Astronomia Britannica exhibiting the doctrine of the sphere, and theory of the planets decimally by trigonometry, and by tables : fitted for the meridian of London ... / by John Newton ...

Astronomia BRITANNICA: The second Book. (Book 2)

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, toge∣ther with their middle motiōs must be sought, but how to state them so exactly as that we may thereby find their true or appa∣rent 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 al∣though 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 necessi∣ty,

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; Al∣though the periodicall revolutions therefore of the planets are the onely proper yeares and first in nature; yet since the civil yeare in every nati∣on 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 Egyp∣tians do also account 12 Months to their yeare, but their months are pro∣portioned 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 dif∣fer from the other onely in this, (that by reason of the Suns excesse in mo∣tion 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 rec∣koned 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 mul∣tiplication 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 Pascha∣lis, upon which the Feast of Easter, and the rest of the moveable Feasts de∣pend. The third Circle which now serves for no other use, then the constitu∣ting 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 revo∣lution 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 re∣puted 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

56142

2987

152

4

900 Turkish years gives

318930

There rests

375072

56142

2987

152

4

150 years gives

53155

There rests

375072

56142

2987

152

4

8 years give

2835

There rests

375072

56142

2987

152

4

Giumadi I. gives

148

There rests

375072

56142

2987

152

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 Pe∣riod 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

70

11

300 yeares give

109575

There rests

109645

70

11

April compleat

59

There rests

109645

70

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 Lon∣don, 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 Fe∣bruary▪

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 Ca∣talogue from that of London. The Letter S notes that the distance is West∣ward, 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 diffe∣rence is to be applied with the contrary title.

And that these Reductions whether in time or motion may be the bet∣ter 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 con∣verting 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 num∣ber 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 en∣tituled, A Table to convert the hours and minutes of a day into Deci∣malls, 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 down∣ward in that column till I come to 8 hours 17 minutes, and in the last co∣lumne 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 a∣gainst 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 remain∣der 004231 being sought amongst the Decimals of the fourths gives me 5 fourths, and this number to be subtracted from it 3858, and the remain∣der 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 revolu∣tions 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, sub∣tract 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 correspon∣dent time in Egyptian yeares from the death of Alexander, which accor∣ding to the former directions is thus.

The Christian Aera began in the 4713 complete yeare of Julian peri∣od, 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

332745

4246

596

231

21

1000 Egyptian yeares give

365000

There rests

697745

332745

4246

596

231

21

900 yeares give

328500

There rests

697745

332745

4246

596

231

21

10 years give

3650

There rests

697745

332745

4246

596

231

21

1 yeare gives

365

There rests

697745

332745

4246

596

231

21

Phamenoth compleate

210

There rests

697745

332745

4246

596

231

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 Annu∣all or Periodicall revolutions of the Planets may be knowne, with their mean or equal motion, for any part of those revoluti∣ons, 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 natu∣rall 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 pla∣nets 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 in∣equality 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 mo∣tion, 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 some∣times swifter in motion, then at other some, yet the cause of this inequa∣lity 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 ap∣parent motions must be lesse then 90 by the quantity of that whole ine∣quality 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 cen∣ter 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 qua∣drant.
• 10 And because the equal motion must not change and that the ap∣parent 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 cen∣ter

• 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, succeed∣ing 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 affecti∣ons 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 Tri∣angle, as that the one shall bisect the other.

An unequal sided Cone being cut through the Axis by a plaine perpen∣dicular 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 Tri∣angle A B and A C be bisected in the point X, and by the point of bi∣section 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.

From the points E V K draw the lines V F, K Q, E D parallel to A I, so shall the segments P R and R S, V Z and Z T be equal, because the right line A I bisects the base B C: and the lines E K and V T being con∣tinued to F and Q, the lines F K and V Q shall be contained between the parallels F V and K Q, and because the lines E K and D Q are contain∣ed between the parallels D E and K Q, and that E X and X K are equal, the triangles E X D and Q K X shall be like and the sides proportionall, that is E X to X D so is X K to X Q. And so is D E to Q K but E X

[illustration]

[illustration]

and X K are equal by construction, and therefore X D and X K are e∣qual, and so are D E & K Q and the perpendiculars Kδ, and Eφ shall be equal also. And because H M is equal to H R, & H R, parallel to Z X, there∣fore as H R, to H M, so is Z X to X M, because E D is parallel to Z M, ther∣fore X M to X E, so is X Z to X D, but X M & X Z are equal, & there∣fore X E and X D are equal, and the double of them E K and D Q are equal also. And because of the parallels V F, K Q, E D and A Z, it is as A Z to Z V so is E D to D V, and also as A Z to Z T so is K Q to Q T, and by composition as the right angle A Z V is to E D V, so is A Z T to K Q T: and because V Z and Z T are equal, and A Z common to both, therefore A Z V is equal to the angle A Z T, and E D V to K Q T, in which because the perpendiculars Kδ and Eφ are equal, T Q and D V are equal, and now if D V and T Q be both added to D T the right line V T shall be equal to D Q, but D Q is equal to E K and therefore also V T and E K are equal, as was to be proved.

