Our next goal will be to determine how,
given the arcs of the periodic [motions] on the eccentre which produces the uniform motion [i.e., the equant] and on the epicycle, one can readily obtain the apparent positions of the planets.
As usual, we’ll start with a new diagram.
In this diagram, we have the three eccentres, $Z$, $D$, and $E$ as usual. We then have the epicycle on the eccentre at $B$ and the planet at $K$. We then draw $\overline{ZB}$ which is extended to the opposite side of the epicycle at $\Theta$ as well as $\overline{EB}$ which is extended to the opposite side of the epicycle at $H$. We also create $\overline{EK}$ and $\overline{BK}$. Lastly, a perpendicular is dropped from $K$ onto $\overline{EH}$ at $L.$
[T]hen, $\angle HBK$ will be given by addition [of the angles $\angle \Theta BK$ and $\angle HB \Theta$], and hence the ratio of $\overline{KL}$ and $\overline{LB}$ to $\overline{BK}$ and also, obviously, [their ratio] to $\overline{EB}.$
Accordingly, the ratio of the whole line $\overline{EBL}$ to $\overline{LK}$ will be given. Hence, $\angle LEK$ will be given and we will have computed $\angle AEK$ which comprises the apparent distance of the planet from apogee.
Let’s break down what Ptolemy is saying here.
First off, he’s presuming we know $\angle HB /Theta$ and $\angle \Theta BK$. The latter we would be able to determine from our mean motion table. It’s not immediately obvious to me how we’d know the former, but we’ll probably find out later.
But, assuming we know these angles, we can add them together to determine $\angle HBK$ which is one of the angles in $\triangle KLB.$ Since we know one of the angles and this is a right triangle, we can solve it (at least, in a demi-degrees context to start) for the sides. This is what Ptolemy is meaning about the ratio of the sides of this triangle.
And within this triangle, $\overline{KB}$ is a radius of the epicycle, which we know in our broader context, so we can then switch those sides to the context in which the radius of the eccentre is $60^p.$
Once we’ve done that, we’ll know $\overline{KL}$ in that context, allowing us to draw a demi-degrees circle about $\triangle EKL.$ Then, we can find the corresponding arc, $arc \; KL$ which we can use, in turn, to find $\angle LEK.$
Next, look at $\angle AEB.$ This is the apparent angle of the center of the epicycle from apogee (i.e., from the point of view of the observer). Ptolemy tells us that we’ll be able to calculate this.
And, if we do, then we can add this to $\angle LEK$ to determine the apparent position of the planet away from apogee.
So it seems there’s a few missing elements here, but I suspect we’ll get more information about them shortly, given the next chapter will be on how the tables of anomalies are constructed.
