Now that we understand how to predict lunar eclipses, we’ll turn our attention towards solar eclipses. However, Ptolemy warns us that these will be
more complicated to predict because of lunar parallax1.
Toomer again provides an example that we can follow along with2. This will be Example $12$ from Appendix A. Surprisingly, nowhere in the Almagest does Ptolemy describe the details of a solar eclipse. As such, Toomer has selected his own example. In this case, we are to determine the details of the solar eclipse of June $16$, $364$ CE (Nabonassar $1112$ in the month of Thoth), which was observed by Theon of Alexandria3. Upon observing the eclipse, Theon then followed Ptolemy’s methods in the Almagest and Handy Tables to compare the predictions against observations and his calculation are what Toomer follows as an example using Ptolemy’s methods4.
We begin by
[determining] the number of equinoctial hours by which the time of true syzygy at Alexandria proceeds or follows noon.
The first step of that is obviously determining the time of true syzygy. And to do that, we’ll again need to refer to VI.3, the Table of Mean Syzygies.
| Years | Days of Thoth | [Increment in] Sun from Apogee | [Increment in] Moon’s [Argument of] Anomaly | [Increment in Argument of Lunar] Latitude |
| $1101$ | $22;41,45$ | $19;11,56$ | $222;53,32$ | $64;41,57$ |
| $11$ | $1;09,39$ | $358;28,11$ | $271;04,19$ | $211;12,03$ |
| Total | $23;51,24$ | $17;40,07$ | $133;57,51$ | $276;54,00$ |
We’ll concentrate on the Days of Thoth column. Converting that to time, this implies the potential eclipse would happen on the $24^{th}$ day of Thoth5 at $8;34$am.
Now, we’ll need to find the anomalies for both the sun and moon. Starting with the sun, we refer to the Table of the Sun’s Anomaly (III.6) and look up the sun’s position from apogee we just calculated of $17;40,07º$. Doing so and interpolating between rows I come up with a value of $0;41,13$ and because it was found in the first column, it is rearwards.
Next for the moon, we’ll use the position of the moon about its epicycle we found ($133;57,51º$) and look that up in the Table of Lunar Anomalies (IV.10). Interpolating between rows, I come up with a value of $3;49,48º$ and since our value came from the first column, it is again rearwards.
Since these two anomalies are not equal and opposite, this is not a true conjunction and we will, indeed, need to work towards the true conjunction by asking how much ahead of or behind the sun the moon is. In this case, both the sun and moon have negative anomalies (meaning they lag their mean positions which are at conjunction), but the moon even more so. As such, the moon will have to catch up to the sun again. This time by
$$3;50º – 0;41º = 3;09º$$
How long will it take to do that? To know, we’ll need to determine the moon’s actual motion around this time. To do so, we’ll increase the position of the moon about its epicycle by $1º$ and recalculate the anomaly. Thus, we’ll look up an argument of anomaly of $134;57,51º$ and find it to be $3;46,08º$, again rearwards. The difference between these two is $0;03,40 \; \frac{º^{_a}}{º^{_m}}$6 so using the same logic as we have previously, we write:
$$0;32,56 \; \frac{º}{hr} + (0;32,40 \; \frac{º^{_m}}{hr} \cdot 0;03,40 \; \frac{º^{_a}}{º^{_m}}) = 0;34,56 \; \frac{º}{hr}$$
Now that we know the speed the moon, we can divide the distance over which the moon must catch up (increased by $\frac{1}{12}$ to account for the motion of the sun) by the speed to determine the amount of time:
$$3;09º \cdot \frac{13}{12} \div 0;34,56 \frac{º}{hr} = 5;52 \; hours$$
So it will take the an additional $5$ hours and $52$ minutes from mean conjunction until true conjunction. This gets added to the time of day
$$8;34 \; am + 5;42 \; hrs = 2;26 \; pm$$
So this eclipse would take place $2$ hours and $26$ minutes after noon7 which is what Ptolemy asked us to find. So what’s next?
