1. In latitude 52° 12′ 42′′ N., in the afternoon, the true altitude of the sun's centre was 39° 5′ 28′′, when its declination was 15° 8' 10"; what was the apparent time of observation? Ans. 2h 56m 428.7.

2. In latitude 24° 30′ N., in the forenoon, the true altitude of the sun's centre was found to be 33° 20', when its declination was 6° 47′ 50′′ S.; required the apparent time of observation. Ans. 8h 46m A.M.

468. PROBLEM XXVII.-Given the latitude and longitude of the place, the right ascension and declination of a fixed star and its altitude, to find the mean time.

'Find the horary distance of the star from the meridian; then find the sun's mean right ascension at the preceding mean noon at the given place, and subtract it from the star's right ascension, increased if necessary by 24 hours; to this interval apply the horary angle by addition or subtraction, according as the star is west or east of the meridian, and the result is the sidereal interval from mean noon, and its corresponding interval of mean time will be the required

As in last problem,

2 L cos H

L sin S+L sin (S-Z) + L cosec P+L cosec C. Let A sun's m. R.A. at preceding noon at place, A'star's right ascension,

d = the difference of A and A',

s the sidereal interval from mean noon, and m, r, the corresponding mean time and retardation. Then d = A'—A, s=d± H, and m=s―r.

EXAMPLE.-At a place in latitude 48° 56′ N., and longitude 66° 12′ W., the true altitude of Aldebaran, which was west of the meridian, was 22° 24' on the 10th of February 1841; what was the mean time of observation?

Star's dec.

Ꮓ

16° 11′ 10′′ N., and R.A. = 4h 26m 498.77.

=

67° 36′ 0′′

L. Sin S

= 9.9998982

Hence, H = 37° 37′ 33′′-82, and H = 5h 1m 0s.5.

Sun's reg. R.A. or sidereal time on 10th 21h
Acceleration for long. 4b 24m 48s W. =+

21m 33s-5

0

0 43.5

Let M be the meridian, A the first point of Aries, MP the meridian of the place, B the place of the star, and S of the mean sun at time of observation. Then,

N

B M

if since noon the sun's increase of mean S right ascension be SS', the sidereal interval at noon between the sun and star is = BS', and as B has passed the meridian by the horary arc BM, the sidereal time from noon till the observation is expressed by S'M S'B+BM=d+H=A-A+H. And when the star is east of the meridian, when observed, as at B', the sidereal time from noon = S'MS'B' - B'M =A'-A-H. And this interval reduced to mean time, gives that required. It is here supposed that the declinations of the sun and stars are the same, but the same reasoning applies when they are different.

EXERCISES.

1. If, at a place in latitude 53° 24′ N., and longitude 25° 18′ W., the altitude of Alphacca when east of the meridian was found to be 42° 8' 50" on the 31st of January 1841, its right ascension being 15h 27m 58s, its declination 27° 14′ 50′′ N., and the registered mean right ascension of the sun 20h 42m 8s. Ans. H = 3h 40m 178.5,

2. Find the hour of observation in mean time at which the altitude of Procyon was 28° 10′ 13′′, when east of the meridian in latitude 7° 45′ S., its declination being 5° 41′ 52" S., its right ascension 7h 29m 30s, and that of the mean sun at mean noon 11h 4m 40s.

Ans. H = 4h 2m 0s-9, and time = 16h 20m 83.

469. The equation of equal altitudes is a correction generally of a few seconds, and seldom exceeding half a minute, that must be applied to the middle time between the instants of two observations at which the sun has equal altitudes in the forenoon and afternoon. It depends on the change of the sun's declination in the interval between the observations.

470. PROBLEM XXVIII.-To find the equation of equal altitudes.

Let L the latitude of the place,

I

the interval of time expressed in degrees, &c., Pthe sun's polar distance,

v variation of declination in 24 hours in seconds,
v = variation of declination in interval I in seconds,
the equation of equal altitudes in seconds.

E
Then, an arc é, called arc first, is such that

tan = L. cot L. + L. cos I -10;

P-8.

and if is another arc, called arc second, then y = And L. EL. cot I+L. cosec + L. cosec P + L. sin p

+L.I+L. '+25-3645,

in which the quantity I in L. I is expressed in minutes. The logarithms require to be carried only to four places. This rule is approximate, but it will give the result correct to a small fraction of a second. The polar distance at the nearest noon may be used, as any small change in it or in the latitude produces a very small effect on the equation.

EXAMPLE. Find the equation of equal altitudes for an interval of 7h 45m 30s, and latitude 46° 30′ S., on the meridian of Greenwich.

Here L 46° 30′ S., I = 7h 45m 30s,
P 97° 10' N., Im 465m-5, v'

1°

58° 11'.

22' 14"

1334".

Hence, the equation of equal altitudes is 18s-9.

It is evident that when the declination of the sun has varied in one direction during the interval between two equal altitudes, that the intervals between the meridian passage or apparent noon, and the instants of the two observations, are different. When it increases, the interval in the afternoon will exceed that of the forenoon, and conversely when it diminishes. For a demonstration of the rule, see Riddle's Treatise of Navigation and Nautical Astronomy. A slight alteration has been made here which improves it a little. Instead of L., where v is the variation due for the interval I, and which requires to be separately calculated, there has been introduced above the constant 5.3645, L. Im and L. v'.

For 24:Iv: v, and hence L. v = L. I + L. v′ - L. 24. And if v', v, are in seconds of space, and 24 and Im in minutes of time, then, since 24h = 1440m, and L. 3v= L. v-L. 30, therefore

L. v=L. Im + L. v' +5·3645.

EXERCISE.—If at a given place when the sun's declination at noon was 17° 54′ N., the sun had equal altitudes for an interval of 5h 40m 6s, the latitude of the place being 57° 10′; what was the equation of equal altitudes?

Ans. 128.1. 471. The middle time for the times of observation of two equal altitudes of the sun, is half the sum of the times.

472. PROBLEM XXIX.-To find the time by equal altitudes of the sun.

"Apply the equation of equal altitudes to the middle time by addition or subtraction, according as the polar distance is increasing or diminishing, and the result is the time shown by the clock at apparent noon; find the mean time at apparent noon, and the difference between it and the preceding time will be the error of the clock."

When a chronometer is used for the times of observation, apply the longitude in time to the mean time at apparent noon, and the result is the mean time at Greenwich at that instant; and the difference between it and the time found