Now that we’ve laid out the models for the planets except Mercury and the special case for Mercury, Ptolemy revisits them to explore some symmetries. We’ll start with the ones for the model for the planets other than Mercury first.
To start, let’s produce a new diagram based on that model. As usual, I’ll break up Ptolemy’s description into several steps to help make it a bit more digestible.
First, we’ll start with just the eccentric circle which carries the epicycle. It will have center $E$ and diameter $\overline{AEG}$.
On that diameter, we’ll add $Z$ which will be the center of the ecliptic (i.e., the observer’s position1), and $H$, which will be the point around which this circle actually rotates (i.e., the equant).
Here, I’ve gone a bit ahead of Ptolemy’s instructions and drawn in the epicycle itself, centered on point $B$ at some distance from apogee (at $A$) about the eccentric which would be $arc \; AB$.
Next, we’ll draw in a reflection of that same epicycle on the other side and call its center $D$. Specifically, by reflection I mean that this epicycle will have the same radius and be the same distance from apogee as the former, but in the opposite direction (i.e., $arc \; AB = arc \; AD$ or $\angle AHB = \angle AHD$).
Then, we’ll create $\overline{BH\Theta}$ which is just $\overline{BH}$ extended until it meets the opposite side of the eccentric at $\Theta$. We can then do the same from the next epicycle forming $\overline{DHK}$ such that $K$ is the extension of $\overline{DH}$ including point $K$ on the opposite side of the eccentric circle.
Next, we’ll add some points to the epicycle such that they lie tangent to them from the point of view of the observer at $Z$, “in the same direction [i.e., both towards the perigee]”:
Here, I’ve added these points at $L$ and $M$. It’s important to note that these points to not necessarily fall on the eccentric circle. I’ve tried to draw it to reflect that although it may be hard to tell.
This step also indirectly had us draw in $\overline{ZL}$ and $\overline{ZM}$ as they help illustrate the points being tangent.
Next, we’ll also join the centers of the epicycles with point $Z$:
At this point, Ptolemy takes a break from setting up the diagram to tell us what he intends to prove:
I say [$1$] that the angles of the equation of ecliptic anomaly [are equal. I.e.,] $\angle ZBH = \angle HDZ$2
[$2$] similarly, that the greatest elongations on the epicycle [are equal. I.e.,] $\angle BZL = \angle DZM$
(For, [if these statements are true], the amounts of the greatest elongations from the mean [position] resulting in the combination [of the hypotheses] will also be equal [on opposite sides of the apsides]).
To prove this, Ptolemy immediately starts adding to the diagram, now connecting the points we’ve set up.
First, we’ll create lines from the center of each of the epicycles to the points on their perimeters that are on the tangent lines. This will create $\overline{BL}$ and $\overline{DM}$ and they will necessarily be perpendicular to $\overline{ZL}$ and $\overline{ZM}$ since a radius is always perpendicular to the tangent it meets.
We’ll also extend lines from $E$ such that they meet $\overline{B \Theta}$ and $\overline{DK}$ perpendicularly.
Then, since $\angle XHE = \angle NHE$3
and the angles at $N$ and $X$ are right
and $\overline{EH}$ is common to the equiangular triangles [i.e., $\triangle NHE$ and $\triangle{XHE}$],
$$\overline{NH} = \overline{XH}$$
and perpendicular $\overline{EN}$ is equal to the perpendicular $\overline{EX}$.
In short, Ptolemy has proven that these two small triangles are not just similar, but equal in all respects, just mirrored.
Therefore, $\overline{B \Theta}$ and $\overline{DK}$ are equidistant from center $E$.
Therefore, , they are equal to one another4 and their halves are equal to one another [i.e., $\overline{BX} = \overline{DN}$5.
Therefore, by subtraction [of $\overline{XH}$ from $\overline{BX}$ and $\overline{NH}$ from $\overline{DN}$],
$$\overline{BH} = \overline{DH}$$
Next, we’ll concentrate on $\triangle BHZ$ and $\triangle DHZ$ in which
$\overline{HZ}$ is common [to both]
and $\angle BHZ = \angle DHZ$
Which we know because these are both supplemental angles to $\angle AHB$ and $\angle AHD$.
Therefore, base $\overline{BZ}$ is equal to base $\overline{DZ}$
and $\angle HBZ = angle HDZ$.
Because, again, these aren’t just similar triangles, but equal triangles since all the angles are the same and at least one side.
But also, $\overline{BL} = \overline{DM}$ ([as they are both] radii of the epicycle),
and the angles at $L$ and $M$ are right.
Ptolemy skips a bit of the logic in this last step, but essentially he’s again proving that $\triangle BLZ = \triangle DMZ$. We know this because two of the three sides are equal ($\overline{BZ} = \overline{DZ}$ and $\overline{BL} = \overline{DM}$) in a right triangle with the equal sides being the hypotenuse and adjacent to the right angle which means we could use the Pythagorean theorem to find the length of the third side in each, which would also be equal. Thus, the two triangles are equal as are the respective angles within them.
Therefore, $\angle BZL = \angle DZM$.
which is what Ptolemy set out to prove.
I’ll leave off there for the actual following along of Ptolemy, but before I finish this post, I think we should ask what we really just proved.
Essentially, what this is stating is that the equation of anomaly is mirrored on both sides of the lines of apsides. This may seem trivial, but needed to be established so we can make some shortcuts and only need to derive the equation of anomaly for one side or the other, and then can simply apply it to both sides with an opposite sign, the same as we’ve done for the solar model.
In the next post, we’ll explore the same for Mercury’s particular model.
- Note that $E$ and $Z$ have flipped their meanings between this figure and the original one in which we set up the general planetary model. There, $Z$ was the center of the eccentric circle and $E$ was the observer at the center of the ecliptic. Quite annoying.
- It’s been awhile since we’ve discussed the equation of anomaly in this sense (i.e., that it’s equal to the angle at the center of the epicycle in a triangle with other points of the observer and the center of the eccentric), so if you need to recall why that’s so, you can find where we originally proved it in this post.
- We know this because we previously stated that $\angle AHE = \angle AHD$ and these angles are vertical angles from them and thus equal as well.
- This comes from Euclid’s Elements, III.14.
- That a line from the center of a circle and perpendicular to a chord of that circle bisects the chord is related to Euclid’s Elements, III.1.