The last planet we’ll need to determine the eccentricity and line of apsides for is Saturn. To do so, we’ll follow exactly the procedures we developed previously. And again, we’ll begin with three observations.
[1] The first of these was observed by us, using the astrolabe instruments, in the eleventh year of Hadrian, Pachon [IX] $7/8$ in the Egyptian calendar [$127$ CE, March $26/27$], in the evening, $1;13º$ into Libra.
[2] The second, in the seventeenth year of Hadrian, Epiphi [XI] $18$ in the Egyptian calendar, [$133$ CE, June $3$]. We computed the time and place of exact opposition from nearby observations as $4$ hours after noon on the $18^{th}$, $9;40º$ into Sagittarius.
[3] We observed the third opposition in the twentieth year of Hadrian, Mesore [XII] $24$ in the Egyptian calendar [$136$ CE, July $8$]. As before, we computed the time of exact opposition as having occurred precisely at noon on the $24^{th}$, and computed the place as $14;14º$ into Capricorn.
The observations here are a bit different than previous ones as, for the second and third, Ptolemy clearly did not observe the opposition. Instead, he notes that they were during the daytime when they could not be seen. Instead, Ptolemy tells us, he says they were determined from “nearby observations,” presumably the night before and after, or something similar. Then the times and positions could be interpolated. Ptolemy doesn’t speak to his method here.
Continuing on, Ptolemy again breaks the observations into pairs to create intervals:
[T]he first to the second [interval] comprises [in time], 6 Egyptian years, $70$ days, $22$ hours, [during which the] apparent motion of the planet [is] $68;27º.$
[T]he second to the third opposition comprises [in time], $3$ Egyptian years, $35$ days, $20$ hours, [during which the apparent motion was] $34;34º.$
If we take those intervals and look up the change in longitude from the mean motion tables, we can calculate the increase in mean longitude. I find them to be $75;43º$ for the first interval and $37;52º$ for the second.
And with that information, we produce a new diagram similar to the previous ones. Ptolemy doesn’t bother explaining this diagram and simply refers readers back to how it was laid out for Mars and Jupiter.
In this diagram, $\angle BDG = 34;34º$ as that was the apparent increase from the point of view of the observer. This is a vertical angle with $\angle EDH$.
Thus, we can enter a demi-degrees context about $\triangle EDH$ in which the hypotenuse, $\overline{DE} = 120^p$ and $arc \; HE = 69;08º.$
We then find the corresponding chord, $\overline{HE} = 68;05^p.$
Before taking our focus off this triangle, we can note that the remaining angle $\angle DEH = 55;26º.$
Next, $arc \; BG = 37;52º$ as this was the mean motion in ecliptic longitude. Thus, the angle it subtends on the opposite side of the eccentre, $\angle BEG = 18;56º$1.
We can add this angle to $\angle DEH$ to determine $\angle BEH = 74;22º$ which means that the remaining angle in $\triangle BEH$, $\angle EBH = 15;38º$2.
We’ll now look at $\triangle EBH$ forming a new demi-degrees context about it in which its hypotenuse, $\overline{BE} = 120^p.$ Additionally, $arc \; EH = 31;16º$ and we can find the corresponding chord, $\overline{EH} = 32;20^p.$
This gives us $\overline{EH}$ in two contexts, so we’ll begin converting everything into the context in which $\overline{ED} = 120^p$ wherein $\overline{EH} = 68;05^p.$
$$\frac{68;05^p}{32;20^p} = \frac{\overline{BE}}{120^p}$$
$$\overline{BE} = 252;41^p.$$
Next, we’ll note that $\angle ADG = 103;01º$ as this is the sum of the two apparent changes in ecliptic longitude. Thus, the supplement, $\angle ADE = 76;59º.$
We can now focus on $\triangle DEZ,$ forming a demi-degrees circle about it in which the hypotenuse, $\overline{DE} = 120^p.$ Then, $arc \; EZ = 153;58º$ and its corresponding chord, $\overline{EZ} = 116;55^p.$
Additionally, we know that $arc \; ABG$ of this eccentre is $113;35º$ as this is the sum of the changes in longitude over the intervals in question. This means that the angle this arc subtends on the other side of the circle, $\angle AEG = 56;47,30º.$
That’s one of the angles in $\triangle ADE$, and we also determined that $\angle ADE = 76;59º$ so the remaining angle, $\angle DAE = 46;13,30º.$
But, instead of focusing on that triangle, we’ll shift our focus to $\triangle ZAE$ which also contains this angle, and we’ll create a demi-degrees context about it in which the hypotenuse, $\overline{AE} = 120^p.$ Then, the arc opposite this angle, $arc \; EZ = 92;27º$ and the corresponding chord, $\overline{EZ} = 86;39^p.$
We also know the length of this segment in our context in which $\overline{DE} = 120^p$, allowing us to convert this piece to that context:
$$\frac{116;55^p}{86;39^p} = \frac{\overline{AE}}{120^p}$$
$$\overline{AE} = 161;55^p.$$
Ptolemy then repeats this process of looking at arcs, next looking at $arc \; AB$ which we calculated to be $75;43º.$ Thus, $\angle AEB = 37;51,30º.$