[I]f the geographical position in question, [i.e.] that of the required place, is different [from that], i.e. if it does not lie beneath the same meridian as Alexandria, we add or subtract the difference in longitude between the two meridians, expressed in equinoctial hours, and [thus] decide how many hours before or after noon the true syzygy occurred at that place too.
In other words, if we’re not in Alexandria, adjust accordingly. For this example we are, so no adjustment is necessary. Moving on,
Then we determine, first, the time of apparent syzygy (which will be approximately the same as mid-eclipse) at the required geographical location, by applying the method of computing parallaxes we explained previously [V.19].
I’ll pause here because there’s been a sneaky shift in language. Previously, we calculated the time of true conjunction in relation to noon. However, now we’re being asked to find the time of apparent syzygy. What does that mean?
Well, it has to do with the difference between mean solar time (based on when the mean sun crosses the meridian) and apparent solar time (when the actual sun crosses the meridian). We’ll need this because we’ll be dealing with parallax shortly and to use that, we’ll be making use of Table of Zenith Distances and Ecliptic Angles. For that, we need to know how much time it has been since the sun actually crossed the meridian whereas the time we computed above is based on the mean sun. This is something we glossed over in the post on calculating lunar parallax because that sample problem gave the point at upper culmination so we were able to calculate the time more directly. Here, we don’t have that information.
So how to go about making that correction?
This is something we haven’t dealt with in a long time but it is precisely what was discussed in this post on the inequality of solar days. Fortunately, towards the bottom of that post, I have a set of simplified instructions that we can follow.
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.
While it might be tempting to use the initial date as the mean conjunction, here, we do actually want to use the beginning of the epoch.
At that time, the mean position of the sun was $0;45º$ into Pisces ($330;45º$ ecliptic longitude) and a true position of $3;08º$ into Pisces ($333;08º$ ecliptic longitude).
To get the positions at true conjunction, we’ll start with mean conjunction. We can get the mean position at mean conjunction quickly from the table at the beginning of this post. The mean sun was $17;40,07º$ past apogee and since the position of solar apogee is $65;30º$ that puts the mean sun at $83;10,07º$ ecliptic longitude. To find the true sun at mean conjunction, we just need to apply the anomaly which we also calculated above of $0;41,13º$ in the rearward direction. So we subtract to find the ecliptic longitude of the true sun at mean conjunction to be $82;28,54º$.
This will allow us to get to true conjunction by adding the additional motion of the sun over the time between mean and true conjunction. We’ve discussed several times in this book how, when the moon is playing catch up to the sun, it has to travel an extra $\frac{1}{12}$ the distance than the separation of the two luminaries at mean conjunction8. That’s because that $\frac{1}{12}$ is the distance the sun travels in that interval. Therefore, in the interval between mean and true conjunction, the sun travelled:
$$3;09 \div 12 = 0;15,45º$$
We can add that to the position of the true sun at mean conjunction to find the position of the true sun at true conjunction of $82;44,39º$ ecliptic longitude.
The anomaly wouldn’t have changed appreciably over this interval so we can add the same anomaly to determine the mean position at true anomaly of $83;25,52º$ ecliptic longitude. Let me make a quick table so we can keep track of that easier:
| Beginning of Epoch | True Conjunction | |
| Mean | $330;45,00º$ | $83;25,52º$ |
| True | $333;08,00º$ | $82;44,39º$ |
That gives us all the positions in question.
2. Use the true position to look up the cumulative rising times at sphaera recta for both dates.
Here, we are asked to use the Rising Time of the Ecliptic Tables (II.8). We’ll be looking up $333;08,00º$ and $82;44,39º$. Doing so I come up with $335,07,02º$ and $82,04,45º$.
