Almagest Book X: Second Iteration for Mars – Part 1

Chapter $7$ of Book X has been a lengthy one. We already have six posts behind us and we’re really just getting started because now we have to repeat all of those posts again, iterating through to better approximate an angle we didn’t know at the outset.

To do a quick review, in the first post in this chapter, we needed $\mathrm{\angle }ZNH$, the apparent change along the equant circle from the point of view of the observer in the diagram below.

However, what we had was $\mathrm{\angle }LNG$ which was the apparent change along the ecliptic from the point of view of the observer.

As you can tell, they’re close, but not quite the same. However, we could get to what we needed from what we had by subtracting $\mathrm{\angle }LNT$ and $\mathrm{\angle }YNM$.

We found both of these angles to be $0;33º$ in this post and $0;50º$ in this post, respectively.

Unfortunately, both of those values were calculated starting from an incorrect value for $\mathrm{\angle }ZNH$. But, the good news is that applying them anyway will (Ptolemy hopes) still get us closer to the correct value! And then we’ll be able to iterate through again to reduce the discrepancy even further.

So let’s get started.

As a quick note, I’m doing these calculations in a Google Sheet. In it, each step references the cells of the previous steps. That way, if I make an error, I can correct it upstream and have the sheet calculate everything downstream, saving myself a lot of work. Because Sheets is saving the full value behind the scenes, it also prevents successive rounding errors. However, I’m only displaying the values to the first sexagesimal place, so sometimes things won’t quite add up as if I were doing each calculation discreetly.

I’ve also color coded the components. Green indicates that the values don’t change in each iteration. Red indicates they do.

We’ll begin by using the same diagram as we did previously:

If you need a reminder on how this diagram is laid out, you can find it in the original post.

As with before, $arcBG=95;28º$. This hasn’t changed since this was determined from the mean motion tables about this circle over the interval given in this post.

Next, we can apply our revisions to $\mathrm{\angle }BDG$.

Previously, we stated that $\mathrm{\angle }BDG=93;44º$. But we need to subtract $\mathrm{\angle }LNT$ and add $\mathrm{\angle }YNM$. Thus,

$$\mathrm{\angle }BDG=93;44º–0;33º–0;50º=92;21º.$$

Its supplement, $\mathrm{\angle }EDH=87;39º$.

We’ll then focus on $\mathrm{△}EDH$. Previously, we created a demi-degrees circle about this to solve it. However, I’ll instead use modern trig to save time:

$$\overline{EH}=120\cdot sin(87;39º)=119;54º.$$

Again, this is in the context where $\overline{ED}={120}^p$.

Next, $\mathrm{\angle }BEG=47;44º$ as it did previously, which again gives us the second angle in $\mathrm{△}BED$. Thus, we can find the remaining angle,

$$\mathrm{\angle }DBE=180º–47;44º–87;39º=44;37º.$$

Focusing next on $\mathrm{△}BEH$, in the context where its hypotenuse $\overline{BE}={120}^p$,

$$\overline{EH}=120\cdot sin(44;37º)=84;{17}^p.$$

We’ve now found $\overline{EH}$ in two contexts, so we can use that to convert $\overline{BE}$ to the context in which $\overline{DE}={120}^p$:

$$\frac{119;{54}^p}{84;{17}^p}=\frac{\overline{BE}}{{120}^p}$$

$$\overline{BE}=170;{43}^p.$$

We’ll now turn to $\mathrm{\angle }ADG$ for which we will also need to apply our corrections. Previously, this angle was

$$\mathrm{\angle }ADG=161;34º.$$

However, we need to add $\mathrm{\angle }KNS$ (which we found to be $0;32º$) and subtract $\mathrm{\angle }YNM$ (which we found to be $0;50º$). Hence:

$$\mathrm{\angle }ADG=161;34º+0;32º–0;50º=161;16º.$$

Next, we can find its supplement, $\mathrm{\angle }ADE$:

$$\mathrm{\angle }ADE=180º–\mathrm{\angle }ADG=180º–161;16º=18;44º.$$

Now we can focus on $\mathrm{△}ZDE$. It’s already in the context we want in which the hypotenuse, $\overline{ED}={120}^p$.

