Now that we’ve determined how much further from the nodes parallax can cause solar eclipses to occur due to the latitudinal parallax, we need to consider the longitudinal effect. As with the last post, Ptolemy is absolutely no help in this. He simply tosses out some values with no explanation or work stating
When [the latitudinal] parallax is $0;08º$ northwards1, [the moon] has a maximum longitudinal parallax of about $0;30º$ … and when its [latitudinal] parallax is $0;58º$ southwards2, it has a maximum longitudinal parallax of about $0;15º$…
Seeking some assistance, I again refer to Neugebauer and Pappus, but immediately run into an issue. Neugebauer minces no words and states
Ptolemy is wrong in stating that $p_\lambda = 0;30º$ and $p_\lambda = 0;15º$ are the greatest longitudinal components of the parallax for locations between Meroe and the Borysthenes. It is difficult to explain how he arrived at this result.
Well… this will be interesting to try to untangle then.
Neugebauer begins by explaining that Ptolemy
knew an important theorem in the theory of parallaxes: for the point on (or very near to ) the ecliptic, the latitudinal component $p_\beta$ is constant if only the zenith distance of the highest point of the ecliptic remains fixed3.
It took me several tries to understand what Neugebauer was getting at here, so let me break it down. In the last post, we determined the maximum latitudinal parallax, $p_\beta$, for the two terrestrial latitude extremes Ptolemy considers. To do that, we simplified our situation, putting the sun and moon on the observer’s meridian. However, to have a component of longitudinal parallax, $p_\lambda$, like we’re after in this post, we’re going to have to move our setup off the observer’s meridian. This may result in a change to the latitudinal parallax we found in the last post, screwing up the work we just did.
What Neugebauer is saying here is that if the summer solstice (i.e., the “highest point of the ecliptic”) is on the meridian, it doesn’t end up changing the latitudinal parallax, even if the sun/moon are elsewhere along their circles. This also works if the summer solstice is on the part of the meridian beneath the horizon (since the meridian is really a great circle) which would mean that the winter solstice is on the visible meridian.
So then, when the first point in Cancer (i.e., the summer solstice) is on the meridian, what point along the ecliptic can we choose to maximize the parallax?
We can recall that the further away from the zenith the greater the parallax so… where the ecliptic intersects the horizon, or $90º$ from the zenith. It doesn’t even matter where along the ecliptic it is; the parallax will always be maximized because that’s as far as you can get from the zenith.
Neugebauer agrees stating,
Ptolemy should have taken the position [along the ecliptic] near sunrise (or sunset).
“Should have”. Indicating he didn’t. So what did he do?
Again, it’s hard to tell because he didn’t show any work, but Pappus may provide some evidence. Neugebauer notes that his work showed
that Pappus computes the components of parallax for [the sun at the first point in Leo] $2$ hours before (or after) noon at Meroe. Since these points have a distance of $\pm 30º$ from [the first point in Cancer], it is clear that $2$ hours before (or after noon, [the first point in Cancer] must be near culmination. Pappus indeed finds that $2;04º$ into cancer is culminating, which is certainly close enough for the application of the above theorem.
Stating this more simply, Pappus used another method to compute both the latitudinal and longitudinal components of parallax for times when he expected that the summer solstice would be on the meridian. This could then be compared to the we value calculated in the last post to double check that it was reasonably close and not pushing the sun further from an eclipse. Pappus also double checked exactly what point would be culminating and found while it wasn’t exactly the summer solstice it was reasonably close.
So, how to calculate both the components of parallax? Neugebauer indicated that he did so by making use of our Table of Zenith Distances and Ecliptic Angles we computed back at the end of Book II. As a reminder, this set of tables gives the zenith distance of the first point of each zodiac constellation for various times relative to that point crossing the meridian as well as the angle between the altitude circle through that point and the ecliptic.
First, when the first point of Leo is $2$ hours before the meridian at Meroe4 we can see that the arc between the zenith and the point in question is $28;42º$.
