Z G P E E B Z E or Z' E' be drawn, and join Z F or Z'F'. Then APB will be the elapsed interval of time between the observations, A PF half the elapsed time, Z P the co-latitude, ZPA the sun's meridian distance when the greater, and Z P B his meridian distance when the less altitude is observed, and ZPF his meridian distance I at the middle time between the observations. And PFA being a right angle, Z FP or Z FE will be the complement of Z FA or ZF B. In the right angled triangle P F A we have by Napier, sin A P. sin sin A P. sin APF APF rad. sin A F, whence rad = sin AF (arc 1,) 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, cos A P.. sec A F rad whence rad. cos A P = cos A F = cos P F (arc 2;) and as A F is always acute, P F will be of the same affection as A P. In the triangle ZF A we have (Prop. 18, p. 97) cos Z FA or sin cos AF. cos F Z ZFE= FZ + sin A F. sin F Z. sin Z FE; and in the triangle BFZ we have cos ZF B, or - cos Z FA= sin A F. sin F Z. cos Z FA= cos A F. cos F Z FZ. sin Z FE. We have then these two equations, viz. = cos A Z = cos AF.cos F Z + sin A F . sin F Z. sin Z FE and cos BZ = cos AF. cos FZ sin A F. sin FZ. sin ZFE. And, by taking their sum and difference, we have cos A Z + cos B Z = 2 cos A F. cos F Z and But in the right angled triangle Z FE, cos Z F = cos Z E. cos EF, and sin ZE = sin Z F. sin ZFE. two preceding equations become Or Hence, by substitution, the cos AZ + cos B Z = 2 cos A F. cos ZE. cos EF and 2 cos A F. cos Z E sec A F. sec Z E, radius being unity; or restoring sin rad 2 sin Z E (arc 3), and Here FZ and Z 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, P F FE PE, otherwise P F + FE, or P F + FE'= PE', or generally P F FE = 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 E P. cos E Z, hence rad = 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 D C, 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 P Z 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, AZ = d, PZ = l', and the latitude 1, 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 being unity; or cos D Z P (= sin D Z C) cosp. sec l sin dec. sec lat rad rad the rule given in p. 234, for comput ing the amplitude. And as I 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. For the azimuth we have (Prop. 19, Spherics,) cos2 AZP 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,) 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 B P, 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, AZ, and the side A B, to find the angle ZA B, the difference between which and PA B is ZAP, or the supplement of PA B+ Z'A B, is Z'A P; hence ZAP 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 Z' 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 Z P, the azimuth, from H, the supplement of which is AZI, the azimuth from I. 1 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. P Then if the polar distances DP 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 sun is on the meridian. But if the polar distances A P and DP 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 Z, 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. F G Now to compute this equation, let BP = D P, and B Z = D Z or Ꭺ Ꮓ ; let A B be part of a parallel of altitude, and BC 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 A PD; and half F G, or half the angle A P B, will be the required equation. But F G C B. cosec P B; and as PB C and Z B A are right angles, if the common angle Z BC 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 A C. cot A B C, is equal A C. cot P B Z, and therefore F G = A C. cot PB Z. cosec P B; or if A C F G A C be in seconds of arc, 2 30 PB Z. 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 DPB, which it may be without sensibly affecting the value of BC, we have cos ZPE. tan P Z = tau PE (arc 1); and PB PE = E B (arc 2,) and sin EP: sin E B :: cot Z PE: cot Z BP = cosec E P. sin (PBPE). cot Z PE, whence the equation of equal altitudes in A C seconds of time is cosec PB. cosec P E. sin (P B‡ PE). 30 cot Z PE. in seconds of time will be = cot 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 |