Almagest Book V: Constructing the Lunar Anomaly Table

In our last post, we showed how it is possible to determine the equation of anomaly by knowing the motion around the epicycle and the double elongation. This, combined with the position of the mean moon¹ gives the true position of the moon. As usual, Ptolemy is going to give us a new table to make this relatively easy to look up. But before doing so, Ptolemy wants to explain what this table is going to look like.

First off, Ptolemy states we won’t just be tossing out the work we did in Book IV in which we created our lunar anomaly table. Rather, we will be providing a supplement. Specifically,

we inserted a third column containing the equation to be added to or subtracted from the anomaly in order to reduce the mean motion counted from $M$ , the mean apogee, to $Z$ , the true apogee.

To understand, let’s bring back the diagram from the previous post:

In that post, where we had a

[double] elongation of $90; 30 º$ , we showed that $a r c Z M = 12; 01 º$ and thus, since the distance of the moon [at $H$ ] from $M$ , the mean apogee, was $333; 12 º$ , we find that its distance from $Z$ , the true apogee, was, obviously, $345; 13 º$ which we must use as argument for the epicyclic equation correcting the mean motion in longitude.

In other words, we can still use the old lunar anomaly table, but we’re going to need to apply a correction in what “apogee” means since the one we used for the old table was based on it being at $Z$ . Thus, we need to apply a correction for it actually being at $M$ .

Then repeating the same methodology as we did in the past few chapters, Ptolemy recalculated for various positions to produce a new table. So let’s take a peek:

Here, the first two columns are labeled as “common numbers.” This is because we’re essentially combining numerous tables all in one for the sake of convenience. As such, what these first two columns are representing can change. For example, if we’re considering column $3$ , that’s the difference between the mean and true anomaly; $a r c Z M$ in the above diagram which we determined is based on the double elongation. Hence, that’s what the first two columns are representing. For column $4$ , which is the equation of anomaly, it’s dependent on the angle around the epicycle from the apogee. And for column $7$ , the latitude is dependent on the angle around the eccentre from the northern limit, so that’s what it’s representing there. Hence why the first two columns are just labeled “common numbers” as opposed to having a more specific definition.

Moving on, the new correction, representing $a r c Z M$ is entered in column $3$ , the Equation for [Mean to True] Apogee. Indeed, as we’ll see in the next post with the full table, when the double elongation is $90 º$ , which is very close to the double elongation we looked at for our example, the value in the third column is $12; 00 º$ . Very close to the $12; 01 º$ we’d calculated.

Next, in column $4$ we have the equation of anomaly we’d given in IV. $10$ .

To that, we’ll need to apply the additional equation of anomaly we’ve come up with in this book and showed how to calculate in the last post. Recall that when we introduced this anomaly, we said that its effect is to amplify the first anomaly. Thus, to determine the true equation of anomaly, we’ll need to add the first anomaly (column $4$ ) to the second (column $5$ ). Ptolemy points out that it will then reach its maximum total anomaly of $7 \frac{2}{3} º$ ². He notes that this is “corresponding to the ratio $\frac{60}{8}$ ” which needs a bit of explanation as we’ll actually be making use of this in a few chapter³.

To understand, consider that at apogee, the distance from the earth to the mean moon we took as $60^{p}$ . In that situation, the radius of the epicycle was $5; 15^{p}$ . In the situation of perigee, the distance between the earth and mean moon is $39; 22^{p}$ while the radius of the epicycle remains the same. In that case, if we take the ratio: $\frac{39; 22}{5; 15}$ it comes out very close to the result of if we took the ratio of $\frac{60}{8}$ . Another way of saying this is that if we instead let the distance of between the earth and moon at perigee be $60^{p}$ instead of $39; 22^{p}$ , then the radius of the epicycle would have to be $8^{p}$ .

Moving on, the sixth column is a bit weird, so we’ll need to walk through that more slowly as well. The purpose is to make calculations between the increments in the first two columns easier, but the way Ptolemy goes about it is tricky as

for each tabulated argument of elongation, the corresponding fraction (given in sixtieths) of the tabulated increment [is given], which must be added to the equation of anomaly tabulated in the fourth column.

Fortunately, Ptolemy walks us through what this means, starting with the following diagram and picking some simple numbers where the elongation is $60 º$ and thus, a double elongation of $120 º$ .

