Now that we have established the periodic motion, anomalies, and epochs of the planet Mars, we shall next deal with those of Jupiter in the same way.
It certainly strikes me that splitting up the planets between books is an odd choice on Ptolemy’s part, but at least the methods will be familiar. As Ptolemy tells us:
Once again, we first take, to demonstrate [the position on] the apogee and [the ratio of] the eccentricity, three oppositions [in which Jupiter is] directly opposite the mean sun.
This will be directly paralleling the work we did in X.7-10 for Mars. So bring on the observations!
[$1$] We observed the first of these by means of the astrolabe instrument in the seventeenth year of Hadrian, Epiphi [XI] $\frac{1}{2}$ in the Egyptian calendar [$133$ CE May $17/18$], $1$ hour before midnight, $23;11º$ into Scorpio.
[$2$] The second in the twenty-first year [of Hadrian], Phaophi [II] $13/14$ [$136$ CE, August $31$/September $1$], $2$ hours before midnight, $7;54º$ into Pisces.
[$3$] And the third in the first year of Antoninus, Athyr [III] $20/21$] [$137$ CE, October $7/8$] $5$ hours after midnight, $14;23º$ into Aries.
As with before, Ptolemy pairs the observations off and breaks them into intervals, first getting the apparent changes from the point of view of the observer.
[T]he first to the second opposition comprises [in time], $3$ Egyptian years, $106$ days, $23$ hours, [during which the] apparent motion of the planet [is] $104;43º$.
[T]he second to the third opposition comprises [in time], $1$ Egyptian year, $37$ days, $7$ hours, and [in true longitude], $36;29º$.
Then, he computes the changes along the circle of mean motion in the interval of time:
For the first interval: $99;55º$. For the second interval $33;26º$.
My own calculations agree with these.
From this information, we’ll create a diagram very similar to the one we started this post with:
As with before, points $A$, $B$, and $G$ are the three observations on the eccentre of mean distance. Point $D$ is the observer on Earth (not at the center of this circle).
We’ve extended $\overline{GB}$ until it hits the other side of the circle at $E$, and then connected $\overline{EA}$, $\overline{EB}$ and $\overline{AB}$.
We’ve also extended $\overline{BD}$ and dropped perpendiculars from $E$ onto the extended $\overline{BD}$ at point $H$, as well as onto $\overline{AD}$ at $Z$. Lastly, a perpendicular was dropped from $A$ onto $\overline{EB}$.
To start, $\angle BDG = 36;29º$1 since this is the angle that describes the apparent change from the point of view of the observer. As is $\angle EDH$ since it’s a vertical angle.
We’ll then create a demi-degrees context about $\triangle EDH$ in which the hypotenuse, $\overline{ED} = 120^p$. In it, $arc \; EH = 72;58º$. We can then find the corresponding chord. Doing so, I find $\overline{EH} = 71;21^p$, in agreement with Ptolemy.
Coutinuing to focus onthis triangle for a moment, we can also find $\angle DEH = 53;31º$ since we know the other two angles in this triangle.
Now, let’s look at $\angle BEG$. This angle is subtended by $arc \; BG$ which is the change in longitude over that interval of time and we stated was $33;26º$ above. Thus, $\angle BEG = 16;43º$.
We can then find $\angle BEH$ by adding:
$$\angle BEH = \angle BEG + \angle DEH$$
$$\angle BEH = 16;43º + 53;31º = 70;14º.$$
Then, focusing on $\triangle BEH$, we can find the remaining angle in it, $\angle EBH$ to be $19;46º$
This allows us to create a demi-degrees context about $\triangle BEH$. In it, the hypotenuse, $\overline{EB} = 120^p$ and $arc \; EH = 39;32º$.
We can then look up the corresponding chord for which I find $\overline{EH} = 40;35^p$, again in agreement with Ptolemy.
This gives us $\overline{EH}$ in two contexts. So we can use that to convert $\overline{EB}$ into the context in which $\overline{EH} = 71;21^p$:
$$\frac{71;21^p}{40;35^p} = \frac{\overline{EB}}{120^p}$$
$$\overline{EB} = 210;58^p.$$
We’ll now look at $\angle ADG$ which is the sum of the apparent change in both intervals or $104;43º + 36;29º = 141;12º$.
Therefore, its complement, along $\overline{EG}$ is $\angle ADE = 38;48º$.
