Continuing on with finding the corrections necessary for the presence of the equant, we’ll produce a diagram similar to the last but with the first opposition switched out for the second at $B$.
Additionally $X$ is on $\overline{ZB}$ where it crosses the circle of mean motion and $N$ has been relocated near perigee on the same circle.
To begin, recall that our first approximation for $\angle XZN$ was $0;35º$ before the perigee. We’ll first look at $\triangle DZH$. In it, the hypotenuse, $\overline{ZD} = 120^p$ and $arc \; DH = 1;10º$. We can then look up the corresponding chord for which I find $\overline{DH} = 1;13^p$.
Similarly, we can take the supplement of $arc \; DH$ to find $arc \; ZH = 178;50º$ and its corresponding chord $\overline{ZH} = 119;59,37º$ which Ptolemy just calls $\approx 120^p$.
We’ll now convert these pieces back to the context in which the diameters of the eccentres is $120^p$:
$$\frac{2;42^p}{120^p} = \frac{\overline{DH}}{1;13^p}$$
$$\overline{DH} = 0;01,39^p$$
which Ptolemy rounds to $0;02^p$1.
$$\frac{2;42^p}{120^p} = \frac{\overline{ZH}}{119;59,37^p}$$
$$\overline{ZH} = 2;41,59^p$$
which Ptolemy rounds to $2;42^p$.
Ptolemy then considers $\triangle BHD$ to calculate $\overline{BH}$ within it. However, because $\overline{DH}$ is so small, he doesn’t show any work and just concludes that it is “negligibly smaller than the hypotenuse, $\overline{BD}$.”
In short, he takes $\overline{BH} = 60^p$ as well.
As we’ve seen before, $\overline{\Theta H} = \overline{HZ}$ and $\overline{E \Theta} = 2 \cdot \overline{DH}$.
Thus, we can subtract $\overline{\Theta H}$ off of $\overline{BH}$ to determine that $\overline{B \Theta} = 57;18^p$. Additionally, $\overline{E \Theta} = 0;04^p$2.
Now, looking at $\triangle E \Theta B$, we have two of the sides, $\overline{B \Theta}$ and $\overline{E \Theta}$, so we can use the Pythagorean theorem to find the remaining side, $\overline{EB}$… but, oh wait. $\overline{E \Theta}$ is still so small that there’s no difference to the first sexagesimal place. So Ptolemy again comes up with $57;18^p$.
We’ll now enter a demi-degrees context about $\triangle EB \Theta$ wherein the hypotenuse, $\overline{EB} = 120^p$. We’ll then need to convert $\overline{E \Theta}$ into this context:
$$\frac{120^p}{57;18^p} = \frac{\overline{E \Theta}}{0;04^p}$$
$$\overline{E \Theta} = 0;08^p.$$
We can then look up the corresponding arc which I find to be $0;08º$ which makes the angle it subtends on the other side of the demi-degrees circle, $\angle EB \Theta = 0;04º$.
We’ll set that aside and check out $\overline{ZX}$ which is a radius with a measure of $60^p$. We can then subtract off $\overline{\Theta Z}$ to find that $\overline{X \Theta} = 54;36^p$.
That gives us two sides of $\triangle E \Theta X$ which means we can find the hypotenuse, $\overline{EX}$ using the Pythagorean theorem:
$$\overline{EX} = \sqrt{54;36^2 + 0;04^2}$$
$$\overline{EX} = 54;36^p.$$
Yet another where the angles are so small that the hypotenuse is indistinguishable from the leg…
Regardless, we’ll use this to enter a demi-degrees context about this triangle:
$$\frac{120^p}{54;36^p} = \frac{\overline{E \Theta}}{0;04^p}$$
$$\overline{E \Theta} = 0;08,48^p.$$
Ptolemy somehow gets $0;10^p$. It’s unclear whether he’s just getting really creative with his rounding or he was carrying over extra precision somewhere.
We can then look up the corresponding arc, $arc \; E \Theta$ which is $0;10º$. Therefore, the angle it subtends on the opposite side of the demi-degrees circle, $\angle EX \Theta = 0;05º$.
Lastly, we can subtract to find:
$$\angle BEX = \angle EX \Theta – \angle EB \Theta$$
$$\angle BEX = 0;05º – 0;04º = 0;01º.$$
Thus,
it is clear that the planet, since its apparent longitude at the second opposition, when it was viewed along $\overline{EB}$ was $7;54º$ into Pisces, would, if it had been viewed along $\overline{EX}$, have had a longitude of only $7;53º$ into Pisces.
That concludes our correction for the second opposition.
- I think these extremely small values do a good job of explaining why Toomer previously stated these points were poorly chosen. When they are so small, the rounding errors become huge as a percentage. The difference between $0;01,39^p$ and $0;02^p$ is a $21\%$ difference!
- Again, we see the problem of rounding small numbers here. If we kept the second sexagesimal place, this should have been $0;03,18^p$ which should have rounded down to $0;03^p$.