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 first Book. (Book 1)

CHAP. 1. To find the Suns greatest Declination, and the Poles Elevation.

THe Declination of a Planet, or other Star is his distance from the Aequator, and as he declines from thence either Northward or Southward, so is his declination nominated North or South. And because that all the Planets (the Sun onely excepted) do move sometimes in and sometimes out of the Ecliptick besides there declination North or South from the Aequa∣tor, they have also latitude North or South from the Ecliptique, while the Planets keep in the ecliptique, one rule will serve to find their Declinati∣on, as well as the Declination, of the Sun, but if they have either North or South latitude, there must another rule be given, in both which rules the greatest Declination of the Sun is supposed to be known; and first there∣fore we will shew how, that may be found instrumentally, and then com∣pute the Declination of a Planet or other Star, with latitude or without.

For the finding the Suns greatest Declination, you must by a Quadrant or other Instrument, take his greatest and his least Meridian altitude; the difference between which altitudes is the distance of the Tropiques, and half the distance of the Tropiques, is the quantity of the Suns great De∣clination, as by the following Figure it doth appear. In which A Z B N represents the Meridian, E F the Eq••inoctial, ♋ ♑ the Zodiack, the

[illustration]

North pole D the South A B the Horizon Z the Zenith, N the Nadir, ♋ G a parallel of the Suns Diurnall motion at ♋, or the Suns greatest Declination from the Equator towards the North Pole. From whence it is apparent that from A to ♋ is the Suns greatest Meridian altitude from A to H his least, if therefore you deduct A H the least Meridian altitude from A ♋ the greatest, the difference H ♋ is the distance of the Tropiques, and because the angles ♋ center F and ♑ center E are equal, therefore the Suns greatest Declination towards the South Pole is equal to his great∣est Declination towards the North, and consequently halfe the distance of the Tropiques, or the arch that is the arch ♋ F is the quantity of the Suns greatest declination. And then if you deduct the Suns greatest declination or the arch A ♋ F from the Suns greatest Meridian altitude or the arch A ♋ the difference wil be A F or the height of the Aequator above the Horizon, the complement whereof to a quadrant is the arch A D equal to C B the height of the Pole.

Example.

The Suns greatest Meridian altitude taken, June 11 at London

61, 99167

The Suns least Meridian altitude, December the 10th.

14, 94167

Their difference is the distance of the Tropiques

47, 05000

Half that difference is the Suns greatest Declination whose difference from the greatest altitude is

23, 52500

The elevation of the Equator and the complement thereof to 90 is the Elevation of the Pole

38, 46667

 

51, 53333

CHAP. 2. The Suns greatest Declination being given, to find his Declination in any point of the Ecliptique.

LEt DFHG represent the Solstitial Colure, D BAG the Equator, F A H the Ecliptique, I the Pole of the Ecliptick, E the pole of the Equator, E C B a meridian line passing from E, through the Sun at C, and falling upon the Equator D A G with right angles at the point B. Therefore in the Rectangle sphaericall Triangle A B C we have known. 1 The Hy∣pothenuse A C the Suns distance from the next Aequinoctiall Point, whe∣ther Aries or Libra and may be supposed to be in 10 degrees of Cancer, and that being neerer unto Libra then Aries, I take his distance from Li∣bra which is two signs and 20 degrees, or 80 degrees.

[illustration]

2. We have known the angle B A C the Suns greatest declina∣tion, which by the accurate Observation of Tycho is found to be 23 deg. 31 minutes and 30 seconds. And in Decimall numbers 23. 525. Hence to find the present decli∣nation the proportion is.

As the Radius is to the sine of the Suns geatest declination. So is the sine of the planets distance from the next Aeqninoctiall point, To the sine of the de∣clination required.

And by this proportion, together with the help of the Canon of artifici∣all sines and tangents, I find the declination of the Sun at the time pro∣posed thus. First, I seek for the sine of Radius or 90 degrees, the measure of the right angle at B, and I find the sine thereof 10, 000000, next I seeke the sine of 80 degrees, and likewise the sine of 23. 5250, these two I adde

together and from their aggregate, I substract the Radius and what re∣maineth is the sine of the declination sought, as in the following work you may perceive.

As Radius
To the sine of B A C	23. 525	9, 6011352
So sine of A C	70.	9, 9729858
To the sine of B C	22. 02910	9, 5741210

CHAP. 3. How to find the Declination of a planet or fixed Star with Latitude.

What the Declination of a planet is, and how to find the same in any point of the Ecliptique hath been already shewed, we will now shew how to finde the same with latitude either north or south, and for demonstration sake. Let G F I H represent the Solstitial Colure, H A F the Aequator, I A E the Ecliptique, K M N a circle of latitude, G M C a cir∣cle of declination, A N the longitude of the star, M C the declination. Then in the Oblique angled spheical▪ Tri∣angle M G K we have limited. 1 The side K G the Suns great∣est declination. 2 The side K M the complement of lati∣tude. 3 The angle M K G the complement of longit. From hence the declination M C may be found at two operations, for first to find K O the pro∣portion is,

[illustration]

As the Radius to the Cosine of the complement of Longitude, so is the tangent of the Suns greatest declinati∣on, To the tangent of the first arch: K O, which being added to or sub∣stracted from the complement of Latitude K M, according to the direction ollowing their aggregate or difference will be the arch O M.

If the Declination sought be in the

• Northerne Signes and
  • North Latitude, substract the arch found from the complement of latitude and what remaineth is the se∣cond Arch O M.
  • South Latitude add the arch found to the comple∣ment of latitude and their aggregate is the second arch if lesse then a quadrant, but if more the com∣plement thereof to a semicircle.
• Southerne Signes and
  • North Latitude, adde the arch found to the comple∣ment of Latitude, and their aggregate is the second arch.
  • South latitude substract the arch found from the com∣plement of latitude, and what remaineth is the second arch.

And then the second proportion is,

As the Consine of the first arch found, Is to the cosine of the Suns great∣est declination, So is the Cosine of the second arch found, to the Cosine of an arch, whose complement is the declination sought: To make it plain we will in each case adde an example.

In the first Quadrant.

We will suppose a starre to be in 10 deg. of Gemini, with 4 deg. of north latitude.

The longitude of a starre so posited is 70 deg. the complement thereof M K G is 20 deg.

The latitude 4 deg. north, the complement thereof 86 M K. And the Suns greatest declination K G 23 deg. 5250.

Then according to the former analogie, first I say,

As the Radius	90.	10, 000000
To the Cosine of M K G	20.	9, 9729858
So is the tangent of K G	23. 5250	9, 6388198
To the tangent of K O	22. 25	9, 6118056
The complement of latitude	00 s. 86 d.	00
The arch K O	22	25
Their aggregate	108	25
Their difference	63	75

Because the latitude is North I take their difference and say,

As the Cosine of K O:	22. ••5	9, 9663954
To the Cosine of K G	23. 5250	9, 9623154
So is the Cosine of O M:	6••. 75	9, 6457058
To the Cosine of G M:	64. 01	9, 6416258

The complement thereof 25. 99 is the declination sought.

To find the declination of the same point with 4 degrees of South lati∣tude: I take the complement of 108. 25. viz. 71. 75.

And say as the Cosine of K O	22. 25.	9, 9663954
To the Cosine of K G	23. 5250	9, 9623154
So the Cosine of O M	71. 75	9, 4957715
To the Cosine of G M	71. 93	9, 4916915

whose complement is the declination sought, viz. 18. 7.

The point of the Ecliptique answering hereunto in the 2 quadrant, is the 20 of Cancer, and hath the same declination North and South. In the third and fourth quadrant the 10 of Sagittary, and the 20 of Capri∣corn, have the same declination, but with contrary latitude.

In the second Quadrant.

We will suppose a starre in 10 deg. of Leo whose longitude is 130. The latitude 4 deg. north.

The complement of longitude M K G is 50, but because the longitude is more then a quadrant, I take the complement of 50, viz. 40 deg. The complement of latitude M K is 86 as before. And the Suns declination fixed, first then I fay,

As the Radius	90	10, 0000000
To the Cosine of M K G	40	9, 8842539
So the tangent of K G	23. 5250	9, 6388198
To the tangent of K O.	18. 44	9, 5230737
The complement of latitude		86. 00
The arch K O		18. 44
Their aggregate		104. 44
Their difference		67. 56
For the north latitude I take their difference and say,
As the Cosine of K O	18. 44	9, 9771084
To the Cosine of K G	23. 5250	9, 9623154
So the Cosine of O M	67. 56	9, 5817400
To the Cosine of G M	68. 35	9, 5669470

whose complement 21. 65 is the declination sought.

For the declination of the same point with South latitude, I take the complement of 104. 44: viz. 75. 56. And say,

As the Cosine of K O	18. 44	9, 9771084
To the Cosine of K G	23. 52. 50	9, 9623154
So the Cosine of O M	75. 56	9, 3968372
To the Cosine of G M	76. 05	9, 3820442

whose complement 13. 95 is the declination with South latitude

The point of the Ecliptique answering hereunto in the first quadrant is the 20 of Taurus, and hath the same declination North latitude North and South.

In the third and fourth Quadrants the points answering thereunto are the 20 of Scorpio and the 10 of Aquarius, and have the same declinati∣on but with contrary latitude.

CHAP. 4. How to find the Right Ascension of any point of the Ecliptique.

