Our goal in this chapter will be to determine a systematic way to the equation of anomaly for the sun at 6º intervals from apogee (and perigee) along the eccentre. In this post, we’ll explore the method for finding this value from apogee for a single value of 30º.
To do so, we’ll start with a diagram very much like the original one we had when we first explored the eccentric hypothesis.
Here, we have the circle of the ecliptic, ABG, centered on the observer at D. The eccentre is circle EZH on center Θ. The sun will be at position Z on the eccentre, so it appears at point B on the circle of the ecliptic. As stated above, we’ll be taking this when point Z is 30º away from the apogee at E. But specifically, that’s in context of the eccentre meaning $arc \; EZ = 30º$ as does its central angle, $\angle{E \Theta Z}$. Our goal will be to find the equation of anomaly, $\angle{DZK}$.
In addition to the points we’ve looked at before, we’ve extended $\overline{\Theta K}$ until a right angle from it creates $\overline{DK}$, which forms right triangle $\triangle{DK \Theta}$. In that triangle, $\angle{D \Theta K}$ is also 30º since it’s an opposite angle to $\angle{E \Theta Z}$ and opposite angles are equal.
Now we’ll want to explore that triangle a bit more through the demi-degrees method we keep running into. So let’s draw in the smaller circle.
Since $\angle{D \Theta K} = 30º$, and it’s on the circumference, that means $\angle{DLK}$, at the center is 60º as is $arc \; DK$. Using our chord table, we get $\overline{DK}$ in the context of the small circle to be 60. Also, in the context of the small circle, $\overline{D \Theta} = 120$, but only 2;30 in the context of the large circles. This lets us set up the scale equation:
$$\frac{\overline{DK}}{60} = \frac{2;30}{120}$$
Solving:
$$\overline{DK} = 60 \cdot \frac{2;30}{120} = 1;15$$
That gets us two sides of the right triangle, $\triangle{D \Theta K}$ so we can use the Pythagorean theorem to get the third, $\overline{\Theta K}$:
$$1;15^2 + \overline{\Theta K}^2 = 2;30^2$$
Solving:
$$\overline{\Theta K}^2 = 2;30^2 – 1;15^2 = 4;41,15$$
$$\overline{\Theta K} = 2;10$$
Recall that $\overline{\Theta K}$ is just an extension of $\overline{Z \Theta}$ which means we can just add it on to get $\overline{ZK} = 62;10$.
And we can use that with the Pythagorean theorem again in $\triangle{DKZ}$, for:
$$62;10^2 + 1;15^2 = \overline{DZ}^2$$
Solving:
$$\overline{DZ} = \sqrt{62;10^2 + 1;15^2} = 62;11$$
Now, we’ll employ the demi-degrees trick again to get at what we’re really after:
In the context of this newer circle, we have that diameter $\overline{DZ} = 120$ whereas in the larger circles it was only 62;11. Also, in context of the larger circles, $\overline{DK} = 1;15$. So we can set up our scaling equation to determine the length of $\overline{DK}$ in the context of the small circle:
$$\frac{\overline{DK}}{1;15} = \frac{120}{62;11}$$
Solving:
$$\overline{DK} = 1;15 \cdot \frac{120}{62;11} = 2;25$$
But that’s the chord, so using the chord tables, we can determine that $arc \; DK = 2;18º$ which is equal to the angle at the center, $\angle{DLK}$. The angle we’re after, $\angle{DZK}$ is half of that, or 1;9º when $\angle{E \Theta Z} = 30º$.
This will be the first entry in our table:
| Angle from Apogee | Equation of Anomaly |
|---|---|
| 30º | 1;9º |
One other thing we can quickly get out of this is the apparent angle from apogee, which is $30º – 1;9º = 28;51º$.
But what if we had that the other way around? That is to say, instead of having the mean motion’s angle (along the eccentre) from apogee being 30º, we had the apparent angle being 30º?
According to Ptolemy,
if any other of the [relevant] angles be given [instead of $\angle{E \Theta Z}$], the remaining angles will be given1
Those other “relevant angles” are the one we just introduced (the apparent angle) and the equation of anomaly itself. Essentially Ptolemy is stating that if we have one, we can solve for the other two.
To prove his claim Ptolemy once again draws the standard eccentric model, but this time drops a perpendicular line, $\overline{\Theta L}$ onto $\overline{DZ}$.
Since we’re now talking about the apparent motion, that’s tracked by $arc \;AB$ which subtends $\angle{ADB}$. According to Ptolemy therefore
the ratio $\frac{\overline{D \Theta}}{\overline{\Theta L}}$ will be given. And since $\frac{\overline{D \Theta}}{\overline{DZ}}$ is also given, $\frac{\overline{\Theta Z}}{\overline{\Theta L}}$ will be given. Hence $\angle{\Theta ZL}$, the equation of anomaly, will be given.
