In the last post, we began the process of determining Mars’ eccentricity and the position of apogee. The technique is fundamentally similar to the three eclipse method used for the moon, but the presence of the equant is throwing a wrench in things. Specifically, in that previous post we showed that an line extended through the observer from one of the observations did not have a length of $120^p$ which affirmed that the observer was still off center.
So in this post, we’ll use that information to determine how far off center so that we can complete the task we set about.
We’ll begin with a modified version of our previous diagram:
Here, we repeat the same circle as before, the “eccentre of mean motion1” on center $K$ with the observer at the center of the ecliptic at $D$2 and $\overline{LM}$ going through both centers.
The observed positions of Mars remain at $A$, $B$, and $G$ with a line extended from $G$ through $D$ to $E$ on the opposite side of the circle.
We’ll now extend a line from $K$ such that it is perpendicular to $\overline{GE}$ at $N$, terminating on the circle at $X$.
In the previous post, we showed that $\overline{GE} = 118;22^p$3. Similarly, we showed that $\overline{ED}$ was $67;50^p$, both in the context in which the diameter of the circle, now drawn in as $\overline{LM}$, was $120^p$.
We can subtract to state:
$$\overline{GD} = \overline{GE} – \overline{ED} = 118;22^p – 67;50^p$$
$$\overline{GD} = 50;32^p.$$
Next, Ptolemy drops the following:
$$\overline{ED} \cdot \overline{DG} = \overline{LD} \cdot \overline{DM}.$$
Toomer notes that this statement is based on Euclid’s Elements III.35 but is more commonly known today as the intersecting chords theorem.
Right now we only know the left side of this equation, so let’s plug in what we know for now:
$$67;50^p \cdot 50;32^p = \overline{LD} \cdot \overline{DM}$$
$$3427;51^p = \overline{LD} \cdot \overline{DM}.$$
So what to do with the right side?
Ptolemy now invokes another of Euclid’s theorems, this time II.5.
This allows him to state the following:
$$\overline{LD} \cdot \overline{DM} + \overline{DK}^2 = \overline{LK}^2.$$
We can solve this for $\overline{LD} \cdot \overline{DM}$ and substitute it into the previous equation. First solving:
$$\overline{LD} \cdot \overline{DM} = \overline{LK}^2 – \overline{DK}^2.$$
And then substituting:
$$3427;51^p = \overline{LK}^2 – \overline{DK}^2.$$
We can now plug in $\overline{LK}$ as it’s the radius of the circle with a measure of $60^p$:
$$3427;51^p = 60^2 – \overline{DK}^2 = 3600 – \overline{DK}^2$$
$$\overline{DK}^2 = 3600 – 3427;51$$
$$\overline{DK}^2 = 172;09$$
$$\overline{DK} = 13;07^p.$$
This is the distance between the center of motion (the equant) and the observer. Presumably, half of this is the eccentricity based on how we’ve defined it previously. However, Ptolemy doesn’t stop here to mention this.
As a brief aside, at the end of the last post, we noted that Ptolemy made a few mathematical errors. Toomer notes that, had the calculations been done correctly, this value would have been $13;02^p$. So there’s definitely a bit of a difference, but not overly substantial.
Next, we’ll notice that $\overline{GN} = \frac{1}{2} \overline{GE}$ since it is a chord at a right angle to a radius. Thus,
$$\overline{GN} = \frac{118;22^p}{2} = 59;11^p.$$
We can then subtract $\overline{GD}$ from this, which we previously showed to be $50;32^p$. This would leave us with $\overline{DN} = 8;39^p$.
This is a second part of right triangle $\triangle DKN$, so we can now create a demi-degrees circle about this triangle and solve it. To do so, we’ll need to convert into that demi-degrees context using $\overline{DK}$, the hypotenuse to scale into it.
$$\frac{120^p}{13;07^p} = \frac{\overline{DN}}{8;39^p}$$
$$\overline{DN} = 79;08^p.$$
We can then use this to determine the corresponding arc which I find to be $82;31º$, but Ptolemy calls $82;30º$. Thus, the angle subtending this on the demi-degrees circle, $\angle DKN$ is half of this or $41;15º$.
If we then zoom out and consider the larger circle in our diagram, we can then state that $arc \; MX = 41;15º$ as well since $\angle DKN$ is the central angle.
While we’re mentally zoomed out, reconsider $arc \; GMX$. This arc is half of $arc \; GMXE$ since $\overline{KX}$ is perpendicular to its chord, $\overline{GE}$.
In the last post, we determined that $arc \; GMXE = 161;07º$, so half of that, $arc \; GMX = 80;34º$.
We can then subtract $arc \; MX$ out of this:
$$arc \; GM = arc \; GMX – arc \; MX = 80;34º – 41;15º = 39;19º.$$
This angle is the angular distance from the point of view of the observer, that Mars was before the perigee for the third observation. Thus, we should now know the position of the line of apsides.
Again, pausing to explore how correcting Ptolemy’s errors at the end of the last post would carry forward, it would change this value to $39;10º$ according to Toomer.
However, Ptolemy again doesn’t pause to acknowledge this and instead works on finding the angles, relative to the line of apsides, of the other two observations.
Ptolemy then reminds us that, between observations, Mars’ position appears to increase by $arc \; BG = 95;28º$.
Thus, we can look at $arc \; LM$ which is $180º$, and then subtract off $arc \; GM$ and $arc \; BG$ to determine:
$$arc \; LB = 180º 39;19º – 95;28º = 45;13º.$$
We also stated that Mars’ position increased by $81;44º$, from the point of view of the observer, from the first observation to the second which is $arc \; AB$. We can then subtract $arc \; LB$ from this to determine:
$$arc \; AL = 81;44º – 45;13º = 36;31º.$$
With this complete, Ptolemy is now ready to start going through the iterative corrections we discussed in the last post to account for the fact that we were playing a bit fast and loose with which center we were taking angles from.
However, I’ll break these into separate posts.
- If you’re following along in Toomer’s translation, you may be a bit confused on where pulled this quote from as it certainly doesn’t appear in the description for this diagram. Rather, it comes from the description of the previous diagram which we explored in the last post. In the paragraph leading up to this new diagram, Ptolemy is referring back to that one, so clearly intends for the circle here to be the same circle as the previous.
- Again, if you’re following along with Toomer, Ptolemy doesn’t specify what point $D$ is for this diagram. Thus, my interpretation comes, again, from the previous diagram in which Ptolemy tells us that $D$ is the “centre of the ecliptic”. While it’s nice of him to have kept $D$ consistent here, it would have been nice if he had been consistent with these three centers through all the diagrams. Sadly, in the first diagram from this chapter, he referred to the center of mean motion as $\Theta$ instead of $K$ as he does here.
- Ignoring the mathematical missteps Ptolemy made towards the end of the derivation.