Not content to simply figure out how long it would take a zodiacal constellation to rise at latitudes other than the equator, Ptolemy sets out to further divide the ecliptic into 10º arcs and he’s promised an easier method than what we’ve done previously. But before we can get there, Ptolemy gives a brief proof which he’ll make use of later.
To start, we begin with the vernal equinox on the horizon:
Here, BED is the horizon for Rhodes, AEG is the celestial equator, ZEH the ecliptic, their intersection E which is on the horizon, is the vernal equinox. L is the south celestial pole.
First, Ptolemy selects an arbitrary point, Θ, somewhere along $arc \; EZ$ and then connects that point to the horizon, parallel to the celestial equator at a point he calls K.
Next, we’ll draw some great circle arcs through these points, beginning at L, and ending on the celestial equator. We’ll also draw one to E.
This also creates two new points on the celestial equator: M and N.
Then it is immediately obvious that the segment EΘ of the ecliptic rises with $arc \; EM$ of the equator at sphaera recta, and with NM at sphaera obliqua, since $arc \; K \Theta$ of the parallel circle, with which segment EΘ rises [at sphaera obliqua], is similar to $arc \; NM$ of the equator and similar arcs of parallel circles rise in equal times everywhere.
Let’s break that down into some smaller chunks. First off, $arc \; E \Theta$ rises with $arc \; EM$ at sphaera recta. As a reminder, sphaera recta is when the observer is on the Earth’s equator. At that point, the horizon would not be BED, but could be any great circle which connects the north and south celestial poles. We have several arcs that are part of such circles. $Arc \; LM$ is one of them. And since both M and Θ fall on it, they would necessarily be on the horizon at the same time, rising until point E, a single point, hits the horizon. Since the arcs Ptolemy described ($arc \; EM$ and $arc \; E \Theta$) have starting points on the horizon at the same time, and the same ending point, they must rise in the same amount of time.
Next up, EΘ rises with $arc \; NM$ at sphaera obliqua. This one we need to do a little backwards thinking. First, let’s notice that $arc \; \Theta K$ rises with $arc \; \Theta E$ at sphaera obliqua, which is to say the horizon BED. The logic of this is backwards of the last one: Θ is the common starting point and K and E are the end points which, as we can directly see from the diagram, are on the horizon at the same time. Then since $arc \; MN$ is formed by the same pair of great circles arcs as $arc \; \Theta K$, the two arcs are similar and since similar arcs rise in the same amount of time, we can then swap out $arc \; MN$ for $arc \; \Theta K$.
Now we’ve looked at the rising times for $arc \; EM$ and $arc \; MN$, the difference of which is $arc \; EN$. Since $arc \; EM$ rose with $arc \; E \Theta$ at sphaera recta and $arc \; MN$ with $arc \; E \Theta$ at sphaera obliqua, $arc \; EN$ is the difference of the rising times of $arc \; E \Theta$ at the two.
Thus we have shown that, for arcs of the ecliptic bounded by point E and the parallel circle through K, in every case, if the great-circle arc corresponding to LKN is drawn, segment EN will comprise the difference between that arc’s rising-times at sphaera recta and at sphaera obliqua.
This value is known as the “ascensional difference” by medieval astronomers. Why is it useful? Because:
$arc \; EN$ is the difference between the rising-times of segment $E \Theta$ at sphaera obliqua and at sphaera recta.
That statement is in the context of having just solved for EN, but let’s recall what we’re really after is $arc \; E \Theta$ as it’s the arc on the ecliptic and that’s what we’ve been working on for the past several posts. But let’s state it a bit more mathematically so we can manipulate it more easily.
$$arc \; EN = arc \; E\Theta_{SR} – arc \; E\Theta_{SO}$$
Since we’re after $arc \; E \Theta$ at sphaera obliqua, let’s solve for that:
$$arc \; E\Theta_{SO} = arc \; E\Theta_{SR} – arc \; EN$$
Thus, we can find the rising time of an arc of the ecliptic by knowing its rising time at sphaera recta and finding $arc \; EN$ which we subtract from it. We’ll be doing an example of this in the next post, so if you’re having trouble following, stay tuned.