In the last post, we introduced three eclipses from Babylonian times which we used to build a couple intervals: The first eclipse to the second, and the second to the third. Using those, we used lunar and solar mean motion tables to figure out the solar position, as well as its change. Since the moon must be opposite the sun in ecliptic longitude for an eclipse to occur, we used the change in solar position to determine the true change in lunar position in these intervals. From that, we could compare that to the mean motion to determine how much of it must be caused by the lunar anomaly. But while we’ve determined this component, we haven’t done anything with them yet.
So in this post, we’ll start using these to answer several questions that will build out the details of the model. Specifically, we want answers to questions like what is the radius of the epicycle? Where, in relation to the ecliptic was the mean moon during these eclipses and what was the equation of anomaly? That’s a lot of information to extract so I’m going to try to break it up a bit and in this post, we’ll only tackle the radius of the epicycle1
To begin, let’s sketch out the epicyclic lunar model2 with the three eclipses drawn on it.
Here we’ve drawn out positions for the three eclipses on the epicycle with the first eclipse being A, the second as B, and the 3rd as G3. It’s important to note that three of these eclipses would not happen with the epicycle in this same position which is obvious since in the last post we saw that the sun was in three very different positions during each. So what we’ve done in this diagram is taken the three different snapshots and stacked them with the centers of the epicycles overlapped so we can easily work on the geometry.
But why did we put them in this order?
Returning to the figures from the last post, we stated that between the first eclipse (A) and the second (B), moved 306;25º around the epicycle4. This means that $arc \; AGB = 306;25º$. In addition, we showed that the change along the ecliptic relative to the mean motion was 3;24º. We won’t use that right now, but keep it in mind.
Similarly, the motion along the epicycle between the second (B) and third (G) eclipse we found to be 150;26º, so $arc \; BAG = 150;26º$. The amount of motion this made in addition to the mean motion for this interval was 0;37º.
Turning to the first of those, we can quickly determine $arc \; BA = 53;35º$ because it is a full circle minus $arc \; AGB$. If we subtract that from $arc \; BAG$ we can determine $arc \; AG = 96;51º$.
Staying on the first eclipse to the second (A to B), in the last post we showed that the moon’s apparent motion was 349;15º but the component of that due to the mean motion was only 345;51º which meant that 3;24º of that motion must have been due to the anomaly.
What does that mean? To visualize, let’s add a few more lines to the diagram from the observer at D.
What this would imply is that the motion of the moon from A to B, should increase5 the observed position along the ecliptic by 3;24º. Thus, $\angle{ADB} = 3;24º$.
Similarly, from the second to the third eclipse, we saw the motion due to the anomaly was 0;37º6. This would similarly imply that $\angle{BDG} = 0;37º$.
If we take the difference, we can determine that $\angle {ADG} = 2;47º$.
Ptolemy’s next statement is rather tricky. He states:
[T]he perigee of the epicycle cannot lie on $arc \; BAG$. This is clear because this arc has an subtractive effect, and is less than a semi-circle, while the greatest speed occurs at perigee.
This explanation is a bit fuzzy, but consider it this way. Starting from point B and going clockwise is there any point on $arc \; BAG$ that you could choose which would produce an increase in ecliptic longitude? Since ecliptic longitude is measured counterclockwise7 and all these points would necessarily lie to the right of B, the answer is no.
You may wonder whether this was simply a result of how I arranged these three points along the ecliptic?
When I drew the diagram, I did take care to get $arc \; BA$ and $arc\; AG$ correct, but I could have rotated it around the center of the epicycle and kept those proportions the same. However, if I rotated it clockwise, $\angle \; BDG$ could cross the point of perigee8, but in doing so, it would have the result of making $\angle{BDG}$ positive instead of the negative we just showed it to be.
I’ll pause here a moment to note that there is a flaw in this drawing such that $\angle{BDG}$ is drawn pretty close to $\angle{GDA}$ which it should be about a fourth of the size. However, as drawn here, it would be impossible to get right because I don’t have the correct ratio of of the epicycle to deferent. I’ve drawn them only a factor of ~3 different which is far too small, as we’ll see. So for now, we’ll just leave the diagram as it is since it gets the relative positions correct.
Jumping back into the flow of things, I hope I’ve convinced you that there’s no way we can get any part of $arc \; BAG$ to lie at perigee. There’s just no way it would let the angles between the eclipses work out from the position of the observer. This means that perigee must lie somewhere on $arc \; GB$ (going clockwise).
