With our corrections for the intervals from Saturn’s oppositions in hand, we’re ready to repeat the calculation. As with before, I’m doing this in a Google Sheet to speed things along. This means that I’ll be using modern trig and that the figures I give her may be subject to some rounding as we go since the Sheet is saving higher precision behind the curtain.
We’ll begin with the same diagram as before:
This time around, we’ll take $\angle BDG$ to be $34;38º$ since that was the apparent change from the point of view of the observer we found in the last post.
Since its vertical angle, $\angle EDH,$ has the same measure, we can enter into a demi-degrees context about $\triangle EDH$ in which the hypotenuse, $\overline{DE} = 120^p.$ We can then find $\overline{EH} = 68;12^p.$ Additionally, the remaining angle in this triangle, $\angle DEH = 55;22º.$
Next, $arc \; BG = 37;52º$ as this was the mean motion in ecliptic longitude which means $\angle BEG = 18;56º$ since it is subtended by this arc on the opposite side of the circle.
This can be added to $\angle DEH$ to determine $\angle BEH = 74;18º.$ Thus, the remaining angle in $\triangle BEH,$ $\angle EBH = 15;42º.$
Next, we’ll focus on $\triangle EBH$ forming a demi-degrees circle about it in which the hypotenuse, $\overline{BE} = 120^p.$
We can then use a bit of trig to determine $\overline{EH}$ in this context which I find to be $15;42^p.$
Since we now have $\overline{EH}$ in two contexts, we can use this to convert $\overline{EB}$ into the context in which $\overline{ED} = 120^p.$
$$\frac{68;12^p}{32;28^p} = \frac{\overline{EB}}{120^p}$$
$$\overline{EB} = 252;02^p.$$
Next, we’ll add the two apparent changes in ecliptic longitude from the view of the observer (post correction) together to determine $\angle ADG = 103;20º.$ The supplement, $\angle ADE = 76;40º.$
We’ll then create a demi-degrees circle about $\triangle DEZ$ in which the hypotenuse, $\overline{DE} = 120^p.$ In that context, we can find that $\overline{EZ} = 116;46^p.$
We also know that $arc \; ABG$ of this eccentre is $113;35º$ as that was the sum of the changes in longitude over the intervals in question. Thus, the angle it subtends on the other side of the circle, $\angle AEG = 56;47,30º.$
This is one of the angles in $\triangle ADE$ and we also determined that $\angle ADE = 76;40º,$ so the remaining angle $\angle DAE = 46;32,30º.$
That is also an angle in $\triangle ZAE$ which we will now create a demi-degrees circle about in which the hypotenuse, $\overline{AE} = 120^p$. This allows us to find $\overline{EZ} = 87;06º$ in this context.
We also know the length of this segment in our context in which $\overline{DE} = 120^p$ so we can use that to convert:
$$\frac{116;46^p}{87;10^p} = \frac{\overline{EA}}{120^p}$$
$$\overline{EA} = 160;52^p.$$
Next, we can note that $arc \; AB = 75;43º$ as this was the mean motion over the interval in question. Thus, $\angle AEB = 37;51,30º.$
We then look at $\triangle AE \Theta$ which contains this angle. We’ll create a demi-degrees circle about it in which the hypotenuse, $\overline{AE} = 120^p.$ We can then determine that $\overline{A \Theta} = 73;39º$ and $\overline{E \Theta} = 94;45º$ in that context. Which we’ll promptly convert into the context in which $\overline{ED} = 120^p$:
$$\frac{160;52^p}{120^p} = \frac{\overline{A \Theta}}{73;39^p}$$
$$\overline{A \Theta} = 98;43^p$$
and
$$\frac{160;52^p}{120^p} = \frac{\overline{E \Theta}}{94;45^p}$$
