Finally, we’ve arrived at the end of Book III where we’ve arrived at a well developed model of the solar motion. But before closing out, Ptolemy has one last chapter to discuss the inequality of the solar day. Ptolemy states the problem as follows:
the mean motions which we tabulate for each body are all arranged on the simple system of equal increments, as if all solar days were of equal length. However, it can be seen that this is not so.
What Ptolemy is really getting at here is that the term “day” is somewhat ambiguous. As such, the different ways by which we might measure a “day” are explored in this chapter.
The first is by measuring the position of the sun “by its return to the horizon, or to the meridian.” This is the definition of a solar day, where time is reckoned by the sun returning to the same position in the sky. I’ve briefly mentioned it in a few posts1, but now we’ll need to explore it in a bit more depth.
Because a solar day is determined with respect to the sun and the sun moves along the ecliptic, it will be different than time measured with respect to the fixed stars2. Specifically, the sun moves rearwards (to the left) along the ecliptic, meaning that it takes just under 4 minutes longer to cross the meridian each day as compared to the background stars. Or as Ptolemy phrases it,
the mean solar day is the period comprising the passage of the 360 time-degrees of one revolution of the equator plus approximately 0;59 time-degrees, which is the amount of the mean motion of the sun during that period.
Note that this is specifically the mean sun and that 0;59º extra is not quite constant for two reasons. The first is that the anomaly is not constant; sometimes it will add a small amount, and sometimes subtract it. Second, because the tilt of the ecliptic with respect to the horizon changes throughout the year resulting in different lengths of the ecliptic rising in different amounts of time. These are both small effects and would be difficult to distinguish for a single day, but are apparent over long periods, so we’ll dig into each of those a bit more as well.
First off, for anomalistic motion, the greatest difference from the mean motion occurs
between the two positions of the sun where its [true] speed equals its mean speed.
I previewed those two points at the end of the post on the table of the anomaly, where we showed that these two points were at 90º and 270º for the mean motion. At each of those two points, the anomaly is at its greatest. But since at 90º, the equation of anomaly is negative 2;23º and at 270º it’s positive 2;23º, that’s a total difference of 4;46º which Ptolemy rounds off to $4 \frac{3}{4}$. As Ptolemy noted, that adds up to quite the difference.
Now let’s turn to the other reason the definition of a solar day varies: The rising time of a section of the ecliptic is not constant. Ptolemy concludes that a section of the ecliptic will rise either the most quickly, or the most slowly, near the solstices. Specifically near the summer solstice, it rises more slowly, and near the winter solstice, it rises more quickly.
This can easily be seen by referring to our table from II.8 in which we determined the rising time for various pieces of the ecliptic. If we add up the the rising times for the two signs on either side of the summer solstice (Gemini and Cancer), we see that 60º chunk rises in 64;32 time-degrees as compared to the 60 time-degrees it would have taken for the mean motion. That’s a difference of about $4 \frac{1}{2}$ time-degrees. If you do the same thing for the two signs on either side of the winter solstice (Virgo and Libra), we find they rise in 55;40 time degrees, which is again about $4 \frac{1}{2}$ time-degrees from the mean motion for a total variance between the two of 9 time-degrees.
As such, Ptolemy notes that using the time of sunrise or sunset is not a good way to define an astronomical day as different latitudes will have different rising and setting times. Instead, we should use the time of the sun on the meridian (i.e. local noon) as this is the same for all latitudes.
So how do these two effects stack up?
The [maximum] subtractive result from both effects occurs over the interval from the middle of Aquarius to [the end of] Libra and the [maximum] additive [result] … from [the beginning of] Scorpio to the middle of Aquarius. Both of these intervals produce a maximum additive or subtractive result which is composed of about $3 \frac{2}{3}º$ due to the effect of the solar anomaly, and about $4 \frac{2}{3}º$ due to the [variation in the time of] meridian-crossing.
Ptolemy states this without proof and so I won’t get into it much further. A full proof is available in Neugebauer’s History of Ancient Mathematical Astronomy on page $62-63$ as Example A. But without going through the proof I think this is relatively to understand by looking at a graph. Unfortunately, Toomer doesn’t provide one, but Neugebauer does:
Here we see the two separate effects by the narrow lines with the dots. The first, the variance due to anomalistic motion, is the one with its first peak in Taurus and a second in Scorpio. The variance due to the uneven rising times of the ecliptic is the one with its peak between Leo and Virgo.
The thick black line is the sum of the two variances at any given point. As we can see, it’s not a simple function. Ptolemy doesn’t dig into the details of it here, calculating it at each point as was done for this graph so I’ll simply note that the combination of these two effects is known modernly as the equation of time3.
In general, we can see that this effect has a fairly large spread from its highest peak, to its lowest point. Ptolemy notes that, if you used a true solar day instead of the mean motion, the motion of the planets is sufficiently slow that the change in position over this period of time would be imperceptible, but the moon (as we’ll be learning about in the next Book) moves sufficiently fast4 that a difference in its motion over that interval would be readily perceptible.
