we
shall
content
our selves
with
the
Trigonometricall
calculation
onely
,
first
,
of
Saturne
for
the
time
before
given
1587.
August
17
:
h
:
18
:
4564
:
and
then
of
the
rest
.
An.
Christ
.
|
Longit.
♄
|
Aphel
.
♄
|
Node
♄
|
1500
|
064.
98279
|
262.
82583
|
109.
79361
|
80
|
258.
76528
|
2.
5●694
|
0.
57611
|
6
|
73.
39056
|
.
19078
|
0.
04306
|
Iuly
▪
|
7.
10111
|
.
01833
|
.
00444
|
D
16
|
.
53183
|
.
00139
|
.
00030
|
H
18
|
.
02500
|
|
|
P
4564
|
.
00063
|
|
|
Meane
Mot.
404.
801
20265.
57327110.
41752
Aphel
.
Subt.
265.
57327
Rests
Anomaly
139.
22793
The
halfe
of
Saturns
first
inequality
,
or
his
eccentricity
supposing
the
Semidiameter
of
Saturns
orbe
to
be
100.
000.
is
5774.
and
the
Semidiameter
of
the
Earths
orbe
10480.
As
they
are
computed
by
Bullialdus
,
but
the
Semidiameter
of
the
Earths
orbe
being
before
supposed
to
be
100.
000
the
Semidiameter
of
Saturns
orbe
will
be
954198
,
for
as
10480
is
to
100.
000.
So
is
100.
000.
to
954198.
and
Saturns
eccentricity
in
the
same
parts
will
be
55145.
for
as
100.
000
is
to
954198.
So
is
5774
to
55145.
whose
double
110200
is
the
side
M
H
in
the
figure
following
;
in
the
triangle
therefore
M
E
H
,
we
have
knowne
,
1.
The
Angle
H
M
E
40.
77207
,
the
complement
of
Saturns
Anomaly
to
a
semicircle
●●9
.
22793
or
the
halfe
sum
of
the
angles
M
E
H
and
M
H
E
viz.
halfe
the
anomaly
69.
61396.
2.
The
side
M
E
1908396
To
finde
the
Angle
M
E
H.
3.
The
side
M
H
110290
To
finde
the
Angle
M
E
H.
As
there
sum
|
2●18686
co.
ar
.
|
3.
694931●
|
Is
to
their
differ
.
|
1798106
|
6.
2548153
|
So
is
tang
.
halfe
sum
|
69.
61396
|
10.
4299016
|
To
tang
.
halfe
diff
.
|
67.
●5375
|
10.
3796482
|
Difference
|
2.
26021
is
the
angle
M
E
H.
|
|
Difference
doubled
4.
52042
is
the
angle
M
B
H
,
or
the
Equation
sought
,
to
be
subtracted
from
Saturns
meane
Longitude
,
the
Anomaly
being
lesse
then
a
semicircle
.
Saturns
meane
Longitude
-
44.
80120
Equation
Subt.
-
4.
52042
Saturns
eccentrick
place
-
40.
28078
2.
To
finde
his
distance
from
the
Sun
.
As
the
sine
M
B
H
|
4.
52042
co.
ar
.
|
1.
1034042
|
Is
to
the
side
M
H.
|
110290
|
5.
0425361
|
So
sine
E
M
H
|
40.
77207
|
9.
8149473
|
To
the
side
B
H
|
913876
|
5.
9608876
|
●
.
From
the
eccentrick
place
of
Saturne
subduct
the
Node
,
there
resteth
the
argument
of
Latitude
:
by
help
whereof
and
the
angle
of
his
greatest
inclination
,
which
according
to
Bullialdus
is
2
d.
50
,
or
4362
,
we
may
easily
finde
his
Reduction
,
but
the
side
E
B
4362
in
the
parts
of
100.
000
,
must
be
reduced
into
the
parts
of
Saturns
semidiameter
954198
,
to
finde
the
curtation
.
As
100.
000
is
to
954198
,
so
is
4●62
to
41622.
Saturns
eccentrick
place
Example
.
-
40.
28078
Node
substract
-
110.
41752
Argument
of
Latitude
-
289.
86326
Whose
complement
is
K
L
-
70.
13674
As
Radius
|
|
|
To
cosine
of
XKL
|
2.
50
|
9.
999586
|
So
is
tang
.
of
KL
|
70.
13674
|
10.
4421682
|
To
tang
.
of
|
70.
12929
|
10.
4417546
|
Whose
difference
|
.
01745
is
the
Reduction
sought
:
|
|
And
to
be
subtracted
from
the
ecceutrick
place
,
if
he
move
from
either
Node
towards
the
limits
of
his
greatest
latitude
,
but
if
he
depart
from
the
limits
and
approach
towards
the
Nodes
the
reduction
is
to
be
added
,
for
so
the
sum
or
difference
will
be
the
place
in
the
Ecliptique
.
As
in
our
example
,
Saturne
is
past
the
limits
of
his
greatest
latitude
,
and
is
approaching
towards
his
Node
,
and
therefore
the
reduction
is
to
be
added
.
Saturns
eccentrick
place
-
40.
28078
Reduction
adde
-
.
01745
The
eccentrick
reduced
-
40.
29823
The
inclination
of
his
orbite
from
the
eccliptique
represented
in
the
second
figure
following
by
the
line
XL
,
may
thus
be
found
.
As
the
Radius
KE
90.
|
|
|
To
the
greatest
in
clination
EB
|
41622
|
4.
6193229
|
So
is
the
sine
of
KL
|
70.
13674
|
9.
9733616
|
To
the
side
XL
|
●9145
|
4
▪
5926845
|
which
is
the
inclination
agreeing
to
the
common
Radius
954
198
,
whereas
the
distance
of
Saturne
from
the
sun
is
to
be
put
for
the
Radius
,
and
then
XL
will
be
but
37491.
As
DL
|
954198
co.
ar
.
|
4.
0203616
|
To
AL
|
913876
|
5.
9608876
|
So
is
XL
|
39145
|
4.
5926845
|
To
XL
|
37491
|
4.
5739336
|
The
distance
of
Saturne
in
his
orbite
from
the
Sun
being
given
with
the
inclination
of
his
orbite
from
the
eccliptique
,
the
distance
corrected
by
curtation
may
thus
be
found
.
To
finde
Saturns
second
inequality
.
Subtract
the
suns
place
from
the
Eccentrick
reduced
,
or
this
from
it
,
so
that
lesse
then
180
degrees
may
remain
,
this
remainer
is
the
Anomaly
of
the
orbe
,
the
complement
whereof
is
the
angle
NAS
,
or
the
halfe
,
is
the
halfe
sum
of
the
opposite
angles
.
Example
▪
Saturns
Eccentrick
reduced
|
|
40.
29823
|
The
Suns
true
place
|
|
154.
07347
|
The
Anomaly
of
the
orbe
|
|
113.
77524
|
The
angle
NAS
|
|
66.
22476
|
The
halfe
Anomaly
|
|
56.
88762
|
As
the
greater
side
AN
|
913107
co
,
ar
.
|
4.
0394781
|
Is
to
the
less
AS
|
100895
|
5.
0038707
|
So
Radius
,
to
the
tang
.
|
6.
30541
|
9.
0433488
|
Adde
|
45.
|
|
As
Radius
|
|
|
To
cotang
.
of
the
summe
|
51.
30541
|
9.
9036304
|
So
is
tang
.
halfe
summe
|
56.
88762
|
10.
1856192
|
To
tang
.
halfe
differ
.
|
50.
84613
|
10.
0892496
|
Summe
|
107.
73375
Angle
ASN
.
|
|
Differ
.
|
6.
04149
Angle
ANS.
|
|
Because
Saturne
eccentrick
reduced
was
subtracted
from
the
Suns
true
place
,
therefore
the
angle
of
his
Elongation
ASN
107.
73375
must
be
subtracted
also
,
and
then
Saturns
place
will
be
125.
19972.
To
finde
Saturns
distance
from
the
Earth
.
To
find
the
latitude
of
♄
from
the
earth
.
|
|
|
As
SX
.
|
877291
Co.
ar
.
|
4.
0568610
|
Is
to
Radius
|
|
10.
0000000
|
So
is
XL
,
|
37491
|
4.
5739336
|
To
the
tang
.
of
XSL
,
|
2.
44711.
|
8.
6307946
|
which
is
the
quantity
of
♄
Southern
Latitude
,
because
the
argument
of
Latitude
was
more
then
6
signes
,
when
it
is
less
then
6
signs
,
the
Latitude
is
North.
CHAP.
14.
Of
the
Motion
of
Jupiter
.
THe
investigation
of
the
place
of
this
and
the
other
Planets
,
is
well
nigh
the
same
with
that
of
♄
,
they
differ
more
in
the
Dimensions
of
their
Orbs
,
then
in
the
manner
of
their
calculation
.
Yet
that
there
may
be
no
mistake
,
we
will
not
onely
give
you
the
Dimensions
of
their
several
orbs
,
but
makes
examples
of
their
calculation
to
the
former
time
given
.
The
meane
motions
of
Jupiter
.
Anno
Christi
.
|
Longitude
♃
|
Aphelion
♃
|
Node
♃
|
Years
1500
|
004.
50000
|
185.
54833
|
097.
93889
|
80
|
269.
04444
|
1.
98000
|
.
54722
|
6
|
182.
13667
|
.
14833
|
.
04111
|
Iuly
|
17.
62333
|
.
01444
|
.
00389
|
Dayes
16
|
1.
33000
|
.
00104
|
.
00030
|
Houres
18
|
.
06222
|
|
|
Parts
4564
|
.
00154
|
|
|
Mean
motion
.
|
474.
69820
|
187.
68414
|
98.
53141
|
Aphelion
subt
.
|
187.
68414
|
Rests
Anomaly
|
287.
01406
|
The
halfe
of
Jupiter
first
inequality
,
or
his
eccentricity
,
supposing
his
Semidiameter
to
be
100.
000
by
the
computation
of
Bullialdus
is
4856
,
and
the
Semidiameter
of
the
Earths
Orbe
19138
▪
and
therefore
to
find
♃
Semidiameter
,
when
the
Semidiameter
of
the
Earths
Orb
is
100.
000
the
proportion
is
,
As
19138.
Is
to
100.
000
so
is
100.
000
to
522520
,
the
Semidiameter
required
,
which
being
doubled
is
the
Diameter
of
the
Ellipsis
,
or
the
side
M
E
1045040.
And
to
find
the
eccentricity
in
the
same
parts
,
say
,
as
100.
000.
is
to
522520
,
so
is
4856
,
to
2537●
.
the
excentricity
,
and
the
double
thereof
.
50746
is
M
H
the
distance
of
the
foces
.
The
complement
of
the
Anomaly
72.
98594
is
the
angle
A
M
E
,
and
the
halfe
compl
:
36.
49297
,
is
the
halfe
sum
of
the
opposite
angles
M
E
H
and
M
H
E.
The
side
M
E
|
1045040
|
|
The
side
M
H
|
50746
|
|
Sum
|
1095786
co.
ar
.
|
3.
9602744
|
Differ
|
994294
|
5.
9975147
|
Tang.
½
sum
|
36.
49297
|
9.
8690974
|
Tang.
½
differ
|
33.
87162
|
9.
8268865
|
Differ
.
|
2.
62135
Angle
M
E
H
|
|
Double
differ
.
5.
24270
Angle
M
B
H
or
the
Equatiō
sought
,
and
to
be
added
to
Jupiters
mean
longitude
,
the
Anomaly
being
above
a
Semi-circle
.
Jupiters
mean
longitude
-
114.
69820
Equation
add
-
5.
24270
Jupiters
eccentrick
place
-
119.
94090
The
Node
subt
.
-
98.
53141
Argument
of
Latitude
.
-
21.
40949
By
help
whereof
,
and
the
angle
of
his
greatest
inclination
1.
36333
,
or
E
B
2379
,
we
may
find
the
reduction
:
but
to
find
the
parts
of
inclination
in
proportion
to
the
given
Radius
522520.
say
,
As
100.
000.
is
to
2379
,
so
is
522520
to
13619.
the
parts
required
.
To
finde
the
Reduction
.
As
Radius
|
|
|
So
Cosine
of
X
K
L.
|
1.
36333
|
9.
9998770
|
To
tang
.
of
K
L.
|
21.
40949
|
9.
5933823
|
To
tang
.
of
X
K.
|
21.
40353
|
9.
5932593
|
Differ
.
|
.
00596
Reduction
.
|
|
Because
the
argument
of
Latitude
is
lesse
then
90.
the
Reduction
must
be
subtracted
from
the
eccentrick
place
.
Jupiters
eccentrick
place
-
119.
94090
Reduction
subt
.
-
.
00596
Eccentrick
reduced
-
119.
93494
To
find
the
Inclination
.
As
Radius
|
|
|
To
the
greatest
inclin
:
E
B
|
13619
|
4.
1341452
|
So
is
the
sine
of
K
L
|
21.
40949
|
9.
5623296
|
To
the
side
X
L
|
4971
|
3.
6964748
|
which
are
the
parts
of
inclination
agreeing
to
the
common
Radius
522520
,
but
the
distance
of
♃
from
the
☉
is
to
be
put
for
the
Radius
,
the
which
distance
may
thus
be
found
.
As
the
sine
of
M
B
H
|
5.
24270.
Co.
ar
.
|
1.
0391741
|
Is
to
the
side
M
H
|
50746
|
4.
7054018
|
So
is
the
fine
of
E
M
H
|
72.
98594
|
9.
9805636
|
To
the
side
B
H
|
531055
|
5.
7251395
|
As
the
common
Rad:
D
L
|
522520
Co.
ar
.
|
4.
2818971
|
To
Jupiter
dist:
B
H
or
A
L
|
531055
|
5.
7251●95
|
So
is
X
L
|
4971
|
3.
69647●8
|
To
X
L
|
5052
|
3.
7035114
|
To
finde
♃
distance
corrected
by
Curtation
.
As
A
L
|
531055
Co.
ar
.
|
4.
2748605
|
Is
to
Radius
|
|
10.
0000000
|
So
is
L
X
|
5052
|
3.
7035114
|
To
the
sine
of
L
A
X
|
54515
|
7.
9783719
|
As
the
Radius
|
|
|
To
A
L
|
531055
|
5.
7251395
|
So
is
the
Cosine
of
L
A
X
|
|
9.
9999831
|
To
A
X
|
531034
|
5.
7251226
|
To
finde
the
second
inequality
of
Jupiter
.
We
must
have
given
,
1.