3. Take the differences in both the mean positions and the rising times. Then the difference between these differences.
The difference in mean positions is:
$$330;45,00º – 83;25,52º = 247;19,08º$$
The difference is rising times is:
$$335;07,02º – 82;04,45º = 253;02,17º$$
Then the difference between the differences:
$$253;02,17º – 247;19,08º = 5;43,09º$$
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.
Straightforward enough:
$$\frac{5;43,09º}{360º} \cdot 1440 = 22;52 \; min$$
I’ll round that off to an even $23$ minutes9.
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.
Here, the difference in rising times was greater so we need to add this to the time for true conjunction. As such, the time of true conjunction was at
$$2;26\; pm + 0;23 \; hrs = 2;49\; pm$$
That gets us over the apparent conjunction hump. Next:
We enter into the Table of Angles [II.13] and the Table of Parallaxes [V.18], using [as arguments] the appropriate latitude, distance in hours from the meridian, point on the ecliptic where the conjunction occurred, and also distance of the moon. We thus find, first, the moon’s parallax along the great circle drawn through the zenith and the moon’s center.
Now we’re actually ready to start dealing with parallax. First, we need to know the position of true conjunction on the ecliptic which we calculated above to be $82;44,39º$ or $22;44,39º$ into Gemini. We also need the time which we just calculated to be $2;49 \; pm$. These will get entered into the Table of Zenith Distances and Ecliptic Angles (II.13), being careful that we’re on the one corresponding as closely to possible to the latitude of Alexandria which is the Lower Egypt table10.
We’re going to need to do quite a bit of interpolation here because we’re between hours on the table and the tables are given for the first points in each constellation and we’re partway through one.
So let’s begin by calculating the arc and west angle11 for $2;49$ hours in both Gemini and the following constellation, Cancer.
Doing so I find the following:
| Arc | West Angle | |
| Gemini $0º$ @ $2;49 \; hr$ | $39;10º$ | $11;08º$ |
| Cancer $0º$ @ $2;49 \; hr$ | $37;57º$ | $19;40º$ |
Now we’ll need to interpolate between each of these for the distance into Gemini ($22;44,39º$).
For the arc, I get $38;14º$12 and for the west angle I find $17;38º$13.
Next, we’ll turn to the Table of Parallax (V.18). Here, we use the arc we just calculated as the argument we look up. First, Ptolemy says to look up the moon’s parallax taking into consideration the moon’s distance. To do so, we start by determining the values from the third and fourth columns.
For column $3$ I get a value of $0;33,34º$ and for column $4$ I get $0;06,32º$. Thus, the parallax at the first limit (from column $3$) is $0;33,34º$ and at the second limit (column $3 + 4$) is $0;40,06º$. However, these limits are based on where the moon is about the epicycle. Which means we need to ask where about the epicycle the moon is.
We calculated the moon’s argument of anomaly initially as $133;57,51$. However, that was for mean conjunction. In the $5;52 \; hrs$ between mean and true conjunction the moon moves an additional $3;12,43º$. Thus, at true conjunction, the moon is at $137;10,34º$ about its epicycle.
We then need to look this value up in column $7$, recalling that the value we’ll look up is always half of the actual argument. So we’ll be looking up $68;35,17º$. Interpolating, I find a value of $0;51,21$. Recall that this is the number of sixtieths from the first limit to the second. Thus, we use this to interpolate between the parallax at the two limits. Doing so I find a total lunar parallax of $0;39,09º$14.
That takes care of finding the lunar parallax. However,
We always subtract from this that solar parallax which is on the same line, and from the result determine, in the way indicated, the component of parallax in longitude by itself, which is computed by means of the angle we found [from the table] between the ecliptic and the great circle through the zenith.
The solar parallax is much easier to calculate since there’s only one column for it. Entering the arc into the table and interpolating, I find a solar parallax of $0;01,45º$.
