Almagest Book IX: Position of Mercury About Epicycle 264 BCE Nov 14/15

Having determined the position of Mercury about its epicycle in Ptolemy’s time, he now sets out to do the same for an ancient observation so that the can determine the rate Mercury moves about the epicycle.

Here’s the observation:

[I]n the ${21}^{st}$ year of Dionysius’ calendar (which was in the ${484}^{th}$ year from Nabonassar), Scorpion $22$, [which is Thoth [I] $18/19$ in the Egyptian calendar [$264$ BCE, Nov. $14/15$], at dawn, Stilbon [i.e., Mercury] was $1$ moon to the rear of the straight line through the northern [star in the] forehead of Scorpius and the middle [star in the forehead], and was $2$ moons to the north of the northern [star in the] forehead. Now, according to our coordinates at that time, the midmost of the stars in the forehead of Scorpius had a longitude for $1\frac{2}{3}º$ into Scorpio, and is the same amount [$1\frac{2}{3}º$] south of the ecliptic, while the northernmost star had a longitude of $2\frac{1}{3}º$ into Scorpio and is $1\frac{1}{3}º$ north of the ecliptic. So, the planet Mercury had a longitude of about $3\frac{1}{3}º$ into Scorpio.

Furthermore, it is clear that it had not yet reached its greatest elongation as morning-star, since, $4$ days later, on Scorpion $26$, it is recorded that its distance from the same straight line towards the rear was $1\frac{1}{2}$ moons; for [by that time] the elongation had become greater, the sun having moved about $4$ degrees, but the planet [only] half a moon. And on Thoth $19$ at dawn, the longitude of the mean sun, according to our tables, was $20\frac{5}{6}º$ into Scorpio., while the longitude of the apogee of the planet was about $6º$ into Libra, since the $400$ or so years between the observations produce a displacement of the apogee of about $4º$.

Here’s the diagram we’ll use this time:

By and large, this one is the same as the previous one. The points are all even given the same lettering, so I won’t repeat all of them except to note that, when we dropped the perpendicular from $H$ onto $\overline{ZG}$ at point $M$, it looks like $B$ is on that line. It’s not. It’s just that $\mathrm{\angle }MHB$ is very small.

To begin, we’ll first determine the angle of the eccentres relative to the line of apsides. In other words, $\mathrm{\angle }ABH$ and $\mathrm{\angle }BGZ$.

This is slightly more complicated than in the last post. There, we were simply able to use Ptolemy’s determination of the solar position to determine where $\overline{GZ}$ pointed. Ptolemy tells us that this is $20\frac{5}{6}º$ into Scorpio, so we might be tempted to state that $\mathrm{\angle }BGZ=40\frac{5}{6}º$ (the $20º$ remaining in Libra from where the line of apsides points, plus the $20\frac{5}{6}º$ into Scorpio for this configuration).

However, we must be careful to recall that Ptolemy determined that Mercury’s line of apsides precesses at the same rate as the sun. This is what Ptolemy is telling us when he reminds us of the “$400$ or so years between the observations”.

Specifically, at that time, the line of apsides did not point to a position of $10º$ into Libra, but would have pointed to a point $6º$ into Libra. Therefore, the measure of $\mathrm{\angle }AGZ=\mathrm{\angle }ABH=44;50º$ (the $24º$ remaining in Libra plus the $20\frac{5}{6}º$ into Scorpio).

We can then state that $\mathrm{\angle }GBH$, the supplement of these angles is $135;10º$.

If we now turn our attention to $\mathrm{△}BGH$, this is an isosceles triangle since $\overline{BG}=\overline{BH}$. In it, we just determined $\mathrm{\angle }GBH$ and, since it is an isosceles triangle, the remaining to angles, $\mathrm{\angle }BGH$ and $\mathrm{\angle }BHG$ must be equal. Therefore, they’re each $22;25º$.

I solved this triangle by bisecting it and then using some trig, coming up with $\overline{GH}=5;{32}^p$. Ptolemy comes up with $5;{33}^p$, in the context in which the eccentricities $\overline{BG}$ and $\overline{BH}$ are both $3^p$.

