Now that we’ve created our table of conjunctions and oppositions, how do we go about using it? As usual, Ptolemy walks through the process in a vacuum, so to help, I’ll follow along with the example Neugebauer does in History of Ancient Mathematical Astronomy on pages $123-124$, although somewhat slimmed down. In particular, I’ll walk through finding the true opposition from the year $718$ in the epoch for the first opposition in the year.
First,
we calculate the number of the year in question in the era of Nabonassar [and then]… determine what combination of 25-year periods (taken from the first or second table, as the case may be1)
So doing this for our example, we’re looking for year $718$ which obviously isn’t in the $25$ year table. Best we can do (without going over) is $701$ years. Keep in mind this example is for an opposition, so we need to look at the right table on the $25$ year interval table wherein we see this will occur $8;40,24$ days into the year. The sun will be $102;37,36º$ from apogee, the moon $192;11,17º$ around the epicycle from its apogee, and the argument of altitude $154;56,32º$.
However, we also need to do this for the additional $17$ years from the single year table.
and single years (taken from the third table2) adds up to that number of years. [We then] add the entries from [each] successive column separately: for conjunctions we add the entries from the first and third tables and likewise for oppositions we add entries from the second and third tables. The sum derives from the entries in the second column will give us the moment of syzygy, counted from the beginning of that year; e.g. if the sum is $24;44$ days, [the syzygy] will be $44$ sixtieths of a day after noon on Thoth $24$; or, if it is $34;44$ days, it will be $44$ sixtieths of a day after noon on Phaophi $4$. The sum derived from the entries in the third column will give us the [mean] position of the sun in degrees counted from apogee; The fourth column, the anomaly of the moon counted from the apogee [of the epicycle]; the fifth column the [argument of] latitude counted from the northern limit.
Looking at the single year table, we see that for $17$ years, we’ll also need to add $25;57,19$ days to the date, $21;26,58º$ to the position of the sun, $47;19,30º$ to the moon’s motion about the epicycle, and $351;29,44º$ to the argument of Latitude.
Thus, the initial mean opposition from these tables alone will occur $34;37,43$ days into the $718^{th}$ year. That means it’s $4;37,43$ days into the second month, Phaophi3. At that time, the mean sun will be $124;04,34º$ from apogee, the moon $239;30,47º$ from its apogee on the epicycle, and the argument of latitude $146;26,16º$.
At the same time, we can readily calculate the subsequent [syzygies of the year in question], either all, or some, as we choose, in logical fashion, by adding the appropriate entries in the fourth, monthly table.
Note that the calculation we just completed gives a syzygy in the second month of the $718^{th}$ year of the epoch. At that point, there’s already been one. And since this example was to find the first, we’ll need to do a bit of math to rewind and find the first. To do so, we look at the last table, for one month intervals and we’ll need to subtract one month.
Subtracting the days, we subtract $29;31,50$ to get that the first opposition happened $5;05,53$ days into year year in question. We’ll also subtract the other columns to determine the sun was $94;58,11º$ from apogee, the moon $213;41,47º$ from apogee on its epicycle, and the argument of latitude, $115;46,02º$.
Now we have all our initial values ready to go.
Once we have derived, by the above procedure, the time of mean conjunction or opposition, and the position of each luminary in anomaly at that time, it will be easy to determine the time and place of the true syzygy, and also the moon’s position in latitude, by comparing the anomalies of the two bodies. For by applying each anomaly in turn, we calculate the true position of the sun, moon, and moon’s latitude, at the moment defined by the mean syzygy in question, by means of the equation thus found, and examine these positions. If we find that the bodies are still at the same longitude [for conjunction], or exactly opposite [for opposition], then the time of true syzygy will be the same [as that of mean syzygy].
