Almagest Book IV: Alexandrian Eclipse Triple – Radius of the Epicycle

Continuing on with Ptolemy’s check on the radius of the epicycle, we’ll produce a new diagram based on the positions of the Alexandrian eclipses. However, instead of doing it piece-by-piece as I did when we explored the Babylonian eclipses, I’ll drop everything into a single diagram since we already have some experience and the configuration for this triple is a bit more for forgiving on the spacing:

As with before, the eclipses in order are at A, B, and G.

From the previous post, we determined that $arcAB=110;21º$ of the epicycle and makes a corresponding angle from the center of the deferent ($\mathrm{\angle }ADB$) of $7;42º$. Meanwhile, $arcBG=81;36º$ on the ecliptic and produces $\mathrm{\angle }BGZ=1;21º$.

Again, we can ask why the diagram was set up this way. Ptolemy explains:

It is clear that the apogee must lie on $arcAB$, since it can lie neither on $arcBG$ nor on $arcGA$, both of which produce an additive effect and are less than a semi-circle.

Again looking at the last post, we noted that the motion from A to B produced a rearwards effect along the ecliptic. Hence, B must be to the right of A. Similarly, B to G had a forwards effect, indicating that G must be to the left of B. Lastly, we could determine the angle between G and A by subtracting $\mathrm{\angle }BDG–\mathrm{\angle }GDA$ and we would find that from A to G should also produce a rearwards effect indicating G should be to the right of A.

The only way we can make these statements true is if apogee lies on $arcAB$. If we rotated it such that A was to the right of apogee, this would force G to be to the left of A. Similarly, if we rotated things so that B was to the left of apogee, then G would be to the right of B.

So with those three points in their relative positions, we again draw lines from D to each, producing point E where $\overline{AD}$ intersects the epicycle. We then connect point E to points B and G. Perpendiculars are dropped from point E onto $\overline{BD}$ and $\overline{GD}$ producing points Z and H respectively and making right triangles $\mathrm{△}BEZ$, $\mathrm{△}EZD$, and $\mathrm{△}EHD$. Lastly a perpendicular is dropped from point G onto $\overline{BE}$ producing point Θ and forming right $\mathrm{△}EG\mathrm{\Theta }$.

With all of that out of the way, back to the math!

As with when we did this for the Babylonian eclipses, we’ll be putting everything in the context of a circle drawn about E, Z, and D. And as we saw last time, point H is also on that circle, so let’s just call it $circleEHZD$ from the outset this time.

In this circle, we have $\mathrm{\angle }EDZ=7;42º$. Thus, in this circle where D is on the circumference, the arc opposite it, $arcEZ=15;24º$. The corresponding chord, $\overline{EZ}=16;4,{42}^p$.

Turning our attention to the epicycle, $arcAB=110;21º$. So the angle it subtends on the circumference of the epicycle, $\mathrm{\angle }AEB=55;11º$¹. We can then take the supplementary angle along $\overline{AD}$ to determine

$$\mathrm{\angle }DEB=180º–55;11º=124;49º$$

Now looking at $\mathrm{△}DEB$ we now know two of the angles ($\mathrm{\angle }DEB$ and $\mathrm{\angle }EDB$) so we can subtract those from 180º to say

$$\mathrm{\angle }EBD=180º–124;49º–7;42º=47;29º$$

Next, we’ll create a circle around B, E, and Z to form $circleBEZ$. On this circle, we have $\mathrm{\angle }EBZ$ which we just found, so the arc opposite it is twice of that which is to say, $arcEZ=94;58º$. Looking up the chord, $\overline{EZ}=88;26,{17}^p$, again in the context of $circleBEZ$.

But again, we already determined $\overline{EZ}$ in the context of our circle we’ll convert everything to, $circleEHZD$. This allows us to set up the conversion ratios, to solve for $\overline{BE}$, which was the hypotenuse in $circleBEZ$, in the context of this circle:

$$\frac{\overline{BE}}{{120}^p}=\frac{16;4,{42}^p}{88;26,{17}^p}$$

$$\overline{BE}={120}^p\cdot \frac{16;4,{42}^p}{88;26,{17}^p}=21;48,{59}^p$$

While we’re looking at the ecliptic, we can notice that

$$\mathrm{\angle }ADG=\mathrm{\angle }ADB–\mathrm{\angle }GDZ=7;42º–1;21º=6;21º$$

So in our $circleEGZD$, $arcEH$ is opposite that meaning it will have twice the measure, so $arcEH=12;42º$ allowing us to head to the chord table to determine $\overline{EH}=13;16,{19}^p$, again in the context of $circleEGZD$.