THe Ascension of the Sun or Starres is the degree of the Aequator that riseth with the same above the Horizno. And the Descension of it is the degree of the Aequator that goeth under the Horizon with the same, both these are either Right or Oblique. The Right Ascension or Descension is the degree of the Aequator that Ascendeth or Descendeth with the Sun or other starre in a Right Spheare, and the Oblique Ascension is the degree of the Aequator, that ascendeth or descendeth with the same in an ob∣lique spheare. The former of these is simple, and of one kind onely; be∣cause there can be but one position of a Right spheare, but the later is va∣rious and manifold according to the diverse inclination of the same. To find the Right Ascension of a planet in the Eclptique, There must be gi∣ven as in the first Chapter, the planets longitude or distance from the next Equinoctiall point: and the Suns greatest declination. Then in the Rect∣angle sphericall Triangle of the first chapter A B C, we have limi∣ted.

• 1 The angle B A C, the Suns greatest declination 23. 31. 30.
• 2 The Hypothenuse A C, the Suns distance from the next Equino∣ctiall point, whose place we will suppose to be in 10 degrees of Gemini, and consequently his distance from Aries is 70 degrees. Hence to find the base A B, the right Right Ascension the point sought, the proportion is.

As the Radius is to the tangent of the planets distance from the next Ae∣quinoctial

point: So is the Cosine of the Suns greatest declination▪ to the tangent of the Right Ascension of the point sought, Example.

As the Radius	90	10, 0000000
To tangent of A C	70.	10, 4389341
So is the Cosine of B A C	23. 5250	9, 9623154
To the tangent of A B	68. 34874	10, 4012495

which is the Right Ascension of the Sun or any other planet without lati∣tude, when they be in the the 10 degree of Gemini.

Note that if the Right Ascension of the point sought be in the second quadrant (is in Cancer, Leo, Virgo.) you must take the complement of the arch found to a semicircle: if in the third Quadrant (as in Libra, Scor∣pio, Sagitarius) you must add a semicircle to the arch found: if in the last quadrant (as in Capricorn, Aquarius, Pisces) you must substract the arch found from a whole circle or 360, and so shall you have the Right Ascension of any point of the Ecliptique, to make it plaine we will in each case add an Example.

In the second Quadrant.

Let the Right Ascension of the point sought be 10 degrees of Virgo, the distance thereof from Libra, which is the next Aequinoctiall point is 20 degrees, according therefore to the former Analogie, I say.

As the Radius	90.	10. 0000000
To the tangent of A C	20.	9. 5610658
So is the cosine of B A C	23. 5250	9. 9623154
To the tangent of A B	18. 45	9. 5233813

whose complement to a semicircle 161. 55 is the Right Ascension thereof.

In the third Quadrant.

Let the Right Ascension of the point sought be in 14 degrees of Scorpio, the distance thereof from Libra, the next Aequinoctiall point is 44 de∣grees. Therefore I say.

As the Radius	90.	10. 0000000
To the tangent of A C	44. deg.	9. 9848371
So is the Cosine of	23. 5250	9. 9623154
To the tangent of A B	4152	9. 9471525

to which if you adde a semicircle or 180 degrees the Right Ascension of of the point sought will be 221. 52.

In the last Quadrant.

Let the Right Ascension of the point sought, be 22 degrees of Aquarius; the distance thereof from Aries, which is the next Aequinoctiall point ••s 38 degrees. Therefore I say.

As the Raduis	90.	10. 000000
To the Tangent of A C	38.	9. 8928098
So is the Cosine of	23. 5250	9. 9623154
To the tangent of A B	35. 62	9. 8551252

which being subtracted from 360 there rests▪ 324. 38 for the Right Ascen∣sion of the point sought.

CHAP. 5. How to finde the Right Ascension of a Planet or other Star with Latitude.

THe Declination being found by the 3 Chapter, we have in the ob∣lique angled Sphericall Triangle of that Diagram G K M all the sides given with the angle M K G, therefore to finde the angle K G M, say. As the Cosins of the declination, is to the Cosine of the Planets distance from the next Aequinoctiall point. So is the Cosine of its latitude to the Cosine of its Right Ascension.

For Example. The Declination of 10 degrees of Gemini, was found to be 25. 99. with 4 degrees North Latitude. Whose complement is G M. 64. 01 the complement of Longitude is the angle M K G 20, the comple∣ment of Latitude is M K 86, hence to finde the Right Ascension, the Analogie is.

As the Sine of G M	64. ••01. co. arith.	0. 0463059
To the sine of M K G	20.	9. 5340516
So is the sine of K M	86.	9. 9984407
To the sine of K G M.	22. 28	9. 5787982

whose complement 67. 72 is the right Ascension of a Star in 10 de∣grees of Gemini, with 4 degrees of North Latitude.

CHAP. 6. The Elevation of the Pole and Declination of the Sun given, to find his Amplitude.

THe Amplitude of the Suns rising or setting is an arch of the Horizon intercepted betwixt the Aequator and the place of the rising and set∣ting of the Sun. And it is either Northern or Southward, the Northern Amplitude is when he sets and riseth on this side of the Equator, toward he North Pole: and the Southern when he sets or riseth on the contrary side. Now when the Sun is in the Aequator, he hath no Amplitude at all: but when he is in the Solstitial points, he hath then the greatest Amplitude. That we may find then the Suns amplitude or distance from the East or West points at the time of his rising or setting; let D P L G F represent the Me∣ridian, F A I the Horizon, D A L the Equinoctial. P the Pole of the Aequator.

[illustration]

Then in the Rectangle Spherical Triangle A B C, let there be given the angle B A C or complement of the Poles elevation, 38. 47 and B C the Suns Declination 23. 15. To find A B the Suns Amplitude, The Ana∣logie is.

As the sine of B A C.	38. 47. co. ar.	0. 2061365
To Radius.	90.	10. 0000000
So is the sine of B C.	23. 15.	9. 5945468
To the sine A B	39. 19.	9. 8006833

CHAP. 7. The Meridian Altitude and Declination of the Sun with the Poles Elevation given to finde his true place in the Zodiac.

IF the Meridian Altitude of the Sun be lesse then the complement of the Poles elevation, subtract the meridian altitude from the height of the Ae∣quator, and what remaineth is the Suns Declination towards the South Pole; But if the Meridian altitude of the Sun be more then the height of the Aequator, subtract the height of the Aequator from the Meridian alti∣tude and what remaineth is the Suns Declination towards the South Pole.

Then in the Diagram of the second Chapter, in the right angle Sphaeri∣call Triangle A B C, we have known, the angle B A C the Suns greatest Declination, and the perpendicular B C the present Declination to finde the Hypothenuse A C, the Suns distance from the next Equinoctiall point, or true place in the Zodiac, for which the Analogie is.

As the sine of B A C.	23. 5250 co. ar.	0. 3988648
To the sine of B C.	23. 15	9. 5945468
So is the Radius or whole sine.	90.	10. 0000000
To the sine of A C	80. 04	9. 9934116

That is in 20 degrees of Gemini, and 4 Centesmes, if the Meridian al∣titude were taken in Summer: But in 20 degrees 04 Centesmes of Capri∣corn, if the Meridian altitude was taken in Winter.

CHAP. 8. Having the Meridian Altitude of an unknown Star and the dist∣ance thereof from a known Star, to finde the Right As∣cension of the unknown Star.

ABout the end of the year 1577 Tycho observed the distance of the litle Star in the breast of Pegasus from the bright Star of the Vultur to be exactly 45 deg. 31 min. or in decimal numbers 45. 51667. And by the Meridian altitude thereof, he found the Declination to be 22 deg. 26 min. North that is 22. 43333. Which given the Right ascension of the said Starre is to be enquired. For the finding whereof in the oblique angled

Sphaericall Triangle (of the annexed Diagram) F O L, there is knowne First, F L the complement of the Declinat. of the bright Star of the Vultur 82. 13333. Secondly F O the complement of the Declination of the Star in the brest of Pegasus, 67. 56667. Thirdly O L the distance of them, 45. 51667, to finde the angle at F or difference of their Right ascen∣sion.

[illustration]

The side L O.	45. 51667.
The side F L.	82. 13333.
The side F O.	67. 56667.
Sum of the sides.	195. 21667.
The halfe Sum.	97. 60833.
Sine of F L.	82. 13333. co. ar.	0. 0041064
Sine of F O.	67. 56667. co. ar.	0. 0341757
Dif. of F L ½ sum.	15. 47500.	9. 4262148
Dif. of F O ½ sum.	30. 04167.	9. 6995164
Quadrat of the sine of halfe the angle.		19. 1640133
Halfe is the sine of	22. 45453.	9. 5820066

Double is the angle L F O. 44. 90906. Equall to the arch D E the diffe∣rence of their Right ascensions, which being added to the Right ascension of the bright Star of the Vulture. 292. 58333. The summe 337. 49239 is the Right ascension of the little Star in the breast of Pegasus.

CHAP. 9. Having the Declination and Right ascension of a star given, to finde the longitude and latitude thereof.

IN the Diagram of the 3 Chapter, having the Right ascension of the little Star in the breast of Pegasus A C. 337. 49239. And the decli∣nation C M. 22. 43333. with the greatest obliquity of the Ecliptique

B A C. 23. 5250. we are to enquire its Longitude A N. and Lati∣tude M N. wherefore in the Triangle A B C. we have the angle B A C. 23. 5250. and the side A C 22. 50761 the complement of the Right ascen∣sion: then I say.