Next Ptolemy fleshes out the diagram a bit more. Specifically, he draws in a new point, E, on $\overline{BD}$ where it intersects the circle of the epicycle and connects this to point A as well as connecting point A to point G. To help keep things clean, I’m temporarily going to redraw this, enlarged, but with point D off the edge of the figure to help keep the part we’re most worried out clear9:
Now, we’ll drop in a line from E onto $\overline{AD}$ such that it’s perpendicular. We’ll call the new point Z.
When you start seeing Ptolemy dropping these perpendiculars, this is a pretty good indication that we’ll be using some demi-degrees since the technique requires right angled triangles inscribed in circles.
We’ll begin by drawing $circle \; {DEZ}$.
In this circle, $\angle{EDZ} = \angle{BDA} = 3;24º$. However, as we’ve seen before with the demi-degrees method, when we convert to demi-degrees, the angle on the circumference gets doubled. As such, $\angle{EDZ} = 6;48ºº$ which is equal to the arc subtending it, $arc \; EZ$, in regular degrees. We could look this up in the table of chords to determine that, in the context of this circle, $\overline{EZ} = 7;7,0^p$, in the context of $circle \; DEZ$
We’ll pause there and turn our attention the the epicycle:
Here, $arc \; BA = 53;35º$. Thus, the angle on the circumference should be half that, but again gets doubled when we switch to the demi-degrees context. Thus, we can say $\angle{BEA} = 53;35ºº$.
Now Ptolemy does something we haven’t seen before. We’re going to temporarily combine contexts to work out some geometry, and then we’ll use the ratio of some lines in each context to help sort it out afterwards.
To illustrate, I’ve highlighted a few triangles:
Here, our first objective will be to solve for $\angle{EAZ}$ in demi-degrees. Once we’ve done that, we can then use another demi-degrees method to solve for $\overline{EZ}$ in a circle about those points. Then comparing the length of this line in each context will allow us to swap between them, translating the pieces we’ve found in this triangle to the context of $circle \; DEZ$.
To get there, we’ll start off in $\triangle DEZ$ in which we can find $\angle{DEZ}$ since we know the other two angles. Thus, $\angle{DEZ} = 180ºº – 90ºº – 6;48ºº = 83;12ºº$10.
We can also state that $\angle{AEZ} = 180ºº – 53;35ºº – 83;12ºº = 43;13ºº$ since the angles along straight line $\overline{BED}$ also add up to 180ºº.
Now we can solve $\triangle{EAZ}$ to state that $\angle{EAZ} = 180ºº – 90ºº – 43;13ºº = 46;47ºº$.
Lastly, we’ll pull ourselves out of the demi-degrees context with another circle:
Here, $\angle{EAZ} = 46;47ºº$, but as we’ve seen from doing the demi-degrees previously, this means that the arc opposite the angle will have the same measure in regular degrees. Thus, $arc \; EZ = 46;47º$ in the context of $circle \; AEZ$.
We can look up that arc in the table of chords to find $\overline{EZ} = 47;38,30^p$, still in the context of this small circle, $circle \; EZA$ where $\overline{EA} = 120^p$. However, this circle isn’t really all that useful. Instead, Ptolemy wants to put it in the context of circle $EZD$ because we’re about to go through this same set of steps again for another line (which we haven’t drawn in yet), and circle $EZD$ will be common to both methods, so it gives a sort of baseline for comparison which we’ll then do a similar join to the epicycle itself, allowing us to solve for what we’re really after.
To do this conversion, Ptolemy looks at $\overline{EZ}$ which we’ve already found in both contexts. In the context of $circle \; AEZ$ it’s $47;38,30^p$, and in the context of $circle \; DEZ$, it’s $7;7^p$. Thus, the ratio of these can be used to convert contexts. In other words, this same ratio can be applied to other lines in $\triangle{AEZ}$ to determine what their length would be in the context of $circle \; DEZ$. Setting this up for $\overline{EA}$:
$$\frac{7;7^p}{47;38,30^p} = \frac{\overline{EA}}{120^p}$$
Solving that we get $\overline{EA} = 17;55,32^p$ in the context of $circle \; DEZ$.
