Now that Ptolemy has worked out his model for the sun and moon in earth radii, we can use this result to calculate parallax for any position in our model. To begin, we will calculate
the parallaxes with respect to the great circle drawn through the zenith and body.
This is essentially the reverse calculation of what we did in this post, so we can reuse the same diagram:
To remind us what we’re looking at, this is a side on view with the small circle in the center being the earth, the middle one being the lunar or solar sphere, and the outer one being the celestial sphere. This is also a slice through those spheres so that what we’re looking at lies in the plane of the meridian and $\overline{KE}$ is the line through the zenith.
Our goal here is going to be able to determine the parallax for either the sun or moon at various distances from the zenith which is $arc \; H \Theta$. Before jumping into the math, Ptolemy lays out what calculations he intends to perform stating,
neither of the luminaries always remain at the same distance. But the resulting difference in the sun’s parallaxes will be very small and imperceptible, since the eccentricity of its circle is small and its distance great. For the moon, however, the resulting difference will be very perceptible, both because of its motion on the epicycle and because the motion of the epicycle on the eccentre, each of which produces quite a large difference in the distance. Therefore, we shall demonstrate the solar parallaxes for a single ratio, namely $1210:1$, but we shall demonstrate the lunar parallaxes for the four ratios which will be most convenient for the methods we shall subsequently develop.
That’s a mouthful, but what Ptolemy is driving at is that the sun is sufficiently far away and the variation in distance is small enough that considering this doesn’t have a significant effect on the parallax. However, because the moon is close and the size of its epicycle a large portion of that distance, it will have a big impact. Thus, Ptolemy will only calculate once for the sun, but at four positions for the moon. What are those four positions? I think it’s easiest to see if we just sketch them out:
For the first two, $1a$ and $1b$, we consider when the moon is at apogee on the eccentre and then at its maximum and minimum distance on the epicycle respectively. We’ve explored these distances in this post. For $1a$, it’s the distance of the mean moon at apogee ($59 \; earth \; radii$) plus the radius of the epicycle ($5;10\; earth \; radii$) for a total distance of $64;10 \; earth \; radii$. For $1b$ it’s the same but subtracted so $53;50 \; earth \; radii$.
Similarly, the distance to the mean moon at perigee for $2$ was $38;43 \; earth \; radii$ again, plus and minus the radius of the epicycle. So the distance for $2a$ is $43;53 \; earth \; radii$ and for $2b$ is $33;33 \; earth \; radii$.
We’ll begin with calculating the parallax for each of these positions for the moon, and the one for the sun when the object is $30º$ from the zenith. So we’ll take $arc \; GD = 30º$ as is the angle angle it subtends, $\angle {GKD}$.
Now let’s look at a demi-degrees circle around $\triangle{AKL}$. In it, we know that $arc \; AL = 60º$ as it’s opposite the angle we just mentioned. Therefore, $arc \; KL = 120º$ as it’s the supplement.
Taking the corresponding chords we get $\overline{AL} = 60^p$ and $\overline{KL} = 103;55^p$ in this context where the hypotenuse, $\overline{AK} = 120^p$.
So if we switch back to the larger context where $\overline{AK}$, the radius of the earth is $1^p$, we get that $\overline{AL} = 0;30^p$ and $\overline{KL} = 0;52^p$.
Now Ptolemy starts bringing in the $5$ cases discussed above. In each of them, we’re considering $\overline{KD}$, the distance from the center of the earth to the object in each of the cases. And from each of them we can subtract out $\overline{KL}$ which we just determined to determine $\overline{LD}$.
If you remember in the previous post where we were working with this diagram, we did a bit of dodgy math, essentially stating that the difference between $\overline{LD}$ and $\overline{AD}$ was sufficiently small due to the actual scale of things, that we could consider them functionally equivalent at the level of precision we’re using. Thus, Ptolemy gives $\overline{AD}$ as the following for each of the cases:
$\overline{AD}_{Sun} = 1209;08^p$
$\overline{AD}_{1a} = 63;18^p$
$\overline{AD}_{1b} =52;58^p$
$\overline{AD}_{2a} = 43;01^p$
$\overline{AD}_{2b} = 32;41^p$
Now, let’s consider another demi-degrees circle about $\triangle{ALD}$ wherein $\overline{AD} = 120^p$. In that circle, we would then determine $\overline{AL}$ to be the following for each of the cases:
$\overline{AL}_{Sun} = 0;02,59^p$
$\overline{AL}_{1a} = 0;56,52^p$
$\overline{AL}_{1b} = 1;07,58^p$
$\overline{AL}_{2a} = 1;23,41^p$
$\overline{AL}_{2b} = 1;50,09^p$
We can then convert those to the relevant arcs:
$arc \; AL_{Sun} = 0;02,50º$
$arc \; AL_{1a} = 0;54,18º$
$arc \; AL_{1b} = 1;04,54º$
$arc \; AL_{2a} = 1;20,00º$
$arc \; AL_{2b} = 1;45,00º$
And the angles these arcs subtend, $\angle{ADL}$ would be half each of these.
$\angle{ADL}_{Sun} = 0;01,25º$
$\angle{ADL}_{1a} = 0;27,09º$
$\angle{ADL}_{1b} = 0;32,27º$
$\angle{ADL}_{2a} = 0;40,00º$
$\angle{ADL}_{2b} = 0;52,30º$
Now recall that $\overline{AZ} \parallel \overline{KD}$. This means that $\angle{ADL} = \angle{ZAD}$ since these are alternate interior angles.
Next, we’ll invoke a bit more of that funny math reasoning. I’ll let Ptolemy state it first in which he says
point $A$ is negligibly different from center $K$, and $arc \; ZH \Theta$ is negligibly greater than $arc \; H \Theta$ (for the whole earth has the ratio of a point to $circle \; EZH \Theta$).
Again, this appeals to the notion that, in comparison to the celestial sphere, the earth has no apparent size, thus $A$ and $K$ being the same point. Similarly, as we shrink the size of the earth in this view, $\overline{AZ}$ converges to $\overline{KH}$ and $arc \; Z \Theta$ converges to $arc \; H \Theta$. As such, the arc subtending $\angle{ADL}$, which we just found describes the parallax for when the angle of the object from the zenith is $30º$.
As is usual, Ptolemy is going to repeat this calculation for other angles which doesn’t get shown and will display all the results in a table in the next chapter. But before getting there,
[w]e need to provide a convenient method of calculating the parallax (corresponding to the appropriate argument) for distances [of the moon] at intermediate positions between apogee and perigee [of the eccentre and epicycle] from the parallaxes tabulated at the above.
But we’ll discuss that more in the next post.
