8. the tropic of cancer, or any parallel of the sun's north declination.

9. the tropic of capricorn, or any parallel of the sun's south declination.

10. On the elevation of the north pole above the horizon, equal to Æ zenith, the latitude of the place.

11. HE the elevation of the equator above the horizon, equai to zenith N, the complement of the latitude.

12. TOW a parallel circle 18 degrees below the horizon, or the boundary between twilight and dark night.

13. Zenith SEC, zenith cd, &c. azimuth, or vertical circles. 14. NSDS, NGcws, &c. meridians or circles of terrestrial longitude.

15. nvxm, ncvm, &c. circles of celestial longitude.

NOTE. The several triangles in this general figure, and the lines which form them, are explained at the head of each problem, to which the triangle is applied.

(L) PROBLEM I. (Plate III. Fig. 1.)

Given the obliquity of the ecliptic and the sun's longitude, to find his right ascension and declination.

In the right-angled spherical triangle AGO, or ABO.

1. Asun's longitude, an arc of the ecliptic from aries. 2. AG, or AB sun's right ascension, an arc of the equinoctial; the point в always comes to the meridian with the sun.

3. GO, or BO sun's declination, an arc of a meridian passing through the sun's place. This arc likewise comes to the meridian with the sun.

4. GA, or BAO the obliquity of the ecliptic, or the angle formed between the equinoctial and the ecliptic.

5. AG, or AB the angle formed by the ecliptic, and the meridian passing through the sun's place.

Any two of these five quantities being given, the rest may be found.

EXAMPLE.

On the 17th of May 1822, when the sun's longitude is 1..25°.57′.25′′; required his declination, right ascension, and the angle formed between the ecliptic and the meridian passing through the sun's place. The obliquity of the ecliptic being 23°.28'.

1. To find the sun's declination, in the triangle ABO.

*

Rad x sine B sine BAO X sine AO.

• (log sine BA +log sine A)-10=log sine B○.

or, Rad sine AO 55°.57'.25":: sine BAO 23°.28' sine BO 19°.16'.

2. To find the sun's right ascension, in the triangle ABO.

Rad x cos BA Ocot AO x tang AB.

(10+log coS BAO)-log cot Alog tang AB. or, Cot A=55°.57′.25" rad::cos BAO=23°.28′ : tang AB= 53°.37′.40′′=3h.34′.30′′.40′′.

3. To find the angle AOB between the ecliptic and meridiant. Rad x cos Acot BAO × cot AB.

*

(10+log cos AO)-log cot BAO=log cot AOB. or, Cot BAO-23°.28': rad::cos A55°.57.25" cot AOB =76°.20′.23′′.

(M) Astronomers reckon the sun's longitude and right ascension from aries quite round the globe. Hence, though at equal distances from the equinoctial points, aries and libra, the sun may have the same quantity of declination and of the same name, viz. from aries to libra, north; and from libra to aries, south; yet the longitudes and right ascensions differ materially at these points where the declinations agree. Hence,

(N) If the sun be in the second quadrant of the ecliptic, subtract the longitude from 180°, use the remainder to find the right ascension; and take the right ascension, when found, from 180°.

(O) If the sun be in the third quadrant, subtract 180° from his longitude, use the remainder to find his right ascension; and add 180° to the right ascension when found.

(P) If the sun be in the fourth quadrant of the ecliptic, subtract his longitude from 360°, use the remainder to find the sun's right ascension; and take the right ascension, when found, from 360°.

(Q) In the first quadrant of the ecliptic between the 20th March and 21st June, the sun's declination is north and increasing; and in the third quadrant between the 22d September and 21st December, the sun's declination is south and increasing. In the second quadrant of the ecliptic from 21st June to 22d September, the sun's declination is north and decreasing; and in the fourth quadrant from 21st December to the 20th March, the sun's declination is south and decreasing.

PRACTICAL EXAMPLES.

1. On the 17th of May 1822, the obliquity of the ecliptic 23°.27.53", the sun's declination=19°.15.57" N. and increas

+ The finding of this angle is of no other use than as an exercise of right-angled spherical triangles.

ing; required his longitude, right ascension, and the angle formed between the ecliptic and the meridian passing through the sun's place.

The sun's longitude

=55°.57′.37′′, or 1$.25°.57′.37".

=76°.20'.

Ans. The sun's right ascension=53°.38', or 3.34.32".

The angle AOB

The answers here would have been ambiguous, had it not been shewn that the sun was in the first quadrant of the ecliptic.

(R) This problem is useful for calculating tables of the sun's longitude. For the latitude of the place being accurately determined, and a quadrant fixed in the plane of the meridian, the observation of the sun's meridian altitude gives his declination, or distance from the equinoctial; whence his longitude for any day is readily obtained.

2. On the 5th of August 1822, the obliquity of the ecliptic= 23°.27.53", the sun's right ascension 134°.54'; required the sun's longitude, declination, and the angle formed between the ecliptic and the sun's meridian.