We can then find that

$$\bar{E}Z=120\cdot sin(18;44º)=38;{32}^p.$$

We then look at $arcAG=177;12º$ indicating $\mathrm{\angle }AEG=88;36º$.

So, in $\mathrm{△}ADE$, we now know $\mathrm{\angle }ADE=18;44º$ and $\mathrm{\angle }AED=88;36º$ which means the remaining angle, $\mathrm{\angle }DAE$ is $72;40º$.

Next, we’ll look at $\mathrm{△}AEZ$. In it, hypotenuse $\overline{AE}={120}^p$.

Therefore,

$$\overline{EZ}=120\cdot sin(72;40º)=114;{33}^p.$$

This allows us to convert $\overline{AE}$ to our context in which $\overline{ED}={120}^p$.

$$\frac{38;{32}^p}{114;{33}^p}=\frac{\overline{AE}}{{120}^p}$$

$$\overline{AE}=40;{22}^p.$$

We’ll put that aside and look at $arcAB=81;44º$ which was the mean motion that occurred between the first and second observation. Therefore, $\mathrm{\angle }AEB=40;52º$ as it is the angle that subtends the arc.

Focusing next on $\mathrm{△}AE\mathrm{\Theta }$, $\mathrm{\angle }AE\mathrm{\Theta }=40;52º$ and the hypotenuse $\overline{AE}={120}^p$. Thus,

$$\overline{A\mathrm{\Theta }}=120\cdot sin(40;52º)=78;{31}^p.$$

We’ll again context switch this into the context where $\overline{ED}={120}^p$:

$$\frac{40;{22}^p}{{120}^p}=\frac{\overline{A\mathrm{\Theta }}}{78;{31}^p}$$

$$\overline{A\mathrm{\Theta }}=26;{25}^p.$$

And we can use the Pythagorean theorem to find $\overline{E\mathrm{\Theta }}$ in this context too which gives me $\overline{E\mathrm{\Theta }}=30;{32}^p$.

We’ll now subtract $\overline{E\mathrm{\Theta }}$ from $\overline{BE}$:

$$\overline{B\mathrm{\Theta }}=170;{43}^p–30;{32}^p=140;{11}^p.$$

Next, we’ll look at $\mathrm{△}AB\mathrm{\Theta }$. We know two sides in it¹, $\overline{B\mathrm{\Theta }}=140;{11}^p$ and $\overline{A\mathrm{\Theta }}=26;{42}^p$. Thus, by the Pythagorean theorem, $\overline{AB}=142;{39}^p$.

Lastly, we’ll convert to the context in which the diameter of this circle is ${120}^p$. As with before, $\overline{AB}=78;{31}^p$ in that context allowing us to find $\overline{ED}$:

$$\frac{78;{31}^p}{142;{39}^p}=\frac{\overline{ED}}{{120}^p}$$

$$\overline{ED}=66;{03}^p$$

and $\overline{AE}$:

$$\frac{78;{31}^p}{142;{39}^p}=\frac{\overline{AE}}{40;{22}^p}$$

$$\overline{AE}=22;{13}^p.$$

We can then find $arcAE$ which I find it to be $21;21º$ using some trig instead of the table of chords.

This, then, is added to $arcABG$

$$arcGBAE=177;12+21;21º=198;33º$$

As we can see, the value here has changed. In the previous iteration, this value was $198;53º$. So this is a notable tweak.

Meanwhile, the arc on the other side of the circle is

$$arcGE=161;27º$$

where it was previously $161;07º$.

This new value for $arcGE$ has a corresponding chord, $\overline{GE}=118;{26}^p$, up slightly from $118;{22}^p$ in the first iteration.

As with before, we’ll break here to make it easier to compare the progress for anyone seeking to do so. In the next post, we’ll use this to calculate the revised line of apsides and eccentricity.

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No update on progress this time since Ptolemy doesn’t show any of his work here!

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1. Again, in the context in which $\overline{ED}={120}^p$.