I stopped my previous post where I did because the material it covered is the end of the example problem Toomer provided. However, Ptolemy still has a few more paragraphs to go because
there is, in fact, a noticeable inequality in these intervals [of immersion/emersion] due, not to the anomalistic motion of the luminaries1, but to the moon’s parallax. The effect of this is to make each of the two intervals, separately, always greater than the amount derived by the above method, and, generally, unequal to each other.
In short, because the parallax changes over the course of the eclipse, it will cause the immersion and emersion durations to be longer than they would otherwise be.
We shall not neglect to take this into account, even if it is small.
Then let’s get to it.
This phenomenon is due to the fact that the effect of the parallax on the moon’s apparent motion is always to produce the appearance of motion which would be in advance (if one were to disregard the moon’s proper motion towards the rear). For, suppose first that the moon’s apparent position is before [i.e., to the east of] the meridian: then, as it gradually rises higher [above the horizon], its eastward parallax becomes continually smaller than at the moment preceding, and thus its motion towards the rear appears slower.
What Ptolemy is getting at here is quite simple to understand if we draw a picture:
Here, I’ve drawn an observer’s view looking south with the moon at two positions. Even though I’ve drawn the total parallax, $p$, the same in both, we can clearly see that as the moon approaches south, the component that is the ecliptic longitude, $p_\lambda$ decreases. Thus, the ecliptic longitude appears to be decreasing2 or having a “forward” direction3.
Or suppose, secondly, that its apparent position is after [ie., to the west of] the meridian: then, again, as it gradually descends [towards the horizon], its westward parallax becomes continually greater than at the moment preceding, and thus, as before, its motion towards the rear appears slower.
In other words, same thing happens on the other side of the meridian. Flipping the above picture horizontally shows this clearly.
For this reason, the intervals in question are always greater than those derived by the simple procedure described.
Our previous discussion had a rearward motion of the moon we calculated. However, since in both cases just described, there is now an induced apparent forward motion, this will slow the total motion meaning it takes longer to go through both immersion and emersion.
Furthermore, the difference between successive parallaxes [at equal intervals of time] becomes greater as one approaches the meridian: hence those intervals [of immersion and emersion] which are nearer the meridian must necessarily become more drawn-out. For this reason, the only situation in which the time of immersion is approximately equal to the time of emersion is when mid-eclipse occurs precisely at noon, for then the appearance of motion in advance resulting from the parallax is about equal on both sides [of mid-eclipse].
Previously we have simply assumed that the immersion and emersion time would be the same because we treated the moon with a single speed throughout the whole of the eclipse. Here, we’re adding a new component to that speed and, unless the moon is directly on the meridian at mid-eclipse, that means that these components will be different for each half.
But when mid-eclipse occurs before noon, then the interval of emersion is closer to the meridian and [thus] longer, while if mid-eclipse occurs after noon, then the interval of immersion is closer to the meridian and longer.
Here, Ptolemy is stating that, whichever is closer to the meridian (immersion or emersion), ends up being longer. This follows because the rate of change of $p_\lambda$ is greatest as you approach the meridian.
So what are we going to do about it?
So in order to correct the time-intervals for this effect, we [first] determine, in the way we explained, the uncorrected length of each of the intervals in question, and the zenith distance at mid-eclipse.
Here, Ptolemy provides his own example:
Suppose, for example, that each interval is $1$ equinoctial hour, and the zenith distance $75º$. In the Parallax Table (V.18) we look for the minutes of parallax corresponding to the argument $75º$ (for, e.g., the moon’s greatest distance, for which one takes the entries in the third column). We find, corresponding to $75º$, [a value of] $0;52º$.
Well, that’s a bit of rounding up as the value is actually $0;51,11º$, but let’s not quibble too much.
Since, by hypothesis, the time-intervals of both immersion and emersion, in the mean, is $1$ equinoctial hour, or $15$ time-degrees, we subtract these from $15º$ from the $75º$ of the zenith distance, and find the minutes of parallax in the same column corresponding to the resulting $60º$ [namely], $0;47º$.
