Despite Ptolemy demonstrating the equivalence of the eccentric and epicyclic hypotheses many times now, he sets out to prove that we can arrive at the same equation of anomaly using the epicyclic hypothesis as we do the eccentric. As we did for the eccentric, we’ll again use the case when the mean motion is 30º from apogee.
Here’s the diagram we’ll be using:
As usual $\angle{A’DA} = \angle{EAZ}$ which, from the problem we’ve stated are both 30º. From there we can also state that their respective arcs, $arc \;A’A = arc \; EZ = 30º$. We’ll immediately extend a line perpendicular to $\overline{ED}$ until it meets $\overline{ZD}$ at Z with point K on $\overline{ED}$.
Now we’ll use the demi-degrees method on $\triangle{AZK}$. Things are sufficiently bunched up and we’ve done this enough times that I’m not going to actually draw in the small circle this time. However, we can state that the diameter of this circle, $\overline{AZ} = 60º$ in the context of circle EHZ but 120º in context of this new smaller circle. That’s a factor of 2, which means $arc\; KZ$ (which is 30º) would need to be scaled up similarly, so in the context of the circle I haven’t drawn, $arc \; KZ = 60º$ and, from the chord table, $\overline{KZ} = 60$.
Let’s briefly recall that the angle this arc subtends on the circumference of the circle is always half that measure, which gives us $\angle{KAZ} = 30º$.
Since we set up $\angle{ZKA} = 90º$, and just stated that $\angle{KAZ} = 30º$, this leaves $\angle{AZK} = 60º$ and the arc subtending it, $arc \; KA = 120º$ and therefore $\overline{KA} = 103;55$1.
So now we’ve determined $\overline{KA}$ and $\overline{KZ}$ which are two sides of the right triangle, $\triangle{KAZ}$. As such, we can use the Pythagorean theorem to solve for $\overline{AZ}$:
$$103;55^2 + 60^2 = \overline{AZ}^2$$
Solving:
$$\overline{AZ} = \sqrt{103;55^2 + 60^2} = 119;59,40$$
Let’s compare that to the context of the overall diagram. Here, $\overline{AZ} = 2;30$.
Thus, we can set up the scale equation to determine the other parts of the diagram in that context: Let’s first do $\overline{KZ}$:
$$\frac{2;30}{119;59,40} = \frac{\overline{KZ}}{60}$$
Solving:
$$\overline{KZ} = 60 \cdot \frac{2;30}{119;59,40} = 1;15$$
Repeating for $\overline{KA}$:
$$\frac{2;30}{119;59,40} = \frac{\overline{KA}}{103;55}$$
Solving:
$$\overline{KA} = 103;55 \cdot \frac{2;30}{119;59,40} = 2;10$$.
We can add $\overline{KA} + \overline{DA}$ to get $\overline{DK} = 62;10$.
And just like that we again have two sides of right triangle $\triangle{DKZ}$, so we can again find the hypotenuse, $\overline{DZ}$ using the Pythagorean theorem:
$$1;15^2 + 62;10^2 = \overline{DZ}^2$$
Solving:
$$\overline{DZ} = \sqrt{1;15^2 + 62;10^2} = 62;11$$
And now we’ll use the demi-degrees method again drawing the circle around $\triangle{DKZ}$ such that the diameter will be $\overline{DZ} = 120$.
We can use this to set up our scaling equation and then apply that scale to $\overline{KZ}$:
$$\frac{120}{62;11} = \frac{\overline{KZ}}{1;15}$$
Solving:
$$\overline{KZ} = 1;15 \cdot \frac{120}{62;11} = 2;25$$
Doing a reverse lookup in our table of chords gets us $arc \; KZ = 2;18º$. But again, that’s only if it’s measured from the center of the circle about $\triangle{DKZ}$. If we take it from point D on the circumference so the angle subtended is the equation of anomaly, it’s half of that, or 1;9º$ when the point defining the mean motion is 30º away from apogee which agrees exactly with what we found previously for the eccentric model.
And again, we can subtract this from the mean motion to determine that the apparent position is 25;51º.
Next, we can demonstrate that we can also derive the equation of anomaly if we know the apparent position directly. To do so, here’s the diagram we’ll use:
I’ve moved the sun a bit further around the deferent because otherwise things get too bunched up2. The diagram is as usual with the addition of $\overline{AL}$ which is perpendicular to $\overline{DZ}$.
This time, we’ll be trying to demonstrate that we can arrive at $\angle{ADZ}$ by starting with $\angle{A’DZ}$ which, as we’ve shown previously, is the same as $\angle{AZD}$3.
Since we know the length of $\overline{AZ}$ in the context of the epicycle to be 60, and the angle stated above, we could use the demi-degrees method to draw a small circle, AZL, with diameter $\overline{AZ}$. From there, $arc \; AL$ would be twice $\angle{AZD}$. We could then use the chord table to figure out length of $\overline{AL}$ in the context of the smaller circle. Dividing those, we can get the ratio $\frac{\overline{AZ}}{\overline{AL}}$.
We can also construct the ratio $\frac{\overline{AZ}}{\overline{AD}}$ because $\overline{AZ}$ is 2;30 and $\overline{AD}$ is 60.
We can again divide those ratios4:
$$\frac{\overline{AZ}}{\overline{AL}} \cdot \frac{\overline{AD}}{\overline{AZ}} = \frac{\overline{AD}}{\overline{AL}}$$
We could then draw another circle, ADL, to determine the length of $\overline{AL}$ directly and in turn, its respective arc. From that, we would be able to determine $\angle{ADL} = \angle{ADZ}$ as it’s half of the arc, which is what we were after.
As with before, we could do these steps backwards to arrive at $\angle{AZD}$ if given the equation of anomaly.
That completes looking at things from the perspective of the angular distance from apogee. In the couple next posts, we’ll repeat these calculations for angles from perigee!
- As usual, from the chord table.
- Otherwise points B & L would be indistinguishable.
- It’s easiest to see it in that post if you look at this image in which the equivalent angles in question are marked in green.
- Which is the same as multiplying by the reciprocal as I’ve done here.