As a reminder, the tables we derived were for mean conjunction. For a true conjunction we’ll also need to take into consideration the anomalies for ecliptic longitude, as well as the moon’s ecliptic latitude4. Fortunately, the tables we derived so far in this book give us all the parameters we need to do this. As such, we can relatively quickly calculate these values. If the anomalies happen to be equal and the moon is directly on the ecliptic with an ecliptic latitude of zero, then it really is the a true conjunction. But if it’s not, we’re going to have to adjust. The same argument can be made for oppositions, although the ecliptic latitudes would need to be $180º$ apart as opposed to equal.
So let’s check our longitudes.
For the sun, its mean position was $94;58,11º$ from apogee. Looking at our table of the sun’s anomaly, we determine this would result in an anomaly of $2;23º$ lagging the mean position since the value we entered was found in the first column. Thus, the solar ecliptic longitude would be $92;35,11º$.
Meanwhile, the moon was $213;41,47º$ from its apogee on the epicycle, so we can look that up in our table of lunar anomalies to determine the anomaly to be $3;00,01º$ which we’ll round to be an even $3;00º$ to keep the number of sexagesimal divisions the same. This time, the anomaly would increase the mean position since we found the value in the second column. Although we didn’t explicitly calculate the lunar mean position, since this was for mean syzygy, by definition, the mean moon is $180º$ from the solar mean position, so $274;58,11º$ in ecliptic longitude. Adding the anomaly, we get the true lunar position to be $277;58,11º$.
Thus we can clearly see that the two do not have the same anomaly, therefore resulting in them not having ecliptic latitudes that are $180º$ apart. So this is not a true syzygy. The two bodies are off in ecliptic longitude by a total of $5;23º$ from true syzygy. This is the elongation at that moment. So what to do?
If not, we take the difference between the bodies in longitude, expressed in degrees, and increase it by a twelfth of the part itself, to account approximately for the additional motion of the sun [between the mean and true syzygy].
Ptolemy does nothing to explain this step so let’s step outside the text for a bit to examine5. What Ptolemy does say here is that we know how far apart the two luminaries are in ecliptic longitude from true syzygy. We just calculated it to be $5;23º$. What he wants to know is how long it will take to close that gap which is ultimately a question of how fast both the sun and moon are moving. So to phrase that mathematically:
$$t = \frac{\eta}{v}$$
where $\eta$ is the elongation or distance to true syzygy and we just calculated, and $t$ is the time until the true syzygy which is what we want to know so we can figure out how much to adjust the time of mean syzygy by. But what is $v$ in this case? Well, it’s not a single value because both the sun and moon contribute. So we can restate the equation to take this into account:
$$t = \frac{\eta}{v_m – v_s}$$
where $v_m$ and $v_s$ are the velocities of moon and sun respectively.
If we do a bit of factoring for no immediately obvious reason we get:
$$t = \frac{\eta}{v_m(1 – \frac{v_s}{v_m})}$$
This is helpful because we now have the ratio $\frac{v_s}{v_m}$ in there. In truth, we’d really want the instantaneous velocities of the sun and moon which would take into account their varying anomalistic motions but Ptolemy settles for the mean motion for each which ends up having a ratio of $\approx{1:13;22,10}$ which Ptolemy evidently rounds off to an even $1:13$. Thus, substituting into the equation we get:
$$t = \frac{\eta}{v_m(1 – \frac{1}{13})}$$
Combining the big in parentheses:
$$t = \frac{\eta}{v_m\cdot \frac{12}{13}}$$
And getting rid of that pesky fraction in the denominator by multiplying the larger fraction by $\frac{13}{13}$ we get:
$$t = \eta \cdot \frac{13}{12v_m}$$
Thus, the total amount was increased by $\frac{1}{12}$ from $\frac{12}{12}$ to $\frac{13}{12}$6.
But we still need to know $v_m$ and this time Ptolemy isn’t going to substitute the mean motion this time. Rather,
The method of finding the moon’s true hourly motion at the syzygy for any position is as follows. We can enter the table of the moon’s anomaly [IV 10] and then determine the size of the increment in the equation [at that point] corresponding to an increment of $1$ degree in anomaly. We multiply this increment by the mean motion in anomaly in $1$ hour, $0;32,40º$, and, if the anomaly [which we entered into the table] as argument is in the lines above the greatest equation, we subtract the product from the mean hourly motion in longitude, $0;32,56º$, but if [the anomaly] is in the lines below [the greatest equation], we add the product to $0;32,56º$. The result will be the moon’s true motion in longitude in one equinoctial hour at that position.