As Radius.
To Tangent of B A C	23. 5250	9▪63••••198
So sine of A C.	22. 50761	9. ••8••••787
To Tangent of B C.	6. 46125	9. 2217985
Adde C M.	22. 43333 The Declination
Sum is M B.	31. 89458

when the Declination is South the arch found must be subtracted from it, and their difference shall be M B.

2. To finde the angle A B C.

As the sine of B C	9. 46125 co. ar.	0. 7841497
To the sine B A C	23. 5250	9. 6011352
So the sine of A C	22. 50761	9. 5829787
To the sine of A B C	68. 36843	9. 9682836

3. To finde the side A B.

As the sine of B A C.	23. 52520. co. ar.	0. 3988648
To the sine of B C.	9. 46125.	9. 2158503
So is Radius.		10. 0000000
To the sine of A B.	24. 31967	9. 6147151

4. The angle A B C is equall to the angle M B N, therefore to finde the latitude M N.

As the sine M N B	90.
To the sine of M B	31. 89458.	9. 722928••
So the sine of M B N.	68. 36843.	9. 9682836
To the sine of M N.	29. 41602.	9. 6912118

Lastly, to finde the arch B N.

As Radins.
To Cotangent of. M B N.	68. 36843.	9. 5983151
So Tangent of M N.	29. 41602.	9. 7511554
To the sine of B N.	12. 92052.	9. 3494705

Which is to be added to A B if the Right ascension be lesse then a semi∣circle, but if the Right ascension exceed 180, as in our example, the Com∣plement of B N. 357▪07948, is the longitude desired.

CHAP. 10. How to finde the Ascensionall Difference.

THe Ascensional Difference, is nothing else but the difference be∣tweene the ascension of any point of the ecliptique in a Right Sphere, and the Ascension of the same point in an oblique Sphaere, as in the annexed Diagram, Let A G E V represent the

[illustration]

Meridian, E M T the Horizon, G D M C V the Aequator, D B part of the Zodiac, A the Pole thereof, D the beginning of Aries, V T the com∣plement of the Poles elevation, B C the Suns Declination, D C the Right ascension, M C the ascensional Difference.

Then in the Right angled sphaericall Triangle B M C we have limited. 1. The angle C M B the complement of the Pole 38. 46667. Secondly, the side B C 19 deg. the Suns Declination, hence to finde, the ascension∣al difference M C the Analogie is.

As the Cotangent of the poles Elevation, is to Radius. So is the tangent of the planets Declination, to the sine of the ascensionall difference.

As the tangent of C M B	38. 46667. co. ar.	0. 09991••6
Is to Radius		10. 000000
So is tangent of B C	22. 0291	9. 6070441
To the sine of M C	30. 61613	9. 7069577

which is the Ascensionall Difference sought.

CHAP. 11. How to find the Oblique Ascension or Descension of any point in the Ecliptique.

OBlique ascension is when a less arch or portion of the Aequator ••iseth, then doth of the Zodiack, or else of that Star may be said to rise obliquely with whom a less portion of the Aequator as∣cendeth above the Horizon, & so the oblike Descension or setting of a Star, is when a less proportion of the Aequator descendeth with it, then doth in a right Sphere.

In the former Diagram, D C represents the right Ascension, M C the Ascensional Difference, D M the oblique Ascension, D B an arch of the Ecliptique above the Horizon, which being greater then D M, a Star in this position of the Sphere, is said to rise obliquely. The quantity whereof is found, by deducting the Ascensional difference C M from the right as∣cension D C, according to the direction following.

If the Declination be

• North
  • Subt. The Ascensional Difference from the right Ascension, and it giveth the oblique Ascension.
  • Adde The Ascensional Difference to the Right Ascension, and it giveth the oblique Descension.
• South
  • Adde The Ascensional Difference to the Right Ascension, and it giveth the oblique Ascension.
  • Subt. The Ascensional Difference from the Right Ascensi∣on, it giveth the oblique Descension.

Right Ascension of ten degrees of Gemini,

68. 348••••

Ascensional Difference

30. 61613

Oblique Ascension of ten degrees of Gemini,

37. 73261

Oblique Descension of ten degrees of Gemini,

99. 08137

CHAP. 12. The Poles Elevation and the Suns Declination being given, to finde his Altitude at any time assigned.

IN this proposition there are three varieties, first, when the Sun is in the Aequator, that is the begining of Aries and Libra, in which case sup∣posing in this Diagram the Sun to be at H two houres or 30 degrees distant from the Meridian A, and the poles Elevation R F equall to A C 51. 53333, the angle at A being right, I say.

[illustration]

As the Radius.
To the Cosine of A H	30	9. 9375••06
So the Cosine of A C	51. 53333	9. 793831••
To the Cosine of C H	57. 40351	9. 7313624

whose complement 32. 59649 is the side L H or altitude sought.

The second varietie is when the Sun is in the Northerne signes Aries, Taurus, Gemini, Cancer, Leo, Virgo. For the solving of the Probleme in this varietie, let A E C represent the Aequinoctiall. F the pole thereof, L E R the Horizon, G the pole thereof. B D a parallel of the Suns decli∣nation. F O the Meridian of the Sun. B H the distance of the Sun from the Meridian. H O the Suns declination North, R F the poles elevation, F G the complement.

Admit the Sun at H be distant from the Meridian B 45 degrees, and in ten degrees of Gemini, where his declination O H is 23. 02910. which

[illustration]

being taken out of the Quadrant F O 90. there remains F H. 67. 97090. which known in the Triangle F G H, we have given these three parts.

First, the side F G 38. 46667. Secondly, the side F H. 67. 97090. and the included angle F G H. 45 degrees to find G H.

[illustration]

As the Radius
To the Tangent of F G	38. 46667	9. 9000864
So the Cosine of F G H	45.	9. 8494850
To the Tangent of F K	29. 32669	9. 7495714
Then from F H	67. 97090
Deduct F K	29. 32669
There rests H K	38. 64421
2. As the Cosine of F K	29. 32669.co. ar.	0. 0595626
To the Cosine of F G	38. 46667	9. 8937451
So Cosine of K H	38. 64421	9. 8926710
To the Cosine of G H	44. 54111	9. 8459787

The complement whereof N H 45. 45889. is the altitude of the Sun above the Horizon.

The third variety is when the Sun is in the southern signes Libra, Scorpio, Sagittarius, Capricorn, Aquarius, Pisces. And in this case supposing the Sun to be in 10 deg. of Sagittarius, and having South de∣clination 22. 02910, and also 45 deg. distant from the Meridian as before in the oblique angled triangle F G H of the annexed Diagram we have gi∣ven 1. The side F G. 38. 46667. 2. The side F H 112. 02910 and the angle G F H 45 deg. Then as before, I say

[illustration]

As Radius
To the tangent of G F	38. 46667	9, 9000864
So Cosine of G F H	45.	9, 8494850
To the tangent of F K	29. 32669	9, 7495714
From F H	112. 02910
Deduct F K	29. 32669
Rest K H	82. 70241

As the Cosine of F K	29. 32669 co. ar.	0, 05956••6
To the Cosine of F G	38. 46667	9, 8937451
So the Cosine of K H	82. 70341	9, 1038818
To the Cosine of G H	83. 43974	9, 0571895

whose compl. H N. 6. 56026 is the Suns altitude required. The like is to be observed of the Moon with the other planets and fixed Starres.

CHAP. 13. Having the Suns greatest Declination, with his distance from the next Equinoctiall point, to find the Meridian angle or intersection of the Meridian with the Ecliptique.

IN the Diagram of the 10 chapter, we have in the triangle D C B. 1. The angle B D C 23. 5250 the Suns greatest Declination. 2 The Hypothenusal D B 70 the distance of the Sun from Aries. 3 The an∣gle B C D 90, to find the angle D B C. The Analogie is,

As the Radius
To the tangent of B D C.	23. 525	9. 6388198
So the cosine of D B	70	9. 5340516
To the Cotang. of D B C	81. 53133	9. 1728714

or the angle of the Ecliptique with the Meridian.

CHAP. 14. To find the angle of the Meridian with the Horizon.

IN the Diagram of the 5 chapter, we have in the triangle B C M, first, the angle B M C 38. 46667 the elevation of the Equator. 2 B C 22. 02910 the Declination of the point given, to find M B C.

As the Cosine of B C	22. 02910 co ar.	0. 0329234
Is to Radius		10. 000000
So is Cosine of B M C	38. 46667	9. 8937451
To the Sine of M B C	57. 63275	9. 9266685

CHAP. 15. The Poles elevation, with the Suns Altitude and Declination gi∣ven, to find his Azimnth.

IN the oblique angled sphaericall triangle G H F, in the second Dia∣gram of the 12 Chapter, we have known H F the complement of the Suns Declination. 2 The side G F the complement of the poles ele∣vation.

3 The side H G the complement of the Suns altitude, to find H G F.

••. The side F H

67. 97090

2. The side G F

38. 46667

3. The side H G

44. 54111

Summe

150. 97868

Halfe summe

75. 48934

Sine of G F	38. 46667 co. ar.	0. 2061682
Sine of H G	44. 54111 co. ar.	0. 1540214
Diff. G F ½ sum	37. 02267	9. 7796909
Diff. H G ½ sum	30. 94823	9. 7111855
Quadrat of the Sine of ½ the angle		19. 8510560
Half is the Sine of	57. 39644	9. 9255280

whose double 104. 79288 is the Suns Azimuth from the North, and the complement thereof 75. 20712 is the Suns Azimuth from the South.

