Almagest Book III: Equation of Anomaly from Perigee using Eccentric Hypothesis

In the past couple Almagest posts, we’ve demonstrated that you can derive the equation of anomaly if you know the angle from apogee of either the mean or apparent motion using either the eccentric or epicyclic hypothesis¹. Now, we’ll do the same using the angular distance from perigee again using 30º as our example.

As usual, we’ll start with a fairly generic drawing of the eccentric model. Again, we’ll be dropping a perpendicular, this time from D (the position of the observer) on to $\overline{\mathrm{\Theta }Z}$.

 

Although we’re not going to actually use point B in this post, I still want to call it out because previously, we’ve always had the actual position of the object on the eccentre further from the observer than we’ve drawn the ecliptic. In this case, the object appears closer. That means to see the projection of the object onto the ecliptic, we’ll need to extend $\overline{DZ}$ until it meets circle AGB. Previously, the intersection with the ecliptic has been interior to the eccentre so we haven’t needed to worry about this because the intersection of the line of sight was already along the line of sight to the object.

Moving along, in this diagram, our objective will be to determine $\mathrm{\angle }DZK$ as this is the equation of the anomaly.

Now to start playing with numbers. As stated before, we’ll be taking the case where the sun is 30º from perigee which is at point H which means $arcHZ=30º$ as it the angle of the eccentre it subtends, $\mathrm{\angle }Z\mathrm{\Theta }H$. Immediately, we’ll use the demi-degrees method to draw in small circle ΘDK.

The result of this is that $\mathrm{\angle }Z\mathrm{\Theta }H$ is on the circumference, which means that the arc that subtends in this small circlet will now be twice that value meaning $arcDK$ is 60º. We can also state that $\mathrm{\angle }\mathrm{\Theta }DK=60º$ since the angles in a triangle add up to 180º and we’ve already got a 30º angle and a 90º angle. Thus, the arc across from it, $arcK\mathrm{\Theta }=120º$.

Translating those arcs into chords using the chord table, we get $\overline{DK}=60$ and $\overline{K\mathrm{\Theta }}=103;55$.

Now let’s scale things back to the full diagram. First, let’s solve for $\overline{DK}$ by comparing it to $\overline{D\mathrm{\Theta }}$:

$$\frac{\overline{DK}}{\overline{D\mathrm{\Theta }}}=\frac{60}{120}=\frac{\overline{DK}}{\overline{2;30}}$$

Solving:

$$\overline{DK}=2;30\cdot \frac{60}{120}=1;15$$

Doing the same for $\overline{\mathrm{\Theta }K}$:

$$\frac{\overline{\mathrm{\Theta }K}}{\overline{D\mathrm{\Theta }}}=\frac{103;55}{120}=\frac{\overline{\mathrm{\Theta }K}}{\overline{2;30}}$$

Solving:

$$\overline{DK}=2;30\cdot \frac{103;55}{120}=2;10$$

We can also determine $\overline{KZ}$ by subtracting $\overline{\mathrm{\Theta }K}$ from $\overline{\mathrm{\Theta }Z}$ to get $\overline{KZ}=57;50$.

Now we have right triangle, $\mathrm{△}DZK$, wherein we know the two sides, allowing us to solve for the hypotenuse, $\overline{DZ}$, via the Pythagorean theorem:

$$\overline{DZ}=\sqrt{1;{15}^2+57;{50}^2}=57;51$$

Now we’ll again use the demi-degrees method drawing small circle DZK with $\overline{DZ}$ as the diameter. We’ll find $\overline{DK}$ by comparing it to $\overline{DZ}$:

$$\frac{\overline{DK}}{\overline{DZ}}=\frac{1;15}{60}=\frac{\overline{DK}}{120}$$

Solving:

$$\overline{DK}=120\cdot \frac{1;15}{60}=2;30$$

Looking up the corresponding arc for that cord, we find $arcDK=2;27º$ and as such, the angle it subtends on the circumference, $\mathrm{\angle }DZK$ would be half of that, or 1;14º which is the equation of the anomaly we were searching for.

We can add that to $\mathrm{\angle }BDG$ to determine the apparent arc along the ecliptic, $arcGB$ to be 31;14º.

As with before, we can walk through a proof of how you would obtain the same information if the arc of the ecliptic were given. In that case, we’ll alter the diagram slightly again, extending $\overline{BD}$ and dropping a perpendicular onto it from point Θ.

In this case, we would start knowing $\mathrm{\angle }GDB$ which is equal to $\mathrm{\angle }\mathrm{\Theta }DL$ as they’re opposite angles.

We would then employ the demi-degrees method drawing in small circle ΘDL so that $\overline{D\mathrm{\Theta }}$ is the diameter. We could then determine $arc\mathrm{\Theta }L$ as it would be twice the angle it subtends, and from that, $\overline{\mathrm{\Theta }L}$.

This would give use the ratio $\frac{\overline{D\mathrm{\Theta }}}{\overline{\mathrm{\Theta }L}}$.

As with before, we’d also know ratio $\frac{\overline{D\mathrm{\Theta }}}{\overline{\mathrm{\Theta }Z}}$.

Dividing these two cancels out the $\overline{D\mathrm{\Theta }}$ leaving us with $\frac{\overline{\mathrm{\Theta }Z}}{\overline{\mathrm{\Theta }L}}$.

This would allow us to determine the length of \overline{\Theta L} using another demi-degrees method with small circle about ΘLZ and consequently the equation of anomaly, $\mathrm{\angle }\mathrm{\Theta }ZL$ which could then be subtracted from the angle from perigee to determine the angle for the arc of the eccentre, $\overline{HZ}$.

That does it for the eccentric model. In the next post, we’ll repeat all this once again for the epicyclic model which will take us out of chapter 5!

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1. And also noted that you can do this backwards if you know the equation of anomaly instead.