Again, let’s take some time to parse that out.
Let’s first consider the part about entering into the table of anomaly the value we have. Our current value for the moon’s motion around the epicycle is $213;41,47º$. As stated, that gives a value of $3;00º$ in anomaly.
Next, Ptolemy asks what happens if we recalculate the anomaly for the argument plus $1º$, so in this case $214;41,47º$. If we do that, we get an anomaly of $3;04º$. If we subtract the latter from the former, the difference is $0;04º$ of anomaly, per degree of motion around the epicycle7. Yes, it’s quite confusing that we’re using two different contexts of “degrees” at the same time so let’s pay extra careful attention here.
Now, let’s look at the the lunar mean motion table for the motion around the epicycle in one hour. It’s $0;32,40º$ of motion around the epicycle. Now let’s multiply that by the increment we just calculated, paying special attention to the units here. I’ll use $º^{_a}$ to denote degrees in anomaly and $º^{_m}$ for degrees of motion about the epicycle. Thus, in one hour we get
$$0;04 \; \frac{º^{_a}}{º^{_m}} \cdot 0;32,40 \; \frac{º^{_m}}{hr} = 0;02,11 \frac{º^{_a}}{hr}$$
Here, we can see that the degrees of motion about the epicycle cancel out and we’re left with units that give the motion in anomaly per hour8. But the motion in anomaly isn’t the only component. The moon also has its mean motion of $0;32;56º$ per hour. Thus, we need to consider that as well.
But we can’t simply add these two components of motion. After all, the motion in anomaly can be either additive or subtractive based on where in the epicycle the moon is. Ptolemy phrases this part somewhat oddly, saying that, if the value we entered is above9 the maximum equation of anomaly, we subtract, and vice versa if below. This falls out very naturally if you just make sure you subtract the original from the original plus $1º$ and then pay attention to the sign. If we do, it is immediately obvious whether to add or subtract the motion in anomaly from the mean motion.
So in our case, the motion in anomaly is additive and we add the $0;02,11º$ to the mean motion of $0;32,56º$ per hour to get a total hourly motion of $0;35,17º$ per hour.
Now that we’ve determined the total speed of the moon,
We then determine how long, in equinoctial hours, the moon in its true anomalistic [i.e. true] motion, takes to cover that interval.
This passage seems to me to be a bit garbled, but if we just remember that we have an angular distance (that we need to increase by $\frac{13}{12}$ and the velocity it takes to traverse that distance, we can calculate the time with simple multiplication. So in our case, the original elongation was $5;23º$. We need to multiply that by $\frac{13}{12}$ to get an adjusted elongation of $5;49,55º$.
We can then divide that by the speed at which the gap is being closed:
$$5;49,55º / 0;35,17^{\frac{º}{hr}} = 9;55 hr$$
This is the discrepancy in time between the mean and true syzygy based on this procedure. But as a final question, we need to understand whether the true syzygy occurs before or after the mean.
If the true longitude of the moon [at mean syzygy] is less than the true longitude of the sun, we add the result to the time of mean syzygy, but if [the moon’s longitude] is greater, we subtract the result from the time of mean syzygy. Similarly, if the true longitude of the moon at mean syzygy is less than the sun’s [true longitude], we add the interval in degrees (increased, again, by a twelfth) to both the longitude and the argument of latitude [at mean syzygy], but if it is greater, we subtract it [from both]. Thus we get the time of true syzygy, and the approximate true position of the moon on its inclined circle.
This passage makes a lot of sense if we’re thinking in terms of conjunction. It’s obviously the context Ptolemy is thinking in given the focus here is really about eclipses. In that context, if the moon’s ecliptic longitude is less than the sun, it has to catch up. So it’s going to take awhile, and we should add that time. If the moon is already past the sun, then true syzygy has already passed and we should subtract.
