= cos A F Z E or Z' E' be drawn, and join Z F or ZF'. Then APB will be the elapsed interval of time between the observations, APF half the elapsed time, Z P the co-latitude, Z PA the sun's P meridian distance when the greater, and Z P B his meridian distance when the less E altitude is observed, and ZPF his meridian distance at the middle time between the observations. And PFA being a right angle, ZF P or ZFE will be the complement of ZFA or ZFB. In the right angled triangle P FA we have by Napier, sin A P.sin sin A P. sin A PF A PF= rad , sin A F, whence = sin AF (arc 1,) rad and as A P F is always an acute angle, A F will also be acute. In the same triangle we have rad .cos A P= cos A F. cos PF, rad . cos A P cos A P.. sec A F whence = cos P F (arc 2 ;) and rad as A F is always acute, P F will be of the same affection as A P. In the triangle Z F A we have (Prop. 18, p. 97) cos Z F A or sin cos A Z cos AF. cos F 2 ZFE= ; or cos A Z= cos A F. cos sin A F. sin FZ FZ + sin A F. sin F Z. sin Z FE; and in the triangle B FZ we cos Z B cos FB . cos F Z have cos Z FB, or cos Z FA= sin FB. sin F Z cos FZ sin AF. sin FZ sin A F. sin FZ.cos ZFA = cos A F . cos F Z sin AF. sin FZ. sin Z FE. We have then these two equations, viz. cos A Z= cos AF.cos F Z + sin AF. sin FZ.sin Z FE and cos BZ = cos A F. cos FZ sin AF. sin FZ. sin Z FE. And, by taking their sum and difference, we have cos A Z + cos BZ = 2 cos A F. cos FZ and cos BZ = 2 sin A F. sin FZ. sin ZFE. But in the right angled triangle Z FE, cos Z F = cos ZE . cos EF, and sin ZE = sin ZF:sin Z FE. Hence, by substitution, the two preceding equations become cos AZ + cos BZ = 2 cos AF. cos Z E.cos E F and cos BZ = 2 sin AF. sin Z E. sin A D sin B C = 2 sin AF. sin 2 E. cos Z B cos AZ From the latter of these equations we have sin AD- sin BC sin ZE = 2 sin AF sin A D + sin BC and from the first, cos E F = 2 cos A F. cos Z E But (Formula 3, p. 73,) A D BC A D + BC sin A D - sin BC= 2 sin 2 2 COS and A D + BC sin A D + sin BC= 2 sin 2 A D BC 2 These expressions substituted in the two preceding equations give AD - BC A D + BC 2 sin 2 sin ZE = 2 AD - BC = sin 2 sin AF 2 COS = sin COS A D + BC AD - BC 2 AD+BC cos EF = 2 cos A F. cos ZE 2 A D - BC sec AF sec Z E, radius being unity; or restoring 2 the radius. A D - BC AD + BC sin cosec A F = sin Z E (arc 3), and COS rad 2 AD - BC sec A F. sec ZE =cos EF,(arc 4.) rad 3 Here FZ and Ž E are always acute, therefore E F is also always acute. Now when Z and P are on the same side of B A G, the great circle passing through the places of the sun at the times of observation, PF FE PE, otherwise PF + FE, or PF + FE= PE', or generally PF FFE=PE, (arc 5.) In the right angled triangle Z PE, rad. cos P Z, or rad, sin lat. cos E Z. cos E P = cos EP.cos E Z, hence = sin lat. Method of computing the amplitude and azimuth. (See p. 235.) In the last figure let Z C represent the prime vertical, and D the place of a celestial object in the horizon, then the angle D ZC, measured by DC, will be the amplitude of the object at D. Let P be the north pole, and A the place of an object above the horizon, then A Z P will be the azimuth from the north, AZ I the azimuth from the south, A PZ the hour angle, or the meridian distance of the object, A D its altitude, A Z its zenith distance, AP its polar distance, and Z P the co-latitude. Put AD= a, A2= a', PZ=l', and the latitude=l, and A P = p. Then to investigate the rule for the amplitude, we have, by Prop. 18, when a' equal 90°, or the object is in the horizon, as at D, radius rad . cos p being unity; or cos D Z P (= sin D Z C) cos 1 cos p. sec 1 sin dec . sec lat - the rule given in p. 234, for computrad rad ing the amplitude. And as l is always acute, DZP will be of the same affection with p, or the amplitude D ZC will be of the same name with the declination. AZP For the azimuth we have (Prop. 19, Spherics,) cos2 2 or Method of computing the hour angle, or the meridian distance of a celestial object. (See p. 241.) Adopting the above notation, we have (Prop. 19, Spherics,) a' + ' P rad 2. sin sin APZ 2 2 sin p. sin l . sin e The direct solution of the preceding problems by spherics is very simple. To find the latitude by double altitudes, we have given first in the isosceles spherical triangle B P A, the two equal polar distances PB, PA, and the included angle BP, the elapsed time, to find A B and the angle P A B. Then in the triangle Z A B are given the two zenith distances BZ, A Z, and the side AB, to find the angle ZÅ B, the difference between which and PAB is ZAP, or the supplement of PA B + Z'AB, is Z' A P; hence Z AP or Z A P is known. We have then Z A and A P, or Z' A and AP, and the included angle in either case, to compute Z P or 2' P, the co-latitude. For the azimuth and hour angle, in the triangle A ZP, all the sides are given to compute A P Z, the hour angle, and A ZP, the azimuth, from H, the supplement of which is AZI, the azimuth from I. D Z C G Method of computing the equation of equal altitudes. (See p. 245.) Let P represent the pole, Z the zenith, A the place of the sun in the morning, D his place in the afternoon of the same day, when his zenith distance D Z is equal to A Z. Then if the polar distances D P and A P are equal, the angles A PZ and D P Z will be equal, and consequently half of A P D, their sum, as measured by a chronometer, added to the time shown by the same chronometer when the sun is at A, will be the time by the chronometer at apparent noon, or when the F sun is on the meridian. But if the polar distances A P and D P are unequal, the angles AP Z and DPZ will also be unequal, one of them exceeding half their sum as much as the other is less, D P Z being greater or less than A P 2, according as D P is less or greater than A P, and half the difference of these two angles is the equation of equal altitudes. Now to compute this equation, let B P = DP, and B Z = D Z or A Z; let A B be part of a parallel of altitude, and B C part of a parallel of declination, and F G the arc of the equator included between P A and P B. Then A C will be the change of the polar distance in the interval measured by APD; and half F G, or half the angle A P B, will be the required equation. But F G=CB. cosec P B; and as P B C and Z B A are right angles, if the common angle Z B C be taken from each, the remaining angles P B Z and A B C will be equal; and the sides of the triangle A B C being necessarily small, it may be considered as a rectilineal one, right angled at C. Hence C B, which is equal to AC. cot A B C, is equal A C. cot PBZ, and therefore F G = AC.cot PBZ.cosec P B; or if AC FG AC be in seconds of arc, in seconds of time will be = cot 2 30 PBZ . cosec P B. Now to compute cot P B Z, let Z E be a perpendicular on P B, and the angle Z PE being taken equal half DP B, which it may be without sensibly affecting the value of BC, we have cos ZPE. tan P Z = tan P E (arc 1); and P B F PE = E B (arc 2,) and sin EP: sin E B :: cot Z PE : cot Z BP = cosec E P. sin (PBF PE). cot Z PE, whence the equation of equal altitudes in AC seconds of time is cosec P B. cosec P E. sin (P B FPE). 30 cot Z PE. On these equations it may be observed, that no great precision is required in the arcs employed in the computation, excepting in the change of declination. T |