Here, we have the observer at the center of the ecliptic at $E$ , the center of the eccentre at $D$ , the mean moon at $B$ , and the moon at its greatest anomaly at $M$ . Point $Z$ is the true apogee and $H$ is the mean apogee. On the eccentre, $A$ is our northernmost point from which we can measure our double elongation.

We’ve extended $\overset{―}{B E}$ until it meets a line dropped from $D$ , perpendicularly at $L$ .

Because we’ve said that the moon is at its greatest anomaly, this means that the line from the observer to the moon, $\overset{―}{E M}$ is tangent to the epicycle and thus forms a right angle with $\overset{―}{B M}$ since it’s a radius.

First, let’s look at $△ D E L$ . In it, we know $∠ D E L$ because it’s the supplement of the double elongation. Thus, it’s $60 º$ and its corresponding arc in a demi-degrees circle about that triangle, $a r c D L = 120 º$ . Since the hypotenuse, $\overset{―}{D E}$ divides the circle in half, we can determine that $a r c E L = 60 º$ in that context, since it’s the supplement.

As usual, the corresponding chords are $\overset{―}{E L} = 60^{p}$ and $\overset{―}{D L} = 103; 55^{p}$ in the context of the demi-degrees circle.

But in the context of the overall diagram, we know that $\overset{―}{D E} = 10; 19^{p}$ instead of the $120^{p}$ it has in the demi-degrees context as the hypotenuse. Thus, we can convert contexts to get $\overset{―}{E L} = 5; 10^{p}$ and $\overset{―}{D L} = 8; 56^{p}$ in the overall context.

Next, look at $△ B L D$ . We just determined $\overset{―}{D L}$ and we know $\overset{―}{B D}$ , the radius of the eccentre is $49; 41^{p}$ . So we use the Pythagorean theorem to show $\overset{―}{B L} = 48; 53^{p}$ . We can subtract of $\overset{―}{E L}$ which we also just determined to get $\overset{―}{B E} = 43; 43^{p}$ in the context of the overall diagram.

Now we’ll do another demi-degrees circle about $△ B E M$ . In it, $\overset{―}{B E}$ the hypotenuse will have a measure of $120^{p}$ . And since we just determined the length of that same line in the overall context, we can use that to context switch $\overset{―}{B M}$ which is normally $5; 15^{p}$ to the context of this demi-degrees circle, wherein it would be $14; 25^{p}$ . Then, looking up its corresponding chord, we would determine $a r c B M = 13; 48 º$ .

So the angle of anomaly, $∠ B E M$ is half of that, or $6; 54 º$ .

Compare that to the maximum equation of anomaly we found from the first model which was $5; 01 º$ : It’s high by $1; 53 º$ when the double elongation is $120 º$ . Stating this in slightly different language, the second anomaly increases the total anomaly by $1; 53 º$ over the first.

Now, we’ll compare that to the maximum equation of anomaly from our new model which is $2; 39 º$ by taking the ratio:

$\frac{1; 53 º}{2; 39 º} = 0; 42, 38$

That $0; 42, 38$ ⁴ is what Ptolemy has entered into the fifth column after recomputing this ratio for each of the double-elongations. This feels like an odd way to do this, but we’ll see how Ptolemy will make use of this shortly as there’s a chapter on how to use this table coming up shortly.

Finally, the last column is the ecliptic latitude. Notice that it is at zero at $90 º$ and $270 º$ , as these are the nodes where the inclined plane of the moon intersects the ecliptic. Ptolemy doesn’t give any detail on how these were calculated other than to say it was the

same procedure as we did to calculate the arcs of the circle through the poles of the equator, between the equator and ecliptic

which was done in I. $14$ , but the exception in this case was that the inclination was $5 º$ instead of $23 \frac{1}{2} º$ for the angle between the ecliptic and celestial equator.

That explains all the columns in the table. In the next post, I’ll give the table itself, and then we’ll explore an example of how to derive a complete solution for the moon using this table.

Which is determined by its motion from epoch.
$7; 39 º$ to be really specific.
See this post [V.17-2].
Note that there’s no units on this as they cancelled out when we took the ratio. This is basically a percentage, but instead of in decimal (base 10) form, it’s in sexagesimal (base 60). So a persexage if you will.