Now that we’ve done our initial determination of the eccentricity for Saturn, and the position of its line of apsides, we’ll work on making the first, and only, iterative correction to these. We’ll again start by finding corrections to the angles for each opposition beginning with a diagram that should look familiar.
Since we’ve been through this process so may times, I don’t think this image needs much explanation. But if you need a refresher, refer to our post where we did this for Mars or Jupiter.
We’ll begin by noting that $\angle NZX = 55;52º$ as this was the angle between the first position and apogee as we calculated in the last post. This means that $\angle DZH$ is as well since this is the vertical angle.
So, we’ll focus on $\triangle DZH$ which contains this angle, creating a demi-degrees context about it in which the hypotenuse, $\overline{DZ} = 120^p$. In this demi-degrees circle, $arc \; DH = 111;44º$ and its supplement, $arc \; HZ = 68;16º.$ We can then find the corresponding chords: $\overline{DH} = 99;20^p$ and $\overline{HZ} = 67;20^p.$
Next, since we know the distance between the eccentres in the context in which the diameter of the eccentres is $120^p$, we can use that to convert to that context1:
$$\frac{3;34^p}{120^p} = \frac{\overline{DH}}{99;20^p}$$
$$\overline{DH} = 2;57^p.$$
and
$$\frac{3;34^p}{120^p} = \frac{\overline{HZ}}{67;20^p}$$
$$\overline{HZ} = 2;00^p.$$
Now, since $\overline{DA}$ is a radius with a measure of $60^p$, we can use the Pythagorean theorem on $\triangle DAH$ to determine $\overline{AH}$:
$$\overline{AH} = \sqrt{60^2 – 2;57^2} = 59;56^p.$$
And as we’ve seen several times, $\overline{\Theta H} = \overline{HZ}$ and $\overline{\Theta E} = 2 \cdot \overline{HD}$ which allows us to determine that $\overline{A \Theta} = 61;56^p.$
We now know two sides of $\triangle AE \Theta$ so we can use the Pythagorean theorem to determine the remaining side:
$$\overline{AE} = \sqrt{61;56^2 + 5;54^2} = 62;13^p.$$
We’ll then consider a demi-degrees circle about $\triangle AE\Theta$ in which the hypotenuse, $\overline{AE} = 120^p.$ We’ll need to convert into that context:
$$\frac{120^p}{62;13^p} = \frac{\overline{E \Theta}}{5;54^p}$$
$$\overline{E \Theta} = 11;23^p.$$
Ptolemy comes up with $11;21^p$. Toomer agrees with my result, but we’ll take Ptolemy’s for consistency.
Next, we’ll find the corresponding arc, for which I find $arc \; E \Theta = 10;51º.$ This means that the angle which this arc subtends on the opposite side of the demi-degrees circle, $\angle EA \Theta = 5;25,30º.$
We’ll jump out of the demi-degrees context and back into the one in which the diameters of the eccentres is $120^p.$ In that context, $\overline{ZX}$ is a radius, so has a measure of $60^p$ to which we can add $\overline{Z \Theta}$ to find $\overline{Z \Theta} = 64^p.$ That gives us two sides of $\triangle E \Theta X$ so we can use the Pythagorean theorem to find the remaining side:
$$\overline{EX} = \sqrt{64^2 + 5;54^2} = 64;16^p.$$
We can now examine $\triangle EX \Theta$, creating a demi-degrees context about it in which the hypotenuse, $\overline{EX} = 120^p.$ We then convert into that context:
$$\frac{120^p}{64;16^p} = \frac{\overline{E \Theta}}{5;54^p}$$
$$\overline{E \Theta} = 11;01^p$$
for which Ptolemy finds $11;02^p.$
If we consider the corresponding arc, we find $arc \; E \Theta = 10;33º$ and the angle this arc subtends on the opposite side of the demi-degrees circle, $\angle EX \Theta = 5;16,30º.$
As with before, we can then subtract to find:
$$\angle AEX = \angle EA \Theta – \angle EX \Theta = 5;25,30º – 5;16,30º = 0;09º.$$
This is our correction for the first opposition.
