Now that we’ve completed the parallax table, we should discuss how to use it. As usual, Ptolemy will now walk us through the steps, but does not provide an explicit example so I’ll follow an example from Neugebauer1, to help illustrate the procedure.
As our goal, Ptolemy states that our goal is to find
the moon’s parallax at any given position, (first) with respect to the great circle drawn through the moon and zenith
As a reminder, we’re operating in a theoretical context here, wherein we’re calculating the parallax for any time in the past or future; not going out and measuring it for a present date. So while Ptolemy has equipped us with an instrument for measuring such a distance, we’ll actually be calculating the lunar distance from the zenith.
To begin, we’ll need the
distance (in equinoctial hours) from the meridian at the latitude in question.
To phrase this differently, we want to know how long before the moon crosses the meridian. Even if the moon isn’t on the meridian, we can calculate how long it will take by knowing the point of the ecliptic at upper culmination2 and the position of the moon. In Neugebauer’s example3, he considers the moon at $5;20º$ ($215;20$ ecliptic longitude) into Scorpio, the first point of Scorpio4 ($210º$ ecliptic longitude) at upper culmination and the observations being done from Rome.
So the distance along the ecliptic between the moon and the point of upper culmination would be the difference between the two: $215;20º – 210;00º = 5;20º$. However, we’re looking for the distance in equinoctial hours which is measured in terms of the celestial equator; not the ecliptic. As such, we’ll need to translate those ecliptic coordinates into equatorial ones.
We can do this from our table of the rising times of the ecliptic. However, we need to be careful because we’re not looking for the arc of the celestial equator cut off by a great circle through the local zenith, but from the celestial poles. As such, we should use the table for sphaera recta5. In doing so, I come up with $212;59º – 207;50º = 5;09º$.
Converting that the equinoctial hours6:
$$\frac{5;09º}{360º} \cdot 24^h = 0;20^h$$
In other words, the moon is about $\frac{1}{3}$ of an hour before being on the meridian.
Now, to get the zenith distance, we’ll need to take this and enter it into
the Table of Angles for the appropriate latitude and zodiacal sign, and take the amount in degrees in the second column corresponding to the hour, interpolating between integer hours if necessary.
As Ptolemy says, we’ll table of zenith distances. I can’t remember having used this table before, so let’s slow this one down a bit and start by reminding ourselves what this table is – It tells us the arc of a great circle between the zenith at that latitude and the first point of a given zodiac sign for a given number of equinoctial hours before that first point hits the meridian.
For this example, we already stated we’re considering the latitude of Rome which Neugebauer adopts as being equal to Hellespont. So we’ll use the appropriate table. Looking at that table, we’ll select our sign in question, Scorpio, and estimate the value from the Arc column $\frac{1}{3}$ of the way between noon and $1$. Doing so, I get $53;12$.
But remember, this was the arc to the first point in Scorpio and the moon is actually $5;20º$ into Scorpio. So to account for that, we’ll repeat the same calculation for the beginning of the next constellation, Sagittarius and then, using those two values, we’ll interpolate between7. When doing so, I come up with $61;57º$. The moon was $\frac{5;20º}{30º}$ the way through that difference giving an interpolated angle from the zenith of $54;45º$. Neugebauer comes up with $54;40º$ which I’ll adopt going forward for clarity.
With this as [the] argument, we enter [it into] the Table of Parallaxes, determine on which line in the first column the argument is to be found, and taking the numbers corresponding to this in the four columns following the column of solar parallaxes, namely the third, fourth, fifth, and sixth columns, write each one down separately.
So we’ll now head over to the parallax table, and look up the value in column one, then record the corresponding values from columns $3 – 6$ since we’re dealing with a lunar parallax calculation. Interpolating, I come up with $0;44;06º$ for column $3$, $0;08,33$ from column $4$, $1;04,27º$ from column $5$, and $0;21,07º$ from column $6$.
Then we take the corrected anomaly (i.e. with respect to the true apogee [of the epicycle]) at that moment: [if it is less than $180º$] we take the anomaly itself, but if it is greater than $180º$, we take ($360º$ minus the anomaly); we halve the amount so obtained, and, entering with this into the same [column of] arguments, determine the number of minutes corresponding to it in both the seventh and eighth columns separately.
Next we’ll use the four values we just wrote down to begin applying some corrections. Based on our previous work, we know that this will have to do with the anomaly on the epicycle as well as the elongation on the eccentre. This example hasn’t given a specific date and time so we can’t calculate them. Rather Neugebauer gives them to us directly as $294;32º$ around the epicycle from apogee and $277;53º$ as the mean elongation from the sun.
We’ll start with the epicycle since that’s all Ptolemy has told us to work with for now. This value is greater than $180º$, so it’s past perigee and we need to consider $360º$ minus that value which is $65;28º$. In addition, we noted that Ptolemy actually uses half that angle as what he considers the argument, so we’ll cut that in half to $32;45º$8. So looking that argument up in columns $7$ and $8$ I get $0;16,34$ and $0;16,03$ respectively.
We take the minutes found from the seventh column, multiply them into the difference found from the fourth column, and (always) add the result to the parallax from the third column. [Likewise] we take the minutes found from the eighth column, multiply them into the difference found from the sixth column, and again (always) add the result to the parallax from the fifth column.
Stating that a bit more clearly: multiply the values from column $7$ and $4$, and then add to the one from column $3$.
$$0;16,34 \cdot 0;08,33º + 0;44,06º = 0;46,28º$$
We then do something similar, multiplying column $8$ by $6$ and then adding to column $5$:
$$0;16,03 \cdot 0;21,07º + 1;04,27º = 1;10,06º$$
Thus, we have obtained two parallaxes; we take the difference between these and write it down.
