Determination of the meridian line.

at 6h 2m 11' P. M., and the second lower transit at 5h 56′′ 4′ A. M. What was the time of the star's passing the meridian the second morning? and what was the azimuth error in the position of the instrument?

Ans. The time of third merid, trans. was 5h 58m 11a A. M.

The azimuth error 1' 8" W.

12. An observer at Boston, wishing to determine his meridian line by means of a clock regulated to solar time, observed the inferior transit of Polaris on April 4, 1839, at 0 A. M., and the superior transit of Ursa Majoris at Oh 53m 59s A. M. What was the azimuth error in the position of his transit instrument?

n

n

The R. A. of Polaris is 1h 0m 50s, that of Ursa Majoris is 13h 41m 14s, and the declination of Ursa Majoris is

50° 7' N.

n

Ans. The azimuth error 7' 18" W.

13. An observer at Boston, wishing to determine his meridian line, in the evening of May 1, 1839, observed by means of a clock regulated to Polaris at 9h 49m 228 P. M., 9h 52m P. M. What was the ment?

The R. A. of Polaris

solar time, the lower transit of and that of a Cassiopeæ at azimuth error of the instru

The R. A. of a Cassiopeæ

1h 0m 56s.

04 31m 228.

The Dec. of a Cassiopeæ = 55° 39′ N,

Ans. The azimuth error = 18' 34" W.

Latitude found by meridian altitudes.

CHAPTER IV.

LATITUDE.

46. Problem. To find the latitude of a place.

Solution. The latitude of the place is evidently, from (fig. 34), equal to the altitude of the pole; so that this problem is the same as to find the altitude of the pole, which would be done without difficulty if the pole were a visible point of the celestial sphere.

First Method. By Meridian Altitudes. [B. p. 166–175.]

Observe the altitude of a star at its transit over the meridian, and let

A''s dist. from point of horizon below the pole ;

then, if the notation of § 28 is used, it is evident, from (fig. 34), that

L= A'p;

(403)

the upper sign being used when the transit is a superior one, and the lower sign when it is an inferior one.

I. Suppose the observed transit to be a superior one; then, if it passes upon the side of the zenith opposite to the pole, we have A' 180°

A, p 90° D.

Latitude found by meridian altitudes.

and (403) becomes

L 90° (AD) = (90°—A)±D=z±D; (404) the upper sign being used when the declination and latitude are of the same name, and the lower sign when they are of different names.

But if the star passes upon the same side of the zenith with the pole, we have

A' A, p = 90°

and (403) becomes

D,

L=(A+D) — 90° — D — (90° —A)=D—z. (405)

:

II. If the transit is an inferior one, we have

A' A, p= 90° — D,

and (403) becomes

=

— D).

(406)

L= (AD) + 90° = A + (90°

Equations (404) and (405) agree with the rule of Case I, [B. p. 166], and (406) with Case II, [B. p. 167.]

III. If both transits are observed, and if A' and A are referred to the upper transits, and

A, the altitude at the lower transit,

1

we have, by (403),

L = A'-p

L = A1+P,

the sum of which is

L= 1⁄2 (A' + A1);

(407)

Latitude found by a single altitude.

so that the latitude is determined in this case without knowing the star's declination.

Second Method. By a Single Altitude.

Observe the altitude and the time of the observation.

I. If the star is considerably distant from the meridian, we have given in the triangle PBZ (fig. 35), PB, BZ, and BPZ to find PZ, which may be solved by Sph. Trig. $60, and gives, by the notation of § 28,

(409)

=cos. PC. cos. z cosec. D,

in which the upper sign is used if the declination and latitude are of the same name, otherwise the lower sign.

90°

L=PZ= PC ± ZC
L= 90° — (PC±ZC) ;

(410)

in which both signs may be used if they give values of L contained between 0° and 90°, and in this case other data must be resorted to, in order to determine which is the true value of L.

Scholium. The problem is, by Sph. Trig. § 61, impossible, if the altitude is greater than the declination, when the hour angle is more than six hours.

II. If the latitude is known within a few miles, it may be exactly calculated by means of (376), or

cos.zcos. [90°-(L+p)]-2 cos. Lcos. D (sin. h)2. (411)

Latitude found by a single altitude.

But if A is the star's observed altitude, and A, its meridian altitude at its upper transit, (403) gives

A1

=L+p, or = 180° — (L+p),

and (411) becomes, by transposition,

sin. A1

=

sin. A +2 cos. L cos. D (sin. h)2; (412)

from which the meridian altitude may be calculated by means of table XXIII, as in the Rule. [B. p. 200.]

III. A formula can also be obtained from (340), which is particularly valuable when the star is, as it always should be in these observations, near the meridian.

In this case we have in (340) applied to PBZ

2s90°-L+p+z=180°-L+p-A (413)

2s-2PZ=L+p—A

=A, -A or = 180° — (A,+4)

(414)

= 180° — (A1+ A) or = A,—A; (415)

1

and if these values are substituted in (340), after it is squared and freed from fractions, they give

(sin. h)2cos. Lcos. D=sin. ¿(A,—A)cos. § (4 ̧+A), (416)

or

1

sin.}(4,−4)=(sin.3h)2cos.Lcos.Dsec.(4,+4); (417) and if, in the second member of this equation, the value of A, is used, which is obtained from the approximate value of the latitude, the difference between the observed and the meridian altitudes may be found at once; and this difference is to be added to the observed altitude to obtain the meridian altitude.