Having set out the above as a preliminary, we can predict lunar eclipses in the following manner.
As Ptolemy states in opening this chapter, we’re finally done with the preliminary work and we’re ready to start diving into how to actually use everything we’ve done to predict eclipses. As usual, Ptolemy walks us through the steps, but does not provide an example, so I will follow my usual procedure of using example $11$ in Appendix A of Toomer’s translation1.In that example, Toomer invites us to examine lunar eclipses around Nabonassar 28, in the month of Thoth (the first month of the Egyptian year).
To begin, Ptolemy states we should
set down the amounts, in degrees, computed for the required opposition at the time of mid-syzygy at Alexandria, of the so-called anomaly, [counted] from the apogee of the epicycle, and the [argument of] latitude, [counted] from the northern limit.
To do this, we’ll begin with VI.3, the Table of Mean Syzygies. There, we’ll refer to the Table of Oppositions (since we’re discussing a lunar eclipse) on the $25$ year Interval tab looking up the value as close to, but below $28$ which would be the second line for $26$ years. There, we want to record the values from the fourth and fifth columns (the anomaly and latitude respectively). Then, we’ll pick up the additional $2$ years from the $1$ Year Interval tables and do the same. While Ptolemy only tells us to write down the Anomaly and Latitude, I’m going to include the other two columns as well because we’ll be making use of them shortly anyway.
| Years | Days of Thoth | [Increment in] Sun from Apogee | [Increment in] Moon’s [Argument of] Anomaly | [Increment in Argument of Lunar] Latitude |
| $26$ | $9;55,35$ | $267;05;38$ | $83;24,29$ | $230;10,05$ |
| $2$ | $8;15,53$ | $7;39,36$ | $285;25,04$ | $46;45,54$ |
| Total | $18;11,28$ | $274;45,14$ | $8;49,33$ | $276;55,59$ |
Next, Ptolemy tells us that we need to
[Correct] the latter by means of the equation [of anomaly].
Here, Ptolemy is indicating here that we so far, we’ve only calculated the mean anomaly. We need the argument of latitude for the true anomaly to know if it will truly fall in an eclipse window. Thus, we next need to adjust this to be the true opposition as described in VI.4. In particular, we need to adjust the Argument of Lunar Latitude above to take into consideration the lunar anomaly.
We therefore take the value of the Argument of Lunar Anomaly we recorded and look it up in the Table of Lunar Anomalies (IV.10). Interpolating between rows for $8;49,33º$, I come up with an anomaly of $0;42,11º$. Being that it was from the first column, it is negative.
Let’s pause and remind ourselves what this is telling us. Mean opposition is the moment when the mean sun and mean moon would be aligned. No anomaly. We’ve just calculated, at that moment, how much anomaly there really is so at the time of mean opposition , the true moon is at $276;13,48º$.
But this still isn’t telling us the true opposition. There’s still more work we need to do to get there. In particular, we need to take the difference of the anomalies because this tells us how much distance the moon will need to or will have covered since true opposition. Thus, we’ll also need to determine the sun’s anomaly (which is why I included the column for the distance of the sun from apogee).
So we’ll turn to the Table of the Sun’s Anomaly (III.6), and calculate the solar anomaly using the distance of the sun from apogee which we calculated above as $274;45,14º$. Plugging that into the table, I find a solar anomaly of $2;21,25º$. And owing to the fact we looked it up from the second column, this is additive.
Thus, if we take the difference of the anomalies:
$$2;21,25 + 0;42,11 = 3;03,36º$$
Since the sun’s anomaly was positive and the moon’s negative, this is how much distance the moon will have to cover from the mean opposition to the true. But, as we’ve seen before, the sun is also moving. We can account for the additional distance this will require the moon to catch up by increasing the difference between their positions at mean opposition by $\frac{1}{12}$. In other words, the total distance the moon will need to travel until true opposition is
$$3;03,36º \cdot \frac{13}{12} = 3;19,33º$$
Since the moon is then moving further, we’ll add this to the argument of latitude at mean opposition:
$$276;13,48º + 3;19,33 = 279;33,21º$$
Now that we know the argument of longitude for true opposition (i.e., corrected for the effects of the anomaly), we can proceed. Ptolemy tells us we should now look up this
corrected [argument of] latitude [in] the tables for lunar eclipses. If it falls within the range of the numbers in the first two columns, we take the amounts corresponding to the argument of latitude in the columns for the [lunar] travel and the column for the digits [of magnitude] in both tables, and write them down separately.
