Autolycus, On a moving sphere Prop. 2©
translated by Henry Mendell (Cal. State U., L.A.)
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Prop. 2: If a sphere turns evenly about its own axis, all the points on the surface of the sphere traverse in equal time similar circular-arcs of the parallel circles on which they move.
(general diagram based on Vat. gr. 204 39r)
(diagram 1)
For let a sphere turn evenly about its own axis, AB, but let there be poles of the sphere, points A, B, and let there be taken some points on the surface of the sphere, G, D. I say that points G, D traverse in equal time similar circular-arcs of the parallel circles along which they move.
(diagram 2) For there be parallel circles, GE, DZ, along which points G, D move, (diagram 3) and let the plane through AB and G be extended. In fact, it will make a section in the sphere, a circle .But let there be a semicircle of it, AGB. (diagram 4) In fact, either it will also go through D or not.
(diagram 5) Let it first go through D and let there be AGDB, and in the rotation of the sphere, let semicircle AGDB move with them in the rotation of the sphere and let it have a position as AEZB. (diagram 6) Since GE, DZ are parallel circles in a sphere, and largest circles through their poles have been described, AGDB, AEZB, circular-arc GE is, therefore, similar to circular-arc DZ. And so, I say that point G comes to be at E and D at Z in an equal time.
(diagram 7) For if not, but if it is possible, let point G come to be at point E but D to H in an equal time. Therefore, with the sphere turning, whenever G comes to be at E and D at H, semicircle AGDB will also have a position as AEHB. (diagram 8) And since each of circles AEZB, AEHB is a largest circle, the straight-line joined from A to E is, therefore, a diameter of the sphere. But also, AB, which is absurd. Therefore, point G does not come to be at E and D at H in an equal time. We will, in fact, similarly show that it will come to be at no other point except point Z.
(diagram 9) Let the semicircle through A, G, B, in fact, not go through D, but through Q, as it holds in the second drawing, (diagram 10) and let there be a parallel circle along which point D moves, DQZ, and let DH be posited similar to GE. But GE is a circular-arc similar to circular arc QZ. Therefore, circular-arc DH is also similar to QZ and is of the same circle. Therefore, circular-arc DH is equal to circular-arc QZ. Therefore, Q also comes-to-be at Z and G at E in an equal time. Therefore, G comes to be at E and D at H in an equal time.