That takes us through this method for the first eclipse pair we’d considered which included the first eclipse (A) and the second (B). Now we’ll repeat this for the next eclipse pair we created in the last post which was the second (B) to the third (G)11. But as with before, we’ll need to create some new right triangles for this to work on, by extending a line from E so that it is perpendicular to $\overline{DG}$ at point H. In addition, we’ll connect point E to point G forming $\overline{EG}$.
I’ll skip drawing in all the circles again, since we just saw the methodology above. The first circle would be circle $EDH$. Since we showed in the last post that $\angle{EDH} = 0;37º$ and this angle is on the circumference, we double it to convert it to demi-degrees. Thus, $\angle{EDH} = 1;14ºº$.
As we stated above, this is then equal to the arc opposite the angle in regular degrees. In other words, $arc \; EH = 1;14º$. This can then be used to find the corresponding chord from the chord table to state, $\overline{EH} = 1;17,30^p$ in the context of circle $EDH$.
At this point, we need to take a time out to think about $circle \; EDH$. It’s actually the same circle as $circle \; DEZ$ above. We know that this is the same circle because of Thales’ Theorem. It states that, if a right angle lies on the circumference of the circle, the hypotenuse must then be the diameter. Since $\angle{EHD}$ and $\angle{EZD}$ are both right angles and have the same hypotenuse, $\overline{ED}$, this means that both circles have the same diameter. Which is the same thing as saying they’re the same circle. So to help remind us of this going forward, I’ll refer to this as $circle \; EHZD$.
Next, we’ll again use the circle of the epicycle to consider $\angle{BEG}$. In the last post, we determined that $arc \; BAG = 150;26º$. And following with the pattern of the demi-degrees method, this means that the angle it subtends must be equal in measure in demi-degrees. On other words, $\angle{BEG} = 150;26ºº$.
Now let’s focus on $\triangle{DEG}$. At this point, we have everything necessary to solve it in the demi-degrees context. We just stated above that $\angle{EDH} = 1;14ºº$. So we just need to find $\angle{DEG}$, which is straightforward since it, plus $\angle{BEG}$ is $180ºº$. And since we knew $\angle{BEG} = 150;26ºº$, this means $\angle{DEG} = 29;34ºº$.
That gives us two of the three angles in the triangle, which again add up to 180ºº, so subtracting out $\angle{DEG}$ and $\angle{EDG}$ we’re left with $\angle{EGD} = 149;12ºº$.
Next, a tiny circle around E, G, and H. Following the rationale of the demi-degrees method, $arc \; EH = 149;12º$ since it’s opposite the angle. And looking at the corresponding chord, $\overline{EH} = 115;41,21^p$12 in the context of this tiny circle.
However, in the context of the $circle \; EHZD$, we showed above that $\overline{EH} = 1;17,30^p$. So again, we’ll set up an inequality to determine the length of $\overline{EG}$ in the context of circle $EHZD$.
$$\frac{\overline{EG}}{120^p} = \frac{1;17,30^p}{115;41,21^p}$$
We can solve this to get $\overline{EG} = 1;20,23^p$ in the context of circle $EHZD$.
So now, we’ve solved for $\overline{EG}$ and $\overline{EA}$ in the context of this circle. And now we’ll do it again. This time for the first eclipse (A) to the third (G). As before, we’ll add a new line. This time, we’ll extend a line from G so that it’s perpendicular to $\overline{AE}$ which gives us a few more right triangles to play with.
With that done, let’s start with the length of $arc \; AG$ which is $arc \; BG – arc \; BA$. So
$$arc \; AG = 150;26º – 53;35º = 96;51º$$
Next up, we’ll start converting things to demi-degrees, this time beginning with the epicycle. There, $\angle AEG$ is subtended by $arc \; AG$ and thus, is equal, in demi-degrees, to the arc in regular degrees. So $\angle AEG = 96;51ºº$.
Ptolemy now draws a small circle around points Θ, E, and G and looks at $arc \; G \Theta$ on this circle, which is opposite $\angle AEG$. And again, the arc in regular degrees is equal to the angle it subtends in demi-degrees. Thus, $arc \; G \Theta = 96;51º$ as well.
Continuing with this small circle, $\overline{GE}$ is the diameter, which means $arc \; EG = 180º$ so we can subtract $arc \; G \Theta$ out to determine $arc \; E \Theta = 83;9º$.