Next, we’ll focus on $\triangle DEZ$, creating a demi-degrees context about it in which the hypotenuse, $\overline{ED} = 120^p$.
In it, $arc \; EZ = 77;36º$, so the corresponding chord, $\overline{EZ} = 75;12^p$.
Now, let’s look at $arc \; ABG$. This is the change in longitude between the two intervals or $99;55º + 33;26º = 133;21º$. Therefore $\angle AEG$, which this arc subtends, is $66;40,30º$.
We’ll now focus on $\triangle EAD$. In it, we have found two of the three angles: $\angle AED = 66;40,30º$ and $\angle ADE = 38;48º$. Thus, the remaining angle, $\angle EAZ = 74;31,30º$.
That’s one of the angles in $\triangle AEZ$ which we’ll now focus on, creating a demi-degrees context about it, in which the hypotenuse, $\overline{AE} = 120^p$.
In it, $arc \; EZ = 149;03^p$ and we can find the corresponding chord, $\overline{EZ} = 115;39^p$.
That now gives us $\overline{EZ}$ in two contexts. So we can use that to convert $\overline{EA}$ into the context where $\overline{ED} = 120^p$:
$$\frac{75;12^p}{115;39^p} = \frac{\overline{EA}}{120^p}$$
$$\overline{EA} = 78;02^p.$$
Next, let’s check out $\angle AEB$. This angle is subtended by $arc \; AB$ which is the change in ecliptic longitude between the first and second opposition or $99;55º$. Thus, this angle is $49;57,30º$.
We’ll now focus on $\triangle AE \Theta$, creating a demi-degrees context about it in which the hypotenuse $\overline{AE} = 120^p$.
In it, $arc \; A \Theta = 99;55º$ and its supplement, $\overline{E \Theta} = 80;05º$. We can then find the corresponding chords: $\overline{A \Theta} = 91;52^p$ and $\overline{E \Theta} = 77;12^p$.
Since we have $\overline{AE}$ in two contexts, we can now use that to convert these two chords into the context in which $\overline{DE} = 120^p$:
$$\frac{78;02^p}{120^p} = \frac{\overline{A \Theta}}{91;52^p}$$
$$\overline{A \Theta} = 59;44^p$$
and
$$\frac{78;02^p}{120^p} = \frac{\overline{E \Theta}}{77;12^p}$$
$$\overline{E \Theta} = 50;12^p.$$
Next, recall that we showed that $\overline{BE} = 210;58^p$ in this context. Therefore, we can subtract to find
$$\overline{B \Theta} = \overline{BE} – \overline{E \Theta}$$
$$\overline{B \Theta} = 210;58^p – 50;12^p = 160;46^p.$$
We’ll now focus on $\triangle AB \Theta$. Since we now know two sides of it in this context ($\overline{A \Theta}$ and $\overline{B \Theta}$), we can use the Pythagorean theorem to find the third:
$$\overline{AB} = \sqrt{\overline{A \Theta}^2 + \overline{B \Theta}^2}$$
$$\overline{AB} = \sqrt{59;44^2 + 160;46^2} = 171;30^p.$$
Ptolemy reminds us that this is in the context in which $\overline{ED} = 120^p$ and $\overline{EA} = 78;02^p$.
However, in the context in which the diameter of the eccentre is $120^p$, we can find the length of $\overline{AB}$ in that context since it is subtended by $arc \; AB$ which was the change in ecliptic longitude between the first and second opposition or $99;55º$. Thus, its chord is $\overline{AB} = 91;52^p$ in that context.
We can now use that to convert a few pieces:
$$\frac{91;52^p}{171;30^p} = \frac{\overline{ED}}{120^p}$$
$$\overline{ED} = 64;17^p$$
and
$$\frac{91;52^p}{171;30^p} =\frac{\overline{EA}}{78;02}$$
$$\overline{EA} = 41;48^p$$
although Ptolemy rounds down to $41;47^p$.
Using $\overline{EA}$ we can now look up the corresponding arc which I find to be $arc \; EA = 40;47º$. Ptolemy finds $40;45º$.
This can then be added to $arc \; ABG$ to find that
$$arc \; EABG = 40;45º + 133;21º = 174;06º.$$
We can immediately see this isn’t dividing the circle in half, and to prove it further, Ptolemy takes the corresponding chord to find $\overline EG = 119;50^p$ where the diameter of the eccentre is $120^p$.
Now, segment $EABG$ is less than a semi-circle, so the centre of the eccentre will fall outside it.
