[21] Trigonometry.
[22] The process may be worth illustrating by means of a simpler problem. A heavy body, falling freely under gravity, is found (the resistance of the air being allowed for) to fall about 16 feet in 1 second, 64 feet in 2 seconds, 144 feet in 3 seconds, 256 feet in 4 seconds, 400 feet in 5 seconds, and so on. This series of figures carried on as far as may be required would satisfy practical requirements, supplemented if desired by the corresponding figures for fractions of seconds; but the mathematician represents the same facts more simply and in a way more satisfactory to the mind by the formula s = 16 t2, where s denotes the number of feet fallen, and t the number of seconds. By giving t any assigned value, the corresponding space fallen through is at once obtained. Similarly the motion of the sun can be represented approximately by the more complicated formula l = nt + 2 e sin nt, where l is the distance from a fixed point in the orbit, t the time, and n, e certain numerical quantities.
[23] At the present time there is still a small discrepancy between the observed and calculated places of the moon. See chapter XIII., [§ 290].
[24] The name is interesting as a remnant of a very early superstition. Eclipses, which always occur near the nodes, were at one time supposed to be caused by a dragon which devoured the sun or moon. The symbols ☊ ☋ still used to denote the two nodes are supposed to represent the head and tail of the dragon.
[25] In the figure, which is taken from the De Revolutionibus of Coppernicus (chapter IV., [§ 85]), let D, K, M represent respectively the centres of the sun, earth, and moon, at the time of an eclipse of the moon, and let S Q G, S R E denote the boundaries of the shadow-cone cast by the earth; then Q R, drawn at right angles to the axis of the cone, is the breadth of the shadow at the distance of the moon. We have then at once from similar triangles
G K - Q M : A D - G K :: M K : K D.
Hence if K D = n. M K and ∴ also A D = n. (radius of moon), n being 19 according to Aristarchus,
G K-Q M: n. (radius of moon)-G K
:: 1 : n
n . (radius of moon) - G K
= n G K - n Q M
∴ radius of moon + radius of shadow
= (1 + 1∕n) (radius of earth).
By observation the angular radius of the shadow was found to be about 40′ and that of the moon to be 15′, so that
radius of shadow = 8∕3 radius of moon;
∴ radius of moon
= 3∕11 (1 + 1∕n) (radius of earth).