So far in this chapter, we’ve explored how to use the table of parallax to calculate the parallax of the moon and sun by knowing its distance from the zenith. But this in and of itself isn’t particularly useful. For example, when we did a sample lunar parallax calculation, we determined in that situation, the moon was about $1;10º$ off from its true position. But $1;10º$ in which direction?
Thus, the next step will be to break that down into its components, determining how far off in both ecliptic latitude and longitude the parallax makes the moon appear. Ptolemy again gives steps, but no example, so we’ll continue the previous example we did for the moon, following Neugebauer1.
To being, Ptolemy states
we again enter, with the same distance of the moon from the meridian in equinoctial hours [as before] into the same part of the Table of Angle, and take the number of degrees corresponding to that hour, in the third column if the moon is to the east of the meridian, or in the fourth column if it is to the west of the meridian.
As instructed, we’ll make use of the table of zenith distances and ecliptic angles, again looking at the one for Hellespont. Recalling from the last post that the moon was $20$ minutes before crossing the meridian and was $5;20º$ into Scorpio, we’ll do some interpolation. First to estimate the first points of both Scorpio and Sagittarius $20$ minutes before the meridian. However, this time, since the moon was before the ecliptic, we’ll look at the column called East Angle instead of the Arc column.
Doing so I get $115;35º$ for Scorpio and $106;42º$ for Sagittarius. Then we estimate for the $\frac{5;20º}{30º}$ that the moon was through Virgo and I get the final east angle as $114;00º$.
We examine the result, and if it is less than $90º$ we write down the number itself; but if it is greater than $90º$, we write down its supplement, since that will be the size in degrees of the lesser of the two angles at the intersection [of the ecliptic and altitude circle] in question.
Our value was greater than $90º$, so we’ll want to do $180 – 114 = 66;00º$.
We double the number written down
Easy enough: $66 \cdot 2 = 132;00º$.
and enter with this [doubled] number, and also with its supplement into the Table of Chords.
Here we go back to the table of chords and look up two values. For $132;00º$ we get $109;37,32^p$ and its supplement, $48;00º$ we get $48;48,30^p$.
I should note that it is actually somewhat curious that we’re doing this because referring to the table of chords carries with it an implicit assumption: That the triangles we’re about to work with are planar as opposed to spherical. This is certainly not true, but it ends up working reasonably well because the parallax is so small2. Ptolemy seems to be aware of this because he states
for circular arcs of such small size are not noticeably different from straight lines.
Continuing on we take the two figures we just determined and
The ratio of the chord of the doubled number to the chord of the supplement will give the ratio of the latitudinal parallax to the longitudinal parallax
The ratio for the doubled to the supplement will be $\frac{109;37,32^p}{48;48,30^p}$. Here, the top will correspond to the latitudinal parallax and the bottom to the longitudinal. We won’t use this directly, but will refer back to it in a moment to help us sort out the components of another few calculations. Instead,
we multiply the amounts of the chords in question by the parallax determined with respect to the altitude circle, and divide the products, each separately, by 120.
In other words, we take the chord of the doubled angle, $109;37,32^p$ and divide it by $120^p$ to get $0;54,49$3 which we then multiply by the total parallax of $1;10º$. So,
$$1;10º \cdot 0;54,49 \approx 1;04º$$
We do the same for the other, dividing $48;48,30^p$ by $120^p$ to get $0;24,24$. This then gets multiplied by the total parallax:
$$1;10º \cdot 0;24,24 \approx 0;28º$$
But which is which the longitudinal piece and which is the latitudinal?
Well, we can refer back to that ratio we looked at above. There, the numerator was the latitudinal and it was clearly the larger number. So the component of the latitudinal parallax is $1;04º$ while the longitudinal is $0;28º$.
We’re not quite done yet as we haven’t stopped to consider whether these are positive or negative. In other words, is parallax moving the moon up or down for the latitudinal, or left or right for the longitudinal? Ptolemy gives the answer:
For the latitudinal parallax, when the zenith is to the north of the point of the ecliptic then culminating on the meridian, the [effect of the] parallax will be towards the south of [the ecliptic]4; but when the zenith is to the south of the culminating point, [the effect of] the parallax in latitude will be towards the north.
In our case, the zenith was certainly more north than the culminating point because we took Hellespont as our location which has a latitude of $40;56º$. As such, the culminating point of the ecliptic will never be more northwards than the zenith. That can only happen between the tropics. As such, the latitudinal component should get a negative sign for a final value of $-1;04º$.
Similarly,
For the longitudinal parallax: the angles tabulated in the [zenith] table represent the northernmost of the two angles cut off to the rear of the intersection of the ecliptic [and altitude circle]. Therefore, when the latitudinal parallax is to the north, if the angle in question is greater than a right angle, the effect of the longitudinal parallax will be in advance [i.e., in reverse order] of the signs, but if the angle is less than a right angle, the effect will be towards the rear [i.e., in order of the signs]. However, when the latitudinal parallax is to the south, the longitudinal parallax will be towards the rear [i.e., in the order] of the signs, but if it is less than a right angle, the longitudinal parallax will be in advance.
Let’s break it down. If we’re just here to follow instructions, we can ignore the first sentence. It’s just hinting at the logic behind the rest of the paragraph.
First, we need to consider the latitudinal parallax we just discussed. In our case, we just stated that the culminating point was south of the zenith, so we need to look at the second portion of Ptolemy’s statement beginning “when the latitudinal parallax is to the south”. In that case, we need to look at the eastern (or western) angle we calculated. Is it greater than $90º$? In our case it is. Which means that the component of the longitudinal parallax will be “towards the rear” which means a higher ecliptic longitude or a positive value. Thus, we can leave our component for longitude as $0;28º$.
We’re still not quite done with this chapter. The reason is that the steps Ptolemy laid out above are actually a bit of a cheat and aren’t actually correct because they make some faulty assumptions5. It was simply done because it was easy. The issue lies with the fact that one of the values used for an angle in this post, wasn’t actually for the angle he pretended it was. However, it’s pretty close, so in the next post we’ll follow as Ptolemy assesses the impact, and then we’ll look at an algorithm for correcting this when necessary. Once that’s done, we’ll be finished with Book V, so we’re almost there!
- This time from page $117$.
- As Neugebauer puts it this
shows a curious primitivity in dealing with a simple problem of spherical trigonometry, to an extent which would lead to serious errors if the parallaxes were not so small.
- Note that we divided parts by parts which leaves this as a dimensionless factor.
- In other words, negative.
- See the previous quote from Neugebauer at the end of this post.