In the last post, we introduced two different models that could potentially explain the anomalous motion of the sun (or other objects). Specifically, the sun sometimes appears to move faster along the ecliptic than at other times1. The first model was the eccentric model in which the observer was placed off center. The second was the epicyclic in which the object would travel around the deferent on an epicycle.
Ptolemy stated that for the simple motion of the sun, either of these models would be sufficient. However, he wanted to demonstrate a key equivalence. Specifically that
for the eccentric hypothesis always, and for the epicyclic hypothesis when the motion at apogee is in advance, the time from least speed to mean is greater than the time from mean speed to greatest; for in both hypotheses the slower motion takes place at the apogee. But [for the epicyclic hypothesis] when the sense of revolution of the body is rearwards from the apogee on the epicycle, the reverse is true: the time from greatest speed to mean is greater than the time from mean to least, since in this case the greatest speed occurs at the apogee.
There’s a lot to unpack here, so let’s break it down.
First off, Ptolemy is saying that the motion of the object is slowest at apogee always for the eccentric model, and in the epicyclic model if the deferent and epicycle rotate in different directions2. If the deferent and epicycle rotate the same direction, then the opposite is true.
The next thing Ptolemy is stating here is that, in these cases where the slowest motion is at apogee, then the time from it’s slowest point, to the time it is moving with the same apparent speed as the deferent, will be when the object has moved an apparent angle of 90º from the point of view of the observer. And he’ll prove it.
Ptolemy starts with the eccentric hypothesis.
Here, E is the true center of the orbit and Z is the position of an observer. Then, $\overline{BZD}$ is drawn so that it forms right angles with diameter $\overline{AEG}$. This then means that B and D have their positions so that each of the arcs cut off ($arc \; AB$, $arc \; BG$, $arc \; GD$, and $arc \; DA$) each appear to be 90º from the position of the observer since the angles they subtend are each 90º.
Our goal will be to demonstrate “that the greatest difference between mean and anomalistic motion takes place at point B and D”. In other words, we want to prove that the position the object appears at is the furthest from where it should be were it undergoing uniform motion from the point of view of the observer. Specifically, at point B it will have lagged behind what it would look like for uniform motion while at point D it would appear to be ahead.
But before getting there, we’ll need to do some intermediate steps. First, I’ll add another set of lines to make things easier to see. Specifically, I’ll draw in $\overline{HE}$ which is parallel to $\overline{BD}$ and $\overline{EB}$3:
First up, notice that $\angle{AEB}$ subtends the $arc \; AB$ from the position of the observer that would observe uniform motion, while $\angle{AZB}$ subtends the same arc from the position of the observer that would see apparent, non-uniform motion. The difference in the two arcs that are actually subtended by those angles is $arc \; HB$ which is subtended by $\angle \; HEB$ which is equal to $\angle \; EBZ$ because alternate interior angles are congruent.
To state that again slightly differently: $\angle EBZ$ is the angle that is the difference between the mean and anomalistic motion. In this section, Ptolemy refers to it as “the difference due to the anomaly” but Toomer notes that, later on, the angle that is the difference between the mean and anomalistic motion will be referred to as “the equation of anomaly.”
Next, Ptolemy states,
the ratio of $\angle{EBZ}$ to 4 right angles equals the ratio of the arc of the difference due to the anomaly to the whole circle
In math terms:
$$\frac{\angle{EBZ}}{360º} = \frac{arc \; HB}{360}$$
Next, Ptolemy takes a quick detour stating we can’t create an angle with endpoints at E and Z that has its vertex on the circumference that is any larger than $\angle{EBZ}$ (which is equal to $\angle{EDZ}$). To prove this, he introduces two new points: Θ and K as test points for their respective quadrants.
To start, Ptolemy quotes I.19 of Euclid’s Elements, which states that if you compare two angles in a triangle, the larger angle will have the larger side across from it. However, Ptolemy intends to use this backwards, noting that $\overline{\Theta Z} > \overline{ZD}$ which means that $\angle \Theta DZ > \angle D \Theta Z$4.
Ptolemy then notes that $\overline{E \Theta} = \overline {ED}$ since they are both radii. Thus, their respective angles are equal, so $\angle E \Theta D = \angle ED \Theta$.
Now we’ll add the unequal angles to the equal ones:
$$\angle \Theta DZ + \angle ED \Theta > \angle D \Theta Z + \angle E \Theta D$$
which forms two new angles but preserves the inequality:
$$\angle EDZ > \angle E \Theta Z$$
We can swap out $\angle EDZ$ for $\angle EBZ$ since they’re equal to get
$$\angle EBZ > \angle E \Theta Z$$
This proves that any angle with its vertex to the upper left or upper right above B or D respectively must be smaller than the angle created with vertex at B or D.