CHAP. 16. How to erect a Figure of Heaven.

AMong the severall wayes for the erecting of a figure, used by the ancient Astronomers, that is held most rationall which divideth the Equinoctiall into twelve parts by circles meeting at the inter∣sections of the Meridian and Horizon, which is according to the following Scheame, in which the line W Z E represents the East and West Azimuth W ♈ E is that halfe of the Equinoctiall above the eareh, and W ♎ E is that half of the Equator under the earth. The arch 7 ♈ 1, doth represent that halfe of the ecliptick above the earth.

And 7 ♎ 1 that part under the earth. The utmost circle N E S W re∣presents the Horizon, N Z S the meridian, N the north, and S the South end thereof. The Eqninoctiall circle W ♈ E ♎ is divided in to 12 equall parts, by which divisions passe arches, from the North and South inter∣sections of the meridian with the Horizon, which cut the Ecliptique at the cuspe of the houses, N 1 S is the cuspe of the Ascendant, N 7 S of the 7 house N 2. 12 S of the second and twelfth, N 3. 11 S of the third and e∣leventh, and so of the rest as you see in the Figure.

[illustration]

To find the Mid-heaven or Tenth House.

When you would erect a figure you must have the true place and Right Ascension of the Sun for the time proposed, let the given time in the me∣ridian of London where the North pole is elevated above the Horizon 51. 53333 be July the 16. After noon 18 hours, 4564 in the yeare 1587. For which time according to our Tables the Suns place is in Vir∣go 4. 07368 which known the Midheaven or 10 house may thus be found.

First, enquire the Right Ascension of the Sun at the time proposed by the 4 Chapter thus

As the Radius
To the tangent of A C	25. 92632	9. 6867627
So consine of B A C	23. 52500	9. 9623154
To the tangent of B C	24. 02427	9. 6490781

whose complement 155. 97573 is the Right Ascension sought.

To the Right Ascension of the Sun

155. 97573

Adde the Right Ascension of time

276. 84600

Right Ascension of the Mid-heaven

432. 82173

Deduct a whole circle

360.

Then rests for the Right Ascension

072. 82173

Then in the Rectangle Sphericall Triangle ♈ 10 F we have given. 1 ♈ F the Right Ascension of the mid-heaven. 72. 82173 2 The Suns greatest Declination F ♈ 10. 23. 5250 to find ♈ 10 the point culminating.

As the Radius.
To the Cotangent of F ♈	72. 82173	9. 4901498
So cosine of 10 ♈ F	23. 5250	9. 9623154
To the cotangent of ♈ F O	74. 17501	9. 4524652

To find the Cuspe of the 11 house.

Unto the Right Ascension of the 10 house

72. 82173

Adde 30 degrees

30. 00000

And then the oblique Ascension of the 11 house

102. 82173

To this adde

30.

Oblique Ascension of the 12 house

132. 82173

 

30.

30.

30.

Oblique Ascension of the Ascendant

162. 82173

 

30.

30.

30.

Oblique Ascension of the 2 house

192. 82173

 

30.

30.

30.

Oblique Ascension of the 3 house

222. 82173

The Oblique Ascension of the houses thus found by a continuall additi∣on of 30 degrees, we must next find the poles elevation upon their severall circles of position thus, In the rectangle sphericall Triangles E A B and E C D we have limited. 1. Their common angle at E the latitude of the place. 2. E B and E D 38 and 60 with the right angles A and C to find the angles B and D, being the angles that the circles of position make with the Equator.

First, then for the 11 and 3 houses.

As the Radius
To the Cosine of E D	60.	9. 69470••0
So tangent of C E D	51. 53333	10. 0999136
To cotangent of C D E	57. 81626	9. 7988836

whose complement 32. 18374 is the height of the pole above those circles of position.

For the 12 and 2 houses.

As the Radius
To the Cosine of E B	30	9. 375306
So tangent of A E B	51. 53333	10. 0999236
To cotangent of A B E	42. 53308	10. 0374442
whose complement	47. 46692 is the height of the pole required.

Having thus found the severall oblique Ascensions of the severall houses together with the elevation of the pole, above their severall circles of po∣sition, in the oblique angled sphericall Triangle ♈ C D we have limited 1. The angle ♈ D C complement to C D E. 2. The angle C♈D the Suns greatest Declination. 3. Their included side ♈ D the oblique Ascension of each house to find ♈ C the point of the Ecliptique. The Analo∣gies are.

For the 1 Operation.

• As the Sine of the halfe summe of the angles.
• To the Sine of halfe their difference.
• So the tangent of halfe the side comprehended.
• To the tangent of halfe the difference of the sides

For the 2 Operation.

• As the Cosine of the half summe of the angles
• To the Cosine of their halfe difference.
• So the Tangent of halfe the side comprehended.
• To the Tangent of half the Summe of the sides.

The Summe of these two arches, shall give you the point of the Eclip∣tique desired.

For the Cuspe of the 11 house.

The oblique Ascension or arch ♈ D is

102. 82173

The halfe thereof is

51. 41086

The complement of C D E 57. 81626 is the angle

♈ D C.	122. 18374.
C ♈ D.	23. 52500.	Or the Suns greatest Declination
Summe	145. 70874	Half Summe	72. 85437
Differ.	98. 65874	½ Differ.	49. 32937
Sine ½ Summe	72. 85437		co. ar.	0. 0197429
Sine ½ Differ.	49. 32937			9. 8799374
Tang. ½ ♈ D	51. 41086			10. 0980083
Tang ½ Differ.	44. 84740			9. 9976886

For the 2 Operation.

Cosine ½ Summe	72. 85437 co. ar.	0, 5304705
Cosine ½ Differ.	49. ••2937	9. 8140541
Tangent ½ ♈ D	51. 41086	10. 0980083
Tangent ½ Summe	70. 15211	10. 44••5329
1 Arch adde	44. 84740

Summe 114. 99951 or ♋ 24. 99951 is the point of the Eclip∣tique for the 11 house.

For the 12 house.

The Oblique Ascension or arch ♈ D is

132. 82173

The halfe thereof is

66. 41086

The complement of C D E 42. 53308 is the angle

♈ D C	137. 46692
C ♈ D	23. 52500
Summe	160. 99192	½ Summe	80. 49596
Differ.	113. 94192	½ Differ.	56. 97096
Sine ½ Summe	80. 49596	co. ar.		0. 0060025
Sine ½ Differ.	56. 97096			9. 9234483
Tang. ½ ♈ D	66. 41086			10. 3598529
Tang. ½ Differ.	62. 81111			10. 2893037

For the 2 Operation.

Cosine ½ Summe	80. 49596 co. ar.	0. 7822080
Cosine ½ Differ.	56. 97096	9. 7364475
Tangent ½ ♈ D	66. 41086	10. 3598529
Tangent ½ Summe	82. 46470	10. 8785084
1 arch adde	62. 81111

Summe 145. 27581 or ♌ 25. 27581 is the point of the Eclip∣tique for the 12 house.

For the Cuspe of the Ascendant.

The Oblique Ascension or ♈ D is

162. 82173

The halfe thereof is

81. 41086

The angle C D E is the same with the height of the Equator 38. 46667 whose complement to a semicircle is the angle.

♈ D C	141. 53333
C ♈ D	23. 52500
Summe	165. 05833	½ Summe	82. 51916
Differ.	118. 00833	½ Differ.	59. 00416

Sine ½ Sum 82. 52916	co. ar.	0. 0037026
Sine ½ Differ. 59. 00416		9. 9330844
Tangent ½ ♈ D 81. 41086		10. 8209024
Tangent ½ Dif. 80. 09004		10. 7576894

For the 2 Operation.

Cosine ½ Summe	82. 52916 co. ar.	0. 8859846
Cosine ½ Differ.	59. 00416	9. 7117867
Tang. ½ ♈ D	81. 41086	10. 8209024
Tang. ½ Summe	87. 81606	11. 4186737
1 Arch add	80. 09004	11. 4186737

Summe 167. 90610. Or ♍ 17. 90610 is the point of the Ecliptique for the Ascendant

For the Cuspe of the 2 house.

The Oblique Ascension is 192. 82173 whose complement is the Arch ♈ D 167. 17827.

The half whereof is 83. 58913. The summe and Difference of the an∣gles are the same with those for the 12 house.

Therefore I say first,

Sine ½ Summe	80. 49596 co. ar.	0. 0060025
Sine ½ Differ.	56. 97096	9. 9234483
Tangent ½ ♈ D	83. 58913	10. 9493879
Tangent ½ Diff.	82. 47036	10. 8788387

For the 2 Operation.

Cosine ½ Summe	80. 49596 co. ar.	0. 7822080
Cosine ½ Differ.	56. 97096	0. 7364475
Tangent ½ ♈ D	83. 58913	10. 9493879
Tangent ½ Summe	88. 05055	11. 4680434
1 Arch adde	82. 47036

Summe 170. 52091 whose complement 189. 47909 or ♎ 9. 47909, is the Cuspe of the 2 house.

For the Cuspe of the 3 house.

The Oblique Ascension is 222. 82173, whose complement is the arch ♈ D 137. 17827. The half whereof is 68. 58913. The summe and dif∣ference of the angles are the same with these for the 11 house.