The
angle
N
A
S
,
which
is
to
be
found
by
subducting
the
☉
place
from
♃
eccentrick
or
reduced
,
or
this
from
it
,
so
that
lesse
then
6
Signes
may
remain
,
this
remainer
is
the
Anomaly
of
the
Orbe
,
and
the
Complement
thereof
is
the
Angle
N
A
S
,
or
the
halfe
is
the
halfe
sum
of
the
opposite
angles
.
Example
.
Jupiters
eccentrick
reduced
-
119.
9●494
☉
True
place
-
154.
07347
The
Anomaly
of
the
Orbe
-
34.
13893
The
angle
N
A
S
-
14.
86107
The
halfe
Anomaly
is
-
17.
0694●
These
given
with
the
sides
N
A
and
A
S
,
I
say
.
As
the
greater
side
A
N
|
531034
co
.
ar
.
|
4.
2748774
|
Is
to
the
Radius
|
|
10.
0000000
|
So
is
the
lesser
side
A
S
|
100895
|
5.
0038707
|
To
the
tang
.
of
|
10.
75783
|
9.
2787481
|
Adde
|
45
|
|
As
Radius
to
Co-tang
.
|
55.
75783
|
9.
8329403
|
So
tang
:
of
½
summe
|
17.
06946
|
9.
4872186
|
To
tang
.
of
½
differ
.
|
11.
80522
|
9.
3201589
|
Summe
|
28.
87468
Angle
A
S
N
|
|
Differ
.
|
5.
26424
Angle
A
N
S
|
|
Because
♃
eccentrick
reduced
was
subtracted
from
the
☉
place
,
therefore
the
angle
of
his
Elongation
A
S
N
28.
87468
must
be
subtracted
also
,
and
so
♃
place
125.
19839.
To
finde
the
distance
of
Jupiter
from
the
Earth
.
As
the
sine
of
A
N
S
|
5.
26424
co.
ar
.
|
1.
0373985
|
To
the
side
A
S
|
100895
|
5.
0038707
|
So
the
sine
of
N
A
S
|
34.
13893
|
9.
7491287
|
To
the
side
S
N
|
603111
|
5.
7803979
|
To
finde
the
latitude
of
♃
from
the
Earth
.
As
the
side
S
N
|
603111
co
.
ar
.
|
4.
2196021
|
To
the
side
X
L
|
5052
|
3.
7035114
|
So
is
the
Radius
|
90
|
10.
0000000
|
To
tang
.
of
X
S
L
|
0.
47998
|
7.
9231135
|
Which
is
the
quantity
of
♃
Northern
latitude
.
|
|
|
CHAP.
15.
Of
the
Motion
of
Mars
.
THere
being
no
other
variety
in
calculating
the
place
of
this
Planet
,
then
what
hath
been
already
shewed
,
in
the
motions
of
Saturn
and
Jupiter
,
we
will
proceed
in
the
same
method
,
and
gather
first
the
middle
motions
for
the
former
time
given
,
and
then
shew
the
Dimensions
of
his
several
Orbs
,
as
we
shall
have
occasion
for
them
.
An.
Christ
.
|
Longit.
♂
|
Aphel
.
♂
|
Node
♂
|
Yeares
1500
|
245.
61611
|
146.
80916
|
45.
40250
|
80
|
193.
32778
|
1.
75139
|
1.
07194
|
6
|
68.
23750
|
.
13139
|
.
08028
|
Iuly
|
111.
10306
|
.
01250
|
.
00778
|
D
16
|
8.
38500
|
.
00096
|
.
00057
|
H
18
|
.
39306
|
|
|
P
4564
|
.
00996
|
|
|
Mean
Mot.
|
627.
07247
|
148.
70540
|
046.
56307
|
Aphel
.
Subt.
|
14.
70540
|
Rest
Anom
.
|
●19
.
36707
|
Supposing
the
Semidiameter
of
♂
his
Orbe
-
100.
000
His
eccentricity
according
to
Bullialdus
is
-
9239
Semidiameter
of
the
Earths
Orbe
-
656●8
The
sine
of
his
angle
of
Inclination
-
3230
And
the
Arch
answering
thereto
-
1.
85111
And
therefore
suppose
the
R.
of
the
Earths
Orbe
-
100.
000
The
eccentricity
of
Mars
will
be
-
14075
The
Semidiameter
of
his
Orbe
-
152350
The
parts
of
his
greatest
Inclination
-
4921
The
Anomaly
119.
36707
is
the
angle
A
M
E
in
the
Ellipsis
of
the
13
Chapter
,
and
therefore
the
halfe
of
it
is
the
halfe
sum
of
the
angles
M
E
H
and
M
H
E
59.
68353
2.
The
side
M
E
|
304700
|
|
3.
The
side
M
H
|
28150
|
|
Summe
|
332850
co.
ar
.
|
4.
4777515
|
Differ
.
|
276550
|
5.
4417726
|
Tang
:
½
sum
|
59.
68353
|
10.
2330382
|
Tang
:
½
diff
:
|
54.
86290
|
10.
1525633
|
Difference
|
4.
82063
Angle
M
E
H
|
|
Difference
doubled
9.
64126
Angle
M
B
H
or
the
Aequation
sought
,
and
to
be
subtracted
from
the
planets
mean
Longitude
,
because
the
Anomaly
is
lesse
then
180.
viz.
119.
36707.
Meane
longitude
of
Mars
-
267.
07247
Aequation
subtracted
-
9.
64126
Mars
his
Eccentrick
place
-
257.
43121
To
finde
his
distance
from
the
Sun
.
As
the
sine
of
M
B
H
|
9.
64126
co.
ar
.
|
0.
7760404
|
To
the
side
M
H
|
28150
|
4.
4494783
|
So
is
the
sine
of
M
B
H
|
60.
63293
|
9.
9402403
|
To
the
side
B
H
|
146473
|
5.
1657590
|
From
the
eccentrick
place
sub
:
Node
.
|
|
119.
36707
|
Argument
of
Latitude
|
|
128.
06614
|
Whose
complement
is
K
L
|
|
51.
93586
|
To
finde
the
Reduction
.
As
the
sine
of
90
|
|
|
To
Cosine
the
great
inclin
:
X
K
L
|
1.
85111
|
9.
9997732
|
So
tang
:
of
K
L
|
51.
93494
|
10.
1061739
|
To
tang
:
of
X
K
|
51.
92039
|
10.
1059471
|
Differ
.
|
.
01455
Reduction
.
|
|
Because
the
argument
of
Latitude
is
more
then
90
,
the
Reduction
must
be
added
to
the
Eccentrick
place
.
The
eccentrick
place
of
Mars
-
257.
43121
Reduction
adde
-
.
01455
Eccentrick
reduced
-
257.
44576
To
finde
the
present
inclination
.
As
Radius
|
|
|
To
the
greatest
inclin
:
E
B
|
9921
|
3.
6920533
|
So
the
sine
of
K
L
|
51.
93494
|
9.
8960878
|
To
the
X
L
|
3873
|
3.
5881411
|
Which
are
the
parts
of
inclination
agreeing
to
the
common
Radius
,
152350.
But
the
distance
of
Mars
from
the
☉
is
to
be
put
for
the
Radius
,
and
then
the
parts
of
inclination
will
be
3724.
For
as
common
Radius
D
L
|
152350
co.
ar
.
|
4.
8171576
|
To
♂
distance
B
H
or
A
L
|
146473
|
5.
1657590
|
So
is
X
L
|
3873
|
3.
5881411
|
To
X
L
|
3724
|
3.
5710577
|
To
finde
the
distance
of
♂
corrected
by
Curtation
.
As
A
L
|
146473
co.
ar
.
|
4.
8342410
|
To
Radius
|
|
10.
0000000
|
So
is
X
L
|
3724
|
3.
5710577
|
To
the
sine
of
L
A
X
|
1.
45703
|
8.
4052987
|
As
Radius
|
|
|
To
A
L
|
146473
|
5.
1657590
|
So
Cosine
of
L
A
X
|
1.
45703
|
9.
9998596
|
To
A
X
|
146426
|
5.
1656186
|
To
finde
the
second
inequality
of
Mars
.
We
must
have
given
,
1
The
angle
N
A
S
,
which
is
to
be
found
by
subducting
the
☉
place
from
the
eccentrick
of
♂
reduced
,
or
this
from
it
,
so
that
lesse
then
6
signes
may
remain
,
this
remainer
is
the
Anomaly
of
the
Orbe
,
and
the
complement
thereof
is
the
angle
N
A
S
,
or
the
halfe
,
is
the
halfe
sum
of
the
opposite
angles
.
Example
.
The
eccentrick
of
♂
reduced
-
257.
44576
The
☉
true
place
-
154.
07347
Anomaly
of
the
Orbe
-
103.
37229
Complement
is
N
A
S
-
76.
62771
Halfe
Anomaly
-
51.
68614
These
given
with
the
sides
N
A
&
S
A
,
I
say
,
As
the
greatest
side
N
A
|
146426
co.
ar
.
|
4.
8343814
|
Is
to
Radius
|
|
10.
0000000
|
So
is
the
lesser
side
S
A
|
100895
|
5.
0038707
|
To
the
tang
.
of
|
34.
56887
|
9.
8382521
|
Adde
|
45
|
|
As
Radius
|
|
|
To
co-tang
.
|
79.
56887
|
9.
2650444
|
So
tang
.
●/●
sum
|
51.
68614
|
10.
1022929
|
To
tang
.
½
diff
.
|
13.
11536
|
9.
3673373
|
Summe
|
64.
80150
angle
A
S
N
|
|
Differ
.
|
38.
57078
angle
A
N
S
|
|
Because
the
Suns
place
was
subtracted
from
the
eccentrick
of
♂
reduced
,
therefore
the
angle
of
elongation
A
S
N
64.
80150
must
be
added
to
the
☉
place
154.
07347
and
then
the
place
of
♂
will
be
218.
87497.
To
finde
the
distance
of
♂
from
the
Earth
.
As
the
sine
of
A
N
S
|
38.
57078
co.
ar
.
|
0.
2051769
|
To
the
side
A
S
|
100895
|
5.
0038707
|
So
sine
of
N
A
S
|
76.
62771
|
9.
9880626
|
To
the
S
N
|
157438
|
5.
1971102
|
To
finde
the
latitude
of
Mars
from
the
Earth
.
As
the
side
S
X
|
157438
co.
ar
.
|
4.
8028898
|
To
the
side
X
L
|
3724
|
3.
5710577
|
So
is
Radius
|
|
10.
0000000
|
To
tangent
of
X
S
L
|
0.
13560
|
7.
3739475
|
Which
is
the
quantity
of
♂
his
Northern
latitude
.
|
|
|
CHAP.
16.
Of
the
Motion
of
Venus
.
HAving
done
with
the
three
superiour
Planets
,
Saturn
,
Iupiter
and
Mars
,
we
come
to
the
two
inferiour
,
Venus
and
Mercury
,
the
investigation
of
whose
places
is
much
after
the
same
manner
with
the
former
,
the
difference
is
in
the
second
inequality
,
occasioned
by
their
motion
under
the
earth
,
the
Orbs
of
the
other
Planets
being
above
it
;
that
this
difference
may
be
the
better
discerned
,
we
have
added
an
example
in
each
for
the
time
before
given
.
The
meane
motions
of
Venus
.
An.
Christ
.
|
Longit.
♀
|
Aphel
.
♀
|
Node
♀
|
Yeares
1500
|
333.
11667
|
303.
97639
|
073.
28944
|
80
|
15.
48472
|
1.
12694
|
.
67028
|
6
|
270.
36028
|
.
08444
|
.
05028
|
Iuly
|
339.
65833
|
.
00806
|
.
00500
|
D
16
|
25.
63472
|
.
00062
|
.
00039
|
H
18
|
1.
20194
|
|
|
P
4564
|
.
03048
|
|
|
Meane
Mot.
|
265.
48714
|
305.
19645
|
074.
01537
|
Aphel
.
Subt.
|
305.
19645
|
Rest
Anom
.
|
320.
29069
|
The
Semidiameter
of
the
orbe
of
Venus
,
in
such
parts
of
which
the
Earths
orbe
is
100.
000
,
by
the
computation
of
Bullialdus
is
72398
,
her
Eccentricity
575.
The
parts
of
her
greatest
inclination
4270.
And
the
angle
it selfe
3.
38111.
In
the
triangle
therefore
of
the
following
Diagram
M
E
H
,
we
have
three
things
given
.
1.
The
halfe
sum
of
the
angles
M
E
H
and
M
H
E
19.
85465
,
viz.
the
halfe
complement
of
the
meane
Anomaly
to
a
circle
.
2.
The
side
M
E
|
144796
|
|
3.
The
side
M
H.
|
1150
|
|
Summe
|
145946
co.
ar
.
|
4.
8358079
|
Difference
|
143646
|
5.
1572934
|
Tang.
halfe
sum
.
|
19.
85465
|
9.
5576273
|
Tangent
halfe
difference
|
19.
56560
|
9.
55072●6
|
Difference
|
.
28905
Angle
M
E
H.
|
|
Differ
.
doubled
|
.
57810
Angle
M
B
H.
|
|
or
the
Equation
to
be
added
to
the
mean
longitude
,
because
the
Anomaly
is
more
then
a
semicircle
.
The
meane
Longitude
of
Venus
-
265.
48714
Equation
adde
-
.
57810
The
eccentrick
place
of
Venus
-
266.
06524
Node
subtract
-
74.
01537
Argument
of
latitude
K
L
-
192.
04987
To
finde
the
distance
of
Venus
from
the
Sun
.
As
the
sine
of
M
B
H
|
0.
57810
co.
ar
.
|
1.
9961373
|
To
the
side
M
H
|
1150
|
3.
0606978
|
So
the
sine
of
H
M
B
|
39.
70931
|
9.
8054279
|
To
the
side
B
H
|
72822
|
4.
8622630
|
To
finde
the
Reduction
.
As
the
Radius
|
|
|
To
the
Cosine
of
the
great
inclina
.
X
K
L
|
3.
38111
|
9.
999●433
▪
|
So
tangent
of
K
L
|
12.
04987
|
9.
329●295
|
To
tangent
of
X
K
|
12.
02950
|
9.
3285728
|
Difference
|
.
02037
|
Reduction
.
|
Because
the
Argument
of
Latitude
is
more
then
180
,
the
Reduction
must
be
subtracted
from
the
eccentrick
place
.
The
eccentrick
place
of
Venus
-
266.
06524
Reduction
subtract
-
.
02037
Eccentrick
reduced
-
266.
04487
To
finde
the
present
inclination
.
As
Radius
|
|
|
To
the
greatest
inclinat
.
E
B.
|
4270
|
3.