Next, we’ll have to demonstrate the same for points below B and D. The process is largely the same. First we note that $\overline DZ > \overline KZ$ and thus, $\angle ZKD > \angle ZDK$. Next, we can again get a few equal angles by noting that $\overline EK = \overline ED$ because they’re both the radius of the circle. Therefore their angles must also be equal so $\angle EDK = \angle EKD$.
We’ll subtract the unequal angles from this:
$$\angle EDK – \angle ZDK \; ? \; \angle EKD – \angle ZKD$$
This time, we’re subtracting angles, so the one with the smaller angle ($\angle ZDK$) being subtracted out is the side that will be the greater than in the resultant inequality. So taking that and looking at what angles the difference creates we get:
$$\angle EDZ > \angle EKZ$$
Again, $\angle EDZ = \angle EBZ$ so we can substitute:
$$\angle EBZ > \angle EKZ$$
This again demonstrates that any point with its vertex in the lower area beneath B and Z must smaller than the angle created with vertex at B and D.
So what does that prove? Recall what we said $\angle EBZ$ was above: It’s the angle that is the difference between the mean and anomalistic motion. And since it cannot be larger anywhere other than points B and D, this proves when Ptolemy stated that the greatest departure from mean is always when the object appears to be 90º from the apogee.
Also, as noted earlier, at point A (the apogee), the object is moving slowest. Yet at point G, it’s moving fastest. At some point in between those two, the speed matches the mean, and that point is at point B.5
Next, let’s compare $arc \; AB$ to $arc \; BG$. Looking at this, we can quickly see which is greater, but let’s prove it. To do so, I’m going to go back to our previous picture.
This means that $\angle AEH$ would be a right angle, and thus, $arc \; AH$ would be too. But we’re interested in $arc \; AB$ which is related to $\angle AEB$ and $\angle AEB$ exceeds $\angle AEH$ by $\angle HEB$ which is equal to $\angle EBZ$6. Or, as Ptolemy put it, “$\angle AEB$ exceeds a right angle by $\angle EBZ$”. We can similarly examine $arc \; BG$ to discover it is short of a right angle by $\angle EBZ$ as well.
So what does that mean? It means that, since $arc \; AB > arc \; BG$ the object takes longer to move from A to B than it does B to C since it moves around the circumference with constant speed with respect to the center. This proves the part of what Ptolemy said above when he said,
for the eccentric hypothesis … the time from least speed [at apogee] to mean [point B] is greater than the time from mean speed to greatest [point G].
Now for the other part regarding the epicyclic hypothesis. Here, we’ll set up the initial diagram much like the one for the outline of the hypothesis although we’ll be changing the letters as usual. We’ll add a few points, starting with H, that is the location of the planet “when its apparent distance from the apogee is a quadrant.”
This single sentence took me several months to figure out and ultimately, the answer was hidden further in the problem when Ptolemy states:
the body transverses the epicycle with the same [angular] speed as the epicycle traverses circle ABG.
To understand this, let’s start with the object at apogee which in this diagram is point E’. I’ve colored it red to make it easier to follow. Also, to help differentiate points, I’m going to use primes7 to denote the before positions of different points. Lastly, I’m making the epicycle super large as we’re going to have some points getting bunched up and this will help out.
As noted when we first introduced the epicyclic hypothesis, we’ll have the epicycle travelling counter clockwise. However, as it does so, the planet will be rotating around the epicycle, clockwise as Ptolemy stated previously. Furthermore, it will travel as far around the epicycle as the epicycle does the deferent until it appears to be 90º from where it started.
Now, we can see that the epicycle has moved around, with it’s center moving from A’ to A. Similarly, the object has moved from point E to H on the epicycle8. Now let’s draw in a few more lines to help us with things.
Here we’ve extended $\overline{DH}$ until it comes to point Θ, which is also on a line, $\overline{A \Theta}$, which is perpendicular to $\overline{ED}$. Point K lies on the intersection of that line and the epicycle while point G lies at the intersection of the previous line and the deferent.
Ptolemy’s first goal here is to demonstrate that $\overline{HD}$ is tangent to $\overline{AH}$, “for that is the position in which the difference between uniform and anomalistic motion is greatest.” In other words, you can’t move H anywhere else on the epicycle and make $\angle ADH$ larger.
To start, let’s recall what Ptolemy said about how much the object moves around the epicycle in relation to the epicycle moving around the deferent (its mean motion): They’re the same. This means that $\angle{EAH} = \angle{A’DA}$.
Next, $\angle{ADH}$ is the deviation of the object’s position from the mean.
