In this next chapter, we’ll be looking at the angle between the ecliptic and horizon. Obviously, this one will depend on the latitude so there will be more complexity than the previous chapter. But as with before, we’ll simplify the process by starting off with some symmetries.
Equinox
In order to calculate these angles also at sphaera obliqua, we must first prove that the points on the ecliptic equidistant from the same equinox produce equal angles at the same horizon.
To do this, we’ll start off with ABGD as the meridian, AEG as the celestial equator, and BED as the horizon. For now, we’ll draw in the ecliptic as MLK.
Remembering that the sun travels counter clockwise around the ecliptic, this means that when it passes the celestial equator at point K, it would be heading south putting K at the autumnal equinox, which is below the horizon.
In this section we’re talking about the angle at the intersection between the ecliptic and horizon. That point is L, and since Ptolemy discusses this in terms of the angle that’s up and right, we’re concerned about $\sphericalangle DLK$.
But we’re trying to prove a symmetry rule here about points equidistant from the same equinox, so we’ll need a second equinoctial point. So, we’re going to let some time pass until the autumnal equinox rises but we’ll rename it Z as not to be confused.
So now we have the autumnal equinox above the horizon at point Z. Again, if we’re looking at the intersection between ecliptic and horizon, it’s now at H so the upper right angle is $\sphericalangle EH \Theta$.
One last thing we’ll need to know is that we’ve drawn this such that $arc \; LK = arc \; ZH$. This means that point L and point H, the vertices of the two angles we’re interested in, are equidistant from the same equinox (at K and Z respectively) and we’re trying to prove that the two angles are equal.
Fortunately, this is pretty simple to do because spherical triangle EHZ and ELK are equal. We can prove this by showing that all three corresponding sides are equal.
We said from the setup that $arc \; LK = arc \; ZH$. Additionally, $arc \; HE = arc \; EL$ as was noted in this post (where they were called $arc \; HE$ and $arc \; EK$). Lastly, $arc \; EZ = arc \; EK$ which is something we showed in this post (where they were called $arc \; EZ$ and $arc \; E \Theta$).
Thus, the triangles are the same as are their respective angles, so $\sphericalangle EHZ = \sphericalangle ELK$.
But those weren’t quite what we’re after. We wanted to show that $\sphericalangle DLK = \sphericalangle EH \Theta$. Fortunately, these are the supplements to the angles we just proved are equal, which means that they too are equal, proving the original statement: That these angles between the ecliptic and horizon equidistant from the same equinox are equal.
Rising and Setting Points
The next thing Ptolemy sets about to prove is that
if two points [of the ecliptic] are diametrically opposite, the sum of the angles [between the ecliptic and the horizon] at the rising-point of one and the setting-point of the other is equal to two right angles.
As with before, the phrase “two right angles” means 180º.
Here, we’ll only need to draw the horizon, ABGD, and the ecliptic, AEGZ, having them intersect at A and G. The goal here will be to prove that $\sphericalangle DGZ + \sphericalangle DAE = 180º$.
Before proving this, let’s take a minute to look at what we’re trying to prove. As can be seen more easily from the picture, we have the ecliptic rising up from the horizon in the east at G and setting in the west at A. The sum of the angle on one side of the ecliptic where it’s rising, and the other side of the ecliptic where it’s setting, always equals $180º$.
This one’s even easier to prove. First off, let’s notice that $\sphericalangle DAZ + \sphericalangle DAE = 180º$ as they’re supplements to the same great circle (AEGZ).
But $\sphericalangle DAZ = \sphericalangle DGZ$ because they’re both formed by the same plane (the one formed by the great circle of the horizon, DABG) intersecting with another (the one formed by the great circle of the ecliptic, AEGZ) which means we can substitute to prove the proposition.
Solstice
Ptolemy doesn’t offer a proof for this, but states that from these two previous theorems, we can create another, stating,
For points equidistant from the same solstice, the sum of the rising-angle at one and the setting-angle at the other will be equal to two right angles.
While we could certainly take Ptolemy’s word for it, The translator offers a proof. Here, we’ll again draw in a horizon and ecliptic. On them, we’ll label the intersection that is rising R, and the setting point S. Since we’re concerned about solstices, we’ll let that be just above the rising horizon at T. Of course 90º along from that will be an equinox which we can label at E. We’ll also sketch in point X is twice the distance along the ecliptic T is from R. We’ll call that distance d.
This means that $arc \; EX = arc \; TE – arc \; TX = 90º – d$.
In addition, $arc \; ES = arc \; RS – arc \; RE = 180º – (90º + d) = 90º – d$
Since both $arc \; EX$ and $arc \; ES$ equal $90º – d$ they must be equal as well. If these are equal, than it means that X and S are equal distances from that equinox, E, which means the angles they form must also be equal by the first of the theorems we proved above.
Next, we can use the other theorem we just proved to look at X and R which are equidistant from solstice T. This means that the angle at R plus the angle at X must equal 180º. But we just demonstrated the angle at S equals the angle at R, we can state that the angle at S + the angle at R also equal 180º.
With these theorems in hand, we’re ready to start the actual calculations for the different signs, but we’ll save that for the next post.