At the end of the last post, we noted that Ptolemy wasn’t quite satisfied with what we did previously because we used some rather faulty assumptions.
As Ptolemy states it:
For lunar parallaxes, we considered it sufficient to use the arcs and angles formed by the great circle through the poles of the horizon [i.e., an altitude circle] at the ecliptic, instead of those at the moon’s inclined circle. For we saw that the difference which would result at syzygies in which eclipses occur is imperceptible, and to set out the latter would have been complicated to demonstrate and laborious to calculate; for the distance of the moon from the node is not fixed for a given position of the moon on the ecliptic, but undergoes multiple changes in both the amount and relative position.
The key phrase here is the “at the ecliptic, instead of the moon’s inclined circle.” This got swept under the rug in that post because Ptolemy didn’t really explain why the algorithm he gave us should work. So to understand, let’s start by taking a harder look at what’s actually going on.
To do so, we’ll follow a new diagram Ptolemy describes1:
Here, $\overline{GA}$ is a section of the ecliptic and $\overline{DA}$ will be a section of the moon’s path. Point $E$ is the zenith and we’ll put the moon’s true position at $D$. Its position due to parallax will be at $H$. A line is dropped from the zenith, through the moon onto the ecliptic at $Z$ and a second line is dropped from the moon such that it is perpendicular to the ecliptic at $B$2. Two lines are drawn from the moon’s apparent position at $H$: one parallel to the ecliptic intersecting the previously drawn line $\overline{DB}$ at $\Theta$ and another, perpendicular to the ecliptic, meeting the ecliptic at $Z$. Now that we’ve got that set up, let’s take a moment to consider what a few of these represent.
We have $\overline{AB}$ which will represent the true distance, measured along the ecliptic, that the moon appears from the node while $\overline{AK}$ is the apparent distance due to parallax3.
The total parallax will be $\overline{DH}$. This gets broken down into its two components: $\overline{D \Theta}$ is the latitudinal component, and $\overline {H \Theta}$ is the longitudinal component.
Take a moment to digest this as this triangle, $\triangle{D \Theta H}$ which contains the parallax and its components is the real crux of the issue here. What Ptolemy suggested in the last post is that we can solve this triangle. This is a right triangle, so if we know a side and one of the angles, this would certainly be true. But do we?
We certainly do know $\overline{DH}$ since this is the total parallax which we have discussed how to calculate. Thus, if we’re truly to solve this triangle, we would need to know one of the non-right angles. And that’s where we run into trouble.
Ptolemy states the issue with what we did previously:
From the preceding theorems, [we know that] parallax $\overline{DH}$ can be found if $arc \; ED$ is given, and both [components of] parallax, $\overline{D \Theta}$ and $\overline{\Theta H}$, if $\angle{GZE}$ is given. But what we determined previously was the arcs and angled formed at given points of the ecliptic by the altitude circle; and the only point on the ecliptic which is given in this situation is $B$. Hence, is clear that we are using $arc \; EB $ instead of $arc \; ED$, and $\angle{GBE}$ instead of $\angle{GZE}$.
Here, Ptolemy first reiterates what I stated above about solving $\triangle{D \Theta H}$ and then admits that we don’t actually know one. The first step in what Ptolemy had us do was to refer to the table of zenith distances and ecliptic angles. But what angle really was that?
The value we plugged in to that table was the time until the moon would cross the horizon which was $20$ minutes in that example. But where did that came from again? It was something we calculated in this post in which we calculated the ecliptic longitude for the true moon. In other words, this was $\angle{GBE}$4. That’s… not helpful because that’s not congruent to anything in the small triangle we’re concerned with. So what we’d really have liked is $\angle{GZE}$ which is equal to $\angle{\Theta HD}$ in our triangle. So let’s ask ourselves what this angle really is? It is the angle between the ecliptic and the great circle drawn between the zenith and the (true or apparent) moon.
So, with that highlighted, the true purpose of this chapter is going to be to determine in which cases we’ll need to actually worry about this, and if we do, how to determine the actual angle we need.