So how to prove $\overline{HD}$ is tangent to the epicycle? We’ll do so by proving it’s tangent to $\overline{AH}$.
First, let’s note that $\angle{A’DA} – \angle{ADH} = 90º$. This is from the setup of the problem when Ptolemy stated that we were setting it up such that the object’s apparent motion, moving from E’ to H, was 90º.
However, above we noted that $\angle{A’DA} = \angle{EAH}$. So we can substitute to get $\angle{EAH} – \angle{ADH} = 90º$.
We can take that even further to demonstrate
$$\angle{EAH} – \angle{ADH} = 90º = \angle{AHD}$$
This wasn’t super obvious to me from looking at the picture, but looking at the picture we can write two equations:
$$\angle{EAH} = 180º – \angle{HAD}$$
$$\angle{AHD} = 180º – \angle{HAD} – \angle{ADH}$$
Subtracting these, the $180º – \angle{HAD}$ cancels from each leaving us with $\angle{EAH} – \angle{ADH} = 90º = \angle{AHD}$. The last part is what we’re really after there which is
$$\angle{AHD} = 90º$$
This is important because $\overline{AH}$ is a radius and this means $\overline{DH}$ is 90º from it, and when a line is tangent to the radius, it’s also tangent to the circle. So again, Ptolemy has demonstrated that when the object’s position is as far away from the mean position, it’s when it’s 90º from the apogee.
We’re still not quite done because there’s still that other part
the time from least speed [at apogee] to mean is greater than the time from mean speed to greatest.
I’ll come back to Ptolemy’s proof of this in just a moment, but I think in this picture, there’s another modern way to prove it without any real math. Specifically, if you were to break the motion up at any other point along $arc \; EH$ it would have a component of motion away from the center of the epicycle. This means that, from point E (which was apogee) to point H, the planet would appear to be moving slower than the mean motion. Similarly, from point H to point Z, there’s a component of motion that’s going the same direction as the center of the epicycle which means it would be moving faster. In addition, at point E, the vector would have all of its motion directly opposite the vector for the center of the epicycle which affirms that it moves slowest at apogee9 and at point Z the opposite is true.
This all means that $arc \; EH$ is the arc that corresponds to the “least speed [at apogee] to mean” where $arc \; HZ$ corresponds to the “mean to greatest”. And since we can clearly see that $arc \; EH$ is greater than $arc \; HZ$, this affirms the proposition given the motion around the epicycle is constant.
So what of Ptolemy’s proof? Ultimately he points out the same thing, stating:
$arc \; EH$ … represents the time from least speed to mean … [and] $arc HZ$ … represents the time from mean speed to greatest.
Ptolemy does this by first looking at $\triangle{AD\Theta}$ and since we just showed that $\overline{AH} \perp \overline{D\Theta}$, we can apply Proposition 8 from Book IV of Euclid’s Elements10, to state that $\angle{KAH} = \angle{ADG}$.
Next, he states that $arc \; KH = arc \; AG$, which makes the mathematician in me hurt because in many ways, they’re not equal since the epicycle is a smaller circle, but because Ptolemy always divides the circumference of a circle into 360 parts, regardless of its size, he gets a pass.
But basically what he’s getting at with this is that $arc \; EH$ (which is the time from least speed to mean) is greater than 90º by $arc \; KH$, since A (which is the point defining the mean motion) is past the 90º point on the deferent by $arc \; AG$. Thus, $arc \; HZ$ (which is mean to greatest speed), is less than 90º by $arc \; KH$ which proves that the time from least speed to mean is greater than mean to greatest speed.
That’s been enough math for one post, so I’ll leave off here for now. But in our next post, we’ll look at what happens when you start combining these two hypotheses!
- It actually moves fastest during the winter when we’re closest.
- As a reminder, the deferent will always rotate counter clockwise which is “rearwards” with respect to the heavens which means that for the above to be true, the epicycle would have to be rotating clockwise which is “in advance”.
- Ptoelmy never has a point H or $\overline{HE}$. I’m only adding this to help illustrate a few points.
- Using the greater side to prove the greater angle really should have been I.18 from the Elements.
- Ptolemy doesn’t appear to prove this. However, as a brief sketch of a modern proof, consider the difference in position to resemble a sine wave: Sometimes it’s ahead of the mean (a straight horizontal line), sometimes it’s behind. Since the angular speed is the integral of the position, this basically means a 90º phase shift. So when it’s the furthest away it can get, that means its difference in speed will be 0 and thus the same as the mean.
- Because alternate interior angles are equal.
- The apostrophes after the letter.
- And from E’ to H in the larger picture.
- Only when the deferent rotates the opposite direction as the epicycle.
- I feel like I also remember this theorem from high school geometry.