In the first post on Book VI, I stated that, while we could calculate the position of the sun and moon every day to determine whether an eclipse was happening, we wanted to rule out as much as possible. To that end, we’ve spent most of our time trying to figure out when we do or do not need to worry about there being an eclipse. First we looked at determining mean conjunctions, then showed how to get from mean to true syzygy, then looked at how far away from a mean syzygy an eclipse could occur, and finally, in the last chapter, we looked at numerous periods to see whether or not they would be possible.
However, we’ve now run out of things that Ptolemy wants to rule out. As such, in what’s left, we’ll need to actually go through at least some of the calculations. Specifically, in this chapter we’re going to work on some tables that, if we input the argument of
the moon’s position in latitude [for a given syzygy, we will know] which of those syzygies will definitely produce an eclipse, as well as the magnitudes and times of obscuration for these eclipses.
Ultimately, we’ll be creating four primary tables:
two for solar eclipses and two for lunar eclipses ([in each case] one for the moon’s greatest distance and one for its least distance.)
In addition, we’ll create an additional table to help with interpolation for the moon at epicyclic distances between greatest and least distance.
For the purposes of this post, we’ll just worry about the solar tables. I’ll split the lunar tables into another post and save much of the discussion of the layout for both until I present the tables so we can have them in front of us to see exactly what he’s talking about.
As indicated above, we won’t be worrying about intervals of time. But rather
the interval which we will establish [between successive entries in the tables] is determined by the amount of obscuration, being $\frac{1}{12} th$ of the diameter of whichever of the luminaries is eclipsed.
This will still be listed by degrees in argument of latitude which is the “input” of the table.
We will also calculate
the distance travelled by the centre of the moon corresponding to each [tabulated] obscuration, without however taking into account either the sun’s additional motion [during the phase of the eclipse] or the moon’s epiparallax [i.e., the change in the moon’s parallax].
To begin, Ptolemy walks us through the first line of the table:
Since the sun’s diameter is $0;31,20º$, and, as was proven, the moon’s diameter at its greatest distance is also $0;31,20º$, it follows that when the moon’s apparent centre is $0;31;20º$ from the sun’s centre on the great circle through both their centres, (and thus is $6º$ from the node along its inclined circle, according to the previous ratio, $11;30 : 1$), that will be the situation in which the moon just touches the sun.
The first part of this is pretty straightforward. Ptolemy, as he has done throughout the discussion of the moon, incorrectly assumes that the moon’s angular diameter at apogee is the same as the sun. Thus, the radius for each is $0;15,40º$ and the total distance between their centers when just touching would be $0;31,20º$
The second part is really a statement of something we talked about in this post. Specifically, we showed that the ratio of the distance between the distance between the true moon and the node to the the ecliptic and the true moon in its circle is about $11;30 : 1$1. We know the separation of the center of the two bodies ($0;31,20º$) , so we can immediately use this ratio to calculate the distance to the node which comes out to $6;0,20º$ which Ptolemy rounds to an even $6º$2.
Since the sun and moon are just touching in this case, the amount of obscuration will be $0$ digits3 and the duration will be instantaneous or $0$.
So the first line of the table for the moon at greatest distance will look as follows:
| Greatest Distance | |||
| Arguments of Latitude | Digits | Minutes of Immersion | |
| $84;00$ | $276;00$ | $0;00$ | $0;00$ |
As a reminder, the argument of latitude is measured from the northern limit, so the nodes are at $90º$ and $270º$ for the descending and ascending nodes respectively. Thus, this row is giving information about what happens $6º$ from the nodes towards the northern limit. There’s obviously a symmetry about the node, so we can also fill in the last line of the table as follows:
| $96;00$ | $264;00$ | $0;00$ | $0;00$ |
The quick example we took above was for the case in which the moon was at its greatest distance and, as we said above, we should also consider what happens with the moon at its closest distance. There, the diameter of the sun remains unchanged, but the moon’s radius will instead be $0;17,40º$. Thus, the total distance between their centers would be $0;33,20º$. Applying the same ratio, we find the distance to the node would be $6;23º$. Ptolemy comes up with $6;24º$. Thus, the first and last lines of the table for the moon at least distance would look as follows:
| Least Distance | |||
| Arguments of Latitude | Digits | Minutes of Immersion | |
| $83;36$ | $276;24$ | $0$ | $0;00$ |
| ⋮ | ⋮ | ⋮ | ⋮ |
| $96;24$ | $263;36$ | $0$ | $0;00$ |
Now, Ptolemy wants to increment the rows of each table, not by the arguments of latitude, but by the digits. So the next row in the table will have a Digits of $1$ meaning we’ll need to calculate the Arguments of Latitude and Minutes of Immersion.
This is quite straightforward for the Arguments of Latitude,
since the amount of the [moon’s] inclined circle which corresponds to $\frac{1}{12} th$ of the solar diameter is about $0;30º$, we increase or decrease the entries in the above-mentioned two columns by that amount, beginning from the lines at both ends and going towards the middle. On the middle line we put $90º$ and $270º$.