Next, we’ll want to turn our attention to $\mathrm{△}GHM$. However, we’ll first need to determine one of the angles within it besides the right angle. So we’ll find $\mathrm{\angle }MGH$. It is the sum of $\mathrm{\angle }AGZ+\mathrm{\angle }BGH$, both of which we know. Thus,

$$\mathrm{\angle }MGH=44;50º+22;25º=67;15º.$$

Now we can create the demi-degrees circle about this triangle. This allows us to state that $arcHM=134;30º$ and looking up the corresponding chord, we find that $\overline{HM}=110;{40}^p$.

We can also determine the remaining angle in this triangle $\mathrm{\angle }MHG$ to be $22;45º$. Thus, the arc subtending this angle, $arcGM=45;30$ and the corresponding chord, $\overline{GM}=46;{24}^p$, all in the context where the hypotenuse, $\overline{GH}={120}^p$.

We just determined $\overline{GH}$ in the context in which the eccentricities all equal $3^p$, so we’ll use that to convert the other pieces of the triangle:

$$\frac{\overline{HM}}{110;{40}^p}=\frac{5;{33}^p}{{120}^p},$$

$$\overline{HM}=5;{07}^p.$$

$$\frac{\overline{GM}}{46;{24}^p}=\frac{5;{33}^p}{{120}^p},$$

$$\overline{GM}=2;{09}^p.$$

Ptolemy comes up with $2;{10}^p$ for the latter of these which I’ll adopt for consistency.

We can now determine $\overline{ZM}$ using the Pythagorean theorem on $\mathrm{△}ZMH$.

$$\overline{ZM}=\sqrt{{\overline{ZH}}^2–{\overline{HM}}^2}$$

Recalling that, $\overline{ZH}$ is the radius of one of the eccentres and thus has a measure of ${60}^p$,

$$\overline{ZM}=\sqrt{{{60}^p}^2–{5;{07}^p}^2}=59;{47}^p.$$

Next, we can add,

$$\overline{ZG}=\overline{GM}+\overline{ZM}=2;{10}^p+59;{47}^p=61;{57}^p.$$

Now, we’ll focus on $\mathrm{△}GDN$ and draw a demi-degrees circle about it.

In it, we know $\mathrm{\angle }DGN=44;50º$ as it’s a vertical angle to $\mathrm{\angle }AGZ$. Thus, the corresponding arc, $arcDN=89;40º$ and thus, the corresponding chord, $\overline{DN}=84;{36}^p$, in this context where the hypotenuse, $\overline{DG}={120}^p$.

As a quick side note, we can determine the remaining angle in this triangle, $\mathrm{\angle }GDN=45;10º$. Ptolemy doesn’t mention this, but it will become helpful later, at least, in the way I’ll approach something that Ptolemy doesn’t fully explain.

We can then use the Pythagorean theorem to determine $\overline{GN}=85;{06}^p$.

Again, we’ll translate these to the context in which the eccentricities are $3^p$:

$$\frac{\overline{DN}}{84;{36}^p}=\frac{3^p}{{120}^p},$$

$$\overline{DN}=2;{07}^p$$

And similarly, we can determine, $\overline{GN}=2;{08}^p$.

Now we’ll add $\overline{GN}$ onto $\overline{ZG}$ to determine $\overline{ZN}=61;{57}^p+2;{08}^p=64;{05}^p$.

We now know two of the sides in $\mathrm{△}ZDN$. Thus, we can use the Pythagorean theorem to state:

$$\overline{ZD}=\sqrt{{\overline{ZN}}^2+{\overline{DN}}^2}$$

$$\overline{ZD}=\sqrt{{64;{05}^p}^2+{2;{07}^p}^2}=64;{07}^p$$

We’ll now create a demi-degrees circle about this triangle and enter into that context, first converting all the sides to a context in which the hypotenuse, $\overline{ZD}={120}^p$:

$$\frac{\overline{DN}}{2;{07}^p}=\frac{{120}^p}{64;{07}^p},$$

$$\overline{DN}=3;{58}^p.$$

While we could determine the remaining side, $\overline{ZN}$, we won’t be needing it.

Instead, we’ll look up the arc corresponding to $\overline{DN}$ from the table of chords, for which I come up with $arcDN=3;47º$. Ptolemy gets $3;48º$ which I will, again, adopt to stay consistent.