The sun's right ascension being greater than 90°, and less than 180°, he is in the second quadrant of the ecliptic,

3. On the 10th of November 1822, the sun's right ascension. 225°.5′.16′′, his declination 17°.5'.18" S. required his longitude, the obliquity of the ecliptic, and the angle formed by the ecliptic and the sun's meridian.

S The sun's longitudem 47°.33′.26", or 7s.17°.33′.26". Ans. Obliquity of the ecliptic= 23°.27'.53". The angle AOB

= 73°.40'.19".

The sun's right ascension being greater than 180°, and less than 270°, he is in the third quadrant of the ecliptic.

4. On the 1st of February 1822, the sun's longitude is 8'.59", or 10.12°.8'.59", his declination 17°.10.15" S.; re

12°.

quired his right ascension, the obliquity of the ecliptic, and the angle formed by the ecliptic and the sun's meridian.

The sun's right ascension=314°.37', or 20h.58'.28". Ans. Obliquity of the ecliptic = 23°.27.55".

The angle AOB

= 73°.46′.

PROBLEM 11. (Plate III. Fig. 1.)

(S) Given the latitude of the place, and the sun's declina

tiont, to find his amplitude, ascensional difference, and the time of his rising and setting.

In the right-angled spherical triangle ABS or AGC.

1. AB or AG=the sun's, or star's, ascensional difference. 2. BS or GC the sun's, or star's, declination.

=

3. AS or AC the amplitude.

4. BAS, measured by the arc oo; or CAG, measured by the arc HÆ,the complement of the latitude.

5. ASB or ACB=the angle formed between the horizon and the meridian passing through the sun.

Any two of the above five quantities being given, the rest may be found.

(T) The same things may be found from the right-angled triangle SON or CHS.

For, in the triangle son, so is the complement of As ; sn is the complement of BS; ON is the complement of the arc go, which measures the angle BAS; and the angle SNO, measured by the arc BQ, is the complement of the ascensional difference. In the same manner the triangle CHS may be compared with the triangle AGC.

EXAMPLE I.

Given the latitude of London 51°.32′ N. on the longest day when the sun has 23°.28′ N. declination; required the sun's amplitude, ascensional difference, time of rising, setting, the length of the day and night, &c.

1. In the triangle ABS to find the amplitude As.

Rad x sine BS=sine BAS X sine as.

(10+ log sine BS)—log sine BAS=log sine as. or, Sine BAS38°.28': rad:: sine BS-23°.28': sine as 39°.48'.

+ This problem may be applied to the rising or setting of a planet, or of a star. But where great exactness is required, the declination of the sun or planet must be calculated as near to the time of rising and setting as possible; especially for the moon on account of her swift and irregular motion. Likewise, the declination of the sun near the equinoxes alters considerably in the compass of an hour. Hence what is generally called a parallel of the sun's declination, and drawn as such in the projection of the sphere, is not strictly a parallel circle, but a spiral line.

When the latitude of the place and the declination of the object have the same name, the right ascension diminished by the ascensional difference leaves the oblique ascension, and increased by the same gives the oblique descension.

But when the latitude of the place and declination have contrary names; the right ascension increased by the ascensional difference gives the oblique ascension; and being diminished by the ascensional difference leaves the oblique descension.

viz. cosine of the latitude, is to radius; as the sine of the sun's declination is to the sine of the amplitude.

(U) This part of the problem is useful in navigation, for finding the variation of the compass.-It is evident, by comparing the triangle ABS with the triangle AGC, that the amplitude is always of the same name with the declination, whether north or south.

2. To find the ascensional difference ab.

*

Rad x sine ABCot BAS X tang BS.

(log cot BAS+log tang BS)-10=log sine AB.

or, Rad: cot BAS38°.28':: tang. BS=23°.28': sine AB=33°.7′. viz. Radius, is to the tangent of the latitude; as tangent of the sun's declination, is to the sine of the ascensional difference.

(W) By comparing the triangle ABS with the triangle AGC, it is evident that when the sun's declination and the latitude of the place are of the same name, the ascensional difference in time subtracted from six hours gives the time of rising, and added to six hours gives the time of sun-setting: but when the latitude of the place and the sun's declination are of contrary names, the ascensional difference added to six hours gives the time of sun-rising, and subtracted gives the time of setting.

In the present case the sun rises at 3h.47.32", and sets at gh.12.28"; double the time of rising gives the length of the night, and double the time of setting gives the length of the day.

(X) When the sun's declination is equal to, or greater than the complement of the latitude; viz. when o is equal to or exceeds go (a circumstance that can never happen in latitudes lower than 66°.32′) the parallel of declination will not cut the horizon Ho, and consequently the sun will never set at these times. The same observations hold true with stars whose co-declination or polar distance CN, is equal to, or less than, the latitude of the place, or elevation of the pole on, and in the same hemisphere.

EXAMPLE II.

Required the amplitude, time of rising, culminating, and setting of Arcturus at Greenwich, latitude 51°.28.40" N. on the first of December 1822, its right ascension being 14b.7.32"," and declination 20°.6'.50" N.

+ See Table VIII. at the end of the Book.