Having determined a correction for the equant for the first opposition, we’ll now repeat the procedure for the second. The diagram to do so is identical to the previous one, but is flipped over the line of apsides.
For this, we’ll be using the the second observation.
If you’re needing a refresher, we discussed three observations in this post and demonstrated the distance of Mars’ apparent position from the position of apogee1, i.e., $arc \; XL$ in this diagram, in this post in which it was described as $arc \; LB$ and we found it to be $45;13º$.
Thus, $arc \; XZ = 45;13º$.
We’ll begin by creating a demi-degrees context about $\triangle X \Theta Z$. In it, the angle the arc we just mentioned is, $\angle X \Theta Z = 90;26º$ as is its vertical angle, $\angle D \Theta F$.
If we then concentrate on $\triangle D \Theta F$, creating another demi-degrees circle about it, we can then state that $arc \; DF$ would have this same measure. We can then find the corresponding chord to be $\overline DF = 85;10^p$ in the context about this triangle in which the hypotenuse, $\overline \Theta D = 120^p$.
We can then use the Pythagorean theorem to determine $\overline{\Theta F} = 84;32^p$.
Now, let’s convert that to our general context in which the eccentricity, i.e., $\overline{D \Theta} = 6;33;30^p$. I’ll start with $\overline{DF}$:
$$\frac{6;33,30^p}{120^p} = \frac{\overline{DF}}{85;10^p}$$
$$\overline{DF} = 4;39^p.$$
And $\overline{\Theta F}$:
$$\frac{6;33,30^p}{120^p} = \frac{\overline{DF}}{84;32^p}$$
$$\overline{\Theta F} = 4;37^p.$$
Ptolemy comes up with $\overline{\Theta F} =4;38^p$ which I’ll adopt for consistency.
Now, recalling that $\overline{DB} = 60^p$ as it’s a radius of the middle of the three circles, we can then use the Pythagorean theorem to solve the last side of $\triangle DBF$:
$$\overline{DF}^2 + \overline{FB}^2 = \overline{DB}^2$$
$$(4;39^p)^2 +\overline{FB}^2 =(60^p)^2$$
$$\overline{FB} = 59;49^p.$$
As we discussed in the last post, $\overline{FQ} = \overline{F \Theta}$, so we can add another $4;37^p$ to $\overline{BF}$ to determine $\overline{QB} = 64;27^p$.
Now, let’s look at $\overline{NQ}$. This is twice the measure of $\overline{DF}$. This is, again, because $\triangle D \Theta F$ is a similar triangle to $\triangle N \Theta Q$ with the latter being twice as large.
Thus, $\overline{NQ} = 2 \overline{DF} = 9;18^p$.
This gives us three of the sides of $\triangle NQB$ in this context, so we can find the remaining side using the Pythagorean theorem:
$$\overline{NB}^2 = (9;18^p)^2+ (64;27^p)^2$$
$$\overline{NB} = 65;07^p$$
for which Ptolemy finds $65;08^p$. Again, I’ll adopt his value.
We can now create a demi-degrees circle about this triangle:
$$\frac{120^p}{65;08^p} = \frac{\overline{NQ}}{9;18^p}$$
$$\overline{NQ} = 17;08^p$$
for which Ptolemy finds $17;09^p$ and I will again adopt.
We can then look up the corresponding arc, $arc \; NQ = 16;26º$ indicating that the angle it subtends, $\angle NBQ = 8;13º$.
Now let’s take a look at $\overline{\Theta Q}$. This is equal to $\overline{\Theta F} + \overline{FQ}$. But, as we stated above $\overline{\Theta F} = \overline{FQ}$. Since we established that $\overline{\Theta F} =4;38^p$, that means that $\overline{\Theta Q} = 9;16^p$.
This can be added to $\overline{\Theta Q}$ which is $60^p$ as it’s a radius, thus giving $\overline{ZQ} = 69;16^p$.
That gives us two sides of $\triangle NZQ$, so we can use the Pythagorean theorem to find $\overline{ZN}$ in this context. Doing so, I find it to be $69;53^p$. Ptolemy comes up with, $69;52^p$ which I’ll adopt.
Now, let’s create a demi-degrees circle about this triangle. In it, the hypotenuse $\overline{ZN} = 120^p$ which we can use to convert the other pieces:
$$\frac{120^p}{69;52^p} = \frac{\overline{NQ}}{9;18^p}$$
$$\overline{NQ} = 15;58^p.$$
Ptolemy rounds this off to an even $16^p$.
Looking up the corresponding arc, I find it to be $arc \; NQ = 15;20º$ in this demi-degrees circle meaning the angle it subtends is half that, or that $\angle NZQ = 7;40º$.
Ptolemy then compares this to $\angle NBQ$ which we found to be $8;13º$.
The difference of these two is equal to $\angle BNZ$ and is $0;33º$.
This gets us to the same place as we ended the previous post.
However, we now have the positions worked out for the first pair of observations, so now we’re able to do a bit more with them.
Ptolemy explains:
Now, since we found the $arc \; KS$ as $0;32º$ for the first opposition2, it is clear that the first interval, taken with respect to the [equant]3, will be greater than the interval of apparent motion by the sum of both arcs, [namely] $1;05º$, and [hence] will contain $68;55º$.
What does Ptolemy mean here?
To answer, let’s do a quick review of what we’ve done since starting on the superior planets.
In the first post, we discussed how the position of a superior planet about its epicycle will be directly tied to the motion of the sun. Most importantly for what we’re discussing now, oppositions with respect to the sun occur when the planet is at the point on its epicycle which is along the line between the Earth and the center of the epicycle which is closer to the Earth.
However, an important point in that post is that the center of the epicycle is carried on the circle of mean distance.
This poses a bit of a problem, because what we’re really looking for is the motion along the equant circle – the circle of mean motion.
To help explain, let’s bring back this diagram from a few posts ago:
We’ve found several things at this point, so let’s recap.
First off, let’s state clearly what we’re after here. It’s $\angle ENZ$. We made a first approximation at it in this post where we said the planet appeared to move $67;50º$ between the first opposition observation and the second.
However, there’s a problem: As discussed in the post where we first introduced the model for superior planets, oppositions occur when the planet is at its position on the epicycle that is closer to Earth, along the line of sight to the epicycle’s center, from the point of view of the observer on Earth.
So, what we really determined wasn’t $\angle ENZ$, it was $\angle ANB$. What we’ve been working at in this post and the last one was a way to get the former from the latter.
And what Ptolemy is telling us to do is add on the missing pieces. In short,
$$\angle ENZ = \angle ANE + \angle ANB + \angle BNZ.$$
Here, $\angle ANE$ we found to be $0;32º$ in the last post and $\angle BNZ$ we found to be $0;33º$ in this post.
Thus, we can add these to $\angle ANB$ (which we’d previously found to be $67;50º$) to determine $\angle ENZ = 68;55º$.
Ptolemy still isn’t telling us what help that is. Instead, we’ll repeat this calculation for the third observation, which I’ll explore in the next post.
- In the direction of $X$ from the observer at $N$
- This arc and value is from the last post. Not this one.
- Ptolemy used the unspecific term “eccentre” here. Toomer notes that Ptolemy here is talking about the equant.