Now that we’ve determined how far away from the nodes a lunar eclipse can occur, we’ll work on doing the same for a solar eclipse1. But before diving in, I want to say that this has been one of the most, if not the most challenging section of the the Almagest so far. One of the primary reasons is that Ptolemy shows no work and gives almost no explanation on how he did this. When such things happen, I often turn to Neugebauer’s History of Ancient Mathematical Astronomy which I did in this case. There, Neugebauer refers to Pappus of Alexandria, a fourth century mathematician who did commentary on the Almagest and walks through a process that arrives at the same values as Ptolemy.
However, there was a very large amount to unpack in just a few pages there and, unlike most cases where I can simply work along with it and see where things are going, this time I had to really understand the whole process before the first steps made any sense. This led me to agonize over what was going on with those first steps, amounting to several days of effort and rewriting this post from scratch several times. The result is twofold. First because I feel this section can only be approached by understanding the methodology before diving into the math, there’s going to be far more exposition than normal and, as a result, this is likely to be one of my longer posts. Second, the struggles I had with trying to understand the method and rewriting this post so many times has left me with a lot of fragments of thoughts in my brain and in the blog editor. I’ve done my best to clean it up, and maybe it’s just those thoughts swirling around in my brain, but this post just doesn’t feel as coherent as I like. Apologies in advance if you struggle to follow. Know I did as well.
Anyway, moving on to the topic at hand.
Normally, I like to start with a quote from Ptolemy to give us some direction, but I think Ptolemy did such a poor job of laying this section out, I’m going to avoid doing so for the majority of the post. Instead, let’s try to understand the process by recalling what we did with the moon and discussing how things will change.
When dealing with the moon we considered the situation in which the edge of the moon’s disc just grazed the earth’s shadow.
In that situation, we knew the radius of the earth’s shadow and the radius of the moon, so we could add them together to get the angular distance between the ecliptic and lunar path and since we knew the angle between the two paths, we could solve that right triangle, approximating it as a planar one.
For the sun, we’re wanting to do the same thing, but there’s a complication: parallax. And to make that even worse, parallax can act both in the latitudinal and longitudinal directions. So we’ll tackle those separately, only exploring the latitudinal effect2 in this post
Since the moon is relatively close to earth, our position on the earth will make a notable difference on the apparent position of the moon in the sky. Indeed, if we look back at our parallax table we constructed in Book V.17, we can clearly see that the lunar parallax can be significantly larger than $1º$. Such parallaxes would mean that they can cause the apparent position to move more than enough that an eclipse doesn’t occur based on location.
And this makes perfect sense. You may recall total solar eclipses happening in your life, such as the August 21, 2017 eclipse, where the eclipse is only visible from a narrow corridor. This is because too far from the the center of the eclipse, the parallax pulls the moon off the sun. Let’s dig into that for a moment. Back in the last book, we determined the angular diameter of the moon at apogee on the epicycle to be $0;31,20º$. Then in this book, we worked out the angular diameter of the moon at perigee on the epicycle as $0;35,21º$, so a radius of $0;17,40º$. For the sun, you may recall Ptolemy has taken the rather odd position that the sun and moon have the same angular diameter when the moon is at apogee on the epicycle. One of the implications of this is that we can therefore use the angular diameter of the moon at apogee, $0;31,20º$ as the diameter of the sun as well or a radius of $0;15,40º$.
Thus, if we consider the moon to be at perigee, its radius is only $\approx 2$ arcminutes larger than the diameter of the sun. Thus, it hardly takes any parallax at all before it’s no longer a total eclipse! Again, if we look at the parallax table, for the first limit (column $3$), we see we hit that limit at around $2º$ in column $1$ which, as a reminder, was the distance of the object from the zenith when using this table for columns $2 – 6$. That translates to a $2º$ shift in latitude on earth. Hence why the eclipse path is so narrow. Take that up to $\approx 34º$ shift in latitude and now the parallax is sufficiently great that the moon would be entirely shifted off the sun and there is no apparent eclipse at all.