[T]he planet at the first opposition, when viewed along $\overline{AE},$ had an apparent longitude of $1;13º$ into Libra. Thus, it is clear that if the epicycle centre were carried, not on [circle] $AL$, but on [circle] $NX,$ it would have been at point $X$ [at the first opposition], and the planet would have been seen along $\overline{EX}, 9’$ in advance of its actual position at $A,$ with a longitude of $1;04º$ into Libra.
Moving on to the second correction, we’ll reproduce the diagram for that situation:
In this new diagram, $\angle NDX = 19;51º$ as that’s the angle we calculated that Saturn appeared after apogee for the second opposition. Thus, its vertical angle, $\angle DZH = 19;51º$ as well.
We then create a demi-degrees circle about $\triangle ZDH$ in which the hypotenuse, $\overline{ZD} = 120^p$. In it, $arc \; DH = 39;42º$ and its supplement, $arc \; ZH = 140;18º.$ The corresponding chords are $\overline{DH} = 40;45^p$ and $\overline{ZH} = 112;52^p.$
This gets converted into the context in which the diameter of the eccentres is $120^p$:
$$\frac{3;34^p}{120^p} = \frac{\overline{DH}}{40;45^p}$$
$$\overline{DH} = 1;13^p$$
and
$$\frac{3;34^p}{120^p} = \frac{\overline{ZH}}{112;52^p}$$
$$\overline{ZH} = 3;21^p.$$
We can then look at $\triangle BDH$ in which we know two sides, allowing us to find the third using the Pythagorean theorem:
$$\overline{BH} = \sqrt{60^2 – 1;13^2} = 59;59^p.$$
And as we’ve seen several times, $\overline{\Theta H} = \overline{HZ}$ and $\overline{\Theta E} = 2 \cdot \overline{HD}.$
Thus, we can add $\overline{H \Theta}$ onto $\overline{BH}$ to determine $\overline{B \Theta} = 63;20^p.$ And since we also know $\overline{\Theta E} = 2;26^p$ we know two sides of $\triangle BE \Theta$ allowing us to find the third using the Pythagorean theorem:
$$\overline{EB} = \sqrt{63;20^2 + 2;26^2} = 63;23^p.$$
We then enter a demi-degrees context about this circle in which its hypotenuse, $\overline{EB} = 120^p$:
$$\frac{120^p}{63;23^p} = \frac{\overline{\Theta E}}{2;26^p}$$
$$\overline{\Theta E} = 4;36^p.$$
The corresponding arc, $arc \; \Theta E = 4;24º$ and the angle this arc subtends in the demi-degrees circle, $\angle \Theta BE = 2;12º.$
We can add $\overline{Z \Theta}$ onto that to determine $\overline{X \Theta} = 66;42^p$ which gives us two of sides of $\triangle XE \Theta$. We find the third with the Pythagorean theorem:
$$\overline{EX} = \sqrt{66;42^2 + 2;26^2} = 66;45^p.$$
We’ll then create a demi-degrees circle about $\triangle XE \Theta$ in which the hypotenuse, $\overline{EX} = 120^p.$
$$\frac{120^p}{66;45^p} = \frac{\overline{E \Theta}}{2;26^p}$$
$$\overline{E \Theta} = 4;22^p$$
which Ptolemy rounds up to $4;23^p.$
We then find the corresponding arc, $arc \; E \Theta = 4;11º$ for which Ptolemy finds $4;12º$. Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EX \Theta = 2;06º.$
Finally, we can subtract:
$$\angle BEX = \angle \Theta BE – \angle EX \Theta = 0;06º.$$
Here too, then, it is clear, since the planet, at the second opposition, when viewed along $\overline{EB},$ had a longitude of $9;40$ into Sagittarius, that if, instead, it were viewed along $\overline{EX},$ it would have a longitude of $9;46º$ into Sagittarius. And we showed that, at the first opposition, it would, on the same hypothesis, have had a longitude of $1;04º$ into Libra. Hence, it is clear that the interval in apparent [longitude] from the first to the second opposition, if it were taken with respect to the eccentre [circle] $NX,$ would be $68;42º$ of the ecliptic.
