64 Rationale of the above rules. (1) In Fig. 4, let S be the Sun (or a point on the ecliptic), SL the perpendicular from S on the plane of the celestial equator, and SM the perpendicular from S on the line joining the centre of the celestial sphere with the first point of Aries. Then in the plane triangle SLM, we have SL = Rsin 8, SM Rsin λ, LSML €, and SLM = 90°. Therefore, or DIRECTION, PLACE AND TIMB = SL/SM Rsin > = and LKAB = $. = KA = Rsin 8, KB = earthsine, LKBA = 90⁰ - $, = M Rsin €/Rsin 90°, Rsin Ex Rsin > R 1397 x Rsin A R Fig. 4 (2) The arcual distance of the diurnal circle from the north pole of the celestial equator-90°-8. Therefore, the day-radius-Rsin (90°-8), or √R²(Rsin 8 )². (3) In Fig. 5, let K be the point of intersection of the diurnal circle and the six o'clock circle, KB the perpendicular from K on the rising-setting line,² and KA the perpendicular from K on the east-west line. Then in the plane triangle KAB, we have A S for Rsin E=1397'.¹ Fig. 5. 1 See LBh, ii. 16. 2. The rising-setting line of a heavenly body is the line joining the point where the heavenly body rises on the eastern horizon with the point where it sets on the western horizon. In other words, it is the line of intersection of the planes of the celestial horizon and the diurnal circle.