How does this play into what we’re going to be doing for solar eclipses? Let me draw a diagram similar to the one we looked at above for the moon:
Here, I’ve drawn in a situation in which the apparent positions of the sun and moon, $S_A$ and $L_A$ respectively, are such that they just barely graze one another. However, we can see that their true positions, $S_T$ for the sun and $L_T$ for the moon, are different. I’ve drawn those in as light gray dashed circles that you may need to enlarge to see, but in such a situation, there would be no eclipse. Thus, parallax can actually help solar eclipses happen by bringing the moon, which in this example would be too far south for an eclipse to occur, northwards and eastwards enough that it could happen. Without any parallax, look how much closer to the node the two would need to be:
While this image isn’t to scale in any manner, it does make it clear that the distance to which some sort of solar eclipse can occur can now happen notably further past where the limit would be were there no parallax.
So, how to deal with all of this?
Let’s sketch out another picture to help us.
This is largely the same picture as above, but instead of trying to draw in the node, I’ve truncated the ecliptic and lunar circle and redrawn them such that they’re parallel. It’s obviously not true, but over short distance such as these, it’s close enough.
This has the advantage of creating a handy little right triangle in the lower left that helps us break down the components of the parallax, $p$, into the longitudinal, $p_\lambda$, and latitudinal, $p_\beta$, components. The same thing is actually happening up by the sun too, but because the solar parallax is small, I didn’t bother drawing it in here. We will take it into consideration when we get to the math however.
As noted, we’re going to break this into two chunks. In the first one, we’ll just concentrate on the latitudinal component. This happens when the objects are on an observer’s meridian3. Redrawing the above diagram for that situation, it will look like this:
What we’re looking for here is the distance between the ecliptic and lunar circle, $arc \; S_T L_T$.
To help us, we know that $arc \; S_A L_A$ as its the sum of the radii or $0;17,40º + 0;15,40º = 0;33,20º$. To that, we can add the lunar parallax, $arc \; L_A L_T$ which gets us $arc \; S_A L_T$ and then we can subtract out the solar parallax, $arc \; S_A S_T$.
However, this begs the question: What are the lunar and solar parallaxes?
Putting that on hold for just a moment, let’s remind ourselves that the big question here is how far from the nodes can this situation occur? To answer that, we’re wanting to figure out how to maximize the total distances between the ecliptic and lunar circle for this diagram which happens, as we can see, when the parallax is largest. As noted before, the parallax is a function of distance from the zenith, so the obvious answer should be to consider the potential parallax at the north/south pole.
However, Ptolemy doesn’t do that. Instead he places other constraints on the latitude, telling us we will only consider
regions stretching from Meroe … up to the mouths of the Bortysthenes…
These are two of the regions that we’ve seen pop up back in Book II. There, Ptolemy discussed $39$ latitude zones, but he has chosen seven of them as the “seven climes” and this is the extremes of those. Thus, Ptolemy choses to limit himself to this region which has terrestrial latitudes between $16;27º$ and $48;32º$4.
Because of this, we won’t be considering the analogous situations beneath the equator. While you may be tempted to think limiting ourselves to a range of latitudes will make things easier, not being able to simply use the symmetry on the other side of the equator means we need to be more careful.
In addition, the image as I’ve drawn it here is specifically for the situation in which the moon is south of the sun. We’ll also need to consider how things work if the moon is north of the sun. As we’ll see, the argument is really the same and we’ll reduce this down quite handily, but as it stands right now, we’re needing to figure out how to maximize the parallax for two cases (the moon north/south of the sun) in two locations (Meroe and the Mouths of the Borysthenes).
Let’s begin by approaching this a bit conceptually so we don’t waste our time doing unnecessary calculations. First, let’s turn to our parallax table. There we can clearly see that the further the object is from an observer’s zenith, the larger the parallax. Thus, we can immediately understand that, to get the sun (and thus the moon) furthest from the zenith, we’re going to be needing the northern and southern extremes for the solar position. In other words, we’ll want to consider the summer and winter solstices.
Next, one of the things that gets left out of the parallax table and the discussion I had in the post surrounding it, is which direction the parallax pushes the apparent position of the object. Very simply, it is always away from the zenith. This means that there are some cases we can immediately ignore. For example, we don’t have to worry about cases in which the true moon is north of the true sun and their position is north of the zenith because that would push the apparent moon further from the sun, preventing an eclipse. Similarly, we can ignore cases in which the moon’s true position is south of the sun and they appear south of the zenith. Again, this would push cause the moon to appear even further away from the sun and prevent an eclipse.