Now we can look up some corresponding chords from the chord table. This gives us:
$$\overline{G \Theta} = 89;46,14^p$$
$$\overline{E \Theta} = 79;37,55^p$$
Both of these being in the context of the small circle we just defined. But we want to convert these to the context of the $circle \; EHZD$. To do so, we again need a line that’s we have solved for in both contexts. We have this in $\overline{GE}$ since we already determined $\overline{GE} = 1;20,23^p$ in the context we’re looking for and it is $120^p$ in the context of the small circle.
Thus, we can set up our ratio to solve for $\overline{G \Theta}$:
$$\frac{1;20,23^p}{120^p} = \frac{\overline{G \Theta}}{89;46,14^p}$$
Solving:
$$\overline{G \Theta} = 89;46,14^p \cdot \frac{1;20,23^p}{120^p} = 1;0,8^p$$
Doing the same for $\overline{E \Theta}$:
$$\overline{E \Theta} = 79;37,55^p \cdot \frac{1;20,23^p}{120^p} = 0;53,21^p$$
So recapping: we’ve now found $\overline{EG}$, $\overline{EA}$, $\overline{G \Theta}$, and $\overline{E \Theta}$ all in the same context of $circle \; EHZD$. We’ll stay in this context for a moment to solve for a few other pieces.
First, we can subtract $\overline{E \Theta}$ from $\overline{AE}$ to determine
$$\overline{A \Theta} = 17;55,32^p – 0;53,21^p = 17;2,11^p$$
We can then solve for $\overline{AG}$ because it’s part of right triangle, $\triangle{AG \Theta}$. We have two of the three sides, so we can use the Pythagorean theorem:
$$\overline{AG}^2 = \overline{A \Theta}^2 + \overline{G \Theta}^2$$
$$\overline{AG} = \sqrt{17;2,11^2 + 1;0,8^2} = \sqrt{291;14,35^p} = 17;3,57^p$$
We’ll hold that thought for a moment and recall that we also know that $arc \; AG = 96;51º$ which means that the chord, $\overline{AG} = 89;46,14^p$ in the context of the epicycle. This gives us another context for $\overline{AG}$ which means that we can now convert everything we’ve solved for in the context of circle $EHZD$ to the context of the epicycle!
We’ll start with $\overline{DE}$. In the context of $circle \; EHZD$ it was the diameter, so we can write the relationship:
$$\frac{\overline{DE}}{120^p} = \frac{89;46,14^p}{17;3,57^p} $$
Solving:
$$\overline{DE} = 120^p \cdot \frac{89;46,14^p}{17;3,57^p} = 631;13,48^p$$
We can do the same for $\overline{GE}$:
$$\frac{\overline{GE}}{1;20,23^p} = \frac{89;46,14^p}{17;3,57^p} $$
$$\overline{GE} = 1;20,23^p \cdot \frac{89;46,14^p}{17;3,57^p} = 7;2,50^p$$
Since we’re in the context of the epicycle, we can now swap that back into the arc length on the epicycle itself to determine $arc \; GE = 6;44,1º$. We can add that to $arc \; BAG$ which we showed previously to be 150;26º to get the arc length from the second eclipse (B) to perigee to be 157;10,1º.
Ptolemy then pauses briefly to consider what this means. Since this isn’t 180º, and thus its chord length isn’t $120^p$, it means that the second eclipse must not have been at apogee. If it was, then we’d basically be done as this would mean that $\overline {EB}$ was the diameter of the epicycle, and we could immediately determine the ratio of the epicycle to that of the deferrent. But it’s not, so we’re close, but not quite there.
To find the diameter of the epicycle, Ptolemy finally introduces a few points that are on the line defined by the center of the deferent through the center of the epicycle:
Now, point M represents perigee and point L apogee.
This allows us to state that:
$$\overline{BD} \cdot \overline{DE} = \overline{LD} \cdot \overline{DM}$$
What Ptolemy is stating here is that, if we have a line from a point external to a circle that cuts through the circle to the point on the far side, the product of the full line to the segment exterior to the circle is always equal. This comes indirectly from Euclid’s Elements III.36, which states that this product for any single line as outlined above is equal to the square of the length of the tangent to that circle. But since the square of the tangent will be the same for any such setup, as long as they share the same external point, we can use the transitive property to ultimately equate the product for two lines directly.
So let’s see how we can use this equation. We’re working in the context of the epicycle right now where $arc \; BE = 157;10,1º$ so it’s chord, $\overline{BE} = 117;37,32^p$ . In this context, we’ve also determined $\overline{ED} = 631;13,48^p$ so we can add these together to get $\overline{BD} = 748;51,20^p$.