Before going forward, I’ll just make a note that Ptolemy was rather inconsiderate in the rest of this chapter. We’ll be drawing some more diagrams and instead of keeping these points labeled in a consistent manner, they will change between diagrams. So regardless of the how we label things, keep in mind that what we’re trying to find here is the angle between the ecliptic the and the great circle drawn through the zenith and moon. This we will compare with the angle used in the approximation used in the previous chapter which was the angle between the ecliptic and the great circle drawn between the zenith and the ecliptic longitude of the true moon.
In addition, consider again that quote above where Ptolemy states:
parallax $\overline{DH}$ can be found if $arc \; ED$ is given.
Thus, as we go through these, we’ll also want to determine if $\overline{ED}$, the distance between the zenith and true moon, can be calculated.
Ptolemy notes that this problem isn’t new to him. It’s something Hipparchus faced as well and he
attempted to correct this kind [of inaccuracy] too, but it is apparent that he attacked the problem in a very careless and irrational way. For firstly, he dos it for [just] a single value of the distance $\overline{AD}$, instead of all [possible] values, or a number of values, as would have been appropriate in a situation where one has chosen to be nicely accurate about small [errors]. Furthermore, without realizing it, he has fallen into a number of [even] stranger errors. Having also [like us] previously demonstrated [the amounts of] the arcs and angles with respect to [intersections of altitude circles with] the ecliptic, and shown that, if $\overline{ED}$ is given, $\overline{DH}$ can be found (he shows this in Bk. I of his On Parallaxes), in order to get $\overline{ED}$ as a given quantity, he assumes $arc \; EZ$ and $\angle{EZG}$ are given (in this way, in Bk. II, he calculated $\overline{ZD}$ and takes $\overline{ED}$ as remainder [of $\overline{EZ} – \overline{ZD}$). However, he was mislead by his failure to notice that the given point of the ecliptic is not $Z$ but $B$, and hence the given arc is not $arc \; EZ$, but $arc \; EB$, and the given angle not $\angle{EZG}$ but $\angle{EBG}$. Yet, it is these [$arc \; EZ$ and $\angle{EZG}$] which are the [necessary] starting points for making even such a partial correction. For in many situations there is quite a noticeable difference between the $arc \; ED$ and the $arc \; EZ$, whereas the difference between $\overline{BE}$ (which really is given) and $\overline{ED}$ is, at most, the amount of the $arc\; BD$ for any given distance [of the moon] from the node.
That block of text isn’t really necessary for the corrections we’re going to work out, but I decided to include it for a few reasons. First, I love seeing the thread of history weaving through this – Ptolemy building on ideas for Hipparchus and other Greek astronomers5. Second, I seem to recall reading somewhere that Toomer took care in this translation to try to break up Ptolemy’s penchant for overly complex run on sentences into something more readable6. This section is a prime example that I think highlights what he was trying to work with7.
Now that this diversion is over, how does Ptolemy suggest we deal with this discrepancy?
To avoid the issue Ptolemy criticized Hipparchus on, we’ll discuss a few different scenarios.
Let’s start by considering one of the simplest scenarios, where the great circle through the zenith and moon is perpendicular to the ecliptic:
Here, we’ll let $\overline{ABG}$ represent the ecliptic with $Z$ as the zenith, and the moon either above the ecliptic at $D$ or below it at $E$. We’ll consider this when $\overline{DBE}$ at right angles to the ecliptic, intersecting at point $B$.
Ptolemy doesn’t describe whether $D$ or $E$ is the true or apparent moon, but it doesn’t particularly matter as it will still fall along $\overline{ZE}$. So lets just say they are the true moon and we’ll draw in an extra point for the apparent moon:
Here, I’ve drawn in the apparent moon at $H$ for a true moon at $D$8. So where is the ecliptic longitude for the true moon? It’s at $B$ on the ecliptic. Where is it for the apparent moon? Again, at $B$. Thus, the angle between the ecliptic and great circle through the zenith at their ecliptic longitudes is the same because it’s the same point for each and is right angles in both cases. Ptolemy phrases this somewhat oddly, saying
the angles at $D$ and $E$ will not differ from at $B$: for [arcs] drawn through these points [from the zenith] are also at right angles to the ecliptic.