We can now look at $\triangle AE \Theta$ which contains this angle, and create a demi-degrees circle about it in which, the hypotenuse, $\overline{AE} = 120^p.$ Then, $arc \; A \Theta = 75;43º$, its supplement, $arc \; E \Theta = 104;17º,$ and its corresponding chord $\overline{A \Theta} = 73;39^p$ and $\overline{E \Theta} = 94;45^p.$
Again, we know $\overline{AE}$ in the context in which $\overline{DE} = 120^p$ so we’ll convert:
$$\frac{161;55^p}{120^p} = \frac{\overline{A \Theta}}{73;39^p}$$
$$\overline{A \Theta} = 99;23^p$$
and
$$\frac{161;55^p}{120^p} = \frac{\overline{E \Theta}}{94;45^p}$$
$$\overline{E \Theta} = 127;51^p.$$
We can then subtract this from $\overline{EB}$ which we showed previously was $252;41^p$ in this context to determine that $\overline{B \Theta} = 124;50^p.$
That gives us two sides in $\triangle AB \Theta$ so we can find the third using the Pythagorean theorem:
$$\overline{AB} = \sqrt{124;50^2 + 99;23^2} = 159;34^p.$$
We’ll then return to $arc \; AB$ which, as we’ve stated, was $75;43º.$ Thus, in the context 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. Doing so we find:
$$\frac{73;39^p}{ 159;34^p} = \frac{\overline{DE}}{120^p}$$
$$\overline{DE} = 55;23^p$$
and
$$\frac{73;39^p}{ 159;34^p} = \frac{\overline{EA}}{161;55^p}$$
$$\overline{EA} = 74;44^p$$
although Ptolemy comes up with $74;43^p.$
We can then determine the corresponding arc, $arc \; EA = 77;01º.$
That can be added to $arc \; ABG$ to determine that $arc \; EABG = 190;36º$ which leaves the arc on the other side of the circle, $arc \; GE = 169;24º.$ Then, we can take the corresponding chord to determine that $\overline{GE} = 119;29^p$ although Ptolemy somehow comes up with $119;28^p.$
We now produce a new drawing:
Again, this diagram is similar to the ones we created for Mars and Jupiter, so I’ll let those descriptions serve for this one as well.
First, Ptolemy reminds us that we determined $\overline{EG} = 119;28^p$ and $\overline{ED} = 55;23^p.$
Thus, we can subtract to determine that $\overline{GD} = 64;05^p.$
We again use the intersecting chords theorem to state:
$$\overline{ED} \cdot \overline{DG} = \overline{LD} \cdot \overline{DM}.$$
Plugging in:
$$55;23^p \cdot 64;05^p = \overline{LD} \cdot \overline{DM}.$$
And again, we can substitute the right side from Euclid’s II.5:
$$55;23^p \cdot 64;05^p = \overline{LK}^2 – \overline{DK}^2.$$
And since $\overline{LK}$ is the diameter of this eccentre with a measure of $60^p$, we can plug in and solve for $\overline{DK}$3:
$$\overline{DK} = \sqrt{3600 – (55;23^p \cdot 64;05^p)} = 7;08^p.$$
Next up, we’ll need to determine the arcs between the oppositions and the line of line apsides.
We’ll first note that $\overline{GN} = \frac{1}{2} \overline{GE} = 59;44^p.$
We can subtract that off of $\overline{GD}$ to determine $\overline{DN} = 4;21^p.$
Then, we’ll look at $\triangle DKN,$ 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:
$$\frac{120^p}{7;08^p} = \frac{\overline{DN}}{4;21^p}$$
$$\overline{DN} = 73;11^p.$$
This allows us to look up the corresponding arc for which I find $arc \; DN = 75;10º$ and thus, the angle which this arc subtends on the opposite side of the demi-degrees circle, $\angle DKN = 37;35º$ which is also the measure of the arc subtending this angle, $arc \; XM$4.
We’ll now recall that $arc \; GX = \frac{1}{2} arc \; GE$ which means that $arc \; GX = 84;42º.$
We can then subtract this arc and $arc \; XM$ off of the half-circle formed by $arc \; LBGXM$ to determine $arc \; LG = 57;43º.$
However, we know that $arc \; BG$ the change in ecliptic longitude along this eccentre, was $37;52º,$ so we can subtract to determine that $arc \; LB = 19;51º.$
We also determined that $arc \; AB = 75;43º$ from which we can subtract $arc \; LB$ to determine that $arc \; AL = 55;52º.$
Thus, we have determined the angular distances from each of the observed oppositions, but we still need to make corrections for the presence of the equant. And as with before, we’ll do that in future posts.
- Ptolemy does most of these calculations in his demi-degrees context in which all angles are doubled.
- Ptolemy doesn’t explain this well and simply treats this as $\angle EDH – \angle BEG$ which is not at all intuitive, but does produce the correct result.
- Toomer notes that even small rounding errors accumulate to a fairly large effect here. Toomer computes with higher accuracy, finding $\overline{DG} = 64;05,21^p$ and $\overline{ED} = 55;23,39^p,$ both of which are in pretty good agreement with Ptolemy’s values. However, when put calculating through to $\overline{DK},$ this results in the final value being $7;03,33^p$ which is notably further off than Ptolemy’s value.
- Toomer again notes that the accumulation of rounding errors produces a significant difference here. He computes with higher accuracy to find $arc \; XM = 38;01º$ – a difference of nearly half a degree!