We’ll continue on to the final opposition for Saturn:
Again, from the previous post, we demonstrated that $\angle NZX = 57;43º$ which means its vertical angle, $\angle DZH$ does as well.
We then create a demi-degrees circle about $\triangle DZH$ in which its hypotenuse, $\overline{DZ} = 120^p.$
In that context, the arc opposite the above angle, $arc \; DH = 115;26º$ and its supplement, $arc \; ZH = 64;34º.$ We then find the corresponding chords: $\overline{DH} = 101;27^p$ and $\overline{ZH} = 64;06^p.$
Next, we’ll convert these to the context in which the diameter of the eccentres is $120^p.$
$$\frac{3;34^p}{120^p} = \frac{\overline{DH}}{101;27^p}$$
$$\overline{DH} = 3;01^p$$
and
$$\frac{3;34^p}{120^p} = \frac{\overline{ZH}}{64;06^p}$$
$$\overline{ZH} = 1;54^p.$$
This gives us two sides of $\triangle DGH$ since $\overline{DG}$ is a radius with a length of $60^p.$ Thus, we find the remaining side with the Pythagorean theorem:
$$\overline{GH} = \sqrt{60^2 – 3;01^2} = 59;55^p$$
which Ptolemy has rounded up to $59;56^p.$
And as we’ve seen several times, $\overline{\Theta H} = \overline{HZ}$ and $\overline{\Theta E} = 2 \cdot \overline{HD}.$
This allows us to add $\overline{H \Theta}$ onto $\overline{GH}$ to find $\overline{G \Theta} = 61;50^p.$ Additionally, $\overline{E \Theta} = 6;02^p.$
We then focus on $\triangle GE \Theta$ which we now have two of the sides for allowing us to find the remaining side via the Pythagorean theorem:
$$\overline{GE} = \sqrt{61;50^2 + 6;02^2} = 62;08^p.$$
We’ll then create a demi-degrees circle about this triangle wherein its hypotenuse, $\overline{GE} = 120^p.$ We’ll need to convert into that context:
$$\frac{120^p}{62;08^p} = \frac{\overline{E \Theta}}{6;02^p}$$
$$\overline{E \Theta} = 11;39^p.$$
Then we can look up the corresponding arc, $arc \; E \Theta = 11;09º$ which means that the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EG \Theta = 5;34,30º.$
We’ll again leave the demi-degrees context and add $\overline{Z \Theta}$ onto $\overline{ZX}$ finding that $\overline{X \Theta} = 63;48^p.$ This gives us two of the three sides of $\triangle EX \Theta$ so we again use the Pythagorean theorem to find the remaining side:
$$\overline{EX} = \sqrt{63;48^2 + 6;02^2} = 64;05^p.$$
Next, we create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{EX} = 120^p$ and convert into it:
$$\frac{120^p}{64;05^p} = \frac{\overline{E \Theta}}{6;02^p}$$
$$\overline{E \Theta} = 11;18^p.$$
We then look up the corresponding arc to find $arc \; E \Theta = 10;48º$ although Ptolemy finds $10;49º.$ This means that the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EX \Theta = 5;24º.$
Finally, we subtract to find:
$$\angle GEX = \angle EG \Theta – \angle EX \Theta = 5;34,30º – 5;24º \approx 0;10º.$$
Hence, since the planet at the third opposition, when viewed along $\overline{EG},$ had a longitude of $14;14º$ into Capricorn, it is clear that, if it had been on $\overline{EX},$ it would have had a longitude of $14;24º$ into Capricorn, and the interval from the second opposition to the third in apparent [longitude], taken with respect to eccentre $NX,$ would have been $34;38º.$
Thus, we’ve derived corrections for all three oppositions as well as redetermined the intervals based on these corrections.
In the next post, we’ll recalculate the eccentricity and line of apsides from this!