Thus, there would be no difference between the two angles we discussed previously so the approximation in this case turns out to be entirely accurate and the entirety of the parallax will be in ecliptic latitude.
Also, can we determine the distance between the zenith and the true moon, which, in this case is $\overline{ZD}$ or $\overline{ZE}$? In both cases, yes.
$$\overline{ZD} = \overline{ZB} – \overline{BD}$$
$$\overline{ZE} = \overline{ZB} + \overline{BE}$$
Those other components being things we can calculate as we’ve seen earlier.
Next, what happens if the zenith lies on the ecliptic as can happen between the tropics:
To illustrate that9, we’ll let $\overline{ABG}$ again be the ecliptic with the zenith at $Z$. Again, we can let the moon be at either $D$ or $E$ which in both cases would have an ecliptic longitude at $B$. Again, for the case of the true moon at $D$, I’ve drawn in an apparent moon at $H$.
Now let’s consider the two angles we’re concerned with. The one Ptolemy used in his previous approximation was the angle between the ecliptic and great circle drawn between the zenith and the ecliptic longitude of the true moon. In other words, the angle between $\overline{AG}$ and $\overline{ZB}$. However, in this case, they’re parallel. So the angle is either $0º$ or $180º$.
The angle we really want to use in this case is the angle between the ecliptic and the great circle drawn between the zenith and moon. This is $\angle{AZD}$ and is very obviously not the same angle. So in this case, we will need to take this discrepancy into consideration. Fortunately, it’s not hard to do in this case because we know $\overline{ZB}$: the distance of the moon’s ecliptic longitude from the zenith, which we can get from the table regarding the zenith and ecliptic as we’ve already seen.
In addition, we know $\overline{BD}$ or $\overline{EB}$ which are the ecliptic latitudes of the true moons, which we showed how to calculate in Book V, Chapter $9$.
These are two pieces of the right triangle, $\triangle{ZBD}$ so we can use the Pythagorean theorem to find the hypotenuse, and then a demi-degrees method to determine the corresponding angles. Specifically, $\angle{DZB}$ or $\angle{EZB}$ which can then be subtracted from $180º$ to find the angle we’re really after.
Now we’ll get a bit more complicated and consider something like the original drawing we had the ecliptic at an incline to the altitude circle. That initial diagram doesn’t have everything we’re going to need to really solve this so we’ll need to do a bit of redrawing.
In this diagram, we’ll take $Z$ as the zenith and $\overline{AG}$ as the ecliptic . As with the past two hypotheticals, we’ll also let $D$ and $E$ be potential positions of the moon. We also draw from each of these lines perpendicular to the ecliptic, both at $B$ for the sake of simplicity. This means that $\overline{DB}$ and $\overline{EB}$ is the ecliptic latitude of each position respectively.
For the moon at $D$ we would also know $arc \; ZB$ and $\angle{ABZ}$ as we’ve already shown how to calculate these for the true moon in previous posts in this chapter, as would $arc \; DB$ since it’s the calculated ecliptic latitude of the true moon.
But, Ptolemy says,
what we need to be given are $arc \; ZD$ and $arc \; ZE$, and angles $\angle{AHZ}$ and $\angle{A \Theta Z}$.
To understand why, let us add in that apparent position of the moon in relation to point $D$ as well as its components:
Here, the apparent moon I’ve added at $M$ and the components of the parallax, $\overline{DM}$. Here, $\overline{MN}$ is the latitudinal component and $\overline{ND}$ the longitudinal. This small triangle, $\triangle{MND}$ is the one we’d need to solve. To do so we’d need either $\angle{NMD}$ or $\angle{NDM}$. The latter of these is equal to $\angle{AHZ}$ so would be sufficient for this purpose. If we did the same for a hypothetical moon at $E$, then we would need $\angle{A \Theta Z}$.
And Ptolemy says we can actually solve for these by drawing in $\overline{DK}$ and $\overline{LE}$, both of which are perpendicular to the great circle drawn through the zenith and the ecliptic longitude of the true moon.
