Almagest Book IV: The Babylonian Eclipse Triple – Solar/Lunar Positions & Epicyclic Motion of Anomaly

So far in this book we’ve covered the ancient Greek values for the various motions of the moon. For the most part Ptolemy has accepted them as authoritative, but to demonstrate some of the methodology, Ptolemy wants to walk us through one: The lunar anomaly.

We shall use, first, among the most ancient eclipses available to us, three [which we have selected] as being recorded in an unambiguous fashion, and, secondly, [we shall repeat the procedure] using among contemporary eclipses, three which we ourselves have observed very accurately. In this way our results will be valid over as long a period as possible, and in particular, it will be apparent that approximately the same equation of anomaly results from both demonstrations, and that the increment in the mean motion [between two sets of eclipses] agrees with that computed from the above periods.

To begin, Ptolemy lays out the general lunar model:

Imagine a circle in the sphere of the moon which is concentric to and lies in the same plane as the ecliptic. Inclined to this, at an angle corresponding to the amount of its [maximum] deviation in latitude, is another circle, which moves uniformly in advance (with respect to the centre of the ecliptic) with a speed equal to the difference between the motions in latitude and longitude. On this inclined circle we suppose the so-called ‘epicycle’ to be carried, with a uniform motion, towards the rear with respect to the heavens, corresponding to the motion in latitude. (This motion, obviously, will represent the pmean[ motion in longitude with respect to the ecliptic). On the epicycle itself [we suppose] the moon to move, in such a way that on the arc near apogee its motion is in advance with respect to the heavens, at a speed corresponding to the period of return in anomaly.

That’s a lot to take in, so let’s draw it out.

Here we can see the lunar sphere. On it, there is the horizontal circle which is in the plane of the ecliptic. Tilted with respect to that is the deferent. The epicycle rides on that deferent in the same plane, which is to say it’s also tilted with respect to the circle in the plane of the ecliptic.

The amount of tilt between the ecliptic plane and the plane of the deferent is determined by how far above or below the ecliptic the moon goes, which is about 5º.

Since we established in the last few posts that the deferent will rotate, this will mean that the nodes (☊, which is the ascending node and ☋ which is the descending node) will slowly rotate with it. While we previously inferred that the deferent must be rotating counter-clockwise, Ptolemy spells this out here giving the same reason I did previously¹.

However, here is where we finally arrive at the statement from Ptolemy that he is treating the epicycle as rotating clockwise.

Although Ptolemy has now laid out the overall geometry for this first model, we’ll immediately discard a portion of it as

for the purposes of the present demonstration we shall suffer no ill consequences if we neglect the advance motion in latitude from the inclination of the moon’s orbit, since such a small inclination has no noticeable effect on the position in longitude.

In other words, for now, we’ll ignore that ~5º tilt of the deferent with respect to the ecliptic because it doesn’t throw things off by very much². So let’s go ahead and get started by looking at some historical eclipses.

The first Ptolemy uses occurred:

in the first year of Mardokempad, Thoth [I] 29/30 in the Egyptian calendar [721 BCE March 19/20³]…. well over an hour after moonrise, and was total.

From this, we’ll first want to figure out exactly where the sun was along the ecliptic at the moment of that eclipse. To do so, Ptolemy starts by noting that the night time was about 12 equinoctial hours long on that date, presumably because this date was within a day or two of the equinox. Thus, since the historical description notes it started “well over an hour after moonrise” Ptolemy calls this an hour and a half. Thus, if the day and night were nearly equal, this would have meant that the sun set at 6:00 pm, and the eclipse started at 7:30 pm, which is $4 \frac{1}{2}$ equinoctial hours before midnight. From there, Ptolemy assumes it should have taken about 4 hours for the total eclipse to occur and thus, the midpoint of totality occurred $2 \frac{1}{2}$ equinoctial hours before midnight.

However, this eclipse was observed in Babylon⁴ and Ptolemy wants to keep things in terms of his home in Alexandria⁵ which is west of Babylon. Thus, the local time in Alexandria would be earlier. Ptolemy adopts a difference of $\frac{5}{6}$ of an equinoctial hour⁶.

So we’ll need to adjust the time at which mid-totality occurred by that amount which gives us that it happened around $3 \frac{1}{3}$ equinoctial hours before midnight, Alexandria time or 8:40 PM.