6304●78
|
So
sine
of
K
L
|
12.
04987
|
9.
3196533
|
To
the
inclinat
.
X
L
|
891
|
2.
9500811
|
Which
are
the
parts
of
inclination
agreeing
to
the
common
Radius
72398
,
but
the
distance
of
Venus
from
the
Sun
,
72822
being
put
for
Radius
,
the
inclination
will
be
896.
As
the
common
Radius
D
L
|
72398
co.
ar
.
|
5.
1402735
|
To
Venus
distance
B
H
or
A
L
|
72822
|
4.
8622630
|
So
is
X
L
|
891
|
2.
9508115
|
To
X
L
|
896
|
2.
9526176
|
To
finde
the
distance
corrected
by
Curtation
.
As
A
L
|
7●822
co.
ar
.
|
5.
1377370
|
To
Radius
,
so
is
X
L
|
896
|
2.
9526176
|
To
the
sine
of
L
A
X
|
0.
70550
|
8.
0903546
|
As
Radius
|
|
|
To
A
L
|
72822
|
4.
8622630
|
So
Cosine
of
L
A
X
|
07055
|
9.
9999671
|
To
A
X
|
72816
|
4.
8622301
|
To
finde
the
second
inequality
of
Venus
.
We
must
have
given
,
1.
The
angle
N
A
S
which
is
to
be
found
by
subducting
the
Suns
place
from
the
eccentrick
of
Venus
reduced
,
or
this
from
it
so
that
less
then
6
signes
may
remain
,
this
remainer
is
the
Anomaly
of
the
orbe
and
the
complement
is
the
angle
N
A
S
,
or
the
halfe
is
the
halfe
sum
of
the
opposite
angle
.
Example
.
The
eccentrick
of
Venus
reduced
-
266.
04487
The
Suns
true
place
-
154.
07347
Anomaly
of
the
orbe
-
111.
97140
Complement
is
N
A
S
-
68.
02860
Halfe
Anomaly
-
55.
98570
These
given
with
the
sides
N
A
and
S
A
the
Analogies
are
.
As
the
greater
side
S
A
|
100895
co.
ar
.
|
4.
9961293
|
Is
to
Radius
|
|
10.
0000000
|
So
is
the
lesser
side
N
A
|
72816
|
4.
8622300
|
To
the
tangent
of
|
35.
81815
|
9.
8583593
|
Adde
|
45.
|
|
As
Radius
|
|
|
To
cotangent
|
80.
81815
|
9.
2085475
|
So
tang
.
halfe
sum
|
55.
98570
|
10.
1707787
|
To
tang
.
halfe
diff
.
|
13.
46926
|
9.
3793262
|
Summe
|
69.
45496
Angle
A
N
S
|
|
Difference
|
42.
51644
Angle
A
S
N
|
|
In
the
superiour
Planets
,
Saturne
,
Iupiter
,
and
Mars
,
the
summe
of
these
angles
is
the
elongation
,
but
in
the
inferiour
Venus
and
Mercury
there
difference
is
the
Elongation
sought
,
and
in
our
Example
is
to
be
added
to
the
Suns
place
,
because
the
Suns
place
was
subtracted
from
the
Eccentrick
of
Venus
reduced
.
Suns
true
place
-
154.
07347
Elongation
A
S
N
add
-
42.
51644
True
place
of
Venus
-
196.
58991
To
finde
the
distance
of
Venus
from
the
Earth
.
As
the
sine
of
A
N
S
|
69.
45496
co.
ar
.
|
0.
0285403
|
To
the
side
A
S
|
100895
|
5.
0038707
|
So
sine
of
N
A
S
|
68.
02860
|
9.
9672296
|
To
the
side
S
N
|
99917
|
4.
9996406
|
To
finde
the
latitude
of
Venus
from
the
Earth
.
As
the
side
S
X
|
9991.
7
co.
ar
.
|
5.
0003594
|
Is
to
Radius
|
|
10.
0000000
|
So
is
X
L
|
896
|
2.
9526176
|
To
the
tangent
of
X
S
L
|
0.
51417
|
7.
9529770
|
which
is
the
South
latitude
of
Venus
.
|
|
|
CHAP.
17.
Of
the
Motion
of
Mercury
.
THe
forme
of
calculating
the
place
of
this
Planet
is
the
same
with
Venus
,
the
Dimensions
of
whose
orbs
we
shall
give
you
,
as
the
learned
Bullialdus
hath
computed
them
,
but
first
we
will
set
down
the
middle
motions
thereof
to
the
former
time
.
The
middle
motions
of
Mercury
.
An.
Christ
.
|
Longit.
☿
|
Aphel
.
☿
|
Node
☿
|
1500
|
352.
53750
|
248.
73556
|
039.
85639
|
80
|
59.
53472
|
2.
31611
|
2.
12417
|
6
|
326.
41889
|
.
17361
|
.
15917
|
Iuly
|
147.
58583
|
.
01694
|
.
01528
|
D
1●
|
65.
47806
|
.
00126
|
.
00117
|
H
18
|
3.
06917
|
|
|
P
4564
|
.
07781
|
|
|
Meane
Mot.
|
234.
70198
|
251.
24348
|
42.
15618
|
Aphel
.
Snbt.
|
251.
24348
|
Rest
Anom
.
|
343.
45850
|
The
proportion
between
the
Earths
orb
,
and
the
orbe
of
Mercury
is
as
100.
000
to
38585
Semicentricity
in
the
same
parts
is
,
8105.
The
parts
or
greatest
inclination
4635.
And
the
angle
it selfe
6.
90.
In
the
triangle
therefore
M
E
H
,
of
the
first
Diagram
of
the
former
Chapter
we
have
known
.
1.
The
halfe
sum
of
the
opposite
angles
M
E
H
and
M
H
E
,
8.
27075
the
halfe
of
16.
54150
which
is
the
complement
of
the
meane
Anomaly
,
343.
4585
to
a
circle
.
2.
The
side
M
E
|
77170
|
|
3.
The
side
M
H
|
16210
|
|
Summe
|
93380
co.
ar
.
|
5.
0297462
|
Differ
.
|
60960
|
4.
7850449
|
So
tang
.
halfe
sum
.
|
8.
27075
|
9.
1628126
|
To
tang
.
halfe
differ
.
|
5.
42532
|
8.
9776037
|
Difference
|
2.
84543
Angle
M
E
H
|
|
Difference
doubled
|
5.
69086
Angle
M
B
H
or
the
Equation
to
be
added
to
the
meane
longitude
,
because
the
Anomaly
is
more
then
a
semicircle
.
|
|
Example
.
The
meane
Longitude
of
Mercury
-
234.
70198
Equation
adde
-
5.
69086
Eccentrick
place
-
240.
39284
Node
subtract
-
42.
15618
Argument
of
Latitude
K
L
-
198.
23666
To
finde
the
distance
of
Mercury
from
the
Sun
.
As
the
sine
of
MBH
|
5.
69086
co.
ar
.
|
1.
0036592
|
To
the
side
MH
|
16210
|
4.
20978●0
|
So
sine
of
EMH
|
16.
54150
|
9.
4544022
|
To
the
side
BH
|
46541
|
4.
6678444
|
To
finde
the
Reduction
.
As
Radius
,
to
cosine
of
XKL
|
6.
90
|
9.
9968431
|
So
tangent
of
KL
|
18.
23666
|
9.
5178453
|
To
tangent
of
XK
|
11.
11322
|
9.
5146884
|
Reduction
|
12344
|
|
And
because
the
argument
of
Latitude
is
more
then
180
,
it
must
be
subtracted
from
the
eccentrick
place
240.
39284
And
then
the
eccentrick
reduced
will
be
.
240.
26940
To
finde
the
present
inclination
.
As
Radius
|
|
|
To
the
greatest
inclination
EB
|
4635
|
3.
6660497
|
So
sine
of
KL
.
|
18.
23666
|
9.
4954646
|
To
the
inclinat
.
XL
|
1450
|
3.
1615143
|
Which
are
the
parts
of
inclination
agreeing
to
the
common
Radius
38585.
But
the
distance
of
Mercury
from
the
Sun
being
put
for
Radius
,
the
inclination
will
be
.
1749
For
as
Radius
DL
|
38585
co.
ar
.
|
5.
41358●5
|
To
Mercury
dist.
BH
or
AL
|
46541
|
4.
6678444
|
So
is
XL
|
1450
|
3.
1615143
|
To
XL
|
1749
|
3.
2429402
|
To
finde
the
distance
corrected
by
Curtation
.
As
AL
|
46541
co.
ar
.
|
5.
3321556
|
To
Radius
|
|
10.
0000000
|
So
is
XL
|
1749
|
3.
2429402
|
To
the
sine
of
LAX
|
2.
15437
|
8.
5750958
|
As
Radius
|
|
10.
0000000
|
To
AL
|
46541
|
4.
6678444
|
So
cosine
of
LAX
|
2.
15437
|
9.
9996929
|
To
AX
|
46509
|
4.
6675373
|
To
finde
the
second
inequality
of
Mercury
.
We
must
have
given
,
1.
The
Angle
NAS
which
is
to
be
found
by
subductiug
the
Suns
place
,
from
the
eccentrick
place
of
Mercury
reduced
,
or
this
from
it
,
so
that
less
then
6
signes
may
remain
,
this
remainer
is
the
Anomaly
of
the
orbe
,
and
the
complement
thereof
is
the
Angle
NAS
,
or
the
halfe
,
is
the
halfe
sum
of
the
opposite
angles
.
Example
.
The
eccentrick
of
Mercury
reduced
-
220.
26940
The
Suns
true
place
-
154.
07347
Anomaly
of
the
orbe
-
96.
19593
Complement
is
NAS
-
83.
80407
Halfe
Anomaly
-
48.
09796
These
given
with
the
sides
NA
and
SA.
the
Analogies
are
As
the
greater
side
SA
|
100895
co.
ar
.
|
4.
9961293
|
Is
to
Radius
|
|
10.
0000000
|
So
is
the
lesser
side
NA
|
46509
|
4.
6675373
|
To
the
tangent
of
|
24.
74799
|
9.
6636666
|
Adde
|
45.
|
|
As
Radius
|
|
|
To
the
cotang
.
of
|
69.
74799
|
9.
5669785
|
So
tang
.
halfe
summe
|
48.
09796
|
10.
0470559
|
To
tang
.
halfe
difference
|
22.
35160
|
9.
6140344
|
Summe
|
70.
44956
Angle
ANS
|
|
Difference
|
25.
74636
Angle
ASN
|
|
Because
the
Suns
place
was
subtracted
from
the
eccentrick
of
Mercury
reduced
,
therefore
the
angle
of
Elongation
ASN
must
be
added
to
the
Suns
place
.
-
154.
07347
Elongation
ASN
adde
-
25.
74636
True
place
of
Mercury
-
179.
81983
To
finde
the
distance
of
Mercury
from
the
Earth
.
As
the
sine
of
ANS
|
70.
44956
co.
ar
.
|
0.
0257891
|
To
the
side
AS
|
100895
|
5.
0038707
|
So
the
sine
of
NAS
|
83.
80407
|
9.
9974556
|
To
the
side
SN
|
106442
|
5.
0271154
|
To
finde
the
Latitude
of
Mercury
from
the
Earth
.
As
the
side
SX
|
106442
co.
ar
.
|
4.
9728846
|
Is
to
Radius
|
|
10.
0000000
|
So
is
XL
|
1749
|
3.
2429402
|
To
the
tang
.
of
XSL
|
0.
94169
|
8.
2158248
|
Which
is
the
south
Latitude
of
Mercury
.
|
|
|
CHAP.
18.
Of
the
Semidiameters
of
the
Sun
,
Moon
,
and
shadow
of
the
Earth
.
THe
angle
of
the
Suns
apparent
Semidiameter
,
in
his
nearest
distance
to
the
Earth
,
Bullialdus
hath
by
observation
found
to
be
16′
45″
,
or
in
Decimall
numbers
27917.
And
by
an
Eclipse
of
the
Moon
,
December
1638
,
he
found
her
Semidiameter
to
be
16′
54″
or
28167
,
and
the
Semidiameter
of
the
Earth
;
shadow
44′
9″
,
or
7
583
,
at
which
time
(
being
the
time
of
incidence
)
her
distance
from
the
Earth
by
his
computation
was
97908
parts
of
the
Semiaxis
of
the
Elipsis
100.
000.
By
this
and
another
observation
in
the
same
Eclipse
,
he
sheweth
how
to
finde
her
apparent
semidiameter
,
in
all
the
other
intervalls
.
The
inferiour
limbe
of
the
Moon
and
the
first
Starre
in
the
foot
of
the
former
Twin
,
(
whose
place
then
according
to
Tycho
was
Gemini
28.
25′
17″
,
or
Gemini
28.
42138
with
South
Latitude
,
0
d.
58′
or
0.
96667.
)
being
in
the
same
Azimuth
,
was
8′
or
13333
higher
then
the
Star
and
the
Altitude
of
the
heart
of
Hydra
then
taken
by
him
at
Paris
was
30
deg.
37′
,
or
30
d.
61667.
From
whence
the
hour
was
found
,
30
h.
40′
,
or
13
h.
66667
and
the
houre
being
given
the
altitude
of
the
Starre
is
also
given
,
deg.
56.
42′
15″
,
or
deg.
56.
70416.
The
apparent
altitude
of
the
center
of
the
Moone
was
deg.
57
7′
9″
,
or
deg.
57
11916
,
but
by
her
latitude
and
place
it
should
have
beene
deg.
57
40′
4″
,
or
deg.
57
66778
and
therefore
her
parallax
of
altitude
32′
55″
,
or
54861.
The
situation
of
the
Moone
and
Azimuth
in
which
her
interiour
limbe
and
the
Stars
were
,
being
given
,
her
aparent
Longitude
was
almost
in
Gemini
,
deg.
28
38′
30″
,
or
Gemini
deg.
28
64167
,
her
parallax
of
longitude
18
min.
or
30000
and
therefore
the
center
of
the
Moon
in
her
true
motion
in
Gemini
28
d.
57
min.
fere
.
or
in
Gemini
28
d.
95000
,
her
parallax
of
Latitude
is
19
min.
or
31667.
to
which
21′
or
35007
,
the
difference
of
the
observed
latitude
of
the
Moon
and
Stars
,
being
added
the
true
difference
is
50
min.
or
83333
min.
and
thence
the
Moons
Latitude
8
min.
or
13333
S.
Now
then
to
finde
the
distance
of
the
Moon
from
the
Earth
,
in
this
Eclipse
,
the
Earths
semidiameter
being
one
degree
,
Let
FEC
represent
the
true
Horizon
,
BDE
,
the
vertical
at
Paris
E
the
center
of
the
earth
,
D
the
City
of
Paris
:
the
Moons
true
altitude
,
AEF
,
deg.
