is 6 hours from the meridian, then its cosine will vanish, and equa. 1, becomes

...

cos n or sin a = sin d sin l..... (3.) from which the altitude, a, becomes known.

In the same circumstances theorem 10, of right angled spherical triangles, gives

cot szp=cot Ps sin pz tan d cos l.... (4.) by which the azimuth in that case may be determined. 8. When the body s is in the horizon, or s and v coincide, we have zs = : 90°, or cos n =0;

hence 0 sin d sin + cos d cos l cos P,

Here it is evident that when the declination and latitude are both of the same kind, cos p is negative, or p is greater than 90°; that is, the time occupied by the heavenly body in passing from the horizon to the meridian, or from the meridian to the horizon, exceeds 6 hours.

If the declination and latitude are of different kinds, cos P will be positive, and the time of passage from the horizon to the meridian, or from the meridian to the horizon, less than 6 hours. Thus, this theorem will serve to determine the times of the risings and settings of the sun, or other heavenly bodies, disregarding the changes of declination, and the effects of parallax and refraction.

To find the azimuth when s is in the horizon, we have from the principles of quadrantal triangles,

9. If the body s be upon the prime vertical, then pzS = 90°, and the formula for right angled triangles give

cos Ptan ZP cot SP cot / tan d (7.)

....

sin d cosec l.... (8.)

10. When the parallel circle s's s" which a heavenly body describes in consequence of the diurnal rotation can have a vertical zsv drawn to touch it, which it may while the declination of that body exceeds the latitude of the place, then the body when at the place of contact of the two circles will move vertically, and with the greatest apparent rapidity. In that case the spherical triangle Psz will be right angled at s, and we shall have from the formulæ for right angled triangles in chap. vi. cot d tan l.... ... (9.)

COS P =

cot d cot l

(10.)

sin

sin a= =sin l cosec d.. (11.)

sin d

11. Returning to the first equation

COS ZS COS PS COS PZ + sin PS sin PZ cos P it may evidently be employed to determine the altitude, or the zenith distance, when the latitude, declination, and horary angle, are known. The same may also be readily effected by an auxiliary angle, as explained in case 2 of oblique spherical triangles, equa. 12 and 13.

If we wish to find the hour from the meridian, by the observed zenith distance, we have from the above equa

tion

COS ZS- COS PS COS PZ

COS P

sin Ps sin PZ

Or, from equa. 2, art. 38, spherical trigonometry,

a theorem fitted to logarithmic use.

A like theorem will obviously serve for the determi❤ nation of the azimuths, viz.

tanz=

sin(ZPPS - sz) sin (Ps + ZS — ZP)

sin & (ZP + Z8 — PS) sin § (PS + ZS + ZP)

(13.)

Or, the angles z, s, and the side zs, may be easily found by Napier's analogies, thus:

12. Since, according to the determinations of astronomers, the crepusculum or twilight commences and terminates when the sun is 18° below the horizon, either previously to his rising, or subsequently to his setting, it follows that equa. 12 will serve to compute the time from noon to the beginning or end of twilight, if zs the zenith distance be assumed equal to 90° + 18°, or 108o. Or, to determine the duration of twilight, regarding the effect of horizontal refraction, we have these formulæ:

sin 18 sec / sec d ·

tan l tan d,

COS P' = for the hour angle for the commencement of the morn ing, or the end of the evening twilight. COS P— - sin 34' sec 7 sec d

tan l'tan d,

for the hour angle from noon to sunrise or sunset, including refraction.

And (PP) = duration of twilight

(15.)

These theorems may be employed in the solution of a variety of questions: a few are subjoined.

Example I.

At Cambridge, in north latitude 52° 12′ 35′′, when does the sun rise and set, on the 4th of May, 1816, and what is the azimuth at rising and setting?

The sun's declination on the proposed day at noon, is 16° 0′ 46′′ north; and the formulæ to be employed are the 5th and 6th. For the first, viz.

COS P- tan d tan l, we have,

log tan d.. 16° 0′ 46′′.... 94578618 tan.. 52° 12′ 35′′.... 10.1104697

The sum is log cos P 111° 43′ 21′′.... 9.5683315

Here since the latitude and declination are both of the same kind, cos P is negative, and belongs to an arc between 90° and 180°. This hour angle converted into time by dividing by 15 or multiplying by, gives 7 26 534 from noon, for the time required.

In cases where great accuracy is needed, the change of declination in the interval must be regarded. To find the azimuth, take the theorem

COS Z sin d secl.

Log sin d.. 16° 0′ 46′′.... 9.4406756

sec /

The sum cos z

52° 12′ 35′′

10.2127004

azimuth from the north S

Example II.

Required the time of the sun's rising and setting at the equator, on the 14th of October, 1816.

It is not necessary in the solution of this example to look out the declination: for, since the latitude is 0, tan is 0, and consequently cos P= 0: therefore P= 90°, the hour angle of 6; that is, the sun rises and sets at 6 o'clock.

Cor. Hence it appears that at the equator all the heavenly bodies are 12 hours above and 12 hours below the horizon; neglecting the effects of refraction.

Example III.

Required the times when the sun is east or west, at

Cambridge, on May 4th, 1816, and his altitude at those times.

The theorems applicable to this example are 7 and 8, viz.

cos P = cot / tan d, and sin a = sind cosec l. Thus, for the hour angle

The

Log cot .. 52° 12′ 35′′.... 9.8895736

tan d.. 16° 0′ 46′′

....9.4578618

sum cos P.. 77° 8'2 hour angle from noon

Then, for the altitude

Log sin d.. 16° 0′ 46′′

cosecl.. 52° 12′ 35′′ .... 10.1022306

The sum sin a.. 20° 25′ 48′′....

9.5429062

Note. By means of the equations to this example, the solar time when the sun is due east or west may be deter mined: and, therefore, in the summer half year, when the sun is above the horizon, when he is in the prime vertical, he will at those moments cast the shadows of opake objects in the east and west direction, and a horizontal perpendicular to such shadows will be a meridian line.

Example IV.

Given the latitude of the place 52° 12′ 35′′ N. the sun's declination 15° 54′ 25′′ N. and the sun's altitude 40°; to find his azimuth and the time from noon.

Here the time and azimuth may be found by means of equa. 12 and 13. Or, the azimuth being found by means of equa. 13, the time may be found by the proportion between the sines of the sides and opposite angles of spherical triangles, when we shall have

sin z sin zs

sin P =

sin SP

"

sin z cos a sec d.