Let m, s, be the observed places of the moon and sun, or of the moon and a fixed star, and let м, s, be their true places. м will be above m, because the moon is depressed by parallax more than it is elevated by refraction; but s will be below s, because the sun is more elevated by refraction than it is depressed by parallax. Observation gives the apparent distance ms, and the apparent zenith distances zm, zs: by applying the proper corrections to these latter we also deduce the true zenith distances zм, zs, and with these data we are to determine the true distance, мs, by computation.

M

ר

Then in the triangle mzs, we have, (Art. 82,) R being 1,

hence, for the determination of D, we have this equation, viz.,

* In observing d with the sextant, it is the nearest point of the limb of the moon which is made to coincide with the other heavenly body, and in observing a with the quadrant, it is the limb also which is made to coincide with the horizon; so that d and a must be corrected for the semidiater of the moon; similar remarks apply to the sun, if he be the other heavenly body.

+ Observe that A and A' are the complements of zм and zs.

from which we immediately get

COS D= (cos d—sin a sin a')

COS A COS A'

cos a cos a'

-+ sin a sin a'

But cos a cos a' - sin a sin a' =cos (a+a') Art. 69; transposing cos a cos a' and substituting the value of — sin ɑ sin a' thus obtained, we have

COS D

cosd+cos(a+a')-cos a cos a'

cos a cos a'

-COS A COS A'+sin a sin a'

A

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 1⁄2 (a+ad) Art. 86, we have

2cos(a+a+d)cos}(a+a'_d)cosacosa'

COSD=

—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 28° 20′ 48′′, a'

a1

= 48° 27′ 32"

= 48° 26' 49" ;

and the computation for D, by formula (1), is as follows:

(Reject 40 from index) 1.536926=log.344292+

A+A' 76 47 37

nat. cos .228460—

True distance 83° 20′ 54"

nat. cos .115832

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 1010, 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 member 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.

Those who are desirous of entering more at large into the problem of the longitude, and of becoming acquainted with the best methods of shortening the computation by the aid of subsidiary tables, may advantageously consult, besides the works already referred to, the Quarto Tables of J. De Mendoza Rios, Lynn's Navigation Tables, Captain Kater's Treatise on Nautical Astronomy, in the Encyclopædia Metripolitana, Kerrigan's Navigator's Guide and Nautical Tables, and Dr. Myers's translation of Rossel on the Longitude.

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.

rical triangle, the three sides being given to find an angle;

the three given sides are the colatitude pz, the zenith distance of the object zs and its polar distance ps and the azimuth being measured by the angle at the zenith z, opposite the polar distance, this is the angle sought. We shall give an example in each of these cases of the problem.

P

Z

EXAMPLES.

1. In January, 1830, at latitude 27° 36′ N., the rising amplitude of Aldebaran was by the compass* E. 23° 30′ N.; i 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. × sin. dec.sin. amp. x cos lat., the computation is as follows.

X

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.

* The compass amplitude must be taken when the apparent altitude of the object is equal to the depression of the horizon.