5766778
,
the
observed
altitude
ADG
,
deg.
5711916.
The
parallacticall
angle
DAE
,
deg.
0.
34861.
Therefore
in
the
Triangle
ADE
we
have
given
all
the
angles
,
and
the
finde
DE
one
Semidiameter
of
the
Earth
,
to
finde
AE
,
for
which
the
anolagy
is
.
As
the
sine
of
DAE
0
d.
|
54861
co.
ar
.
|
2.
0188745
|
To
the
side
DE
1
|
|
0.
0000000
|
So
is
the
sine
of
ADB
32
d.
|
88084
|
9.
7347147
|
To
the
side
AE
|
56.
70
|
1.
7535892
|
This
foundation
being
laid
,
he
proceedeth
to
the
rest
:
and
to
shew
how
we
may
possibly
fall
into
some
absurditie
,
he
supposeth
the
Moons
distance
from
the
Earth
in
this
Ecclipse
to
be
but
55
semidiameters
,
or
the
side
BC
in
the
following
figure
,
the
apparent
angle
of
the
semidiameter
of
the
Earths
shadow
CHI
,
0.
73583
AEF
represents
the
Sun
,
his
semidiameter
AE
,
the
angle
of
his
apparent
Semidiameter
when
he
is
Perig.
AGE
16.
45
,
or
in
decimalls
27916
BHG
represents
the
Earth
.
BG
the
Semidiameter
thereof
,
hence
to
finde
HI
in
the
triangle
HIC
the
proportion
is
.
As
the
sine
of
HIC
|
89.
26417
co.
ar
.
|
0.
0000358
|
To
the
side
HC
|
54
|
1.
7323937
|
So
is
the
Radius
HCI
|
90
|
10.
0000000
|
To
the
Hypothenusal
HI
|
54.
004
|
1.
7324295
|
2.
In
the
triangle
HBI
we
have
given
the
sides
BH
.
1.
and
HI
54.
004
with
the
angle
BHI
179.
26417
,
hence
to
finde
the
Angle
BHI
,
the
Analogie
is
.
As
the
greater
side
H
I
|
54.
004
co.
ar
.
|
8.
2675705
|
Is
to
the
less
H
B
|
1.
|
0.
0000000
|
So
is
Radius
|
|
10.
0000000
|
To
the
tangent
of
|
1.
06082
|
8.
2675705
|
Adde
|
45.
|
|
As
Radius
|
|
10.
0000000
|
To
the
Cotang
.
of
|
46.
06082
|
9.
9839145
|
So
Tang.
halfe
summe
|
36791
|
7.
8075980
|
To
tang
.
halfe
differ
.
|
35454
|
7.
7915125
|
Angle
B
I
H
|
01337.
And
the
angle
C
B
I
|
72246
|
3.
In
the
triangle
C
B
I
,
we
have
given
the
angles
and
the
side
B
C
55
to
find
C
I.
Therefore
say
,
As
Radius
To
the
tangent
of
C
B
I
|
72246
|
8.
1007064
|
So
is
B
C
|
55
|
1.
7403627
|
To
C
I
|
0.
6935
=
to
B
K
|
1.
8410691
|
and
therefore
K
G
3065
,
and
the
angle
K
I
G
0.
31857.
For
,
●●
I
K
|
55
co.
ar
.
|
8.
2596373
|
To
Radius
|
|
10.
0000000
|
So
is
K
G
|
0.
3065
|
1.
4864305
|
To
the
tangent
of
K
I
G
|
0.
31857
|
7.
7450678
|
and
the
angle
B
D
G
is
equal
thereunto
,
but
so
the
angle
of
the
Suns
appar●n●
Semidiameter
A
G
E
27916
by
observation
,
is
lesse
then
the
angle
A
D
E
,
which
is
absurd
,
and
therefore
some
part
assumed
is
false
.
The
Semidiameters
of
the
Sun
and
Moon
must
not
be
changed
,
constant
experience
agreeing
with
these
observations
.
In
this
Eclipse
therefore
Bullialdus
doth
take
for
the
distance
of
the
Moone
from
the
earth
,
B
C
57.
85
Semidiameters
of
the
earth
,
and
the
Semidiameter
of
the
earths
shadow
,
C
B
I
75111.
Hence
to
find
C
I
,
the
analogie
is
.
As
Radius
To
the
side
B
C
|
57.
85
|
1.
7623034
|
So
the
tang
.
of
B
C
|
0.
75111
|
8.
1176019
|
To
C.
I.
|
0.
75841
|
1.
8799053
|
Let
B
K
be
equal
to
C.
I.
So
is
K
G
24159.
The●
As
I
K
|
57.
85
co.
ar
.
|
8.
2376066
|
To
Radius
|
|
10.
0000000
|
So
is
K
G
|
24159
|
1.
3830789
|
To
the
tang
.
of
K
I
G.
|
23928
or
B
D
G
|
7.
6207755
|
equal
to
E
D
A
,
and
the
Suns
apparent
Semidiameter
being
given
A
G
E
,
27916
,
the
angle
G
A
B
,
or
the
difference
between
the
angles
A
G
E
,
and
E
D
A
shall
be
given
also
,
viz.
03988.
the
Suns
Horizontal
parallax
when
he
is
Perigaeon
.
And
the
Moones
Perigaeon
distance
from
the
earth
,
in
Syzigiis
,
56.
50
Semidiameters
of
the
earth
.
For
,
as
|
97908
co.
ar
.
|
5.
0091819
|
To
|
57.
85
|
1.
7623034
|
So
is
|
95638
|
4.
9806304
|
To
B
C
|
56.
50
|
1.
7521157
|
Hence
to
find
the
Moones
Horizontal
parallax
when
she
is
perigaeon
,
the
analogie
is
,
in
the
preceding
Diagram
.
As
E
G
or
B
C
|
56.
50
co
,
ar
.
|
8.
2478843
|
Is
to
Radius
:
So
is
D
E
|
1.
|
0.
0000000
|
To
the
sine
of
E
G
D
|
1.
01399
|
8.
2478843
|
The
Horizontal
parallax
of
the
Sun
when
he
is
perigaeon
or
the
angle
B
A
G
was
found
to
be
-
.
03988
The
Moones
Horizontal
parallax
is
-
1.
01399
Their
aggregate
-
2.
05387
Semidiameter
of
the
Sun
subtract
-
.
27916
There
rests
the
angle
C
B
I
-
.
77471
or
the
apparent
semidiameter
of
the
earths
shadow
in
loco
transitus
Lun●
,
Perig.
In
the
triangle
therefore
B
C
I
,
we
having
the
angles
and
the
side
B
C
given
,
C
I
shall
be
also
given
.
For
,
As
the
sine
of
|
90
deg.
|
|
Is
the
side
B
C
|
56.
50
|
1.
7521157
|
So
tang
.
of
C
B
I
|
77472
|
8.
1310339
|
To
the
side
C
I
|
7640
=
B
K
|
1.
8831496
|
And
therefore
K
G
|
●360
|
|
And
in
the
triangle
A
G
B
having
the
angles
and
B
G
given
the
side
A
B
is
also
given
,
for
As
the
tang
.
of
B
A
G
|
03988
co
.
ar
.
|
3.
1585620
|
Is
to
B
G
1.
So
is
Radius
|
|
10.
0000000
|
To
A
B
|
1440.
66
|
3.
1585520
|
which
is
the
distance
of
the
Earth
from
the
Sun
,
when
he
is
Perihelion
.
And
because
the
Suns
eccentricity
is
1784
,
his
Apogaean
distance
is
101784
,
hence
to
find
his
distance
,
in
Semidiameters
of
the
earth
,
say
,
As
his
Perigaean
distance
|
98216
co.
ar
.
|
5.
0078178
|
Is
to
his
distance
|
1440.
66
|
3.
1585620
|
So
is
his
Apog
.
distance
|
101784
|
5.
0076794
|
To
his
Apog
.
dist.
|
1493.
03
|
3.
1740692
|
Then
as
B
S
|
1493.
03.
or
E
G
|
6.
8259308
|
To
Radius
,
so
is
E
D.
|
1
|
10.
0000000
|
To
the
sine
of
E
G
D
|
0.
3855
|
6.
8259308
|
The
Suns
Horizontal
parallax
when
he
is
Apogaeon
.
As
Radius
,
to
A
B
:
|
1440.
66
|
3.
1585620
|
So
is
tang
.
of
A
B
E
|
27916
|
7.
6877120
|
To
A
E
|
70189
|
0.
8462740
|
Then
as
B
S
|
1493.
03
co.
ar
.
|
6.
8259308
|
To
Radius
,
so
is
S
T
|
70189
▪
☉
Semid
.
|
0.
8462740
|
To
the
tang
.
of
S
B
T
|
26936
|
7.
6722048
|
The
apparent
Semidiameter
of
the
Sun
when
he
is
Apogaeon
.
The
Sun
being
Perigaean
,
we
have
given
B
G
1.
K
B.
75841.
KG
.
24159
and
B
C.
56.
50
,
the
distance
of
the
Moon
from
the
Earth
when
she
is
Perigaean
;
from
whence
the
longitude
of
the
earths
shadow
may
thus
be
found
.
As
K
G
|
2360
co
.
ar
.
|
10.
6270880
|
To
K
I
|
56.
50
|
1.
7521157
|
So
is
C
I
|
7640
|
1.
8831496
|
To
C
D
|
182.
93
|
2.
2623533
|
Add
B
C
56.
50
then
is
B
D
239.
43.
the
longitude
of
the
earths
shadow
.
Let
B
S
be
the
Apogaean
distance
of
the
Sun
,
-
1493.
03
The
angle
of
the
Suns
apparent
Semidiameter
S
G
T
-
26936
The
Perigaean
Semidiameter
or
the
angle
A
G
E
-
27916
Their
difference
is
the
angle
Z
G
E
-
00980
Let
TG
be
produced
to
N
,
then
shall
the
angle
I
G
N
be
equal
to
the
angle
Z
G
E
,
but
the
Sun
being
Perigaean
,
the
angle
B
D
G
was
found
to
be
o.
degrees
239●8
.
whose
complement
is
the
angle
B
G
D
89.
76072
therefore
when
the
Sun
is
Apogaean
,
it
shall
be
89
▪
77052
,
therefore
B
X
G
0.
22948
,
equal
to
K
N
G.
Hence
to
find
K
G
the
analogie
is
.
As
Radius
|
90
|
10.
0000000
|
To
K
I
|
56.
50
|
1.
7521157
|
So
tang
.
of
K
I
G
|
0.
22948
|
7.
6006035
|
To
K
G
|
22632
|
1.
3547192
|
And
K
B
|
77368.
Then
to
find
C
B
N.
Say
.
|
|
As
B
C
|
56.
50
co.
ar
.
|
8.
247●843
|
To
Radius
|
90
|
10.
0000000
|
So
C
I
or
rather
C
N.
|
77368
|
1.
8885613
|
To
the
tang
.
of
C
B
N
|
7844●
|
8.
1364456
|
The
Sun
being
Apogaean
;
and
the
angle
C
B
I
,
the
Sun
being
Paerigaean
,
was
before
found
to
be
77471
,
and
therefore
the
difference
of
the
earths
shadow
between
the
Suns
Apogaean
and
Perigaean
is
,
00971.
Then
,
As
K
G
|
22632
co
.
ar
.
|
10.
6452808
|
To
K
I
|
56.
50
|
1.
7521157
|
So
is
C
N
|
77368
|
1.
8885613
|
To
C
X
|
19●
.
18
|
2.
2859578
|
Add
B
C
|
56.
50.
Then
is
B
X
|
249.
68.
|
The
semidiameter
of
the
Moon
,
when
she
is
Perigaean
,
is
greater
then
the
semidiameter
of
the
Sun
,
being
Apogaean
,
and
therefore
Bullialdus
doth
make
it
17.
or
28333
,
and
because
the
eccentricity
of
the
Moon
is
given
4362
,
her
Apogaean
distance
in
Syzygiis
104362
,
the
Moon
being
Perigaean
her
distance
from
the
earth
is
,
95638
,
and
in
semidiameters
of
the
earth
56.
50
and
therefore
her
Apogaean
distance
in
semidiameters
of
the
earth
,
by
the
analogy
following
,
As
|
95638
co.
ar
.
|
5.
0193696
|
To
|
56.
50
|
1.
7521157
|
So
is
|
104362
|
5.
0185423
|
To
|
61.
66
|
1.
7900276
|
As
her
Apogaean
dist.
|
61.
66
Co.
ar
.
|
8.
2099724
|
To
the
Moons
Perig.
semid
.
|
28333
|
1.
4522925
|
So
is
the
Moon
Perig.
dist.
|
56.
50
|
1.
7521157
|
To
the
Apog
.
semid
.
|
25964
|
1.
4143806
|
We
have
the
semidiameter
of
the
Cone
C
I
76400
,
and
her
Perigaean
distance
56.
50
,
and
D
C
182.
93
,
but
when
the
Moon
is
Apogaean
,
D
C
will
be
no
more
then
177.
77.
found
by
abating
K
I
or
K
N
61.
66.
from
B
D
239.
43.
Hence
to
find
C
I
or
C
N
in
the
same
parts
say
.
As
D
C
|
182.
93
co.
ar
.
|
7.
7376467
|
To
C
I
|
7640
|
1.
8831496
|
So
is
C
D
|
177.
77
|
2.
2498584
|
To
C
I
|
7424
|
1.
8706547
|
Then
as
B
C
|
61.
66
co.
ar
.
|
8.
2099724
|
Is
to
Radius
|
90
|
10.
0000000
|
So
is
C
I
|
7424
|
1.
8706547
|
To
the
tang
.
of
C
B
I
|
.
68981
|
8.
0806271
|
Here
then
we
have
determined
|
|
|
Apogaeon
|
26936
|
The
Suns
Semidiameter
|
|
|
|
Perigaeon
|
27916
|
|
Apogaeon
|
1493.
03
|
His
distance
from
the
earth
|
|
|
|
Perigaeon
|
1440.
66
|
|
Apogaeon
☉
|
249.
68
|
The
Axis
of
the
earths
shadow
|
|
|
|
Perigaeon
☉
|
239.
43
|
The
Semidiameter
of
the
shadow
,
when
the
Sun
is
Apogaeon
.
In
loco
●ransitus
Lunae
,
Apog
78442
Perig.
77471.
|
Apogaeon
|
25964
|
The
Semidiater
of
the
Moone
in
Syzygiis
|
|
|
|
Perigaeon
|
28333
|
|
Apog
.
|
61.