This gives us a date and time to determine where the sun was located along the ecliptic. Ptolemy doesn’t work through this, but we haven’t done any reckoning of the solar position since the last book, so let’s do it ourselves.

To do so, we’ll refer to this post in which Ptolemy explained the method and we walked through the method. To sum it up, we’ll need to first break the time period between the beginning of the epoch and the date and time in question down into chunks of time that can be looked up in the solar mean motion table. Then we’ll add that to a fixed number we derived here to get the angular distance along the ecliptic from apogee. This allows us to look that result up in the table of the sun’s anomaly. From there, we can add another fixed number to account for the angular distance between the apogee and the vernal equinox. Lastly, we’ll add in the anomaly we just looked up.

To get started, recall that Ptolemy’s era for everything began in Nabonassar I, which is 747 BCE. So if we’re in 721, that’s a difference of 26 years. There is no 26 year reckoning in our mean motion table, so we’ll need to add a few together to come up with one. We can do 18 + 8 years.

Next, we would tackle the months⁷, but since we’re still in the first month, we haven’t completed a month yet. So that’s a nice easy zero.

Ptolemy lists the eclipse as occurring on the 29th, which means we’re actually completed 28 days.

Lastly, recall that date reckoning happens from noon in the Almagest, so if the eclipse took place $3 \frac{1}{2}$ hours before midnight, then that’s at 8:40 PM, which is $8 \frac{1}{3}$ hours after noon. However, the amount the sun moves along the ecliptic in 20 minutes is very little. Somewhat less than the accuracy to which Ptolemy will be reporting the final number, so I’ll go ahead and just round that off to 9 hours.

So let’s get started with looking all those up in the mean motion table:

	º	‘	”	”’	””	””’	”””
18 years	355	37	25	36	20	34	30
8 years	358	3	18	2	49	8	40
0 months	0	0	0	0	0	0	0
28 days	27	35	52	2	9	50	28
9 hours	0	22	10	36	27	17	11
TOTAL	21	38	46	17	46	50	49

So after all that time, the position has advanced 21;38,46,17,46,50,49º from the point it was at the beginning of the epoch. To get the total distance from apogee, we next add that to 265;15,0,0,0,0,0.

	º	‘	”	”’	””	””’	”””
Advance from Epoch	21	38	46	17	46	50	49
To Apogee	265	15	0	0	0	0	0
TOTAL	286	53	46	17	46	50	49

This gives us the distance from apogee the sun is on the date in question, which we’ll use to look up the solar anomaly. For convenience, we’ll round this number off to an even 287º which isn’t in the table either, but given the increments in degrees are an interval of 6 and the difference in arc minutes between 282 and 288 is also 6, it allows us to quickly come up with a value for the anomaly as 2;15º.

We’ll set that aside for now and instead add in the distance between apogee and the vernal equinox to determine the ecliptic longitude of the mean sun.

	º	‘	”	”’	””	””’	”””
Advance from Apogee	286	53	46	17	46	50	49
To Vernal Equinox	65	30	0	0	0	0	0
TOTAL	352	23	46	17	46	50	49

Now, we’ll add in the value we just determined for the anomaly. However, before doing so we must remember that if the value we looked up was in the first column, the sun was lagging behind the mean sun, so we would subtract. If it was from the second column, it was ahead so we would add. In this case, we looked up 287º which was in the second column, so we’ll add:

	º	‘	”	”’	””	””’	”””
Advance from Equinox	352	23	46	17	46	50	49
Equation of Anomaly	2	15	0	0	0	0	0
TOTAL	354	38	46	17	46	50	49

Thus the true longitudinal position of the sun at the time of mid-eclipse is 354;38,46º which Ptolemy rounds off to 354;30º, stating it’s $24 \frac{1}{2}º$ into Pisces. Since the moon would be directly opposite the sun in ecliptic longitude, this would mean the moon should be at 174;30º ecliptic longitude.

That does it for the first eclipse.

The second eclipse occured in the 2nd year of the Mardokempad reign, Thoth [I], 18/19 (720 BCE, March 8/9).

However, Ptolemy notes that this eclipse was not full. Rather,

The [maximum] obscuration… was 3 digits from the south at exactly midnight.

