Almagest Book V: Second Anomaly Eccentricity

So far, we’ve stated that the effect of the second anomaly is to magnify the first anomaly. In the last chapter, we worked out how much larger. Since this second model works by bringing the moon physically closer and further by offsetting the center of the lunar orbit with an eccentric and having that eccentre orbit the Earth, we can determine how far that center must be. In other words, the eccentricity of this second anomaly.

Here we have the Earth, at the center of the ecliptic (not drawn in) at $E$ . The eccentric, centered on $D$ is circle $A B G$ , which goes about its orbit around $E$ on the dashed circle. So for the moment depicted, $G$ is the perigee of the mean moon and $A$ is the apogee.

The moon’s epicycle is centered on $G$ and represented by circle $B H Z$ . We’ll take the moon’s position at its maximum combined anomaly to be at position $Θ$ which is where $\overset{―}{E Θ}$ is tangent to the epicycle since this is when it would be at its largest value. Therefore, using the maximum size of the anomaly from the last chapter, $∠ G E Θ = 7; 40 º$ .

Our goal here will be to get an understanding of $\overset{―}{D E}$ in relation to $\overset{―}{D G}$ . To do so, Ptolemy starts with his usual demi-degrees method, drawing a circle about right triangle, $△ D G Θ$ ¹.

In such a circle, $a r c G Θ$ would be double $∠ G E Θ$ , so $15; 20 º$ . Looking up the corresponding arc from our chord table tells us the chord, $\overset{―}{G Θ} = 16^{p}$ in the context of this small circle where $\overset{―}{G E} = 120^{p}$ .

This is the radius of the epicycle which we have already found twice. Ptolemy does a bit of rounding on its value and says that, in the context where the radius of the deferent is $60^{p}$ it should be $5; 15^{p}$ ². So we can use this as our scale factor to also convert $\overset{―}{G E}$ back to the same context using the following equation:

$\frac{5; 15^{p}}{16^{p}} = \frac{\overset{―}{G E}}{120^{p}}$

$\overset{―}{G E} = 120 \cdot \frac{5; 15^{p}}{16^{p}} = 39; 22^{p}$

However, in the context in which $\overset{―}{G Θ} = 5; 15^{p}$ we also had that the distance from the Earth to the mean moon was $60^{p}$ . That obviously doesn’t work anymore since the distance is constantly changing. As such, we’re going to have to pick a moment and decide it to be 60 at that time. Ptolemy picks the apogee. So in this, he’s defining things such that $\overset{―}{A E} = 60^{p}$ . This means, that is we add $\overset{―}{E G}$ on, we determine the radius of the eccentre to be $99; 22^{p}$ .

Half of this would be the radius, so $\overset{―}{A D} = \overset{―}{D G} = 49; 41^{p}$ .

We can subtract $\overset{―}{E G}$ from this to determine $\overset{―}{D E} = 10; 19^{p}$ which is what we were after.

But let’s take a moment to consider what this means: it means that at closest, the mean moon would have a distance of $39; 22^{p}$ . Add in the epicycle and it could be as close as $34; 7^{p}$ .

Conversely, at its furthest, the mean moon could be up to $60^{p}$ and if at the apogee of the epicycle too, then $65; 15^{p}$ . That’s a huge variation in distance that should have been readily apparent as the moon’s apparent size should vary proportionally. That it doesn’t should have been a gigantic red flag that this model is incorrect. But Neugebauer notes, “the longitudes are so well represented by the new theory that it was not replaced by another model before the late Islamic Period [ $13^{t h}$ to $14^{t h}$ centuries] and then again by Copernicus. Ptolemy himself never mentions this difficulty although he cannot have overlooked it.”

We know this is a right triangle at $Θ$ because a tangent to a circle is always at a right angle to the radius.
As a reminder, with more careful math, it was $5; 13^{p}$ or $5; 14^{p}$ .