Ptolemy concludes by giving instructions for how to convert any interval given in true solar days to mean solar days:
Determine the ecliptic position of the sun in both mean and anomalistic motion at the beginning and end of each interval
Take the increment, in degrees [from the first] anomalistic (i.e. apparent) position to [the second] apparent position, enter with it into the table of rising-times at sphaera recta, and [thus] determine the time taken by the apparent distance [of the sun between the first and second positions] to the meridian, measured in degrees of the equator.
We then take the difference between this number of time-degrees and the mean distance [of the sun from first to second positions], measured in degrees, and convert this difference, which is in time-degrees, to a fraction of an equinoctial hour.
We add the result to the number of [true] solar days given if the amount of the time-degrees [corresponding to the rising-time of the apparent motion] was greater than the mean motion, or subtract it if less.
The interval we arrive at will be corrected for expression in mean solar days.
We’ll be using this in the next book when we start looking at the mean motion of the moon, so before finishing out this post, let’s do an example.
To start, we’ll need two dates to define an interval of time.
We’ll use Hadrian 17 (Nabonassar 880) Pauni [X] 20/21, 11:15pm as our first time and Hadrian 19 (Nabonassar 882) Choiak [IV] 2/3, 11:00pm as our second5.
To begin, we’ll need to determine the position of the mean and anomalistic sun at both times. The method by which to do so was discussed in this post. So let’s go through it as practice, at least for the first of these.
First, we need to sum break the interval of time since the beginning of the epoch down into pieces we can look up on the mean motion table. Let’s begin with years. As stated above, we’re in year 880, which isn’t complete yet, so 879 years have actually elapsed. We can break this down into 810 + 54 + 15, all of which are in our mean motion table.
Next, the months. Our first period is in Pauni which is the 10th month, so we’re through 9, which we can look up on the table as well, recalling that each month in the Egyptian calendar has 30 days. So that’s 270 days.
Then, we’re on day 20, so 19 are complete which we can find on the table as well.
Lastly, hours. A reminder these are reckoned from noon, so 11:15 pm is only 11 hours and 15 minutes into that day. However, I’m going to round this down to 11 hours even because the change of solar position in one hour is only ~2 ;30 arcminutes. So if we were to divide that by 4, it’s a little over 30 arcseconds. Since we’ll only be reporting to the nearest arcminute, this may throw us off by a rounding of 1, but we’ll deal with that shortly.
Now let’s lay that all out in a way that’s a bit more easy to digest:
| º | ‘ | ” | ”’ | ”” | ””’ | ””” | |
| 810 years | 163 | 4 | 12 | 15 | 25 | 52 | 30 |
| 54 years | 346 | 52 | 16 | 49 | 1 | 43 | 30 |
| 15 years | 356 | 21 | 11 | 20 | 17 | 8 | 45 |
| 270 days (9 months) | 266 | 7 | 17 | 29 | 26 | 19 | 30 |
| 19 days | 18 | 43 | 3 | 27 | 10 | 57 | 49 |
| 11 hours | 0 | 27 | 6 | 17 | 53 | 33 | 14 |
| TOTAL | 71 | 35 | 7 | 37 | 15 | 35 | 18 |
This has determined the advance from the point the sun was at at the beginning of apogee. So now, we’ll need to determine the distance to apogee by adding the distance in ecliptic longitude the point at the beginning of the epoch was from apogee:
| º | ‘ | ” | ”’ | ”” | ””’ | ””” | |
| Advance | 71 | 35 | 7 | 37 | 15 | 35 | 18 |
| To Apogee | 265 | 15 | 0 | 0 | 0 | 0 | 0 |
| TOTAL | 336 | 50 | 7 | 37 | 15 | 35 | 18 |
This is all working towards the position of the mean sun, but we do this step to pause for a moment and use this value to determine the anomaly from our table of the solar anomaly. It’s in between values, so we’ll do some rough estimation and call it 0;55º.
We’ll set that aside until we finish the mean sun calculation which we’ll continue now by adding in the distance between apogee and equinox:
| º | ‘ | ” | ”’ | ”” | ””’ | ””” | |
| Advance from Apogee | 336 | 50 | 7 | 37 | 15 | 35 | 18 |
| To Vernal Equinox | 65 | 30 | 0 | 0 | 0 | 0 | 0 |
| TOTAL | 42 | 20 | 7 | 37 | 15 | 35 | 18 |
So the position of the mean sun on the first date is 42;20º. Toomer has this as 42;21º which would follow if I’d included the extra 30 seconds of arc from the 15 minutes I rounded off earlier, so let’s adopt this.
If we want the position of the true sun, we need to take the anomaly into consideration. Since the value we looked up came from the second column, we need to add it, which gets the true position as 43;15º.
After all that work, that was only the mean and true position for the first of the two dates. If we repeated this for the second, we’d find the mean position as 206;42º and the true as 205;10º.
And that got us through the first step. Now for the part about time-degrees. For that, we’ll need to use our rising-times table. Specifically, we’ll use it at sphaera recta, plugging in the values for the position of the true sun.