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 El∣lipsis 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 pro∣portionall

between V X and X T, because it cuts the line V T at right an∣gles in the point X. And therefore the square of X O is equal to the rect∣angle 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 that we may now see how the aforesaid affections of the planets are represented in this figure, let us suppose the Ellipsis following to be exactly made according to these directions. Let X be the centor, M the, umbilique point agreeing with the Axis of the Cone and center of the El∣lipsis, whose plane parallel to the plane of the base is represented by the line F M. The conjugate Diameter let be the right line O X N divi∣ding the Ellipsis into two equal parts at the center X, Z is the center of the circle V Z T, the plane and Diameter of which circle passe by X the center of the Ellipsis, now because the center is Z and the Diameter V T

[illustration]

their V Z and Z T are equal, and the plane passing by the Axis doth in like manner divide all the Diameters of all the other circle•• E D, F G L Q and P K the which are also parallel to the base of the Cone.

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 to∣wards 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 mo∣ved 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 mid∣dle 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 A∣phelion, 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 Ellip∣sis, the arch Y O, or M X, and so the center of the Ellipsis is in X, the mid∣dle 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 notwith∣standing 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 grea∣ter circles, by lesser lines in lesser circles, & because the lesser circles are pla∣ced 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 cir∣cles comprehended between E D and P K, these circles are contained in one plain Superficies, and by these circles the planets describing an Ellip∣sis doe continually passe, and yet they are all of them excentricall in re∣spect 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 obser∣vation; 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 a∣bout 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 a∣bout 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 draw∣ing an Ellpsis, doth if not demostrate, yet at least illustrate that method.

[illustration]

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, keep∣ing 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 draw∣ing 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 Geometri∣cally 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 Dia∣meter, 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 fol∣lowes.

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 conveni∣ently

reckon the middle motion in any more circles then one, and there∣fore 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 pla∣nets 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.

[illustration]

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 mo∣tion 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 in∣crease 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 semi∣diameter 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.

[illustration]

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 cir∣cle 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 e∣quated 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 ac∣celeration 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 pro∣portionall of the planets acceleration; and therefore also (faith Bullial∣dus) 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 diffe∣rence answering to that distance. From hence and the two following pro∣blemes 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 dia∣meter▪ 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.

In the circle B C D whose center is A, let the points given be B, C, D, and let it be required to find the diameter of that circle upon which per∣pendiculars

[illustration]

being let fall fall from the points B, C, D, the segments of the Diameter shall be proportional to another proportion given: let the pro∣portion of the segments intercepted by the perpendiculars let fall from B and C be required to be as S to R.

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 there∣fore 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.

[illustration]

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 cir∣cle 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.

[illustration]

And because E F and D C are pa∣rallel, B D shall cut E F at right an∣gles 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 be∣ing given, there is found the Dia∣meter E F, upon which the perpen∣diculars 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.

[illustration]

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 inter∣valls 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 dif∣ference 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 pro∣portion 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 corre∣cted by Tycho, 1584 be they found; Between the Autumne and the Vern∣al 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.

[illustration]

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 A∣phelion 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 Ex∣centricity, 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.

[illustration]

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••duct∣ed 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 deduct∣ing 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 be∣tween 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 Tau∣rus 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 pla∣nets 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

As the sine of M B H	2. 05136 co. ar.	1. 4461743
Is to the side M H	3590	3. 5550944
So is the Sine of B M H	86. 76916	9. 9993071
To the side B H	100132	5. 0005758
The side X H	1795

[illustration]

[illustration]

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 ex∣actly 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 ad∣ding 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 with∣out 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 quan∣tity 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,

As the side H L	99938 co. ar.	5. 0002727
Is to Radius, so is K L	99019	4. 9957185
To the angle K H L	82. 22718	9. 9959912

