Now that we’ve sorted out the angles around the circle of mean motion, Ptolemy sets about checking to ensure that they return the observed angles from apogee from the point of view of the observer.
To do so, he begins with the following diagram:
In this diagram, the point $A$ is the position of Mars at the time of the first opposition.
In the previous set of posts, we demonstrated that $arc \; AE = 41;33$. Thus, in this diagram, $\angle A \Theta E = 41;33$ as well, as does its vertical angle, $\angle D \Theta F$.
We’ll now focus on $\triangle D \Theta F$, creating a demi-degrees context about it. In that context, the hypotenuse, $\overline{D \Theta} = 120^p$. We can then look at $arc \; DF$ opposite $\angle D \Theta F$. Since that arc will be double the angle, it will be $83;06º$.
In addition, we can look at its supplement, $\overline{F \Theta}$ which would be $96;54º$.
Ptolemy then looks up the corresponding chords. I find them to be $79;36^p$ and $89;48^p$ respectively. But Ptolemy comes up with $79;35^p$ and $89;50^p$ respectively. As usual, I’ll adopt Ptolemy’s values for consistency.
We then convert back to our context in which the diameter of the circle is $120^p$. In that context, $\overline{D \Theta} = 6^p$. Since we know that piece in both contexts, we can use it to do our conversion:
$$\frac{6^p}{120^p} = \frac{\overline{DF}}{79;35^p}$$
$$\overline{DF} = 3;58,45^p$$
by my reckoning. However, Ptolemy comes up with $3;58,30^p$.
Ptolemy also converts $\overline{F \Theta}$. I find it to be $4;29,30^p$ and Ptolemy evidently rounds to $4;30^p$.
We can then look at $\triangle DFA$. In it, $\overline{DA} = 60^p$ since it’s a radius. And since we just found $\overline{DF}$ we can use the Pythagorean theorem to find $\overline{FA} = 59;52^p$. Ptolemy evidently finds it to be $59;50^p$.
As we’ve seen several times in such diagrams, $\overline{F \Theta} = \overline{FQ}$ and that $\overline{QN} = 2 \cdot \overline{FD} = 7;57^p$.
Since we just found $\overline{FA}$ we can add on $\overline{QF}$ to find $\overline{QA} = 64;20^p$.
Next, we can look at $\triangle QNA$. We know two sides of this, we we can determine the remaining side, $\overline{NA}$ using the Pythagorean theorem.
Doing so, I find $\overline{NA} = 64;49^p$. Ptolemy is a bit further off at $64;52^p$ which I will, again, adopt.
We can now create a demi-degrees context around $\triangle NAQ$ and convert into it, to find $\overline{NQ}$ in that context:
$$\frac{120^p}{64;52^p} = \frac{\overline{NQ}}{7;57^p}$$
$$\overline{NQ} = 14;42^p.$$
Ptolemy is again, slightly different at $14;44^p$.
We can then find the corresponding arc, $arc \; NQ = 14;06º$.
This means that the angle it subtends on the opposite side of the circle, $\angle NAQ = 7;03^p$.
Lastly,
$$\angle ANE = \angle AOE – \angle NAQ$$
$$41;33º – 7;03º = 34;30º.$$
This is the amount by which the planet was in advance of apogee at the first opposition.
This isn’t enough for us to check on its own. But we’ll now do the same for the second opposition. Then we can take the angular increase and compare that to what we found for the angular increase in this post to ensure it matches.
So we’ll create a very similar diagram for the second opposition:
In this diagram, the points are all the same as they were in the previous diagram with the addition of $B$ which is the position of Mars along the circle of mean distance at the second opposition.
Thus, $\angle B \Theta E = 40;11º$ as does its vertical angle, $\angle D \Theta F$.
We’ll again enter a demi-degrees context about $\triangle D \Theta F$ wherein the arc opposite that angle, $arc \; DF$ has twice the measure or $80;22º$. Therefore, its supplement, $arc \; F \Theta = 99;38º$.
And we can look up their corresponding chords:
$$\overline{DF} = 77;26^p$$
which is what Ptolemy also arrives at.