Similar logic can be applied for oppositions, except we’ll need to add $180º$ to the solar longitude. So, in our case, the solar ecliptic longitude was $92;35,11º$. Adding $180º$ to that we get the point opposite the sun to which we need to compare to be $272;35,11º$.
Meanwhile, the moon’s position $277;58,11º$. It is already past the anti-solar point indicating that true syzygy happened $9;55$ hours prior to when we initially calculated.
Before closing out this post, we can discuss briefly how accurate this ends up being. While the example I worked was only for the first opposition in this year, Neugebauer works through the calculations for all oppositions in it. He then compares these to modern calculated values from Goldstine, New and Full Moons (1973). Here’s the full $13$ oppositions that happened that year shown as a discrepancy against the calculated value:
Here, we can see the discrepancy, at worst is off by about an hour and $18$ minutes. For the example we just calculated, it was an hour and $15$ minutes early. As a reminder, we showed the moon’s velocity at that time was a little over half a degree per minute, so this would imply that, at the moment Ptolemy predicted, the moon was $2-3$ full moon widths away from the sun – enough that eclipse timing might be off. But is that really enough to cause the eclipse not to happen at all?
As a quick back of the envelope calculation, we can consider a similar process as to what Ptolemy did above to find the lunar velocity in ecliptic longitude, but instead do it for lunar latitude. Near the nodes, where the lunar latitude changes most rapidly and the only places where eclipses can take place, the latitude changes by about $0;05,20º$ degrees in latitude per degree of argument in latitude. Thus, if we look up the change in argument in latitude in one hour ($0;33,04º$ per hour), we estimate a change of about $0;08,49º$ in latitude per hour. So in an hour and a quarter, that would be a discrepancy of $0;11,01º$. This would certainly change the nature of an eclipse, but only in the case when the eclipse was already only $\approx \frac{1}{3}$ of the full moon would it prevent it.
Meanwhile, if we consider that discrepancy as a fraction of a full lunar month, it’s only off by about $0.2\%$. Ultimately, the results of these calculations are quite impressive!
- By which Ptolemy means either the Conjunction or Opposition table.
- Which I have separated into a table on the 1 year interval tab.
- Again, refer to this post if you need a refresher on the Egyptian months.
- But not the sun’s since it is always $0$ by definition.
- A big thanks to Neugebauer who explains this.
- If you’re following along in other texts, you may or may not see $1;05$ pop up as it is the same as $\frac{13}{12}$ but in sexagesimal.
- This value is obviously not constant as it will depend where on the epicycle the moon is. However, for small intervals of time and thus motion on the epicycle like we’re dealing with here, approximating it as a constant works well enough.
- A few thoughts about this step of the calculation come to mind. The first is that, if we think back to the larger picture of what we’re doing, we’re trying to find the lunar motion over a distance so we can calculate how long it will take to traverse an angular distance. While this method calculates a reasonable speed for the moment of mean syzygy, if the gap between mean and true syzygy is large enough, the time to bridge it may take several hours. Thus, the speed during that time will change. Ptolemy obviously doesn’t think this is a large enough concern to warrant mention, but Neugebauer notes “Later works, e.g. of the Byzantine period, would explicitly calculate the [solar and lunar ecliptic longitudes] for the moment thus obtained and iterate the process until no elongation remains.” This is undoubtedly a better solution.
Second, we might also notice that the tables we reference were both from Book IV, before we calculated the effect of the second anomaly. We may again wonder if that has any impact. Pedersen explores this in Survey of the Almagest (p 226) and notes that it would have an impact, but Toomer, perhaps somewhat chidingly, points out that doing so with so much mathematical rigor is unnecessary since the effect of the second anomaly is zero at syzygy.
- Meaning literally above in the table. So $0º$ to $96º$ in the first column and $264º$ to $360º$ in the second column.