$$\overline{E \Theta} = 127;00^p.$$
We can then subtract $\overline{E \Theta}$ from $\overline{EB}$ to determine $\overline{B \Theta} = 125;01^p.$
We now know two sides in $\triangle AB \Theta$ so we can find the remaining side using the Pythagorean theorem:
$$\overline{AB} = \sqrt{98;43^2 + 125;01^2} = 159;18^p.$$
Next, we’ll return to $arc \; AB$ which was $75;43º.$ Thus, in the contxt in which the diameter of the eccentre is $120^p,$ the corresponding chord, $\overline{AB} = 73;39^p$ allowing us to convert into that context:
$$\frac{73;39^p}{159;18^p} = \frac{\overline{DE}}{120^p}$$
$$\overline{DE} = 55;29^p$$
and
$$\frac{73;39^p}{159;18^p} = \frac{\overline{EA}}{160;52^p}$$
$$\overline{EA} = 74;22^p.$$
The corresponding arc, $arc \; EA = 76;36º.$
We add this to $arc \; ABG$ to determine that $arc \; EABG = 190;11º.$
From that, we can determine the chord, $\overline{EG} = 119;32^p.$
We’ll now switch to our second diagram:
First, we’ll find that $\overline{DG} = 64;03^p$ as it’s $\overline{DE}$ subtracted off of $\overline{EG}.$
Then, we use the intersecting chords theorem. We’ve done this enough times that I’ll skip straight to the final calculation:
$$\overline{DK} = \sqrt{3600 – (55;29^p \cdot 64;03^p)} = 6;50^p$$
which is in exact agreement with Ptolemy.
We then set about finding the arcs between the oppositions and the line of line apsides.
We’ll first note that $\overline{GN} = \frac{1}{2} \overline{GE} = 59;46^p.$
This gets subtracted off of $\overline{GD}$ to determine $\overline{DN} = 4;17^p.$
We’ll then focus on $\triangle DKN,$ creating a demi-degrees circle about it in which the hypotenuse, $\overline{DK} = 120^p$ and use that to convert $\overline{DN}$ into that demi-degrees context:
$$\frac{120^p}{6;50^p} = \frac{\overline{DN}}{4;17^p}$$
$$\overline{DN} = 75;13^p.$$
Then, we can determine that $\angle DKN = 38;49º$ which is the same as $arc \; XM$ since this is the central angle that arc subtends.
Next, we’ll recall that $arc \; GX = \frac{1}{2} arc \; GXE$ which means that $arc \; GX = 84;55º.$
We’ll now look at $arc \; LGM$ which is $180º$ and subtract off $arc \; XM$ and $arc \; GX$ to determine $arc \; LG = 56;16º.$
Also, we know that $arc \; BG = 37;52º,$ which we can subtract off from $arc \; LG$ to determine $arc \; LB = 18;24º$.
Lastly, we know that $arc \; AB = 75;43º,$ allowing us to subtract off $arc \; LB$ to determine $arc \; AL = 57;19º.$
To sum up, we find that:
From the first opposition to the apogee: $arc \; AL = 57;19º$
From the apogee to the second opposition: $arc \; LB = 18;24º$
From the apogee to the third opposition: $arc \; LG = 56;16º.$
Ptolemy’s values are, again, slightly different:
From the first opposition to the apogee: $arc \; AL = 57;05º$
From the apogee to the second opposition: $arc \; LB = 18;38º$
From the apogee to the third opposition: $arc \; LG = 56;30º.$
And again, I don’t have any good explanation for why as I can’t spot any mistake and I’m rather surprised that I’m in exact agreement on the elongation but drift so significantly in the last few steps and neither Toomer nor Neugebauer call anything out here. So if you spot a mistake, please let me know!
Toomer does note that a further iteration finds corrections of $0;9.28º,$ $0;5,36º,$ and $0;9;40º$ which are pretty minimal and why Ptolemy doesn’t bother with another iteration.
We still have to validate these numbers by showing they can reproduce the original observations, but we’ll save that for the next post.
