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This is presentation (not a translation) of the Ptolemy's construction of the solar model, which he attributes to Hipparchus. He uses only the time intervals for spring (vernal equinox to summer solstice) and summer (summer solstice to autumnal equinox). The other assumptions come from reflections on the model. Chord functions are only used where they are most natural. Elsewhere, the reader may employ whatever pleases her.
This argument may be generalized to show that if the mean motions are known, only three points are required to establish an eccentric model.
The season lengths, according to Hipparchus are:
|Spring 94 1/2 days||Fall: 88 1/8 days|
|Summer: 92 1/2 days||Winter: 90 1/8|
Ptolemy assumes a simple eccentric model. He also repeats the derivation using an epicycle model.
We assume that the sun moves with a constant regular motion
and need to find the distance of the center of the deferent from
the earth, ZE, and the angle of ZE to some fixed line, e.g., to
the summer solstice, ZEX (analysis).
(diagram 1) Let the earth E be the center of the zodiac ABGD. Since the seasons are unequal, the center of the sun's motion is not at E. Let it be at Z. (diagram 2) Clearly, Z is in the quadrant of the longest season, since the sun must move through the greatest angle to appear to have moved mearly 90 degrees. (diagram 3) Draw circle NPOS with center Z as the circle of the sun. It is also obvious that the sun moves in spring through 94 1/2 days / (365 1/4) * 360 degrees = 93;9 degrees. In summer, it moves through 92 1/2 days / (365 1/4) * 360 degrees = 91;11 degrees. Hence, the total motion is 184;20 degrees.
(diagram 4) Construct through Z and parallel to AG and BD, respectively NO and PS. Hence, arcs OL + NQ + 180 degrees = arc QPL (the arc of spring and summer). And OL = NQ. Hence, NQ = (QPL - 180) / 2 = 2;10 degrees.
(diagram 5) Draw chord QTU. QTU = CHORD(QNU) = CHORD(2 NQ) = CHORD(4;20). For a radius ZN = 60 points, QTU4;32, so that TQ2;16. TQ =ZR. Hence ZR2;16.
Similarly, arc PK = arc PS - arc NQ - 90 degrees = 93;9 - 2;10 - 90 = 0;59. Draw chord CFK. Arc KPC = 2PK = 1;58 degrees. Hence, CK = CHORD(1;58) = 2;4 points for radius PZ = 60 points. RE = PK = 1/2 C1;2.
(diagram 6) Hence, ZE2 = ZR2 + RE2 = 5;8,15 + 1;4,4 = 6;12,19. ZE = 2;29,30. That is, ZH : ZE= 60 : 2;29,30. Or the deferent is about 24 * the eccentricity.
Therefore, since ZE and ZX are known, ZEX24;30 (the angular distance preceeding the summer solstice).
Since arc SM = arc PK and arc OL = arc NQ, Fall arc LM = 90 - arc OL - arc SM = 90 - 2;10 - 0;59 = 86;51. 86;51/360 * 365 1/4 days88 1/8.
Winter arc MQ = 90 - arc NQ + arc SM = 90 - 2;10 + 0;59 = 88;49. 88;49/360 * 365 1/4 days90 1/8.