And the angle H L K	07. 77282	And L F K	168. 50353
And the arch H K	15. 54564	And L F	84. 25176
The arch H I L add	175. 95083	The place of the Aphelion at F
The arch L I H B	191. 49647

[illustration]

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 de∣ducted 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 A∣lexander 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

08. 0278834536

Gemini

24. 2516600000

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 di∣stant 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 me∣ridians between Uraniburge and Alexandria, it will be at Uraniburge; Mechir 28. 83828. And the vernall Equinox observed by Tycho at Ura∣niburge 1588, was March the 9. 86458, and the Earths aphelion then was in Cancer 5. 39377; and therefore the arch answering to the ex∣centricity 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 E∣quinoctial 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 Zodi∣acks 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 Ano∣maly 985603599, these things premised we will now determine the E∣pochaes of the middle motions.

The middle Equinoctiall Anno Christi 1588, March 11. 9••966, is from the Aera Nabonassari 2336 Pharmuthi ••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 Nabonas∣sari. 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 Lon∣don, 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 Zodi∣ack, and distance from the earth may thus be found.

1 Write out the Epocha next before the given time and seve∣rally 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 Febru∣ary,

the dayes must be increased by an Unite) then adding them alto∣gether, 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.

[illustration]

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 Ae∣quations 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 di∣stance 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 longi∣tude and his Right Ascension; to be the absolute Aequation of naturall dayes, the which is also clearly demonstrated, according to the Coper∣nican 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 lon∣gitude from the Equinoctiall point in the ecliptick, ♎ F the like arch pro∣jected 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 me∣ridian, 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 dif∣ference 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 E∣quation of time) composed. In which you must enter with the signe and de∣gree of the Suns place either in the uppermost and left hand columnes de∣scending,

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.

[illustration]

CHAP. 9. Of the Theory and Motion of the Moon.

THe Moon according to our Hypothesis is a secondary planet, mo∣ving 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 E∣quation 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 revoluti∣on, 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 op∣position, 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 me∣thod we follow, in our calculation of this inequality of the Moon, ma∣king 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 Anoma∣ly, found by subtracting the former Equation from the simple Anomaly, which is 46. 11125

Aequation subtract

3. 49552

Aequated Anomaly

42. 61573

[illustration]

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, e∣qual to the angle A H G, which being subtracted from the Equated A∣nomaly 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 sub∣tracted 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 e∣vection, 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 e∣quated

Anomaly which suppose 25 degrees, the side M H 8724 and the side H E ••00000, to find the angle at E.

[illustration]

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 di∣stance 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 Bullial∣dus . 97805 which is so little differing from what we have found, that I have taken his Table and converted it into Deci∣mall numbers.

And for the finding the part proportial between the quadrature and the Zyzygia, Bullialdus whom we follow, hath annexed scruples of pro∣portion in this manner.

As Radius to 60 minutes or one degree; so is the fine of halfe the de∣grees 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 Reflecti∣on 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 some∣times 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 sub∣tracted, 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 o∣ther 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 dili∣gent and accurate observations, whose method of calculation is as fol∣loweth.

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 di∣stance 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,

[illustration]

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 re∣maines 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 inclina∣tion 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 Incli∣nation 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 In∣clination 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 la∣itude

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.

[illustration]

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 di∣stance 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 Longi∣tude 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 Lati∣tude.

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 where∣of 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 oppo∣site 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, name∣ly of the distance of the Moon from her Umbilique▪ 5. 0133408 and of the Eccentricity of her Evectioon 3. 6024365. And because the double di∣stance 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 dou∣ble 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 ano∣maly 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 Ano∣maly 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 an∣swering 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 there∣fore 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 accord∣ing to the title.

Example.

Suppose the distance of the Sun and Moone were 293. 39161, that is, de∣ducting 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 ap∣ply 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 agre∣gate 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 E∣quated, 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 ex∣cesse 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 an∣cients 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 Deci∣malls 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 Semidia∣meter 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.

[illustration]

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 Equa∣tion sought, to be subtracted from Saturns meane Longitude, the Anoma∣ly 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 rest∣eth the argument of Latitude: by help whereof and the angle of his great∣est 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 to∣wards 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 se∣cond 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, where∣as 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.

As AL	913876 co. ar.	4. 0391124
Is to Radius	90	10. 0000000
So is LX	37491	4. 5739••36
To the tang. of LA••	2. 35121	8. 6130460
As Radius
To AL	913876	5. 9608876
So cosine of LAX	2. 35121	9. 9996343
To AX	913107	5. 9605219

[illustration]

[illustration]

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.

