To find the semidiurnal arc, or the time between a known celestial object's rising or setting, and its passing the meridian, neglecting the effects of dip, refraction, and parallax.

Let O (see the last figure) be the object at rising or setting; then in the quadrantal triangle O P Z, are given O Z a quadrant, ZP the co-latitude, and O P the object's polar distance, to find Z PO, the angle whose measure is the semidiurnal arc. Or in the triangle E QO, right angled at Q, are given Q E O the co-latitude, and QO the declination, to find E Q, the time between the object's rising or setting, and its passing the six o'clock hour circle; and the sum or difference of E Q and six hours, according as the latitude and declination are of the same or different names, is the semidiurnal arc.

By this problem the time of the sun's rising and setting, and the length of the day are found; for the semidiurnal arc shows the time of setting, and deducted from twelve hours, leaves the time of rising; and the semidiurnal arc of any object, added to the time of its passing the meridian, shows the time of its setting; and, subtracted, shows the time of its rising.

When the declination is equal to or exceeds the co-latitude, the object will be always above, or always below the horizon of the place, according as the declination and latitude are of the same or of contrary denominations.

EXAMPLES FOR EXERCISE.

In each of the following examples the, semidiurnal arc is required?

PROBLEM X.

To find the apparent time at which the sun's centre rises and sets, at a given place, on a given day, allowing for the effect of dip, refraction, and parallax.

If from the sum of the dip and the horizontal refraction the sun's parallax be subtracted, the remainder, added to 90°, will be his distance from the zenith when his centre appears on the horizon. Hence, in an oblique angled triangle, as O Z P, we have O Z the corrected zenith distance, OP the polar distance, and Z P the co-latitude, given, to compute Z PO the meridian distance, or the time from noon at which his centre rises or sets.

Note. The declination used in the computation should be that which the sun has at the time of rising or setting. If therefore the semidiurnal arc be computed from the declination at the noon of the given day, by the last problem, the approximate time of rising or setting may be found from it; and thence, by applying the longitude, the corresponding Greenwich time will be obtained. The declination may then be corrected for that time, and the true time of rising or setting found as above.

EXAMPLES FOR EXERCISE.

In each of the following examples the apparent time at which the sun's centre rises and sets is required?

Given the latitude and the sun's declination, to find the beginning and end

of twilight.

In the last figure, let R be the place of the sun on the twilight circle; then in the triangle R Z P are given R Z 108°, RP the sun's polar distance, and Z P the co-latitude of the observer, to find Z PR, the time from noon when twilight begins or ends.

Remark. If the latitude and the sun's polar distance differ only 18°, or less, there will be continual twilight while the sun is below the horizon.

EXAMPLES FOR EXERCISE.

In each of the following examples the times of the beginning or the end of twilight are required?

Given the altitude of a known star, when another known star is on the same vertical with it, to find the latitude.

Let G and F (see the last figure) be the places of the two stars, G being that whose altitude G K, or zenith distance G Z, is given. Then in the triangle G P F, are given G P and PF, the polar distances, and G P F the difference of the right ascension of the two stars, to find the angle P GF; and in the triangle Z GP are then given ZG and PG, the zenith distance and polar distance of G, and the included angle Z G P, to find Z P the co-latitude.

EXAMPLES FOR EXERCISE.

In each of the following examples the true latitude is required?

Star whose Observed Star on the
True
Height of Latitude
Alt. is obsvd. Altitude. same Vertical. the Eye. by Acct. Latitude.

Given the meridian distance of a known celestial object and its altitude,

to find the latitude.

Let F (see the last figure) be the place of the object; then in the triangle Z P F are given F P the polar distance, F Z the zenith distance, and Z P F the meridian distance of F, to find Z P the co-lati tude; an example of the ambiguous case in spherics. Let F Y be a

perpendicular from F on PZ, or P Z produced; then cot FP: rad: : cos Z PF: tan P Y; and cos FP: cos P Y cos F Z: cos Z Y. The sum or difference of P Y and Z Y is P Z the co-latitude; and from the latitude by account it will generally be obvious whether the sum or the difference ought to be taken.

It will be found advisable in practice to apply this problem only when the object is near the meridian, as a small mistake in the apparent time may under other circumstances produce a considerable error in the result.

If the object be the sun, the apparent time, or the difference between the apparent time and twenty-four hours, is his meridian distance; in the case of a star, the difference between its right ascension and the right ascension of the meridian, is its meridian distance.

EXAMPLES FOR EXERCISE.

In each of the following examples the true latitude is required?

From the altitude of two known stars, observed at the same instant, to find

the latitude.

In the figure, p. 275, let S and M represent the true places of the stars; then in the triangle M P S are given M P, P S the polar distances, and M P S the difference of the right ascensions of the two stars, to find their distance M S, and the angle PSM; and in the triangle M S Z there are there given M Z, S Z the true zenith distances of the stars, and M S their distance, to find the angle Z SM; the difference between which and P S M is ZS P. Hence there are given Z S and S P, the zenith distance and polar distance of S, and the included angle Z S P, to find Z P the co-latitude.

If one of the altitudes be increasing and the other decreasing, Z will fall between S P and M P; and if Z and P are on different sides of M S, the sum of M S P and Z S M will be Z SP; but this may be avoided in practice, by selecting two stars whose polar distances are greater than the co-latitude.

If the altitudes cannot both be obtained at the same instant, add

the interval between the observations, reduced into sidereal time, (by adding one second for every six minutes) to the right ascension of the star first observed, and the sum will be its reduced right ascension, with which, the altitude and polar distance of the star, and the altitude, right ascension, and polar distance of the other star, make the computation as above.

EXAMPLES FOR EXERCISE.

The latitude is required in each of the following examples, the altitudes of both stars being taken at the same time?

In each of the following examples the altitudes of two stars, and the interval between the observations, are given to find the latitude.

From the observed altitude, and angular distance of the sun and moon on a given day, to find both the latitude and longitude of the place of observation.

Take from the Nautical Almanac the sun's semidiameter for the given day, and the semidiameter and horizontal parallax of the moon for the hour which corresponds most nearly with the given distance, and correcting the altitudes and distance for semidiameter, &c. clear the distance in the usual manner from the effects of parallax and refraction, and find from the Nautical Almanac the Greenwich time which corresponds to it. If this time differ much from the time before taken, let the semidiameter and parallax of the moon be taken out for the time thus found, the distance re-computed, and the corrected time at Greenwich found from it; for which time also let the declinations of the sun and moon be taken, and thence their polar