PTOLEMY DAY 10 PREPARING TO FIND OTHER ARCS BETWEEN THE ECLIPTIC AND EQUATOR. You might have noticed that in Day 9, when we found the arc

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1 PTOLEMY DAY 10 PREPARING TO FIND OTHER ARS BETWEEN THE ELIPTI AND EQUATOR You might have noticed that in Day 9, when we found the arc D between the tropics, we made no use of our table of chords. We simply observed the arc, watching the Sun s latitude throughout the year. But Ptolemy will next show us how to H find other arcs intercepted between the ecliptic and equator. For example, if E is the place where the ecliptic and equator G E intersect, and arc E and arc ED are each 90, then the arc of the great circle through and D will be half the arc between the tropics, or about 23½. But what if arc EG on the equator is something less than 90, and we draw a great circle through the celestial poles and through G, intercepting the ecliptic at S H? How great will arc GH be? Will we be able to tell, if we are given arc EG in degrees? We will. And one can see how this kind of power of getting the values of arcs on a great circle given other arc-values on the same sphere could be very useful in astronomy. And to do this, we will need to use our table of chords and arcs. But first we will need to develop a little more geometry! We will need some Lemmas about spherical geometry, and in order to prove those, we will need first a few Lemmas concerning plane geometry. And since we will need to compound ratios in order to understand these Lemmas, and since that mathematical operation is extremely common in Ptolemy (and in opernicus and in Kepler, and also in Apollonius, whose geometry Kepler presupposes to some extent), we will have to begin with a refresher on what it means to compound ratios. So here is our agenda for today: (1) Review what it means to compound ratios. (2) Prove the Menelaos Theorems as Lemmas preliminary to the spherical proofs. (3) Prove the Spherical Menelaos Theorems as Lemmas preliminary to determining the values of certain arcs on the celestial sphere. (4) Apply these Menelaos Theorems to an astronomical problem, namely finding the length of the longest day of the year where you live. We will cover Step (1) today, Step (2) on Day 11, Step (3) on Day 12, and Step (4) on Day 13. N 70

2 (1) OMPOUNDING RATIOS. Suppose you have two ratios, ab : bc & ef : fd. There is any number of ways one could unite these terms so as to produce a new ratio. For example: ab + ef : bc + fd That is a new ratio, formed out of the terms of the original ratios. And if one liked, one could call this new ratio the ratio compounded of the original ratios. But there is a reason we don t do this. If you start with two ratios and go through this process, and if I start with two ratios that are the same as yours (but perhaps expressed in different numbers), and then I go through the same process, then your result and my result might not be the same! The sameness of my starting ratios with yours, and the sameness of the process, does not guarantee the sameness of the result, which makes the process not very useful. For example, let the starting ratios be 1 : 2 & 3 : 4 Then the new ratio, formed by adding the antecedents and adding the consequents, is 4 : 6 But now suppose we start with the same ratios as before, but this time expressed in different numbers, such as 2 : 4 & 9 : 12 Then the new ratio, formed the same way, will be 11 : 16 which is certainly not the same as 4 : 6. But there is another way of producing a new ratio from two given ratios which always gives the same result, regardless of the terms in which the original ratios are expressed. Moreover, this way of combining ratios (which I am about to describe) is important, because in mathematics and in nature there are many interesting truths which involve this way of producing new ratios. Hence this way of producing ratios deserves to be named, and so we name it compounding ratios. The definition of compounding ratios is this: Given any two ratios, a : b & c : d, find another pair of ratios the same as these but in which the consequent of the first ratio is the same as the antecedent in the second, and then form the ratio of the extremes. The new ratio is said to be the compound of the original two ratios. 71

