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the body. M'S' is the apparent distance, and MS the true distance; which can be calculated by applying the rules of solution of oblique spherical triangles to the two triangles MZS and M'ZS'.
For the apparent altitudes of M and S are known, and hence their true altitudes can be found (443-445); and therefore their apparent and true zenith distances can be found. Also, in the triangle M'ZS' are known the apparent distance M'S', and the apparent zenith distances M'Z, S'Z; hence (349) angle Z can be found. Then, in triangle MZS, are known the angle Z, and the true zenith distances MZ, SZ; hence the true distance MS can be calculated (355 or 331.) The formulas given above, however, afford a more concise solution of the problem. Their demonstration is given in various works.*
If the registered horizontal parallax of the moon (438) is not reduced for the latitude when computing a lunar distance, that is, if the ellipticity of the earth is not taken into account, there will be an error on the computed distance, seldom amounting, however, to more than 10", though at its maximum it is about 14"; but this error within the tropics is scarcely sensible. When the parallax is reduced, this error, though only partially corrected, will be considerably diminished, and the correction at its maximum, under the latitude of 45°, will amount only to 12", or to 5'-2 on the longitude, and diminishes towards the equator and the poles, where the error is nothing.t
In the first three exercises, the logarithms are carried only to five places of decimals.
1. Given the apparent altitudes of the centres of the sun and moon 32° 0′ 1′′ and 24° 0′ 8′′, their true altitudes 31° 58′ 38′′ and 24° 51′ 48′′, and their apparent distance 68° 42′ 15′′; required their true distance. Ans. 68° 19' 34".
* Woodhouse's Astronomy, p. 858; Delambre," Abrégé d'Astronomie," p. 632; Francœur, "Geodesie," p. 452, and "Astronomie Pratique," p. 248.
For an exposition of this subject, the reader may consult " Mackay on the Longitude," and a paper by Professor Henderson in the 40th number of the "London Journal of Science."
2. Suppose that on the 6th April 1821, in latitude 47° 39′ N., and longitude 57° 16′ W., by account at 3h 56m P.M. per watch, it was found that the apparent altitudes of the centres of the sun and moon were 26° 9′ 7′′ and 46° 34′ 44′′, their true altitudes 26° 7′ 19′′ and 47° 14' 19", and the apparent distance of their centres 76° 0′ 7′′; required the true distance and the true apparent time of observation, the sun's declination at the time being 6° 32′ 12′′ N. Ans. The distance = 75° 45' 44", and time
=3h 51m 24s. 3. If in longitude 11° 15′ W. by account at 3h 45m A.M. per watch, the apparent altitude of the centre of the moon was 24° 29′ 33′′, and that of Regulus 45° 9′ 12′′, and their true altitudes 25° 16′ 50′′ and 45° 8′ 15′′, and the apparent distance of Regulus from the moon's centre 63° 35′ 4′′; required the true distance. Ans. 63° 4' 40".
4. On the 9th of April 1837, at 5h 29m 36s-8 mean time, suppose that the nearest limbs of the sun and moon were 54° 30′ 12′′ distant; that the apparent height of the sun's centre was 21° 50′ 14′′, and that of the moon's = 61° 10' 10"; the moon being east of the sun, and both west of the meridian; the latitude by account was 41° 47′ N., and the longitude 2h 10m W.; the horizontal equatorial parallax was 55′ 31′′-1, and the semi-diameters of the sun and moon 15′ 59′′ and 15′ 7′′-8; required the true distance.* Ans. 55° 17′ 20′′.
5. On the 6th of May 1840, at a place in latitude 36° 40' N., and longitude by account 39° W., at 7h 40m A.M., the apparent altitudes of the centres of the sun and moon were 30° 33′ 0′′-2 and 53° 15' 40"-9, their true altitudes 30° 31′ 27′′ and 53° 50′ 31′′·3, and the apparent distance of their centres 62° 0′ 9′′-1; required their true distance. Ans. 61° 52′ 35′′.4.
