It’s been awhile since I’ve done an Almagest post so quick recap. In Book IV, we worked out a first model for the moon, which was a simple epicycle model inclined to the plane of the sun’s sphere. In this book, Ptolemy showed us that this model was insufficient as the moon’s speed varies more than should predict and so we added an eccentric as well as having the center of the eccentre rotate around the observer. Finally, we introduced the concept of a “mean apogee” which is the position on the epicycle we’ll need to measure from in order to do calculations.
That’s been a lot, but with all of this completed, we should now be able to use it to calculate lunar positions which Ptolemy walks us through in this chapter. Unfortunately, he does this in the form of a generic prescription of steps instead of a concrete example. Fortunately, Toomer provides an example1 that I’ll use to supplement Ptolemy’s narration.
For the example, we’ll attempt to calculate the position of the moon on Nabonassar 466, Thoth [I] 7/8, 2 equinoctial hours after midnight2.
The first step is to
take, for the moment in question at Alexandria, the mean motions of the moon in longitude, elongation, anomaly and latitude, in the way we explained.
First off, Toomer notes that Ptolemy didn’t actually explain this. At least, not in the context of the moon. But it’s basically the same as we did for the sun. What it means is that we need to first determine the time since epoch, and then look up the corresponding values from the respective tables.
Toomer provides the interval as 465 years, 6 days, and 14 hours.
The values Ptolemy has mentioned here all come from the lunar mean motions table we derived in the last book. As usual, we’ll need to break the period up to intervals that are on the table which can be done as shown below:
| Period | Longitude | Elongation | Anomaly | Latitude |
| Epoch Position | $41;22,00º$ | $70;37º$ | $268;49º$ | $354;15º$ |
| $450$ years | $260;46,44º$ | $10;11,03º$ | $323;26,05º$ | $320;54;06º$ |
| $15$ years | $140;41,33º$ | $144;20,22º$ | $250;46,52º$ | $70;41,48º$ |
| $6$ days | $79;3,30º$ | $73;8,40º$ | $78;23,24º$ | $79;22,34º$ |
| $14$ hours | $7;41,10º$ | $7;6,41º$ | $7;37,16º$ | $7;43,02º$ |
| TOTAL | $169;35$ | $305;24º$ | $209;03º$ | $112;57º$ |
Here, I’ve included the starting position at epoch in the first line. Ptolemy oddly skips this for the longitude as it is included in a later step but I felt it better to include here. The total line subtracts out any full revolutions of $360º$ and is rounded to the nearest minute.
Next, we
double the figure computed for the elongation, and (after subtracting $360º$ if necessary), enter with this into the table of anomaly and take the corresponding amount in the third column. If the double elongation is less than $180º$ we add the amount to the mean anomaly, but if the double elongation is greater than $180º$ we subtract the amount from the mean anomaly.
From above, we showed the average elongation to be $305;24º$. Doubling that we would have $610;48º$ from which we can subtract out a full circle to get $250;48º$.
We’ll take that, and look it up in the complete table of lunar anomaly. There isn’t a line specifically for $250;48º$, but fortunately, it’s bracketed on either side by $13;03º$ so we can use that as the value3.
Since the value of the double elongation is greater than $180º$, we subtract this from the mean anomaly:
$$209;03º – 13;03º = 196;00º$$
This gives us the true anomaly which
we enter …. into the same table, and take the corresponding equation in the fourth column, and also the corresponding increment in the fifth column, and write down [both] separately.
So for the epicyclic equation (column $4$), we look up $196;00º$ which is half way between two values so fairly easy to estimate as $1;33º$4.
For the increment in the epicyclic equation (column $5$), I come up with $0;57º$5.
Next we enter the double mean elongation into the same table, take the sixtieths corresponding to it in the sixth column, multiply the increment which we wrote down separately by that number of sixtieths, and always add the result to the previously computed equation from the fourth column.
Again, our double elongation was $250;48º$ so that’s what we’ll look up from the sixth column to find the sixtieths. When doing so, I get $0;36,52$.
Now do to the math:
$$0;35,52º \cdot 0;57 + 1;33º = 2;07º$$
If the true anomaly is less than $180º$, we subtract this sum from the mean longitude and mean [argument of] latitude, but add it to them if the true anomaly is greater than $180º$.
In our case, the true anomaly we calculated to be $196;00º$ which is greater than $180º$, we’ll add this result to our figures for the mean longitude and mean latitude to get a true longitude of $171;42º$ and an argument of latitude of $115;04º$.
Thus, we have [two] numbers: we add the one for the longitude to the position [of the mean moon] at epoch: the result will be the true position of the moon [in longitude].
This is the step I mentioned I did already when we calculated the movement during the interval from epoch so we don’t need to do it again here.
With the one for the [argument of ] latitude, counted from the northern limit, we enter into the same table: the number corresponding to it in the seventh column will be the distance of the moon’s centre from the ecliptic, measured along the great circle through the poles of the ecliptic. If the argument falls within the first 15 lines, it will be to the north [of the ecliptic], but if it falls below the first 15 lines, it will be to the south.
So now we take the $115;04º$ (the argument of latitude) and look it up in the latitude column (column $7$). I get a value of $2;07º$. Since the value we looked up is below the fifteenth line of the table, this indicates the moon is below the ecliptic.
This completes how we can compute the position of the moon. However, we’re still a long way from the end of this book. The moon is close enough to the earth that being at different positions on the surface of the earth can produce effects of parallax, which we’ll explore in future chapters!
- Appendix A, Example 9 wherein Toomer begins a calculation from Book VII, Chapter $3$. We’ll continue that calculation in Chapter $18$ of this book, again as an example.
- This is actually an upcoming solution from Book VII, chapter $3$.
- The fact that the values for $252º$ and $249º$ both have a value of $13;03º$ suggests these lie on either side of a local maximum, so it’s likely that our value of $250;48º$ might actually be slightly higher. Indeed, Toomer comes up with a value of $13;04º$ which implies he went through the calculations instead of relying on the table.
- Toomer comes up with a slightly lower value of $1;30º$, likely both because of the disagreement I mentioned in the last comment, as well as likely going through the full calculations again.
- Toomer gets $0;55º$.