As Ptolemy indicates, we take the difference of the lunar and solar parallax which is:
$$0;39,09º – 0;01,45º = 0;37,24º$$
This then needs to get broken down into components using the west angle. In this case, the longitudinal parallax can be found with the following:
$$cos(17;38º) = \frac{p_\lambda}{0;37,24º}$$
Solving, I find the longitudinal component of parallax to be:
$$p_\lambda = 0;35,39º$$
We always add to this [longitudinal parallax] the increment of ‘epiparallax’ corresponding to the number of equinoctial hours represented by the longitudinal parallax.
Here, Ptolemy is telling us we’re going to need to adjust our time of apparent conjunction again, due to the longitudinal parallax. To find out by how much, we divide the distance the parallax moved the moon by the hourly mean motion in longitude:
$$0;35,39º \div 0;32,56 = 1;05 \; hrs$$
This gets added to the previous estimation for the time of apparent conjunction:
$$2;50\; pm + 1;05 \; hr = 3;55 \; pm$$
And now we have a new problem because this changed the time of apparent conjunction which means all the work we just did to calculate the parallax is wrong because they were calculated for a different time. It’s looking like this is going to be an iterative process. Thus, we’ll now work on making what many other authors look to call a “second approximation”.
As a quick note on terminology, while the term “epiparallax” just got used above, we should note that it was really the “increment of epiparallax”. I point this out because Ptolemy now tells us the method of calculating the epiparallax:
This epiparallax is determined as follows. We take the difference (as determined from the same table) between the parallax corresponding to the original zenith distance and the parallax corresponding to the zenith distance after the pass of the number of equinoctial hours [represented by the longitudinal parallax].
First, we’ll need to recalculate the parallax for the same point now at $3;55 \; pm$. This requires first getting the new zenith distance (arc) and west angle from the Table of Zenith Distances and Ecliptic Angles (II.13). Again, we need to take care that we’re on the table for Lower Egypt.
| Arc | West Angle | |
| Gemini $0º$ @ $3;55 \; hr$ | $53;22º$ | $10;51º$ |
| Cancer $0º$ @ $3;55 \; hr$ | $52;09º$ | $21;01º$ |
Then we interpolate between each of the constellations for the distance into Gemini ($22;44,39º$).
For the arc, I get $52;27º$ and for the west angle I get $18;33º$.
This arc then gets plugged into the Table of Parallax (V.18), again looking up the lunar parallax from columns $3$ and $4$. For column $3$ I get a parallax of $0;42,54º$ and from column $4$ I get an additional $0;08,19º$ for a total parallax at the second limit of $0;51,13º$.
However, now we need to interpolate between these two limits based on the motion about the epicycle for the moon which we will need to again adjust for the additional time difference. In $1;05 \; hr$, the moon will have moved $0;35,23º$ about the epicycle. We will add this onto the position we previously calculated15:
$$137;10,34º + 0;35,23º = 137;45,57º$$
This then gets divided by two and looked up in column $7$ to determine the proportion between the two limits. Thus we look up $68;52,59º$ in the table and interpolating between rows I get $0;51,34$.
This is the proportion between the two limits above. Applying it, I find that the final parallax is $0;50,03º$.
Although Ptolemy doesn’t state it directly, we must also find the solar parallax and subtract that16. Again, this is done for an arc of $52;27º$ for which I find a solar parallax of $0;02,17º$.
Subtracting this from the lunar parallax I come up with a difference in parallax of $0;47,46º$.
We take the longitudinal component of this by itself, plus an additional amount (if it is significant) which is the same fraction of the latter as the later is of the original [longitudinal] parallax.
The first half of this sentence is straightforward: We again need to take the longitudinal component of parallax using the west angle found above. Here, I’ve denoted this parallax with a prime mark ($’$) to denote this is the second approximation parallax as this will come into play shortly.
$$cos(18;33º) = \frac{p’_\lambda}{0;47,46º}$$
$$p’_\lambda = 0;45,17º$$
But what of the second half of the sentence17?
Essentially what Ptolemy is telling us is that since we have again adjusted our parallax, this will again add time and we’ll have to go through this process again… That sounds less than fun. So instead, we’ll more quickly approximate the results of that further calculation until there are diminishing returns.