Therefore I say first,

Sine ½ Summe	72. 85437 co. ar.	0. 0297429
Sine ½ Differ.	49. 32937	9. 8799374
Tangent ½ ♈ D	68. 58913	10. 4065869
Tangent ½ Diff.	63. 71032	10. 3062672

For the 2 Operation.

Cosine ½ Summe	72. 85437 co. ar.	0. 5304705
Cosine ½ Differ.	49. 32937	9. 8140541
Tangent ½ ♈ D	68. 58913	10. 4065869
Tangent ½ Sum	79. 94186	10. 7511115
1 Arch adde	63. 71032

Summe 143. 65218 whose complem. 216. 34782 or ♏ 6. 34782 is the point of the Ecliptique for the Cuspe of the third house.

The six Orientall houses being thus found, the other 6 are also found, by consequence, being the same degrees and parts of the opposite signes.

And thus we have not onely erected a figure for the time given, but composed a Table, for the generall erecting a figure in this latitude, for by adding together the first and second numbers, in every of these proportions, there is composed 2 numbers for every house, to each of which the Artificiall tangent of halfe the oblique Ascension being added, their aggregates are the tangents of two arches, which added together is the distance of the Cuspe of the house from the first point of Aries. One∣ly note that if the oblique Ascension be more then 180 deg. you must take the tangent of halfe the complement to a whole circle; And to find the Cuspe of the house, you must also take the complement of the added arches, as shall be cleared by example.

A New Table of Houses for the Latitude of London.

	11 & 3 hous.	Ascendent	12 & 2 hous.
1 Oper.	9. 8996803	9. 9367870	9. 9294508
2 Oper.	10. 3445246	10. 5977713	10. 5186555

Having found the oblique Ascensions of the houses as before was shew∣ed, take out the tangent of half thereof from a Table of Tangents, whose Radius is 10. 0000000, and set them down twice on your paper as you see in the following example, then seek in the top of the Table, the house whose cuspe you would find; and under the same you have two numbers, the first of which being added to one of the tangents in the paper, and

the second to the other, will give you the tangents of two arches, whose aggregate is the Cuspe desired.

Example.

The Oblique Ascension of the 11 house was before found to be 102. 82173. And the tangent of the halfe thereof 51. 41086 is 10. 0980083 the first number in the Table under the 11 house is 9. 8996803 their same rejecting Radius 9. 9976886, is the tangent of 44. 84740. The second number in the Table.

Tangent	51. 41086	10. 0980083
1 Number		9. 8996803
Tangent	44. 84740	9. 9976886

is ••0. 3445246, which being added to the same tangent, there summe 10. 4425329 is the tangent of 70. 25211 and the aggregate of these two arches.

Tangent	51. 41086	10. 0980083
2 Number		10. 3445246
Tangent	70. 15211	10. 4425329
1 Arch	44. 84740
Summe	114. 99951

gives the point of the Ecliptique in Cancer 24. 999

2 Example.

The Oblique Ascension of the 2 house was before found to be 192. 82173 The tangent of halfe the complement 83. 58913 is 10. 9493879

Tangent	83. 58913	10. 9493879
1 Number		9. 9294508
Tangent	82. 47036	10. 8788387

Tangent	83. 58913	10. 9493879
2 Number		10. 5186555
Tangent	88. 05055	11. 4680434

1 Arch. 82. 47036 Summe 170. 52091 whose complement ••••••••47909▪ is the Cuspe desired.

The Cuspe of your houses being thus found, draw your figure after this manner, and then place the Signes and Degrees on the Cuspes of their proper houses, beginning with the 10 house.

[illustration]

A Figure of the 12 Houses of Heaven.

10 house ♊ 14. 17.	The opposite houses are in the opposite Signes.	4 house ♐ 14. 17.
11 house ♋ 24. 99.	The opposite houses are in the opposite Signes.	5 house ♑ 24. 99.
12 house ♌ 25. 27.	The opposite houses are in the opposite Signes.	6 house ♒ 25. 27.
Ascendent ♍ 17. 90.	The opposite houses are in the opposite Signes.	7 house ♓ 17 90.
2 house ♎ 09. 47.	The opposite houses are in the opposite Signes.	8 house ♈ 9. 47.
3 house ♏ 06. 34.	The opposite houses are in the opposite Signes.	9 house ♉ 6. 34.

The 12 signes being placed in the figure we must next enquire for the true places of the Planets, with the Dragons head, the which according to the doctrine of the second Book at the time proposed are as followeth.

Suns place	♍ 4. 07.	☊ ♏ 1. 40.	♃ ♌ 5. 19.	♀ ♏ 16. 58.
Moons place	♊ 26. 24.	♄ ♌ •• 5. 19.	♂ ♑ 8. 87.	☿ ♎ 19. 81.

The places of the Planets being thus found, consider under what signe any of them falleth, which being observed, note which of the houses con∣tain the signe in which the planet is, observing the number of Degrees, and in due order place the planet in that house.

Right Ascension of the Mid-heaven, 72. 82.

[illustration]

A Scheme of Hea∣ven for July 16, Hour 18, parts 4564, AD. 1587. In the Meridian of London, whose Lati∣tude is 51. 53.

And when a Planet contains a lesser number of Degrees then the Cuspe of the house, he is to be placed before the house, in the house preceding, but if greater, he is be placed beyond the Cuspe, according to the sequel, and consequence of the signes.

As in the figure, Sol is in 9 degrees and 7 parts of Virgo, and the signe is placed upon the Cuspe of the Ascendent, but because the Cu••pe thereof containeth more degrees of that signe then the Sun, I place the Sun in the preceding house, that is in the twelfth.

CHAP. 17. To find the Angle of the Ecliptick with the Horizon, or the Altitude of the Nonagesime deg. together with its distance from the Mid-heaven.

BY the rules delivered in the last Chapter, find the point culmina∣ting, whose declination being ended to the altitude of the Equator ••n the Northerne signes, or subtracted in the Southern, gives you the altitude of the Mid-heaven, suppose 10 degrees of Gemini were

in the Mid-heaven, the declination thereof by the 2 Chapter is 22. 02910. which being added to the altitude of the Equator 38. 46667, because the de∣clination is North, their sum 60. 49577 is the Altitude of the Mid-heaven, and the meridian angle of the same point by the 13 Chapter is 81. 53133, hence in the triangle 7. 10. S of the Diagram in the last Chapter, we have given 10. S. 60. 49577 the altitude of Mid-heaven. The angle 7. 10. S 81. 53133. the Meridian angle to find the angle 10. 7. S.

As the Radius

To the sine of 7. 10. S.	81. 53133	9. 9952385
So Cosine of 10. S	60. 49577	9, 6923956
To Cosine of 10. 7. S	60. 84861	9. 6876341

And to find 7. 10 or the distance of Mid-heaven from the Nonagesime degree, the Analogie is

As Radius

To the Cosine of 7. 10. S	81. 53133	9. 6677126
So is the cotangent of 10. S	60. 49577	9. 7527167
To the tangent of 7. 10	14. 75047	9. 4204293
Mid-heaven add	70. 00000

Summe 84. 75047 is the Nonagesime degree. And note that arch found is to be added to the Mid-heaven from Capricorn to Cancer, to be subtracted from Cancer to Capricorn.

CHAP. 18. To find the Parallactical angle, or angle of the Ecliptique with the Verticall circle.

THe angle of the Ecliptique with the verticall circle, is an angle made by the oblique cutting of the circle of altitude, with the E∣cliptique, which is a right angle, when the said circle passeth through the 90 degree of the Ecliptique, but falling without the same it is oblique; as in the following figure, D denotes the Zenith, D C B the verticall circle, D E H V T the Meridian, T A B H the Ho∣rizon, V A E the Ecliptique, C the angle of the intersection of the E∣cliptique with the Vertical.

In the 16 Chapter is shewed how to find the point of the Ecliptique Ascending, and the Suns altitude at any time in the 12 Chapter, which being obtained we may speedily find the parallactical angle.

Admit the Sun at C be in ♍ 4. 07368 distant from the Meridian hours 5. 5436 or 83. 1540. Ad ortum. His Declination will be found by the 2 Chapter to be 10. 05037, and his altitude by the 12 Chapter 12. 10189. The point Ascending by the 16 Chapter is ♍ 17. 90610,

[illustration]

which known, in the triangle A B C we have. 1. A C 13. 90610. Se∣condly, B C 12. 10189, to find the angle A C B. I say then,

As the tangent of A C	13, 9061 co. ar.	0. 6062707
Is to Radius		10. 0000000
So tangent B C	12. 10189	9. 3312569
To Cosine of A C B	30. 00067	9. 9375276

CHAP. 19. The elevation of the Pole and Declination of the Sun given, to find the time when he will be due East and West.

IN the second variety of the 12 Chapter, the complement of the Suns Declination F H i•• 67. 97090 and the complement, of the poles eleva∣tion F G 38. 46667, hence the angle G F H equal to the Arch of the Aequator A O is to be sought, therefore I say.

[illustration]

As Radius

To Cotangent F H	67. 97090	9. 6070621
So tangent F G	38. 46667	9. 9000864
To Cosine of G F H	71. 24779	9. 5071485

whose comple. O F E 18. 75221 being converted into time giveth 1 hour 25147 parts of an hour, and so long it is after 6 in the morning when the Sun will be due East, and before 6 at night when he will be due West.

