The fourth chapter in book two is a very short one. In fact, it’s a single paragraph so I almost didn’t dedicate an entire post to it but ultimately decided to as it didn’t really fit with either the previous or next chapter.
In this, Ptolemy says,
it is a straighforward computation to determine for what regions, when, and how often the sun reaches the zenith. For it is immediately obvious that for those beneath a parallel which is farther away from the equator than the 23;51,20º (approximately), which represents the distance of the summer solstice, the sun never reaches the zenith at all, while for those beneath the parallel which is exactly that distance, it reaches the zenith once a year, precisely at the summer solstice.
While Ptolemy hints at some sort of calculation, I don’t find that any such calculation is necessary. It immediately follows from an understanding of the celestial sphere.
Here we have the celestial sphere drawn about as simply as we can get it. Around the center is the celestial equator, and tilted with respect to it is the ecliptic. Towards the top, I’ve drawn in a circle parallel to the equator that represents the highest one that can still intersect with the ecliptic, which it does, at only one point, to the right. Because the angle of the ecliptic is 23.5º1, so too must the circle be elevated that much from the equator. Thus defining the Tropic of Cancer (and conversely the Tropic of Capricorn).
However, if we draw the circle lower, we can clearly see there are two points of intersection:
Thus, points between the tropics will see the sun reach the zenith twice a year, once at the ascending node, and once at the descending, which Ptolemy notes as well stating,
It is furthermore clear that for those beneath a parallel less far than the above-mentioned amount the sun reaches the zenith twice.
But he continues:
The time when this happens is readily supplied from the Table of Inclinations … For we take the distance from the equator, in degrees, of the parallel in question (which must, obviously, lie within the [parallel of the] summer solstice), and enter with it the second set of columns; we take the corresponding argument, in degrees from 1º to 90º, in the first set of columns; this gives us the distance of the sun from each of the equinoxes towards the summer solstice when it is in the zenith for those beneath the parallel in question.
Upon first reading, that’s a lot to break down. So let’s try an example: The easiest one in my estimation is the first situation we considered here where the parallel intersects with the ecliptic only at one point: On the solstice. There, we know what the answer is so we can check to make sure we applied the method correctly.
Since Ptolemy uses a value of 23;51,20º, we’ll adopt that. Looking that up in the Table of Inclinations, we see that the corresponding arc of the ecliptic is 90º. That’s 1/4th of 360º, which is correspondingly 1/4th of the year after the equinox. That’s the solstice, so we’ve applied it correctly.
But what of another latitude. Let’s take 15º. That doesn’t happen to quite fall in the column of degrees, so let’s adopt the closest one: 15;4,4º. We see the corresponding arc is 40º. This would imply that the sun crosses the zenith $\frac{40}{360} = \frac{1}{9}$ of 365.25 which is about 40.5 days past the equinox as well as before the next one. Since the spring equinox is around March 21, this would be around April 30. Since the autumnal equinox is around September 21, the other date would be around August 12.
We can check this by popping into Stellarium:
Here, I’ve put in the location as 15ºN Latitude, 0º Longitude, set the date to April 30, and fiddled with the time to get the sun as close to local noon as possible. While the zenith marker is turned on, it’s washed out by the sun, but we can see from the info on the left, the sun is at 89;50,36.6º which is quite close to the zenith. Likely a bit of error is introduced as Stellarium doesn’t have sub second resolution and Ptolemy was using slightly incorrect values for the angle between the ecliptic and equator.
But let’s also check the other date:
Again, we can see that the sun is indeed very close to the predicted position.
It should be noted that this method also works for positions south of the equator, but a small adjustment is necessary: Instead of going towards the summer solstice from each equinox, one should move towards the winter solstice.