In other words, if we take $\frac{1}{12}$ of the sun’s diameter we get:
$$\frac{0;31,20º}{12} = 0;02,37º$$
We then multiply that by the ratio above to get the distance associated with that measured from the node:
$$0;02,37º \cdot 11;30 = 0;30,02º \approx 0;30º$$
Thus, for each additional digit of obscuration, the Argument of Latitude will increment by $\approx 0;30º$. This holds true for both tables since the diameter of the sun (which is the object being obscured) does not change.
Now for the Minutes of Immersion. For a bit of clarity, this isn’t minutes as in a measure of time. Rather, this is arcminutes, and as we’ll see, it’s only measuring the time from the first moment of the eclipse to mid-eclipse.
But before jumping in, Ptolemy tells us he’s going to make a few simplifying assumptions:
We computed the travel of the moon tabulated for each obscuration geometrically, but as if [the problem were confined to] as single plane and straight lines, since such small arcs do not differ sensibly from the corresponding chords, and furthermore the moon’s motion on its inclined circle is not noticeably different from its motion with respect to the ecliptic.
In short, we’re again assuming planar geometry instead of spherical and, over these short distances, that the moon’s path is parallel to the ecliptic. However, Ptolemy does give one warning:
the moon’s motion in longitude is affected by the use of arcs of the inclined circle instead of arcs of the ecliptic, and also that it does not follow that the time of syzygy is exactly the same as the time of mid-eclipse.
Here’s what he means:
I’ve sketched the sun at point $B$ with the node at $A$. When the moon is at point $D$ it is mid-eclipse. However, at this point in time, the sun and moon do not actually have the same ecliptic longitude (i.e., when syzygy occurs). That happens when the moon is at $G$. Thus,
the time of syzygy will differ from the time of mid-eclipse by $arc \; GD$.
So is this a big deal? Ptolemy says no as
the differences they cause are small and imperceptible. While it would be absurd not to recognize any of these effects, on the other hand, when one considers the resulting complication in the methods necessary to deal with each of them, deliberate neglect of effects small enough to be overlooked both in theory and observation evokes [in the reader] a strong feeling of the advantage of the greater simplicity, and no regret, or little, for the resulting error in representing the phenomena. In any case, we find that the arc corresponding to $arc \; GD$ does not , in general, exceed $0;05º$.
Ptolemy briefly walks through how to demonstrate this which I’ll omit.
Finally, Ptolemy is ready to illustrate how the final column is calculated. First, we’ll let the sun be at $A$ and then consider the moon at three positions, $B$, $G$, and $D$ in succession4.
In this, $B$ is the point where the lunar disk first touches that of the solar disc, $G$ is mid-eclipse, and $D$ is the last moment where the lunar disc touches that of the sun. In this figure we can state that:
$$\overline{AD} = \overline{AB} = 0;31,20º$$
We can also determine $\overline{AG}$ if we know how many digits are being obscured. For his example, Ptolemy uses a value of $3$ digits. Thus, $\overline{AG}$ is less than $0;31,20º$ by $\frac{3}{12}$ of the solar diameter (which is also $0;31,20º$) or $0;07,50º$. Thus, $\overline{AG} = 0;23,30º$. We can then calculate $\overline{GB} = 0;20,44º$ by the Pythagorean theorem. Ptolemy comes up with $0;20,43º$. Instead of expressing this in degrees, he instead expresses it in arcminutes as $20;43$ and this is what he enters into the respective column of the table. So our table is now looking like this:
| Greatest Distance | |||
| Arguments of Latitude | Digits | [arc]Minutes of Immersion | |
| $84;00$ | $276;00$ | $0;00$ | $0;00$ |
| ⋮ | ⋮ | ⋮ | ⋮ |
| $85;30$ | $274;30$ | $3$ | $20;43$ |
| ⋮ | ⋮ | ⋮ | ⋮ |
| $96;00$ | $264;00$ | $0;00$ | $0;00$ |
The same sort of calculation can be done when the moon is at least distance. However, in this case, $\overline{AB} = 0;33,20º$ and $\overline{AG} = 0;25,30º$5.
That’s all there is for the construction of the solar eclipse tables. In the next post, we’ll explore the lunar tables (which aren’t dissimilar) and then the tables for adjustment, before presenting the tables in their entirety.
- In that post we actually showed it was really $11;28,25 : 1$, but we’ve left that behind and are adopting the rounded figure.
- If you’re curious, using the precise value for the ratio is only $10$ arcseconds off instead of $20$.
- Recalling that a digit is $\frac{1}{12}$ of the diameter of whichever body is being obscured.
- This is drawn as if viewed from inside the celestial sphere and I have not included the ecliptic.
- Which is the total separation minus the obscuration or $0;33,20º – 0;07,50º$.