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limb be 230.48', and the eye 18 feet above the level of the sea; required the moon's true central altitude.
The moon's semidiameter and horizontal parallax at noon (Naut. Alm.), being 16'.39% and 61'.8”; and at midnight, 16.38" and 61.4".
Answer. The time at Greenwich is 65.30'; D's true semidiameter 16'.44", horizontal parallax 61'.6"; apparent altitude of the ) 's centre 23°.27'.13"; the D's correction 53'.52", and the true altitude of her centre 240.21'.5'.
PROBLEM XII. (I) Given the sun's meridian altitude to find the latitude of the place of observation.
RULE. Reduce the sun's declination to the meridian of the place of observation* (B. 270.)
Subtract the sun's corrected altitude (G. 277.) from 90°, the remainder is the zenith distance.
If the sun be south of the observer when the altitude is taken, call the zenith distance north ; but, if north, call it-south.
Then, if the zenith distance and declination have the same name, their sum is the latitude; but, if they have contrary names, their difference is the latitude, and it is always of the same name with the greater of the two quantities.
EXAMPLE I. On the 17th of October 1822, in longitude 52°W., suppose the meridian altitude of the sun's lower limb to be 28o.40', the observer at sea, the sun to the south of him, and the eye 14 feet above the surface of the water. Required the latitude of the place of observation ? Obser. alt. O's lower limb 28°,40' Time at given place
Oh. O Semidiameter
+ 16'. 6"
-3h.28' 289.56'. 6" Time at Greenwich
=Sh. 28 Refraction (Table IV.) - 1'.44"
O's declination at noon 9o. 9', 20's.
28°.54'.22" O's declination 18th at noon 9°.31',20" O’s paral. in alt. (Tab. VI.)
Increase in 24 hours
22. 0" 28°.54'.50" Dip for 14 feet (Table V.)= - 3'.44"
24h. : 22'::3h.28' : 3.10"
O's decline. 17th at noon = 99.9'.20's. Truc altitude O's centre=280.50ʻ.56"
O's true declination 90.12.30's. Zenith distance =61o. 9'. 4 n. l Zenith distance
=61° 9' 4"N.
* The same rule, with a little variation, will answer for a planet; and for the moon, correcting for the declination as at C. 271. See Dr. Mackay's Navigation, 1st Edition, page 151 to 155.
2. On the 18th of March 1822, in longitude 106° E, suppose the meridian altitude of the sun's lower limb, when north of the observer, to be 74.56', and the height of the eye 16 feet above the level of the sea, required the latitude.
The sun’s semidiameter (Naut. Alm.) being 16'.5"; the declination at noon on the 17th 19.26'.59'S., and on the 18th 10.3'.17" S.
Answer. The sun's true altitude -75°.8'.3"; time at Greenwich 16h.56' on the 17th, sun's corrected declination 10.10.19"S. latitude 16o.2'.16" S.
3. On the 28th of September 1822, suppose the meridian altitude of Arcturus* to be 36o.18' north of the observer, its declination 20°.6'.50" N, and the eye 20 feet above the level of the sea, required the latitude.
Answer. 33o.40'.44" S.
THE APPLICATION OF RIGHT-ANGLED SPHERICAL TRIANGLES
TO ASTRONOMICAL PROBLEMS.
(K) The celestial sphere is represented by Plate III. fig. 1.
i. Let the circular space, South, Zenith, North, Nadir, represent the brazen meridian † of a celestial, or terrestrial globe, having its north pole elevated above the horizon.
2. Imagine the globe to be cut in halves by the brass meridian, and the semi-globe to be of transparent glass with the circles of the sphere drawn on it. Now if a sheet of paper be put upon the section, and a light be placed in the point aries (the eye being in libra), the shadows of all the most useful circles of the sphere, will form a plane figure similar to Figure I. Plate III.
3. nas will present the axis of the globe, n the north pole, s the south pole.
4. Ang the equator.
6. Zenith A the quadrant of altitude screwed on the zenith, and passing through aries. Or, Zenith A Nadir, the prime vertical passing through aries A.
7. is OAOT the ecliptic, n its north pole, m its south pole. 8. Os the tropic of cancer, or any parallel of the sun's north declination.
* See the note on D. 273.
+ It is here presumed that the learner has some knowledge of the globes, and of the circles described thereon. It would be a very good exercise to solve the suc
9. W vs the tropic of capricorn, or any parallel of the sun's south declination.
10. On the elevation of the north pole above the horizon, equal to Æ zenith, the latitude of the place.
11. HÆ the elevation of the equator above the horizon, equal to zenith n, the complement of the latitude.
12. TOw a parallel circle 18 degrees below the horizon, or the boundary between twilight and dark night.
13. Zenith sec, zenith cd, &c. azimuth, or vertical circles.
14. NSDS, NGO cws, &c. meridians or circles of terrestrial longitude.
15. nvxm, ncvm, &c. circles of celestial longitude.
NOTE. The several triangles in this general figure, and the lines which form them, are explained at the head of each problem, to which the triangle is applied.
(L) PROBLEM 1. (Plate III. Fig. 1.) Given the obliquity of the ecliptic and the sun's longitude, to find his right ascension and declination.