That gives us the product:
$$\overline{BD} \cdot \overline{DE} = 748;51,20^p \cdot 631;13,48^p = 472700;5,32^p$$
Ptolemy now digs out another one of Euclid’s Elements. This time, II.6, which allows him to state:
$$\overline{LD} \cdot \overline{DM} + \overline {KM}^2 = \overline{DK}^2$$
What Euclid is saying here is that if a line ($\overline{LM}$) is bisected (which it is at point K since that’s the center) and another line ($\overline{DM}$) is added to it, then the product of the whole line ($\overline{LD}$) times the line added on ($\overline{DM}$) plus the square the square of half the bisected line (which could be either $\overline{LK}$ or $\overline{KM}$) is equal to the square of the sum of the added line ($\overline{DM}$) plus half of the bisected line (again, $\overline{LK}$ or $\overline{KM}$).
This gives us another equation involving the product $\overline{LD} \cdot \overline{DM}$ which we stated previously was equal to $\overline{BD} \cdot \overline{DE}$ which we just determined to be $472700;5,32^p$.
In addition, $\overline {KM}$ is a radius in the context we’re currently working so it’s $60^p$. So we can plug that into the equation too to get:
$$472700;5,32^p + {60^p}^2 = 476300;5,32^p = \overline{DK}^2$$
We can then take the square root of that to determine
$$\overline{DK} = 690;8;42^p$$
Again, we’re still in the context of the epicycle, where $\overline{KM} = 60^p$. Just doing a bit of back of the envelope math (which Ptolemy doesn’t do), this would mean that the radius of the deferent is about 11.5 times the size of the epicycle.
But what we really want to get to is the size of things in the context of the deferent where $\overline{DK} = 60^p$, so let’s context switch one last time to determine the radius of the epicycle in that context:
$$\frac{\overline{KM}}{60^p} = \frac{60^p}{690;8,42^p}$$
$$\overline{KM} = 60^p \cdot \frac{60^p}{690;8,42^p} = 5;13^p$$
Thus, we’ve answered one of the big questions we set out at the beginning of this post: What is the radius of the epicycle? It’s $5;13^p$ when the radius of the deferent is $60^p$.
That was a ton of work for this one value. Fortunately, in deriving this, we’ve calculated a bunch of values that will be useful going forward and thus, for the answers to these other questions, we will be able to make use of them and the next post shall be relatively short. In it, we’ll be determining the equation of anomaly which we can then use to determine the position of the mean moon.
- Just walking through that first part of the proof already makes this one of the longer posts I’ve yet done.
- Ptolemy draws out the eccentric model as well and notes that the same general proof can be done with it. But Ptolemy solidly works in the epicyclic model here, so I will as well.
- In the past, I’ve always drawn in the full deferent. Here, I’ve obviously only drawn in half of it to help the scale a bit.
- Recall that Ptolemy has stated for this model that the motion on the epicycle is clockwise.
- Keeping in mind that positive directions are counter-clockwise.
- In the last post we actually determined this as being negative since the motion along the ecliptic is clockwise from B to G. But if we are just careful to consider this as we go, we can do away with the negatives for now as I noted in the footnote in the previous post.
- With respect to the ecliptic.
- Or at least more leftwards than point B would then be.
- I’m also dropping the deferent at this point as it’s not necessary.
- I should be clear that I’m following Ptolemy’s use of demi-degrees here. Or, more accurately, Toomer’s translation as Ptolemy didn’t actually call things demi-degrees or use this notation. Regardless, I’m making special note that I’m following Ptolemy’s method because constantly switching to demi-degrees seems to me to be absolutely unnecessary. We’ll be repeating this exact same set of procedures for another set of three eclipses in a few posts, in which I’ll skip the demi-degrees all together, although we will still have to be careful about the context of which circle we’re working in. I’m still going to go through the unnecessary math here to make it easier for readers using this blog as a supplement to Toomer to follow along. If you’re doing so, you might try reading that post first as I believe it to be a much cleaner presentation. However, I go through things much faster and reference some things we learned in this post, so you may still find yourself a bit lost.
- This brings to mind the phrase I often hear in SCA dancing: “We had so much fun with that, we’ll do it again.”
- Toomer notes that this should actually be $115;41,24^p$, but this is such a small difference, it’s not going to matter by the end of all this since we’ll be rounding off.