Now, let’s consider the angles involved specifically, the east angle since we’re considering the angle before it crosses the meridian. First, let’s consider what that would look like from inside the celestial sphere5:
Here, I’ve highlighted the angle in question. It’s the one that is rearwards (i.e., east) and north of the ecliptic. From the table, we can find it should be $15;28º$. We’ll use this value shortly. That’s everything we need from that table.
Now we’ll turn to our parallax table and plug in the distance of this point from the zenith we just found of $28;42º$. Interpolating between rows, I come up with a lunar parallax of $0;31,09º$6.
Now, let’s pause so I can draw things out a bit more schematically:
Here, I’ve sketched out the same configuration as the image above, but added the effect of the parallax. We are considering the true moon at $L_T$7.
Since parallax always works along the altitude circle8 and away from the zenith, at $Z$, this means the apparent lunar position would be at $L_A$.
Thus the length of $\overline{L_T L_A} = 0;31,09º$ which is the total parallax, $p$. Now what we need to do is break it down into the longitudinal parallax, $p_\lambda$, latitudinal parallax, $p_\beta$. Fortunately, this little triangle, $\triangle{p p_\beta p_\lambda}$ is a right triangle since $p_\lambda$ is perpendicular to $p_\beta$ by definition. In addition, we know the angle in that triangle at $L_T$ as it’s the east angle we found above to be $15;28º$.
Now, this isn’t a planar triangle, but we’ll approximate it as one to solve9. In that case, I get10 that,
$$p_\beta = 0;08,18º$$
$$p_\lambda = 0;30,01º$$
We can compare the the lunar latitudinal parallax to the value we found in the last post of $0;08,22º$ which confirms that we haven’t impacted it significantly11. Thus, we’re free to use the longitudinal parallax as needed. So what exactly are we trying to do with it?
It’s actually quite easy. In the last post, we determined how far from the nodes an eclipse could occur due to latitudinal parallax. All the longitudinal parallax does is push that further out. And since the longitudinal parallax acts in the direction along the ecliptic (which we treated as parallel to the ecliptic over these short distances12), we simply need to add it on.
Thus, all we need to do is add this on to the distance from the nodes we calculated in the last post. As a reminder, we did this for Meroe so far, so we need to be careful about which case we’re applying this to. For Meroe, we found that the true moon could be $7;55,09º$ from the node. So adding on another $0;30,01º$ we get that the moon could be up to $8;25,10º$ which is in rough agreement with Ptolemy’s value of $8;22º$13.
So where did Ptolemy go wrong again?
His issue was that choosing the situation in which the sun and moon are at the first point in Leo and the summer solstice is culminating isn’t the largest longitudinal parallax. That’s only a point $2$ hours from the meridian. As noted above, we should be choosing a point of the ecliptic on the horizon. Neugebauer notes this is especially true
in the case of Meroe, where the ecliptic stands almost vertical to the horizon.
That means that we can have points on the ecliptic that are $6$ hours from the meridian, which would be around the first point of Libra if the summer solstice was still on the meridian. Thus, looking at the Table of Zenith Distances we see that it would have a zenith distance of $75;39º$ which translates into a total lunar parallax of $1;02,10º$ with an east angle of $7;09º$.
Breaking that down into components:
$$p_\beta = 0;07,44º$$
$$p_\lambda = 1;12,28º$$
Here, we’re starting to drift a ways from $p_\beta$ being $0;08,22$ from the last post, but it’s only a little over half an arcminute. Still a relatively small amount. Regardless, we can see that the longitudinal parallax can be significantly higher than the previous case which Ptolemy said was the maximum. By $\frac{3}{4}$ of a degree!
Neugebauer doesn’t explain any reason that Pappus or Ptolemy erroneously chose the first case we showed above, and I can’t come up with one either. And a similar error is made for the Mouths of the Borysthenes.
