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Almagest Book III: On the Calculation of the Solar Position

This chapter is easily the shortest one in Book III. It literally consists of a paragraph (which I’ll quote in its entirely but break into two for ease of reading) that gives a very quick description of how one calculates the position of the sun at any given time from the epoch derived in the last chapter.

So whenever we want to know the sun’s position for any required time, we take the time from epoch to the given moment (reckoned with respect to the local time at Alexandria), and enter with it into the table of mean motion. We add up the degrees [and their subdivisions] corresponding to the various arguments [18-year periods, years, months, etc.], add to this the elongation [from apogee at epoch], 265;15º, subtract the complete revolutions from the total, and count the result forward from Gemini 5;30º rearwards through [i.e. in the order of] the signs. The point we come to will be the mean position of the sun.

Next we enter the same number, that is the distance from apogee to the sun’s mean position, into the table of anomaly, and take the corresponding amount in the third column. If the argument falls in the first column, that is if it is less than 180º, we subtract the [equation] from the mean position; but if the argument falls in the second column, i.e. is greater than 180º, we add it to the mean position. Thus we obtain the true or apparent [position of] the sun.

That’s the entire chapter. But let’s walk through an example1.

In this example, we’ll attempt to determine the position of the sun on Nabonassar 548, Mecchir 9th, at $1 \frac{1}{3}$ equinoctial hours after midnight2.

We add up the degrees [and their subdivisions] corresponding to the various arguments [18-year periods, years, months, etc.]

First, we’ll tackle the years. Remember that the era started with year 1. As such, we’re really only 547 years and change beyond the beginning of the Nabonassar era and thus the epoch. We can break that up into 540 years and 7 years, both of which are in the table of mean motion. The change in apogee over that time for 540 years is 228;42,48º and for 7 years is 358;17,53º.

Next, Mechir is the 6th month, but this date isn’t through it yet. So, since there’s always 30 days per month, the previous 5 months contributed 150 days. That’s another 147;50,43º.

Similarly, the 9th day isn’t complete yet, so we add 8 days, which is 7;53,6º.

Now recall that Ptolemy defines his days from noon when discussing the epoch. So if we’re $1 \frac{1}{3}$ hours after midnight, that’s actually $13 \frac{1}{3}$ hours past the start of the day. We’ll break that into 13 hours (0;32,2º) and 20 minutes (divide 1 minute by 3 to get 0;0,49º).

The sum of all this: 743;17,21º.

add to this the elongation [from apogee at epoch], 265;15º, subtract the complete revolutions from the total

I’ll do these two steps backwards3. First, we can subtract out 720º of that since that’s two full rotations, which leaves us with 23;17,21º that the sun has advanced along the eccentre in that amount of time relative to where it was at the epoch.

Thus, we add that increase to the distance after apogee which we calculated in the last post, of 265;15º to get 288;32,21º as the position of the sun on the eccentre.

count the result forward from Gemini 5;30º

Note from above that that 265;15º wasn’t actually the position from the vernal equinox4, but apogee. So to convert to true ecliptic longitude, we need to add in the distance of apogee after the vernal equinox which we determined back in this post to be 65;30º. So we need to add 65;30º to get 354;2,21º5.

Next we enter the same number, that is the distance from apogee to the sun’s mean position, into the table of anomaly, and take the corresponding amount in the third column. If the argument falls in the first column, that is if it is less than 180º, we subtract the [equation] from the mean position; but if the argument falls in the second column, i.e. is greater than 180º, we add it to the mean position.

So far, we’ve only gotten the position of the mean sun on the eccentre. We want to know the apparent position. So we apply a value from our table of anomaly, corresponding to the value before we added the 65;30º which was 288;32,21º. Unfortunately, the table isn’t quite that specific since, in this range, it’s broken down into 6º intervals. So we’ll need to do some extrapolation.

No explanation is given on how to do this, but my method would be to determine the percent 0;32,21º into that 6º interval6, which is 8.98%. Then we can apply that percentage to the difference between the equation of anomaly for 288º and 294º which is a difference of 0;6º which is 0;0,32º. Note that the equation from of anomaly from 288º to 294º is falling, so we need to subtract this from the value of the equation of anomaly for 288º which was 2;14º, giving the equation of anomaly to be 2;13,28º.

Next, we need to determine whether to add or subtract this from the value of the mean sun. Here, Ptolemy is quite specific. If the position of the mean sun is in the first column, we subtract. If it’s in the second, we add. Since we found this value of the mean sun’s position in the second column, we add.

$$354;2,21º + 2;13,28º = 356;15,49º$$

That’s almost all the way back around the full circle to aries, but not quite. It’s 26;15,49º into pisces, which would be the apparent position of the sun on the given date at the given time.

That sure was a lot to take in, so let me summarize the method:

1. Determine the number years, days, and hours since the beginning of the epoch.

2. Use this to determine the total increment in mean motion from the table of mean motion.

3. Add to that, 265;15º to get the difference in ecliptic longitude from the apogee. Set this number aside until step 5.

4. Add another 65;30 to get the true ecliptic longitude of the mean sun.

5. Use the total at the end of step 3, to look up the equation of anomaly from the table of the sun’s anomaly.

6. If the value from step 3 was found in the first column, subtract this from the total in step 4. If it was in the second column, add it.

This will give the true longitude for the sun for a given date and time.

With that out of the way, there’s only one chapter left in Book III, and we’ll cover it in the next post!

 

  1. Example taken from Toomer’s Appendix A.
  2. This is actually a date we’ll be looking at in Book IV, so it’s quite convenient we’re getting it out of the way now!
  3. It doesn’t matter since these operations are commutative.
  4. Which is our zero point in ecliptic longitude
  5. It should be apparent that we could combine these last two steps: Instead of adding the distance after apogee and then the distance between the vernal equinox and apogee, we could just note that the actual position of the mean sun at the beginning of the epoch was, as we noted in the last post, 330;45º, or 0;45 into Pisces. However, we’re going to need that figure we just calculated that was the distance from apogee in just a moment to calculate the anomaly, which is probably why Ptolemy breaks this down into separate steps.
  6. Since 288º is in the table.