Our next step in calibrating Saturn’s model, is to verify that the steps we’ve taken so far is able to reproduce the original observations.
We begin with a familiar diagram.
As we demonstrated in the last post, $\angle AZL = 57;05º$ which means that its vertical angle $\angle DZH$ is as well.
We immediately create a demi-degrees circle about $\triangle DZH$ in which its hypotenuse, $\overline{DZ} = 120^p.$ In it, $arc \; DH$ will be $114;10º$ and its supplement $arc \; ZH = 65;50º.$ The corresponding chords are $\overline{DH} = 100;44^p$ and $\overline{ZH} = 65;13^p.$
We then convert this into the context in which the diameter of the eccentres is $120^p$ and the distance between the centers is $3;25^p.$
$$\frac{3;25^p}{120^p} = \frac{\overline{DH}}{100;44^p}$$
$$\overline{DH} = 2;52^p$$
and
$$\frac{3;25^p}{120^p} = \frac{\overline{ZH}}{65;13^p}$$
$$\overline{ZH} = 1;51^p.$$
We can then look at $\triangle ADH$ in which we know two sides and determine the remaining side using the Pythagorean theorem:
$$\overline{AH} = \sqrt{60^2 – 2;52^2} = 59;56^p.$$
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 $\overline{A \Theta} = 61;47^p$ and $\overline{E \Theta} = 5;44^p.$
That gives us two sides of $\triangle AE \Theta$ so we can again use the Pythagorean theorem to determine
$$\overline{AE} = \sqrt{61;47^2 + 5;44^2} = 62;03^p.$$
We can then convert this into a demi-degrees context about this triangle in which the hypotenuse, $\overline{EA} = 120^p$
$$\frac{120^p}{62;03^p} = \frac{\overline{E \Theta}}{5;44^p}$$
$$\overline{E \Theta} = 11;05^p.$$
We can then find the corresponding arc, $arc \; E \Theta = 10;36º.$ This means the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EA \Theta = 5;18º.$
Jumping back to our starting point, we said that $\angle AZL = 57;05º.$ We can subtract $\angle EA \Theta$ from this to find $\angle AEL = 51;47º.$
That [$51;47º$], then was the amount by which the planet was in advance of the apogee at the first opposition.
We still need the position at the second opposition to be able to determine the interval so we’ll do that now.
From our last post, we determined that $\angle BZL = 18;38º.$ Again, the vertical angle $\angle DZH$ is the same, allowing us to create a demi-degrees circle about $\triangle DZH$ in which the hypotenuse, $\overline{DZ} = 120^p.$
In that context, $arc \; DH = 37;16º$ and its supplement, $arc \; ZH = 142;44º.$
The corresponding chords are, $\overline{DH} = 38;20^p$ and $\overline{ZH} = 113;43^p.$
We’ll convert these back into the context in which the diameters of the eccentres is $120^p$:
$$\frac{3;25^p}{120^p} = \frac{\overline{DH}}{38;20^p}$$
$$\overline{DH} = 1;05^p$$
and
$$\frac{3;25^p}{120^p} = \frac{\overline{ZH}}{113;43^p}$$
$$\overline{ZH} = 3;14^p.$$
We can then look at $\triangle DBH$ in which we now know two sides. So, we find the remaining side using the Pythagorean theorem:
$$\overline{BH} = \sqrt{60^2 – 1;05^2} = 59;59^p.$$
Next, we’ll use the fact that $\overline{\Theta H} = \overline{HZ}$ and $\overline{\Theta E} = 2 \cdot \overline{HD}$ to determine $\overline{B \Theta} = 63;13^p$ and $\overline{E \Theta} = 2;10^p.$ This gives us two sides of $\triangle EB \Theta$ so we can use the Pythagorean theorem to determine:
$$\overline{EB} = 63;15^p.$$
We then create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{EB} = 120^p$:
$$\frac{120^p}{63;15^p} = \frac{\overline{E \Theta}}{2;10^p}$$
$$\overline{E \Theta} = 4;07^p.$$
Next, we find the corresponding arc, $arc \; E \Theta = 3;56º.$ Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EB \Theta = 1;58º.$
Finally, we can subtract this from $\angle BZL$ to determine that $\angle BEL = 16;40º.$
Therefore, at the second opposition, the apparent position of the planet was $16;40º$ to the rear of the apogee. And we showed that, at the first opposition, it was $51;47º$ in advance of the same apogee. Therefore, the interval in apparent [longitude] from the first opposition to the second is computed as the sum of the above amount, $68;27º$, in agreement with the distance found from the observations.