66
|
The
distance
of
the
Moon
from
the
earth
in
Syzygiis
|
|
|
|
Perig.
|
56.
50
|
|
Perig.
|
23928.
|
Semiangle
of
the
Cone
|
|
|
|
Apog
.
|
22948.
|
CHAP.
19.
Of
the
Proportion
and
Magnitude
of
the
three
great
b●dies
,
the
Sun
,
Moon
and
the
Earth
.
THat
it
is
a
hard
matter
exactly
to
determine
the
true
Magnitude
of
the
coelestial
bodies
,
is
not
I
beleeve
denied
by
any
,
it
will
be
therefore
sufficient
if
we
shall
determine
them
so
,
as
that
there
be
no
sensible
errour
in
them
;
and
to
such
exactnesse
,
we
may
traine
by
the
rules
and
proportions
following
.
As
the
Semidiameter
of
the
Earths
shadow
C
B
I.
Is
to
the
Semidiameter
of
the
shadow
in
parts
of
the
Earths
Semidiameter
C
I
=
B
K
:
So
is
the
apparent
Semidiameter
of
the
Moon
.
To
the
Semidiameter
of
the
Moone
in
parts
of
the
Ear●hs
Semidiameter
,
that
is
As
C
B
I
|
77471
co.
ar
.
|
10.
1108609
|
To
C
I
|
7640
|
1.
8831496
|
So
is
the
Moones
semid
.
|
28333
|
1.
4522925
|
To
the
Moones
semid
.
|
27945
|
1.
4463030
|
And
Sphears
being
in
triplicated
proportion
of
their
diameters
,
the
proportion
of
the
earth
to
the
Moone
will
be
as
1.
00000.
00000.
00000.
the
Cube
of
the
earths
Semidiameter
to
02182.
28939.
33625.
the
Cube
of
the
Moones
semid
.
27945.
and
therefore
dividing
the
earths
Semidiameter
by
the
Moons
,
the
quotient
will
be
45.
823
,
and
so
many
times
is
the
body
of
the
Moon
contained
in
the
Earth
.
The
proportion
between
the
Semidiameter
of
the
Earth
and
the
Semidiameter
of
the
Sun
,
may
be
found
by
this
analogy
,
As
Radius
|
90
|
|
To
A
B
|
1440.
66
|
3.
1585614
|
So
is
tang
.
A
B
E
|
27916
|
7.
687706●
|
To
A
E
|
7.
0189
|
0.
8462683
|
But
if
to
A
B
|
1440.
66
|
|
You
adde
B
D
|
239.
43
|
|
Their
summe
is
A
D
|
1680.
09
|
|
And
then
,
As
Radius
|
90
|
|
To
the
side
A
D
|
1680.
09
|
3.
2255415
|
So
is
tang
.
of
A
D
E
or
K
I
G
|
23928
|
7.
6207755
|
To
A
E
|
7.
0197
|
0.
8463170
|
And
now
if
you
take
the
lesser
semidiameter
of
the
Sun
,
the
Cube
thereof
will
be
●45
.
781
,
but
taking
the
Semidiameter
of
the
Sun
to
be
but
7
semidiameters
of
the
earth
,
the
Sun
will
be
343
times
bigger
then
the
Earth
.
The
proportion
of
the
Semidiameter
of
the
earth
and
the
Moone
is
as
1
to
27945
,
of
the
Sun
and
the
Earth
as
7
to
1
,
and
therefore
of
the
Sun
and
the
Moon
as
7
to
27945.
The
Cube
of
7
is
343
,
the
Cube
of
27945
is
02182
,
&c
,
by
which
dividing
the
Cube
of
the
Suns
semidiameter
the
quotient
will
be
15717.
47
and
so
many
times
is
the
Moone
contained
in
the
Sun
.
CHAP.
20.
Of
the
proportion
between
the
Orbs
of
the
superiour
and
inferiour
Planets
,
and
the
Orb
of
the
Earth
.
WHat
proportion
the
Orbs
of
these
Planets
have
to
the
earths
Orb
,
we
have
set
down
in
those
Chapters
,
in
which
we
have
shewed
the
manner
of
computing
their
places
,
and
by
what
meanes
the
truth
of
those
proportions
may
appeare
,
we
shall
set
downe
in
this
,
and
because
we
have
used
those
proportions
which
Bullialdus
hath
with
great
diligence
computed
;
we
shall
exhibit
here
an
Example
in
Saturne
according
to
which
method
the
proportions
between
the
orbe
of
the
earth
and
the
orbes
of
the
other
planets
are
also
to
be
found
.
And
Saturns
proportion
to
the
earths
orbe
,
as
Bullialdus
hath
determined
it
,
and
which
we
have
used
,
Chap.
13.
is
as
100.
000
to
10480.
The
observation
from
whence
this
proportion
is
gathered
was
made
Anno
Christi
1587.
January
the
9th
.
Hour
9th
.
75
parts
,
at
which
time
Saturn
was
observed
to
be
in
Aries
26
,
13333.
with
South
latitude
,
deg.
2.
46667.
The
Suns
true
place
then
was
in
♉
deg.
29
,
41778.
and
his
distance
from
the
earth
,
98374.
Saturns
true
place
from
the
Sun
by
calculation
was
in
♉
deg.
2
,
31416.
whose
difference
from
his
observed
place
,
deg.
6
,
18083.
is
the
parallax
of
the
orbe
or
the
angle
A
N
S
,
and
the
angle
N
A
S
,
87.
●0362
,
is
found
by
deducting
Saturns
place
from
the
place
of
the
Sun
,
which
with
his
distance
from
the
Sun
or
side
A
N
95596
being
given
,
the
side
A
S
will
be
found
to
be
10310.
Now
as
the
Suns
distance
from
the
earth
,
-
98374
Is
to
the
distance
,
-
10310
So
is
the
Semidiameter
of
♄
orbe
,
-
100.
000
To
the
Semidiameter
of
the
earths
orbe
,
-
10480
By
a
second
Observation
made
Anno
Christi
1590
,
February
8
,
about
8
of
the
clock
in
the
evening
;
Saturn
was
in
Gemini
,
deg.
7.
53333.
with
South
latitude
,
deg.
1.
50.
The
true
place
of
the
Sun
at
the
same
time
was
in
Pisces
,
deg.
0.
02805.
and
his
distance
from
the
earth
,
98953.
And
Saturns
place
from
the
Sun
by
calculation
was
in
Gemini
,
deg.
13.
82167.
from
which
deducting
his
place
taken
by
observation
,
their
difference
,
6.
28834
is
the
parallax
of
his
orbe
,
represented
by
the
angle
A
I
S.
And
subtracting
Saturns
place
73.
82167
from
the
Suns
place
330.
02805
their
differences
256.
20638
reject
a
semicircle
is
the
angle
,
I
A
S
76.
20638
and
Saturns
distance
from
the
Sun
represented
by
A
I
94338
and
hence
the
side
A
S
10423.
And
now
as
the
Suns
distance
from
the
earth
-
9893
Is
to
the
distance
A
G
-
10423.
So
is
the
Semidiameter
of
Saturns
orbe
-
100.
000
To
the
Semidiameter
of
the
earths
orbe
-
10533
By
a
third
observation
made
in
the
same
year
of
Christ
1590
Septemb.
7
at
midnight
,
Saturns
place
was
in
Gemini
deg.
28.
1
with
South
latitude
deg.
1.
18333.
The
Suns
true
place
at
the
same
time
was
in
Virgo
deg
24.
49833.
And
his
distance
from
the
earth
100300.
Saturnes
place
from
the
Suns
by
calculation
,
was
in
Gemini
deg.
21.
76722
,
which
being
deducted
from
his
place
taken
by
observation
,
their
difference
is
the
parallax
of
his
o●●e
,
or
the
angle
A
K
L
6.
33278
,
and
deducting
Saturns
place
from
the
place
of
the
Sun
the
angle
A
L
K
is
9●
.
73111
,
and
therefore
the
side
A
L
10415.
Now
as
the
Suns
distance
from
the
earth
-
100300
Is
to
the
distance
A
L
-
10415
So
is
the
Semidiameter
of
Saturns
Orbe
-
100000
To
the
Semidiameter
of
the
earths
orbe
-
10383
But
Bullialdus
whom
we
follow
doth
retaine
the
first
of
these
10480
as
being
the
meane
,
and
most
agreeable
to
Tycho's
observation
;
And
from
these
three
observations
the
inclination
of
Saturns
orbe
may
thus
be
found
.
The
Triangles
L
D.
N.
I
D
O
,
and
D
G
M
of
the
following
Diagram
,
have
their
sides
and
angles
equal
with
the
triangles
,
N
A
S.
I
A
S
and
L
A
K
in
the
Diagram
preceding
,
being
drawne
from
the
same
observations
;
in
every
of
which
we
are
to
compute
♄
distance
from
the
earth
;
for
which
in
the
triangle
L
D
N
by
the
first
of
these
observations
we
have
given
the
angle
L
N
D
86.
71556.
The
angle
L
D
N
87.
10362
and
♄
distance
from
the
Sun
L
D
95596
to
find
L
N.
As
the
sine
of
L
N
D
|
86.
71556
co.
ar
.
|
0.
0007130
|
Is
to
the
side
D
L
|
95596
|
4.
9804397
|
So
the
sine
of
L
D
N
|
87.
10362
|
9.
9994447
|
To
the
side
L
N
|
95630
|
4.
9805974
|
Then
in
the
right
angled
triangle
L
K
N
right
angled
at
K
,
we
have
given
the
angle
of
South
latitude
L
N
K
2.
46667
,
and
the
side
L
N
95630
,
to
find
L
K.
As
Radius
|
|
10.
0000000
|
Is
to
L
N
|
95630
|
4.
9805974
|
So
is
the
sine
of
L
N
K
|
2.
46667
|
8.
6338534
|
To
the
side
L
K
|
4116
|
3.
6144508
|
Hence
to
find
the
angle
of
Latitude
at
the
Sun
say
,
As
the
side
L
D
|
95596
co
.
ar
.
|
5.
0195603
|
Is
to
Radius
|
|
10.
0000000
|
So
is
the
side
L
K
|
●116
|
3.
6144508
|
To
the
Sine
of
L
D
K
|
2.
46756
|
8.
6340111
|
By
the
second
observation
in
the
triangle
I
D
O
we
have
given
the
angle
I
O
D
97.
50528
whose
complement
is
82.
49472
the
angle
I
D
O
76.
20638
,
and
♄
distance
from
the
Sun
I
D
,
to
find
his
distance
from
the
earth
I
O.
As
the
sine
of
I
O
D
|
82.
49472
co.
ar
.
|
●
.
00373●●
|
Is
to
the
side
I
D
|
94338
|
4.
974686●
|
So
is
the
sine
of
I
D
O
|
76.
20638
|
9.
9872910
|
To
the
side
I
O
|
92409
|
4.
9657145
|
Then
in
the
right
angled
Triangle
H
I
O
right
angled
at
H
,
we
have
given
the
angle
of
Saturns
South
latitude
H
O
I
1.
50
,
and
the
side
I
O
to
find
H
I.
By
the
third
observation
in
the
triangle
GMD
we
have
given
the
angle
M
D
G
87.
26889
the
angle
M
G
D
86.
39833
the
side
G
D
94239
to
find
M
G.
As
the
sine
of
G
M
D
|
86.
39833
co.
ar
.
|
0.
0008588
|
To
the
sine
of
M
D
G
|
87.
26889
|
9.
9995064
|
So
is
the
side
G
D
|
94239
|
4.
9742306
|
To
the
side
M
G
|
94318
|
4.
9745958
|
Then
in
the
right
angled
triangle
G
M
F
right
angled
at
F
we
have
given
the
angle
of
Saturns
South
latitude
G
M
F
1.
18333
,
and
the
side
G
M
to
find
G
F.
As
Radius
|
|
10.
0000000
|
Is
to
the
side
M
G
|
94318
|
4.
9745958
|
So
is
the
sine
of
G
M
F
|
1.
18333
|
8.
3149535
|
To
the
side
G
F
|
1948
|
3.
2895493
|
Hence
to
find
the
angle
of
latitude
at
the
Sun
,
As
the
side
D
G
|
94239
co
.
ar
.
|
5.
0257694
|
Is
to
Radius
|
|
10.
0000000
|
So
is
the
side
G
F
|
1948
|
3.
2895493
|
To
the
sine
of
G
D
F
|
1.
18434
|
8.
3153187
|
These
things
premised
the
places
of
the
Nodes
and
the
angle
of
inclination
of
the
planes
may
thus
be
found
.
In
the
following
Diagram
let
the
place
of
the
first
observation
be
at
A
,
and
the
angle
of
latitude
at
the
Sun
D
G
2.
46756.
The
second
at
B
,
and
the
latitude
G
E
1.
4693●
.
The
third
at
C
,
and
the
latitude
G
F
1.
18434.
The
Arches
of
apparent
motion
from
the
First
observation
to
the
second
A
B
-
41.
50751
Second
to
the
the
third
B
C
-
7.
94555
First
to
the
third
A
C
-
49.
45306
The
halfe
of
these
arches
are
the
measure
of
the
angle
B
A
C
|
3.
97277
|
whose
sines
are
the
sides
A
B
|
35435
|
of
the
angle
B
C
A
|
20.
75375
|
whose
sines
are
the
sides
B
C
|
6929
|
|
|
whose
sines
are
the
sides
A
C
|
41828
|
The
angles
of
Latitude
at
the
Sun
found
are
L
D
K
|
2.
46756
|
whose
sines
are
the
sides
G
D
|
4306
|
I
D
K
|
1.
46932
|
whose
sines
are
the
sides
G
E
|
2617
|
G
D
F
|
1.
18434
|
whose
sines
are
the
sides
G
F
|
2064
|
The
difference
between
G
E
and
G
F
is
F
E
-
553
The
difference
between
G
D
and
G
F
is
F
D
-
2242
The
difference
between
G
E
and
G
D
is
E
D
-
1689
Hence
to
find
the
side
A
L
in
the
triangle
A
B
L
use
this
analogie
As
F
D
|
2242
co
ar
.
|
6.
6493644
|
Is
to
C
A
|
41828
|
4.
621467●
|
So
is
E
D
|
1689
|
3.
2276296
|
To
A
L
|
31511
|
4.
4984611
|
Then
to
find
the
angle
A
B
L
We
have
given
the
side
A
B
|
35435
|
and
B
A
L
|
3.
97277
|
We
have
given
the
side
A
L
|
31511
|
and
B
A
L
|
3.
