Sidebilder
PDF
ePub
[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

which gives a.

We have therefore found the altitude and azimuth.

Again, let z, P, - (Fig. 116,) be the places of the zenith, pole, and star; QQ', CC' the equator and ecliptic intersecting in the first point Aries R, and cutting the meridian in the points Q', C. Then it is evident that Q'R = A - H. Also the L R = 23° 28', and Q' is a right 2. Consequently tan. Q'C' = sin. (A – H) tan. 23° 28'.

sin. 23° 28 sin. Z Q'C'R = sin. (A-H) x

sin. QC

sin. Q'C' and sin. C'R =

sin. 23° 28' which give Q'C' and L 2C'z' Hence in the Azz'C', we have

zC =

90° - L - Q'C' LzC'ź = 180° Q'C'R and C'zz' = 180° which will give, when resolved, the < C'z'z and c'é. Consequently,

C'z'z and Rz' (= C'z'- RC') are known, which are the other quæsita, of the problem.

891. Since the day of the year is given, the sun's rightascension (a) is known from the tables. Let also the given co-declinations and right-ascensions of the stars be denoted by

D, D'; A, A.

a.

COS. Z

Moreover, let H be the hour L or distance of the sun from noon at the instant the stars are on the same azimuth ; then the angular distances of the stars from the meridian are evidently h = H + A

- a, and h' = H + A' Hence, if L = the co-latitude of the place, and z, z' the zenith distances from the A (z, L, D), (z', L, D'), we get

cos. L. cos. D cos. (H + A - a) =

sin. L. sin. D.

cos. ¿ cos.L. cos. D cos. (H,+ A' a)

sin. L. sin. D Also from the A (D, D', z' – z) we get cos. (A' – A) = cos. (z' — 2) – cos. D. cos. D

sin. D. sin. D D+D

D+D = sin.

+ M). sin. M being such that

A'- A sin. M'= sin. D. sin. D'.

2 Hence z' - z is known (= m), which therefore gives

cos. L. cos. D cos. (H + A

sin. L. sin. D and cos. (H + A' – a) =

cos. (z + m) - cos. L. cos. D'

sin. L. sin. D' by which two equations, z being eliminated, we shall have H expressed in terms of A, A', a, L, D, D'; which will therefore be known.

:: sin.8z'—2

[ocr errors]
[ocr errors]

2

COS.

[ocr errors]

892. When the hour angle from noon = azimuth from the south, the zenith distance is evidently equal to the co-declination (D). Hence if I be the co-latitude, and H the required hour L, we have

cos. D - cos. D. cos. L cos. H=

sin. D. sin. L

= cot. D.

1-cos. L

[ocr errors][merged small][merged small][merged small][merged small][merged small]

893. Let L, D, z be the co-latitude of the place, the sun's co-declination and zenith distance; also let a be his azimuth, and H the hour angle from noon ; then from the A (L, D, z), we get

sin. (S-L). sin. (S-2) sin,

sin. L. sin. z where SE

L+D+z

2

[blocks in formation]

Moreover, since the sun's declination is given, the day of the year is known from the tables. Consequently the azimuth and hour L are found.

Again, since the co-latitude of the sun is 90°, and the obliquity of the ecliptic, or distance between the poles of the equator and ecliptic, is 23° 28' from the A (90°, 23° 28', D) we have

cos. P = cot. 23°, 28', x cot. D, P being the angle of position.

894. Since the day is given, the earth's longitude (L) is known from the tables. Let L' and a be the given longitude and latitude of the star, and 8 the required inclination; then from the right-angled'a (90° + L - L', a, 6) we get

cos. 0 = cos. a. cos. (90 – L' – L)

= cos. a. sin. (L' -L)

895. At the moment the star and sun are seen rising together, let the time from noon be H, and let L be the co-latitude

cos. L.

of the place, A the right-ascension of the sun, D his co-declination, A', D', H' the right-ascension, co-declination, and angular distance from the meridian of the star. Then from the quadrantal A (90°, L, D), (90°, L, D'), we have

COS. 90

cos. D cos. H=

sin. L. sin. D

cot. L. cot. D cos. H' = cot. L. cot. D'. But H' - H= A' - A

io cos. (H + A' – A) = -cot. L cot. D'..... (1) But since the obliquity of the ecliptic is 23° 28', we have

cot. D = sin. A. tan. 23° 28'
.. cos. H = cot. L. sin. A. tan. 23° 28' (2)
Now, from equation (1) we get

H-A +A

' and .. H - A is known. Hence, by substituting in equation (2), we get both H and A, which give the hour of the day, and (by the tables) the day of the year.

896. Let 0, o be the true anomalies corresponding to the distances po ř' ; then from the equation to a parabola (s=.

cos.

a

2

[merged small][ocr errors][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

897. Let L, L', L", be the longitudes, , 1, 2" the co-latitudes of the three places 2, z', z". Also let d, d', d", be the distances between

2, z'; z',z" ; 2, 2". Then from the A

(d, a, a), (d', a', a), (ď", a, a) we get

- cos. a. cos. a'
cos. (L-L') =

sin. a, sin, a'
cos. (L'-L') = cos. d' — cos. a' cos. a."

sin. a. sin. a"

cos. d

[blocks in formation]

whence

sin.2

2

[ocr errors]

siu, ed'

home = sin. (^++ M) sin. (– M)
= sin. ( + M). sin. (*'**" - M)

( + M). sin. (*^' – M)

[ocr errors]

2

ata"

sin.24

= sin.

2

M, M, M" being such that

L sin.2M = sin. a. sin, a'. cos.

L'

2

&c. &c. which therefore give the three distances d, d' d".

Hence if D, D, D' denote the angles respectively opposite 19 d, d', d", we have

cos. d. cos. D=

cos. d. cos. ď sin. d. sin. d"

cos. d' - cos. d. cos. d"
cos. D' =

sin. d. sin. d"
cos. d' cos. d. cos, d'

sin, d, sin. ď

cos. D'':

which give

[ocr errors]
« ForrigeFortsett »