CHAP. 20 The Elevation of the Pole, with the Suns Declination and Al∣titude given, to find his distance from the Meridian.

IN the Oblique angled Spherical triangle G H F in the 2 Diagram of the 12 Chapter, we have known H F the complement of the Suns Declination, G F the complement of the poles elevation, H G the complement of the Suns altitude, to find G F H the angle of the Suns distance from the Meridian.

1 The side H G	44. 54111
2 The side F H	67. 97090
3 The side G F	38. 46667
Summe	150. 97868
••alf Summe	75. 48934
••••ne of F H	67. 97090 co. ar.	0. 0329234
••••ne of G F	38. 46667 co. ar.	0. 2061682
Differ. G F ½ Summe	37. 02267	9. 7796909
Differ. F H ½ Summe	07. 51844	9. 1167578
Quadrat of the sine of half the Angle		19. 1355403
Which bisected, is sine of	21. 69295	9. 5677701

And the double therof is 43. 38590. The Suns distance from the Meridian, And converted into time, gives two houres, 89259 parts.

CHAP. 21. To find the time of the Suns rising and setting, with the length of the Day and Night.

THe Ascensional difference of the Sun being added to the Semi∣diurnal arch in a Right Sphere, that is to 90 degrees in the Nor∣thern signes, or substracted from it in the Southern, there summe or difference will be the Semidiurnal arch, which doubled is the day Arch, and the Complement to 360 is the night Arch, which bisected is the time of the Suns rising, and the day Arch bisected is the time of his setting.

As when the Sun is in ten degrees of Gemini, his Ascensional difference is found to be

30. 61613

The Quadrant Add

90.

The Semidiurnal Arch

120. 61613

The diurnal arch

241. 23226

Whose Complement

118. 76774

Converted into time gives 7 houres 91078 parts, which bisected gives the time of the Suns rising. 3 hours 95539, parts; or a little before 4 of the clock.

CHAP. 22. To find the distance of a star from the Meridian.

IF a Starre be between the Mid-heaven and the Horoscope deduct the Right Ascension of the Mid-heaven from the Right Ascension of the Starre, what remaineth is the distance from the Meridian. If a starre be between the Mid-heaven and the 7 house, deduct the Right Ascen∣sion of the starre from the Right Ascension of the Mid-heaven, and what remaineth is the distance as before.

IF a starre be between the 7 house and the Imum Coeli or fourth house, deduct the Right Ascension of the Imum Coeli from the Right Ascension of the starre, and what remaineth is the distance from the Meridan. If a star be between the Ascendant and the Imum Coeli deduct the Right Ascensi∣on of the star from the Right Ascension of the Imum Coeli, and what re∣maineth is the distance from the Meridian as before.

For Example. In the preceding figure, the Right Ascension of the Mid-heaven is 072 deg. 82 parts. The Sun is in the 12 house and his

Right Ascension

155. 97

From which deduct the Right Ascension of the M. C.

72. 82

The distance of the Sun from the Meridian is

83. 15

CHAP. 23. To find the Elevation of the Pole above any circle of position.

A Circle of position, is as it were a certaine Horizon (upon which the point or star proposed doth arise) passing by the two intersect∣ions of the Horizon with the Meridian, and may be either above or under the Earth, in respect of the place for which the figure is erected.

A star posited in the

• 1, 2, 3, 4, 5, 6, house is Under the earth
• 7, 8, 9, 10, 11, 12, house is Above the earth

Thus in the annexed Diagram A H C is a circle of position passing by the Horizontal point of the Significator at H, and the two intersections of the

Horizon of the place at A and C, and L M is the elevation of the pole a∣bove this Horizon of the star or circle of position.

[illustration]

To find which there must be known. 1. The latitude of the place. 2. The Declination of the star or point proposed. 3. The distance thereof from the Meridian. Hence to find the angle of Inclination of the circle of po∣sition with the meridian, the proportions are as followeth.

1. As the Radius, To the tangent of the complement of the stars decli∣nation: so is the Cosine of the stars distance from the meridian, To the tangent of the first-arch. To which the pole of the place being added, or subtractod from it according to the following direction, their summe or difference is the second arch.

If the distance of the star from the meridian, be more then 90 and the declination South under the earth or north above it, subtract the first arch from the poles elevation, and what remaineth is the second arch.

If the distance of a star from the meridian be lesse then 90, and the de∣clination south under the earth, or north above it, adde the poles elevation, to the first arch, and their agg••gate if lesse then 90 is the second arch, if more then ••0 the complement thereof.

If the distance of a star from the meridian, be either more or lesse

then 90, and the Declination North under the earth, or South above it. Substract the elevation of the Pole from the first arch, and what remaine•••• ••s the second arch.

If the distance of a star from the Meridian be a just quadrant, the angle of inclination may be found at one operation, as in the fourth example.

2 As the sine of the first arch found: Is to the cotongent of the Stars distance from the Meridian: So is the sine of the second arch found; To the cotangent of the angle of inclination. Then to find the elevation of the Pole above the circle of position, the analogie is.

3 As the Radius, To the sine of the Pole of the place: So is the sine of the angle of inclination; to the sine of the Pole of the Circle.

1 Example.

Let the distance of a Star from the Meridian, be more then 90, viz. 97 deg. And the Declination of the Star 31 deg. North above the earth: the Pole of the place 45. Then in the oblique Spherical Triangle H M C we have limited. 1 The side M C the Poles elevation 45 degrees. 2 The side H M the complement of the Stars declination 59 degrees. 3 The an∣gle H M C, the Stars distance from the Meridian 97, or instead thereof the acue angle I M H 83, the complement of the other to a Semicircle.

Hence to find I M the proportion is.

As the Radius	90	10. 000000••
To the tangent of H M	59	10. 2212262
So is the Cosine of I M H	83	9. 085894••
To the tangent of I M	11. 47	9. 3071206

which being substracted from 45 the Poles elevation, there resteth 33. 53 the second arch.

2 As the sine of I M	11. 47	9. 2985361
To the Cotangent of I M H	83	9. 089143••
So is the sine of I C	33. 53	9. 742232••
To the Cotangent of H C M	71. 17	9. 5328404

The angle of inclination.

3 As the Radius	90	10. 0000000
To the sine of C M	45	9. 8494850
So is the sine of H C M.	71. 17	9. 9761116
To the sine of L M.	42. 01	9. 8255966

The height of the Pole above that Circle of position.

2 Example.

Let the distance of a Star from the meridian be less then a quadrant,

viz. 44. 7. The Declination of the Star, 14. 51. North above the Earth, the Pole of the place 45. then as before.

1 I say, As the Radius	90	10. 0000000
To the tangent of H M.	75. 15.	10. 5765162
So the Cosine of I M H	44. 7.	9. 8574210
To the tangent of I M.	69. 74.	10. 4329372

To which add the Pole, 45. 00. There aggregate is 114. 74. whose Com∣plement 65▪ 26. is the second Arch.

2 As the sine of I M.	69. 74.	9. 9722634
To the cotangent of I M H.	44. 7.	10. 0141010
So the sine of I C	65. 26.	9. 9581892
To the Cotangent of H C M	45	10. 0000267

The angle of Inclination.

3 As the Radius	90	10. 0000000
To the sine of C M	45	9. 8494850
So is the sine of H C M	45	19. 8494850
To the sine of L M	30	9. 6989700

The height of the Pole above that Circle of position.

3 Example.

Let the distance of a Star from the Meridian be 22. 82. the declination 13. 53. South above the Earth: And the Pole 49.

1 Then as the Radius	90	10. 0000000
To the tangent of H M	76. 47	10. 6186455
So is the Cosine of I M H	22. 82	9. 9646026
To the tangent of I M	75. 37	10. 5832481

Subtract the Pole 49. 0••. There rests 26. 37. for the 2d arch

2 As the sine of IM	75. 37	9. 9856855
To the Cotangent of I M H	22. 82	10. 3759530
So is the sine of I C	26. 37	9. 6475454
To the Cotangent of H C M	42. 51	10. 0378229

The angle of Inclination.

3 As the Radius	90.	10. 0000000
To the sine of C M	49.	9. 8777798
So is the sine of H C M	42. 51	9. 8297661
To the sine of L M	30. 66	9. 7075459

The height of the Pole above that Circle of position.

4 Example.

If the distance from the Meridian be a just Quadrant, or 90 degrees

then omitting the two first proportions, the angle of Inclination may be found at one operation, by this analogie.

As the tangent of the Complement of declination, is to Radius.

So is the sine of the Pole, to the cotangent of the angle of inclination.

Let then the declination be 23, and the Pole 45. I say,

As the tangent of H M	67	10. 372148••
Is to Radius or the angle H M C	90	10. 0000000
So is the sine of M C	45.	9. 8494850
To the Cotangent of H C M	73. 30	9. 4773369
Then as Radius, to the sine of M C	45.	9. 8494850
So is the sine of H C M	73. 30	9. 9812850
To the sine of L M	42. 63	9. 8307700

The height of the Pole above that Circle of position.

CHAP. 24. Of the Ark of Direction, what it is, and how to finde it.

AStrologers use to fore-tel the general Fortune of any Native by the consideration of the 12 Houses, but the particular time, in which we may expect, what is promised by the position of the Heavens at the time of the Birth, they measure out by the arke of Direction. That is, by the distance of the Significators from there Promittors reckoned in the Aequator, by Significators usually meaning, the Ascendent, Mid∣heaven, Sun, Moon, and part of Fortune: And by Promittors, the seve∣ral Aspects of these Significators to the Planets, or the twelve Houses.

