Finally we’re ready to calculate our final eccentricities and line of apsides. And instead of breaking it into two posts, I’m going to do it all at once so we can be done with it1.
Here we go!
We begin again with this diagram.
In it, $arc \; AB$ and $arc \; BG$ both remain unchanged as these were the increases in mean motion on this circle.
We now need to apply the second correction to $\angle BDG$.
My initial expectation was that this correction would be applied in addition to the previous correction. However, Ptolemy clearly applies the correction derived in the second iteration to the original $\angle BDG$, indicating that the first correction we derived overshot how much correction would be needed.
Our original, knowingly incorrect, value we used for $\angle BDG$ was $93;44º$. We again subtract the angles on either side of this: $\angle ANE = 0;28º$ and $\angle BNZ = 0;27º$ to find that the updated $\angle BDG = 92;38º$.
Then, the supplement, $\angle EDH = 87;22º$.
In $\triangle EDH$, if the hypotenuse, $\overline{ED} = 120^p$, then,
$$\overline{EH} = 120 \cdot sin(87;22º)$$
$$\overline{EH} = 119;52º.$$
Next, $\angle BEG = 47;44º$ as it’s half of the measure of $arc \; AB$.
That gives us two of the angles in $\triangle DBE$ which allows us to determine the remaining one, $\angle DBE = 44;54º$.
We can then determine $\overline{EH}$ in $\triangle BEH$ (in which $\overline{BE} = 120^p$ since it’s the hypotenuse) using a bit of trig:
$$\overline{EH} = 120^p \cdot sin(44;54º)$$
$$\overline{EH} = 84;42^p.$$
We’ll then switch $\overline{BE}$ into the context in which $\overline{ED} = 120^p$ using $\overline{EH}$ as our conversion piece:
$$\frac{119;52^p}{84;42^p} = \frac{\overline{BE}}{120^p}$$
$$\overline{BE} = 169;50^p.$$
Next, we’ll need to use $\angle ADG$ which will again get corrections applied to it. This time, I find it to be $161;22º$.
It’s supplement, then, is $\angle ADE = 18;38º$.
We can then focus on $\triangle DZE$ in which the hypotenuse, $\overline{ED} = 120^p$, already putting us in the context we want to be in for now.
Using the angle we just found, we can then determine $\overline{EZ}$:
$$\overline{EZ} = 120 \cdot sin(18;38º)$$
$$\overline{EZ} = 38;20º.$$
We’ll now use $arc ABG = 177;12º$ to state that the angle it subtends, $\angle AED = 88;36º$, unchanged from previous iterations.
That means we now know two of the three angles in $\triangle DAE$, so we can subtract to find the third:
$$\angle DAE = 180º – 88;36º – 18;38º$$
$$\angle DAE = 72;46º.$$
We’ll then focus on $\triangle ZAE$ which has it’s hypotenuse $\overline{AE} = 120^p$ in a demi-degrees context about it. We can then determine,
$$\overline{EZ} = 120^p \cdot sin(72;46º)$$
$$\overline{EZ} = 114;37^p.$$
We then convert $\overline{AE}$ into our context in which $\overline{ED} = 120^p$:
$$\frac{38;20^p}{114;37^p} = \frac{\overline{AE}}{120^p}$$
$$\overline{AE} = 40;08^p.$$
As with before, $arc \; AB = 81;44º$ so the angle it subtends on the circumference, $\angle AEB = 40;52º$.
And within $\triangle AE \Theta$, $\overline{A \Theta}$ remains unchanged since previous iterations as $78;31^p$ where its hypotenuse $\overline{AE} = 120^p$.
That too gets converted into the context in which $\overline{ED} = 120^p$:
$$\frac{40;08^p}{78;31^p} = \frac{\overline{A \Theta}}{120^p}$$
$$\overline{A \Theta} = 26;16^p.$$
Then, concentrating on $\triangle AE \Theta$, we can determine $\overline{E \Theta}$ in this context since we know two of the three sides in this triangle. Thus, we can use the Pythagorean theorem:
$$\overline{E \Theta} = \sqrt{40;08^2 – 26;16^2}$$
$$\overline{E \Theta} = 30;21^p.$$
We can then determine $\overline{B \Theta}$ by subtracting:
$$\overline{B \Theta} = \overline{BE} – \overline{E \Theta} = 169;50^p – 30;21^p$$
$$\overline{B \Theta} = 139;29^p.$$
This gives us two of the three sides in $\triangle {BA \Theta}$ so we can use the Pythagorean theorem to find the remaining side, $\overline{AB}$:
$$\overline{AB} = \sqrt{\overline{A \Theta} + \overline{B \Theta}} = \sqrt{26;16^2 + 139;29^2}$$
$$\overline{AB} = 141;56^p.$$
This is the piece that we then use to convert into the context in which the diameter of the circle is $120^p$ since, in that context, we know that $\overline{AB} = 78;31^p$ as that’s it’s chord length (for $arc \; AB$) in that context.
