Exam. 2. Required the moon's longitude, latitude, equatorial parallax, semi-diameter, and hourly motions in longitude and latitude, on the 9th of April, 1838, at 8h. 58m. 19sec. P. M. mean time at Washington.

Ans. Long. 6. 19° 45′ 31′′.2; lat. 36' 21".9 S.; equat. par. 54' 36".3; semi-diameter 14′ 52".7; hor. mot. in long. 30′ 15".2; and hor. mot. in lat. 2' 47".0, tending south.*

PROBLEM XV.

The Moon's Equatorial Parallax, and the Latitude of a Place, being given, to find the Reduced Parallax and Latitude. With the latitude of the place, take the reductions from Table LXIV, and subtract them from the Parallax and Latitude.

Exam. 1. Given the equatorial parallax 55′ 15′′, and the latitude of New York 40° 42′ 49" N., to find the reduced parallax and latitude.

2. Given the equatorial parallax 60' 36" and the latitude of Baltimore 39° 17′ 13′′ N., to find the reduced parallax and latitude.

Ans. Reduced par. 60' 32", and reduced lat. 39° 5′ 59′′. 3. Given the equatorial parallax 57' 22", and the latitude of New Orleans 29° 57′ 45′′ N., to find the reduced parallax and latitude.

Ans. Reduced par. 57' 19", and reduced lat. 29° 47' 50".

The smaller equations were omitted in working this example.

PROBLEM XVI.

To find the Longitude and Altitude of the Nonagesimal Degree of the Ecliptic, for a given time and place.

For the given time reduced to mean time at Greenwich, find the sun's mean longitude and the argument N from Tables XVIII, XIX, XX, and XXI. To the sun's mean longitude, apply according to its sign the nutation in right ascension, taken from Table XXVII with argument N; and the result will be the sun's mean longitude, reckoned from the true equinox.

Reduce the mean time of day at the given place, expressed astronomically, to degrees, &c., and add it to the sun's mean longitude from the true equinox. The sum, rejecting 360°, when it exceeds that quantity, will be the right ascension of the midheaven, or the sidereal time in degrees.

Next, find the reduced latitude of the place by Problem XV; and when it is north, subtract it from 90°; but when it is south, add it to 90°; the sum or difference will be the reduced distance of the place from the north pole.

Also take the obliquity of the ecliptic for the given year from Table XXII.*

These three quantities having been found, the longitude and altitude of the nonagesimal degree may be computed from the following formulæ :

log. cos. † (H — w) — log. cos. } (H + w) = A (1);

log. tang (Hw) + 10 — log. tang } (H+ w) = B. . . (2); log. tang E = A + log. tang (S-90°).

log. tang F = log. tang E + B . . . (4);

NE+F+90° ... (5);

(3);

If great precision is required, the apparent obliquity is to be used in place of the mean. (See Prob. X.)

in which,

H = the reduced distance of the place from the north pole;

w the Obliquity of the Ecliptic;

ω

=

S the Sidereal Time converted into degrees;

=

N = the required Longitude of the Nonagesimal;

h = the required Altitude of the Nonagesimal;

E and F are auxiliary angles.

We first find the logarithmic sums A and B. With these we determine the angles E and F by formulæ (3) and (4), and with these again N and h by formulæ (5) and (6).

The angles E, F are to be taken less than 180°; and less or greater than 90°, according as the sign of their tangent proves to be positive or negative.

Note 1. In case the given place lies within the arctic circle, we must take, in place of formula (5), the following:

NE F + 90°.

Note 2. As the obliquity of the ecliptic varies but slowly from year to year, the values which have once been found for the logarithms A, B, and C, will answer for several years from the date of their determination, unless very great accuracy is required.

Note 3. The angle h derived from formula (6), is the distance of the zenith of the given place from the north pole of the ecliptic. This is not always equal to the altitude of the nonagesimal. Throughout the southern hemisphere, and frequently in the northern near the equator, it is the supplement of the altitude. In employing this angle in the following Problem, it is, however, for the sake of simplicity, called the altitude of the nonagesimal in all cases.

Exam. 1. Required the longitude and altitude of the nonagesimal degree of the ecliptic at New York, on the 18th of September, 1838, at 3h. 52m. 56sec. P. M. mean time.

The sun's mean longitude taken from the tables, for the given time, is 5. 27° 19' 17", and the argument N is 987. The nutation taken from Table XXVII with argument N is-1". Hence, the sun's mean longitude from the true equinox is 5. 27°

19' 16". The given time of day, expressed astronomically, is 3h. 52m. 56sec.; which in degrees is 58° 14' 0".

The reduced latitude of New York, found by Problem XV, is 40° 31' 20", and this taken from 90° leaves the polar distance 49° 28′ 40′′. The obliquity of the ecliptic, derived from Table XXII, is 23° 27' 37".

2. Required the longitude and altitude of the nonagesimal degree of the ecliptic at New York, on the 10th of May, 1838, at 11h. 33m. 56sec. P. M. mean time.

Ans. Long. 200° 12′ 23′′, and alt. 37° 0' 34".