PTOLEMY DAY 26 DETERMINATION OF STATIONS EPICYCLIC PROGRESSION PROOF:

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1 PTOLMY Y 26 TRMITIO O STTIOS PIYLI PRORSSIO PROO: iven: Prove: is our eye : > epicycle speed : speed and hence we can draw secant that ½ : = epicycle speed : speed is a random point on arc When the goes through arc, it is in PRORSSIO epicycle Join through to L (only for regression part later). Join,,. ow, > [closer to center] : > : [Lemma, ay 25] ½ : > : 2 i.e. epicycle speed : speed > : ut if we give ourselves an angle through which the epicycle moved in the same time that the moved through angle, then epicycle speed : speed = : hence >. (To adjust the disproportion to a proportion.) So, while the epicycle goes from to, the goes from to. So to our eye, the goes from to, which is PRORSSIO. Q

2 Incidentally, if 1 : 3 = : 60 [i.e. if the epicyclic speed is to the speed as 1 to 3, and = 60 ], then = 20, and we can t construct this without conics. nd if the epicycle speed is to the speed in me irrational ratio, then maybe we cannot construct at all. ut that s okay, because is not a math construction, but the angle swept out by point on the s epicycle in a given time, i.e. it is a physical reality. TRI PRORSSIO PROO: iven: Prove: H is our eye (and arc = arc, H joined) : > epicycle speed : speed ½ : = epicyclic speed : speed but ½ : = ½ H : H (by Lemma, ay 25) ½ H : H = epicyclic speed : speed = L : ½ H + H : H = L + : ½ : H = L + : i.e. ½ : H = eccentric speed : speed is a random point on arc. When the goes through arc, it is in PRORSSIO. L M H (eye) eccentric Join through to L, H through to M. Since H is the harmonic point for secants from (since arc = arc), thus arc L = arc M thus arc L = arc M. ow < : > : [Lemma, ay 25] 192

3 + : > + : i.e. : > L : [uc. 1.32] H : H > L : [: = H:H, Thm p.394] or H : H > M : [arcl = arcm] H : H > M : [, on same arc ] H + H : H > M + : i.e. : H > H : [uc. 1.32] ½ : H > H : 2 ½ : H > H : [angle at center is double] ut, by our givens, ½ : H = eccentric speed : speed and therefore eccentric speed : speed > H :. So if it follows eccentric speed : speed = H :, H > H. ut the eye is at H. Therefore, while the goes from to on the eccentric, in the same time the eccentric circle itself goes from to, and thus the appears to go from to, which is PRORSSIO. Q... L M H (eye) eccentric ' OT: The outer circle is drawn just to indicate the motion of the eccentric LM around point H, i.e. from to. 193

4 PIYLI & TRI RRSSIO PIYLI RRSSIO PROO: L iven: is our eye : > epicycle speed : speed and hence we can draw secant L that ½ L : = epicycle speed : speed is a random point on arc epicycle Prove: When the goes through arc, it is in RRSSIO i.e. it retrogrades through during the time it goes through arc Well, > L : < : L [Lemma, ay 25] ½ L : < : 2 L ½ L : < : thus epic.spd : spd. < : [½L : is as speeds] So if we replace with as to make the ratios equal, then and > epic.spd : spd. = : So in the same time that the epicycle progresses through angle, the will sweep out angle around the center of the epicycle, that the s net motion will be from to, i.e. angle to our eye, which is RRSSIO. Hence, too, if we replace with X to make the ratios equal (p. : Star = X : ), then X <, and in time goes from to, epicycle goes less than, hence the is regressing. Q... Q: Why give that : > ep. :? (Otherwise we can t draw the secant.) Q: What if spd. = zero? (The planet is always in progression.) Q: What if ep. spd. = zero? ( Stations are at the tangent points!) Q: What if : = ep. :? (One station at.) Q: What if : < ep. :? (o stations; all progression.) xercise: In a diagram, correlate (with letters) the planet s zig-zag (or loop) pattern in the sky with the places the is on the epicycle. 194

5 TRI RRSSIO PROO: iven: Prove: H is our eye ½ M : H = eccentric spd. : spd. (the unseparated ratio corresponding to ½ L : for the epicyclic) is a random point on arc When the goes through arc, it is in RRSSIO. Well, L : < : L [as before, for epicycle, above] L + : < + L : L L : < L : L [uc.1.32] MH : H < L : L [p.394] MH + H : H < L + L : L M : H < M + M : L i.e. M : H < H : L [uc.1.32, r HM] L ½ M : H < H : 2 L ecc.spd : spd < H : H M So if we replace with as to make the ratios equal then and > : ecc.spd : spd. = H : eccentric So in the same time that the eccentric progresses through angle H, the will sweep out angle around the center of the eccentric, that the s net motion will be from to, i.e. angle H to our eye, which is RRSSIO. Q