Doing so and interpolating between rows I get2:
| Distance | Argument of Latitude | Digits | Minutes of Immersion |
| Greatest | $279;33,21º$ | $2;30,42$ | $26;08,31$ |
| Least | $279;33,21º$ | $4;39,58$ | $39;25,33$ |
So now we have things worked out for greatest and least distance. Now we need to estimate between the two. To do so, we’ll use the
anomaly as argument, enter into the correction table, and take the corresponding number of sixtieths.
To remind ourselves what he’s referring to, take a look back at this post in which we discussed how the table in question was derived. In particular, this table is designed to tell us how to tell us how to interpolate between the above values based on how far from apogee on the epicycle the moon is.
However, what Ptolemy leaves out of his instructions is that we need to be careful about what argument of anomaly we look up. We can’t use the one we calculated above ($8;49,33º$) because this is for the mean opposition. We need to adjust for the true opposition again.
To do so, we’ll first determine how long of a time it will be between mean and true opposition. And to do that, we’ll need the speed of the moon at this time as the mean speed doesn’t cut it since the speed of the moon varies due to its anomalistic motion. This was covered in the same post covering VI.4.
In it, we first asked what the anomaly was for the position of the moon on its epicycle for mean opposition. In this post, it was negative $0;42,11º$. Next, we recalculated the anomaly for a $1º$ increment about the epicycle. Doing so I find an anomaly of $0;46,51º$. This means that a $1º$ increase in anomaly led to a $0;04,40º$ increase in anomaly rearwards.
Thus, to calculate the true hourly motion, we take the moon’s mean speed per hour about the epicycle, $0;32,40 \; \frac{º^{_m}}{hr}$3, multiply it by the increment per hour, $0;04,40 \; \frac{º^{_a}}{º^{_m}}$ and then subtract this from the moon’s mean motion in longitude per hour (since the increase in anomaly is rearwards).
$$0;32,56 \; \frac{º}{hr} – (0;32,40 \; \frac{º^{_m}}{hr} \cdot 0;04,40 \; \frac{º^{_a}}{º^{_m}}) = 0;30,24 \; \frac{º}{hr}$$
Thus, the moon’s true speed at this time is $0;30,24 \; \frac{º}{hr}.$
Lastly, we need to take this speed and multiply it by the distance the moon needs to catch up to be opposite the sun which we calculated to be $3;03º$ also recalling that we need to increase that value by $\frac{1}{12}$ to account for the extra distance the moon will need to catch up since the sun is moving too. Thus we write:
$$3;03º \cdot \frac{13}{12} \div 0;30,24 \; \frac{º}{hr} = 6;31 \; hours$$
Therefore, the true opposition will take place $6$ hours and $31$ minutes after mean opposition. Taking this amount of time and multiplying it by the mean motion of anomaly from the Lunar Anomaly Table for a period of $1$ hour:
$$6;31 \cdot 0;32,40º = 3;32,53º$$
This is how much the moon moved around the epicycle in that amount of time. Thus, we’ll need to add it to the position on the epicycle from mean opposition:
$$8;49,33º + 3;32,53º = 12;22,26º$$
This is what we’ll need to enter into the correction table. Doing so and interpolating between rows, I find a correction value of $0;00,46$. This is basically a percent4 how far from greatest to least distance we are, and Ptolemy then tells us how to use this:
We then take this fraction of the difference between the [two sets of ] digits, [derived from] the two tables, which we wrote down, and also of the difference between the [two sets of] minutes of travel, and add the results to the amounts derived from the first table.
Here, Ptolemy is basically telling us we will interpolate between the values for greatest to least distance using this number to tell us how far from the values for greatest we are towards the values for least.
So starting with the digits, the difference between greatest and least is $2;09,16$. And $0;00,46$ of that is $0;01,39$. This is then added to the amount for greatest distance to get a total digits of $2;32,21$.
Similarly for the minutes of immersion, the difference between the two is $13;17,02$. And $0;00,46$ of that is $0;10,11$ which gets added to the duration at greatest distance to get a total immersion time of $26;18,42$ minutes. But hold that thought a moment because we’ll return to it after this slight detour because while this is all well and good for this eclipse, where the argument of latitude was in both tables, what happens when it has a value in the table for least distance, but not at greatest? Ptolemy tells us how to deal with it then:
If, however, it happens that the argument of latitude falls within the range of the second table [i.e., the one for least distance] only, we take [as a final result] the appropriate fraction (determined by the number of sixtieths found [from the correction table]) of the digits and minutes [of travel] corresponding [to the argument of latitude] in the second table alone. The number of digits which we find as a result of the above correction will give us the magnitude of the obscuration in the lunar diameter at mid-eclipse.
