Almagest Book X: The Equant of Venus

Now that we’ve sorted out the size of the epicycle, Ptolemy next step will be to determine the point about which the eccentre rotates as he doesn’t want to assume it rotates about the center of the eccentre¹.

To start, we’ll set up a new diagram²:

Here, $\overline{AG}$ is the line of apsides where $A$ points towards the point $25º$ into Taurus (apogee) and $G$ the point $25º$ into Scorpio (perigee).

Point $E$ is the center of the epicycle with points $Z$ and $H$ on its perimeter at greatest elongations as viewed from point $B$, the observer.

Point $D$ will be the point about which rotation actually takes place and $\overline{DE}$ will be taken such that it’s perpendicular to $\overline{AG}$.

Point $\mathrm{\Theta }$ is not yet defined.

Now, Ptolemy introduces another set of observations:

[$1$] We observed the first in the eighteenth year of Hadrian, Pharmouthi [VIII] $2/3$ in the Egyptian calendar [February $17/18/$ $134$ CE]. In this, Venus was at greatest elongation from the sun as morning-star, and when it was sighted with respect to the star called Antares, its longitude was $11\frac{11}{12}º$ in Capricorn, at which time the longitude of the mean sun was $25\frac{1}{2}º$ into Aquarius. So the greatest elongation from the mean as morning-star was $43\frac{7}{12}º$.

[$2$] We observed the second in the third year of Antoninus, Pharmouthi [VIII] $4/5$ in the Egyptian calendar [February $18/19$ $140$ CE], in the evening. In this, Venus was at its greatest elongation from the sun, and when it was sighted with respect to the bright star in the Hyades, its longitude was $13\frac{5}{6}º$ into Aries, while the longitude of the mean sun was, again, $25\frac{1}{2}º$ in Aquarius. Hence, in this case, the greatest elongation from the mean as evening-star was $48\frac{1}{3}º$.

From these two observations, we can determine that $\mathrm{\angle }ZBH=43\frac{7}{12}º+48\frac{1}{3}º=91;55º$ as that’s the sum of the elongations³.

This can then be divided in half to determine $\mathrm{\angle }ZBE=\mathrm{\angle }EBH=45;57,30º$.

We’ll now consider a demi-degrees circle about $\mathrm{△}BEZ$.

In this, $arcEZ=91;55º$ and its corresponding chord, $\overline{EZ}=86;{16}^p$ in this context in which the hypotenuse, $\overline{BE}={120}^p$.

However, in the previous post, we determined the radius of the epicycle, $\overline{EZ}$ in this case, to be $43;{10}^p$. We can use this to change contexts.

Doing so, we determine that $\overline{BE}=60;{03}^p$ in that context.

Next, let’s take a look at the difference between the two elongations:

$$48\frac{1}{3}º–43\frac{7}{12}º=4\frac{3}{4}º.$$

Half of this, or $2;22,30º$, is equal to $\mathrm{\angle }DEB$⁴.

We can then create a demi-degrees circle about $\mathrm{△}DEB$ in which we now know $\mathrm{\angle }DEB=2;22,30º$. Thus, its corresponding arc $arcDB=4;45º$ and the chord, $\overline{DB}=5;{00}^p$ in the context where the hypotenuse, $\overline{BE}={120}^p$. Ptolemy evidently rounds down to $4;{59}^p$.

However, we just determined $\overline{BE}=60;{03}^p$, so we’ll convert to that context. Doing so, we find that $\overline{BD}=2;{30}^p$.

And this is the point at which $\mathrm{\Theta }$ comes into play. Because we previously showed that the distance between the earth and the center of the eccentre was $1\frac{1}{4}º$ which is exactly half of $\overline{BD}$. Thus, $\mathrm{\Theta }$ is the actual center of the eccentre and $D$ as the off-center point about which it rotates, known as the equant, at twice that distance.

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1. Point $D$ in the previous post.
2. This diagram, and the procedure we’re about to go through, is nearly identical to the one we did for Mercury, albeit with a slightly different diagram.
3. If you need a reminder on how, refer back to the post where we did this for Mercury. Specifically, the diagram where I add the position of the sun at $S$.
4. Again, see the post where we did this for Mercury for justification.