In the right-angled spherical triangle ago, or ABO 1. Asun's longitude, an arc of the ecliptic from aries.
2. AG, or as=sun's right ascension, an arc of the equinoctial; the point B always comes to the meridian with the sun.
3. GO, or BO =sun's declination, an arc of a meridian passing through the sun's place. This arc likewise comes to the meridian with the sun.
4. GAO, or BAO=the obliquity of the ecliptic, or the angle formed between the equinoctial and the ecliptic.
5. AOG, or A OB=the angle formed by the ecliptic, and the meridian passing through the sun's place.
Any two of these five quantities being given, the rest may be found.
On the 17th of May 1822, when the sun's longitude is 16.250.57'.25"; required his declination, right ascension, and the angle formed between the ecliptic and the meridian passing through the sun's place. The obliquity of the ecliptic being 230.28'. 1. To find the sun's declination, in the triangle ABO.
Rad x sine BO=sine BAO x sine AO.
or, Rad : sine AO= 550.57'.25"::sine BAO = 230.28' : sine BO=19o.16'. 2. To find the sun's right ascension, in the triangle ABO.
Rad x cos BA O =cot A o x tang AB.
(10+log cos BAO )– log cot A 0 = log tang AB. or, Cot AO=559.57'.25" : rad::COS BAO=230.28' : tang AB 530.37.40" = 3h.34.30".40"". 3. To find the angle A B between the ecliptic and meridiant.
Rad x cos AO=cot BAO Xcot A OB. :: (10+log cos ao)- log cot Bao=log cot AOB. or, Cot BAO =230.28': rad::cOS AO = 55°.57'.25" : cot AB -76o.20ʻ.23".
(M) Astronomers reckon the sun's longitude and right ascension from aries quite round the globe. Hence, though at equal distances from the equinoctial points, aries and libra, the
have the same quantity of declination and of the same name, viz. from aries to libra, north; and from libra to aries, south; yet the longitudes and right ascensions differ materially at these points where the declinations agree. Hence,
(N) If the sun be in the second quadrant of the ecliptic, subtract the longitude from 180°, use the remainder to find the right ascension; and take the right ascension, when found, from 180°.
(O) If the sun be in the third quadrant, subtract 180° from his longitude, use the remainder to find his right ascension; and add 180° to the right ascension when found.
(P) If the sun be in the fourth quadrant of the ecliptic, subtract his longitude from 360°, use the remainder to find the sun's right ascension; and take the right ascension, when found, from 360°.
(Q) In the first quadrant of the ecliptic between the 20th March and 21st June, the sun's declination is north and increasing; and in the third quadrant between the 22d September and 21st December, the sun's declination is south and increasing.-- In the second quadrant of the ecliptic from 21st June to 22d September, the sun's declination is north and decreasing; and in the fourth quadrant from 21st December to the 20th March, the sun's declination is south and decreasing.
1. On the 17th of May 1822, the obliquity of the ecliptics 230.27.53", the sun's declination=190.15'.57" N. and increas
† The finding of this angle is of no other use than as an exercise of right-angled
ing; required his longitude, right ascension, and the angle formed between the ecliptic and the meridian passing through the sun's place.
The sun's longitude 550.57.37", or 188.8.131.52". Ans. The sun's right ascension=539.38', or 3h.34.32". The angle A OB
=760.20'. The answers here would have been ambiguous, had it not been shewn that the sun was in the first quadrant of the ecliptic.
(R) This problem is useful for calculating tables of the sun's longitude. For the latitude of the place being accurately determined, and a quadrant fixed in the plane of the meridian, the observation of the sun's meridian altitude gives his declination, or distance from the equinoctial ; whence his longitude for any day is readily obtained.
2. On the 5th of August 1822, the obliquity of the ecliptic= 239.27.53", the sun's right ascension = 1340.54'; required the sun's longitude, declination, and the angle formed between the ecliptic and the sun's meridian.
The sun's longitude= s 12o.25'.53", or 45.120.25'.53". Ans. Declination The angle AOB
739.40'.34". The sun's right ascension being greater than 90', and less than 180°, he is in the second quadrant of the ecliptic,
3. On the 10th of November 1822, the sun's right ascension I 2250.5'.16", his declination 170.5'.18" S. required his longitude, the obliquity of the ecliptic, and the angle formed by the ecliptic and the sun's meridian.
The sun's longitude=m 47o.33.26", or 184.108.40.206". Ans. Obliquity of the ecliptic=
230.27.53". The angle AOB
73° 40'.19". The sun's right ascension being greater than 180°, and less than 270o, he is in the third quadrant of the ecliptic.
4. On the 1st of February 1822, the sun's longitude is12o. 8'.59", or 105.12°.8'.59", his declination = 170.10'.15" S.; required his right ascension, the obliquity of the ecliptic, and the angle formed by the ecliptic and the sun's meridian.
The sun's right ascension=3149.37', or 209.58' 28". Ans.
Obliquity of the ecliptic = 230.27'.55".
PROBLEM II. (Plate III. Fig. 1.) (S) Given the latitude of the place, and the sun's declina