In the last post, we determined the basic properties of the eccentric model to predict the motion of the sun. Now, we’ll use these properties in conjunction with the model itself to be able to predict “the greatest difference between mean and anomalistic motions” by referring back to our original model.
So let’s draw a simple diagram:
Here, circle ABG is the eccentre on center D and the observer lies at E. The sun will be located at B where $\overline{BE} \perp \overline{AG}$ at E since, as we showed a few posts ago, this is where the equation of anomaly is greatest. Thus, our goal will be to determine $\angle{DBE}$ as this is the angle between mean and anomalistic motion.
In our last post, we determined $\overline{DE}$ to be 2;29:30 which Ptolemy now rounds off to 2;30. The length of $\overline{BD}$ is 60 as it’s the radius of one of Ptolemy’s circles. That’s all we need to solve this using the demi-degrees trick we’ve seen a few times now1. So we’ll draw in a circle around those points:
This makes $\overline{DE}$ a chord of $arc \; DE$. However, we’ll need to convert the length of $\overline{DE}$ in the context of the larger circle, to the context of the new circle.
Previously, $\overline{BD}$ had a length of 60. But as a diameter of the new circle, it’s now 120. That’s a scale factor of 2x. So we’ll need to scale $\overline{DE}$ up by 2x as well, which means its new length is 5 units. From that, we can use the table of chords to determine the length of $arc \; DE \approx 4;46º$, or as Ptolemy puts it, 4;46ºº because that’s the angle at the center of the small circle, subtended by the arc. However, we’re looking for the one on the circumference which is half that, or 2;23º which is the greatest equation of anomaly.
Now, where does that occur?
Let’s remove the small circle and draw in $\overline{DZ}$ which is parallel to $\overline{BE}$.
This means that point B is $\angle{ADB}$ away from apogee. However, this angle is $\angle{ADZ} + \angle{BDZ}$. And that’s easy to find because $\angle{ADZ} = 90º$ and $\angle{BDZ} = \angle{DBE} = 2;23º$2. Thus, $\angle{ADB} = 92;23º$ away from apogee.
Similarly, based on the results we found in this post, the equation of anomaly directly across from E will also be equal. Thus, since point B is the greatest on the left semi-circle, the point directly opposite it will be equal and therefore the greatest there as well. However, it will occur 2;23º before the 270º mark3, so at 267;37º from the apogee (again going counter-clockwise).
That’s all we really needed to show, but for good measure, Ptolemy decides to repeat the process for the epicyclic model just to drive home their equivalency.
So let’s start with the basic model again:
Here, I’ve essentially reused the diagram from this post, but with a few points renamed and a few removed as they won’t be necessary. This diagram was used because it puts the sun at Z, which is 90º away from apogee at E’. As we showed in that post, this is again the point of the greatest equation of anomaly. So to demonstrate that this produces the same results as the other model, we’ll need to show the angle from apogee of the mean motion, $\angle{A’DA}$ to be the same as what we found above for $\angle{ADB}$ in the other diagram which was 2;23º.
One of the requirements for the two hypotheses being equal is that their ratios be the same, so we’ll use $\overline{AZ} = 2;30$. Again, we’ll use the demi-degrees method and draw a circle with the points of $\triangle{ADZ}$ on its perimeter.
In the context of the deferent where $\overline{AD} = 60$, $\overline{AZ} = 2;30$ but since, in this new circle, ADZ, $\overline{AD} = 120$, $\overline{AZ}$ again equals 5. Using the same logic above of $arc \;AZ = 4;46º$ from the chord table, the angle it subtends on the other circumference, $\angle{ADZ}$, must again be half of that or 2;23º, which agrees with our previous result for the maximum of the equation of anomaly.
And again, we can determine what angular distance this takes place from apogee by adding this angle to $\angle A’DZ$ which is $90º + 2;23º = 92;23º$ which agrees with the previous model.
That was a nice easy finish to chapter 4. In the next chapter, we’ll begin the process of constructing a table to determine the anomaly at numerous points in the sun’s path so we’ll have a handy reference instead of having to do the math every time we need to determine something.
But before we go, here’s how today’s progress changes the progress report:
So close to 20%!
- If we were using modern math, this could easily be solved with the sine function, but given that wasn’t developed until a few hundred years after Ptolemy, I’m starting to feel like this is a pretty cool trick.
- Since alternate interior angles are congruent.
- From the point of view of D. It’s at the 270º mark for the observer at E.