Now back to the epicycle. Here,

$$arcABG=arcAB+arcBG=110;21º+81;36º=191;57º$$

Thus, $\mathrm{\angle }AEG$, being on the circumference is half that measure, or $95;59º$. We can then determine $\mathrm{\angle }DEG=84;01º$ since it’s the supplement along $\overline{AD}$. That allows for us to solve $\mathrm{△}EGD$ for $\mathrm{\angle }EGD$:

$$\mathrm{\angle }EGD=180º–6;21º–84;01º=89;38º$$

Now we’ll create $circleEGH$ since this puts the angle we just found on the circumference. Thus, in the context of this circle, $arcEH$ is twice that angle or $179;16º$. The arc that subtends this angle would then be $\overline{EH}=119;59,{50}^p$ in the context of this circle.

But we want to convert back to $circleEHZD$. Fortunately, we’ve around found $\overline{EH}$ in that context, so we can now set up the conversion ratio to help us find $\overline{GE}$:

$$\frac{\overline{GE}}{{120}^p}=\frac{13;16,{19}^p}{119;59,{50}^p}$$

$$\overline{GE}={120}^p\cdot \frac{13;16,{19}^p}{119;59,{50}^p}=13;16,{20}^p$$

Recapping, we’ve now figured out $\overline{EB}$ and $\overline{GE}$ in the context of $circleEHZD$.

Again, we’ll return to the epicycle where $arcBG=81;36º$. Thus, $\mathrm{\angle }BEG=40;48º$.

Next, we’ll draw $circleEG\mathrm{\Theta }$. In the context of this circle, $arcG\mathrm{\Theta }=81;36º$ and the chord, $\overline{G\mathrm{\Theta }}=78;24,{37}^p$.

We can also determine

$$\mathrm{\angle }EG\mathrm{\Theta }=180º–90º–40;48º=49;12º$$

Therefore the arc opposite this angle, $arcE\mathrm{\Theta }=98;24º$ and its chord, $\overline{E\mathrm{\Theta }}=90;50,{22}^p$.

Now we’ll switch the context from this circle back to $circleEHZD$ by using the chord we’ve found common to both $\overline{EG}$². First let’s do $\overline{G\mathrm{\Theta }}$:

$$\frac{\overline{G\mathrm{\Theta }}}{78;24,{37}^p}=\frac{13;16,{20}^p}{{120}^p}$$

$$\overline{G\mathrm{\Theta }}=78;24,{37}^p\cdot \frac{13;16,{20}^p}{{120}^p}=8;40,{20}^p$$

Repeating for $\overline{E\mathrm{\Theta }}$:

$$\frac{\overline{E\mathrm{\Theta }}}{90;50,{22}^p}=\frac{13;16,{20}^p}{{120}^p}$$

$$\overline{E\mathrm{\Theta }}=90;50,{22}^p\cdot \frac{13;16,{20}^p}{{120}^p}=10;2,{49}^p$$

Since everything is now in the context of $circleEHZD$, we can now subtract

$$\overline{B\mathrm{\Theta }}=\overline{BE}–\overline{E\mathrm{\Theta }}=21;48,{59}^p–10;2,{49}^p=11;46,{10}^p$$

Now we’ll check out $circleBG\mathrm{\Theta }$. We’ve determined two of the sides of this right triangle ($\overline{B\mathrm{\Theta }}$ and $\overline{G\mathrm{\Theta }}$), so we can use the Pythagorean theorem to solve for the third ($\overline{BG}$)³:

$$\overline{BG}=\sqrt{{11;46,{10}^p}^2+{8;40,{20}^p}^2}=14;37,{12}^p$$

This is the chord we’ll use to finally switch context to the epicycle. For, in the epicycle, $arcBG=81;36º$ and thus its chord, $\overline{BG}=78;24,{37}^p$.

