Now that we’ve demonstrated that the equation of anomaly is symmetric about the line of apsides for the general model, we’ll demonstrate the same for Mercury’s model. Or, as Ptolemy puts it,
we must prove that in this situation too the angles of the equation of ecliptic anomaly [are equal].
We’ll start by producing a diagram based on Mercury’s particular model.
Again, I’ll build it as we go to make it easier to digest.
Quick note before I do, the eccentric circles start getting a bit complicated here. While the primary one (which had center $D$ in our previous post) will be holding still, the other one gets dragged around in a small circle. Thus, if we’re going to be discussing the position of that second one at two different times, it will be in two different places. I suspect that for this reason, Toomer elected to ignore the circles all together and only draws in the line of apsides, the important points, and the epicycles. I’ve elected to leave the circles in, but have deemphasized them by giving them a lighter coloration, and also trying to color code them for the different positions.
Anyway, getting started:
[L]et $\overline{ABG}$ be the diameter through the centres and apogee of the [eccentric] circles, and let $A$ be taken as the centre of the ecliptic, $B$ as the centre of the eccentre producing the anomaly, and $G$ as the point about which rotates the centre of the eccentre carrying the epicycle.
Draw, again on both sides [of the apogee], lines $\overline{BD}$ and $\overline{BE}$, representing the uniform motion of the [center of the] epicycle towards the rear, and lines $\overline{GZ}$ and $\overline{GH}$ representing the revolution of the eccentre in advance with a speed equal [to the center of the epicycle’s]. (So it is clear that the angles at $G$ and $B$ must be equal and $\overline{BD}$ must be parallel to $\overline{GZ}$, and $\overline{BE}$ to $\overline{GH}$). On $\overline{GZ}$ and $\overline{GH}$ take the centres of the [moving] eccentres – let them be $\Theta$ and $K$ – and let the eccentres drawn on those centres (on which those centres (on which the epicycles are located), pass through the points $D$ and $E$. On points $D$ and $E$ draw epicycles (again equal)…
This is a sizeable chunk and pretty poorly phrased. Let me start by drawing this out and then I’ll discuss the issues.
There’s two major problems that I find with Ptolemy’s description above.
The first is that he describes the lines we’ve added from $G$ as connecting to points $Z$ and $H$, but doesn’t tell us where they are. Some of the steps he will describe later will make it evident, but if you’re trying to draw this diagram by following along from his description, it’s presently impossible. I’ve used the later information to position things correctly for now but hasten to note that, while these points may look like they’re falling on the moving eccentres, they’re not.
The second problem is that he insists that these lines should be parallel to the ones coming from $B$, but doesn’t explain why. However, this relates to the poorly chosen language I referred to when Ptolemy first described the model for Mercury. This is another point that makes it clear he was not describing the second eccentre as having the same speed as the epicycle, but instead meant it as the same speed as the first eccentre which drives the center of the epicycle.
As I described in that post, this means that the lines driving each of these two eccentres will always have an angle equal and opposite the line of apsides. That means $\overline{ZG}$ is the one mirroring $\overline{EB}$ and $\overline{GH}$ mirrors $\overline{DA}$. Thus, the lines Ptolemy describes as parallel must be so.
I’ve also drawn in the points $\Theta$ and $K$ which are the centers of the respective eccentres that move in advance. I’ve color coded them to match their respective circles and drawn the epicycles on them.
[J]oin $\overline{AD}$ and $\overline{AE}$ and draw $\overline{AL}$ and $\overline{AM}$ tangent to the epicycles on the same side [of the epicycles].
Getting this far, Ptolemy states that the angles we’re wanting to prove are equal is:
$$\angle ADB = \angle AEB$$
and that the greatest elongation on the epicycle:
$$\angle DAL = \angle EAM$$
To demonstrate this, Ptolemy continues developing the diagram:
Join $\overline{B \Theta}$, $\overline{BK}$, $\overline{\Theta D}$, and $\overline{KE}$, and drop perpendiculars $\overline{GN}$ and $\overline{GX}$ from $G$ on to $\overline{BD}$ and $\overline{BE}$, perpendiculars $\overline{DZ}$ and $\overline{EH}$ from $D$ and $E$ on to $\overline{AL}$ and $\overline{AM}$.
And here’s the portion we find out where $Z$ and $H$ must be located: i.e., positioned such that a line from $D$ falls perpendicularly on the line extended from $\Theta$ and similarly on the other side.
This has gotten rather busy, so let me remove the eccentric circles to make things easier to see:
Now we can start into the proof proper.
The, since $\angle GBN = \angle GBX$ [by hypothesis] and the angles at $N$ and $X$ are right, and $\overline{GB}$ is common to [$\triangle GBN$ and $\triangle GBX$], $\overline{GN} = \overline{GX}$ and $\overline{DZ} = \overline{EH}$.