In short, for greatest distance, we would simply use zero for the digits and immersion time. This doesn’t seem like an entirely correct solution, but is an edge case anyway, so I’m not overly concerned about it and won’t say anything further.
But returning to the case in which we do have values for both greatest and least distance and closing out
As for the minutes [of travel] resulting from the same correction, we always increase them by $\frac{1}{12}th$ to allow for the sun’s additional motion [during the phase of the eclipse], and divide the result by the moon’s anomalist [i.e., true] hourly motion at that point. The results of the division will give us the duration of each phase of the eclipse in equinoctial hours.
In other words, we need to take the value for the duration we’d gotten started on above, increase it by $\frac{1}{12}$ and then divide by the speed of the moon we came up with:
$$26;18,42 \cdot \frac{13}{12} \div 0;30,24 = 56;15 \; min$$
This give the time from the beginning of the eclipse to mid-eclipse or, as Ptolemy calls it, the “duration of the immersion” as doing this calculation for the fourth (Minutes of Immersion) column
will give the duration of the immersion (and also that of the emersion likewise); and the result derived from the fifth column will give the duration of half of the totality.
In the case we’re considering here, the eclipse is not total, so we cannot apply the later half of this statement.
The times of entry and exit at the beginning and end [of the various phases] can be derived immediately by adding or subtracting the individual durations to or from the time of the middle of totality, that is, approximately, the time of true opposition.
For this, we’ll need to first calculate the time of true opposition. This is done using the Days of Thoth I included in the first table. This tells us how far into the month the mean opposition occurs: on the $18^{th}$ day at $4:35$pm5. We then add to the time of mean conjunction, the time to true conjunction:
$$4;35 + 6;31 = 11;06 pm$$
Therefore, mid-eclipse is at $11:06$pm.
We can then add or subtract $57$ minutes to determine when the eclipse begins ($10:09$pm) and ends ($12:03$am).
We can also immediately find the area digits by enter with the digits of the diameter into the final small table and taking the corresponding amount in the third column.
Toomer’s example doesn’t include this, but it’s a straightforward interpolation in the Table for Magnitudes of Solar and Lunar [Eclipses] table.
And that’s how to calculate everything Ptolemy covers for lunar eclipses.
The chapter doesn’t quite end there. Ptolemy reminds us that, because of the changing speed of the lunar and solar anomalies,
reason informs us that the time interval from the beginning of an eclipse to its middle is not always equal to the time interval from mid-eclipse to the end.
However, he’s not going to be bothered with it since,
as far as the senses are concerned, no noticeable error with respect to the phenomena would result from supposing these intervals [to be] equal in times. For, even when [the luminaries] are near mean speed, where the change [in speed] resulting from an [equal] increment [in the argument] is greater [than elsewhere], the motion of the number of hours represented by the whole duration of [even] the maximum possible eclipse does not exhibit the least noticeable difference [in duration] due to the change [in speed].
TL:DR – It’s too small to care about.
Moving on,
we can [now] see, by examining the matter on the above basis, that we were quite right to reject as erroneous the period for the moon’s [return in] latitude which Hipparchus demonstrated.
This seemingly comes out of nowhere, but is referring to the discussion of Hipparchus’ value in the mean motion of latitude that was discussed in this post. Specifically, both analyzed a period between eclipses that spanned $211,438$ days and $23$ hours. Hipparchus’s value for the mean motion was $3$ arcminutes past a full revolution, but Ptolemy found it to be $12$ arcminutes.
Ptolemy then goes into detail about Hipparchus’ error but I’ll shorten it by stating that Ptolemy finds the error to be Hipparchus not properly accounting for the anomaly at each eclipse as well as failing to make any correction needed due to the epicycle bringing the moon closer. However, these two effects nearly cancelled each other out, which resulted in Hipparchus very nearly having the correct answer.
- I chose this as opposed to the example from Neugebauer’s History of Ancient Mathematical Astronomy because Ptolemy’s methodology is meant to calculate a “required opposition” (i.e., a specific opposition) whereas Neugebauer’s example looks at an entire year and asks for what oppositions we should expect eclipses.
- My values are differing by several minutes from Toomer’s here. Likely because I was doing calculations in Excel, so less rounding happened along the way.
- This comes from the Lunar Mean Motions table.
- Except we’re presenting this in sexagesimal so a “persecond”? “Persexagesimal?”
- Recalling that the Egyptian day begins at noon.