So we’ll convert $\overline{DE}$ using another set of conversion ratios:

$$\frac{\overline{DE}}{{120}^p}=\frac{78;24,{37}^p}{14;37,{12}^p}$$

$$\overline{DE}={120}^p\cdot \frac{78;24,{37}^p}{14;37,{12}^p}=643;35,{11}^p$$

Next, we’ll do the same for $\overline{GE}$⁴:

$$\frac{\overline{GE}}{13;16,{20}^p}=\frac{78;24,{37}^p}{14;37,{12}^p}$$

$$\overline{GE}=13;16,{20}^p\cdot \frac{78;24,{37}^p}{14;37,{12}^p}=71;11,{55}^p$$

From this chord, now in the context of the epicycle, we can determine the corresponding arc, $arcGE=72;47,10º$⁵.

We’ve already shown that $arcABG=191;57º$, so the remainder, $arcGA$ (going clockwise) will be $168;3º$.  From this, we can subtract, $arcGE$ to determine $arcEA=95;15,50º$. Again converting to the corresponding chord, $\overline{EA}=88;39,{35}^p$.

As with the last set of eclipses, Ptolemy pauses here to note $arcAE$ wasn’t a semi circle and $\overline{EA}$ wasn’t ${120}^p$ so the center of the epicycle isn’t falling on this line, indicating this A and E weren’t at apogee/perigee. As such, we’ll need to redraw our diagram, now including apogee (L), perigee (M), and the center of the epicycle (K).

 

We’ll again create the equal products based on Euclid’s Elements III.36, but this time we’ll chose the first eclipse at A instead of point B:

$$\overline{AD}\cdot \overline{DE}=\overline{LD}\cdot \overline{DM}$$

Here, we can calculate $\overline{AD}$:

$$\overline{AD}=\overline{AE}+\overline{DE}=88;39,{35}^p+643;35,{11}^p=732;14,{46}^p$$

This gives us everything we need for the left side of this equation. So we can rewrite⁶:

$$732;14,{46}^p\cdot 643;35,{11}^p=471263;37,{49}^p=\overline{LD}\cdot \overline{DM}$$

And again based on Elements II.6 we can write:

$$\overline{LD}\cdot \overline{DM}+{\overline{KM}}^2={\overline{DK}}^2$$

So plugging in and rearranging:

$${\overline{DK}}^2=471263;37,{49}^p+{{60}^p}^2=474863;37,{49}^p$$

$$\overline{DK}=689;6,{12}^p$$

And now we’ll do one last context swap to determine the radius of the epicycle on the context of the deferent:

$$\frac{\overline{KM}}{{60}^p}=\frac{{60}^p}{689;6,{12}^p}$$

$$\overline{KM}={60}^p\cdot \frac{{60}^p}{689;6,{12}^p}=5;13,{27}^p$$

Ptolemy’s value ends up being higher by just enough that he justifies rounding this up to $5;{14}^p$, but either way, this is in excellent agreement with the radius we found from the Babylonian eclipse triple.

So I’ll leave things here, and again save the equation of anomaly and position of the mean moon for the next post!

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I think today may have been a record day for progress. The four posts that went up today covered almost 8 pages of material! The first one was definitely the hardest and I’d been working on it for a few days, but from these four posts, that was 1.11% of the Almagest right there! However, it was also a marathon day, spending somewhere around 12 hours working today.

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1. If you’re following along from Toomer’s translation, he starts switching into demi-degrees. Since this is unnecessary, I’m not going to follow along, but this will result in some minor differences in the values due to rounding differences.
2. Which was ${120}^p$ since it’s the diameter.
3. Ptolemy comes up with a slightly different value of $14;37,{10}^p$ which isn’t a lot here, but does start making a difference when we calculate $\overline{DE}$ in the context of the epicycle here in a moment, by $0;0,{32}^p$! However, since we’ll be rounding to only the first division in the end, this still won’t make much of a difference.
4. Due to the aforementioned minor difference, my value for $\overline{GE}$ is now going to be $0;0,{45}^p$ different than Ptolemy’s.
5. Given my values are fairly discrepant from Ptolemy’s now, I couldn’t rationalize adopting his value for this arc, and didn’t feel like looking it up in the chord table, so for full transparency, this is based on modern trig and we’re a full $0;1º$ off from Ptolemy’s values now.
6. Due to the large multiplication, the small variances I’ve been picking up from Ptolemy are getting pretty pronounced, but we’ll be taking square roots shortly so they’ll get quite small again.