In our last post, we walked through Ptolemy’s method for finding the angular distance between the mean and true conjunction.
When I initially wrote the post, I followed Neugebauer’s explanation as Ptolemy’s was quite difficult to parse and although they contain a lot of the same key elements, one important piece is left out of Neugebauer’s solution. Specifically, the part where Ptolemy’s iterative method leads naturally to the increase over the anomaly of $\frac{1}{12}$. Writing a later post, I realized that this was an important piece of information since it pops up later and thus made the effort to more completely understand Ptolemy’s method and rewrote the post to explain it. However, I didn’t want to lose the original work, following Neugebauer since readers may appreciate some explanation of Neugebauer’s work as it too is quite dense. Thus, I’ve included that original text beneath the fold as a separate post.
To determine $x$ in the above diagram we’ll first determine how long each luminary takes to move… whatever distance is is that they move. Recall that
$$velocity = \frac{distance}{time}$$
Which we can rearrange to be:
$$time = \frac{distance}{velocity}$$
So let’s start by determining the distance each object travelled. For the moon1, it moved $x + 5;01º$.
Meanwhile, the sun2 moved $x – 2;21º$.
So let’s plug those into their respective equations but for now, we’ll just leave the motions as unknown using $v_s$ for the sun and $v_m$ for the moon.
$$\frac{x – 2;21}{v_s} = t \; \; \; \; \frac{x + 5;01}{v_m} = t$$
For the two of these two objects to end up at the same equatorial longitude at the same time, the times they moved from the first picture must be equal. Hence why I did not put subscripts on $t$. Therefore, we can directly equate these equations:
$$\frac{x – 2;21}{v_s} = \frac{x + 5;01}{v_m}$$
Now, I’m going to briefly rearrange this equation to prove a quick point:
$$\frac{x – 2;21}{x + 5;01} = \frac{v_s}{v_m}$$
Here, I’ve just moved thing around temporarily to show that hidden in here is the ratio of speeds of the sun and moon which Ptolemy called $1:13$ in this post3. So we can substitute that in.
$$\frac{x – 2;21}{x + 5;01} = \frac{1}{13}$$
Now we’re down to just the variable we’re looking for, so let’s solve for $x$:
$$\frac{x – 2;21}{1} = \frac{x + 5;01}{13}$$
$$(x – 2;21) \cdot (13) = (x + 5;01) \cdot (1)$$
$$13x – 30;33 = x + 5;01$$
$$12 x = 35;34$$
$$x = 2;57,50º$$
This is… a bit odd. To remind ourselves, at the beginning of this post, we calculated the distance along the lunar circle that a true conjunction could happen and still have an eclipse. However, this value was done in the context of the ecliptic. So I have a feeling that Ptolemy did one final step here and attempted to translate this back to the corresponding arc of the moon’s circle by constructing a right triangle and pretending it’s planar. If I do so, this gives a corresponding arc of the lunar circle of $2;58;30º$. As you can see, not particularly different, but does help justify Ptolemy’s round off to an even $3º$ in the math he didn’t show.