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this example by the second method than by the first, in which the variation for only 1 hour is employed.

461. PROBLEM XX.-To find the time of culmination of a planet at any given meridian.

The rule is exactly the same as that in last problem, only, as the increase in right ascension of a planet for 24 hours is small, the right ascension is given, not for every hour, but only for every noon; and the meridian passages are given to the tenth of a minute. In the first formula of last problem, therefore, when adapted to this one, v and vare the daily variations in right ascension of the sun and planet. Hence,


24 — (v'—v): T'v'-v:r, and TT'+r. When vv, r is negative, and TT'—r. When ' is negative, that is, when the motion of the planet is retrograde, then r is also negative, and

24 — (v' + v): T′ = v' + v : r, and T = T’— r.

EXAMPLE.-Find the time of the meridian passage of Mars, at a place in longitude 45° 30′ W., on the 22d of May 1841.


By the second Method. Longitude in time 3h 2m. Time of registered meridian passage on 22d Time of registered meridian passage on 23d = +9 2·7

9h 6m.9

Acceleration in 24h

Acceleration in 3h 2m

Time of mer. passage at given place on 22d =

By the first Method.

Registered R.A. of Mars on 22d
Registered R.A. of Mars on 23d

Decrease in 24h
Decrease in 3h 2m

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0 4.2

0 0.5

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R.A. at noon at given place,
Registered R.A. of sun on 22d
Increase or acceleration in 3h 2m


A' =



4 0

R.A. of sun at noon at given place, A =
Approximate time = =T' A'—A = 9 7
The motion of the planet being retrograde,
Om 218-3m 56s.6 —— 4m 178.6,
24 — (v' — v) = 24h 4m 17s-6.*



P.L. 24h 4m 178.6 = ·00129= - P.L. 23h 55m 42s.4
P.L. 9 7 58-9
P.L. 0 4 17.6

= ⚫41959


= 2.04412

P.L. 0 1 T 9 7

37.6 58-8

T 9 6 21.2

13h 8m 148.33

3 59


0 0


= T', the required time.




1. Find the time of Jupiter's transit over the meridian in longitude 160° E., on 10th January 1841, the registered times of meridian passage, on the 9th and 10th, being 21h 15m.6 and 21h 12m.5. Ans. At 21h 13m.9.

2. Find the time of the meridian passage of Jupiter on the 10th of January 1841, at a place in longitude 160° E., its registered right ascension, on the 9th and 10th, being 16h 33m 468.27 and 16h 34m 368-37, and that of the sun, on the 10th, being 19h 19m 208.23. Ans. At 21h 13m 54s.

462. PROBLEM XXI.-To find the meridian altitude of a celestial body at a given place, the declination of the body and the latitude of the place being given.

'Find the declination of the body at its meridian passage at the given place; then take the sum or difference of the colatitude and declination, according as they are of the same or of different names, and the result will be the meridian altitude."

*When vv, the first term, 24— (v'— v), exceeds 24 hours. But it will never exceed 24 hours by more than 10 minutes, and L 1440 - L 1430 L 1450 — L 1440, when carried only to 5 figures. Hence, · L [24 — (vv)] may be taken instead of +L [24+v+v], and it must be added to the logarithms of the second and third terms.

Let L, C= the latitude and colatitude,

D, P= the declination and polar distance,
A, A' the meridian altitudes at upper and lower
C± D.

then A

When the declination exceeds the latitude, the altitude then would exceed 90°, and the supplement is to be taken, which is the altitude from the opposite point of the horizon below the pole; or in this case, A =L+P.

When the declination exceeds the colatitude, the lower meridian passage will be above the horizon, and the altitude then is A' D — C, or A' — L—P.



In all these formulas, the latitude and declination are of the same name, except in A = C — D.

EXAMPLE. Required the meridian altitude of the moon, at a place in latitude 56° 20′ 10′′ N., and longitude 40° 45′ W., on 2d May 1841.

By the example to article 458, the time of transit is 9h 52m 28s, and the declination must be found for this time, that is, for the reduced time 2h 43m+ 9h 52m 28s = 12h 35m 28s, on 2d May.

