Ptolemy starts the chapter reminding us that we found distance to the moon at its maximum distance was $64;10^p$. This was comprised of the distance from earth the mean distance at syzygy being $59^p$ and the radius of the epicycle being an addition $5;10$. Using that,
let us see the size of the sun’s distance which results.
To get started, let’s look at the following diagram1:
Here, we’ve drawn out the various bodies all lined up for the situation of a solar eclipse2 wherein all three bodies lie in the same plane. The sun is the largest circle at the top, the moon, the small one in the middle, and the earth as the lowest.
The lines connecting the various bodies are all drawn such that they are tangent which has the effect of meaning that the horizontal lines through each of the bodies are parallel but do not quite lie through the centers of their respective bodies3. The effect is exaggerated due to scale here because at large distances they would be much closer to the true diameter so Ptolemy calls them “sensibly equal to diameters.”
Another thing this means is that, in this situation, the sun and moon would have precisely the same angular diameter since they fit in the same cone. Recall that Ptolemy previously stated4 that this only happens at the moon’s maximum distance.
Lastly, we’ve drawn in another line, $\overline{OPR}$ which is the same distance from the center of the earth as $\overline{E \Theta HS}$ and thus point $P$ represents the distance of the moon at its furthest distance as well. This is a quantity which we established previously so we can quickly note that that $\overline{\Theta N} = \overline{NP} = 64;10^p$ where $\overline{NL} = 1^p$. The overall line, $\overline{OR}$ is the width of the earth’s shadow at that distance.
We have also previously derived
that the moon’s diameter at the distance in question, namely the greatest distance at syzygies, subtends $0;31,20º$
which tells us that $\angle{ENH} = 0;31,20º$. Now, we’ll divide that angle in half and instead look at $\angle{\Theta NH}$ which would then be $0;15,40º$. And that’s part of the right triangle, $\triangle \Theta NH$ which we will apply a demi-degrees circle about in which the arc subtending that angle, $arc \; \Theta H$ would have double the measure bringing us right back to $arc \; \Theta H = 0;31,20º$. We can then quickly determine that $arc \; \Theta N = 179;28,40º$ as it’s the supplement.
Next, we determine the length of the chords in that context and get $\overline{H \Theta} = 0;32,48^p$ and $\overline{N \Theta} \approx 120^p$.
Since we know $\overline{N \Theta}$ in the context of the larger drawing, we can use this to determine the length of $\overline{H \Theta}$ in that same context as $0;17,33^p$.
Another thing we determined in that same post is the ratio of the radius of earth’s shadow to the radius of the moon. In terms of this diagram that would be $\frac{\overline{PR}}{\overline{\Theta H}}$ and we determined that ratio to be $2;36$. From this, knowing $\overline{H \Theta}$, we can determine $\overline{PR} = 0;45,38^p$.
I’m going to pause here because Ptolemy’s next step is a bit confusing unless we do another short proof. Consider the following right triangle:
This triangle has a base, $C$, and then some distance above it, there’s a line, $B$ parallel to $C$. At twice that distance, there’s another line, $A$, again parallel. I’ve dropped some vertical lines from them to help highlight two smaller triangles:
These two triangles I’ve highlighted are both similar to the overall triangle5, but in addition, are also equal to one another which we know because those verticals I dropped are the same length. That means that the horizontal pieces of each of those highlighted triangles is also equal. Let’s call it $x$ and take $B$ as the “default” length. That would mean $A = B – x$ and $C = B + x$. If we solve each of those for $x$ and then set them equal:
$$x = B – A$$
$$x = C – B$$
$$B – A = C – B$$
Rearranging a bit:
$$2B = C + A$$
That’s what Ptolemy is going to make use of here. In this case, the triangle is upside-down, $\overline{PR} = A$, $\overline{NM} = B$, and $\overline{\Theta S} = C$. Right now, we already know the length of $\overline{PR} = 0;45,38^p$ and $\overline{NM} = 1$6. Thus,
$$2 = 0;45,38 + \overline{\Theta S}$$
$$\overline{\Theta S} = 1;14,22^p$$
We can then subtract out $\overline{\Theta H}$ from that to determine $\overline{HS} = 0;56,49^p$7
Now let’s consider a few other triangles, starting with $\triangle{NMG}$ and $\triangle{HSG}$. Again, we have a triangle inside a triangle with the bases parallel, so these are again similar triangles. This allows us to relate their respective pieces as ratios and then equate them:
$$\frac{\overline{NM}}{\overline{HS}} = \frac{\overline{NG}}{\overline{HG}}$$
And we could do the same thing with $\triangle{NDG}$ and $\triangle{N \Theta H}$ which gives us the ratio relationships of:
$$\frac{\overline{ND}}{\overline{\Theta D}} = \frac{\overline{NG}}{\overline{HG}}$$
But notes that the latter of both of these is the same. So we can ultimately relate all three:
$$\frac{\overline{NM}}{\overline{HS}} = \frac{\overline{NG}}{\overline{HG}} = \frac{\overline{ND}}{\overline{\Theta D}}$$
In our current context, $\overline{NM} = 1^p$ and $\overline{HS} = 0;56,49^p$, so we can plug that into our inequality and regardless of what context we switch to, that ratio will remain the same.
So Ptolemy sets up another context in which the distance from the center of the earth to the center of the sun, $\overline{ND} = 1^p$. In that context, $\overline{D \Theta}$, the distance from the moon at its maximum distance from earth, would be $0;56,49^p$. Doing a bit of subtraction, we can then state $\overline{\Theta N}$, the distance from the earth to the moon at its greatest distance, would be $0;3,11^p$. However, in the context, where that same segment, $\overline{\Theta N} = 64;10^p$ that would mean that the distance between the sun and earth, $\overline{ND} \approx 1210^p$8.
Next, Ptolemy turns his attention towards finding the distance from the center of the earth, to the tip of the shadow cone at $X$, by again creating some ratios of related parts of similar triangles. This time,
$$\frac{\overline{NM}}{\overline{PR}} = \frac{\overline{NX}}{\overline{PX}}$$
If we temporarily assign $\overline{NX}$ a value of $1^p$, we can then determine the length of $\overline{PX} = 45;38^p$ in that context. Again subtracting we determine that $\overline{PN} = 0;14,22^p$.
Thus, context switching back to the context in which $\overline{PN} = 64;10^p$ and $\overline{NM} = 1^p$, then $\overline{XP} \approx 203;50^p$. If we then add on $\overline{NP}$9, we determine $\overline{NX} = 268;00^p$.
So to sum up:
If we consider the earth’s radius to be $1^p$:
- The distance to the mean moon at syzygy is $59;00^p$
- The distance to the sun is $1210^p$
- The distance to the apex of the shadow cone is $268^p$
We’re nearing the end of Book V. There’s four chapters left in which we’ll discuss the volumes of the relative bodies and some further consideration of parallax!
Happy New Year!
- Obviously not drawn anything close to scale.
- This is actually one of my favorite diagrams we’ve seen so far if for no other reason than we can clearly see the moon’s shadow here and see exactly why a solar eclipse can’t be viewed from everywhere on earth at the same time; the shadow is limited to a very narrow portion of the surface.
- If it helps to understand why, think back to when we were looking at the tangents of the epicycles and noted that they are not at $90º$ from the apogee.
- Incorrectly.
- Because all of their angles are equal.
- Since it’s effectively the radius.
- Ptolemy reminds us that this is in the context in which the radius of the earth is $1$.
- The correct answer is $23,455$.
- Which is again the distance to the moon at its maximum distance or $64;10^p$.