Similarly, we can find $\overline{F \Theta} = 91;41^p$. Again, both of these are in the context of this demi-degrees circle in which $\overline{D \Theta} = 120^p$.
And since we know $\overline{D \Theta} = 6^p$ in the context in where the diameter of the circle is $120^p$, we can use that to context switch the parts we just found:
$$\frac{6^p}{120^p} = \frac{\overline{DF}}{77;26^p}$$
$$\overline{DF} = 3;52^p.$$
Similiarly,
$$\frac{6^p}{120^p} = \frac{\overline{F \Theta}}{91;41^p}$$
$$\overline{F \Theta} = 4;35^p.$$
Next, focusing on $\triangle DFB$, we know two sides: $\overline{DB} = 60^p$ since it’s a radius, and $\overline{DF}$ which we just found.
Thus, we can use the Pythagorean theorem to find the remaining side, $\overline{BF} = 59;53^p$.
We can then determine $\overline{BQ}$ by adding $\overline{FQ} + \overline{BF} = 64;28^p$.
Additionally, $\overline{NQ} = 2 \cdot \overline{DF} = 7;44^p$.
This gives us two sides of $\triangle BNQ$, so we can use the Pythagorean theorem to find the remaining side, $\overline{BN} = 64;56^p$.
We’ll now create a demi-degrees context about that triangle in which the hypotenuse, $\overline{BN} = 120^p$ which we can use as our conversion factor:
$$\frac{120^p}{64;56^p} = \frac{\overline{NQ}}{7;44^p}$$
$$\overline{NQ} = 14;18^p.$$
Ptolemy comes up with $14;19^p$ and Toomer notes that this is from him having carried out the series of computations with more sexagesimal places than he shows in the text, which is quite common.
Regardless, we can look up the corresponding chord. Doing so, I find $arc \; NQ = 13;42º$. Thus, the angle opposite this arc in this circle is half that measure, which is to say, $\angle NBQ = 6;51º$.
And as with before,
$$\angle ENB = \angle E \Theta B – \angle BNQ$$
$$\angle ENB = 40;11º – 6;51º = 33;20º.$$
This is the angle by which Mars was past apogee from the point of view of the observer at the second opposition.
Thus, if we add this to what we found in the first part of this post, we should get the same increase in what was observed, if we did everything right. So,
$$34;30º + 33;20º = 67;50º.$$
If we look back at this post, we can see this exactly matches the observed increase. This should tell us that our line of apsides and eccentricity are correct.
But, as a final check, we can repeat this calculation for the third opposition and ensure the increase from the second to the third also matches our initial observation.
So, we’ll again create a similar diagram:
In this, $\angle G \Theta Z = 44;21º$ as we determined in this post.
We’ll begin by looking at a demi-degrees circle about $\triangle D \Theta F$ in which $\angle F \Theta D$ is the same as $\angle G \Theta Z$. Therefore, in this context where the hypotenuse $\overline{D \Theta} = 120^p$, $arc \; DF = 88;42º$ and it supplement, $arc \; F \Theta = 91;18º$
We can then look up the corresponding chords, $\overline{DF}$ and $\overline{F \Theta}$. I find them to be $83;53^p$ and $85;49^p$ respectively, in agreement with Ptolemy.
Now we’ll convert to the context in which the diameter of the eccentre is $120^p$.
$$\frac{6^p}{120^p} = \frac{\overline{DF}}{83;53^p}$$
$$\overline{DF} = 4;11,39^p.$$
Ptolemy rounds this to $4;11,30^p$.
$$\frac{6^p}{120^p} = \frac{\overline{F \Theta}}{85;49^p}$$
$$\overline{F \Theta} = 4;17,27^p.$$
Ptolemy rounds this to $4;17^p$.
Next, we’ll use the Pythagorean Theorem on $\triangle DGF$.