Ptolemy again uses this to indicate that we will need to iterate to get a better approximation, which we will come to shortly. However, for the remainder of this post, we’ll finish out finding our first approximation for the apogee and eccentricity.
To continue, though, we’ll need a new diagram:
In this figure, $\overline{LM}$ is the line of apsides with $K$ as the center of this eccentre and $D$ as the observer on Earth. Points $A$, $B$, and $G$ remain the three observations Ptolemy has given and $E$ remains the extension of $\overline{GD}$. Lastly, we’ve dropped a perpendicular onto $\overline{EG}$ from $K$ at $N$, extending it to the edge of the eccentre at $X$.
In this, the diameter, $\overline{LM} = 120^p$ and $\overline{EG}$ as we just showed, is $119;50^p$. We also showed that $\overline{ED} = 64;17^p$.
Thus, we can subtract to determine $\overline{DG} = 55;33^p$.
We’ll then use the intersecting chords theorem again to state:
$$\overline{ED} \cdot \overline{DG} = \overline{LD} \cdot \overline{DM}.$$
We can immediately plug in the left side of this equation from what we have:
$$64;17^p \cdot 55;33^p = \overline{LD} \cdot \overline{DM}.$$
And for the right side, we can again make use of Euclid II.5 which tells us
$$\overline{LD} \cdot \overline{DM} + \overline{DK}^2 = \overline{LK}^2.$$
So doing a bit of rearranging and substituting in, we get:
$$64;17^p \cdot 55;33^p = \overline{LK}^2 – \overline{DK}^2.$$
And since $\overline{LK} = 60^p$, being the radius of the eccentre, we can solve for $\overline{DK}$:
$$\overline{DK} = \sqrt{3600 – 3570;56}$$
$$\overline{DK} = 5;23^p.$$
This is the distance between the observer on Earth and the center of mean motion2.
Moving on to the line of apsides, $\overline{GN} = \frac{1}{2} \overline{GE}$ meaning $\overline{GN} = 59;55^p$.
But we’ve also shown that $\overline{GD} = 55;33^p$, so we can subtract that off to find $\overline{DN} = 4;22^p$.
We can now concentrate on $\triangle DKN$, entering a demi-degrees context 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 since we know it in both contexts:
$$\frac{120^p}{5;23^p} = \frac{\overline{DN}}{4;22^p}$$
$$\overline{DN} = 97;20^p.$$
This allows us to find the corresponding arc. Looking this up in the chord tables, I find $arc \; DN = 108;25º$, but Ptolemy finds $108;24º$.
Thus, the angle this arc subtends, $\angle NKD = 54;12º$. This is a central angle for this eccentre, so the arc subtending it, $arc \; XM = 54;12º$ also.
Let’s turn back to $arc \; EABG$. We previously said that this was $174;06º$. However, because $\overline{KX}$ is perpendicular to its chord, $\overline{KX}$ bisects both the chord and the arc. Thus, $arc \; XG = 87;03º$.
We can then subtract off $arc \; XM$ to determine $arc \; MG = 32;51º$3.
Furthermore, we know that $arc \; BG = 33;26º$, we can subtract $arc \; MG$ to determine $arc \; BM = 0;35º$4.
And since $arc \; AB$ was $99;55º$, we can subtract this and $arc \; BM$ off of the $180º$ semi-circle defined by $\overline{LG}$ to determine that $arc \; LA = 79;30º$.
Taking stock quickly, we’ve now determined the position of the three observations relative to the line of apsides along the circle of mean motion. The first was $79;30º$ away from apogee, the second $0;35º$ before perigee, and the last $32;51º$ after perigee.
However, we still need to go through our iterations to improve this since we didn’t have the right angles at the outset. But we’ll tackle that in a future post.
- Ptolemy performs this series of calculations in the demi-degrees context in which all angles are doubled.
- Toomer notes that this is slightly high. As we’ve seen going through this calculation, Ptolemy has a few minor rounding errors. Their accumulated effect is that this value is high by $0;03^p$, so a better value would be $5;20^p$.
- Toomer notes that Ptolemy has accumulated some serious rounding error here. A more precise value would have been $32;21º$, so we’re off by half a degree!
- Toomer notes that the “smallness of the corrections for this and the next opposition shows that these oppositions have been badly chosen. To display the greatest difference between the simple eccentric and equant models, all three oppositions should be near the octants (as they are for Mars).