As the final post for this chapter, we’ll examine the equation of anomaly for angles measured from perigee in the epicyclic model. Again, we’ll start with a basic diagram for the epicyclic model and this time, we’ll drop a perpendicular onto $\overline{DA}$ from H to form $\overline{HK}$.
To get us started, we’ll notice that the angular distance of the point of mean motion, at A, is $arc \; AG$ away from perigee, at G on the deferent1. This subtends $\angle{ADG}$, which we’ll again be taking as 30º. From there, we can point out that this is the same angle as $\angle{HAK}$ because $\overline{AH} \parallel \overline{GD}$ making the two angles alternate interior angles. Thus, the arc that represents the angular distance of the mean from perigee can also be the arc that subtends this angle, $arc \; H \Theta$. So this arc too is 30º in the context of the overall diagram.
We’ll again jump into the demi-degrees method, drawing small circle AKH with diameter $\overline{AH}$. Here, $\angle{KAH}$ retains its measure, but in the context of this smaller circle, it will mean that the arc subtending at angle is twice that. So $arc \; KH = 60º$ which has a chord length of 60.
We can again find $\angle{KHA} = 60º$2 which means $arc \; AK = 120º$ and its chord 103;55. As a reminder, $\overline{AH} = 120$ in the context of the small circle as it’s the diameter.
Now let’s put things back in terms of the the overall diagram, staring with $\overline{KH}$ which we’ll compare to $\overline{AH}$:
$$\frac{\overline{KH}}{\overline{AH}} = \frac{60}{120} = \frac{\overline{KH}}{2;30}$$
Solving:
$$\overline{KH} = 2;30 \cdot \frac{60}{120} = 1;15$$
Next up, $\overline{AK}$:
$$\frac{\overline{AK}}{\overline{AH}} = \frac{103;55}{120} = \frac{\overline{AK}}{2;30}$$
Solving:
$$\overline{AK} = 2;30 \cdot \frac{103;55}{120} = 2;10$$
And now that we’re back in the context of the big circle, we can subtract $\overline{KA}$ from $\overline{DA}$ to determine $\overline{DK} = 57;50$
Now we’ll focus on triangle $\triangle{DHK}$. Again, this is a right triangle and we know two sides, so we can determine the hypotenuse, $\overline{DH}$ via the Pythagorean theorem:
$$\overline{DH} = \sqrt{1;15^2 + 57;50^2} = 57;51$$
We’ll do one more demi-degrees method, this time about circle DKΘ on radius $\overline{DH}$ with the goal of determining $\overline{KH}$.
$$\frac{\overline{KH}}{\overline{DH}} = \frac{1;15}{57;51} = \frac{\overline{KH}}{120}$$
Solving3:
$$\overline{KH} = 120 \cdot \frac{1;15}{57;51} = 2;34$$
This is the chord, so we can use the chord table to reverse lookup and determine that $arc \; KH = 2;27º$, which would be twice the value of the subtended angle if that angle is on the circumference which means $\angle{HDK} = 1;14º$ which was the equation of the anomaly and in agreement with our figure from the eccentric hypothesis.
This can be added to the 30º of actual motion to determine that the apparent position along of the sun in angular distance from perigee is 31;14º.
Lastly, if we know the apparent position instead of the position of the mean, we can again calculate the equation of anomaly from the following diagram.
Here, we’ve dropped perpendicular from A onto $\overline{DZ}$ forming $\overline{AL}$4. In this setup, the apparent angular position from perigee is $arc \; BG$ which subtends $\angle{BDG}$ which is equal to $\angle{LHA}$. We can see that these are the same angle because they both make use of $\overline{DZ}$ as one side, and $\overline{AH} \parallel \overline{GD}$.
We can use the demi-degrees method around $\triangle{LHA}$ to determine the ratio $\frac{\overline{AH}}{\overline{AL}}$. Also, we know the ratio of $\frac{\overline{AH}}{\overline{AD}}$. We again divide these to get the ratio $\frac{\overline{AD}}{\overline{AL}}$. Using this with the demi-degrees method again on $\triangle{ADL}$ will again allow us to solve for $\angle{ADL}$ which is the equation of anomaly.
That completes the math portion of this chapter. There’s still another page in this chapter, but essentially Ptolemy is just talking about how he’s going to lay out the table that is the entirety of the next chapter, derived from the methods we’ve explored in this chapter. As such, I’ll discuss that portion of the text with the table itself in the next post!
- At least as I’ve drawn it. If you’re following along with the translation, you’ve surely noticed I draw my epicyclic diagrams different than Toomer. His are always rotated so that the epicycle with point E is upright whereas I find it much easier to understand when I keep the deferent fixed. Typically this doesn’t result in the diagrams being different with the exception that I usually leave the “ghost” of where the epicycle started and mine would be rotated. However, in this case, G is positioned on the deferent towards perigee whereas Toomer extended $\overline{ED}$ across the circle and placed G there. But since G isn’t used in this, it doesn’t particular matter.
- Again, this is is the supplement of the previous angle.
- I actually come up with 2;35,34 on this which would round to 2;36, but it may be an issue of I’ve been continually using rounded numbers. While Ptolemy only displayed his values to the first division, if he kept higher precision throughout, it’s feasible that he could have come up with 2;34.
- Things got pretty scrunched up here so to be clear, point L is just a bit closer to H than B.