In the last post, we looked at how Menelaus’ theorem could be used to tell us which point on the ecliptic would culminate with a star. In this post, we’ll do something similar, but instead look at
the points on the equator and ecliptic which rise or set simultaneously with a fixed star.
To do so, Ptolemy stars with the following diagram:
In this diagram, circle $ABGD$ is now the meridian and circle $AEG$ the celestial equator with $Z$ as one of its poles.
We’ll let circle $BED$ be the horizon with a star on it at point $H$.
We’ll then draw a great circle through the star and the celestial poles intersecting the celestial equator at point $\Theta$.
Lastly, we’ll draw in point $K$ which is equidistant as $E$ from $\Theta$.
Again, what we’re wanting to understand here is what point on the celestial equator is on the horizon at the same time as $H$, which would be $E$.
As in the previous post, we’ll use a Menelaus configuration which I’ve already highlighted:
$$\frac{Crd \; arc \; 2ZB}{Crd \; arc \; 2BA} = \frac{Crd \; arc \; 2ZH}{Crd \; arc \; 2H \Theta} \cdot \frac{Crd \; arc \; 2 \Theta E}{Crd \; arc \; 2AE}$$
Let’s go through the pieces:
$Arc \; ZB$ can be found by subtracting $90º – arc \; AB$ since $arc \; AZ$ is $90º$ by definition (since it’s a great circle to its pole).
But what is $arc \; AB$? It’s the latitude of the observer or, as Ptolemy puts it, the “elevation of the pole”1.
For $arc \; ZH$, this is again, $90º – arc \; \Theta H$.
Where does $arc \; \Theta H$ come from? Ptolemy tells us that we can get this from the calculations we did in the previous post in which we looked at the simultaneous culmination. While $H$ won’t be on the horizon in that case, it does maintain its position relative to the celestial equator, and thus its value will be unchanged between these two scenarios.
Lastly, we have $arc \; AE$ which is $90º$ since it’s between a point on the horizon and a point on the meridian.
Thus, we have everything we need to solve for the missing piece, $arc \; \Theta E$, recalling that $\Theta$, in this diagram, is the Right Ascension of the star in question. Since that is a known quantity2, we could then figure out where $E$ lies along the celestial equator.
You may notice that we never actually used point $K$ in this situation. Well, that one comes in to play if we were to consider the simultaneous setting. As Ptolemy puts it,
For the simultaneous settings, too, it can easily be seen that if we cut off an arc, $\Theta K$, in advance of $\Theta$ equal to $arc \; Theta E$, the star will set together with point $K$ of the equator. For in that situation the setting takes place on an arc [of the horizon measured from the meridian] equal to $arc \; BH$, and cuts off an angle in advance of the meridian equal to that enclosed to the rear [of it] by $arc \; AZ$ and $arc \; Z \Theta$ in the present situation.
In other words, the rising point on the horizon will trail the Right Ascension of the star where as the setting point will precede it in a symmetrical manner. This is quite easy to see if I just let the sphere rotate and carry the star around to the other horizon:
Here, we can clearly see that $K$ is the point on the celestial equator that hits the horizon simultaneously with the star.
But what about the point of the ecliptic that rises and sets with the star?
For that, Ptolemy tells us that
from the arcs of the equator and ecliptic which rise and set together, which we have computed for each clima [II.8], there will immediately be given that point on the ecliptic which rises together with point $E$ of the equator and the star, and that point which sets together with point $K$ and the star.
In other words, once we know the arc of the celestial equator, we can use the rising time table to determine the related section of the ecliptic.
Lastly, Ptolemy states,
It is clear that, at the moment when the sun is exactly in those points of the ecliptic, there will come to pass the risings, culminations, and settings of the fixed star [in question] taken with respect to the sun’s centre which are called ‘true simultaneous cardinal positions.’
This is really more of a definition than anything else, but is how Ptolemy ends Chapter $5$.
In the next post, we’ll follow along as Ptolemy explores the first and list visibilities of a given star.
- This is fairly easy to see if you consider the opposite side between $D$ and the unmarked north celestial pole. This is clearly the elevation of the pole, as Ptolemy describes, and $arc \; AB$ is merely its reflection across the sphere.
- I.e., it could be converted to from the equatorial coordinates which we certainly have.