This means that the angle it subtends, $\mathrm{\angle }DZN=1;54º$.

We can then determine the other non-right angle in this triangle, $\mathrm{\angle }ZDN=88;06º$.

Next, we can determine

$$\mathrm{\angle }ADZ=\mathrm{\angle }ZDN–\mathrm{\angle }GDN.$$

As a quick note, Ptolemy specifies that we’ll find $\mathrm{\angle }ADZ$ “by subtraction” but doesn’t mention what we should subtract. The above is my interpretation.

In contrast, Toomer suggests it’s $\mathrm{\angle }AGZ–\mathrm{\angle }DZN$. This certainly works out just fine and fits well with the pieces that Ptolemy just derived. However, to me at least, it’s less intuitive as it requires a separate proof whereas my suggestion follows directly from the geometry. However, it requires we know $\mathrm{\angle }GDN$ which I calculated above, but noted that Ptolemy didn’t investigate. Thus, I suspect Toomer is correct about the method in which Ptolemy intended this comment. My version is simply offered as a slightly simpler method.

Resuming the calculation:

$$\mathrm{\angle }ADZ=88;06º–45;10º=42;56º.$$

We’ll now turn back to the observation in which Mercury was observed, at $L$ to be $27;20º$ from the apogee (the $24;00º$ remaining in Libra + $3\frac{1}{3}º$ into Scorpio). Thus, $\mathrm{\angle }ADL=27;20º$.

Thus, we can subtract:

$$\mathrm{\angle }ZDL=\mathrm{\angle }ADZ–\mathrm{\angle }ADL=42;56º–27;20º=15;36º.$$

We’ll now construct a demi-degrees circle about $\mathrm{△}ZDX$. In this triangle, $\overline{ZD}={120}^p$, as it’s the hypotenuse.

Then, $arcZX=31;12º$ and its corresponding chord, $\overline{ZX}=32;{16}^p$.

We’ll now convert back to our overarching context:

$$\frac{\overline{ZX}}{32;{16}^p}=\frac{64;{07}^p}{{120}^p},$$

$$\overline{ZX}=17;{14}^p.$$

Ptolemy comes up with $17;{15}^p$ which I will, again, adopt.

We’ll now focus on $\mathrm{△}ZLX$, forming a demi-degrees circle about it. In it, the hypotenuse, $\overline{ZL}={120}^p$. But in our previous context, this has a length of $22;{30}^p$¹, which we can use to convert into this context.

$$\frac{\overline{ZX}}{17;{14}^p}=\frac{{120}^p}{22;{30}^p},$$

$$\overline{ZX}=91;{55}^p.$$

Ptolemy inexplicably rounds this off to an even ${92}^p$.

Ptolemy then calculates the corresponding arc, $arcZX$. He comes up with a value of $100;08º$. Toomer notes that, while there’s various methods by which you might calculate this from the table of chords, none of them quite return this value.

When I calculated it, I came up with $100;05º$². The closest Toomer was able to get with any method was $100;07º$ although he doesn’t offer any insight on what method was used. Regardless, I will again adopt Ptolemy’s value as we’re almost at the end here anyway.

This means that $\mathrm{\angle }ZLX=50;04º$. Then, the remaining angle in this triangle, $\mathrm{\angle }XZL=39;56º$.

Let’s zoom out a bit and consider $\mathrm{△}ZXD$ again. In it, we’ve already shown that $\mathrm{\angle }ZDL=15;36º$.

We can then use this to determine that $\mathrm{\angle }XZD=74;24º$.

If we then subtract $\mathrm{\angle }XZL$, which we just determined, from this, we get:

$$\mathrm{\angle }LZD=\mathrm{\angle }XZD–\mathrm{\angle }XZL=74;24º–39;56º=34;28º$$

We can then subtract from this, $\mathrm{\angle }DZN$, which we previously established was $1;54º$, to determine, $\mathrm{\angle }KZL=32;34º$.

Here, $\mathrm{\angle }KZL$ is the angle of Mercury after perigee (recalling that Mercury moves about the epicycle counter-clockwise). Thus, this indicates that Mercury was $212;34º$ after apogee at the time of this ancient observation.