Ptolemy walks through this in a very abstract sense:
For since $\angle{ABZ}$ is given10, and $\angle{ABE}$ is always a right angle, the right-angled triangles $\triangle{BKD}$ and $\triangle{BLE}$ are given, and so is the ratio of $\overline{ZB}$ to the sides containing the right angle, since [the ratio of $\overline{ZB}$] to the hypotenuses of $\overline{DB}$ and $\overline{BE}$ is given. Hence, there will be given $\overline{ZD}$, the hypotenuse [of the right-angled triangle, $\triangle{ZDK}$, of which sides $\overline{ZK}$ and $\overline{KD}$ are given], and $\overline{ZE}$, the hypotenuse [of right angled triangle, $\triangle{ZLE}$, of which sides $\overline{ZL}$ and $\overline{LE}$ are given], and also the angles $\angle{DZK}$ and $\angle{EZL}$, which are the differences from the required angles.
That was a lot of math in one brick of text, so lets walk through that. The first part said that we know $\angle{ABZ}$. This is the east or west angle we showed how to calculate previously.
Then, since $\angle{ABE}$ is a right angle, we can determine $\angle{ZBE}$ as it’s just $90º – \angle{ABZ}$. This is one of the angles in $\triangle{BEL}$ and is a vertical angle with $\angle{DBK}$ so we would know that angle in $\triangle{BDK}$.
Knowing that, Ptolemy states that this allows us to solve for either one of these triangles, but doesn’t detail how. Fortunately, it’s not hard to figure out how. Recall $\overline{DB}$ and $\overline{BE}$. Again, these are the ecliptic latitudes of the true moon which we know how to calculate. Thus, we know the hypotenuse of these right triangles, as well as one of the non-right angles. From there, we can use a demi-degrees triangle to solve for the remaining sides in each.
Ptolemy states is that this also gives us
the ratio of $\overline{ZB}$ to the sides containing the right angle
In other words, he’s saying we can get $\frac{\overline{ZB}}{\overline{DK}}$ and $\frac{\overline{ZB}}{\overline{BK}}$ in that triangle and $\frac{\overline{ZB}}{\overline{LE}}$ and $\frac{\overline{ZB}}{\overline{BL}}$ in that triangle. This is certainly true, since $\overline{ZB}$ is the angle between the zenith and the ecliptic longitude of the true moon ($B$) which we also said we can get in the previous case. And since we just stated we can calculate all the denominators, so from there, it’s just division.
What I’m not sure of is why this is particularly helpful. Ptolemy states that these ratios allow us to determine $\overline{ZD}$ and $\overline{ZE}$. Perhaps there’s some way I’m just not seeing, because the simpler way to me seems to be to add $\overline{BK}$ to $\overline{DB}$ to get side $\overline{ZK}$. And since we said we can determine $\overline{DK}$, that’s two sides of right triangle $\triangle{DZK}$ so we can use the Pythagorean theorem to determine $\overline{ZD}$ that way. The same can be done via subtraction of $\overline{BL}$ from $\overline{ZB}$ to get $\overline{ZL}$ in right triangle $\triangle{ZLE}$, and again solve for $\overline{ZE}$ in that triangle, thus knowing all sides of $\triangle{DZK}$ and $\triangle{ZLE}$.
And knowing the sides in the triangle, we can use the demi-degrees method to work out the angles. This would allow us to determine $\angle{DZK}$ and $\angle{LZE}$ in particular. And those become useful for what we ultimately need since11
$$\angle{AHZ} = \angle{ABZ} + \angle{DZB}$$
$$\angle{A \Theta Z} = \angle{ABZ} – \angle{EZL}$$
And these were the true angles we were after.
Ptolemy talks a bit about these three cases, discussing when the difference between the two angles we’ve considered are the largest. Turns out and it’s when the zenith lies on the ecliptic. In that case, we already saw that the ecliptic and great circle between the zenith and true lunar ecliptic longitude is $0º$. Meanwhile, the angle between the ecliptic and great circle through the zenith and moon becomes whatever the lunar latitude is. Recall that the moon’s circle is tilted by about $5º$ with respect to the ecliptic, which means that the two angles we’ve been discussing can vary up to $5º$ as well.