This has the effect of completely discounting the case of needing to consider the position of the moon south of the ecliptic at the Mouths of the Borysthenes because its terrestrial latitude is $48;32º$ and its zenith will always fall $48;32º$ above the celestial equator. Since the sun never ventures more than $23;51º$ above the celestial equator, this means that the parallax at the Mouths of the Borysthenes will always cause the apparent position of the moon to be further south.
This directional effect of parallax also means that we need not worry about the opposite case, when the moon is north of the ecliptic for Meroe. This is because the maximum amount of parallax making the apparent sun appear more south happens when the sun is at the winter solstice. Since the amount of parallax is directly related to the angle between the object’s position and the zenith, we can immediately understand that the parallax will be larger at the Mouths of the Borysthenes simply because its zenith will be further from the winter solstice than at Meroe.
Lastly, we need not worry ourselves with the terrestrial latitudes between the two locations given as they will fall between the extremes we’ve already set. We can simply find the maximums for the distance between the ecliptic and celestial equator for north and south at the limits of the two terrestrial latitudes and everything else will fall between them.
Thus, thinking through things, we have taken the myriad of possible cases we’d need to check and distilled them down to two:
1) When the sun is at the summer solstice and the moon is south of the ecliptic from Meroe and
2) When the sun is at the winter solstice and the moon is north of the ecliptic from the Mouths of the Borysthenes.
Finally, we’re ready to start doing some math, beginning with the first case. And almost immediately I run into an issue. In particular, following Neugebauer’s walkthrough of Pappus, I believe there’s some fudging going on. Specifically, to make use of the parallax table for the moon we need to know the moon’s distance from the zenith, which we don’t know. We do know the position for the sun. It’s the difference between the distance of the sun from the celestial equator5 and the distance of zenith from the celestial equator. In other words:
$$23;51º – 16;27º = 7;24º.$$
Popping that into the table and finding the sun’s parallax I come up with $0;00,23º$ after interpolating between rows. If we go ahead and use that same value for the moon in columns $3$ and $4$ as discussed above, we get
$$0;07,00º + 0;1,22º = 0;08,22º$$
Again, as we saw above, we need to add the distance between their apparent centers and subtract out the solar parallax so6,
$$0;08,22º + 0;33,20 – 0;00,23º = 0;41,19º.$$
This agrees, within rounding, of the value Ptolemy gives of $0;41º$. Thus, I suspect Ptolemy, and thus Pappus, were doing a bit of a cheat here using the solar distance from the zenith instead of the lunar one.
What is the impact?
Let’s use this result as a first approximation. The moon’s distance south of the ecliptic if there were no parallax and this grazing eclipse would be $0;33,20º$. If we add on that first approximation, it could be as much as $0;41,20º$ in which case its distance from the zenith would have been
$$7;20,00º – 0;41,20º = 6;38,40º$$
in which case the lunar parallax would be
$$0;06,17º + 0;01,22º = 0;07,39º.$$
Again, we would subtract out the solar parallax to get a distance between the ecliptic and lunar circle of $0;07,16º$7. Thus, Ptolemy’s estimation seems a bit high, but by less than a single arcminute. Not measurable in Ptolemy’s time. Thus, this method seems reasonable enough.
We can now do the same for the second case in which the sun is at the winter solstice and the moon is northwards of the ecliptic as viewed from the Mouths of the Borysthenes. There, the distance of the sun from the zenith will be
$$23;51º + 48;32º = 72;23º.$$
The solar parallax is then $0;02,42º$ and the lunar8 is
$$0;51,17º + 0;09,54º = 1;01,17º.$$
Adding the distances between the apparent centers and subtracting out the solar parallax we get the distance between the ecliptic and lunar path to be, $1;31,49º$9.
Again, we can attempt to estimate the effect of the use of the sun’s zenith distance instead of the moon. There, we would instead take the moon further north of the sun (i.e., towards the zenith) by $1;34,37º$ (the sum of their diameters plus our first approximation for parallax) in which case the distance from the zenith is $70;58,23º$ in which case I get a lunar parallax of
$$0;50,53º + 0;09,49º = 1;00,42º.$$
Again, Ptolemy’s value seems to be high, this time by about half an arcminute.
