Now that we’ve worked out corrections for the change in angles caused by the equant, we’ll apply them and recalculate the eccentricity and line of apsides.
To get us started, Ptolemy reminds us of how the angles have changed by giving their corrected values:
For the first opposition: $23;14º$ into Scorpio
For the second opposition: $7;53º$ into Pisces
For the third opposition: $14;30º$ into Aries.
This has the effect of changing the intervals:
From [the] first to second oppositions: $104;39º$
From [the] second to third oppositions: $36;37º$
Ptolemy stops showing his work there, but we will walk through it here using the same diagram we did previously.
If you need a reminder on the setup for this diagram, refer back to this post.
As a quick note, I’m again doing these calculations in a Google Sheet1 which will allow for the preservation of higher levels of accuracy to avoid rounding errors. Thus, remainders past the first sexagesimal place may cause figures not to add correctly if you’re following along.
To start, $\angle BDG = 36;37º$ as this is the apparent change from the point of view of the observer. Additionally, $\angle EDH$ is the same since it is a vertical angle.
We can then create a demi-degrees context about $\triangle EDH$ to determine $\overline{EH}$ using a bit of trig:
$$\overline{EH} = 120 \cdot sin (36;37º) = 71;35^p.$$
Being that we’ve created a demi-degrees context, this is in the context in which the hypotenuse, $\overline{ED} = 120^p.$
Staying in this triangle for a moment, we can also determine $\angle DEH = 53;32º$ as it’s the remaining angle in this right triangle.
Next, recall that $arc \; BG$ is the motion along the eccentre of mean motion between the second and third opposition (not the apparent motion along the ecliptic), so is $33;26º$. This means that the angle it subtends on the other side of the circle, $\angle BEG = 16;43º.$
Additionally, we can find $\angle BEH$:
$$\angle BEH = \angle BEG + \angle DEH$$
$$\angle BEH = 16;43º + 53;23º = 70;06º.$$
We can then focus on $\triangle BEH$, creating a demi-degrees context about it in which the hypotenuse, $\overline{EB} = 120^p.$ We’ll use a bit of trig to find $\overline{EH}$:
$$\overline{EH} = 120 \cdot cos(70;06º) = 40;51º.$$
We now have $\overline{EH}$ in two contexts so we can use that to convert $\overline{EB}$ into the context in which $\overline{EH} = 71;35^p$:
$$\frac{71;35^p}{40;51^p} = \frac{\overline{EB}}{120^p}$$
$$\overline{EB} = 210;17^p.$$
Next, we’ll focus on $\angle ADG$ which is the sum of the apparent change in both intervals or $104;39º + 36;37º = 141;16º.$
Therefore, its supplement, along $\overline{EG}$, $\angle ADE = 38;44º.$
We’ll now take a look at $\triangle DEZ$, creating a demi-degrees circle about it in which the hypotenuse, $\overline{ED} = 120^p$.
In it, $\overline{EZ}$ can be found with a bit of trig:
$$\overline{EZ} = 120 \cdot sin(38;44º) = 75;05^p.$$
We’ll then look at $arc \; ABG$ which is the change in longitude over the two intervals and is unchanged from the first iteration at $133;21º$, Thus, the angle it subtends, $\angle AEG = 66;40;30º.$
Now we’ll look at $\triangle EAD.$ In it, we know two of the three angles: $\angle AED = 66;40;30º$ and $\angle ADE = 38;44º$. Thus, the remaining angle, $\angle EAZ = 74;36º.$
This allows us to concentrate on $\triangle AEZ$, creating a demi-degrees context about it in which the hypotenuse, $\overline{EA} = 120^p$.
We can then find $\overline{EZ}$ using trig:
$$\overline{EZ} = 120 \cdot sin(74;36º) = 115;41^p.$$
That now gives us $\overline{EZ}$ in two contexts. So we can convert $\overline{EA}$ into the context in which $\overline{ED} = 120^p$.
$$\frac{75;05^p}{115;41^p} = \frac{\overline{EA}}{120^p}$$
$$\overline{EA} = 77;53^p.$$
We’ll now focus on $\angle AEB.$ This angle is subtended by $arc \; AB$ which is the change in ecliptic longitude between the first and second opposition and remains unchanged from the first iteration at $99;55º$. Thus, $\angle AEB = 49;57,30º.$
This allows us to focus on $\triangle AE \Theta$ which we’ll create a demi-degrees context about in which the hypotenuse, $\overline{AE} = 120^p.$
We can then find the other two sides:
$$\overline{A \Theta} = 120 \cdot sin(49;57,30º) = 91;52º$$
$$\overline{E \Theta} = 120 \cdot cos(49;57,30º) = 77;12º$$
And since we have $\overline{AE}$ in two contexts, we can use that to convert these two pieces into the context in which $\overline{DE} = 120^p.$
$$\frac{77;53^p}{120^p} = \frac{\overline{A \Theta}}{91;52º}$$
$$\overline{A \Theta} = 59;38^p$$
and
$$\frac{77;53^p}{120^p} = \frac{\overline{E \Theta}}{77;12º}$$
$$\overline{E \Theta} = 50;06^p.$$
Next, we’ll recall that we showed that $\overline{BE}$ was $210;17^p$ in this context. Therefore, we can subtract off $\overline{E \Theta}$ to find $\overline{B \Theta} = 160;10^p.$
We can now focus on $\triangle AB \Theta.$ Since we now know two sides of it in this context, we can use the Pythagorean theorem to find the remaining side, $\overline{AB}$:
$$\overline{AB} = \sqrt{59;38^2 + 160;10^2} = 170;55^p.$$
Again, this is in the context in which $\overline{ED} = 120^p$.
