Imágenes de páginas
PDF
EPUB

The Rational Way of Regiomontanus; as if, forfooth, there were any Reafon 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 confequently the Cufpides of the several Houses are eafily known by Calculation of a Right-angled Triangle in the three first Schemes; and of a Right and Oblique One in the laft. But in the following Problems, I fhall fhew how to find what Degree and Minute of the Ecliptic, conftitutes the Cufp of every House or Manfion 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 at 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 precifely.

[ocr errors]

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 Afcenfion of Medium Cali or Cufp of the 10th House.

5. Laftly, by the continual Addition of 30 Degrees to the Cufp of the 10th Houfe, and the confequent Sums, we hall obtain the Oblique Afcenfion of the 11th, 12th, 1ft, 2d, 3d Houfes; and fo. this Work, according to the Time and Place in this Problem, will ftand in the following Order.

VOL. II.

[blocks in formation]

Anno Dom. 1735, May
Equation of Time fubftract
Remains the Equal Time
Sun's true Place then
Sun's Right Afcenfion
Ap. Time from Noon add.
Right Afc. of Med. Cali
Add

Ob. Afc. of the 11th House
Add

Ob. Afc. of the 12th House
Add

Ob. Afc. of the Afcendant
Add

Ob. Afcenf. of the 2d Houfe
Add

Ob. Afcenf. of the 3d House.

D. H. M. S.
12 10 36 40 Ap. Time.
oo 00
3 44
12 10 32 56

II 2 II 22
60 6 00

159 10 00

[blocks in formation]

Having proceeded thus far, we may now construct

the Scheme; and firft let it be according to the Method of Regiomontanus, as is here reprefented in the Margin. Wherein HZON is the Meridian of the Place; 2, the Equinoctial; whose Poles are P and S; 10 2 4, the Ecliptic; HAO, HBO,

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

&c. are the Circles of Pofition; PFSL, the Elevation of the Pole above the Circles of Pofition of the

11th, 9th, 3d, and 5th Houfes, to which the Complement of the Angle HA Æ is equal; fo PG = S K is the Height of the Pole above the Circles of Pofition of the 12th, 8th, 2d, 6th Houses, which is equal to the Complement of the Angle HB E. Now we fhall eafily find the Elevation of the Pole above the Circles of the Houses, and the Cufps of the feveral 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.

Practice.

In the Right-angled Triangle HÆ A, there is given H E 39° 04', the Co-Latitude; and E A = 30° 00', the Distance of the Circle of Pofition in the Equinoctial from the Meridian; to find the Angle at A, whofe Complement is the Arch P F, the Height of the Pole required.

[blocks in formation]

Whose Complement is 31° 38' = PFSL, the Elevation of the Pole above thofe Circles of Pofition, 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.

Practice.

In the Right-angled Triangle HB Æ, there is known HÆ 39° 04', E B 60° oo'; to find the Angle at B, whofe Complement is equal to P G the Height of the Pole required.

The Proportion.

As the Sine of В Æ = 60° 00' Is to Radius

So is the Tangent of Æ H = 39° 04′

To the Tangent of HB = 43° 08′
Whofe Complement is 46° 52'

9.9375306

10.0000000

9.9094022

9.9718716

PG

SK the

Height of the Poles above the Circles of Pofition, as required,

[blocks in formation]

of the Pole above the Cir. II, 9, 3, 5 cles of Position of the feve-12, 8, 2, 6 ṛal Houses ftands thus | 1, 7,

[blocks in formation]

Note; Thefe Elevations will be the fame in every Scheme erected in the Latitude of Chichester 50° 56′ for any Time of the Year.

Thus having found the Oblique Afcenfion of the Houses, and the Elevation of the Pole above their Circles of Pofition; we proceed now to find the Cufps of the Houses, as follows.

PROBLEM IV.

To find the Cufps of the 10th and 4th House.

Practice.

In the Right-angled Triangle 24, there is given the Side 2=39° 16', the Right Afcenfion of Imum

[ocr errors]

Cali, as being the oppofite Point in the Equinoctial to 219° 16' the Right Afcenfion of Med. Cali; there is alfo given the Angle at Y= 23° 29', the Obliquity of the Ecliptic; to find the Side 24, the Longitude of the Cufp of the fourth House in the Ecliptic. The Proportion.

As Radius
Is to the Co-fine of 2 = 23° 29′
So is Co-tangent of r2= 39° 16'
To the Co-tangent of Y 4 = 41° 43'

So the Cufp of the is in the Sign Taurus to which is Scorpio m Houfe or Medium Cali;

10.0000000

9.9624527 10.0875019

10.0499546

fourth Houfe, or Imum Cali, 11° 43', the Oppofite Point 11° 43', the Cufp of the 10th as required.

PROBLEM V.

To find the Cufps of the 1ft House, Horoscope, or Ascendant; and of the 7th House, or Defcendant.

Practice.

In the Oblique Triangle 1 C Y, there is given the Angle at r=23° 29'; the Angle at C 140° 56'; and the Side included C Y= 50° 44', the Complement of the Oblique Afcenfion of the Horoscope to a whole Circle; to find the Side Y 1, in the Ecliptic.

[blocks in formation]
« AnteriorContinuar »