EXAMPLE III. Suppose the distance from an unknown star at T was measured to be 25°.42'.10", from a star at x, whose latitude is 39°.33'.30′′ South, longitude 3.11.13'.44"; and 51°.6'.56", from another known star at c, whose latitude is 22°.51'.57′′ North, and 2.18°.57.57" longitude; required the latitude and longitude of the unknown star. First. In the triangle xmc, you will have given xm=the complement of x's latitude, cm=the distance of c from the same pole of the ecliptic, and the angle xmc=the difference between their longitudes vo. Hence xc will be found 65°.47'.42", and the angle mxc= 153°.17.21". Secondly. In the triangle XTC, are given XT Now mxc cxT=104°.3′.21′′=mxт. and TC Thirdly. In the triangle mxT, are given xm ΧΤ To find Tm 74°.1′.46", the distance of the star T from the pole of the ecliptic, and the angle xmT=11°.41′.58", the difference between the longitude of x and T. (0) By the assistance of the foregoing problem, and a knowledge of any one fixed star's true situation, the places of all the rest may be determined. Indeed, it is by this problem principally, that astronomers have rectified the places of all the fixed stars; and thence, by a similar mode of proceeding found the true places of the planets. Leadbetter tells us in his Astronomy, page 230, that he made use of this method for constructing his astronomical tables. The preceding examples contain every variation that can happen with respect to the stars latitudes and longitudes, but the same things may be determined from their right ascensions and declinations, as in the following example: EXAMPLE IV. (Plate III. Fig. 9.) Suppose the distance of a star or planet at Y, was observed to be 650.47.42", from Sirius the Dog star at x, whose right ascension is 990.0'.21", and declination 16°.26'.35′′ South; and to be 51°.6'.56" distant from Procyon at T in the Little Dog, whose right ascension is 112°.6′.47", and declination 5°.45'.3" North; it is required to find the right ascension and declination of this unknown star or planet. First. First. In the triangle XNT, we have The angle XNT 13°. 6′.26′′ the difference between the right ascensions of x and T, measured by the arc RQ. pole. XN=106.26.35 =x's distance from the pole N. Hence two sides and the included angle are given, to find the third side XT=25°.42′10′′, and the angle NXT=31°.22′. Secondly. In the triangle YXT, we have given YX=65°.47′.42" To find the angle YXT=49°.14'. YT 51. 6.56 XT 25. 42. 10 } Then YXT 2 NXT=17. 52. =YXN. The YXT being greater than the NXT shews the right ascension of the unknown star at Y, to be less than the right ascension of that at X. Thirdly. In the triangle XYN, are given XY= 65°.47′.42′′ LYXN= 17.52. XN=106.26. 35 You will find To find YN the complement of the declination BY, and the angle YNX the difference between the right ascensions of y and x,=BR. YN=44°.14′ consequently BY=45°.46′ N. You will findYNX 23°.38'.55", hence the right ascension of Y=AB=75°.21′.26". (P) In the same manner we may find the distance between any two places on the earth; suppose x and T to be two places whose latitudes and longitudes are known, then in the triangle XNT, XN is the distance of x from the pole, TN the distance of T from the same pole, and the angle XNT, measured by the arc RO, the difference of longitude between the two places, hence the distance XT is easily found. In a similar manner the distance between T and y, or y and x may be obtained. EXAMPLE V. What is the extent of Africa, from Cape Verd, latitude 14°.46' N., longitude 17°.47′ W. to Cape Guardafui, latitude 11°.47′ N., longitude 51°.35' E.? Answer. 67.19' the arc of a great circle contained between the places, or 4678 English miles, of 69 to a degree, or 4039 geographical miles, of 60 to a degree. PROBLEM XIII. (Plate III. Fig. 1.) (Q) Given the latitude of the place, the sun's declination and altitude, to find the azimuth. EXAMPLE EXAMPLE I. In latitude 51°.32' North the sun's altitude was observed to be 50°, and his declination 23°.28' North; required his azimuth from the North. In the triangle szN, we have given SN 66°.32′ the co-declination, or sun's polar distance; for DS is the sun's declination North. sz=40. 0 the co-altitude of the sun; for cs is his altitude above the horizon Ho. ZN=38.28 the co-latitude of the place, or what on the elevation of the pole wants of 90°. To find the angle szN, measured by the arc of the horizon co, the sun's distance from the North. Call the co-latitude ZN the base, and draw the perpendicular sf Then, (M. 200.) tangzN=zt=19°.14' : tang (SN+sz) 2 ::tang (SN-sz) : tang ft Take the difference latitude. Then, Radius, sine of 90° tang ft 53°16′ 13°.16' 9.54269 10.12710 9.37250 9.95691 between ft and half the complement of zt=22°.56' :: tang of the sun's altitude=cs=50° 10.00000 9.62645 10.07619 9.70264 : sine of azimuth from the east or west=50%. 