The Toomer translation I’m using has a flurry of footnotes that go with this, pointing to another book of Euclid’s that we haven’t encountered before: Data. I don’t find them particularly helpful so I’ll discuss them but ultimately present a solution to the problem in a slightly different manner. To do so, let’s take the statements and their footnotes piece by piece:
the ratio $\frac{\overline{D \Theta}}{\overline{\Theta L}}$ will be given
The footnote that goes with this points to Euclid’s Data 40. Much like the Elements, Data is organized into a series of propositions, but not separated into different books as Data is much shorter. Since Data is also harder to find a copy of, Toomer provides translations of the propositions. In this case, he has translated it to state,
if the angles of a triangle are given, its sides are given in form.
My first comment is that it’s somewhat hard to validate this statement simply because the numbering of Data appears to be inconsistent. For example, here’s a copy of Data printed 1756. There, we can find the 40th (Roman numeral XL) proposition on page 56 which generally agrees with the above statement2. However, in this copy, dated 1848, that same proposition is numbered 43. However, the one numbered 40 in that copy doesn’t seem to have any resemblance to the one Toomer has quoted.
But regardless of the numbering, let’s try to parse what’s going on here. What this proposition seems to me to be about is similar triangles. As long as the angles in the triangle stay the same, the ratio of their sides is the same. But what triangles is Ptolemy talking about here?
It’s obvious from the two sides given that $\triangle{DL \Theta}$ has to be one of them. My suspicion is that $\triangle{DBA}$ is the other despite Ptolemy never stating that $\overline{AB}$ should be drawn in, as it would indeed be a similar triangle. We could get $\overline{AB}$ from the chord table easily and $\overline{BD}$ and $\overline{AD}$ are both 60 since they’re radii. So we can easily find the ratio of their sides. Those same ratios would then apply to the smaller triangle, $\triangle{DL \Theta}$, as well, and since we know $\overline{D \Theta}$ to be 2;30 from this post, we could easily determine the lengths of the other segments, $\overline{DL}$ and $\overline{D \Theta}$. I’m not sure we needed one of Euclid’s theorems for that…
The next bit:
And since $\frac{\overline{D \Theta}}{\overline{DZ}}$ is also given
We know this because, as we just stated, $\overline{D \Theta} = 2;30$ and $\overline{DZ} = 60$.
On to the last bit:
$\frac{\overline{\Theta Z}}{\overline{\Theta L}}$ will be given.
Here again, Toomer cites Euclid, this time, Data 8:
magnitudes having a given ratio to the same magnitude have a given ratio to each other.
If you’re wanting to see the full proposition, it can be found here (correctly numbered this time). But without worrying about Euclid, we can simply multiply the two ratios we just looked at3.
$$\frac{\overline{D \Theta}}{\overline{\Theta L}} \cdot \frac{\overline{\Theta Z}}{\overline{D \Theta}} = \frac{\overline{\Theta Z}}{\overline{\Theta L}}$$
So on to the last bit:
Hence $\angle{\Theta ZL}$, the equation of anomaly, will be given.
Here, Toomer cites Data 43 which states:
if, in a right angled triangle, the sides about one of the acute angles have a given ratio, the triangle is given in form
Again, I’m not too big on this proposition and we can certainly solve this using the demi-degrees method we’ve seen in the past several posts.
So we haven’t walked through a numerical example of this, but we’ve laid out the logic that affirms we can arrive at the equation of anomaly from the apparent angle from apogee.
Lastly, Ptolemy states that:
Or suppose, secondly, that the equation of anomaly, i.e. $\angle{\Theta ZD}$ is given: we will get the same results in reverse order.
I think this is pretty self explanatory. If we want to arrive at the arcs of either the eccentre or the arc of the ecliptic, we can use the respective diagram and method in reverse to arrive at the appropriate arc. As such, I won’t say anything more about it.
So now we’ve shown how to get the equation of anomaly from either the arc of the ecliptic or eccentre or vice versa if we’re using the eccentric hypothesis. But despite Ptolemy stating that we’re going to use the eccentric model for the sun, he’ll repeat this for the epicyclic model which we’ll explore in the next post.
- When Ptolemy says “given” he means “solvable”.
- It uses the odd term “species” here which is defined on page 1, where it states that a figures is “given in species, when all their angles are given in magnitude, and the ratios of all their sides are given, each to each.”
- I have a suspicion that Euclid was cited because doing this is rather algebraic and algebra as we think of it didn’t really develop many of its formulisms until several centuries after Ptolemy. However, because those formulisms, like cancelling fractions, are simply algebraic statements of things like Euclid’s propositions, I don’t feel overly horrible doing them.