We’ve finally hit the last chapter in Book I. In this chapter our objective is to “compute the size of an arc of the equator”. At first pass, that doesn’t seem to have much to do with the title. Arcs of equator vs rising times?
However, Earth is a clock, rotating once every 24 hours. Thus, if we know the length of an arc, we know something about when an object following that arc through the sky will rise and set because it’s a certain proportion of 360º per 24h. Notice that if you actually complete that division, it comes out to an even 15º/hr. That’s not a coincidence.
Fortunately, to work on this problem, we won’t even need a new diagram. We can recycle the one from last chapter. Again this time we’ll be wanting to determine all sorts of arc lengths, but we’ll start with the one where $arc \; EH = 30$º.
Again, we’ll assume that $arc \; EH$ is a given and now our objective is it find $arc \; E \Theta$ which lies along the celestial equator.
This time we’ll use the other Menelaus’ Theorem to state:
$$\frac{Crd \; arc \; 2ZB}{Crd \; arc \; 2BA} = \frac{Crd \; arc \; 2ZH}{Crd \; arc \; 2H\Theta} \cdot \frac{Crd \; arc \; 2 \Theta E}{Crd \; arc \; 2EA}$$
As with last time, we have an equation with 6 variables, one of which we’re trying to solve for, so we’ll need to know all other five. Here we go!
I’ll differ here slightly from Ptolemy and start with $arc \; 2EA$ because this one’s a freebie. Since E is the equinox and A is 90º away, $arc \; 2EA = 180$º and $Crd \; arc \; 2EA = 120$ parts.
Next, let’s do $arc \; 2AB$ since this one’s easy given we discussed it in the last post where we used a value of $\frac{11}{83}$ (parts of 360º). That came out to 47;42,36 º with a corresponding chord length of 48;31,55. So that one’s out of the way.
Take a look at $arc \; ZB$. That’s the distance from the pole to the summer solstice. If we continued all the way to A to get $arc \; ZA$ that would be from the pole to the equator which is defined as 90º. Which means that $arc \; ZB$ is 90º – $arc \; AB$. But we’re after $arc \; 2ZB$ so what we need is really 180º – $arc \; 2AB$. We just discussed $arc \; 2AB$ above, so we can easily determine $arc \; 2ZB = 132;17,21$1. We can look this up in our chord table to determine the chord length. Again, it’s not exactly on a half degree increment, so we’ll again need to apply some of the sixtieths to better approximate the correct answer which is that $Crd \; arc \; 2 ZB = 109;44,53$.
Again, let’s steal another freebie from the last post where we determined $arc \; 2 H \Theta = 23;19,59$º. It’s corresponding chord from the table and applying some sixtieths is $Crd \; arc 2 H \Theta = 24;15,57$ parts. However, let’s be careful to remember that this arc was specifically for $arc \; EH = 30$º. Any other value, and we’d need to use a different figure. However, the values are all recorded in the Table of Inclinations.
Next let’s tackle $arc \; 2ZH$. We can see $arc \; ZH = arc \; Z \Theta – arc \; H \Theta$. But $arc \; Z \Theta$ is from the pole to the equator so it’s 90º. Which means $arc \; 2ZH = 180 – arc \; 2 H \Theta$. We just discussed $arc \; 2 H \Theta$ above so applying that gives us $arc \; 2ZH = 156;40,1$º. Looking up it’s chord in the table and again applying the necessary sixtieths we get $Crd \; arc \; 2 ZH = 117;31,15$ parts. Again, this arc (and its chord) are dependent on what value is taken for $arc \; EH$, so this will also change as different angles are calculated.
At this point, we have all five variables except the one we’re trying to calculate, so let’s plug them in:
$$ \frac{109;44,53}{48;31,55} = \frac{117;31,15}{24;15,57} \cdot \frac{Crd \; arc \; 2 \Theta E}{120}$$
Rewriting that to solve for our variable:
$$Crd \; arc \; 2 \Theta E = 120 \cdot \frac{109;44,53}{48;31,55} \cdot \frac{24;15,57}{117;31,15}$$
I don’t mind doing addition and subtraction in sexagesimal, but I’m not a fan of doing multiplication and division, so let’s convert that to decimal degrees:
$$Crd \; arc \; 2 \Theta E = 120 \cdot \frac{109.749}{48.532} \cdot \frac{24.266}{117.521}$$
That feels better. Plugging that into a spreadsheet and converting back to sexagesimal I get:
$Crd \; arc \; 2 \Theta E = 56;1,53$ parts. Doing a reverse lookup from our Table of Chords we find the corresponding arc is somewhere between 55;30º and 56º. Applying sixtieths until we get as close as is possible we come up with $arc 2 \Theta E = 55;40$º and thus, dividing it in half we get $arc \Theta E = 27;50$º.
Again, we can repeat this for as many chords as we like. But here we actually selected 30º to start with for a special reason. What happens if we multiply 30º by 12? It’s 360º; a full circle.
Why 12? Because that’s how many constellations are in the Zodiac2.
Thus, it has been shown that the first sign of the ecliptic… rises in the aforementioned manner [i.e. at sphaera recta] in the same time as 27;50º of the equator.
If we repeated the above calculation for 60º of the ecliptic3, we would determine that $arc \Theta E = 57;44$º and thus, subtracting the first sign from that, the second one would rise in the same amount of time as 29;54º of the equator. Similarly, the third sign would rise in the same amount of time as 32;16º of the equator. And we can stop there because now we’ve covered 90º and the pattern will repeat itself, backwards as we’ve hit the solstice and are moving back towards the equinox.
If that’s a bit hard to visualize, here’s a handy diagram that may help (read counter-clockwise):
Ptolemy does break that down a bit more into 10º intervals.
| Arc | Rising Time | Cumulative |
| 1st 10º | 9;10º | 9;10º |
| 2nd 10º | 9;15º | 18;25º |
| 3rd 10º | 9;25º | 27;50º |
| 4th 10º | 9;40º | 37;30º |
| 5th 10º | 9;58º | 47;28º |
| 6th 10º | 10;16º | 57;44º |
| 7th 10º | 10;34º | 68;18º |
| 8th 10º | 10;47º | 79;5º |
| 9th 10º | 10;55º | 90º |
In addition, we should note that each 30º comprises one of the zodiac constellations. As such, Ptolemy also considers the total for each:
| Constellation | Rising Time |
| 1st | 27;50º |
| 2nd | 29;54º |
| 3rd | 32;16º |
- Ptolemy seems to have rounded this off to 132;17,20 without explanation which is what I’ll continue with.
- There’s actually a 13th constellation whose modern border dips into the Zodiac as well: Ophiuchus. However, it’s generally not considered.
- Ptolemy does this calculation but I have chosen to omit it.