For that one, recall that we had the case in which the winter solstice was on the meridian. Thus, we should consider the first point of constellations around the winter solstice: Aquarius, Pisces, and Aries. For no reason, Ptolemy and Pappus evidently chose the first point in Pisces, when it’s $4$ hours from the meridian. Let’s again take a look at how that looks from inside the celestial sphere via Stellarium:
Again, I’ve drawn in the east angle. So now let’s redraw that to include the parallax:
Here again, we have the position of the moon being moved from $L_T$ to $L_A$ along the altitude circle which we can break into components of ecliptic longitude and latitude. So let’s get started looking things up in our tables.
First, from the Table of Zenith Distances and Angles14 we find the distance from the zenith is $80;03º$ and the east angle is $104;28º$.
We then turn to the Parallax Table where I then get a total lunar parallax of $1;02,44º$ after interpolation.
But before we can simply start doing trig, we need to think a bit more about the angle as the east angle is not in the triangle this time. Rather, it’s the supplement along $\overline{Z L_T}$. Thus, the angle within the triangle at $L_T = 75;32º$.
Now we can go ahead and break that triangle into its components:
$$p_\beta = 1;00,45º$$
$$p_\lambda = 0;15,41º$$
Again, this shows pretty reasonable agreement with the previous post for $p_\beta$ where we found that the latitudinal component of the parallax was $1;01,17º$ – Off by about half an arcminute. This time, we come up with a longitudinal parallax of $0;15,41º$ which gets added to the distance from the node we calculated previously of $17;35,54º$ for a total distance from the node of $17;51,35º$.
This is higher than Ptolemy’s value of $17;41º$ by a notable amount. This is due to how Ptolemy calculated the distance from the node that we covered in the previous post where he rounded the distance between the ecliptic and lunar circle down from $1;31;49º$ to an even $1;31º$.
However, we could still get further from the zenith, thereby increasing the longitudinal parallax, if we chose a point nearer to the horizon, which would be nearer to the first point in Aries, but the impact is not nearly as large as it was for the previous case – only a few arcminutes.
Thus, we’ve now shown how to include the longitudinal parallax effects when determining the eclipse limits. So we’re almost there, but there’s one final piece to consider, which I’ll cover in the next post: the effect of the lunar and solar anomalies.
- I.e., the first case we considered in the last post, calculated for the sun at the summer solstice when the moon is south of the ecliptic as viewed from Meroe.
- I.e., the second case we considered in the last post, calculated for the sun at the winter solstice and the moon north of the ecliptic from the Mouths of the Borysthenes.
- He refers to p122 of another of his works for a proof.
- In case you missed it, the various latitudes are tabs along the bottom of the Google Sheet.
- It is incredibly helpful to take a look at this in Stellarium. In an earlier version of this post, I’d drawn the diagrams entirely wrong!
- Recalling, of course, that we must add the interpolated values for columns $3$ and $4$.
- Which was really the position of the true sun, but as we showed in the last post, we can consider it at the same distance from the zenith as the sun without too much issue
- It’s been awhile since we used this term, so as a reminder of what it means, it’s the great circle that goes through the object or point in question such that it makes the shortest distance to the horizon. In other words, it’s part of the great circle along which you’d measure the altitude of the object in your night sky.
- I used trig here for convenience.
- If you’re comparing to Neugebauer, you’ll notice that he comes up with notably different values here. I cannot say for certain since he didn’t show much of his work either, but I suspect that he subtracted out the solar parallax before solving for the components. When I do that, I still don’t get quite the same value, but I get much closer. This is actually something that he did in the steps I covered for the last post as well, and called the result the “net parallax”. I didn’t like how he rolled that together, so I kept it separate.
- As another quick sanity check here, we should also confirm that the effect of the parallax is northwards as predicted. Indeed, we can easily see from the diagram that the position of $L_A$ is north of the ecliptic since north is to the left as drawn above (which can clearly be seen by referring to the screenshot from Stellarium).
- That really aren’t short.
- In the last post I had a footnote that indicated Ptolemy did some arbitrary rounding that caused him to underestimate. So this result lines up well.
- Remember to switch to the tab on the Table of Zenith Distances for Borysthenes. I almost forgot and wondered why my numbers didn’t make sense!