Now we’ll turn our attention to finding the size of Venus’ epicycle. Fortunately, the lack of the extra sphere that Mercury had will make this much easier.
Ptolemy starts with the following diagram:
In this image, circle $ABG$ is the eccentre carrying the epicycle with its center at $D$ and the observer, on earth, at $E$.
We’ll take point $A$ as the point $25º$ into Taurus (apogee) and $G$ as the point $25º$ into Scorpio (perigee). Points $Z$ and $H$ are tangent to these circles (i.e., at greatest elongation).
We’ll first create a demi-degrees circle about $\triangle AEZ$. In this, $\angle AEZ = 44;48º$, which is the greatest elongation Ptolemy reported at this position in the previous post.
We can then look at the subtending arc, $arc \; AZ$ which would then have a measure of twice this, or $89;36º$. This is then looked up in the table of chords to find the corresponding chord, $\overline{AZ} = 84;33^p$ in this demi-degrees context in which $\overline{AE} = 120^p$.
We can then do the same for $\triangle EHG$ in which $\angle HEG = 47;20º$. We then find that $\overline{GH} = 88;13^p$ in the context where $\overline{EG} = 120^p$.
However, these are two different contexts. We’ll temporarily pick one to standardize the other to.
Ptolemy picks the first one in which $\overline{AE} = 120^p$ and the radius of the epicycle has a measure of $84;33^p$.
To convert the other triangle, we’ll again set up a demi-degrees circle about it, and let $\overline{GH} = 84;33^p$ instead, so we can determine the length of $\overline{EG}$.
We’ll then convert the other triangle to this context. Doing so, I find $\overline{EG} = 114;59^p$. Ptolemy comes up with a slightly different value of $115;01^p$.
He then adds $\overline{AE} + \overline{EG}$ to determine $\overline{AG} = 235;01^p$ in this context, half of which ($\overline{AD}$ or $\overline{DG}$) is $117;30,30^p$.
We can then subtract to determine $\overline{DE}$:
$$\overline{AE} – \overline{AD} = \overline{DE} = 120^p – 117;30,30^p = 2;29,30^p.$$
Ptolemy rounds this down to $2;29^p$.
We can then convert this to the larger context we’ve been using for all of the models in which $\overline{AD} = 120^p$ and $\overline{DE} \approx 1 \frac{1}{4}^p$ in which case we find that the radius of the epicycle, $\overline{AZ} = 43 \frac{1}{6}^p$.