In other words, $GZDN$ and $GHEX$ are parallelograms that are both equal.
And also $\overline{\Theta D} = \overline{KE}$.
Ptolemy states this without an actual proof. Toomer doesn’t provide one either, but notes that the actual proof is “quite intricate” and points to Manitus’ translation of the Almagest into German1. I don’t think I’ll follow up on it and will take their word on it2
[A]nd the angles at $Z$ and $H$ are right.
That certainly follows since they were created as perpendiculars.
So $\angle D\Theta Z = \angle EKH$.
This comes from the fact that Ptolemy has established $\triangle DZ \Theta = \triangle EHK$. This was done by showing that two of the sides and the included angle were equal.
And because [in triangles $\triangle G \Theta B$ and $\triangle GKB$] $\overline{\Theta G} = \overline{GK}$ (by hypothesis), and $\overline{BG}$ is common, and $\angle \Theta GB = \angle KGB$, hence $\angle G \Theta B = \angle GKB$.
In this we first know that $\overline{\Theta G} = \overline{GK}$ because $\Theta$ and $K$ are both centers of their respective eccentres which are equidistant from $G$.
We also know that $\angle \Theta GB = \angle KGB$ because $\overline{GZ}$ is parallel to $\overline{BD}$ and $\overline{GH}$ is parallel to $\overline{BE}$. And since $\overline{BD}$ and $\overline{BE}$ are equal and opposite, this means that the angles at $G$ must also be. Then, the angles in question are their supplementary angles which must be as well.
This proves that these triangles are equal so $\angle G \Theta B = \angle GKB$ as they are the related angles within them.
Therefore, by subtraction, $\angle B \Theta D = \angle BKE$
This one isn’t particularly straight forward. To follow, look at $\triangle B \Theta D$. As always, the sum of the angles is $180º$ and within it, we actually know $\angle \Theta DB$ because it’s an alternate interior angle to $\angle D \Theta Z$. Similarly, we know $\angle \Theta DB$ as it’s an alternate interior angle to $\angle G \Theta B$.
Thus we can state:
$$\angle B \Theta D = 180º – \angle D \Theta Z – \angle G \Theta B$$
Here, I’ve used the angles we know instead of the ones in the triangle, but since they’re equal it’s the same thing.
We can then make the same argument on the other side:
$$\angle BKE = 180º – \angle EKH – \angle GKH$$
Then, recall that we established $\angle D \Theta Z = \angle EKH$ and $ \angle G \Theta B = \angle GKH$ which means that, if we subtract these two equations from each other, they’ll all cancel, along with the $180º$ and thus, $\angle B \Theta D = \angle BKE$.
[A]nd base $\overline{BD}$ is equal to base $\overline{BE}$.
This again looks at those same two triangles. We’ve now proven that the angles are all equal (making them similar triangles at the least) and one side which means that the remaining sides must be as well.
Now we’ll look at triangles $\triangle BAD$ and $\triangle BAE$.
$\overline{BA}$ is common, and $\angle DBA = \angle EBA$ (by hypothesis).
The second portion of this is again due to $\angle GBD$ being equal but opposite to $\angle GBE$ and thus, the angles Ptolemy just described, being their supplemental angles, must be as well.
So base $\overline{AD}$ is equal to base $\overline{AE}$.
This comes from side-angle-side similarity since we’ve shown that $\overline{BD} = \overline{BE}$, $\angle DBA = \angle EBA$, and $\overline{BA}$ is common to both.
[A]nd $\angle ADB = \angle AEB$.
Since we’ve shown these are similar triangles, this is true, which is the first proposition Ptolemy wanted to prove.
By the same reasoning [as before], since $\overline{DL} = \overline{EM}$ [epicycle radii], and the angles at $L$ and $M$ are right, $\angle DAL = \angle EAM$.
Again, arguments of similar triangles in which we know two sides of the right triangle to be equal, so the respective angles must be as well. Thus, in both positions, the equation of anomaly will be the same.
That’s it for this chapter.
Now that we have a handle on the model itself, we’ll start looking at the parameters and setup of the model itself in the next chapter.
- p. $435$. Here’s an online copy of the text although a quick check in an online translator indicates the page numbers probably don’t match up as that link is to an early edition and Toomer cites the 2nd edition, edited by Neugebauer (1963).
- It is worth noting that the reason for the intricacy Toomer gives is that “the radii of the deferent in its two positions are not $\overline{\Theta D}$ and $\overline{KE}$, but $\overline{KD}$ and $\overline{\Theta E}$.” We’ve essentially crammed two diagrams on top of each other, following Ptolemy’s direction. But the reality is that this is two separate diagrams – one for the epicycle in each of its positions. They should really be separated and worked independently.