Continuing on with our second iteration in which we calculate the revised line of apsides and eccentricity of Mars’ model.
As with before, I’m doing my calculations in a Google Sheet to make it easier to correct should I make an error. This also means that the Sheet will be keeping much higher precision to prevent rounding errors. However, as I’m only showing rounded values, things may look a bit off. So if something seems funny, I encourage checking out the aforementioned Sheet.
Getting started, we’ll make use of the same diagram as we did the first time around:
From the previous post, we determined that $\overline{GE} = 118;26^p$ and that $\overline{ED} = 66;03^p$, both in the context in which $\overline{LM} = 120^p$.
We can then write
$$\overline{GD} = \overline{GE} – \overline{ED} = 118;26^p – 66;03^p$$
$$\overline{GD} = 52;23^p.$$
Again, we can use the intersecting chords theorem to write
$$\overline{ED} \cdot \overline{GD} = \overline{LD} \cdot \overline{DM}.$$
We have the left hand side of this, so we’ll plug in:
$$66;03^p \cdot 52;23^p = \overline{LD} \cdot \overline{DM}$$
And then we’ll use Euclid’s II.5:
$$\overline{LD} \cdot \overline{DM} = \overline{LK}^2 – \overline{DK}^2.$$
Substituting in the left hand side of the equation:
$$66;03^p \cdot 52;23^p = \overline{LK}^2 – \overline{DK}^2.$$
And then plugging in $\overline{LK} = 60^p$:
$$66;03^p \cdot 52;23^p = 3600 – \overline{DK}^2$$
$$\overline{DK} = 11;50^p.$$
This is the distance between the center of motion (the equant) and the observer and has gone down since the first iteration which was $13;07^p$.
And good news, we’re on the right path here. Although Ptolemy didn’t show all the steps I’ve shown, he does confirm that for this first iteration, this was the value at which he arrived1. Toomer also arrived at this value.
Next,
$$\overline{GN} = \frac{1}{2} \overline{GE} = \frac{118;26^p}{2} = 59;13^p.$$
We can then subtract $\overline{GD}$ from this, which we previously showed to be $52;23^p$, leaving us with $\overline{DN} = 6;50p$.
This is a second part of right triangle, $\triangle DKN$.
We’ll then switch into a demi-degrees context about that triangle:
$$\frac{120^p}{11;50^p} = \frac{\overline{DN}}{6;50^p}$$
$$\overline{DN} = 69;17^p.$$
Using a bit of trig to find $\angle DKN$:
$$sin(\angle DKN) = \frac{69;17^p}{120^p}$$
$$\angle DKN = 35;16º.$$
This is the same measure as $arc \; MX$.
Jumping back to the last post, we’d determined that $arc \; GE = 161;27º$. Again, since $\overline{KX}$ is perpendicular to chord $\overline{GE}$ it bisects both the chord and the arc meaning that $arc \; GMX = 80;44º$.
We can subtract $arc \; MX$ from this to determine:
$$arc \; GM = arc \; GMX – arc \; MX = 80;44º – 35;16º$$
$$arc \; GM = 45;28º.$$
This gives us the amount prior to the line of apsides Mars was at the third observation.
Toomer reports finding this same value. Ptolemy gives a value that is slightly different at $45;33º$. I suspect the difference is accumulated rounding errors and that he did not use a spreadsheet as I did.
Next, we’ll find the values for the other two observations with respect to the line of apsides.
Again, $arc \; BG = 95;28º$.
So we can look at $arc \; LM$, which is $180º$ and subtract off $arc \; GM$ and $arc \; BG$:
$$arc \; LB = 180º – 45;28º – 95;28º = 39;04º.$$
We also know that $arc \; AB = 81;44$. So we can subtract off $arc \; LB$ to find
$$arc \; AL = arc \; AB – arc \; LB = 81;44º – 39;04º$$
$$arc \; AL = 42;40º.$$
Toomer does not report his values for these, but since they’re straightforward calculations, I have no doubt his numbers would match mine given the first of these three values did.
With those values in hand, we’re ready to calculate the corrections for the three angles. And once that’s complete, we can go through one more iteration and finally be done with this!