For the clearer understanding of what the arke of Direction is, in the Diagram of the last Chapter, Let A D C represent the Meridian D V S E the Aequator, A C the Horizon, M the North-pole, A H C a Circle of position above the Earth, H B and N R two parallels of Declination. H the Significator, D O his right Ascension. H O his Declination. R the Promittor D S his right Ascension, R S his declination. Now when the Promittor at R comes to N, it is in the same Circle of position with the Significator at H, and the Circle of Declination M R S will be changed into the Circle of Declination M N V, and then the arch of the Aequator, D V is the Right Ascension of the Promittor at N, and therefore the arch of the Aequator V S is the arke of Direction sought. And the manner of finding thereof is as various, as the position of the Significator may be in the figure, which is threefold, viz. Either in the Meridian, in the Signes Ascending, or in the Signes Descending.

CHAP. 25. How to direct the Mid-heaven, and the Imum Coeli.

A Star posited in the meridian, that is, either in the mid-heaven or Imum Coeli, must be directed to his Promittors, by the right As∣censions of the Significator and Promittor. If a Significator po∣sited in the mid-heaven be to be directed. Substact the Right As∣cension of the mid-heaven, from the Right Ascension of the Star or Pro∣mittor, taken with its latitude if it have any, and what remaineth is the ark of Direction.

For Example. Let the mid-heaven of the preceding figure in the 16 Chapter, be directed to the 12 degree of Capricorn.

The Right Ascension of the 12 degree of Capricorn is

283. 03

From which substract the Right Ascension of the mid-heaven

072. 02

There rests for the Ark of Direction

211. 23

In like manner: If the Imum Coeli or fourth House, or a Star posited upon the Cuspe thereof be to be directed, you must substract the Right As∣cension of the Imum Coeli, or fourth House, from the Right Ascension of the Promittor, and what remaineth is the ark of Direction.

CHAP. 26. How to direct the Ascendent, or Significator posited in the Signes Ascending.

THe Horoscope or Ascendent, or a Significator posited in the signes ascending, that is, in the 12, 11, 10, 1, 2, or 3 houses, must be dire∣cted to Promittors, by the oblique Ascensions answering to the ele∣vation of the Pole above the Circle of position of the Significator. The elevation of the Pole above the ascendent is the same with that of the place for which the figure is erected. The Poles elevation above the Circle of position of any other Significator must be found as hath been shewed in 22 Chapter. Then if you deduct the oblique Ascension of your Signifi∣cator, from the oblique Ascension of your Promittor, what remaineth is the arke of direction.

For example. Let the Ascendent of the preceding figure in the 16 Chap∣ter be to be directed to the 26 deg of Taurus. The elevation of the Pole of the Circle is the same with that of the place. viz. 51. 5••.

And therefore the oblique Ascension of 26 deg. of Taurus,

27. 50

To which add a Circle that substraction may be made

360

And then the Oblique Ascension is

387. 50

The Oblique Ascension of the Ascendent substract

162. 82

There rests for the ark of Direction

224. 68

Another Example.

Let us suppose a Significator to be posited in the signes Ascending but not upon the Cuspe of the Ascendent, in this case though the Arke of Direction must be found out by the Oblique Ascensions as be∣fore; yet the elevation of the Pole above the Circle of position must be first found, because it is not given, as in the last Example.

In the Diagram of the 23 Chapter, let the distance of the Significator from the meridian be 44. 07. The declination thereof North above the Earth H O 14. 85. Hence the Pole L M is 30 degrees. Let this Significa∣tor be posited in 10 degrees of Taurus, the Right Ascension thereof D O 37. 58. His Ascensional difference under the elevation of 30 degrees F O 8. 80. Let the Promittor be in 25 of Gemini, the Right Ascension thereof D S. 84. 55. And the Ascensional difference when this Promittor comes to N. is represented by the letters F V. 14. 47. Now then if you substract the Ascensional Difference of the Significator F O, 8. 80. from the Ascensio∣nal Difference of the Promittor F V. 14. 47. There will remaine O V 5. 67. we call it the Aequation of the Arke of Direction, which be∣ing added to the Right Ascension of the Significator D O. 37. 58. There aggregate is the arch D V 43. 25. Then if you deduct the arch D V 43. 25 from the Right Ascension of the Promittor, D S. 84. 55. There difference is V S 41. 30. The Ark of Direction sought.

Or thus, the Oblique Ascension of the Promittor is 70. 08. The Oblique Ascension of the Significator is 28. 78. There difference 41. 30, is the Ark•• of direction sought, as before.

CHAP. 27. How to direct a Significator posited in the Signes Descending.

THe Descendent or seventh House or Significator there posited, or in any signe descending, that is, in the 4, 5, 6, 7, 8, or 9 Houses, must be directed to his Promittors, by the oblique Descensions answering to the Elevation of the Pole of the Circle of position of the Significator. As suppose the mid-heaven be in 22. 33 of Gemini. And the Right Ascension thereof, 81. 65. Let the Significator be in 10 de∣grees of Taur••s. The declination thereof 14. 85. North. His distance from the meridian 44. 07. Hence the Elevation of the Pole above that Circle of position is 30, as before.

Let the Promittor be in 25 degrees of Gemini, the declination 23. 40 North: Now the declination of the Significator and Prommittor being the same with the former, and the Pole of the Circle the same, the Ascen∣sional

Differences as well of the Significator as of the Promittor, must needs be the same with the former Example, and consequently the same Aequation of the ark of Direction: if then you would find this ark of Direction, by the Right Ascensions, Adde this Aequation of the ark of Direction, to the Right Ascension of the opposite point of the Promit∣tor: and from their aggregate, subtract the Right Ascension of the opposite point of the Significator, and what remaineth is the ark of Direction.

As the opposite point to this Promittor in the 25 of Gemini, is the 25 of Sagittarius, the Right Ascension whereof is

264. 55

The former Aequation of the ark of Direction

5. 67

Their Aggregate

270. 22

The opposite point to the Significator in the 10 of Tau∣rus is the 10 of Scorpio, whose Right Ascension is

217. 56

Which subtracted from the former aggregate leaveth

52. 64

For the ark of Direction sought.

Or if you will by the oblique Ascensions of these opposite points thus:

The oblique Ascension of the 25 of Sagittarius is

279. 01

The oblique Ascension of the 10 of Scorpio, is

226. 38

There Difference is

52. 63

The ark of Direction as before.

Or lastly, by the oblique Descensions, according to the intention of this Chapter, if you subtract the oblique Descension of the Significator, from the oblique Descension of the Promittor, what remaineth will be the ark of Direction.

As in this Example the oblique Descension of the Promittor, under the elevation of 30 degrees is

99. 01

The oblique Descension of the Significator,

46. 38

Which being subtracted from the oblique Descension of the Promittor, there resteth

52. 63

The ark of Direction sought as before.

CHAP. 28. How to find the Arch of the Aequator, whereby is made the general Table of Positions.

FOr finding this arch of the Aequator, there must be given the angle of Inclination of the Circle of position with the Meridian, & the height of the Pole above that Circle, both which may be found by the 9 Chap∣ter:

but •••• to our present purpose, we are not tied unto such a ••••dious calcu∣lation, because the Pole of the Circle may be supposed, and then the angle of Inclination may be found at one Operation, and this arch of the Ae∣quator at another.

And first, the Elevation of the Pole above any Circle of position being given together with the latitude of the place or Countrey, the angle of In∣clination may be found, by this analogy.

As the sine of the Pole of the place, is in proportion to Radius: So is the sine of the Pole of the Circle, to the sine of the angle of Inclination.

For Example. In the Rectangle Spherical Triangle of the 9th Chapter L M C right angled at L. Let M C the Elevation of the Pole of the place be 45. And the Pole of the circle LM 42, hence to find the angle LCM. I say,

As the sine of M C	45.	9. 8494850
Is to the Radius	90.	10. 0000000
So is the sine of L M	42.	9. 8255108
To the sine of L C M	71. 13	9. 9760258

Then to find the arch of the Aequator, the proportion is: As the Radius to the sine of the Complement of the Pole of the place: So is the tangent of the angle of Inclination, to the tangent of the arch of the Aequator. For Example, In the triangle A D F of the afore-said Diagram, Let there be given the side A D the Comple. of the Pole of the place 45. The angle of Inclinat. DAF 71. 9. Hence to find the arch of the Aequator. DF, I say,

As the Radius	90	10. 0000000
To the sine of AD	45	9. 8494850
So is the tangent of D A F	71. 13	10. 4662••85
To the tangent of D F	64. 20	10. 3157275

Which is the arch of the Aequator sought.

CHAP. 29. How by the general Table of Positions, to make a particular Table for any Latitude there exprest.