In it, $\overline{DG} = 60^p$ since it’s a radius, and we just found $\overline{DF}$. Therefore, the remaining part,
$$\overline{GF} = \sqrt{\overline{DG}^2 – \overline{DF}^2}$$
$$\overline{GF} = \sqrt{60^2 – 4;17^2} = 59;51^p.$$
As with before, $\overline{F \Theta} = \overline{FQ}$ and $\overline{QN} = 2 \cdot \overline{DF}$ = 8;23^p$.
We can therefore subtract $\overline{QF}$ from $\overline{GF}$ to find that $\overline{GQ} = 55;34^p$.
That gives us two of the sides of $\triangle GQN$, so we can use the Pythagorean theorem to find $\overline{GN}$:
$$\overline{GN} = \sqrt{55;34^2 – 8;23^2} = 56;12^p.$$
We’ll now enter a demi-degrees context about $\triangle GNQ$.
$$\frac{120^p}{56;12^p} = \frac{\overline{QN}}{8;23^p}$$
$$\overline{QN} = 17;54^p.$$
Ptolemy find it to be $17;55^p$.
Looking up the corresponding arc, I find it to be $17;10º$ which is $arc \; NQ$.
Therefore, $\angle \Theta GN$ is half that or $8;35º$.
As stated previously, $\angle G \Theta Z = 44;21º$. We can add $\angle \Theta GN$ to that to determine $\angle GNZ = 52;56º$ which is
the amount by which the planet was in advance of the perigee at the third opposition.
We can now combine that with the result from the second observation to do another check:
[W]e also showed that at the second opposition, it was $33;20º$ to the rear of apogee. So we have found $93;44º$ between the second and third oppositions, computed by subtraction [of the sum of $52;56º$ and $33;20º$ from $180º$], in agreement with the amount observed for the second interval.
Again, the $93;44º$ between observations comes from this post.
Furthermore, since the planet, when viewed at the third opposition along line $\overline{GN}$ had a longitude of $2;34º$ into Sagittarius according to our observation, and $\angle GNZ$ at the center of the ecliptic was shown to be $52;56º$, it is clear that the perigee of the eccentre, at point $Z$, had a longitude of [$2;34º$ in Sagittarius $+ 52;56º =$] $25;30º$ into Capricorn, while the apogee was diametrically opposite $25;30º$ into Cancer.
Thus, Ptolemy fixes the points of apogee and perigee along the ecliptic.
But before Ptolemy closes out the chapter, he brings our attention back to the epicycle which we have neglected for quite some time with a new diagram:
In this diagram, we’ve placed the epicycle’s center at $G$ and extended $\overline{G \Theta}$ across the epicycle to point $M$, also drawing in point $L$ on the near side. Similarly, we’ve added point $K$ on the nearside of the epicycle on $\overline{GN}$.
In this,
[the] mean motion of the epicycle, counted from apogee of the eccentre [is] $135;39º$ (for its supplement, $\angle G \Theta Z$ was shown to be $44;21º$.
In other words, $\angle G \Theta E$ = $135;39º$.
Next,
[the] mean motion of the planet from the epicycle apogee, $M$ (i.e., $arc \; MK$ [is] $171;25º$ (for $\angle \Theta GN$ was shown to be $8;35º$ [above], and since it is an angle at the center of the epicycle, the $arc \; KL$ from the planet at $K$ to the perigee at $L$ is also $8;35º$. Hence the supplementary arc from the apogee $M$ to the planet at $K$ is, as already stated $171;25º$).
Pretty self explanatory here as $\overline{ML}$ sections off $180º$ of the epicycle. Take out $arc \; LK$ from that (which is the same as $\angle \Theta GN$, and you find $arc \;MK = 171;25º$.
Thus, we have demonstrated, among other things, that at the moment of the third opposition, i.e., in the second year of Antoninus, Epiphi $12/13$ in the Egyptian calendar, $2$ equinoctial hours before midnight, the mean positions of the planet Mars were:
in longitude (so-called) from the apogee of the eccentre: $135;39º$
in anomaly from the apogee of the epicycle: $171;25º.$
And at long last, that concludes chapter $7$.
In the next chapter, we’ll turn our attention to finding the side of the epicycle.
