places, has led to the introduction of other planes and circles into the science, independent of the position of the observers, and even of the figure of the earth. Thus, when the situation of the celestial equator, and the manner of valuing the angles between meridians by the measure of time, became known, they were employed to determine the position of the heavenly bodies by means of their right ascension and their observed declination. Afterwards, as it was found that a considerable portion of the celestial phenomena relative to the planetary system, occur in the plane of the ecliptis, or in planes but little inclined to it, it was found expedient to refer the position of the stars to the same plane, that is, to determine their latitude and longitude (chap. viii. art. 17, 18). These, and many other branches of astronomical inquiry, which we shall not here be able to touch, depending upon the mutual relations and intersections of different circles of the sphere, fail necessarily within the department of trigonometry. A few only will here be selected. ProBLEM I. 2. Given the obliquity of the ecliptic, and either the right ascension and declination of a star, or its latitude and longitude, to find the other two, and the angle of position. Let Ec in the annexed figure be a portion of the ecliptic, EQ a portion of the equa'tor, the two circles intersecting in r: the first point of Aries, in an angle, i, of 23°27'49". Let P be the elevated pole of the ecliptic, F that of the equator, PP' a portion of the solstitial colure. From P' and P let quadrants P's L, PSR, of E great circles, be drawn through s, the place of the star. Then Ek will be the right ascension, a, of the star, SR its declination, d, or Ps, its co-declination, EL its longitude, l, sL its latitude, L, or Ps its co-latitude, PSP’ = p, the angle of position; and PP' is given = 23° 27' 49", the present measure of the obliquity of the ecliptic. It is farther evident that sp? is the complement of the longitude, and P'Ps = P'PE + EPR = 90° -- right ascension; as indicated at the poles of their respective eircles. Now, if they are the right ascension and declination which are supposed known, in addition to the obliquity, we shall have, from the triangle spp", (see chap. vi. equa. 2 and 4), cos P's = sin PP'sin SP cos P'Ps + cos P'P cos SP, sin P'Ps Adopting the preceding literal representatives of these sides and angles, and remembering that cos P'Ps. = cos (90° + a) = — sin a, these become sin L = — sin icos d sin a + cos i sin d .... (1.) tand sin i + sin a cos i tan l = − coS at (2.) = tand sin i sec a + tan a cos i } These two formulae may be accommodated to loga-rithmic computation, by taking a subsidiary angle 4 such that tan 4. - sin a and * for then, exterminating sin a from the former, and tan d from the latter, we shall have - in d sin i = ****".... (3.) cos p 3. If, on the contrary, the latitude and longitude of the star are given, we shall have the declination and sin d = sin icos L sin l + cos i sin L.... (5.) _ – tan L sin i + sin l cos i tan a = cos l (6.) = — tan L sin i sec l + tan l cos i These equations are obviously analogous to those we have just deduced for the latitude and longitude, the only difference being that here the obliquity enters the formulae negatively. Hence by taking another subsidiary angle o', so that sin l and eliminating as before, there will result, / --- r - - - 4. As for the angle of position PSP’ = p, it is easily found from the relation between the sines of angles and the sines of their opposite sides; for from hence we have - sin icos a - sin icos t 5. We have also, from the same consideration, cos a cos d = cos L cos l . . . . (10.) And when L = 0, as is always the case with the sun, we have cos l tan a = tan l cost .... (12.) sin d = sin l sin i .... (14.). sin l = sin d cosec i ... (15.) The preceding formulae have been deduced upon the supposition that the heavenly body has not gone beyond the first quadrant of right ascension from the vernal equinox. But they are applicable to all positions by simply regarding the mutations of signs in the several sines, cosines, tangents, &c. according to the arcs to which they refer. The right ascensions and longitudes, being reckoned from the first point of Aries through their respective circles, will at once indicate by their attendant signs, + or -, to which quadrant they belong. And as to the declinations and latitudes, they being regarded as positive when toward the elevated pole, will be regarded as negative when towards the contrary pole. Thus, with us, north latitudes and declinations will be positive, south latitudes and declinations, megative. Example I. The right ascension of Aldebaran being 67° 40' 30", its declination 16°8' 20" N. Required the longitude, latitude, and angle of position. in equa. 3, 4, and 9, they become sin L = sin d cos (4 + i) sec p, tan l = tan a sin (4 + i) cosec 4, and sin p = sin icos l sec d. the log. operations corresponding to which will be as follows: From log sin a .. 67° 40' 30".... 9.9661625 Take ... tan d ... 16° 8' 20" . . . . 9:46.14544 Here, because cos (4 + i) being in the second quadrant is negative (chap. iv. art. 4), and the other terms are positive, the product is negative, and therefore the latitude is south. Next, to find l the longitude, add together, Log tan a ... . . . 67° 40' 30".... 10.3865391 sin (4 + i)..96° 5' 35".... 9.9975396 cosec p . . . . 72° 37' 46".... 10-0203305 The sum tan l...... 68°29'28".... 10.4044092 Here all the terms being positive, their product is positive; conseq. the longitude is in the first quad. Lo: for p the angle of position, add together, og sin i ... .23° 27' 49" .... 9-6000647 cos / . . . . 68°29′ 28” . . . . 9:5642090 seca. . . . . 16° 8' 20" .... 10-0174616 When the longitude of the moon is 1' 7° 41' 23", and the latitude 3° 49' 57° S., what are the right ascension and declination ? Ans. Right ascension 36° 36', declination 10°28' N. |