C=166 19 8=PZO

which is the azimuth from the north.

(161.) Def. Twilight is that faint light which appears in the horizon before the sun rises and after he sets. It is generally supposed to begin in the morning, and end in the evening, when the sun is 18 degrees below the horizon.

(162.) Problem XIII. Given the latitude and the sun's declination, to determine the time when the twilight begins and ends.

In the spherical triangle Z P O (fig. 31.) we have given ZP co-latitude, P O co-declination when north (or 90° + declination when south) and Z O

108°, to find the angle Z PO, the horary angle.

Example. At London, in latitude 51° 32', when

the sun's declination is 11° 46' s., at what time did

the twilight begin?

Answer 3m 12s after five o'clock.

(163.) Problem XIV. To find the time when the apparent diurnal motion of a star is perpendicular to the horizon.

Let xy (fig. 31.) be the parallel described by the star, and suppose the vertical circle Zh to be drawn touching it at S: now when the star arrives at this point, its motion is perpendicular to the horizon, being at that instant in the direction of the vertical Zh. Join PS, then Z SP is a right angle, and hence we have,

cos Z P Scot Z P. tan P S.

=tan 1. tan polar dist. of star.

Now the time when the star is upon the meridian, being supposed to be known, the angle ZPS converted into time shews the interval to that time.

(164.) Definitions. Let S (fig. 32.) represent the position of a star, AQ the equator, and A L the ecliptic; draw the great circles S B, SC, perpendicular to AQ, AL, intersecting them respectively in B, C; then AB is called the right ascension of the star,

SB its declination, A C the star's longitude, and SC its latitude.

(165.) Problem XV. Given the right ascension and declination of a heavenly body, and the obliquity of the ecliptic, to find the latitude and longitude.

Referring to figure 32, draw the great circle A S; then in the right angled spherical triangle AS B, we have,

hence the angle SA B becomes known, consequently also the angle S A C.

Again,

cos A S=cos A B . cos S B...... (2.)

... A Sis given.

Hence, in the triangle SA C right angled at C,

therefore AC the longitude, and S C the latitude of

the body become known.

=

Example. The right ascension of the fixed star Aldebaran, at midnight (Greenwich time) on Feb. 10th, 1834, 66° 35′ 55′′· 1, its declination=16o 10′ 6′′ 2 N., and the obliquity of the ecliptic=23° 27′ 40′′. Required the latitude and longitude of the star.

Here, A B=66° 35′ 55′′ · 1, SB=16° 10′ 6′′ · 2, and B A C 23° 27′ 40′′.

By equation (1.) we have,

10+ log tan 16° 10′ 6′′ 2......19.4622912

log sin 66° 35′ 55′′ · 1...... 9·9627221

log tan SA B

.........

9.4995691

... SA B 17° 31′ 55′′, and

... the star S' is situated between the ecliptic and the equator, or in south latitude; hence the angle SA C =5° 55′ 45′′.

log cos A S............... 95814497

.•. A S=67° 34′ 33′′.

Again, by equ. (3.), (4.)

tan A C-tan AS. cos SAC

log tan A C....... 10 3820741

... A C-67° 28′ 2' the 's longitude.

Hence, S C=5° 28′ 46′′=the 's latitude south.

The right ascensions and declinations of the stars are ascertained by observation, and hence the tables of the latitudes and longitudes have been computed.

(166.) As the obliquity of the ecliptic and the position of the vernal equinox are subject to continual change, the right ascensions, declinations, and longitudes of all the fixed stars are perpetually varying, principally from these causes. The longitude increases by a quantity nearly equal to the annual precession of the equinoxes; but the variations of the right ascensions, and declinations, owing