Suppose the sun's right ascension, on the 17th May, to be 53° 38′ and his declination at the same time 19° 15' 57", then cos 53° 38′ x cos 19° 15′ 57′′

sum rejecting 109.747989 = log. cos. 55° 57′ 43′′ Hence q or the longitude required is 55° 57′ 43′′.

EXAMPLE II.

90. The same being given as in the last example, required the obliquity of the ecliptic.

The required part is the angle E in the figure. Of E, s and e, since the three are contiguous leaving out the right angle, s is the middle part, hence applying the rule of Napier,

R sin stan e cot E

We put cot E instead of tan E because, according to the directions before given, the complements of the oblique angles are to be employed. Taking the value of cot E from the above equation, we have

91. Had the angle s been required, it might be found by observing that of the three parts, s, e and s,the latter two of which

obliquity of the ecliptic is continually, though very slowly, changing. It given with the minutest attainable accuracy in the Nautical Almanac,

are given, e is the middle part, and s and s adjacent. Hence, using the complement of the oblique angle s;

Applying logarithms as before, the value of s will be found. We leave this as an exercise for the learner.

In the solution of the above triangle, it will be observed that we have found each of the unknown parts in terms of the two given, and have not employed one of those first calculated to obtain another. This is agreeable to the principle laid down at Art. 41, of plane trigonometry, and the reason is the same. Such a method of proceeding is always practicable in the solution of right angled spherical triangles.

In the examples which we have taken above, we have supposed the base and perpendicular of a right angled spherical triangle given. Any other two parts being given, each of the unknown parts may be calculated by the aid of Napier's rules, in a manner entirely similar to what has been just exhibited.

18

EXAMPLE III.

92. Given the sun's declination to find the time of his rising and setting at any place whose latitude is known.

Let nEsa represent the meridian of the place, z being the zenith, and нo the horizon, and let s's" be the apparent path of the sun on the proposed day, cutting the horizon in s. Then the arc EZ will be the latitude of the place, and consequently EH, or its equal ao, will be the colatitude, and this

Ꮓ

E

Ꭺ

H

measures the angle oAQ; also RS will be the sun's declination, and AR, expressed in time, will express the time of sunrise from 6 o'clock, for nas is the six o'clock hour circle.

Hence, in the right-angled triangle SAR, we have given RS and the opposite angle A to find AR, the time from 6 o'clock.

EXAMPLE L..

Required the time of sunrise at latitude 52° 13' N., when the sun's declination is 23° 28'.

By Napier's rule,

Rad. sin ARcot A. tan Rstan. lat. tan. dec.

* Degrees are converted into hours by multiplying by 4 and dividing by

60, which is equivalent to multiplying by 15.

SCHOLIUM.

It should be here remarked that the time thus determined is apparent time, which is that which would be shown by a clock so adjusted as to pass over 24 hours during one apparent revolution of the sun, or from its leaving the meridian to its return to it again, the index pointing to 12, when the sun is on the meridian. But it is impossible that any clock can be so adjusted, because the interval between the successive return of the sun to the meridian is continually varying, on account of the unequal motion of the sun in its orbit, and of the obliquity of the ecliptic; each of these varying intervals is called a true solar day, and it is the mean of these during the year which is measured by the 24 hours of a well regulated clock, this period of time being a mean solar day; hence, at certain periods of the year, the sun will arrive at the meridian before the clock points to 12, and at other periods the clock will precede the sun; the small interval between the arrival of the index of the clock to 12 and of the sun to the meridian is called the equation of time, and it is given in page ii. of the Nautical Almanac for every day in the year; this correction, therefore, must always be applied to the apparent time determined by trigonometrical calculation to obtain the true time, or that shown by a well regulated clock or chronometer.

Another circumstance too must be taken into account, in order to determine the apparent time with rigorous accuracy, viz. the change in the declination of the sun from sunrise to noon. In the Nautical Almanac, the declination of the sun is given for every day at noon, and if this be used in the computation we shall assume that the declination has not varied from sunrise to noon, which is not the case; hence it will be necessary to compute the declination for the time of sunrise, as determined above, and then to resolve the problem with this corrected declination. The correction is obtained by taking from the Nautical Almanac the variation of declination in 24 hours, and then finding by proportion the variation for the time required.

2. Required the time of sunrise at latitude 57° 2′ 54′′, when the sun's declination is 23° 28'?

3h 11' 49"

PROBLEM III.

93. Given the latitude of the place, and the declination of a heavenly body, to determine its altitude and azimuth when on the six o'clock hour circle.

Let HZPO be the meridian of the place, z the zenith, нo the horizon, s the place of the object on the six o'clock hour circle psp, which of course passes through the east and west points of the horizon, and ZsB the vertical circle passing through the sun. Then, in

S

P

the right-angled triangle SBA, the given quantities are as, the declination, and the arc or, or angle SAB, the latitude of the place, to find the altitude вs, and the azimuth Bo from the north point of the horizon; or to find the complement AB of this azimuth, that is, the sun's bearing from the east.

EXAMPLES.

1. What was the altitude and azimuth of Arcturus, when upon the six o'clock hour circle of Greenwich, lat. 51° 28' 40′′ N., on the 1st of April, 1822; its declination on that day being 20o 6' 50" N.?

By Napier's rule we have

Rad. sin BS = sin a sin as

Rad. cos A

Rad. cos atan AB cot As...* cot Bo=

cot As

This sign.. signifies therefore.