Fig. 20.—The eclipse method of connecting the distances of the sun and moon.it with the known angular diameters of the sun and moon, he obtained, by a simple calculation,[1] a relation between the distances of the sun and moon, which gives either when
- ↑ 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 :: I : n
n. (radius of moon)—g k = n g k—n q m
∴ radius of moon + radius of shadow
= (i + in) (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 = 83 radius of moon;
∴ radius of moon = 311 (I + In) (radius of earth).
But the angular radius of the moon being 15', its distance is necessarily about 220 times its radius,
and ∴ distance of the moon
= 60 (I + In) (radius of the earth),
which is roughly Hipparchus's result, if n be any fairly large number.