IF thou wouldst make a particular table of positions, first, divide your paper or book into as many Columnes as the largness of the page will bear, then in the head of your table write the particular latitude, for which you would have the table, and under this title write in the third Co∣lumn of your page. 1 In the fourth, 2, and so forward till you have fild the Columnes of your first page, do so likewise in the 2d. page, and so for∣ward till you have written twice over the several Elevations of the Pole a∣bove the circle of position from one unto that degree for which your table

is intended; then in the first Columne of your left hand page write, North Declination under the earth and South above it, in the second columne thereof write orderly the several degrees of declination, beginning with a cypher or nought, and then 1, 2, 3, and so forwards till you have writ∣ten 32, and in the two first columnes of your right hand page write the contrary, that is, in the first thereof write South Declination under the earth and north above it; and in the second the severall degrees of Decli∣nation beginning with 32, and so downward 31. 30. 29. until you come to nought: having thus prepared your book according to that Table in this book, enter your general table of positions with one degree of Ele∣vation, and in a straight line directy under the latitude of your place, you shall find the arch of the Aequator answering thereunto, then look for the Ascensional differences answering to every degree of Declination, under one degree of the Poles Elevation, the which being substracted from the arch of the Aequator, write the remainder in a direct line under one de∣gree of Elevation, in that page which must serve for North Declination under the Earth: but for the South Declination under the Earth adde the Ascensional difference of every degree of Declination to the former arch of the Aequator, and write their aggregate under one degree of Elevation and right against that degree of Declination, whose Ascensional difference was added thereunto, and so shall you have one Column of your table fi∣nished, to make it plain we will add an Example.

In the latitude of 51 degrees 33 hundred parts, the arch of the Aequa∣tor answering to one degree of the Circle is 0 deg. 79. and the Ascensional difference for one degree of Declination under one degree of Elevation is 0. 1. of which being subtracted from 79, there remaineth 78, which I write against one degree of Declination in that page of the Table serving for North Declination.

Again, to the same arch of the Aequator, I add 0. 01. and their aggre∣gate is 80, and this I write against one degree of Declination, and under one degree of Elevation in that page of the Table which serveth for South Declination under the Earth, and thus must you also add or subtract the Ascensional differences of all the other degrees of Declination, according to this Example. And so shall you have a particular Table of positions for your particular latitude. The use of this Table is to finde the Elevation of the Pole above any Circle or position in that particular latitude, for which the Table is framed, as shall be shewed in his proper place.

CHAP. 30. Of the Doctrine of the Sphere in Tables.

ALthough in the former book there is plainly shewed you, how to find the Declination, Right Ascension, Ascensional Difference, Oblique Ascension, Cuspes of the 12 Houses, and the height of Pole above any Circle of position, by Trigonometrical calculati∣on: yet considering that, that way is not altogether so expedite and ready for practice, as some may desire, wee will also shew you how to finde the same by those Tables, that are hereunto annexed.

Precept 1.

The Declination of the Sun or other Planet is found by the signe in the head or foot, and the degree on the right side, if the signe be in the foot, or on the left side if it be in the head: for the common angle gives the Decli∣nation sought, if you have respect, to your Planets proper latitude, and the proper part proportional.

For Example: Suppose the Moon were in 19 degrees 56 parts of Leo, and her declination were required, having North latitude 3 degrees. Her Declination in that latitude, in 19 degrees of Leo will be found 18. 02

In 20 degrees it is		17. 70
There difference is		0. 32
Now then if one degree give	0. 32	1. 5051499▪
What shall	0. 56	1. 7481880
They give 18 ferè		1. 2533379

Which being subtracted from 18. 02 because the Declination doth de∣crease, the Moons Declination will be found to be 17. 84

Precept 2.

The Right Ascension by the Table following is had by the signe in the head and degree in the left side, and in the common angle is the Right Ascension.

Example. The Right Ascension of a Planet in Aries 1•• 251, and 2 degrees of South latitude be sought.

The Right Ascension of 13 Degrees

12. 73

The Right Ascension of 12 Degrees

11. 82

Difference

91

If one Degree give	91	1. 9590413
What shall	25	1. 3979400
They give	22	1. 3569813

Which being added to 11. 82, the R. ascension of the point sought will be 12. 04.

But if the R. ascen. of the point sought be in Southern signes, you must add 180 to the arch found in the table, and you have your desire, thus the R. ascens. of 12 deg. 25 parts of Libra, with two degrees of North Latitude by adding 180 to the former arch will be 192. 04.

Precept 3.

The ascensional difference of any part of the Zodiack is found by the degree of declination in the left side, and the degree of the Poles elevation in the head, the common angle gives the ascensional difference ••ought. Ex∣ample. Let a planets Declination be 4. 43. whose ascensional difference is sought under the elevation of 52 deg. The ascensional difference in that Elevation answering to 5 degrees of Declination is 6. 43. to 4 degrees is 5. 13. there difference is 1. 30.

If one degree

Give	1. 30	2. 1139433
What shall	4••	1. 6334684
They give	56 ferè	1. 7474117

Which being added to ••. 13, the ascensional difference of the point sought will be 5. 69.

Precept 4.

The oblique ascension of any part of the Zodiack is found by the signe and degree in the first Column in the left hand, and the Poles Elevation on the head of the Table, the common angle will give you the oblique Ascension sought.

Example. Let the Oblique Ascension of Virgo 09. 23. in the elevati∣on of 42 degrees be sought. In that elevation.

The Oblique Ascension of Virgo 10 degrees is

15••. 87

Virgo 6 degree is

155. 80

Their Difference

4. 07

If 4 degrees	4. 00	9. 3979400
Give	4. 07	0. 6095944
What shall	3. 23	0. 50920••5
They give	3. 28	0. 5167369

Which being added to 155. 80. the obli•••• Ascension sought will be 159. 08

Precept 5.

These things premised, the Right Ascension of M. C, and thereby the Mid-heaven it self will easily be found for the houres from Noon (convert∣ed into Aequinoctial degrees, by the table for that purpose) being added to

the Suns Right Ascension, do make the Right Ascension of the Mid hea∣ven, which sought in the Area of the Table of Right Ascensions gives the mid-heaven it self.

Example.

Let the Sun be in 4 deg. 07 parts of Virgo, as in the 16 Chapter, the time from Noon, Houres

18. 4564

The Suns Right Ascension in 4 degrees of Virgo

1••5. 9••

His Right Ascension in 5 deg. of Virgo,

156. 85

Difference

••0. 95

If one degree give	95	0. 9777236
What shall	07	1. 8450980
They give	06	1. 8228216

Which being added to 155. 90. The Suns Right Ascension is 155. 96. The Aequinoctial degrees answering to 18 houres, viz. 270 degrees. And the degrees answering to 4564 parts of an houre are 06. 84, and therefore the Right Ascension of time is 276. 84, which being added to the Suns Right Ascension, their aggregate 432. 80. or rejecting 360 degrees, the remainer 72. 80, is the Right Ascension of the Medium Coeli, whereto answers in the Table of R. Ascensions; 74. 75. ••••, the point of the Ecliptick for the Mid∣heaven it self.

Precept 6.

The Oblique Ascensions of the other Houses are found by a continual addition of 30 degrees to the Right Ascension of the Mid-heaven, as hath been also shewed in the 16 Chapter, and thus the Oblique Ascension of the 11 House is 102. 80. 12 House is 132. 80. Ascendent is 162. 80. 2 House is 192. 80. 3 House is 222. 80.

The Poles of Elevation above the Circle of Position of the Ascendent is always the same with that of the place, for which the figure is erected, the Poles Elevation for the 11, 12, 2 and 3 Houses you may find in the Table for that purpose annexed to the Table of Oblique Ascensions.

Example. By that Table under our Elevation of 51. 53. the Poles Ele∣vation above the Circles of position of the 11 and 3 Houses, is 32. 18. And the Poles Elevation above the 12 and 2 Houses is 47. 46.

Now then to find the Cuspe of the 11 House, look the Oblique Ascen∣sion thereof, 102. 80. in the Table of Oblique Ascensions answering to 32. 18, of Elevation, and the point of the Ecliptick answering thereunto is Cancer 24. 98.

In like manner, if you look the oblique Ascension of the 12 house, 132. 80 under the elevation of 47. 46. the Cuspe thereof will be Leo, 25, 26.

If you look the oblique Ascension of the Ascendent 162. 80. under the Elevation of 51. 43 the Cuspe thereof will be Virgo 17. 94.

If you look the Oblique Ascension of the second House 192. 80. under the Elevation of 47. 46. the Cuspe will be Libra 9. 44.

Lastly, if you look the Oblique Ascension of the third House, 222. 80 under the Elevation of 32. 18. the Cuspe thereof will be Scorpio, 6. 35. The other 6 Houses, are in the same Degrees and parts of the opposite signes.

Precept 7.

The Poles Elevation above the Horizon of a Star is found by the de∣gree of his Declination, in the second Column of the right hand page of your Table, if the Declination thereof be North under the Earth, or South above it; And in the second Column of the left hand page, if the declinati∣on be South under the Earth, or North above it, and his distance from the Meridian in a straight line, for in the head of your Table in a direct line (using a double proportional part if need be) you shall have the degree of the Poles Elevation above the Circle of position.

Example.

In the preceding figure of the 16 Chapter, the Sun is in Virgo, 4. 07 and therefore his Declination is 10. 05. North above the Earth; his di∣stance from the Meridian by the 22 Chapter, was found to be 83. 15. Now then to find the Poles Elevation, I look in that Table for the Suns declina∣tion in the second Column of the left hand page, and his distance from the Meridian in a straight line, and in the last folio of that Table, I finde a∣gainst 10 degrees of Declination, the distance of a Star from the Meridi∣an neerest to my number to be 83. 37. and over the head thereof 50, which is the degree of the Poles Elevation above that Circle of Position.