In the last post we followed along as Ptolemy discussed the construction and use of his parallactic instrument, which he would use to measure the lunar parallax. To do so, Ptolemy waited for the moon to
be located on the meridian, and near the solstices on the ecliptic, since at such situations, the great circle through the poles of the horizon and the center of the moon very nearly coincides with the great circle through the poles of the ecliptic, along which the moon’s latitude is taken.
That’s pretty dense, so let’s break it down with some pictures, First, let’s draw exactly what Ptolemy has described above:
It’s been awhile since we’ve looked at a drawing of the celestial sphere, so let’s review. First off, around the center, I’ve drawn the celestial equator which is just the extension of Earth’s equator. Straight up from that is the north celestial pole. From our point of view, that’s the point around which everything appears to rotate.
Second, we have the ecliptic, which is the path of the sun. Here, I’ve drawn it with the winter solstice to the left and summer solstice on the right. This too has a pole which I’ve labeled as the north ecliptic pole. As we just stated, everything appears to rotate around the north celestial pole including the point of the north ecliptic pole, so I’ve drawn its path in which will help us1 shortly.
Lastly, we have the horizon. I’ve drawn this in for a latitude of something around $48º$ which we can tell because the north celestial pole is that angular distance above the northern point on the horizon. I didn’t choose this for any fancy reason but simply because it separated the planes nicely. I’ve also labeled the cardinal points along the horizon.
Now let’s consider the meridian. By definition, this is the line across the sky from north on the observer’s horizon, through the zenith, to south. I’ll highlight that in red:
As Ptolemy proscribed, I’ve placed the moon on the ecliptic and as we know about the moon, it sticks fairly close to the ecliptic, bouncing above and below by about $5º$. Here, I’ve drawn it slightly above the ecliptic so it didn’t overlap with the summer solstice.
Now let’s dig into the great circles to which Ptolemy was referring. The first was “the great circle through the poles of the horizon and the center of the moon”. What’s being described here is simply the meridian we just looked at since technically the meridian is a great circle and continues under the ground through the nadir.
The other great circle he refers to is “the great circle through the poles of the ecliptic, along which the moon’s latitude is taken.” Let me draw that one in blue:
As you can see, it’s the same circle in this setup. That means that, in this position, we’re easily able to discern any differences to the moon’s position that might be caused by our position. To get a better understanding of what goes wrong when the moon is not near a solstice, let’s redraw our this last diagram with the moon near one of the equinoxes. I’ll leave the horizon in the same position:
Here, you can clearly see that the great circle through the ecliptic pole is no longer aligned with the great circle through the meridian. It’s hard to tell what effect that is going to have, so let’s rotate this picture around a bit so the moon is towards us:
Now you can see the impact pretty clearly: The misalignment of the two great circles means there’s a little spherical triangle right beneath the moon with one side being the meridian, one the great circle through the ecliptic poles, and the third being the ecliptic. Recall that the parallactic instrument measures along the meridian (the red circle). And that when we determine the latitude of the moon, we’re measuring angle above/below the ecliptic (the blue circle). If they’re aligned, then when we measure one, we’re measuring directly relative to the other. When they’re misaligned, so too are the angular measures and they can no longer be compared.
So we can see why measuring the moon at the solstices is important. In addition, Ptolemy states that we should also take the measurement when the moon is
near the northern limit of its inclined circle. For in the region of those points, the moon’s latitude remains sensibly the same over a considerable interval.
This is quite easy to understand if you think in a bit of calculus terms. The moon bounces up and down above the ecliptic on its inclined path. So its angular distance above or below the ecliptic essentially forms a $sin$ wave. If you consider the derivative of it, which is the rate of change, it is zero at the peaks and troughs, which means that it moves the least in latitude around those extrema. This means that misjudging the date of the solstice slightly, or observing a day later due to poor weather would not significantly impact the observation.
So what did the observations end up being?
[T]he distance of the center of the moon from the zenith was always about $2 \frac{1}{8}º$. Hence, by this method too, the moon’s greatest latitude either side of the ecliptic was is shown to be $5º$. For the zenith distance of the equator at Alexandria has been shown to be $30;58º$; if we subtract this from the $2 \frac{1}{8}º$, the result [$28;50 \frac{1}{2}º$] is about $5º$ greater than the distance from the equator to the summer solstice, which was shown to be $23;51º$
Again, let’s turn the to our picture where I’ve drawn out the angles in question:
Here, we know the angle of the zenith above celestial equator is always the latitude of the observer2 so $30;58º$ for Alexandria. Second, we know the distance between the celestial equator and the ecliptic to be $23;51º$. Lastly, Ptolemy observed the distance from the zenith to be $2 \frac{1}{8}º$ ($2;7,30º$). So looking at the picture, we can clearly see the little gap is the angle of the moon above the ecliptic.
So to calculate that we take:
$$30;58º – 23;51º – 2;7,30º = 4;59,30º$$
which is right about $5º$ as Ptolemy stated. This can then be compared against the calculated angle of the moon from the ecliptic to determine if there is an effect due to parallax, and if so, how much.
Ptolemy did this several times and in the next chapter, Ptolemy will walk us through an example.
- Or at least help me with my drawing.
- I’m pretty sure I’ve proven this somewhere in the blog previously but I’m not finding it now. However, it’s easy enough to convince yourself. In this picture, imagine the north point of the horizon on the north celestial pole (which would be the case for an observer on the earth’s equator). The zenith would then be on the celestial equator. As you rotate the horizon to account for latitude, the zenith moves as well always maintaining the same angle from the celestial equator as the north point on the horizon from the north celestial pole.