Pedersen explains in Survey of the Almagest, stating that a total eclipse is

equal to 12 digits, one digit being $\frac{1}{12}$ of the apparent diameter of the Moon. [For example if] the magnitude is e.g. 5 digits [this] means that at the maximum or central phase $\frac{5}{12}$ of the diameter is inside the shadow of the Earth.

So what Ptolemy is stating here is that only $\frac{3}{12}$ of the diameter moon was covered which is $\frac{1}{4}$, and that the covered part of the moon was the “south” or bottom⁸.

Ptolemy doesn’t say immediately what we need to do with the digits for now and instead moves on with calculating the position of the sun. Again, he converts the Babylonian time (midnight) to Alexandrian time ($\frac{5}{6}$ of an hour before midnight), and gives the position of the sun, without walking through the math, as $13 \frac{3}{4}º$ into Pisces which is $343 \frac{3}{4}º$ ecliptic longitude.

For the third of the Babylonian eclipses Ptolemy states it

is recorded as occurring in the (same) second year of the Mardokempad, Phamenoth [VII] 15/16 in the Egyptian calendar [720 BCE Sept 1/2]. The eclipse began, it says, after moonrise, and the [maximum] obscuration was more than half from the north.

With the previous two eclipses, they happened quite close to the vernal equinox which means the number of hours in day and night were close enough to equal as not to worry about it. However, in early September, we know they are not. Thus

the length of night at Babylon was about 11 equinoctial hours, and half the night was about $5 \frac{1}{2}$ [equinoctial] hours. Therefore the beginning of the eclipse was about 5 equinoctial hours before midnight (since it began after moonrise), and mid-eclipse about $3 \frac{1}{2}$ hours before midnight (for the total time for an eclipse of that size must have been about 3 hours).

The purpose of the above explanation is to determine the timing of the eclipse since the Babylonian account is given in relation to moonrise instead of midnight. Since eclipses only happen at opposition, moonrise is at the same time as sunset, hence the interest in knowing the division of day and night.

So what this is really saying is that sunset/moonrise happened about $5 \frac{1}{2}$ hours before midnight or around 6:30 PM Babylonian time. And since the eclipse began a little after moonrise, Ptolemy gives a half hour to account for this which places it at 5 hours before midnight or 7:00 PM.

But that was only when it began. To estimate when it was at its maximum, Ptolemy uses his knowledge of eclipses to state this one’s full duration⁹ should have been about 3 hours. Since the maximum occurs half-way between that, that’s 1.5 hours after it started which is 8:30 PM or 3.5 hours before midnight.

Again, that’s Babylon time. To convert to Alexandrian time, we need to subtract 50 minutes as we did before which puts us at 7:40 PM Alexandria time or $4 \frac{1}{3}$ hours before midnight.

Using the same methods as above, Ptolemy determines the sun at this time was $3 \frac{1}{4}º$ into Virgo, which is $153 \frac{1}{4}º$ ecliptic longitude.

Next, Ptolemy starts paring up the eclipses we just looked at. Specifically, he wants to find the changes in ecliptic longitude between them. As I showed in the example I walked through above, we could determine the position of the moon by adding or subtracting 180º from the solar position, since we’d be doing that for each one, it doesn’t make a difference. So long as we’re talking about the change in position, we can use the sun’s position just as easily which is what Ptolemy does. I’ll follow suit.

Comparing the first eclipse to the second: The sun in the first was at 354;30º. In the second 343;45º. Subtracting the first from the second we get a change in position of 349;15º.

Similarly, comparing the second to the third: The sun during the second eclipse was again at 343;45º and during the third was at 153;15º. Subtracting the third from the second we get a change in position of 169;30º.

Next, Ptolemy looks at the time intervals. From the first to the second, it happened in 354 days, 2 hours and 30 minutes if you take a difference of the two times given. However, this could also be expressed in terms of mean solar time by converting as we explored in this post. It’s only about 4 minutes different: 354 days, 2 hours and 34 minutes which is the value we’ll use.

Similarly, from the second to the third eclipse, it’s a period of 176 days, 20 hours, and 30 minutes reckoned simply, and 174 days 20 hours, and 12 minutes if converted to mean solar days.

So having those periods in mind, Ptolemy then turns to the change in ecliptic longitude due to the mean motion of the moon over said periods using our lunar motion tables. Let’s walk through it with him. To do so, we’ll again need to break the period up into pieces we can find on the mean motion table. For the first period, I’ll use 330 days + 24 days + 2 hours + 34 minutes (which we’ll have to calculate¹⁰).