To do so, we’ll start by first taking the difference of the initial parallax ($p_\lambda$) and the second parallax ($p’_\lambda$) which Toomer denotes as $d$, presumably for “difference”18:
$$d = 0;45,17º – 0;35,39º = 0;09,38º$$
We then consider what proportion of the original parallax this increment was. In other words:
$$\frac{d}{p_lambda} = \frac{0;09,38º}{0;35,39º} = 0;16,13$$
We then predict that, if we were to go through the whole iterative process of finding the parallax again, the next increment (which Toomer calls $f$) would be the same proportion of $d$. In other words:
$$f = 0;09,38º \cdot 0;16,13 = 0;02,36º$$
Thus, we can see how rapidly this diminishes and that it is unlikely we’ll need to estimate additional iterations. We’ll then add that onto the previous increment.
$$d + f = 0;09,38º + 0;02,36º = 0;12,14º$$
This is what Toomer labels the epiparallax which we’ll add to the first longitudinal parallax we calculated19
$$0;35,39º + 0;12,14º = 0;47,53º$$
This is now the total longitudinal parallax.
To the total parallax in longitude, computed in this way, we add $\frac{1}{12}$ of itself, to account for the additional motion of the sun, and convert the total to equinoctial hours by dividing it by the moon’s true hourly motion at the conjunction.
Ah yes… the sun’s motion. We’ll include that now to get our final parallax:
$$0;47,53º \cdot \frac{13}{12} = 0;51,52º$$
And to find the time it took since our original apparent conjunction, we divide this by the moon’s hourly motion at conjunction which we calculated way up towards the beginning of this post to be $0;34,56 \; \frac{º}{hr}$.
Thus, the time added to true conjunction due to parallax is:
$$0;51,52º \div 0;34,56 \; \frac{º}{hr} = 1;29 \; hrs$$
If the longitudinal parallax we found is towards the rear [i.e., in the order] of signs (we explained previously how to determine this), we subtract the amount in degrees which we had converted into equinoctial hours from the moon’s position, as previously determined, at the moment of true conjunction, in longitude, latitude, and anomaly (each separately): this gives us the [corresponding] true positions of the moon at the moment of apparent conjunction, while the number of hours itself [resulting from the above computation] tells us by how much the apparent conjunction precedes the true.
Here, Ptolemy is telling us that we need to take into consideration the direction of the parallax. If it moves it rearwards in the signs (i.e., towards the west), then we need to subtract. We determine this by considering the angle which, in our case was the west angle. From the linked post where this was explained Ptolemy tells us that, when the parallax is to the south (as it is in this case), then the angle being less than a right angle (which it is) means that the effect of the parallax will be in advance.
This is quite easy to understand if we simply draw a picture. Here’s one that I used for a paper I’m writing:
Here, the effect of the parallax is southwards and the angle20 is greater than a right angle. This has the effect of making the longitudinal parallax push the apparent position rearwards with respect to the motion of the sky, which means an increase in ecliptic longitude. If the angle were less than a right angle, the small triangle formed would be on the other side of the altitude circle pushing the effect of the parallax westwards which is forwards in terms of the motion of the sky and lower in ecliptic longitude.
Since our angle is this latter case, the effect of the parallax is not towards the rear so Ptolemy’s statement here does not apply. But he next tells us what to do in that case:
But, if the longitudinal parallax we found is in advance [i.e., in the reverse order] of the signs, contrariwise, we add the amount in degrees to the position, as previously determined, at the moment of true conjunction, in longitude, latitude and anomaly (each separately); and the number of hours will give us the amount by which the apparent conjunction is later than the true.
