With the previous theorem about the ascensional differences complete, it’s time to move on to determine how to figure out the rising time of arcs of the ecliptic for 10º segments at various latitudes using what Ptolemy promises to be a shortcut in the math. In the modern sense it really doesn’t seem to be much of a shortcut, but that’s because with the assistance of calculator’s, the equations we were using previously seem much less daunting. If it had to be done by hand, I’m sure it would be far more tedious.
Instead, Ptolemy reduces the number of calculations by going through the proof regarding ascensional differences as well as making use of some previously calculated values to avoid having to do other calculations.
To get started, Ptolemy revises the previous drawing, making it a bit simpler by removing the ecliptic and renaming a few of the points, as well as changing a few definitions.
We’re no longer worrying about the ecliptic. Also, the is no longer at point E. Instead, Ptolemy takes point H to be the winter solstice, which, being on BED is on the horizon. We’ll take K to be some point between Capricorn and the end of Pisces1.
Before diving in, let’s take a moment to consider what we’re after. The funny thing is that it’s not actually represented on this diagram. We’re trying to figure out the rising times of segments of the ecliptic (specifically at sphaera obliqua), which Ptolemy just removed. However, from our last post, we learned that we can determine this by knowing two pieces of information: The segment’s rising time at sphaera recta (which we derived at the end of the last book) and what in this diagram is $arc \; EL$. So let’s start by going after $arc \; EL$
Here again, we have a Menelaus configuration. To make it more obvious, I’ll highlight it:
Thus, using Menelaus’ Theorem II, we can state2:
$$\frac{Crd \; arc \; 2 H \Theta}{Crd \; arc \; 2 HZ} = \frac{Crd \; arc \; 2KL}{Crd \; arc \; 2KZ} \cdot \frac{Crd \; arc \; 2 E \Theta}{Crd \; arc \; 2EL}$$
As usual, we’ll take stock of what we know.
First off, we know $arc \; 2H \Theta$. The reason for this is that $arc \; H \Theta$ is the angular distance from the H, the winter solstice, to the celestial equator. That’s the obliquity of the ecliptic. As we’ve seen previously, Ptolemy uses a value of $\frac {11}{83}$ for this which is 47;42,40º for a $Crd \; arc$ of 48;31,55.
From that, we can easily get $arc \; HZ$ as $arc \; Z \Theta$ is 90º. Thus we subtract out $arc \; H \Theta$ and then double to get $arc \; 2HZ = 132;17,20$º and $Crd \; arc \; 2HZ = 109;44,53$.
Next, let’s look at $arc \; KL$. This is the angle between K, which is a point on our horizon, and the celestial equator. As we’re interested in the value for a 10º arc3, we can look this value up in our table of inclinations. Thus, when the arc is 10º, $arc \; KL = 4;1,38$º, $arc \; 2KL = 8;3,16$º, and $Crd \; arc \; 2KL = 8;25,39$.
As with before, we can then get $arc \; KZ$ because $arc \; LZ$ is 90º, and we can subtract out $arc \; KL$ to show $arc \; KZ = 85;58,22$º. Thus $arc \; 2KZ = 171;56,44$º, and $Crd \; arc \; 2KZ = 119;42,14$.
Now let’s pause for a moment and take a moment to think about what we’ve been working on. The ultimate goal here is to be able to determine the rising times for every 10º arc. This means we’re going to repeat this calculation several times. To make life easier, Ptolemy takes a break before moving on to see if we can’t find a pattern that will make the repeated calculations a bit easier. So to see what he does, let’s plug in our variables and see where we stand:
$$\frac{48;31,55}{109;44,53} \cdot \frac{119;42,14}{8;25,39} = \frac{Crd \; arc \; 2 E \Theta}{Crd \; arc \; 2EL}$$
Let’s take a moment to think about each of the three fractions in this equation. The first one was composed of values that are constant. Recall they had to do with the obliquity of the ecliptic and the compliment. So no matter how many times we repeat this, we won’t have to worry about that one.
The second fraction wasn’t quite as easy as the root came from our table of inclinations and its compliment. So we’ll need to look at this independently each time, which Ptolemy does now.
| Degrees | $arc \; 2KL$ | $Crd \; arc \; 2KL$ | $arc \; 2KZ$ | $Crd \; arc \; 2KZ$ |
| 20º | $15;54,6$º | $16;35,56$ | $164;5,54$º | $118; 50,47$ |
| 30º | $23;19,58$º | $24;15,56$ | $154;40,2$º | $117;31,15$ |
| 40º | $30;8,8$º | $31;11,43$ | $149;51,52$º | $115;52,19$ 4 |
| 50º | $36;5,46$º | $37;10,39$ | $143;54,14$º | $114;5,44$ |
| 60º | $41;0,18$º | $42;1,48$ | $138;59,42$º | $112;23,57$ |
| 70º | $44;40,22$º | $45;36,18$ | $135;19,38$º | $110;59,47$ |
| 80º | $46;56,32$º | $47;47,40$ | $133;3,28$º | $110;4,16$ |
Ptolemy now pauses for a moment to notice something more about the terms we’ve collected on the left side of the equation: They’re independent of latitude. Which is actually handy because not only does Ptolemy want to develop this list of rising times for each 10º arc, he also wants to do it for every latitude. So if latitude doesn’t come into play, this will be a huge time saver.
