When writing the post on finding the latitudinal parallax as part of determining the limits for eclipses, I commented in a footnote that I’d developed a different method for determining this. While Ptolemy’s methods are reasonably accurate, I figured I should go ahead and share the one I came up with using the first case (when the sun is at the summer solstice and the moon is south of the ecliptic from Meroe) as an example. To do so, let’s consider again the configuration of the sun and moon in that instance.
This is essentially the same diagram we we looked at previously, except now I’ve gone ahead and left the zenith distance in.
For this situation, there’s several knowns that I’ve also added. We know that $arc \; S_T Z = 7;24,00º$ because that’s the distance of the sun from the zenith. From that, we can use our Table of Parallaxes to determine $arc \; S_A S_T = 0;00,23º$ since that’s the solar parallax for that distance from the zenith.
We also know $arc \; S_A T_A$ as that’s the sum of the solar and lunar radii for the configuration we’re considering and is thus $0;33,20º$. We can subtract $arc \; S_A S_T$ from that to determine $arc \; S_T L_A = 0;32,57º$.
This leaves us with two pieces that are unknown: $d$, the distance of the true moon from the zenith, and $p$, the lunar parallax which is what we’re really after.
From this, we can write the equation
$$d + p + 0;32,57º = 7;24,00º.$$
Rearranging slightly we get:
$$d + p = 6;51,03º.$$
That wouldn’t immediately seem to be helpful since it’s one equation with two unknowns. Normally, we can’t solve that. However, in this case, the parallax, $p$, is a function of $d$.
Thus, turning to that Table of Parallaxes let’s consider what this means. The distance between the zenith and true moon, $d$ is what we’d input into column $1$ in that table. The lunar parallax for this case is columns $3 + 4$. So what this equation means in terms of this table, is if we were to sum columns $1$, $3$, and $4$, when would it equal $6;51,03º$?
Fortunately, since this table was in a handy google sheet, I was able to copy/paste it into another one and have it do that SUM for me1. Here’s a small screenshot of that:
We can easily see that the value of $d + p$ falls between the values for an angle from the zenith of $6$ and $8$. Thus, we can interpolate between those two rows and find it would be when the parallax has a value of around $0;07,45º$.
Although the configuration is a bit different for the other case we considered (when the sun is at the winter solstice and the moon is north of the ecliptic from the Mouths of the Borysthenes), the same method works there, again giving results that agree reasonably well with Ptolemy’s.