97277
|
whose
complement
is
176.
02723
,
and
the
halfe
thereof
88.
01361
is
the
halfe
summe
of
the
opposite
angles
.
As
the
summe
of
A
B
and
A
L
|
66946
co.
ar
.
|
5.
1742754
|
Is
to
their
difference
|
3924
|
3.
3937289
|
So
the
tang
.
halfe
summe
|
88.
01361
|
11.
4598852
|
To
the
tang
.
halfe
differ
.
|
59.
38668
|
10.
22889●
|
The
summe
|
147.
40029
is
the
angle
A
L
B
|
|
The
difference
|
28.
62693
is
the
angle
A
B
L
|
|
whose
double
57.
25386
is
the
arch
A
M
,
●o
which
if
you
adde
the
arch
A
B
41.
50751
their
summe
is
the
Arch
B
A
M
98.
76137
,
whose
complement
to
180
gives
the
arch
B
H
and
M
I
81.
23863
,
and
the
halfe
thereof
is
the
arch
B
H
40.
61931.
Now
then
the
point
B
is
in
73.
82167
To
which
if
you
adde
B
H
-
40.
61931
The
Node
ascending
is
-
114.
44098
And
deducting
B
C
from
B
H
|
40.
61931
|
whose
natural
sines
are
G
E
|
65098
|
Their
difference
is
C
H
|
32.
67376
|
whose
natural
sines
are
G
F
|
53980
|
The
summe
of
H
B
and
B
A
is
H
A
|
82.
12682
|
whose
natural
sines
are
G
D
|
99057
|
To
find
the
angle
of
Inclination
by
the
first
observation
,
say
As
G
D
|
99057
co
.
ar
.
|
5.
0041149
|
To
G
D
|
4306
|
3.
6340740
|
So
is
G
K
Radius
|
|
10.
0000000
|
To
G
K
the
sine
of
deg.
|
2.
48908
|
8.
6381889
|
To
find
the
angle
of
Inclination
by
the
second
observation
,
say
As
G
E
|
65098
co
.
ar
.
|
5.
1864324
|
To
G
E
|
2617
|
3.
4178037
|
So
G
K
Radius
|
|
10.
0000000
|
To
G
K
the
sine
of
D.
|
2.
30397
|
8.
6042361
|
To
find
the
angle
of
Inclination
by
the
third
observation
,
say
As
G
F
|
53980
co.
ar
.
|
5.
2677672
|
To
G
F
|
2064
|
3.
3147096
|
So
is
G
K
|
100.
000
or
Radius
|
10.
0000000
|
To
G
K
the
sine
of
deg.
|
2.
19128
|
8.
5824768
|
But
by
reason
of
this
varietie
,
we
may
well
suspect
the
truth
of
one
or
other
,
of
the
observations
given
;
and
therefore
Bullialdus
hath
a●●igned
another
place
for
the
point
of
the
Node
ascending
;
viz.
deg.
110.
42333
and
the
angle
of
Inclination
somewhat
more
then
that
found
by
the
first
observation
.
viz.
deg.
1.
50
that
so
he
might
make
these
observations
to
agree
as
neere
as
might
be
.
And
now
if
you
suppose
the
Node
ascending
to
be
-
110.
42333
And
from
thence
deduct
the
place
of
the
2d
.
observation
-
73.
82167
Their
difference
shall
be
the
arch
H
B
36.
60166
and
the
sine
thereof
is
G
E
59624.
Againe
deducting
B
C
7.
94555
from
B
H
,
their
difference
will
give
C
H
28.
65611
,
and
the
sine
thereof
G
F
47954.
And
by
adding
A
B
41.
50751
to
H
B
their
summe
shall
be
H
A
78.
10917
and
the
sine
thereof
G
D
97854.
For
the
latitude
agreeing
to
the
first
observation
As
G
K
|
100.
000
|
5.
0000000
|
Is
to
G
K
deg.
|
2.
50
|
8.
6396795
|
So
is
G
D
|
9.
7854
|
4.
9905785
|
To
G
D
or
the
latitude
|
2.
44632
|
8.
6302580
|
whereas
it
should
have
been
2.
46667.
For
the
latitude
agreeing
to
the
second
observation
As
G
K
|
100.
000
|
5.
0000000
|
Is
to
K
G
deg.
|
2.
50
|
8.
6396795
|
So
is
G
E
|
59624
|
4.
7754210
|
To
G
E
or
the
latitude
|
1.
49029
|
8.
4151005
|
whereas
it
should
have
been
1.
56667.
For
the
latitude
agreeing
to
the
third
observation
As
G
K
|
100.
000
|
5.
0000000
|
Is
to
G
K
or
the
angle
of
inclin
.
|
2.
50
|
8.
6396795
|
So
is
G
F
|
47954
|
4.
6808248
|
To
G
F
or
the
sine
of
latit
.
|
1.
19856
|
8.
3205043
|
whereas
it
should
have
been
but
1.
19
CHAP.
21.
To
find
the
mean
Conjunction
and
Opposition
of
the
Sun
and
Moon
.
FOr
this
purpose
the
Table
which
Shakerley
transcribed
from
Bullialdus
,
we
have
here
exhibited
in
Decimall
numbers
,
the
use
whereof
,
as
he
hath
explaind
the
same
is
this
:
Set
down
first
the
Epoch●
next
preceding
the
yeare
given
,
then
the
yeares
and
moneths
compleate
having
a
care
of
the
yeare
Bisse●tile
,
and
to
every
one
set
downe
the
time
answering
in
the
Table
;
then
adde
them
altogether
,
and
the
summe
subtract
from
the
next
greater
in
the
Canonian
,
under
the
title●
,
if
you
●ee●
a
Conjunction
,
or
☌
,
if
an
opposition
,
the
remainer
sheweth
the
time
required
compleat
from
the
beginning
of
the
moneth
current
.
Example
.
I
would
know
the
time
of
the
meane
opposition
of
the
Sun
and
Moon
in
March
,
1652.
The
worke
is
this
.
-
Hou
.
Parts
The
Epocha
1640
-
701.
96639
Yeares
compleat
11
-
20.
14722
February
compleat
Bissextile
-
022.
53539
The
summe
subtract
-
744.
64500
From
the
Opposition
next
greater
-
1063.
10139
Rests
the
meane
opposition
-
318.
45639
that
is
13
dayes
6
houres
and
45639
parts
.
CHAP.
22.
To
find
the
true
Opposition
or
Conjunction
of
the
Sun
and
Moon
.
FOr
the
time
of
the
meane
Conjunction
or
Opposition
given
,
find
the
true
place
of
the
Sun
,
and
the
eccentrick
place
of
the
Moon
,
and
compare
them
;
if
they
either
be
precisely
the
same
or
precisely
opposite
,
the
time
of
the
true
Conjunction
or
Opposition
agrees
with
the
meane
;
but
if
they
differ
take
the
difference
,
by
subtracting
the
lesse
from
the
greater
,
and
that
call
the
distance
of
the
Sun
and
Moon
.
2
Out
of
the
Table
of
Semidiameters
,
and
hourly
motions
;
with
the
meane
Anomalies
of
the
Sun
and
Moon
,
take
out
their
hourly
motions
,
and
subtract
the
hourly
motion
of
the
Sun
,
from
the
hourly
motion
of
the
Moon
,
by
the
remainer
(
which
is
the
hourly
motion
of
the
Moon
from
the
Sun
)
divide
the
distance
of
the
Sun
and
Moone
before
kept
,
the
quotient
gives
the
time
,
which
must
be
added
to
the
mean
time
of
Conjunction
or
Opposition
,
if
the
excesse
be
in
the
Suns
place
,
or
subtracted
,
if
in
the
Moones
place
.
3
At
this
time
thus
corrected
,
find
againe
the
true
place
of
the
Sun
and
●●centrick
place
of
the
Moon
,
together
with
their
distance
,
and
repeat
your
former
work
,
till
you
find
them
absolutely
to
concurre
,
and
the
time
thus
found
shall
be
the
true
time
of
Conjunction
or
Opposition
.
As
in
the
Example
,
|
D.
|
Hou
.
parts
|
At
the
time
of
the
meane
☍
March
|
13
|
6.
45639
|
The
true
place
of
the
Sun
is
|
|
4.
85039
|
The
Eccentrick
place
of
the
Moone
|
|
180.
38631
|
The
Distance
of
the
Sun
from
☍
Moon
|
|
4.
46408
|
Mean
Anomaly
of
the
Sun
|
|
266.
40860
|
His
hourly
motion
|
|
.
04112
|
Meane
Anomaly
of
the
Moone
|
|
30.
26681
|
Her
hourly
motion
|
|
.
51827
|
Hourly
motion
of
the
Moone
from
the
Sun
|
|
.
47715
|
By
which
dividing
the
distance
|
|
4.
46408
|
The
quotient
gives
Hours
9.
3558
to
be
added
.
|
|
|
So
the
time
first
corrected
March
|
13
|
15.
81279
|
The
true
place
of
the
Sun
is
|
|
5.
23520
|
The
eccentrick
place
of
the
Moone
|
|
1●5
.
15798
|
The
distance
of
the
Sun
from
☍
of
the
Moon
|
|
.
07722
|
Which
divided
by
the
hourly
motion
of
the
Moon
from
the
Sun
|
|
.
48114
|
Gives
in
time
to
be
added
|
|
.
16049
|
So
the
time
secondly
corrected
|
13
|
15.
97328
|
The
true
place
of
the
Sun
is
|
|
5.
24180
|
The
eccentrick
place
of
the
Moon
|
|
185.
●4010
|
The
distance
of
the
Sun
from
☍
the
Moon
|
|
.
00170
|
Which
divided
by
the
hourly
motion
of
the
Moon
from
the
Sun
|
|
.
48121
|
Gives
in
time
to
be
added
|
|
.
00353
|
So
the
true
time
|
13
|
15.
97681
|
The
true
place
of
the
Sun
is
|
|
5.
24194
|
The
eccentrick
place
of
the
Moon
|
|
185.
24190
|
4
For
this
time
find
out
the
true
motion
of
the
Moons
Latitude
,
and
thereby
the
Reduction
,
which
divide
by
the
hourly
motion
of
the
Moon
from
the
Sun
,
and
the
quotient
contrary
to
the
title
of
Reduction
,
apply
to
the
last
corrected
;
so
have
you
the
true
time
.
In
our
Example
.
The
true
motion
of
Latitude
|
|
174.
04881
|
The
Reduction
Adde
|
|
.
02398
|
The
quotient
Subt.
|
|
.
04983
|
So
the
true
opposition
|
13
|
15.
92698
|
5
Lastly
,
apply
the
equation
of
time
to
this
equal
time
to
make
it
apparent
.
The
true
time
of
the
☍
|
13
|
15.
92698
|
The
equation
of
time
Subt.
|
|
.
02884
|
The
apparent
time
of
the
☍
▪
|
13
|
15.
89814
|
CHAP.
23.
To
find
whether
there
will
be
an
Eclipse
or
not
.
THere
are
two
wayes
to
know
this
,
of
which
the
one
is
more
easie
,
the
other
more
certain
,
The
first
is
this
;
At
the
true
conjunction
,
if
the
true
motion
of
latitude
be
within
17
degrees
backward
or
forward
of
6
or
12
signes
,
or
at
the
opposition
within
12
degrees
,
there
is
a
possibility
of
an
Ecclipse
,
otherwise
not
.
In
our
Example
the
Moons
true
motion
of
latitude
is
174
deg.
04881
,
which
being
not
fully
6
degrees
distant
from
6
signes
,
shewes
the
necessity
of
an
Eclipse
.
The
other
way
is
this
.
If
at
the
visible
conjunction
the
visible
latitude
of
the
Moon
be
lesse
then
the
aggregate
of
the
Semidiameters
of
the
Sun
and
Moon
,
there
must
be
an
eclipse
or
else
not
.
2.
If
at
the
true
opposition
the
true
latitude
of
the
Moon
,
be
lesse
then
the
summe
of
the
Semidiameters
of
the
Moon
and
the
earths
shadow
,
there
must
be
an
Eclipse
otherwise
not
.
This
latter
way
is
most
certain
,
onely
subject
to
this
inconvenience
,
that
a
great
part
of
the
calculation
is
performed
before
we
come
to
the
〈◊〉
〈◊〉
〈◊〉
〈◊〉
〈◊〉
,
or
power
to
judge
of
the
possibility
.
CHAP.
24.
To
find
the
Quantity
of
a
Lunar
Ecclipse
.
BY
the
true
motion
of
the
Moons
Latitude
,
find
her
true
latitude
,
according
to
the
former
directions
:
this
in
our
example
is
0.
51496
North
Descendant
.
2
Find
out
the
Semidiameter
of
the
Moon
by
her
meane
anomaly
out
of
the
Table
,
as
also
h●●
Horizontall
parallax
;
and
with
the
meane
anomaly
of
the
Sun
,
take
out
the
Semiangle
of
the
Cone
of
the
shadow
;
and
this
subtract
from
the
Moons
Horizontall
parallax
,
there
rests
the
Semid
.
of
shaddow
.
3
Adde
together
the
Semidiameter
of
the
shadow
,
and
Semidiameter
of
the
Moon
,
and
from
the
summe
subtract
the
Latitude
of
the
Moon
,
the
remainder
is
the
scruples
of
the
Moons
diameter
ecclipsed
.
Example
.
Horizontall
parallax
of
the
Moon
-
.
94409
Semiangle
of
the
Cone
Subt.
-
.
23455
Semidiameter
of
the
shaddow
-
.
70954
Semidia
meter
of
the
Moon
-
.
26431
Summe
of
the
Semidiameters
-
.
97385
Latitud●
of
the
Moon
,
Subt.
-
.
51496
Scruples
deficient
-
.
45889
4
Convert
these
Scruples
into
digits
or
parts
,
whereof
the
Moones
body
containes
12
,
thus
,
As
the
Moones
diameter
52862
co.
ar
.
-
5.
276857
Is
to
the
Scruples
deficient
45889
-
4.
661708
So
is
12
digits
-
1.
079181
To
the
digits
ecclipsed
10.
417
-
1.
017746
Yet
note
that
Lunar
Eclipses
are
of
three
sorts
.
-
1
Partiall
,
when
the
Scruples
deficient
are
lesse
then
the
diameter
.
-
2
Totall
without
continuance
when
they
are
equall
.
-
3
Totall
with
continuance
when
the
scruples
deficient
are
greater
then
the
diameter
,
and
in
these
the
digits
eclipsed
are
more
then
12
,
which
are
so
to
be
understood
,
as
that
they
shew
how
far
the
ecclipse
is
over
the
body
of
the
Moon
.