6. At a place in latitude 10° 1′ 50′′ N., and longitude by account 30° 5' W. of Paris, on the 17th December 1823, at 14h 59m 485-8 P.M., the apparent altitudes of the moon's centre and of Regulus were 48° 0′ 49′′ and 70° 34′ 9′′, their true altitudes 48° 40′ 38′′ and 70° 33′ 49′′, and their
* This example is taken from Francœur's "Geodesie," and the next two from his "Astronomie Pratique."
apparent distance was 58° 25′ 36′′; what was their true distance? Ans. 57° 47′ 12′′-4.
7. Required the distance between the centres of the sun and moon, from these data :
Altitude of upper limb of moon
Distance of nearest limbs of the two bodies 83° 26′ 46′′
0 5 27
= 0 42 19
Semi-diameter of moon, including augmen. Correction of sun's altitude, including dip Correction of moon's altitude, including dip The latitude = 10° 16′ 40′′, and longitude by account 149° E., the observations being taken on 5th June 1793, about 1h 30m P.M. Ans. 83° 20′ 55′′.
THE LONGITUDE BY LUNAR DISTANCES.
480. PROBLEM XXXVI.-Given the true angular distance between the moon and the sun or a star, and the time of observation, to find the longitude.
The time, when not previously known, can be calculated by means of the altitude of one of the bodies, as in article 467.
The time at Greenwich can be found thus:- Take from the Nautical Almanac the two distances to which the given distance is intermediate; then the difference between the registered distances is to that between the first registered distance and the given distance, as three hours is to a fourth term, which is to be added to the time of the first registered distance, to give the required time.
Then the difference between the time at the place and that found at Greenwich will be the longitude."
the difference between the first registered distance and the given distance, tthe interval of time corresponding to d, Tthe time required at Greenwich,
time of the first registered distance,
the difference between the two registered distances, that is, for intervals of three hours,
Or, L. t
d': d3ht; hence t=
L. 3+ L. d-L. d', or P.L. t P.L. d-P.L. d'.
Since the moon moves over 360° in about 30 days, therefore an error of 10" in measuring the lunar distance will cause an error of about 5' on the longitude. For
When only the first differences of the lunar distances are taken, the result will be a few seconds of time wrong. Thus, in the fourth of the following exercises, the correction found, when the second differences are used, is 5-8, or about 1'4, the correct longitude being 2h 0m 16s.9.
To find the time at Greenwich.
Distance at Oh = 33° 58′ 7′′
EXAMPLE. Find the time of observation and the longitude of the place of observation, from the data of the example to the preceding problem.
2)192 38 52.6
S96 19 26.3 S-Z= = 34 42 40.3
33° 58′ 7′′ D33 56 48.4
d' 1 28 4
d = 0 1 18.6
and d'd=3h: t, or 1° 28′ 4′′: 1' 18"-63h: 2m 40s.6, and Tt'+t=0h+2m 408-60h 2m 408-6.
To find the time at the place.
By art. 468 (the declination of star being 12° 50' 53"-4), Z=61° 36′ 46′′
L. Sin S
L. Cos H
9.7536188 = ⚫0110020 = ⚫0926862
2)19-8546563 = 9.9273281
And H 32° 2′ 15′′, and H = 4h 16m 18s
= 9 58 43.3
Longitude of place
= 11 49 27·7 W.
The time at the place could also be found from the observed altitude of the moon.
The principle on which the rule is founded is so simple as to require no explanation. It proceeds, however, on the hypothesis that the moon's motion is uniform, which it so nearly is for three hours, that the error arising from this assumption amounts at most only to a few seconds. When extreme accuracy is required, what is called the equation of second differences, which depends on the differences of the first differences, is used as a correction.*
1. Find the true longitude for the data in the third exercise of last problem, supposing the true time of observation to be 24th January 1813, at 3h 45m A.M., the true distance 63° 4' 40", and that the distance of the centre of the moon from Regulus on 23d at 15h = 62° 20′ 8′′, and at 18h = 63° 48′ 54′′. Ans. 11° 19' 30" W.
2. Find the longitude from Paris for the data in the fourth exercise of the last problem, the exact mean time of observation at the place being 5h 26m 368.8, the true lunar distance 55° 17′ 20′′, and the registered distance at 6h in the Connaissance des Temps being = 54° 11′ 36′′, and the difference for 3h = 1° 25′ 55′′. Ans. 2h 48m 6s-1 W. 3. Required the longitude west of Paris for the fifth exercise in last problem, the exact mean time of observation
*See Nautical Almanac.