So far in this book, we’ve refined our lunar model, shown how to use it to calculate the lunar position1, discussed a new instrument suitable for determining lunar parallax, as well as an example of the sort of observation necessary to make the calculation.
Now, Ptolemy walks us through an example of how to calculate the lunar distance using an example entirely unrelated to the one we saw in the last post.
In the twentieth year of Hadrian, Athyr [III] $13$ in the Egyptian calendar [135 CE, Oct. $1$2], $5 \frac{5}{6}$ equinoctial hours after noon, just before sunset, we observed the moon when it was on the meridian. The apparent distance of its center from the zenith, according to the instrument, was $50 \frac{11}{12}º$. For the distance [measured] on the thin rod was $51 \frac{7}{12}$ of the $60$ subdivisions into which the radius of revolution had been divided, and a chord of that size subtends an arc of $50 \frac{11}{12}º$.
So this tells us everything we need to know to calculate the true lunar position based on our model. As usual, we must start by determining the time since epoch which Toomer gives as $882$ Egyptian years, $72$ days and $5 \frac{1}{3}$ equinoctial hours.
Ptolemy skips the math here but as we’re still new to it, I think it’s beneficial to walk through it, at least for the moon. To recall the steps necessary we’ll refer to V.10. First, we use the time since epoch to determine the total motion in all of the different motions which comes from the lunar mean motions table:
| Period | Longitude | Elongation | Anomaly | Latitude |
| Epoch Position | $41;22,00º$ | $70;37º$ | $268;49º$ | $354;15º$ |
| $810$ years | $37;24,07º$ | $234;19;55º$ | $222;10,57º$ | $217;37,22º$ |
| $72$ years | $315;19,29º$ | $332;49,46º$ | $267;44,58º$ | $267;20,39º$ |
| $60$ days | $70;34,59º$ | $11;26,41º$ | $63;53,56º$ | $73;45,40º$ |
| $12$ days | $157;07,00º$ | $146;17,20º$ | $156;46,47º$ | $158;45,08º$ |
| $5 \frac{1}{3}$ hours | $2;55,40º$ | $2;42,32º$ | $2;54,12º$ | $2;56,23º$ |
| TOTAL | $265;43,17$ | $78;13,14º$ | $262;19,50º$ | $354;40,12º$ |
The next step is to double the elongation as this will give us the distance from apogee which we find as $156;26,28º$. That gets looked up in the third column of our complete table of lunar anomaly. The value $156;26,28º$ is not in the complete table of lunar anomaly, so we must estimate between the value of $156º$ and $159º$. At this point, I believe Ptolemy made a minor mathematical error. To explain, I’ll walk through my math a bit more slowly than usual when making these estimations.
There is a difference of $3º$ separating these two values, so if we make a linear approximation between the two, we are $\frac{0;26,28º}{3º}$ of the way from $156$ to $159$. Expressing this as a percent, that’s $14.7$%. Now, the difference in the longitudes (expressing it in decimals since it’s easier to work with) is $0.867º$. Taking $14.7$% of that we get that we’re $0.127º$ of the way from the value of longitude at $156º$ to the value at $159º$. And since it’s decreasing, we need to subtract.
Doing so, and converting back to sexagesimal, I get the equation for [mean to true] apogee to be $7;40º$. However, Ptolemy came up with $7;26º$. The only way I can explain this is that Ptolemy forgot that the two lines in the table were $3º$ apart and did not divide. If I do that, I get $7;25º$ which is suspiciously close to Ptolemy’s value.
To be fair, the approximation I’ve just given is a linear approximation between these two values, and the overall shape is decidedly not linear. So it is possible that this was another case of Ptolemy going through the full calculation again? To be extra sure, I plotted the data in Excel to get a better read on it:
You’ll want to enlarge the image as the thumbnail makes this impossible to read, but what we can clearly see from the graph is, even taking the nonlinearity into account, the value at $156;26,28º$ is clearly above $7 \frac{1}{2}$, matching the value I’ve come up with and disagreeing with the value of Ptolemy’s. However, to continue to follow Ptolemy’s work, we’ll adopt his value.
Continuing on, the value we looked up was less than $180º$ so we’ll add that to the mean anomaly to get the true anomaly:
$$262;19,50º – 7;26º = 254;53,50º$$
This then gets entered back into the same table and we look to columns $4$ and $5$ to get their values. From that, I get the epicyclic equation to be $4;57º$3 and for the increment I get $2;39$.
Next, we again look up the double elongation and take the value from the sixth column. I get $0;56,41$. This gets multiplied by the increment and then added to the epicyclic equation:
$$2;39º \cdot 0;56,41 + 4;57 = 7;27º$$
We then check our true anomaly to see if it’s greater or less than $180º$. Since ours is greater, we add it to the mean longitude and mean [argument of] latitude. Thus, our true longitude is $273;10º$ which is $3;10º$ into Capricorn. For the true [argument of] latitude, we then get $2;7º$4 from the northern limit.
Now, that value once again gets looked up in the same table, now looking at column $7$ to get the angle above or below the ecliptic of the moon. This gets a bit odd as there is no row for $0º$ or $360º$. However, we know it should be the same as the value for $180º$ which is $5º$. Thus, we can estimate the distance from the ecliptic as $4;59º$. And because this is in the first fifteen lines of the table (i.e., $<90º$ or $>270º$) this is north of the ecliptic.
However, consider again this image from our last post:
Here, we can see that one of the angles we’re going to need is the angular distance between the ecliptic and celestial equator. It’s been awhile, but we actually created a table for this way back in Book I. This table only includes values through $90º$ as it starts from the vernal equinox and goes to the summer solstice. After that, the values repeat but going down until the autumnal equinox. Then they repeat again, going up, but below the celestial equator, until the winter solstice then back down again completing the cycle.
For this example, $273;10º$ is $3;10º$ past the winter solstice, so we need to look up a value of $90º – 3;10º = 86;50º$ for which I get an estimate of $23;49º$ below the celestial equator. Let’s redraw that picture for this situation:
As with before, the distance from the zenith to the celestial equator is the latitude of the observer ($30;58º$ at Alexandria). We just determined the angle between the ecliptic and equator as $23;49º$, and the angle from the zenith to the moon was observed to be $50;55º$. Thus, we can calculate the observed angle of the moon above the ecliptic:
$$30;58º + 23;49º – 50;55º = 3;52º$$
However, the true angle based on our calculation was $4;59º$. As such, there’s a discrepancy of $1;07º$ due to parallax.
We could look at this the other way around and ask what the distance from the zenith should have been if there were no parallax which is really the route Ptolemy takes in which he performs the following calculation:
$$30;58º + 23;49º – 4;59º = 49;48º$$
Either way, we’re stating the same thing in different ways.
I’ll break off here and in the next post, we’ll explore how to use this observation to calculate the lunar distance.
- In ecliptic coordinates.
- In the last post, we stated that the observations must be taken near a solstice. As such, you might panic and think we’ve done this wrong as Oct $1$ is clearly nowhere near the winter solstice. However, the moon need not be taken on the date of a solstice, but near the solstice point along the ecliptic. On this date, it was a week or so after the autumnal equinox, so the sun was shortly after the autumnal equinox point. So the moon being on the meridian would put it very close to the point of the winter solstice.
- I’ve founded here to the nearest sixtieth. A more complete estimate would be $4;56,50º$, but since the other values we’re working with here are only to the precision of sixtieths, I think it’s appropriate to stay the same.
- Ptolemy comes up with $2;6º$.