CHAP.
25.
To
find
the
duration
of
a
Lunar
Ecclipse
,
or
the
continuance
of
the
totall
darknesse
,
where
the
Ecclipse
is
totall
.
FInd
the
scruples
of
Incidence
thus
:
Take
the
Logarithmes
of
the
summe
and
difference
,
of
the
Moones
latitude
and
the
summ
of
the
Semidiameters
of
the
Moon
,
and
the
shadow
,
halfe
the
summe
of
the
two
Logarithmes
shall
be
the
Logarithme
of
the
Scruples
of
incidence
required
.
Example
.
Summe
of
the
Semidiameters
|
97385
|
|
Latitude
of
the
Moone
|
51496
|
|
Their
summe
|
1.
48881
|
5.
172836
|
Their
difference
|
.
45889
|
4.
661308
|
Summe
of
the
Logarithmes
|
|
9.
834544
|
Scruples
of
incidence
82656
|
|
4.
917272
|
2
Divide
the
Scruples
of
incidence
by
the
hourly
motion
of
the
Moon
from
the
Sun
,
the
quotient
gives
the
time
of
incidence
or
halfe
duration
of
the
Eclipse
.
This
subtracted
from
the
true
time
of
the
opposition
▪
gives
the
beginning
of
the
Eclipse
;
or
added
to
it
gives
the
ending
.
Example
.
The
Scruples
of
incidence
82656
-
4.
917272
Divided
by
the
hourly
motion
of
the
Moon
from
the
Sun
48121
-
4.
682326
Time
of
incidence
,
hours
1.
7177
-
0.
234946
The
true
time
of
the
opposition
-
13
d.
15.
89814
Time
of
incidence
subt
.
-
1.
71770
The
beginning
of
the
Eclipse
-
14.
18044
Time
of
incidence
added
,
gives
the
end
-
17.
61584
The
whole
duration
-
3.
43540
3
If
the
Eclipse
be
total
and
you
desire
to
know
the
continuance
of
total
darkenesse
,
take
the
difference
of
the
Semidiameters
instead
of
the
sum
,
and
thereby
worke
as
you
are
directed
in
the
first
example
of
this
Chapter
,
and
you
have
the
halfe
tarrience
in
the
shadow
,
whose
double
is
the
thing
sought
.
CHAP.
26.
To
find
the
Moons
Latitude
at
the
beginning
and
end
of
the
Eclipse
.
MUltiply
the
Suns
hourly
motion
,
by
the
time
of
incidence
,
the
product
being
added
to
the
scruples
of
incidence
,
gives
you
the
motion
of
the
Moone
agreeing
to
the
time
of
incidence
.
2
From
the
true
motion
of
latitude
at
the
true
opposition
subtract
,
this
motion
of
the
Moon
;
there
rests
the
true
motion
of
latitude
at
the
beginning
of
the
Eclipse
;
or
if
you
adde
it
you
have
the
motion
of
latitude
at
the
ending
,
with
which
out
of
the
Table
of
latitude
you
may
find
out
the
latitude
,
answering
to
the
beginning
and
end
,
as
in
our
Example
.
The
time
of
incidence
-
1.
7177
The
Suns
hourly
motion
-
.
0411
Their
product
is
-
.
07059
The
scruples
of
incidence
-
.
82656
The
summe
-
.
89715
Motion
of
Latitude
at
true
☍
-
1●4
.
04882
Motion
of
Latitude
at
beginning
-
173.
15166
Latitude
at
beginning
North
Desc.
-
.
59223
Motion
of
Latitude
at
ending
-
174.
94596
Latitude
at
ending
North
Desc.
-
.
43746
CHAP
27.
To
find
the
middle
of
the
Eclipse
or
greatest
darkenesse
.
THe
time
of
the
true
Conjunction
or
Opposition
the
most
received
,
is
when
the
Sun
and
Moon
are
in
one
line
perpendicular
to
the
Ecliptique
,
to
find
this
with
the
Moons
true
Latitude
at
the
true
opposition
,
enter
the
little
Table
of
the
difference
of
the
true
Conjunction
or
opposition
from
the
greatest
obscuration
,
and
you
shall
find
the
difference
with
the
title
,
which
divide
by
the
hourly
motion
of
the
Moon
from
the
Sun
,
and
the
quotient
according
to
the
title
,
apply
to
the
time
of
the
true
opposition
;
so
have
you
the
time
of
the
greatest
darkenesse
,
or
middle
obscuration
.
Example
.
The
Moons
Latitude
,
North
Descend
.
|
|
.
51496
|
The
difference
adde
|
|
.
04489
|
Which
divide
by
the
hourly
motion
of
the
Moon
|
|
.
48121
|
Gives
the
difference
in
time
to
be
added
|
|
.
09328
|
To
the
true
opposition
March
13.
15
|
|
.
89814
|
So
the
middle
of
the
Eclipse
|
13.
|
15h.
99142
|
The
Calculation
of
the
forementioned
Ecclipse
according
to
the
preceding
directions
.
|
d.
|
ho.
|
Meane
opposition
March
|
13
|
6.
45639
|
Interval
Adde
|
|
9.
35580
|
True
opposition
|
13.
|
15.
97681
|
True
place
of
Sun
|
|
5.
24194
|
Eccentrick
place
of
the
Moone
|
|
185.
24190
|
Meane
Anomaly
of
the
Sun
|
|
266.
79954
|
Meane
Anomaly
of
the
Moone
|
|
35.
44378
|
True
motion
of
Latitude
|
|
174.
04881
|
True
Latitude
North
descend
.
|
|
.
51496
|
Reduction
Adde
|
|
.
02398
|
Hourly
motion
of
the
Sun
|
|
.
04111
|
Hourly
motion
of
the
Moon
|
|
0.
52232
|
Hourly
motion
of
the
Moon
from
the
Sun
|
|
.
48121
|
Reduction
in
time
Subt.
|
|
.
04983
|
True
opposition
corrected
|
13
|
15.
92698
|
Equation
of
time
Subt.
|
|
.
02884
|
T●uest
opposition
|
13.
|
15.
89814
|
Horizontall
Parallax
of
the
Moone
|
|
.
94409
|
Semiangle
of
the
Cone
|
|
.
23455
|
Semidiameter
of
the
shadow
|
|
.
70954
|
Semidiameter
of
the
Moone
|
|
.
26431
|
Sum
of
the
Semidiameters
|
|
.
97385
|
Scruples
deficient
|
|
.
45880
|
Digits
eclipsed
|
|
10.
4197
|
Scruples
of
incidence
|
|
.
82656
|
Time
of
incidence
|
|
1.
71770
|
Beginning
of
the
Eclipse
|
13
|
14.
18044
|
End
of
the
Eclipse
|
13
|
17.
61584
|
The
whole
duration
|
|
3.
43540
|
Latitude
of
the
Moone
at
the
beginning
North
D.
|
|
.
59223
|
Latitude
of
the
Moone
at
the
ending
North
D.
|
|
.
43746
|
Difference
from
the
middle
,
Added
|
|
.
09328
|
The
middle
of
the
Eclipse
|
13.
15
|
.
99142
|
CHAP.
28.
Of
the
Calculation
of
the
Suns
Eclipse
.
THis
Eclipse
is
not
properly
,
the
Eclipse
of
the
Sun
,
but
of
the
earth
in
regard
it
is
not
the
Sun
,
but
the
earth
which
looseth
light
the
Sun
being
only
apparently
darke
,
the
earth
in
truth
,
we
will
how
ever
use
the
name
,
that
others
have
given
it
,
and
shew
you
the
manner
of
the
Calculation
.
Find
the
meane
conjnction
,
and
from
thence
the
true
,
which
correct
by
the
Reduction
and
Equation
of
time
in
all
things
as
in
the
Moon
.
Example
of
a
Solar
Eclipse
,
which
happened
March
28.
1652.
|
D
|
|
Mean
Conjunction
March
|
28
|
0.
82333
|
Suns
place
|
|
19.
36150
|
Eccentrick
place
of
the
Moon
|
|
20.
89832
|
Distance
of
the
Moon
from
the
Sun
|
|
1.
53682
|
Meane
Anomaly
of
the
Sun
|
|
280.
96132
|
His
hourly
motion
|
|
.
04090
|
Meane
Anomaly
of
the
Moon
|
|
223.
17513
|
Her
hourly
motion
|
|
.
60109
|
Hourly
motion
of
the
Moon
from
the
Sun
|
|
.
56019
|
which
dividing
the
distance
|
|
1.
53682
|
Gives
in
the
quotient
subt
.
|
|
2.
74336
|
So
the
time
first
corrected
|
●7
|
22.
07997
|
True
place
of
the
Sun
|
|
19.
24962
|
The
Eccentrick
place
of
the
Moon
|
|
19.
29533
|
The
distance
of
the
Sun
from
the
Moon
|
|
.
04571
|
which
divided
by
the
hourly
motion
of
the
Moon
from
the
Sun
|
|
.
56137
|
Gives
in
time
subtract
|
|
.
08143
|
So
the
time
secondly
corrected
|
27
|
21.
99854
|
True
place
of
the
Sun
|
|
19.
24631
|
Eccentrick
place
of
the
Moon
|
|
19.
24766
|
Distance
of
the
Sun
from
the
Moone
|
|
.
00135
|
which
divided
by
the
hourly
motion
of
the
Moon
from
the
Sun
▪
|
|
.
56140
|
Gives
in
time
subtract
|
|
.
00240
|
So
the
true
time
of
Conjunction
|
27
|
21.
99614
|
The
true
pl●ce
of
the
Sun
|
|
19.
24621
|
Eccentrick
place
of
the
Moon
|
|
19.
24625
|
True
motion
of
Latitude
|
|
8.
80745
|
The
Reduction
Subtract
|
|
0.
03536
|
The
quotient
|
|
.
06298
|
So
the
true
Conjunction
|
.
27
|
22.
05912
|
Equation
of
time
subtract
|
|
.
09970
|
The
apparent
time
of
the
☌
|
.
27
|
21.
95942
|
CHAP.
29.
To
find
the
Parallaxes
of
Longitude
and
Latitude
.
BY
the
rules
delivered
in
the
former
part
,
find
at
the
true
Conjuction
the
Midheaven
,
with
its
altitude
and
the
Meridian
angle
.
Example
.
The
Suns
place
-
♈
19.
246
The
Suns
Right
Ascension
-
17.
749
Time
in
Degrees
-
329.
391
Right
Ascension
of
Midheaven
-
347.
140
Midheaven
-
♓
16.
022
Meridian
angle
-
67
,
078
Declination
of
Midheaven
-
5.
533
Altitude
of
the
Equator
at
London
-
38.
467
Altitude
of
Midheaven
-
32.
834
2
The
angle
of
the
Ecliptique
and
Horizon
,
or
altitude
of
the
Nonagesime
degree
,
and
his
distance
from
the
Midheaven
is
thus
found
,
by
the
17
Chapter
of
the
first
part
.
As
the
Radius
To
the
sine
of
the
Meridian
angle
|
67.
07
|
9.
9642509
|
So
is
Cosine
of
the
Altitude
of
M
C
|
32.
83
|
9.
9244255
|
To
the
Cosine
of
the
Angle
,
&c.
|
39.
30
|
9.
8886764
|
Then
as
Radius
To
the
Cosine
of
the
Meridian
angle
|
67.
07
|
9.
5906259
|
So
is
cotang
.
of
the
altitude
of
the
M.
C.
|
32.
83
|
10.
1903074
|
To
the
tang
.
of
the
distance
of
the
M.
C.
from
the
Nonagesime
degree
|
16.
02
|
9.
7809333
|
This
M
C
falling
betweene
Capricorn
and
Cancer
this
distance
is
to
be
added
to
the
Midheaven
|
|
♓
16.
02
|
And
the
Nonagesime
degree
will
be
in
|
|
♈
17.
14
|
3
Find
the
Node
Ascendent
and
Subtract
it
from
the
Nonagesime
degree
,
with
the
remainder
enter
the
Table
of
the
Moones
latitude
,
which
if
North
adde
to
the
angle
of
the
Ecliptique
and
Horizon
;
if
South
subtract
it
from
it
,
so
have
you
the
altitude
of
the
Nonagesime
degree
of
the
Moones
orbe
.
Example
.
The
Node
Ascendent
Subt.
-
10.
439
The
Nonagesime
degree
-
17.
140
There
rests
-
6.
701
Which
gives
the
Moones
Latitude
Adde
-
.
570
The
angle
of
the
Ecliptick
and
Horizon
-
39.
300
Altitude
of
the
Nonagesime
degree
of
the
Moones
orbe
-
39.
870
4
Take
the
distance
of
the
S●n
from
the
Nonagesime
degree
,
which
in
our
Example
of
the
true
Conj●ction
is
●
d.
10.
5
Out
of
the
table
of
Horizontall
parallaxes
,
take
the
Horizontall
Parallax
of
the
Sun
and
Moon
,
the
difference
of
them
is
the
Horizantall
parallax
of
the
Moon
from
the
Sun
.
Example
.
The
Horizontall
parallax
of
the
Sun
is
-
.
03912
The
Horizontall
parallax
of
the
Moon
-
.
99396
Horizontall
parallax
of
the
Moon
from
the
Sun
-
.
95484
6
Adde
the
Logarithme
of
the
Horizontall
parallax
of
the
Moon
from
the
Sun
,
the
sine
of
the
Altitude
of
the
Moones
orbe
,
and
the
sine
of
the
distance
of
the
Sun
from
the
Nonagesime
,
their
summe
subtracting
twice
Radius
,
is
the
Logarithme
of
the
parallax
of
longitude
.
Example
.
Horizontall
parallax
of
the
Moon
from
the
Sun
|
95484
|
1.
9799306
|
Altitude
of
Nonagesime
in
the
Moones
orbe
,
sine
|
39.
87
|
9.
8068904
|
Distance
of
the
Sun
from
the
Nonagesime
,
sine
|
2.
10
|
8.
5639994
|
Parallax
of
longitude
02243
|
|
2.
3508204
|
Here
note
that
whensoever
the
Suns
place
is
lesse
then
the
Nonagesime
degree
,
the
Parallax
of
Longitude
,
makes
the
luminaries
appeare
more
west
than
the
truth
,
and
in
the
occidentall
Quadrant
,
when
more
then
in
the
orientall
.
7
Adde
the
Logarithme
of
the
Horizontall
parallax
of
the
Moon
from
the
Sun
to
the
Cosine
of
the
Nonagesime
in
the
Moones
orb
;
the
summe
rejecting
Radius
is
the
Logarithme
of
the
parallax
of
Latitude
.