We haven’t really been dealing with “time-degrees” lately and I feel like I never really addressed them properly back in books I and II, so as a reminder, “time-degrees” are just time expressed in degrees instead of hours, minutes, and seconds. Since there’s $24$ hours in a day each hour is $\frac{1}{24}$ of $360º$ or $15$ “time-degrees” which corresponds to a $15º$ rotation of the sky4.
Thus, Ptolemy is proposing we subtract the duration of this motion from the zenith distance. This is an… odd choice as the rotation of the sky is not necessarily aligned along the altitude circle. As such, simply lopping $15º$ off makes little sense! However, we’re here to play along. Not re-engineer. So we’ll play along.
Hence, the displacement in advance resulting from the parallax at the (average) position nearer the meridian comes up to $0;05º$.
So from the beginning of the eclipse to mid eclipse, the moons motion is $0;05º$ less than the motion it would otherwise have based on what we calculated in the last post.
We also add the [$15º$] to $75º$, and find the minutes of total parallax corresponding to the resulting $90º$ in the same column, [which is] $\approx 0;53,30$. Thus, here the displacement in advance resulting from the [parallax at] the position nearer the horizon is $0;01,30º$.
Here, Ptolemy instructs us to do the same thing for the second half of the eclipse and, as predicted, the resulting discrepancy is smaller.
We take the longitudinal components of these increments we have found, and convert each [separately] into a fraction of an equinoctial hour by means of the moon’s true motion, as described, and then add each result to the appropriate mean interval, calculated simply, of immersion or emersion; that is, we add the greater to the interval bounded by the position nearer to the meridian, and the lesser to the interval bounded by the position nearer the horizon.
We don’t actually have a true hourly motion of the moon in the example Ptolemy has given us, but if we adopt the one from the previous post, we can still see how this calculation would play out:
$$0;05º \div 0;34,56 \frac{º}{hr} = 8;30 \; min$$
$$0;01,30º \div 0;34,56 \frac{º}{hr} = 2;34 \; min$$
So from this we can see that the immersion would be $\approx 8 \frac{1}{2} \; min$ longer and the emersion $\approx 2 \frac{1}{2} \; min$ longer.
It is obvious that the difference between the two interval in the above example is $0;03,30º$ or about $\frac{1}{9} of an equinoctial hour, which is the time taken by the moon in mean motion to traverse that distance.
I suppose that Ptolemy here is using the mean hourly motion since we don’t have a true hourly motion in this example, in which case, his argument is reasonable here.
That being said, this whole procedure really hung on the $15º$ adjustment in zenith distance which just doesn’t work. As Toomer puts it:
Ptolemy’s procedure here is, to say the least, crude. Instead of computing the actual zenith distances of the bodies at the beginning and end of the eclipse, he simply applies the $15º$ of one hours motion of the heavens to the zenith distance at mid-eclipse. Finding the total parallax from the zenith distance, he applies it if it were the longitudinal parallax. The procedure is perhaps explicable as illustrating the maximum possible effect of this factor: possible solar eclipse is about $2$ hours; to get the maximum parallactic difference between the two intervals we have to take the zenith distance as great as possible. Allowing $15º$ hourly motion, $75º$ is the maximum zenith distance which permits the whole the eclipse to be visible. The total parallax is the maximum possible value of the longitudinal parallax. To be consistent, however, Ptolemy should have taken the moon at least distance (for which the difference between parallaxes is greater), i.e., col $3$ + col $4$ in V.18.
So that closes out this chapter. And we’re very nearly to the end of Book VI. There’s only one topic left in which we’ll be discussing the inclination of eclipses. We’ll discuss, build a table, show the table, and then discuss it’s use and we’ll be done with the sun and moon!
- Toomer: “I.e., to the fact that the true speed of both the sun and moon does not remain constant over the course of the eclipse.
- Since ecliptic longitude is measured right to left from this point of view.
- I.e., motion in the same direction as the motion of the stars.
- There’s some fuzziness with the whole solar/sidereal day here, but let’s not think too hard at this time about it right now.