	º	‘	”	”’	””	””’	”””
330 days	28	12	22	4	17	45	0
24 days	316	13	59	25	24	12	0
2 hours	1	6	8	48	19	4	43
34 minutes	0	18	39	59	32	57	28
TOTAL	345	51	10	17	33	59	11

So overall, in the period in question, the mean moon advanced 345;51º.

Now let’s repeat that for the anomaly, again using the mean motion table.

	º	‘	”	”’	””	””’	”””
330 days	351	26	39	38	15	54	30
24 days	313	33	34	31	8	47	36
2 hours	1	5	19	29	41	29	20
34 minutes	0	18	30	31	24	45	19
TOTAL	306	24	4	10	30	56	45

So in the same interval of time, we calculate the moon moves 306;24º on the epicycle (i.e., “in anomaly”). Ptolemy comes up with a value of 306;25º. My assumption is that he rounded to the nearest arcminute based on the arcseconds as this would produce the value he arrived at and save a lot of arithmetic.

Doing the same for the other interval, from the second to the third eclipse, we get the moon would have advanced 170;7º in longitude and 150;26º in anomaly.

Now, let’s break away from the math for a moment and consider the position of the sun and moon necessary for these eclipses to happen. Since eclipses only occur when the moon is opposite, the distance of the sun travelled on a given interval must equal the distance the moon travelled during the same interval. However, the moon’s motion will be comprised of both the mean motion and an anomaly¹¹. Writing that as an equation¹²:

$$\Delta \lambda_{sun} = \Delta \lambda_{MeanMoon} + \Delta \lambda_{LunarAnomaly}$$

We’ve already found the first two pieces of this equation which means we can easily solve for the 3rd. For the first eclipse to second:

$$\Delta \lambda_{LunarAnomaly} = 349;15º – 345;51º = 3;24º$$

Stating this more plainly: in the time between these first and second eclipse, the anomaly accounted for 3;24º of the motion of the moon.

Similarly, we can do the same for the second to third eclipse¹³:

$$\Delta \lambda_{LunarAnomaly} = 169;30º – 170;7º = -0;37º$$

I’ll wrap this post up for now since walking through the math, it’s already gotten quite long. In the next post, we’ll start using these values we just derived to start building out the actual physical model.

That the overall motion of the moon with respect to the background stars is “rearwards.”
Toomer notes that it leads to a 6 arcminute error. The moon is about 30 arcminutes in diameter.
There seems to be some discrepancy here on the date in the Julian calendar for this eclipse as well as several others we’re about to discuss. The date I have given here comes from Pedersen’s Survey of the Almagest. Toomer lists the date as being in 720 BCE. In this case, I have elected to use the Pedersen date because I have found other sources that attest that the reign of Mardokempad (also known as Merodach-baladan) began in 721 BCE.
Which is in modern day central Iraq.
Modern day northern Egypt.
Toomer notes that this is too low as it corresponds to 50 minutes difference. The true value should be $58 \frac{1}{2}$ minutes. In his later book, Geography, Ptolemy updated the difference but overshot be even further!
Which are always defined as 30 days in the Egyptian calendar.
The value here is likely taken from the Babylonian records and is overestimated. This is unsurprising since the Earth does not leave a crisp shadow, with makes such things hard to estimate. Toomer notes Ptolemy’s own calculations, which we’ll see later, estimate this eclipse as only being 2.5 digits and modern calculations give 1.5-1.6 digits.
Being less than full.
Which I’ve done by taking $\frac{34}{60} = 56.6…% which I multiplied by the mean longitudinal motion of 1 hour.
In truth, the sun’s motion is also a function of the mean motion + the anomaly, but we’ve already taken that into consideration when we calculated the change in solar position above.
Here, I’m using modern math/astronomy notations so I figure I should explain: Δ indicates a change in λ, the ecliptic longitude.
If you’re following along with the Neugebauer translation, there is a minor typo here in that he did not include the negative sign on the final result of -0;37º although it is clearly negative from the equation shown. However, this really doesn’t make a difference since, if we’re talking about the arc length, it is always positive, and if we’re comparing two points in ecliptic longitude, we would need to compare whether they’re clockwise or counter-clockwise anyway. So treating this as an absolute value is perfectly reasonable.