NT In the triangle Tnc. Let the star at T represent Procyon, that at c, Capella. пс = 67o. 8'. 3" =comp. Capella's lat. oc. =105°.58'.14" -90°. + Procyon's lat.=NT. LTNC = 330.57'.45"=diff . of their longitudes=0s. Here we have two sides and the contained angle given, to find the third side cT=51°.6.39" Answer. PROBLEM XII. (Plate III. Fig. 8.) (X) The places of two stars being given, and their distances from a third star, to find the place of this third star. EXAMPLE I. Suppose the distance of a comet or new star c, was measured by a sextant to be 65°.47'.42" from Sirius the dog star, lat. 39o.33'.30" South, longitude 38.110.13'.44"; and 519.6'.56", from Procyon the Little Dog, lat. 15°.58'.14" South, longitude 38.220.55'.42"; it is required to find the latitude and longitude of this comet or star. Let x represent the place of Sirius, or one of the known stars, T that of Procyon the other known star, having the greatest longitude, and c the place of the comet or unknown star. First. In the triangle xnt, we have given xn=129o.33'.30" Sirius's distance from the pole of the ecliptic nearest the comet or star; mn=1059.58'.14" Procyon’s distance from the same pole. _XNT= 11°.41.58" the diff. of longitude between the given stars. Secondly. In the triangle xtc, we have given Tc=51°. 6'.56" (any of the methods(W.231. or P. 248,) and xc=65o.47.42" Syou will find cxt=49o.14'. This angle being greater than nxt shews the longitude of the star at c to be less than at x. Hence cxT~NXT = 22°.31'.21" Thirdly. In the triangle xnc, we have given xn=129o.33.30" To find the side nc the comet or star's xc= 65°.47'.42" (co-lat, and xnc the diff. between the dnxc= 22°.31'.21") star's long, at c and Sirius's long. nc=670.8.26" the star's co-latitude, hence its latitude oc= 22°.51'.34", and xnc=22o.16'.42", and since the star's longitude has been shewn to be greater than the comet's; from Sirius's longitude=1010.13'.44" take 22o.16'.42", the remainder 780.57'.2" is the unknown star's longitude. EXAMPLE II. Suppose the distance of an unknown star at x was measured to be 65°.47'.42", from a known star (Capella) at c having 22°. 511.57" North latitude, and 28.180.57'.57" of longitude; and its distance from another known staxat t (Procyon)=25°.42.10", this star having 150.58'.14" South latitude, and 35.220.55'.42" of longitude; required the latitude and longitude of the unknown star? First. In the triangle cmt, we shall have cm=motoc=112°.51.57"=Capella's distance from the pole of the ecliptic nearest the comet, or unknown star. CMT=330.57'.45' = the diff. long. between Capella and Procyon. Tm = 74o.1'.46"-Procyon's distance from the pole of the ecliptic. To find tc=519.6'.56", the distance between the known stars; and the angle mcT=43°.37'.41". Secondly. In the triangle xct, we have given By any of the methods (W. 231. or P. 248.) xct will be found=240.58'; this angle being less than mct shews the longitude of the star at x to be greater than that at c. Hence mcT~XCT=18°.39'.41"=xcm. Thirdly. In the triangle xcm, we have To find xm the unknown star's coCX = 65° 47'.42'' xcm= 180.39'.41" (tween the star's longitude at x, and Capella's longitude at c. xm=50°.26' the star's co-lat., hence its latitude vx=390.34', and xmc=22°.15'.6", and since the given star's longitude has been shewn to be less than the required star's longitude, this difference must be added to Capella's longitude, viz. to 29.18°. 57.57" add 22°.15'.6", the sum is 38. 11°.13.3" the longitude required. cm=1120.51':57", latitude, and xmc the difference be EXAMPLE III. XC XT xm Suppose the distance from an unknown star at T was mea-sured to be 250.42.10", from a star at x, whose latitude is 399.33.30" South, longitude 35.11o.13.44"; and 51°.6.56" from another known star at c, whose latitude is 229.51.57" North, and 2.189.57'.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=650.47'.42", and the angle mxc= 153o.17.21". Secondly. In the triangle xtc, are given To find the angle cxt=490.14 and Tc Now mxc^cxt=1049.3.21"=MXT. To find tm=74o.1.46", the distance of the star xmt=110.41'.58", the difference between the lonLmxT gitude of x and T. (Y) 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 999.0'.21", and declination 16°26'35" South; and to be 5106'.56" distant from Procyon at r in the Little Dog, whose right ascension is 112°.6'.47", and declination 50.45'.3" North; it is required to find the right ascension and declination of this unknown star or planet XT =YXN. First. In the triangle xnt, we have The angle xxt=13°.6'.26''=the difference between the right ascensions of x and t, measured by the are RQ. XN=106°.26'.35"=x's distance from the pole n. TN= 84°.14'.57"=T's distance from the same pole. Hence two sides and the included angle are given, to find the third side xt=250.42'.10", and the angle nxt=31°:22'. Secondly. In the triangle yxt, we have given YX=650.47'.42" To find the angle ext=49o.14'. YT=51. 6. 56 Then YxT ~ NXT=17: 52. XT= 25. 42.10 The Zyxt being greater than the L 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 clination by, and the angle ynx the XN=106. 26.35 difference between the right ascenLyxn= 17. 52. sions of y and x, =BR. YN=44°.14', consequently by=450.46' N. You will find ynx=239.38'.55", hence the right ascension (of y=AB=75°.21'.26". (Z) In the same manner we may find the distance between any two places on the earth; suppose x and 1 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 RQ, 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 14o. 46' N., longitude 170.47'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 XI. (Plate III. Fig. 1.) (A) Given the latitude of the place, the sun's declination and O EXAMPLE I. In latitude 51°.32' North the sun's altitude was observed to be 50°, and his declination 230.28' North; required his azimuth from the North. In the triangle szy, we have given sn=660.32'. 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 (W. 231.) tang 4 ZN =z=190.14 9.54269 : tang į (SN +sz) = 53o.161 10.12710 :: tang 4 (SN~sz) = 13o.16' 9.37250 : tang ft = 420.10 9.95691 Take the difference between ft and half the complement of latitude. Then Radius, sine of 90°. 10.00000 : tang ft zt=220.56 9.62645 :: tang of the sun's altitude CS= 50° 10.07619 ; sine of azimuth from the east or west=30°.17' 9.70264 Hence 90° +30.17'=120°.17' the sun's azimuth from the North, (B) 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°. (C) 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. (B. 270.) 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. |