We’ll now note that in the context in which the diameter of the eccentre is $120^p$, $\overline{AB} = 91;52^p$ as it’s the chord of $arc \; AB$ which remains unchanged in this iteration.
Thus, we can now use $\overline{AB}$ to help us convert a few other pieces:
$$\frac{91;52^p}{170;55^p} = \frac{\overline{ED}}{120^p}$$
$$\overline{ED} = 64;30^p$$
and
$$\frac{91;52^p}{170;55^p} = \frac{\overline{EA}}{77;53^p}$$
$$\overline{EA} = 41;52^p.$$
We can then use this to find $arc \; EA = 40;50º$.
This, then, can be added to $arc \; ABG$:
$$arc \; EABG = 177;12 + 40;50º = 174;11º.$$
We can then determine the corresponding chord, $\overline{EG} = 119;51^p.$
With that worked out, we’ll switch to the second diagram:
First, we’ll recall that we just determined $\overline{EG} = 119;51^p$ from which we can subtract $\overline{ED}$ which we found to be $64;30^p$. Thus, $\overline{DG} = 55;21^p.$
We then use the intersecting chords theorem to state:
$$\overline{ED} \cdot \overline{DG} = \overline{LD} \cdot \overline{DM}.$$
Plugging in the left side:
$$64;30^p \cdot 55;21^p = \overline{LD} \cdot \overline{DM}.$$
And for the right side, we’ll again swap in the relevant portion from Euclid II.5 giving us:
$$64;30^p \cdot 55;21^p = \overline{LK}^2 – \overline{DK}^2$$
where $\overline{LK} = 60^p$.
Thus, we can solve for $\overline{DK}$:
$$\overline{DK} = \sqrt{3600 – (64;30^p \cdot 55;21^p)}$$
$$\overline{DK} = 5;30^p.$$
Thus, we have the distance between the observer on Earth and the center of mean motion which agrees exactly with what Ptolemy derived.
We can now focus on the line of apsides.
First $\overline{GN} = \frac{1}{2} \overline{GE} = 59;55^p.$
But, we’ve also shown that $\overline{GD} = 55;21^p$, so we can subtract that off to find $\overline{DN} = 4;35^p$.
We’ll now look at $\triangle DKN$ and create a demi-degrees circle about it in which the hypotenuse, $\overline{DK} = 120^p$. We’ll first convert $\overline{DN}$ into that context using $\overline{DK}$ for the conversion:
$$\frac{120^p}{5;30^p} = \frac{\overline{DN}}{4;35^p}$$
$$\overline{DN} = 99;56^p.$$
We can then find $\angle NKD$:
$$\angle NKD = 120 \cdot sin(99;56º) = 56;23º.$$
This central angle is subtended by $arc \; XM$ so $arc \; XM = 56;23º$ as well.
Turning back to $arc \; EABG$, we determined it was $174;11º$. Half of it, $arc \; XG = 87;06º$.
Subtracting off $arc \; XM$ we find that $arc \; MG = 30;43º.$
Furthermore, we know that $arc \; BG = 33;26º$. So we can subtract $arc \; MG$ off to determine $arc \; BM = 2;43º.$
And since $arc \; AB = 99;55º$, we can subtract this and $arc \; BM$ off of the $180º$ semi-circle defined by $\overline{LG}$ to determined $arc \; LA = 77;22º.$
So to sum up. We find that:
From the apogee to the first opposition: $arc \; LA = 77;22º$
From the second opposition to perigee: $arc \; BM = 2;43º$
From the perigee to the third opposition: $arc \; MG = 30;43º.$
Ptolemy’s values are slightly different:
From the apogee to the first opposition: $arc \; LA = 77;15º$
From the second opposition to perigee: $arc \; BM = 2;50º$
From the perigee to the third opposition: $arc \; MG = 30;36º.$
I don’t have a good explanation on why our values differ here since Ptolemy doesn’t show any work and nether Toomer or Neugebauer have any commentary. Thus, I’ll chalk it up to accumulated differences in rounding errors, although I admit I’m surprised that I could agree exactly with Ptolemy for the eccentricity, and then get such large discrepancies in the last handful of steps.
Hopefully someone will spot the error if I’ve made one and let me know.
Ptolemy, thankfully, does not go through a further iteration. Instead, he states,
The above quantities have been accurately determined by this method, for the differences in the intervals [as measured along deferent and equant], when calculated from these data, are very nearly the same as the previous set.
Toomer and Neugebauer both agree with this. Toomer notes that a further iteration would produce a change in the eccentricity of less than $0;01^p$ and about $0;10º$ in the line of apsides.
Next, Ptolemy will demonstrate that these figures are good enough by showing they can reproduce one of the oppositions observed.