I’ll start with the number of hours. Again, we’ll add this time on to our initial time (adjusted for apparent conjunction) of true conjunction (not the one adjusted for the first estimation).
$$2;49 \; pm + 1;29 \; hrs = 4;18 \; pm$$
Next we add the final parallax we just calculated to the longitude, latitude, and anomaly as suggested. However, before doing so, there is one other adjustment we need to make that Ptolemy must have forgotten to tell us21. Specifically, in regards to the argument of latitude, when we calculated this at the very beginning of the post, it was for mean position of the mean conjunction and we still haven’t corrected it to be for either the true conjunction.
First, we’ll apply the anomaly. We already calculated this to be $3;49,48º$, and should be subtracted22. Our initial calculation of the moon’s argument of anomaly, was $276;54,00º$ so we’ll subtract the anomaly to find it’s true position at mean conjunction to be $273;04,12º$.
Next, we need to add the extra motion between mean and true conjunction. We can do this by recalling that the distance between the luminaries at mean conjunction was $3;09º$ and following our standard procedure, we multiply this by $\frac{13}{12}$ to determine the distance it must traverse to be $3;24,45º$:
$$273,04,12º + 3;24,45º = 276;28,57º$$
So now that we know where the true moon was at true syzygy, we can go ahead and do as Ptolemy suggests and add the final anomaly to everything.
First, the ecliptic longitude:
$$82;44,39º + 0;51,52º = 83;36,31º$$
Which is $23;36,31º$ into Gemini.
Next, the argument of latitude:
$$276;28,57º + 0;51,52º = 277;20,49º$$
And finally, the argument of anomaly recalling that, after we adjusted for the time between mean and true anomaly, it increased to $137;10,34º$:
$$137;10,34º + 0;51,52º = 138;02,26º$$
Whew. Lots of work so far, but we’re still not finished. According to Toomer, we’re now set to determine the “circumstances of the eclipse”.
Next, using the same methods, we determine from the distance in equinoctial hours of the apparent conjunction from the meridian, first, what the moon’s parallax is measured along the great circle through the moon and zenith. From the result, we subtract the solar parallax for the same argument, and use this to determine, as before (by means of the angle formed between the circles [of ecliptic and altitude] at that moment), the latitudinal parallax [i.e., the parallax] along a circle orthogonal to the ecliptic.
So now we’re starting on the latitudinal parallax. We’ve waited to calculate this because we want to have a final apparent position of the moon after the shifts due to longitudinal parallax. As we just determined, the moon should appear at $83;36,31º$ ecliptic longitude at $4;18 \; pm$. So we’ll calculate the latitudinal parallax at that time.
And so it’s back to the Table of Zenith Distances and Ecliptic Angles (II.13):
| Arc | West Angle | |
| Gemini $0º$ @ $4;18 \; hr$ | $58;18º$ | $11;30º$ |
| Cancer $0º$ @ $4;18 \; hr$ | $57;02º$ | $22;00º$ |
Then we estimate between these based on how far into the constellation the point is. For the arc, I get $57;19º$ and for the west angle I find $19;46º$.
We’ll pop that arc into the Table of Parallax (V.18). Doing so, I find a value of the first limit of $0;45,28º$ and an adjustment at the second limit of $0;08,49º$ so a total at the second limit of $0;54,17º$.
Now we need to estimate between the first and second limit. We’ll use the argument of anomaly we just calculated of $138;02,26º$ again dividing this in half so we can look it up in the table ($69;01,13º$). Interpolating, I find a value of $0;51,55$ as the sixtieths. Using that to interpolate between limits, I find a total parallax of $0;53,06º$.
As usual, we’ll need to subtract out the solar parallax which I find to be $0;02,25º$ for a difference in parallax of $0;50,41º$.