Registered dec. of moon on 2d at 12h
Registered dec. of moon on 2d at 13h

Increase of dec. of moon on 2d for 1h
Inc. of dec. of moon on 2d for 35m 28s


Hence declination at transit,

=8° 17′ 29′′-6 S. =8 31 59.6





D 8° 26'


Meridian altitude,

C-D = 25 13 46.2

A If the apparent altitude were required, this just found altitude would require to be corrected for refraction and parallax. But the observed altitude can more readily be corrected, as in article 445; and the true altitude thus found can be compared with that calculated above. When the meridian altitude of a star is required, its declination may be considered the same for a day, or even for a year, for most nautical purposes.

3"-8 S.






Let NS be the horizon, EQ the equator, NZSR a meridian, and B, B', B", celestial bodies in the meridian, and Bb, B'b', and B"b", parallels of declination. Then ZEL.; and hence ES C, and b ABS: C+D. So when B" is south of the equator, B'E=D, and B"SA; hence A C—D. For the body B', DL., and B'S = C + D is 90, and A = B'N 180° — (C + D) = L+P. and of the same name, then A'b'N




Let PEQ be the solstitial colure, EQ the equator, CC' the ecliptic, PRP' a meridian through the sun's centre S. Then A is the first point of Aries, C of Cancer, C' of Capricorn, and the point diametrically opposite to A is the first point of Libra; also,


When DL., D— C.


1. Find the meridian altitude of Castor, on the 11th of May 1841, at a place in latitude 28° 30′ 25′′, its declination being 32° 13′ 54′′-9 N. Ans. 86° 16' 30"-1.

2. Find the meridian altitude of Jupiter, on the 10th January 1841, at a place in latitude 46° 35′ 28′′ N., and longitude 160° E., its registered declination, on the 10th and 11th, being 21° 19′ 58"-7 and 21° 21' 40"-7 S. (See 2d exercise, art. 461, for time of transit.) Ans. 22° 3' 48"-4.

463. PROBLEM XXII.-Of the obliquity of the ecliptic, the sun's longitude, declination, and right ascension, any two being given, to find the other two.



ASL is the sun's longi-

ARA is the sun's right ascension,
SRD is the sun's declination;
angle SAR O is the obliquity of the ecliptic.


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Now, the triangle ARS is right-angled at R, and any two parts of it, except the right angle, being given, the other two can be found by the rules of right-angled spherical trigonometry.

EXAMPLE. The sun's registered longitude, on the 11th May 1841, will be 50° 34' 21"-4, and the obliquity of the ecliptic 23° 27′ 42′′-15; find the sun's declination and right ascension at mean noon at Greenwich.

In the triangle ARS are given ASL=50° 34′ 21′′-4, and angle SAR = 0 23° 27′ 42′′-15.

To find AR = A.

L. Cot L
L. R
L. Cos O

= 9.9149829 = 10.


To find RSD.
L. R
= 10.
= 9.6000316

L. Sin O
L. Sin L


L. Tan A


L. Sin D


And A = 48° 7′ 47′′-2 = 3h 12m 31s∙14, and D = 17° 54' 38".

When the longitude exceeds 90°, so will the right ascension. Since right ascension is reckoned from the vernal equinox, and since the equator and ecliptic intersect at two points diametrically opposite, it is evident that, to any particular declination, there belongs four different right ascensions, and of these the one required must be determined by the time of the year.


1. When the sun's longitude, on the 10th of June 1841, will be 79° 21′ 23′′, what will be its declination and right ascension, the obliquity of the ecliptic being 23° 27′ 41′′-53? Ans. D 23° 2′ 3′′-8 N., and A = 5h 13m 418.9.

2. The obliquity of the ecliptic, on the 20th of July 1841, being 23° 27′ 41"-42, and the sun's declination 20° 40′ 38"-I N.; required his right ascension and longitude.


Ans. A 7h 58m 21s 15, and L = 117° 30′ 48′′-2. 3. The sun's right ascension and declination, on the 8th of September 1841, will be 11h 7m 2s-5, and 5° 40′ 35′′.7 N.; what will be its longitude and the obliquity of the ecliptic? Ans. L 165° 36′ 53′′-3, and O = 23° 27′ 41′′·8.

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