Example
.
Horizontall
parrallax
of
the
Moon
from
the
Sun
|
95484
|
1.
979930●
|
Altitude
of
Nonag
.
in
the
Moones
orb
,
Cos●ne
|
39.
87
|
9.
8850789
|
Parallax
of
latitude
|
0.
73284
|
1.
8650095
|
CHAP.
30.
To
find
the
visible
motion
of
the
Moon
from
the
Sun
for
any
time
assigned
.
AT
the
beginning
and
end
of
the
time
proposed
find
the
parallax
of
the
Moone
from
the
Sun
in
Longitude
,
and
then
observe
these
rules
.
-
1
If
during
all
the
time
proposed
,
the
luminaries
be
in
the
orientall
quadrant
,
and
the
parallax
of
longitude
increase
or
be
greater
at
the
end
of
the
time
given
then
at
the
beginning
,
adde
the
differences
of
the
two
parallaxes
of
longitude
unto
the
true
motion
of
the
Moone
from
the
Sun
agreeing
to
the
time
given
,
or
if
it
decrease
,
subtract
it
and
you
have
what
you
desire
.
-
2
If
during
all
the
time
the
luminaries
be
in
the
occidentall
quadrant
,
and
the
parallax
of
longitude
increase
,
subtract
the
said
difference
from
the
true
motion
,
if
it
decrease
,
adde
it
,
and
you
have
the
visible
motion
.
-
3
If
at
the
beginning
of
the
time
the
luminaries
be
in
the
orientall
quadrant
,
and
at
the
ending
in
the
occidentall
,
subtract
the
said
difference
from
the
true
motion
,
and
you
have
the
visible
motion
during
that
time
.
Example
.
Let
it
be
proposed
to
find
the
visi●le
motion
of
the
Moon
from
the
Sun
,
for
one
halfe
hour
before
the
true
Conjunction
.
In
our
Example
,
the
true
halfe
hourly
motion
is
-
.
28070
Parallax
of
Longitude
at
the
beginning
-
.
07628
Parallax
of
Longitude
at
the
end
-
.
02243
Their
difference
is
Subtract
-
.
05385
from
the
true
halfe
hourly
motion
Rests
-
.
22658
the
visible
halfe
hourly
motion
before
the
Conjunction
.
CHAP.
31.
To
finde
the
time
of
the
visible
Conjunction
of
the
Sun
and
Moon
.
USe
this
Analogie
,
As
the
apparent
motion
in
any
time
assigned
(
found
by
the
former
Chapter
)
is
to
the
time
assigned
;
so
is
the
parallax
of
longitude
at
the
true
Conjunction
to
the
difference
in
time
between
the
true
and
visible
Conjunction
.
This
difference
in
the
orientall
quadrant
must
be
subtracted
from
the
time
of
the
true
Conjunction
,
in
the
occidentall
quadrant
added
thereto
;
so
have
you
the
visible
Conjunction
.
Example
.
As
the
visible
halfe
hourly
motion
22685
co.
ar
.
-
0.
644262
Is
to
the
time
assigned
50′
-
1.
698970
So
parallax
longitude
02243
-
2.
350829
To
the
difference
betwixt
the
true
and
visible
Conjunction
04943
-
2.
694061
The
true
Conjunction
-
27
21.
95942
Difference
subtract
-
.
04943
The
visible
Conjunction
-
27
21.
90999
At
this
time
finde
out
the
true
distance
of
the
Moon
from
the
Sun
,
as
also
the
parallax
of
longitude
,
if
they
agree
it
is
a
signe
that
the
visible
Conjunction
is
truly
found
,
otherwise
repeat
the
former
worke
till
there
be
a
concurrance
.
Example
At
the
visible
Conjunction
,
March
-
27
21.
90999
The
true
distance
of
the
Moone
from
the
Sun
-
.
0●847
The
Parallax
of
Longitude
-
.
02775
Their
difference
-
.
00072
which
being
so
small
sheweth
that
the
visible
Conjunction
is
precisely
enough
found
.
CHAP.
32.
To
finde
the
visible
Latitude
of
the
Moon
,
at
the
time
of
the
visible
Conjunction
.
IN
these
Northerne
regions
which
we
inhabit
,
the
parallax
of
latitude
allwayes
makes
the
Moon
to
appeare
more
South
then
indeed
she
is
to
find
the
visible
latitude
therefore
observe
these
rules
.
1
At
the
time
of
the
visible
conjunction
find
out
the
true
latitude
of
the
Moon
thus
.
If
the
Eclipse
happen
in
the
orientall
quadrant
,
adde
the
parallax
of
longitude
to
the
motion
of
the
Sun
,
agreeing
to
the
difference
between
the
true
and
visible
Conjunction
,
and
the
summe
subtract
from
the
true
motion
of
latitude
at
the
time
of
the
true
Conjunction
;
or
if
the
Eclipse
happen
in
the
occidentall
quadrant
adde
the
said
summ●
thereto
,
and
you
have
the
true
motion
of
latitude
at
the
visible
Conjunction
,
by
which
as
formerly
taught
,
finde
out
the
true
Latitude
of
the
Moone
.
Example
.
Motion
of
the
Sun
agreeing
to
04943
-
.
00202
Parallax
of
Longitude
at
visible
☌
-
.
02775
The
Summ
Sub●
.
-
.
02977
Motion
of
Latitude
at
true
☌
-
8.
80745
Motion
of
Latitude
at
visible
☌
-
8.
77768
True
Latitude
at
visible
☌
North
-
.
75808
At
the
same
time
find
the
parallax
of
latitude
,
and
compare
it
with
the
true
latitude
.
If
the
latitude
be
South
,
adde
them
together
,
the
summe
is
the
South
visible
latitude
of
the
Moone
,
but
if
North
,
subtract
the
lesse
from
the
greater
;
there
remaines
the
visible
latitude
of
the
Moon
,
which
shall
be
North
when
the
latitude
is
greater
then
the
parallax
,
otherwise
South
.
Example
.
The
true
latitude
of
the
Moon
,
North
-
0.
75808
Parallax
of
Latitude
-
0.
73633
The
visible
latitude
North
-
.
02175
CHAP.
33.
To
find
the
quantity
of
a
Solar
Eclipse
.
THis
differs
very
little
from
that
in
the
24
Chapter
,
for
finding
the
quantity
of
a
lunar
Eclipse
,
for
it
with
their
meane
Anomalie●
you
enter
the
Table
,
and
thence
take
out
the
Semidiameter
of
the
Sun
and
Moone
,
and
adde
them
together
;
and
from
the
summe
subduct
the
visible
latitude
of
the
Moone
,
at
the
visible
Conjunction
,
there
rests
the
Scruples
of
the
Suns
body
deficient
,
which
as
in
the
Moon
,
so
here
in
the
Sun
convert
into
digits
.
Example
.
Semidiameter
of
the
Sun
-
.
27386
Semidiameter
of
the
Moon
-
.
27815
Summe
of
the
●emidiameters
-
.
55201
Visible
latitude
Subtracted
-
.
02175
Scruples
deficient
-
.
53026
So
the
digits
eclipsed
-
11.
61500
CHAP.
34.
To
find
the
beginning
and
ending
of
the
Suns
Eclipse
.
BY
the
visible
latitude
of
the
Moon
,
and
the
summe
of
the
Semidiameters
of
the
Sun
and
Moon
,
find
the
Scruples
of
incidence
,
as
in
the
Moones
Eclipse
Chap.
25.
2
For
one
hour
before
the
visible
Conjunction
,
find
by
the
30
Chapter
,
the
visible
hourly
motion
of
the
Moon
from
the
Sun
,
by
which
divide
the
Scruples
of
incidence
,
the
quotient
is
the
time
of
incidence
,
which
subtracted
from
the
time
of
the
visible
Conjunction
,
leaves
the
beginning
of
the
Eclipse
3
For
one
hour
after
the
visible
Conjunction
,
finde
the
visible
hourly
motion
of
the
Moon
from
the
Sun
,
by
which
divide
the
Scruples
of
incidence
,
the
quotient
is
the
time
of
Repletion
:
which
added
to
the
time
of
the
visible
Conjunction
,
gives
the
end
of
the
Eclipse
.
Example
.
Summe
of
the
Semidiameters
-
.
55201
Visible
latitude
-
.
02175
Scruples
of
Incidence
-
.
55158
At
1
ho.
before
the
visible
☌
March
27
-
20.
90999
Parallax
of
longitude
Orient
.
-
.
13209
True
hourly
motion
of
the
Moon
from
the
Sun
-
.
56140
Visible
hourly
motion
-
.
45174
Time
of
incidence
-
1.
22150
Beginning
of
the
Eclipse
March
27
-
20.
68849
At
1
ho.
after
the
visible
☌
27
-
22.
90999
Parallax
of
longitude
Occid
.
-
.
10119
Visible
hourly
motion
of
the
Moone
from
the
Sun
-
48796
Time
of
repletion
1
-
h.
.
1303●
End
of
the
Eclipse
27
-
23.
04029
The
whole
duration
-
2.
35180
CHAP.
35.
To
find
the
Visible
latitude
of
the
,
Moon
at
the
beginning
and
end
of
the
Suns
Eclipse
.
FOr
the
beginning
,
adde
to
the
minutes
of
Incidence
the
motion
of
the
Sun
agreeing
to
the
time
of
Incidence
,
and
the
summe
subtract
from
the
true
motion
of
latitude
at
the
time
of
the
visible
Synod
,
so
have
you
the
true
motion
of
latitude
at
the
beginning
,
by
which
find
the
true
latitude
,
and
by
these
according
to
the
second
rule
of
the
32
Chapter
,
may
be
had
the
visible
latitude
.
Example
.
The
Scruples
of
incidence
-
.
5515●
Motion
of
the
Sun
answering
to
the
time
-
.
05016
The
summe
subt
.
-
.
60174
Motion
of
latitude
at
visible
☌
-
8.
77768
Motion
of
latitude
at
beginning
-
8.
17594
True
latitude
North
-
.
70648
Parallax
of
latitude
-
.
82004
Visible
latitude
South
-
.
11353
2
For
the
end
,
adde
to
the
minutes
of
incidence
the
motion
of
the
Sun
agreeing
to
the
time
of
repletion
and
the
sum
adde
to
the
true
motion
of
latitude
at
the
time
of
the
visible
Conjunction
;
so
have
you
the
true
motion
of
latitude
at
the
end
;
by
which
proceed
as
before
,
to
find
the
visible
Latitude
.
Example
.
Scruples
of
Incidence
-
.
55158
Motion
of
the
Sun
agreeing
to
the
time
of
repl
.
-
.
04642
The
Summe
Adde
-
.
59800
Motion
of
Latitude
at
the
visible
☌
-
8.
77768
Motion
of
Latitude
at
the
ending
-
9.
37568
True
Latitude
North
-
.
80925
Parallax
of
Latitude
-
.
65218
Visible
Latitude
North
-
.
15707
CHAP.
36.
To
Delineate
the
Eclipses
▪
of
the
Sun
and
Moon
.
FOr
the
Moon
draw
the
lines
AC
and
BD
to
intersect
one
another
at
right
angles
in
E
,
which
point
of
intersection
is
the
place
of
the
Ecliptique
where
the
Eclipse
happens
:
upon
which
as
a
Center
draw
the
Peripherie
ABCD
,
of
the
quantity
of
the
summe
of
the
Semidiameters
of
the
Moon
and
the
earths
shadow
(
which
may
be
done
by
helpe
of
a
Scale
or
Sector
of
equal
divisions
)
also
to
the
quantity
of
the
Semidiameter
of
the
earths
shadow
,
draw
upon
the
same
center
another
Peripherie
.
Then
because
the
Moones
Eclipse
begins
on
the
east
part
of
her
body
,
you
must
upon
the
west
side
of
your
plane
,
note
downe
the
latitude
of
the
Moon
in
the
arch
BCD
,
which
here
▪
represents
the
west
part
▪
and
may
be
thus
done
,
From
E
upon
the
line
BD
prick
out
the
latitude
at
the
beginning
;
towards
B
,
if
the
Latitude
be
North
,
towards
D
if
South
,
and
it
terminat●s
at
G
,
from
which
draw
a
parallel
to
AC
,
and
in
the
arch
BC
it
marks
out
F.
Also
for
the
end
of
the
Eclipse
proceed
in
like
manner
on
the
other
side
,
and
you
have
the
latitude
terminated
at
I
,
and
the
parallel
falling
at
H.
Then
draw
a
line
between
F
and
H
and
where
it
intersects
BD
marke
it
with
K.
Lastly
,
upon
the
centers
F
,
K
and
H
,
draw
three
equal
circles
,
having
for
Radius
the
Semidiameter
of
the
Moone
,
and
the
worke
is
done
.
Typus
Eclipseos
Lunae
praedictae
.
Example
of
the
forementioned
Eclipse
of
the
Moon
March
15.
1652
Summe
of
the
Semidiameters
EB
-
.
97385
Semidiameter
of
shadow
EM
-
.
70954
Initial
latitude
of
the
Moon
EG
,
North
-
.
59223
Final
latitude
of
the
Moon
EI
North
-
.
43746
Semidiameter
of
the
Moon
MB
-
.
26431
2
For
the
Eclipse
of
the
Sun
,
it
differs
nothing
at
all
from
this
of
the
Moon
,
but
onely
that
instead
of
the
Semidiameter
of
the
shadow
of
the
earth
you
use
the
Semidiameter
of
the
Sun
;
and
the
visible
latitude
for
the
true
.
Example
of
the
forementioned
Sol●●
Eclipse
March
28.
●652
.
Summe
of
the
Semidiameters
EB
-
.
5520●
Semidiameter
of
the
Sun
EM
-
.
27386
Initiall
visible
latitude
EG
South
-
.
11353
Finall
visible
latitude
EI
North
-
.
15707
Semidiameter
of
the
Moon
MB
-
.
27815
Typus
Eclipseos
Solis
pr●dict●
.
CHAP.
37.
The
use
of
the
Table
of
Refractions
.
ALthough
the
Table
of
Refractions
,
belongs
not
to
the
calaulation
of
these
Tables
,
yet
will
it
not
be
amisse
to
shew
its
use
in
comparing
of
observation
with
calculation
.
Know
then
that
Refraction
,
causeth
the
stars
to
appeare
higher
then
really
they
are
.
Therefore
with
an
observed
altitude
enter
this
Table
and
take
out
the
Refraction
,
which
subtract
from
the
observed
altitude
,
and
you
have
the
true
altitude
,
or
having
the
true
altitude
,
the
apparent
is
found
by
adding
the
Refraction
thereto
.
FINIS
.