We still need to find the second interval and thus, the third opposition.
Again, in our last post, we determined that $\angle GZL = 56;30º$ which is the same as its vertical angle $\angle DZH$.
We enter the demi-degrees circle about $\triangle DZH$ in which the hypotenuse, $\overline{DZ} = 120^p.$ In it, $arc \; DH = 113;00º$ and the supplement $arc \; ZH = 67;00º.$ The corresponding chords are $\overline{DH} = 100;04^p$ and $\overline{ZH} = 66;14^p.$
We now convert back to the context in which the diameters of the eccentres is $120^p$:
$$\frac{3;25^p}{120^p} = \frac{\overline{DH}}{100;04^p}$$
$$\overline{DH} = 2;51^p$$
and
$$\frac{3;25^p}{120^p} = \frac{\overline{ZH}}{66;14^p}$$
$$\overline{ZH} = 1;53^p.$$
We then find $\overline{GH}$ using the Pythagorean theorem:
$$\overline{GH} = \sqrt{60^2 – 2;51^2} = 59;56^p.$$
And then we can $\overline{\Theta H} = \overline{HZ}$ and $\overline{\Theta E} = 2 \cdot \overline{HD}$ to determine $\overline{G \Theta} = 61;49^p$ and $\overline{E \Theta} = 5;42^p.$
That’s two of the sides of $\triangle GE \Theta$ so we’ll again solve for the remaining side via the Pythagorean theorem:
$$\overline{GE} = \sqrt{61;49^2 + 5;42^2} = 62;05^p.$$
We’ll then create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{GE} = 120^p.$ We then convert into that context:
$$\frac{120^p}{62;05^p} = \frac{\overline{E \Theta}}{5;42^p}$$
$$\overline{E \Theta} = 11;01^p.$$
Next, we look up the corresponding arc, $arc \; E \Theta = 10;32º$ which means that the angle this arc subtends on the opposite side of the demi-degrees circle, $\angle EG \Theta = 5;16º.$
We subtract this from $\angle GZL$ to determine $\angle GEL = 51;14º.$
That [$51;14º$], then, is the amount by which the planet was to the rear of the apogee at the third opposition. And we showed that at the second opposition, it was $16;40º$ to the rear of the same apogee. So, the distance in apparent [longitude] from the second opposition to the third, is computed as the difference, $34;34º$, which is, again, in agreement with that derived from the observations.
So the model checks out.
But still remaining in this chapter is to determine the position of Saturn about its epicycle at one of the oppositions.
It is immediately clear, since the planet at the third opposition had a longitude of $14;14º$ into Capricorn, and was shown to be $51;14º$ to the rear of apogee, that the apogee of its eccentre had, at that moment, a longitude of $23;00º$ into Scorpio while its perigee was diametrically opposite at $23;00º$ into Taurus.
We’ll now draw a new diagram:
As we demonstrated previously, $arc \; LG = 56;30º.$ Additionally, $\angle EG \Theta = 5;16º$ (as we showed above) which means that $arc \; \Theta K$ of the epicycle is as well.
Thus, we can immediately determine that $arc \; H \Theta$ of the epicycle is
$$180º – 5;16º = 174;44º$$
which is the position of Saturn about its epicycle at the time of third opposition when the mean longitude was $56;30º.$
And that’s it for this chapter.
In the next one, we’ll determine the size of the epicycle. After that, we’ll again look at the period of mean motion, and then determine the epoch positions. That will catch our work for Saturn up to that of Mars and Jupiter.