So9
$$1;10,06º – 0;46,28º = 0;23,38º$$
Next, we take the mean elongation of the moon from the sun, or else the mean elongation of the moon from the point opposite the [mean] sun, whichever of these two distances is the lesser, and entering with this too into the the arguments in the first column, take the minutes corresponding to it in the ninth and last column.
This part is phrased rather poorly in my opinion so before just applying the steps, let’s take a moment to explore.
Let’s first remind ourselves of what the argument was for column $9$. When we laid out the table, I reminded us that it was either half the angular distance from apogee or the mean elongation which is what we have here.
Next, let’s think about the arguments for the other columns. Specifically, the table only goes up to $90º$. In the cases of columns $2-6$ this was because these are all angular distances from the zenith so as long as the object is above the horizon, that’s all we should need. In columns $7$ and $8$, the values can actually range anywhere from $0-360º$, but there is a symmetry, in which everything from $180-360º$ is a reflection of the $0-180º$, so the table only needed to show half the values. Still, the table only goes through $90º$ so, as I noticed when we laid out the table, Ptolemy squeezed into the same table by using half the actual angles.
Now to column $9$ in which symmetries again exist which is why Ptolemy was able to get away with fewer rows in the table. However, this time, the symmetries aren’t every $180º$, they’re every $90º$.
When the elongation is $0-90º$, we can use the table directly. From there, we’re off the table, but symmetries again exist. If the elongation falls between $90-180º$, we would then determine the elongation, not from the mean sun, but the point opposite the mean sun. In other words, $180º$ minus the elongation. If we get into the range of a true elongation of $180-270º$, then it would be the elongation minus $180º$. Lastly, if we’re between $270$ and $360º$ elongation, it would be $360º$ minus the elongation.
Neugebauer lays this out a bit more neatly10:
So as long as we’re using the mean elongation, we never actually have to worry about the half angles used in columns $7$ and $8$ because the value we’ll use is never greater than $90º$.
Let me sketch that out based on the situation we’re considering here.
Here, I’ve drawn in the moon with an elongation from the mean sun as we were given of $277;53º$. Since this falls in the to $270-360º$ range, we’ll take $360 – 277;53 = 82;07º$ which is what we’ll enter into the table and for which I get a value of $0;59,00$ after a bit of interpolation.
Lastly,
We multiply these into the difference between the two parallaxes which we wrote down, and (always) add the result to the smaller (that is, the one derived from the third and fourth columns).
In other words, we’ll take the piece from columns $3$, $4$, and $7$, and add that to the product of column $9$ and what we got from columns $5$, $6$, and $8$ keeping in mind our order of operations.
So:
$$0;46,28 + 0;59,00 \cdot 0;23,38 = 1;09,42º \approx 1;10º$$
This will give us the moon’s parallax as measured along the great circle through the moon and the zenith.
I’ll close out this post here, but we’re still not done with this chapter. There’s a few more things we’ll still want to cover to set us up for the next book which will be all about eclipses. So we’ll first look at how to use this result to determine the solar parallax. Then, we need to look at how to adjust the result we just obtained and get the parallax with respect to the ecliptic since that’s the context we really care about for eclipses.
This post took far longer than I’d have liked to get up. A big part of the reason was what I covered in the introduction of this post – I was trying to follow Toomer’s example which has some bad math in it I couldn’t reproduce and I wasted a lot of time trying to figure out what he did. So I ended up having to start all over with the Neugebauer example and by then, I had forgotten about the distinction between the double elongation and distance from apogee in the penultimate step. I got frustrated and had to put it down for a few weeks. It didn’t help that real life work got crazy this past month. Around the same time, lots of recruiters started calling me, and I’m starting a new job next week!
- This example begins on page $114$ of History of Ancient Mathematical Astronomy. Toomer also has an example in his translation of the Almagest, in Appendix A, example $10$. However, Toomer’s example will incorporate corrections that will come later in this chapter so not particularly useful at this point. There’s also a step that is very poorly estimated, so I want to have a good example first before trying to tackle the bad one.
- We haven’t used this term in quite awhile, so as a reminder, the culminating point is the current point of the ecliptic that’s on the meridian.
- In this example, we’re being given the point of upper culmination. In Toomer’s example, he did not but instead gave the time and location which would allow us to calculate the point of upper culmination as this was one of the applications of the rising time table from back in Book II.
- He actually gives it at Libra $30$, which is the same thing, but given it’s displayed in the rising time table as Libra $30$, I can understand why.
- Where the great circle cutting off the arcs of the ecliptic and celestial equator through the poles is the great circle through the zenith.
- I come up with a marginally higher value after rounding of $0;21^h$ than Neugebauer, but for consistency, we’ll use his numbers.
- It is in this step that I find issue with Toomer’s example. In it, the moon is actually below the horizon. Thus, the table does not contain the necessary values needed to execute this step. I have tried estimating based on the interval for the previous hour and came up with a value over a degree higher than Toomer who ultimately adopts a situation in which the moon is directly on the horizon, at $90;00º$.
- Tiny bit of rounding here as it is more accurately $32;44º$.
- Here I diverge from Neugebauer’s example slightly as he saves this step until the corrections from elongation is complete and simply does this as part of the final calculation. I do so because I’m trying to follow Ptolmey’s instructions as they are written.
- Here, $\eta$ is the elongation so $\overline{\eta}$ is the mean elongation (elongation from the mean sun) and $\eta ‘$ is the elongation we’ll be using to look up in the table.