17′ Hence 90° +30°.17' 120°.17′ the sun's azimuth from the North. (R) A slight review of Figure I. Plate III. will naturally point out to us the following observations: If the latitude of the place and the declination of the sun be of the same name, the sun is less than 90° from the elevated pole; therefore the co-declination is the polar distance. If the sun or star be in the equinoctial, it is 90° from the pole, and forms a quadrantal triangle. If the declination be of a contrary name with the latitude, the distance of the object from the elevated pole will be equal to the declination increased by 90°. (S) The preceding solution is on a supposition that the sun's declination, as taken from the Nautical Almanac, remains the same in all places, and at all times of the day. But this cannot be strictly true, for the declination is subject to a daily variation, and is adapted to the meridian of Greenwich. An azimuth will not be materially affected by this change, but it will make a considerable difference when we want to find the time of the day. EXAMPLE EXAMPLE II. Suppose the sun's altitude at London, latitude 51°.32' North, to be 10.14' when the declination is 20°.11' South, what is the azimuth from the North? Answer. 125°.34'. EXAMPLE III. Given the latitude of the place 51°.32′ North, the sun's altitude 25°, his declination 4°.47′ South; required the azimuth from the North? Answer. 137°.16'. EXAMPLE IV. Suppose in latitude 51°.32′ North, the altitude of the star Arcturus in Boötes was 44°.30', when its declination was 20°.16′ N.; required its azimuth from the North at that time. Answer. 117°.8'. EXAMPLE V. In latitude 48°.51' North, when the sun's declination is 18°.30' North, and his altitude 52°.35'; what is his azimuth from the North? Answer. 134°.36'.8". (T) This problem is not only useful for finding the variation of the compass, but it will be found very convenient for determining the true meridian, and the four cardinal points. For, if with a line representing the vertical on which the sun is, an angle be made equal to the supplement of the azimuth from the North, the meridian will be obtained: provided care has been taken to correct the sun's altitude by refraction. (U) An azimuth may be constructed by a scale of chords, sufficiently accurate for Nautical purposes; Thus, With the chord of 60° describe a circle (Plate IV. Fig. 2.) set off the complement of the latitude from z to P, and the complement of the altitude from z to a, draw zon through the centre c and aoa at right angles to it. Set off the polar distance from P to p, draw pp through the centre c, and from the point I where it intersects aoa draw IR parallel to zon, and through c the centre draw cs parallel to aoa. With c as a centre, and the distance oa, cross the line IR in m; through m draw cn, then sn, measured on a scale of chords will be the azimuth from the south (in North latitude) if the point I fell on the left-hand of ZN, and from the north if it fall on the right-hand. See the Mathematical Repository, vol. I. page 283, second edition, edition, 1799, and vol. II. page 26 and 27. See also Dr. Mackay's Navigation, first edition, 1804, page 200. PROBLEM XIV. (Plate III. Fig. 1.) (W) Given the latitude of the place, the sun's declination and altitude, to find the hour of the day. EXAMPLE I. At London, latitude 51°.32′ North, the altitude of the sun's centre was 38°.19', and his declination 19°.39′ North, required the hour of the day? In the triangle szN, we have given SN=70°.21' the sun's distance from the elevated pole. sz=51.41 the sun's co-altitude. ZN=38. 28 the co-latitude, To find the angle zNs, measured by the arc #D, the distance between the meridian which the sun is on at the time of ob servation, and the meridian of the place of observation. Since Az is equal to the latitude ON; it follows, that if the latitude and declination be both north or both south, the dif ference between them is equal to the meridional zenith distance; if the one be north and the other south, their sum is the meridional zenith distance. Hence the following general method of solution: 4.59230 Num. in col. ris., Requisite Tables, 3.30′ Ör, Increase the index by 5+ Log sine of half the hour angle 26°.14.40" 9.64563 2 The hour angle 52.29. 203h.29'.57".20" (X) This |