3 For example, given 1 : 2 & 3 : 4 we can take two ratios the same as these in which the consequent of the first ratio will be the same as the antecedent in the second, like this: 3 : 6 & 6 : 8 and the ratio 3 : 8 is said to be the compound of the original two ratios. One notation for this is as follows: 3 : 8 :: (1 : 2) comp (3 : 4) or 3 : 8 = (1 : 2) c (3 : 4) Notice that the compound ratio is the same regardless of the numbers in which the original two ratios are expressed. We could express them thus: 2 : 4 & 9 : 12 and if we go through the process again, compounding these ratios, we can still use 3 : 6 & 6 : 8 and thus get 3 : 8 as the compound ratio. Or, if we like, we can use 18 : 36 & 36 : 48 so that the compound result is 18 : 48, which, simplifying, is the same ratio as 3 : 8. If we are dealing with integers, then there is an easy, always-works method for forming the compound ratio. All we have to do is take the ratio of the product of the original antecedents to the product of the original consequents. For example, given 4 : 5 & 2 : 3 the compound ratio must be (4 2) : (5 3). The reason for this is that we can get the consequent of the first ratio to be the same as the antecedent of the second (which is the key step to compounding) by multiplying the terms of the first ratio by 2 and the terms of the second ratio by 5, giving us 8 : 10 & 10 : 15 But this means the new extremes will be 4 2 and 5 3, and hence their ratio will be the compound of the original ratios, by the definition. 72

4 From this, one can begin to see why people say that compounding ratios is the equivalent of multiplying fractions. If the fractions are ratios of integers, then the new fraction resulting from multiplying the original two will be the same as the compound ratio. For example: = 8 15 and (4 : 5) c (2 : 3) = 8 : 15 I say begin to see rather than just see, however, because it is not clear how to multiply things such as 2 and 3 (our given ratios might consist of such irrational terms), although there is no difficulty compounding ratios with terms such as these. To multiply 2 by 3, one would first have to learn a new definition of multiplication besides taking 2 three times and adding them all up (as one does for 2 3), since there is no clear sense to taking 2 the square-root-of-three-times. (As it turns out, the way to define the multiplication of irrationals will be by completing a proportion among straight lines. So compounding ratios is more basic and intelligible in itself than multiplying fractions, since one can define compounding ratios without thinking of multiplying fractions, but not the reverse, except when the fractions happen to have integral terms. But this is a digression.) Note also that it is possible to compound ratios which are not of comparable things. If the original ratios to be compounded are a : b & c : d, and a : b is a ratio of volumes, while c : d is a ratio of areas, then we can still compound these so long as we can find two other ratios, the same as these, in which the terms are comparable. For instance, suppose we can find a pair of straight lines A : B which have the same ratio as a : b, and we can also find a pair of straight lines : D which have the same ratio as c : d. Then we can compound A : B and : D the normal way, and the resulting ratio is the compound of the original ratios a : b and c : d as well. There is more than one way to present an image of compounding ratios to make it more memorable. One way is by the accompanying diagram. L If our two original ratios are ab : bc & ef : fd, let abc and efd be drawn parallel to one another, and draw any third line L parallel to them also. We will use this line L to form the ratio compounded of the two originals. p a b A B c e q f 73 d

5 Pick some point P not in line with abc and join pa, pb, pc and extend these to points A, B, on L. learly AB : B = ab : bc. Now join Be, f, and extend these until they meet at a point, q. Join qd and extend it till it cuts L at a point D. learly B : D = ef : fd. But by the definition of compounding, AB : D = (AB : B) c (B : D) And therefore, by the sameness of these ratios with the originals, AB : D = (ab : bc) c (ef : fd) 74

6 Another image which helps to understand the compounding of ratios is as follows. Suppose you have two maps, Map 1 and Map 2, and each one depicts roads in the correct ratios (i.e. the same ratios that the roads themselves have in reality), but neither one tells you how much real distance an inch on the map represents. There is no indication of scale, in other words. Now you see that on Map 1 AB : B = 3 : 2 which you determine with a ruler, measuring the relative lengths of the roads as depicted on that map. Again on Map 2 B : D = 12 : 4 an you conclude, then, that AB : D = 3 : 4? By no means. The scales of the maps are clearly different, since one and the same road, road B, is represented as 2 inches on Map 1, but as 12 inches on Map 2. To adjust for this difference of scale, in order to find the ratio of AB : D from the two given ratios, you must compound the ratios: AB : D = (AB : B) c (B : D) [by def. of compounding] so AB : D = (3 : 2) c (12 : 4) so AB : D = 36 : 8 so AB : D = 9 : 2 B D A MAP 1 MAP 2 B 75