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Method of computing the altitude of a celestial object. (See p. 252.).

In fig. p. 269, let Z be the zenith, F the object, F Pits polar distance, F Z its zenith distance, Z P the co-latitude; and let Z E be a perpendicular from Z upon F P, or on F P produced, if Z P E be an obtuse angle. Then rad . cos ZPE = cot ZP . tan P E. Or rad . cos Z PE cos Z PE. cot, lat

= tan PE (arc 1;) and PFF cot ZP

rad PE=FE (arc 2 ;) and cos PE :cos FE:: cos ZP (sin lat): cos ZF

cos F E. sin lat cos F E. sin lat. sec PE = (sin altitude) =

rad 2

cos P E

To find under what circumstances, in a given latitude, a small mistake in

observing or correcting the altitude of a celestial object will produce the smallest error in the time computed from it.

S

Let Z be the zenith distance, P the pole, r the supposed place, and m the true place of the object. Let m s be a parallel of altitude, join the points m and r, and let p q be the

z arc of the equator contained between the meridians P m and Pr.

P Then as P m and P 9 are equal, mr may be considered as a small portion of a påral

PC lel of declination, rs will be the error in altitude, and p q the measure of the required error in time. And as the sides of the triangle msr will necessarily be small, that triangle may be considered as a rectilineal one right angled at s; and because the angle Prm is also a right angle, the angles s m r and Pr Z being each the complement of mrs, are equal to each other. Now we have rs:mr :: sin s m (Z r P) : rad and mr: p9:: cos qr :

rad Hencers: P9 :: COS IT

sin ZrP : rad”, or p q = rs. rad 2

but sin » P, or cos qr : sin r ZP :: sin ZP cos 9 r. sin ZP (cos lat): sin Zr P. Whence cos qr. sin Zr P = cos lat. sin r ZP,

rad 2 and consequently by substitution we have p q =rs.cos lat. sin r ZP Therefore when the latitude and error in altitude are given, the corresponding error in time will be the least when the sine of r Z P, the azimuth, is the greatest; that is, when the azimuth is 90°, or when

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or correcting an altitude for the time will be of the least consequence in the result, when the object observed is on the prime vertical.

Methods of clearing the distance of the moon from the sun or a star, from

the effects of parallax and refraction. (See p. 254.)

METHOD I. Let m be the apparent, and M the true place of the moon, s the apparent, and S the true place of the sun; then m A, s B will be the apparent altitudes, m Z, sZ the apparent

Z zenith distances, MA, S B the true

P altitudes, and M Z, S Z the true zenith distances; m M will be the correction of th moon's altitude, s S that

M of the sun's, m s the apparent distance, and M S the true distance.

Put d = ms, D = MS, h = m A,

= Bs, H = MA, H' = BS. Then (Prop. 18, Spherics,) cos m Z s = cos A A

B

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Then cos D = (cos d + cos h + h') . 2 cos N cos H + H' 2 cos d , cos N + 2 cos N. cos h + h - cos H + H'.

ܘܘ

- vers

But (Form. 2, p. 73) ? cos d . cos N = cos d + N + cos do N, and 2 cos N. cos h + k = cos (N + h + h). + cos (N ~ h + h').

Therefore cos D= cos d + N + cos do N + cos (N + h +h') + cos (Noh + h) cos H + H'; or 1 – ços D=1+ cos H + H' – cos d + N cos d ON cos (N + h + h) cos (N wh + h), or vers D = suvers H + H'

(1

vers d + N) (1 do N) (1

vers (N + h + ')) (1 vers (Noh + ',)) or vers D = suvers H + H + vers d + N + vers d « N + vers (N + 1 + 1) + vers (No + 2) - 4; the formula which, in words, is the first method given for clearing the distance.

Table 25 gives N, and Table 28 gives the versed and suversed sine of any arc, the first figure being omitted as unnecessary in the solution of this problem ; for as d and D cannot in any case differ much more than a degree, vers D will always be found either in the same column with vers d, or in one of the adjoining columns, and therefore the leading figure in the versed sine can never be required to determine the arc to which vers D corresponds.