But recall that this angle was just a piece of the overall process to determine the apparent lunar position. So we need to consider what that impact is in the final calculation. Ptolemy skips the math12 but determines the maximum impact it would have is about $10$ minutes of arc. That’s not negligible, but not too bad.
And furthermore, we’ve been working up to calculations of eclipses and in those never happen when the moon is at that maximum ecliptic latitude. The highest latitude the moon can have for an eclipse is about $1 \frac{1}{2}º$
13 in which case the maximum difference in that would have on the difference in lunar parallax would be about $1 \frac{1}{2}$ arc minutes. Quite small.
Thus, for most purposes, the first approximation we worked out previously is fine in most cases. But to close out this book, Ptolemy is going to walk us through how to go ahead and calculate the correction should we choose to. However, I’ll save that for another post.
- When drawing this out, I was tempted to go ahead and try to draw it on the celestial sphere as seeing all these angles, arcs, and circles always makes things easier to wrap my head around. But in this case, I elected not to because, as we saw in the last post, Ptolemy is approximating spherical triangles with planar ones, so this two-dimensional diagram is actually more appropriate. However, Ptolemy substituting planar triangles for spherical ones means we’ll be playing a bit fast and loose with the language here.
- The length of this line, $\overline{DB}$, would then the the true moon’s ecliptic latitude
- I’ve included these here because they’re part of Ptolemy’s description of the diagram and in Toomer’s image, but these aren’t particularly important to our discussion.
- You might be wondering if that means that we calculated $\angle{GBD}$ as I did for several minutes. However, remember that the table we’re using calculates the angle between a great circle through the zenith and the ecliptic. The true moon only comes into play because its ecliptic longitude decides the vertex of these two great circles.
- That being said, Toomer notes that no one has actually figured out what Ptolemy is talking about here because we don’t have any extant copies of Hipparchus’ work. Toomer suggests that there has been some attempts to do so (although he doesn’t elaborate), but does refer to a work by Pappus of Alexandria (4th century) which contained a supposed reconstruction of Hipparchus’ method in Pappus’ commentary on the Almagest. But Toomer deems it “entirely fictitious” and refers to Neugebauer. Following that thread through to History of Ancient Mathematical Astronomy (page 324), Neugebauer states that this is because one of the values Pappus used in this supposed calculation in incorrect and appears to be a scribal error in particular lineage of the Almagest. Thus, it is unreasonable to assume that this error persisted back to Hipparchus’ works and thus, the calculation cannot be faithful to Hipparchus’ actual work.
In addition, Neugebauer notes that “single steps are in part obviously meaningless” and suggests that Pappus created the method himself “in order to obtain some procedure which Ptolemy could have rightly criticized.” He goes on to highlight several nonsensical calculations in Pappus’ work, insinuating Hipparchus would not have engaged in such calculations. Based on what I’ve seen from Ptolemy’s commentary on Hipparchus, I’m inclined to agree if for no other reason that I feel Ptolemy would have highlighted the same issues Neugebauer does.
- I’m trying to look through the introduction to find this again, but not finding it at the moment and it’s not worth dwelling on.
- I also note how inconsistent the mathematical notation is here where we seem to casually flip between lines and arcs likely due to the aforementioned approximation I’ve discussed in which Ptolemy is approximating the spherical triangles with flat ones. My feeling is that when arcs can be calculated, Ptolemy is describing these as arcs and is not as lines and this is not an accident on Toomer’s part but I’m not certain.
- Ignoring the case of the true moon at $E$ for now.
- I’m deviating slightly from both Ptolemy and the diagram in Toomer here as they felt rather confusing to me.
- Meaning, we know how to calculate it.
- Ptolemy doesn’t give a proof for this following statement, but it’s quite easy to prove. For the first, simply draw a line through $H$ parallel to $\overline{ZK}$. Then you can see that the angle between the ecliptic and this line at $H$ is the same as the one between the ecliptic and the original at $B$ and the excess is simply the alternate interior angle to $\angle{DZB}$. The same sort of proof can be done for the second case.
- Toomer has a footnote (pg. 271) that walks through the steps, but I’ll omit it.
- Which would happen for a lunar eclipse.