So now that we know the distance between the ecliptic and lunar circle for these limits, we can then estimate the distance to the nodes as we did for the lunar limits – by approximating the triangle as a planar right triangle where the angle at the node is $5º$. However, instead of approaching this calculation head on with his usual demi-degrees method, Ptolemy instead takes a more round about approach stating,
the ratio between the arc from the node to the [true moon] and the arc [between the ecliptic and lunar circle] is about $11 \frac{1}{2} : 1$ for distance between the eclipse limits as can easily be seen from our previous demonstration of the inclination of the lunar orbit.
The reference to the “previous demonstration of the inclination of the lunar orbit” is referring to where he came up with the $5º$ inclination of the lunar orbit in V.12 and was covered in this post.
Here, what Ptolemy is trying to do is save us a bit of work. Instead of going through the demi degrees method each time he does it once using $1º$ as the arc we’ve just calculated between the ecliptic and lunar circle. In that case, you would get a distance to the node from the true lunar position of $11;28,25º$ which Ptolemy evidently rounds off to $11;30º$. Thus, the ratio of the distance between the distance between the true moon and the node to the the ecliptic and the true moon in its circle is $11;30 : 1$.
That’s a bit of a mouthful, so here’s a picture that may make it clearer:
What Ptolemy is saying is that the ratio $\frac{n}{s} = \frac{11;30}{1}$. We obviously can’t extend this out too far since these are great circles and not linear, but for the scales we’re talking about, Ptolemy deems it good enough.
Since our triangle would be a similar triangle, we can simply apply this ratio. In other words, we can take the results of our two scenarios and multiply by $11;30$ to get the distance to the nodes.
So, for the first case, in which the moon was south of the ecliptic at the summer solstice at Meroe, we get a distance between the true moon and the node10 of
$$11;30 \cdot 0;41,19º = 7;55,09º.$$
In the second case, in which the moon was north of the ecliptic at the winter solstice as viewed from the Mouths of the Borysthenes, we get a distance between the true moon and the node11 of
$$11;30 \cdot 1;31,49º = 17;35,54º.$$
Neugebauer notes that Ptolemy’s “arbitrary way of rounding made the [values of the distance to the nodes too small by $0;01,29º$ and $0;08,25º$ respectively.” But overall, not a horrible result.
However, we still need to consider the longitudinal effects of parallax which we’ll tackle this in the next post!
- As mentioned in my last post, I’m doing things slightly out of order here. In this book, Ptolemy first derived some needed parameters for the moon, then sun, then worked out the eclipse limits for the moon and sun. Here in the blog, I’ve paired the sections on the moon, skipping over the sun and coming back to it now as I think it makes things flow a bit more nicely. However, if you’re trying to follow along in the Almagest, you may have felt a bit lost.
- Specifically, I mean the impact to ecliptic latitude.
- If you’ve forgotten, the fancy word for this is “culmination.”
- Presumably because he later calls this range the “inhabited world” and thought there wasn’t any need to calculate other latitudes since he wasn’t aware that anyone actually lived there.
- This is known as the declination, but since we don’t tend to make much use of the RA/Dec system in this blog, I’m trying to avoid using the terminology associated with it.
- Ptolemy evidently does this in two steps, first subtracting out the solar parallax to get $0;07,59º$ which he rounds off to an even $0;08º$ and (incorrectly) refers to as the latitudinal parallax.
- During my frustration trying to understand the Neugebauer/Pappus method presented here, I ended up working out an entirely different approach to this problem. Using that method, I come with a total lunar parallax of $0;07,45º$. As this post is already getting quite long, I’ll save that for another post.
- Again, using the sun’s distance from the zenith.
- Again, if we first subtract the solar parallax, we get a value of $0;58,30º$ which Ptolemy calls $0;58º$.
- Ptolemy comes up with a value of $7;52º$ due to rounding to him rounding the distance between the great circles to $0;41º$.
- Ptolemy comes up with a value of $17;26º$ due to rounding in previous steps leading him to use a value of $1;31º$ instead of $1;31,49º$ as I’m using here.