6 OT: If : is not greater than e : s, then either : = e : s, in which case you get one station at, or : < e : s, and we get no station at all, because, given the foregoing proofs, progression will always produce the greater angle, and therefore we get nothing but progression. L M eccentric H 196

7 OSTRUTIO O TH STTIO POIT. P an we construct the station points? iven our eye is at, and given an epicycle with center, radius, and given that : > epicycle speed : speed can we construct the line that ½ : = epicycle speed : speed? We could employ me rt of edekindian postulate, and say to ourselves the ratio of the half of the intercepted segment to the external segment decreases continuously as goes away from, there must be a place in between where the proportion holds good. ut it turns out we can actually construct it. H iven: To do: e : s is a given ratio (expressed in numbers or lines) ind a secant that ½ : = e : s irst cut at H that H : H = e : s [uc.6.10] ext, describe a circle on H as diameter, cutting circle at. escribe a circle on as diameter. ow join, cutting the circle on at P, and cutting circle at. Join P. Join H. ow, P is 90 [angle in a semicircle] and H is 90 [angle in a semicircle] P is parallel to H P : = H : H P : = e : s [we cut H : H = e : s] ut P is perpendicular to P, and is the center, hence P = P, hence P = ½ ½ : = e : s Q

8 PORISM: Since the construction is possible only if the circle on H cuts the circle, H must be inside circle, and therefore : > H : H, i.e. : > e : s, and hence this is a condition for the possibility of the construction. WIT SO WHR S TH QUT? Ptolemy appears to have ignored the equant, and even the eccentricity of the deferent, for these initial demonstrations. That is, he assumes that the center of the epicycle makes its mean movement about, our eye, when in fact it does not. This is because he is far only considering the matter abstractly and in the simplest case. It is only in h. 2 that he begins to study the particular planets, and to apply the general theorem to the model for Saturn (for example), to find out where it makes stations, how long it should spend in regression (depending on the place in the zodiac etc.). irst he takes it at its mean distance, where the mean movements in longitude and anomaly will be very nearly the same as the apparent movements. So that is like the ideal case already done, but with the particular numbers for Saturn s speeds in L and. He finds the angle about the center of the epicycle subtending the semi-arc of the regression, and from knowing the mean speed in anomaly (by the tables generated from the periodic joint returns), he knows the time it takes Saturn to move through double that angle, and figures the regression, when Saturn s epicycle is at its mean distance from arth, should take about 138 days. HOW O RTH I THY IUR THIS OUT? How did Ptolemy s predecesrs discover this rule for where station occurs? One way could be by observations. We can observe the at station () at a given time. nowing when it is at station, by our tables of planetary motion, we can al know how far the has gone from its last time at apogee, i.e. we know. y planetary theory, we know the magnitude of the epicycle, i.e. we know :. Since we know, we know its supplement,. Hence we know and the ratio :, and therefore r is lved, and all its sides and angles are known. Therefore we al know, and too its supplement,. ut =, and hence =, and we know, i.e.. Thus we know [from r ] and we know [by uc.1.32] and we know the ratio : [the magnitude of the epicycle] and therefore r is lved, and all its sides and angles are known. Therefore we know : [from r ] and we know : [from r ] that, putting in the same terms for each ratio, we know : we know : as well, i.e. : is given. P 198

9 omparing this value (for each planet) to the ratio of the epicycle s speed to the s speed, we notice in each case that ½ : = epicycle speed : speed IM LITR. ut since it is very difficult (or impossible) to observe a station accurately, i.e. to say exactly when a is at station (since it is really just an instant in time, and just seems longer because the planet is moving slowly for a while on either side of station), this is not likely to be the way the ancients discovered the rule. Perhaps they discovered it by beginning with a simple case, e.g. when there is only 1 station, namely at P, perigee, and by thinking (if this is not anachronistic to say) ewton-style. We know that if a station occurs at P, then the speed of the and of the epicycle must be such that, as we take equal arcs P, P smaller and smaller, T P arc ep. speed : speed u = : sp. speed : speed u = : P sp. speed : speed u = : P P P P sp. speed : speed u = P : arcp P (chords ult. as arcs) ( : = P :, and P = P) (ratio of inverses) ep.speed : speed = P : P since is ultimately equal to P, and since (the ratio P : P being fixed, and that of the speeds al being a fixed ratio) there is no longer any rean to say ultimately. ut P is half of PT. So this might lead to suspicion that this is an instance of a general rule, which could easily be verified by a procedure like the one above, looking to a pair of symmetrical stations, instead of just one of them at P. QUSTIO: oes the TULLY sit still in space (i.e. have an instantaneous velocity of zero)? Or does it just appear to be still, when in fact it is heading right for us or straight away from us? In the simple case of one station, where it happens at perigee, it seems to be a real standstill. ut maybe not in the more general case. 199