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METHOD II. Resuming the equation cos d

sin h . sin h' cos D - sin H . sin H' cos h . cos h'

we have

cos H . cos H' cos d — sin h . sin, h' 1

sin H . sin H' cos h . cos h'

, or

cos H. cos H'
cos h
cos h' + sin h. sin h'

cos d
cos h. cos h'

cos D

=1

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cos n

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- cos d cos H NH cos h . cos h'

cos H.cos H' Whence cos D='cos H~H'

cos H.cos H' (cos hoh cos d).

cosh.cos h' cos H. As before, put

= 2 cos N. cos h . cos h'

cos H'

Then cos D = cos H H' 2 cos N. cos h uh + 2 cos N. cos d, or cos D = cos H »H' cos (hwh + N) - cos (h wh' « N) + cos

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(hook + N) + cos (hol' « N) - cos N + d - cos N «d, or 1 - cos D=1. cos H - H' +1+ cos (h h' + N) + 1 + cos ( g h o N) + 1 cos N + d +1 cos Nod 4, or vers D= vers H ~ H' + suvers (h oh + N) + suvers (h woh ~ N) + vers N + d + vers Nod 4.

This formula, in words, gives the second practical method of clear. ing the distance.

Remark. The method of computing M S directly by spherical trigonometry is obvious ; for in the triangle m s Z, all the sides are given, viz. m s, the apparent distance, and m Z, s 2, the apparent zenith distances, to compute the angle m Zs; and in the triangle MZ S there are given M Z and S Z, the true zenith distances, and the included angle M Z S, before found from the triangle m Z s, to find MS the true distance.

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USEFUL MISCELLANEOUS NAUTICAL PROBLEMS SOLVED

BY THE DIRECT APPLICATION OF SPHERICAL TRIGONOMETRY.

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А

In the annexed figure let A represent the first point of Aries, AD the first quadrant of the ecliptic, E its north pole, A C the first quadrant of the equator, and P its north pole; let S be the place of the sun in the ecliptic, S H a perpendicular from Son A C, and M, I, and G the places of any other celestial objects. Let E M F, P M B be great circles drawn to the ecliptic and

D

MY equator from the poles E and P, making of course the angles at B and F

ht angles; and let great circles be drawn, as

H B in the figure from I and G, to E and P. Then D AC, measured by D C, or its equal EP, is the obliquity of ecliptic; A S is the sun's longitude, A H his right ascension, and SH his north declination; A F the longitude, M F the latitude, M E the co-latitude, A B the right ascension, B M the declination, and MP the polar distance of M; MEI is the difference of longitude, and MPI the difference of right ascension of M and I; M E G is the difference of longitude, and MPG the difference of right ascension of M and G; GEI is the difference of longitude, and GPI the difference of right ascension of G and I.

The first quadrants only of the ecliptic and equator are drawn in the figure ; but these circles may be conceived to be produced round

the globe, forming two great circles, of which the poles are E and P. The longitudes and right ascensions of celestial objects are reckoned from A, in the direction A D, A C, quite round to A again ; the latitude is north when the object and E are on the same side of the ecliptic, and south when they are on different sides; the declination is north when the object and P are on the same side of the equator, and south when they are on different sides.

The figure appropriate to any given problem will be readily conceived, by refering the data to a celestial globe.

PROBLEM I. Given the obliquity of the ecliptic, and the sun's longitude, to find his

right ascension and declination. Here in the spherical triangle A S H, right angled at H, are given AS, and the angle SA H to find A H and SH.

If the longitude exceed six signs, or 180° there will be formed on the other side of A C, a right angled spherical triangle, whose base on the equator is the sun's right ascension from 1, the hypothenuse, on the ecliptic, the sun's distance from 2, or the excess of his longitude above 180°, the included angle being the obliquity of the ecliptic, and the perpendicular from the sun's place on the equator, the south declination. If 180° be added to the right ascension, computed from *, the sum will be the right ascension from r.

ExamPLES FOR EXERCISE. The obliquity of the ecliptic being 23° 27' 50", it is required to compute the sun's right ascension and declination in each of the following examples :

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PROBLEM II. From the obliquity of the ecliptic, and the latitude and longitude of any

celestial object, to find its right ascension and declination. Let M (see the last figure) be the object; then in the right angled spherical triangle AF M, are given A F the longitude, and MF the

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