points, and the circle of altitude passing through the place of the body; or it is the corresponding angle at the zenith between the celestial meridian, and the circle of altitude passing through the body: thus SO or NO, or the angles OZS or OZN is the azimuth of X. The amplitude of a heavenly body is the distance from the east at which it rises, or the distance from the west at which it sets, these arcs or distances being measured on the horizon: thus the amplitude of X is the arc WD or ED, (the dotted line D, XD being the arc of a parallel of declination described by X from rising at D, to setting at D.) The hour angle of a heavenly body is the angle at the pole between the celestial meridian and the circle of declination passing through the place of the body: thus ZPX is the hour angle of X. If L be the place of the sun, west of meridian, its hour angle ZPL is called apparent time: but when the sun is east of the meridian as at C, then apparent time is found by subtracting the hour angle ZPC from 24 hours.

If Y be the place of any heavenly body when on the prime vertical, V its place when its hour angle ZPV is 6 hours, and D its place when setting, then if we put the latitude of spectator ZQ

hour angle ZPD when body is setting.. hour angle ZPY when body is due west. altitude VD, at six o'clock..

altitude WY when due west.. azimuth ND from north at setting.. amplitude WD...........

azimuth PZV at six o'clock...

= l

=h

=h,

=α

Ξα

=2

=m

=Z,

and the declination of the heavenly body

d

the following formula are easily deduced from Napier's Rules for quadrantal and right angled triangles.

(88) Given the sun's meridian altitude = 70° (zenith north of the sun) and its declination 20° N.: required the latitude of the place. Ans. 40° N. (89) Civen the sun's meridian altitude = 70° (zenith north), and its declination=5° S required the latitude. Ans. 15° N.

:

(90) Given the sun's meridian altitude 70° (zenith south), and declination 25° N.: required the latitude. Ans. 5° N.

(91) Give the sun's meridian altitude = 30° (zenith south), and declination 10° N.: required the latitude. Ans. 50° S.

(92) Given the altitudes of a circumpolar star at its inferior transit 20°, and at its superior transit = 70°: required the latitude. Ans. 45° N.

(93) A circumpolar star passes the zenith of a place and its altitude at the inferior transit is 20°: required the latitude. Ans. 55° N.

(94 The altitude of a circumpolar star at its inferior

* In these formula, both the latitude and sun's declination are of the same name. In the problems, these quantities are considered north, unless the contrary is expressed.

transit is equal to its zenith distance at its superior transit required the latitude. Ans. 45° N⚫

(95) Given the sun's altitude (a), and hour angle (h) when the declination is nothing: to find the latitude (1).

Er. a 22° 56', h=3h. Ans. 1 56° 33′ 30′′ N. (96) Given the sun's amplitude m, and declination d: to find the latitude 7.

Ex. m-E 37° 30′ N., d=15° 12′ N.

Ans. 1=64° 29′ 15′′ N.

(97) Given the sun's altitude a at six o'clock, and declination d: to find the latitude.

Er. a=18° 45', d=20° 4′ N.

Ans. 1 69° 31′ 40′′ N.

(98) Given the latitude l, and sun's declination d; to

find his altitude and azimuth at six o'clock.

Ex. 1-50° 48′ N., d=23° 27′ 45′′,

Ans. a 17° 58′ 15′′, az.=N 74° 39′ 30′′ E.

(99) Given the sun's declination d, and latitude of to find his altitude and the time when he

the place

is on the prime vertical.

Ex. d=23° 27′ 45′′ N., l=50° 48'.

Ans. Alt.

=

30° 55', time = 4h 37m 4s.

(100) Given the latitude of the place, and the sun's declination, to find his amplitude, time of rising or setting, and the length of the day and night.

Ex. 1-50° 48′ N. d=18° 28' N.

Ans. Ampl.-E. 30° 4′ 30′′N.; sun rises at 4b 23m 19s A. M.; length of day = 15h 13m 22s.

a

(101 The altitude of a star when due east was 20° and it rose E.b.N.: required the latitude.

(102) The altitude of a star whose declination was 20° N. when due west was 30°: required the latitude. Ans. Lat. 43° 9′ 15′′ N.

(103) Given the sun's altitude (west of meridian) and declination, and the latitude of the place: to find the azimuth and the apparent time of observation.

Ex. a=46° 20′, d=23° 27′ 45′′, l=50° 48′.

Ans. Az.-N.111° 51' W; app. time=2h 57m 163. (104) Given the sun's altitude, declination and azimuth to find the latitude. Ex. a=18° 45′, d=22° 10′ N.; az.=S. 57° 45′ W. Ans. Lat=59° 4′ N. (105) Given the sun's altitude (a), hour angle h, and decl. d: to find the latitude (zenith being north of the sun).

Ex. a=37° 20′, 4=2h 15o, d=22°30′ N.

=

Ans. Lat. 71° 31′ N. (106) Given the right ascension of the sun, RA, and obliquity of the ecliptic w: w to find his longitude and declination.

Ex. (1). R. A. 4b 10" 20", w=23° 27′ 45′′.

Ans. Long. 64° 33′ 15′′, d=21° 4′ 15′′ N. Ex. (2). R. A.=17h 10m, w=23° 27′ 45′′.

Ans. Long.=258° 30′ 15′′, d=22° 58′ S. (107) Given the right ascension and declination of a heavenly body, and the obliquity of the ecliptic w: to find its latitude and longitude.

Ex. (1). R.A. = 2h 59m 37s, d = 21° 27' 48" N. w=23° 27' 45".

Ans. Lat. 4° 15′ N., long. =48° 37′ 30′′.

=

Ex. (2). R. A. = 16h 14m, d = 25° 51′ N., w=' 23° 27' 45".

Ans. lat.

46° 6′ 15′′ N., long. = 234° 36′ 30′′.

(108) Two places have the same latitude, namely, 45° N., and their difference of longitude is 10° 36′ : required their distance.

Ans. 449.25 nautical miles. (109) Required the distance from Portsmouth to Buenos Ayres lat. of Portsmouth 50° 48' N., long. 1° 6' W. lat. Buen. Ayres 34° 37' S. long. 58° 24'W. Ans. 5949.8 nautical miles or 6847.2

:

Eng. miles (69 to a degree). (110) Required the distance of the moon from a Leonis (Regulus): the right ascen. former being 0° 32′ 45′′ and 5° 19′ S., 148° 18′ 45′′, and 13° 10′ 15′′ N.

and decl. of the

and of the latter

Ans. 148° 36'.

(111) Required the sun's azimuth and depression below the horizon at 7h P. M. (app. time) the decl. being 10° 15' S., and the latitude of the place 50° 48' N. Ans. 17° 23′ 30′′. (112) Given the sun's longitude 202° 24′ 15′′, and the moon's latitude and longitude 4° 54′ 30′′ N. and 89° 25' 30" required their distance.

Ans. 112 53′ 30′′.

(113) If a ship from latitude 50° 10′ N., start on a S. W. course, and sail 100 miles on a great circle: what will be her last course. Ans. S. 45° 38′ W.

(114) The latitude of a place A is 40° N. of B 50° N. and their distance from each other 20°: the longitude of A is 15° E. required the latitude and longitude of