We’ll take that and use the west angle of $19;46º$ to determine the latitudinal parallax:
$$sin(19;46º) = \frac{p_\beta}{0;50,41º}$$
$$p_\beta = 0;17,08º$$
We convert the result to a distance along [the moon’s] inclined circle, i.e., we multiply it by $12$.
$$0;17,08º \cdot 12 = 3;25,36º$$
If the effect of the latitudinal parallax is northwards with respect to the ecliptic, we add the result to the previously determined true position in [argument of] latitude at the moment of apparent conjunction, when the moon is near the ascending node, but subtract it when the moon is near the descending node. Contrariwise, if the effect of the latitudinal parallax is southwards with respect to the ecliptic, we subtract the distance derived from the parallax from the previously determined position in [argument of] latitude at the moment of apparent conjunction, when the moon is near the ascending node, but add it when the moon is near the descending node.
First, let’s discuss the direction of the parallax. The time of this eclipse we’ve calculated to be $4;18 \; pm$. While the moon can rise or set north of east or west from Alexandria, we’re not nearly late enough that it would have crossed into the northern half of the sky. And since the effect of parallax is always away from the zenith, this means the latitudinal parallax will be southwards.
Next, we have to consider which node we’re near. We’ve calculated the argument of latitude to be $277;20,49º$ which places it just after the ascending node. Thus, the effect will be negative and we get the final position of the moon from the northern limit of:
$$277;20,49º – 3;25,36º = 273;55,13º$$
We thus obtain the amount of apparent [argument of] latitude at the moment of apparent conjunction. With this argument, we enter the solar eclipse tables, and in our argument falls within the range of numbers in the first two columns, we can say that there will be a solar eclipse, and that its middle coincides approximately with the moment defining the apparent conjunction.
Now we’ll head over to Eclipse Tables (VI.8) making sure we’re on the tab for solar eclipses and check to see if $273;55,13º$ is in the range of numbers in the first two columns for either table. And it’s very clearly in both which indicates there will indeed be an eclipse!
Now we’ll try to work out the details.
So we set down, separately, the amounts of the [magnitude in] digits and the minutes of immersion and emersion corresponding to the argument of latitude, as derived from each of the two tables, then enter, with the distance of the moon in anomaly from the apogee [of the epicycle] at the apparent conjunction, into the table of correction , take the corresponding number of minutes, and take the corresponding fraction of the difference between each [pair of] results we wrote down. In every case, we add the result to the number derived from the first table [at greatest distance].
Now, in the Eclipse tables, we’ll look up the digits and minutes of immersion for both greatest and least distance, interpolating as necessary.
| Digits | Minutes of Immersion | |
| Greatest Distance | $4;09,34$ | $23;47,53 \; min$ |
| Least Distance | $4;57,34$ | $26;21,36 \; min$ |
Next, we need to use the tab for the Table of Correction to determine where between the greatest and least distance we fall using the argument of anomaly we found earlier ($138;02,26º$) as our value we’re looking up. Doing so, I find the sixtieths to be $0;51,40$. Applying that to the digits I come up with $4;50,54$ and for the minutes of immersion I come up with $26;00,15 \; min$ of immersion.
We increase the minutes of travel [found by this procedure] for both [stretches, i.e., immersion and emersion] by $\frac{1}{12}$, to account for the sun’s additional motion and convert the result into equinoctial hours [by dividing] by the moon’s true [hourly] motion.
Increasing the minutes of immersion by $\frac{1}{12}$:
$$26;00,15 \cdot \frac{13}{12} = 28;10,17 \; min$$
Next we divide this by the moon’s hourly velocity which we determined earlier to be $0;34,56 \frac{º}{hr}$. However, before we do, we should note there’s a slight mismatch in our units. The immersion time is in minutes and the hourly motion is in hours. Thus, we’ll need to convert the minutes of immersion to hours so the units cancel properly. Fortunately this is quite easy to do in sexagesimal as we just move each of the places down one:
$$0;28,10 \; hrs \div 0;34,56 \frac{º}{hr} = 0;48,23 \; hrs$$
Thus, we have the length of both the immersion and emersion; this, however, is on the assumption that the [change in] parallax has no effect on these time intervals.
At this point we’re done. The mid-eclipse is at $4;18 \; pm$. According to this calculation, it should have first begun $\approx 48 \; min$ before then at $3;30 \; pm$ and ended $\approx 48 \; min$ after around $5;06 \;pm$ with a maximum coverage of $4;50,54$ digits.