As the sine of AXS	6. 04149 co. ar.	0. 9777838
To the side AS	100895	5. 0038707
So the sine of XAS	66. 22476	9. 9614845
To the side SX	877281	5. 9431390

[illustration]

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 La∣titude 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 Di∣mensions 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 El∣lipsis,

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 excentri∣city, 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 cōmon 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 Pla∣net, then what hath been already shewed, in the motions of Sa∣turn and Jupiter, we will proceed in the same method, and ga∣ther 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 Aequa∣tion sought, and to be subtracted from the planets mean Longitude, be∣cause 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 Radi∣us, 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 ♂ redu∣ced, 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 Mer∣cury, the investigation of whose places is much after the same manner with the former, the difference is in the second inequali∣ty, occasioned by their motion under the earth, the Orbs of the other Pla∣nets 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 an∣gle 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.

[illustration]

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 Radi∣us, 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 sub∣ducting 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.

[illustration]

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 ad∣ded to the Suns place, because the Suns place was subtracted from the Eccen∣trick of Venus reduced.

[illustration]

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 trian∣gle 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 A∣nomaly, 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 E∣quation 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 sub∣tracted 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 sub∣ductiug 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 di∣stance 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 shew∣eth 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 for∣mer 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 Alti∣tude 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 there∣fore 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 E∣clipse, 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.

[illustration]

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 semidiame∣ter AE, the angle of his apparent Semidiameter when he is Perig. AGE 16. 45, or in decimalls 27916 BHG represents the Earth. BG the Semi∣diameter 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 ap∣par••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 Pe∣rigaean; 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 sha∣dow.

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 an∣gle 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 Semidia∣meter 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 pro∣portion 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 Semidiame∣ter 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 Semidi∣ameter 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 there∣of 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 determi∣ned 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 di∣stance from the earth, 98374. Saturns true place from the Sun by calcula∣tion 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 an∣gle 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 be∣ing 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 subtract∣ing

Saturns place 73. 82167 from the Suns place 330. 02805 their diffe∣rences 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

[illustration]

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 lati∣tude 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 be∣ing 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 Sa∣turns 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 observa∣tions; 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 an∣gle 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.

As Radius		10. 0000000
To the side I O	92409	4. 9657145
So is the sine of H O I	1. 50	8. 4179190
To the side H I	2419	3. 3836335

Hence to find the angle of Latitude at the Sun

As the side I D	94338 co. ar.	5. 0253134
Is to Radius		10. 0000000
So is the side H I	2419	3. 3836335
To the sine of I D H	1. 46932	8. 4089469

[illustration]

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 gi∣ven 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 incli∣nation 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

[illustration]

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 Bulli∣aldus, 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 shew∣eth the time required compleat from the beginning of the moneth cur∣rent.

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 op∣posite, 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 quoti∣ent 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 re∣peat 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 ap∣parent.

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 con∣junction, if the true motion of latitude be within 17 degrees backward or forward of 6 or 12 signes, or at the opposition within 12 de∣grees, 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 neces∣sity 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 opposi∣tion the true latitude of the Moon, be lesse then the summe of the Semidi∣ameters of the Moon and the earths shadow, there must be an Eclipse o∣therwise 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 〈 in non-Latin alphabet 〉〈 in non-Latin alphabet 〉, 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 ano∣maly 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 durati∣on 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 Chap∣ter, 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 pro∣duct 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 recei∣ved, 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 Conjucti∣on 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 Nona∣gesime 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 de∣gree, with the remainder enter the Table of the Moones latitude, which if North adde to the angle of the Ecliptique and Horizon; if South sub∣tract 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 Pa∣rallax of the Sun and Moon, the difference of them is the Horizantall pa∣rallax 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 Nonage∣sime 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 ori∣entall 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 diffe∣rence 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 diffe∣rence 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 Con∣junction, 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, o∣therwise 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 Semidia∣meters 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 inci∣dence, 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 sub∣tract from the true motion of latitude at the time of the visible Sy∣nod, 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 moti∣on 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 ano∣ther at right angles in E, which point of intersection is the place of the Ecliptique where the Eclipse happens: upon which as a Cen∣ter 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 quan∣tity of the Semidiameter of the earths shadow, draw upon the same cen∣ter 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 be∣ginning; 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 man∣ner 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 in∣tersects 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.

[illustration]

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••.

[illustration]

CHAP. 37. The use of the Table of Refractions.

ALthough the Table of Refractions, belongs not to the calaulati∣on 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 ad∣ding the Refraction thereto.

FINIS.