This is then compared with the position we explored in the previous post from Ptolemy’s time at which point Mercury was $99;27º$ from apogee.

Ptolemy states the interval between these observations as $402$ Egyptian years, $283$ days, and $13\frac{1}{2}$ hours.

This interval contains $1268$ complete returns of the planet in anomaly (for $20$ Egyptian years  produce very nearly $63$ returns, so $400$ years produce $1260$, and the remaining $2$ years, plus the additional days, another $8$ complete returns).

We can get the first of these ($63$ returns in $20$ years) by looking back at the stated number of returns from IX.3. There, Ptolemy told us that Mercury makes $145$ returns in $46$ [solar] years. Thus,

$$\frac{145returns}{46years}=\frac{xreturns}{20years}$$

$$x=63.04returns$$

which we can round down to $63$ complete returns.

For the $400$ years, there would be $20$ times this amount or $1,260$ returns³

Then, we’ll repeat for the additional $2$ years, $283$ days, and $13\frac{1}{2}$ hours, which I calculate as $2.46,48$ years.

$$\frac{145returns}{46years}=\frac{xreturns}{2;46,48years}$$

$$x=8;30,45returns$$

Again, we only want the complete returns, so that rounds down to $8$⁴.

Thus, we have shown that in $402$ Egyptian years, $283$ days, and $13\frac{1}{2}$ hours, the planet Mercury moved in anomaly, beyond $1,268$ complete revolutions, $246;53º$, which is the amount by which the position at our observation is beyond the previous one.

In this post, we determined the historic position of Mercury to be $212;34º$ about the epicycle. Thus, to complete the revolution, it still had $147;26º$ to go. If we then add the $97;22º$ from the previous post (Ptolemy’s observation in his time) we then get $246;53º$.

So, if we incorporate the full revolutions, the total motion in anomaly is:

$$1,268\cdot 360º+246;53º=02,06,52,06;53º.$$

This motion was done in a period of

$$365\frac{days}{year}\cdot 402years+283days+0;33,45days$$

$$=40,50,13;33,45days.$$

Taking the ratio of these to find the degrees per day, I come up with a value of $3;06,24,06,58,39,48\frac{º}{day}$.

Ptolemy then directs us to compare this value to what we get out of the mean motion table.

And just about the same increment [in anomaly] results from the tables we set out before: for it was on the basis of these very same calculations that we made our correction to the periodic motions of Mercury, by reducing the above interval to days, and the above revolutions in anomaly plus the increment to degrees. For when the total of degrees is divided by the total of days, there results the mean daily motion in anomaly which we set out for Mercury in our previous discussion.

As Ptolemy states, this value is very similar to the one that is laid out in his table of mean motion, matching up to the third sexagesimal place exactly. Toomer notes this discrepancy and explores in in Appendix C of his translation.

There, he suggests that Ptolemy is falsely claiming that these calculations were used to derive the mean motion table. Rather, he suggests, Ptolemy took these mean motions from another source which he likely trusted, and simply performed these calculations as a check on it finding that three sexagesimal places of agreement is enough to affirm it.

Alternatively, he suggests that Ptolemy did indeed use the above equations as the basis for his calculations, but then applied some small correction which he did not explain. If that was the case, then the actual methodology would likely be unrecoverable.

Regardless, that ends this chapter and we only have one more, very short chapter left to go in Book IX.

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1. As demonstrated in this post.
2. I did this by using the chord table and, starting at an arc of $100º$ (as this was the one closest to a chord of ${92}^p$, successively added the sixtieths until I was as close to ${92}^p$ as possible.
3. If we were more careful with the math, we could do the following:

   $$\frac{145returns}{46years}=\frac{xreturns}{400years}$$

   $$x=1,260;52,12returns.$$

   Again, we would need to remove the fractional interval and only keep the complete rotations of $1,260$ returns. However, this does make it seem that there is some fractional component with which Ptolemy doesn’t grapple. However, including this makes the final result further from the values given in the Mean Motion tables than the one we’ll be ending up with shortly.

4. It does strike me that we’ve left $0;52,12$ and $0.30,45$ returns on the table here, so I feel like there should be one more return. Ptolemy does not explain why he does not include it.