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The Rational Way of Regiomontanus ; as if, forsooth, there were any Reason for dividing the Heavens into twelve Parts this way, rather than by a thousand others we might invent.

The four Schemes above are general, adapted to the Lat. 50° 56', the Horoscope in each being the first Min. of Libra ; and consequently the Cuspides of the several Houses are easily known by Calculation of a Right-angled Triangle in the three first Schemes ; and of a Right and Oblique One in the last. But in the following Problems, I shall shew how to find what Degree and Minute of the Ecliptic, constitutes the Cusp of every House or Mansion in the Figure, for any particular Time and Place.

PROBLEM I. Let it be required to erect a Scheme or Figure of the Heavens at Chichester, Lat. 50° 56', for May the 12th Day, at 36' 40" paft 10 ac Night, A. D. 1735, the apparent Time.

In order to this, we must first find the Equation of Time, and thereby reduce the Apparent to the Equal or Mean Time.

2. The Place of the Sun must be found to the Mean Time precisely

3. The Sun's Right Ascension at the fame Moment.

4. Time from Noon must be reduced to the Degrees and Minutes of the Equator, by which means we shall have the Right Ascension of Medium Cæli or Cusp of the roth Høuse.

5. Lastly, by the continual Addition of 30 Degrees to the Cusp of the roțh House, and the consequent Sums, we shall obtain the Oblique Ascension of the 11th, 12th, ist, 2d, 3d Houfes, and so this Work, according to the Time and Place in this Problem, will stand in the following Order. VOL. II.

Anno

Nn

D. H. M. S. Anno Dom. 1735, May 12.10 36 40 Ap. Time. Equation of Time substract oo oo

3 44 Remains the Equal Time 12 10 32 56 Sun's true Place then

II 2 II 22 Sun's Right Ascension

60 6 oo Ap. Time froni Noon add.

159 10 00 Right Asc. of Med. Cæli

219 16 oo Add

30 Oo oo Ob. Asc. of the 11th House

249 16 oo Add

30 Oo oo Ob. Asc. of the 12th House

279 16 oo Add

30 00.00 Ob. Asc. of the Ascendant

309 16 oo Add

30 Co oo Ob. Ascens. of the 2d House

339 16 oo Add

30 00 00 Ob. Ascens, of the 3d House 9 16 oo

101

A
B

12

C.

K

Having proceeded thus far, we may now construct the Scheme; and first let it be ac

P cording to the Me

Æ А.
thod of Regiomon-
tanus, as is here re-
presented in the
Margin. Where- u
in HZO N is the

r
Meridian of the
Place ; Æ 2. the
Equinoctial; whose
Poles are P and S;

S 10 2 4, the Ecliptic ; HAO, HBO, &c. are the Circles of Position ; PF=SL, the Elevation of the Pole above the Circles of Position of the

4

IIth, 9th, 3d, and 5th Houses, to which the Complement of the Angle H A Æ is equal ; fo P G = S K is the Height of the Pole above the Circles of Position of the 12th, 8th, 2d, 6th Houses, which is equal to the Complement of the Angle HB Æ. Now we shall easily find the Elevation of the Pole above the Circles of the Houses, and the Cusps of the several Houses, by the following Problems.

PROBLEM II.

To find the Elevation of the Pole above the Circles of the 11th, 9th, 3d, and 5th Houses.

Praktice. In the Right-angled Triangle H Æ A, there is given H Æ = 390 04', the Co-Latitude ; and Æ A = 30° oo', the Distance of the Circle of Position in the Equinoctial from the Meridian ; to find the Angle at A, whose Complement is the Arch P F, the Height of the Pole required.

The Analogy As the Sine of A Æ = 30l oo' 9.6989700 Is to Radius

10.0000000 So is the Tangent of Æ H= 39° 04", 9:9094022 To the Tangent of H AE = 58° 22' 10.2104322 Whose Complement is 31° 38' = P F=SL, the Elevation of the Pole above those Circles of Position, as required.

PROBLEM III.

To find the Elevation of the Pole above the Circles of the 12th, 8th, 2d, and 6th Houfes. VOL. II.

Nn 2

Practice.

Pratice, In the Right-angled Triangle HBÆ, there is known H Æ = 39° 04', Æ B = 60° oo'; to find the Argle at B, whose Complement is equal to PG the Height of the Pole required.

The Proportion. As the Sine of = 600 00' 9.9375306 Is to Radius

10.0000000 So is the Tangent of Æ H = 39°04' 9.9094022 To the Tangent of HB Æ = 43° 08' 9.9718716 Whose Complement is 46° 52' = PG= SK the Height of the Poles above the Circles of Position, as required.

Houses The Elevation therefore ( 10, 4, of the Pole above the Cir. 111, 9, 3, 5 cles of Position of the seve. 112, 8, 2, 6 = 46 52 ral Houses stands thus

| 1, 7,

= 50 56 Note; These Elevations will be the same in every Scheme erected in the Latitude of Chichester 50° 56' for any Time of the Year.

Thus having found the Oblique Ascension of the Houses, and the Elevation of the Pole above their Circles of Position ; we proceed now to find the Cusps of the Houses, as follows.

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00 00

31 38

PROBLEM IV,

To find the Cusps of the ioth and 4th House.

Praktice. In the Right-angled Triangle 2? 4, there is given the Side Y Q=39° 16', the Right Ascension of Imum

Cæli, as being the opposite Point in the Equinoctial to 2199 16' the Right Ascension of Med. Cæli ; there is also given the Angle at 1 = 23° 29', the Obliquity of the Ecliptic ; to find the Side Y 4, the Longitude of the Cusp of the fourth House in the Ecliptic.

The Proportion. As Radius

10.0000000 Is to the Co-line of 1= 2329' 9.9624527 So is Co-tangent of Y Q=39° 16'

10.0875019 To the Co-tangent of 1 4 = 41° 43' 10.0499546

So the Cusp of the fourth House, or Imum Cæli, is in the Sign Taurus 8 11° 43', the Opposite Point to which is Scorpio m 11° 43', the Cusp of the roth House or Medium Cæli ; as required,

PROBLEM V.

To find the Cusps of the ift House, Horoscope, for Ascendant ; and of the 7th House, or Descendant.

Prazlice. In the Oblique Triangle 1 C Y, there is given tite Angle at Y = 239 29'; the Angle at C = 140° 56'; and the Side included CT = 50° 44', the Complement of the Oblique Ascension of the Horoscope to a whole Circle ; to find the Side ? 1, in the Ecliptic.

164 25

Then

The Sum of the Angles is
The Difference of the same
Half Sum
Half Difference
The Half of the Side

117 27 82 12 ! 58 43

25 22 Opera

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