In our last post, we explored how to find the angular distance around the horizon from the ecliptic and celestial equator. In this chapter, we explore another value that can be derived from knowing the length of the longest day (i.e. on the summer solstice): the elevation (or altitude) of the celestial pole (which is also the latitude).
Once again, we’ll start with the same diagram we’ve been using for awhile now:
Again I’ve highlighted a specific portion of the figure to help make it more apparent that this is a Menelaus configuration so we’ll be using Menelaus’ Theorems. Last time we used Menelaus’ Theorem I, so this time we’ll be using Menelaus’ Theorem II. This allows us to state:
$$\frac{Crd \; arc \; 2E \Theta}{Crd \; arc \; 2A \Theta} = \frac{Crd \; arc 2EH}{Crd \; arc 2BH} \cdot \frac{Crd \; arc \; 2BZ}{Crd \; arc \; 2AZ}$$
Quickly, let’s remind ourselves that BED is our horizon and AEG is the celestial equator and we’re wanting to know the angle of the celestial pole above the horizon. Ptolemy pulls a bit of a fast one here because in the last post we specifically noted that Z is the south celestial pole which, for an observer above the tropic of cancer as the setup for this problem is assumed to be, it not visible. However, the angle of one pole below the horizon is necessarily the same as the other above the horizon.
But looking at our diagram we see there are two lines that connect Z to the horizon (BED): $arc \; HZ$ and $arc \; BZ$.
So which one are we looking for? Will they be the same?
What we’re really looking for is the altitude above the horizon and that’s defined as the shortest great circle arc to the horizon. That may make you think that we should solve for both, and see which one’s shorter, but that excludes the possibility that some other arc that we may not have considered is even shorter. So we need to think a bit deeper.
In much of this discussion, we’ve been working these problems as if we could stand outside the celestial sphere. But let’s put ourselves back in the view of a stargazer. We stand with the sky a dome above us. For any star, the shortest distance to the horizon will always be straight down from the star. This doesn’t particularly help us for any general point, but since the point we’re concerned with in this exercise is a celestial pole, that means it will be either due north or due south.
The line that runs across the celestial sphere connecting the north and south poles has a special name: the meridian. So is either of the arcs we’re questioning lie on the ecliptic?
Yes. $Arc \; BZ$ does as we defined it in our last post. So that’s the one we’re looking for. And, lucky us, BZ is one of the components in the form of Menelaus’ theorem we chose.
So now we get to play the game of determining all the other variables. Fortunately, some of them we got in our last post. So instead of discussing them again here, I’ll simply refer the interested reader to refer there.
We found that $arc \; E \Theta = 18;45$º so $arc \; 2E \Theta = 37;30$º and $Crd \; arc 2E \Theta = 38;34,22$.
In addition we found that $arc \; 2A \Theta = 142;30$º so $Crd \; arc \; 2A \Theta = 113;37,54$.
The end goal of the entire last post was $arc \; EH$ which we found to be 30º, so $arc \; 2EH = 60$º and $Crd \; arc \; 2EH = 60$.
We also found $arc \; BH$ along the way to be 60º, so $arc \; 2BH = 120$º and $Crd \; arc \; 2BH = 103;55,23$.
The only piece of the equation we need that we didn’t get in the last post was $arc \; AZ$. However, this one is straightforward since A is on the celestial equator and Z is a pole, which means the arc between them, $arc \; AZ = 90$º. So $arc \; 2AZ = 180$º and thus $Crd \; arc \; 2AZ = 120$.
Time to plug things in.
$$\frac{38;34,22}{113;37,54} = \frac{60}{103;55,23} \cdot \frac{Crd \; arc \; 2BZ}{120}$$
We can reduce this slightly before moving on:
$$\frac{38;34,22}{113;37,54} = \frac{1}{2(103;55,23)} \cdot Crd \; arc \; 2BZ$$
Moving the denominator to the other side:
$$Crd \; arc \; 2BZ = \frac{2(103;55,23) \cdot 38;34,22}{113;37,54} $$
$$Crd \; arc \; 2BZ \approx 70;33$$
Doing a reverse lookup from our chord tables informs us:
$$arc \; 2BZ = 72;1º$$
Which means:
$$arc \; BZ \approx 36º$$
If you look back at what we stated in the setup of the last problem, Ptolemy stated that we were taking our example as the Greek city of Rhodes and did indeed give this value.