It’s been awhile since we’ve used this table, so let’s walk through it more slowly. First, we’ll break the 43;15º into 40º + 3;15º. The 40º is right on the table and has a rising time of 37;30 time-degrees.
So we just need to figure out the 3;15º. Since the table is in 10º intervals, that’s $\frac{3;15}{10} = 32.5\%$ of the way through the interval. Since the interval (shown in the second column) is 9;58º, that an additional 3;14 time degrees for a total of 40;44 time-degrees. Repeating that for the second time we get a rising time of 203;17 time-degrees.
Now take the difference of the two: $203;17 – 40;44 = 162;33$.
Next he says to take the difference in time-degrees (which we just calculated) with the difference in position of the mean sun. We’ve calculated the positions, but not their difference. So let’s do that now:
$$206;42º – 42;21º = 164;21º$$
Now we can take the difference between the two:
$$164;21º – 162;33 = 1;48º$$
Next, we need to convert this to the proportion of the equinoctial hour. Really, this just means take this number as the proportion of 360º to the length of a day (24 hours = 1440 minutes). So writing that out:
$$\frac{1;48º}{360º} = 0.5\% = \frac{x}{1440}$$
Solving for x:
$$0.5\% \cdot 1440 = 7;12 \; minutes$$
Now, what to do with this?
Ptolemy says we add this if difference of the rising times was greater than the difference in mean motion and subtract otherwise.
In this example, the difference in rising times was $162;33º$ and the difference in mean motion was $164;21º$. The mean motion is greater so we subtract this from our original interval was 1 year, 166 days, 23 hours and 45 minutes. Thus, if we subtract 7;12 minutes from that we get the interval in mean solar days to be 1 year, 166 days, 23 hours, 37 minutes, and 48 seconds.
That was a lot of work, so let’s summarize the steps:
1. Calculate both the mean and true positions for the sun at both dates. If no initial date is given, use the beginning of the epoch.
2. Use the true position to look up the cumulative rising times at sphaera recta for both dates.
3. Take the differences in both the mean positions and the rising times. Then the difference between these differences.
4. Convert this to a time period by taking this as a proportion of 360º and multiplying by the number of minutes in a day.
5. If the difference of the rising times was greater than the difference of the mean motion, add that from the total interval. If vice versa, subtract.
So that closes out Book III, having to do with the sun. In the next book, we’ll be starting on the moon which seems to be a natural progression as the phases of the moon depend on the sun.
UPDATE: I’ve been having to revisit this post nearly $2$ years after writing it and feel there’s quite a bit more that needs to be said here as dealing with a particular use case of this conversion has proven rather difficult.
In part, it’s because I was rather lazy as, when Ptolemy has made use of this conversion, I made a poor habit of not checking his figures and simply accepted his adjustment. Had I bothered to check a few, I would have realized my gap in understanding far sooner. Sadly, beyond this sample problem I have had no practice with this method and my ignorance was allowed to fester.
This is particularly inexcusable on my part because this sort of calculation is frequent in Book IV. Nearly any time Ptolemy gives a date and time for a position of the moon he gives two values. The first is “reckoned simply” (i.e., unadjusted using this method). The second is often stated as “reckoned accurately” (i.e., with an adjustment for the equation of time applied).
However, when working through an example of Toomer’s in which he did not present a time that was already adjusted and only gives an adjustment with only a partial calculation I was forced to revisit this procedure. After several hours of struggling with it, I realized that I had failed to understand a key fact: When making this sort of adjustment, typically two dates are not given. Rather, the first date is to be taken as the beginning of the epoch.
This was something I initially missed when writing this post, but in retrospect is reasonably apparent. Indeed, there is a final sentence of Ptolemy’s in this chapter that I initially omitted as I failed to understand its importance:
At our epoch, that is, Year 1 of Nabonassar, Though 1 in the Egyptian calendar, noon, the position of the sun was, in mean motion as we showed just above, $0;45º$ into Pisces [$330;45º$ ecliptic longitude], and in anomalistic motion about $3;08º$ into Pisces [$333;08$ ecliptic longitude].
A further note I’d add to this is that for the epoch, the rising time at sphaera recta for the true position is $335;07,02º$
A further footnote from Toomer makes this even more apparent wherein he states:
Ptolemy gives the data for era Nabonassar because they will be required every time needs to compute the lunar position accurately.
The example originally done in this post is still entirely valid, as it’s giving the adjusted interval between two dates. But if we’re considering a more general adjustment, then using the beginning of the epoch is the appropriate starting date.
However, for a further example where I do use the beginning of the epoch as the initial date, please see my post on calculating solar eclipses (I hope to provide a link when that post goes live).
- Specifically it got a mention in the two posts relating to converting Alt-Az to RA-Dec which are unrelated to the Almagest, and this post where I discussed equal hours in relation to a solar day without defining it further.
- Typically when teaching this, I say “distant stars”, but since I’m trying to communicate this in the way Ptolemy understands it, I’ll avoid doing so.
- Pedersen notes that the later Latin phrase for this is the equatio dierum
- About half a degree per hour!
- These dates come from an upcoming calculation we’ll see in the next book which are used as Example 8 in Appendix A of the Toomer translation.