The next demonstration Ptolemy does is actually a reverse of what we did in the past 2 posts. Now, given latitude Ptolemy asks what the difference in length between the longest (or shortest) day and the daylight on the equinox would be. Again, Ptolemy has actually already given us the answer for the case we’re considering of the Greek city of Rhodes which is at 36º N latitude for which the longest day (the summer solstice) is 14.5 hours. Since the length of the day on the equinox is 12 hours, the answer will be 2.5, but Ptolemy simply wants to demonstrate that this can be achieved mathematically.
Again, we’ll use Menelaus’ Theorem II.
Here again is our diagram:
However, this time, Ptolemy flips the Menelaus configuration. Pulling it off the sphere, it would look like this:
This allows us to write:
$$\frac{Crd \; arc \; 2BZ}{Crd \; arc \; 2AB} = \frac{Crd \; arc 2HZ}{Crd \; arc 2H \Theta} \cdot \frac{Crd \; arc \; 2E \Theta}{Crd \; arc \; 2AE}$$
This time we’re after $arc \; 2E \Theta$. As always, here’s the variables we know:
$Arc \; 2 BZ = 72$º which means $Crd \; arc\; 2 BZ = 70;32,3$
For $arc \; AB$ we’ll need to remember that $arc \; AZ = 90º$ (angular distance from celestial equator to pole) and $arc \; BZ = 36º$ (altitude of said pole which is our given). So $arc \; AB = arc \; AZ – arc \; BZ = 54º$. Therefore $arc \; 2AB = 108º$ and $Crd \; arc \; 2AB = 97;4,56$.
$Arc \; 2 HZ= 132;17;20$º (as shown in this post) and $Crd \; arc \; 2 HZ = 109;44,53$.
$Arc \; 2H \Theta$ is that one we keep seeing come up; it’s twice the angle between the celestial equator and ecliptic, which Ptolemy uses as $\frac{11}{83}$ of a full circle or 47;42,40º which gives us $Crd \; arc \; 2H \Theta = 48;31,55$.
$Arc \; AE$ again is the angular distance form a solstice (A) to an equinox (E), and thus 90º, so $arc \; 2AE = 180$º and $Crd \; arc \; 2AE = 120$.
That gives us all our variables to plug in so let’s do it:
$$\frac{70;32,3}{97;4,56} = \frac{109;44,53}{48;31,55} \cdot \frac{Crd \; arc \; 2E \Theta}{120}$$
There’s not any simplification here, so I’ll skip straight to the end of all the algebra which gives us $Crd \; arc \; 2E \Theta = 38;34$. From the chord table we then look up the corresponding arc to get $arc \; 2E \Theta \approx 37;30$º.
Converting that to hours instead of degrees gives us 2.5 hours, as expected.
That could easily be a stopping point, but before moving on, Ptolemy shows us that we can derive $arc \; EH$ (which was the subject of a previous post) via another formula now that we’ve derived a few other things. Specifically, he again uses Menelaus’ Theorem I. Since that’s the same version of the equation Ptolemy originally used, obviously he’s applying it differently.
It turns out he’s again using the flipped version described above which allows us to state:
$$\frac{Crd \; arc \; 2AZ}{Crd \; arc \; 2AB} = \frac{Crd \; arc \; 2Z\Theta}{Crd \; arc \; 2H\Theta} \cdot \frac{Crd \; arc \; 2EH}{Crd \; arc \; 2BE}$$
As usual the rundown of variables:
We can see $arc \; AZ$ is from the equator to a pole as is $arc \; Z\Theta$. So they’re 90º and $Crd \; arc \; 2AZ = Crd \; arc \; 2Z\Theta =120$º.
We discussed $arc \; AB$ above so $Crd \; arc \; 2AB = 97;4,56$.
Similarly, $arc \; H\Theta$ was discussed above so $Crd \; arc \; 2H \Theta = 48;31,55$.
For $arc \; BE$, it’s 90º so again we get $Crd \; arc \; 2BE = 120$.
Plugging in:
$$\frac{120}{97;4,56} = \frac{120}{48;31,55} \cdot \frac{Crd \; arc \; 2EH}{120}$$
A bit of quick simplification:
$$Crd \; arc \; 2EH = \frac{120 \cdot 48;31,55}{97;4,56}$$
Going through the calculations we get:
$$Crd \; arc \; 2EH = 59;59,19$$
That’s suspiciously close to 60 and because a fair bit of rounding goes on in these calculations, I suspect that if we were a bit more rigorous about the math it would be 60. So looking that up in our chord tables, we see that the corresponding arc would also be 60º. Thus, half of it would be 30º as expected.