Thus, we can now convert $\overline{ED}$:
$$\frac{78;31^p}{141;56^p} = \frac{\overline{ED}}{120^p}$$
$$\overline{ED} = 66;23^p.$$
And we’ll also convert $\overline{AE}$:
$$\frac{78;31^p}{141;56^p} = \frac{\overline{AE}}{40;08^p}$$
$$\overline{AE} = 22;12^p.$$
This allows us to determine the corresponding arc, $arc \; AE = 21;19º$2.
This can then be added to $arc \; GBA$ to determine $arc \; GBAE = 198;31º$ which means the arc going around the other side, $arc GE = 161;29º$. The length then, of the chord, $\overline{GE}$ is $118;26^p$.
Now we need to change diagrams:
Here, we can determine $\overline{GD}$ by subtracting:
$$\overline{GD} = \overline{GE} – \overline{ED} = 118;26^p – 66;23^p$$
$$\overline{GD} = 52;03^p.$$
We now use the intersecting chords theorem again:
$$66;23^p \cdot 52;03^p = \overline{LD} \cdot \overline{DM}$$
as well as Euclid’s II.5 which we’ll substitute the left side of this into and then solve:
$$66;23^p \cdot 52;03^p = 60^2 – \overline{DK}^2$$
$$\overline{DK} = 12;02^p.$$
That gives us our new distance between the center of motion and observer.
Ptolemy gives his value as $\approx 12^p$. His sudden fuzzy rounding makes it impossible to get a good idea of how well we match. But needless to say, mine does indeed round down to $\approx 12^p$ as well.
Toomer does provide a more exact value of $11;59,50^p$, which also rounds to $12^p$. I’m uncertain as to why our values did not match better, being nearly two minutes different whereas, in the previous iteration, we had identical values. And since I’ve been doing these calculations in a Google Sheet, I simply copy/pasted the calculations for the next iteration, which updated the references meaning that it’s highly unlikely I made an error in the calculations. I’m tempted to chalk it up to accumulated rounding errors which mine shouldn’t have since Sheets preserves the higher accuracy through the calculations.
Regardless, we now turn to finding the arcs from each of the observations with respect to the line of apsides.
Again, $\overline{GN} = \frac{1}{2} \overline{GE}$ since $\overline{KX}$ on which $N$ lies is perpendicular to it, therefore bisecting the arc and the chord. Thus, $\overline{GN} = 59;13^p$.
We can then subtract $\overline{GD}$ from this to determine $\overline{DN} = 7;10^p$.
Then, we can use a bit of trig to determine $\angle DKN$:
$$\angle DKN = sin^{-1}(\frac{7;10^p}{12;03^p})$$
$$\angle DKN = 36;35^p.$$
This is the same measure as $arc \; MX$.
As we stated before, since $\overline{KX}$ is perpendicular to $\overline{GE}$ it bisects the arc, meaning
$$arc \; GMX = \frac{1}{2} 161;29º = 80;44º.$$
We can then subtract off $arc \; MX$ to determine:
$$arc \; GM = 44;09º.$$
Then,
$$arc \; LB = 180º – 44;09º – 95;28º = 40;23º.$$
And lastly,
$$arc \; AL = 81;44º – 40;23º = 41;21º.$$
Ptolemy reports somewhat different values again.
| Me | Ptolemy | Toomer3 | |
| $arc \; GM$ | $44;09º$ | $44;21º$ | $44;18,45º$ |
| $arc \; LB$ | $40;23º$ | $40;11º$ | $40;13,15º$ |
| $arc \; AL$ | $41;21º$ | $41;33º$ | $41;30,45º$ |
Toomer’s values are obviously closer to Ptolemy’s here than mine.
I don’t have an immediate explanation on the discrepancy and, after all this calculation, I don’t feel like going through it all again by hand to investigate compiled rounding errors. Toomer’s footnote also indicates that he adopted “Ptolemy’s elements.” It’s unclear whether he meant from the previous iteration or this one. So, that may have something to do with it as well.
But ultimately, I feel our values are all in decent enough agreement for us to move on.
Ptolemy does not go through any further iterations4. Instead, he uses the rest of the chapter to “show by means of the same [configurations] that the observed apparent intervals between the three oppositions are found to be in agreement with the above quantities.”
In other words, he is going to show that the values he’s come up with are good enough.
- At least, until we have to do this all again for Jupiter and Saturn.
- I did so using trig but the chord table agrees.
- Toomer only reports a value for the first arc here. The remaining values I have calculated since they are dependent only on static values and the first arc.
- And he’s right not to. Both Toomer and Neugebauer agree that a further iteration would result in changes less than a minute of difference in the eccentricity and only a few minutes in the direction of the line of apsides. Thus, they would be below the accuracy to which Ptolemy is dealing. Neugebauer goes so far as to note that Ptolemy’s direction for the line of apsides is in remarkable agreement with modern calculations for the period.