Now, there are a few more assumptions that were hidden in this post and Ptolemy still has another few paragraphs to dedicate to discussing them, but as this is the end of his detailed procedure, I’ll leave off here and save that discussion for the next post.
- And boy howdy is he not kidding. This is not only my first post which has broken the $4,000$ word mark, but also passed the $5,000$ word mark! I’m very tempted to break it up into multiple posts, but as this is one long procedure, it really doesn’t make sense to.
- Neugebauer and Pedersen are both conspicuously missing any further discussion pertaining to solar eclipses and neither has any example.
- This eclipse happens to be important to modern astronomers because there are very few eclipses from the first few centuries CE which have reliable observations. This eclipse was used by astronomers to help understand the tidal interactions between the earth and moon which case the moon’s orbit to speed up and expand and the earth’s rotation to slowly decrease. While better methods now exist thanks to the reflectors left on the moon by Apollo astronauts which give precise measurements of the moon’s motion away from us, it is still historically significant. For more information see The Eclipse of Theon and Earth’s Rotation by J.M. Steele.
- If you are following along with Toomer, you may notice that I have done some steps in a slightly different order. This is because I attempted to follow Ptolemy’s instructions as closely as possible.
- Again, recalling that Egyptian days start at noon and that if the Days of Thoth had a value in the whole number place of $0$, it would still be the first day, so it’s always $1$ day later.
- If you’re following along with the Toomer example as I am, you may notice an issue here. Specifically, in previous steps, Toomer has rounded to the first minutes place on previous figures. By doing so, the rounding comes to $0;03$ on this step. Only by including the seconds place in previous steps (as I have here) can we see how he arrived at $0;03,40$ which he expresses as $3 \frac{2}{3}’$.
- It should be noted that this pushed the eclipse into the next Egyptian day since these days start at noon, but as the precise date of the eclipse does not factor into the calculations, we don’t need to consider it further.
- In truth, I feel like this is a fairly crude approximation as it relied on the ratio of the motion of the sun and moon and was somewhat rounded. I feel like a better value here could be derived from the Table of the Mean Motion of the Sun, applied over the interval of time between mean and true conjunction. Doing so I find a value of $0;14,27º$.
- Toomer came up with $24$ minutes.
- Alexandria’s latitude is about $31º$ north, but we’ve consistently used the Lower Egypt table previously.
- Ptolemy hasn’t yet called for the west angle, but we’re going to need it soon, so I’m going to grab it while we’re looking at this table. And as a reminder, we’re using the west angle since the position is after noon and thus west of the meridian.
- Toomer comes up with a value of $38;28º$. I am curious as to whether he also interpolated between the table for Lower Egypt and Rhodes since the Lower Egypt table is about $0;38º$ off from the latitude Ptolemy considers for Alexandria. But not curious enough that I care to calculate it right now.
- Toomer finds $17;35º$.
- Toomer has a notably higher value of $0;39,35º$ here likely due to his higher value for the arc.
- The one adjusted for the previous $5;52 \; hrs$.
- Recall the admonishment of “always” when we did this above.
- Toomer’s example that I’m following along with has a line inserted here that implies this is the actual beginning of the computation of the epiparallax. Seems a bit of an odd take to me.
- The step of taking the difference was actually indicated in the quote above this most recent one. Ptolemy’s instructions are a bit jumbled here and we needed to wait until after we took the longitudinal component.
- We’re adding it to the first since it incorporates the additional parallax from the second estimate.
- Always the “upper left” one in the four angles made by the intersection of the ecliptic and the altitude circle/
- Seriously. There’s no mention of the argument of latitude prior to this!
- This is a bit of a cheat because the anomaly is measured along the ecliptic and this argument of latitude is measured along the moon’s inclined circle, but over these short distances, it shouldn’t matter much.