To make things even simpler, Ptolemy does a bit of simplification, reducing the four terms on the left side to a fraction. Unfortunately, the details of how he did this aren’t explained, but we can certainly check the math to verify that they’re correct.
| Arc | Value |
| 10º | $\frac{60}{9;33}$ |
| 20º | $\frac{60}{18;57}$ |
| 30º | $\frac{60}{28;1}$ |
| 40º | $\frac{60}{36;33}$ 5 |
| 50º | $\frac{60}{44;12}$ |
| 60º | $\frac{60}{50;44}$ |
| 70º | $\frac{60}{55;45}$ |
| 80º | $\frac{60}{58;55}$ |
That joy of the equation being independent of latitude is short lived because latitude does come into play on the right side of the equation. Specifically in the term $Crd \; arc \; 2E \Theta$. However, we’ve actually gone through how to get that term previously. As Ptolemy states:
It is immediately obvious that for each latitude we will have $arc \; 2E \Theta$ as a given arc since it is, in degrees, the difference in time-degrees of the equinoctial day from the shortest day. Hence, from $Crd \; arc \; 2E\Theta$ and the ratio ($\frac{Crd \; arc \; 2E\Theta}{Crd \; arc \; 2EL}$), $Crd \; arc \; 2EL$ will be given, and [hence] $arc \; 2EL$. We will subtract half of this, namely $arc \; EL$, which comprises the above-mentioned difference [between rising-times at sphaera recta and sphaera obliqua], from the rising-time of the ecliptic arc in question at sphaera recta, and thus obtain the rising-time of the same arc at the given latitude.
That’s pretty tough to parse, but what he’s saying is that we have a method for getting $arc \; 2E \Theta$ and that once we have it, we use it to get $arc \; EL$ which we can then subtract it from the rising time of the arc we’re looking for at sphaera recta to get what we’re looking for.
But let’s walk through the example, Ptolemy provides, again using Rhodes. As we’ve seen previously, $arc \; 2E \Theta = 37;30$º so $Crd \; arc \; 2E \Theta = 38;34$. Let’s pop that into our equation using the simplified left sides I’ve listed above, starting with a 10º arc:
$$\frac{60}{9;33} = \frac{38;34}{Crd \; arc \; 2EL}$$
And solve:
$$Crd \; arc \; 2EL = \frac{38;34 \cdot 9;33}{60}$$
$$Crd \; arc \; 2EL = 6;8$$
Next6, we’ll find the arc that chord subtends from our chord table, which tells us $arc \; 2EL = 5;52$º. Therefore $arc \; EL = 2;56$º.
Thus, we only need one more thing for this arc of 10º. We need to know how long it would take to rise at sphaera recta.
Back at the end of the first book, we looked at the rising times of arcs of the equator at sphaera recta, and at the very end of that post it was even broken down into 10º arcs. From there, we see that the rising time of the first 10º arc is 9;10º.
Thus, the rising time for a 10º arc of the ecliptic at Rhodes is7:
$$arc \; E\Theta_{Rhodes} = arc \; E\Theta_{SR} – arc \; EL$$
$$arc \; E\Theta_{Rhodes} = 9;10º – 2;56º = 6;14º$$
Thus, the first 10º of the ecliptic rises in 6:14º. However, this will become slightly more involved as it did when we were doing larger segments, such that to get the next 10º, we’ll need to do 20º and subtract out the previous 10º. So let’s walk through that.
First, we’ll need to use the simplified form of the left side of the equation for 20º:
$$\frac{60}{18;57} = \frac{38;34}{Crd \; arc \; 2EL}$$
And solve:
$$Crd \; arc \; 2EL = \frac{38;34 \cdot 18;57}{60}$$
$$Crd \; arc \; 2EL = 12;11$$
Finding the arc that subtends this we get
$$arc \; 2EL = 11;40º$$
Which means:
$$arc \; EL = 5;50º$$
Again, we look up the time it took 20º of the celestial equator to rise at sphaera recta which was ($9;15º + 9;10º = 18;25º$). And subtracting $arc \; EL$ from that, we get the rising time for 20º of the ecliptic at Rhodes to be $18;25º – 5;50º = 12;35º$.
But again, this is for 20º. So we need to subtract out the previous 10º:
$$12;35º – 6;14º = 6;21º$$
Thus, the second 10º arc of the ecliptic to rise at Rhodes takes 6;21º.
This same method can be applied to the remaining 10º arcs all the way up to 90º. Which is all we will need to do as we’ve previously shown that the pattern repeats itself based on the rules about arcs of the ecliptic equidistant from the same equinox and solstice.
While Ptolemy does complete the calculations for each 10º interval I’ll forego doing so as there’s no new math, and the results of doing so are going to appear in the table derived from all of this in the next chapter.
- Ptolemy specifically defines this as the last point in Pisces (which is also the first point in Aries), but as he never makes use of that I don’t want readers to get confused, hence why I’m putting that in the footnote.
- We’ve actually flipped the equation over (i.e., taken its inverse), and flipped the configuration left-to-right for this equation
- This 10º arc is specifically the one from the spring equinox back towards the winter solstice.
- The translation I’m using makes a note here that this value is slightly off and should be 115;52;26. However, he and I leave it here since this value will be used more in the future and changing it will result in a cascade of changes that would need to be made.
- As noted previously, the value in the last table which gets propagated through causes this value to be slightly off as well as it should be 36;31;40 which would round to 36;32. Thus, it’s off by 1 minute of arc, which is still relatively small and not worth changing.
- I’m going to deviate somewhat from Ptolemy’s treatment here in two ways. The first is that he opts to pop that back into the equation and do a bit of rearranging stating: $\frac{60}{38;34} = \frac{9;33}{6;8}$. I do not find this to be helpful, and have thus omitted it. Secondly, he then repeats this for every other 10º segment, but to make things a bit easier to read, I’ll finish this one 10º arc before presenting all the results as it gets rather cluttered.
- This is the same equation as at the end of the last post, but looks slightly different as some of the variables have changed.