Dividing the last term of the numerator by the denominator, the quotient is 1; then observing that cos A cos A'+ sin a sin A′ = cos (A + A') and that cos d+cos (a + a') = 2 cos(a + a' + d) cos (a+a' ~ d) Art. 83, we have COS D= 2 cos(a + a + d) cos (a + a'~d) cos ▲ cos a' cos a cos a' -COS(A+A)(1) ( EXAMPLE. 1. Suppose the apparent distance between the centres of the sun and moon to be 83° 57′ 33′′, the apparent altitude of the moon's centre 27° 34' 5", the apparent altitude of the sun's centre 48° 27' 32", the true altitude of the moon's centre 28° 20′ 48′′, and the true altitude of the sun's centre 48° 26' 49"; then we have d = 83° 57′ 33′′, a = 27° 34' 5'', a' = 48° 27' 32'' and the computation for D, by formula (1), is as follows: By glancing at the formula (1), we see that 30 must be rejected from the sum of the above column of logarithms, to wit, 20 for the two ar. comp. and 10 for R, which must be introduced into the denominator, in order to render the expression homogeneous, so that the logarithmic line resulting from the process is 9.536926. Now, as in the table of log. sines, log. cosines, &c., the radius is supposed to be 101o, of which the log. is 10, and in the table of natural sines, cosines, &c., the rad. is 1, of which the log. is 0; it follows that when we wish to find, by help of a table of the logarithms of numbers, the natural trigonometrical line corresponding to any logarithmic one, we must diminish this latter by 10, and enter the table with the remainder. Hence the sum of the foregoing column of logarithms must be diminished by 40, and the remainder will be truly the logarithm of the natural number represented by the first term in the second number of the equation (1). If this natural number be less than nat. cos (A + A'), which is to be subtracted from it, the remainder will be negative, in which case D will be obtuse. VARIATION OF THE COMPASS. 114. We shall conclude this part of our subject by briefly considering the methods of finding the variation of the compass, or the quantity by which the north point, as shown by the compass, varies easterly or westerly from the north point of the horizon. The solution of this problem merely requires that we find by computation, or by some means independent of the compass, the bearing of a celestial object, that we observe the bearing by the compass, and then take the difference of the two. The problem resolves itself, therefore, into two cases, the object whose bearing is sought being either in the horizon or above it in the one case we have to compute its amplitude, and in the other its azimuth. are the colatitude PZ, the zenith distance of the object zs, and its polar 1. In January, 1830, at latitude 27° 36' N., the rising amplitude of Alde baran was by the compass* E. 23° 30' N.; required the variation. From the Nautical Almanac, it appears that the declination of Aldebaran at the given time was 16° 9' 37" N., therefore since, by Napier's rule, Rad. X sin dec. = sin. amp. x cos lat., the computation is as follows: As the object is farther from the magnetic east than from the true east, the magnetic east has therefore advanced towards the south, and therefore the magnetic north towards the east; hence the variation is 5° 11' 43" E. 2. In latitude 48° 50′ north, the true altitude of the sun's centre was 220 2', the declination at the time was 10° 12′ S., and its magnetic bearing 161° 32' East. Required the variation. *The compass amplitude must be taken when the apparent altitude of the object is equal to the depression of the horizon. The variation is west, because the sun's observed distance from the north, measured easterly, being greater than its true distance, intimates that the north point of the compass has approached towards the west. 3. In latitude 48° 20′ north, the star Rigel was observed to set 9° 50' to the northward of the west point of the compass; required the variation, the declination of Rigel being 8° 25' S. Variation, 22° 33' West. |