Having demonstrated the conceptual model, Ptolemy now turns to determining
in what part of the ecliptic Mercury’s apogee lies by the following method.
This is done by using
observations of greatest elongations in which the distance [of Mercury] as a morning-star from the mean longitude of the sun (i.e., from the mean longitude of the planet) is equal to its distance as an evening-star. For once we had found such a situation, it necessarily follows from our [previous] demonstrations that the point on the ecliptic halfway between the two positions [of Mercury as a morning-star and evening-star] occupies the apogee of the eccentre.
First off, we should remember that Mercury and Venus are inferior planets. In modern astronomy, this means they are closer to the sun than us which limits the angular distance they can have from the sun (i.e., their greatest elongation).
For Mercury, this distance is actually rather complicated as its orbit is notably elliptical1 which means that, unless you’re capturing it when it’s actually at its greatest distance from the sun, even if it’s in the right position2, then you won’t observe the true maximum elongation.
So what is the absolute best case scenario?
Let’s sketch it out.
In this diagram, $\overline{SM} = 70$ million km and $\overline{SE} = 150$ million km. Also, $\angle SME$ is a right angle since $\overline{EM}$ is tangent to the radius. Thus, we can solve for $\angle SEM$ with a bit of trig. Doing so, I find the maximum elongation to be $27.8º$.
But again, that’s an absolutely ideal scenario in which Mercury is at aphelion and tangent in its orbit while earth is at perihelion. A rare event which Ptolemy acknowledges stating,
The observations which we used for this purpose are few in number, because precisely such combinations [of planet/sun positions] rarely occur.
And even with this admission, he doesn’t quite achieve the ideal scenario, as we’ll see3.
But before getting there, let’s think a bit more about how this technique works.
Essentially, it makes use of the symmetries we explored in the lasts two posts. In them, we showed that the anomaly is mirrored on both sides of the apsidal line.
We discussed above what elongation means in the context of modern astronomy, but now consider what it means in the Ptolemaic model – it’s when the anomalies are at their maximum in such a way that they add to (in the case of Mercury being visible in the evening) or subtract from (in the case of Mercury being visible in the morning) the ecliptic longitude4. Thus, the elongation and anomaly are directly related and if we can find two positions where the observed elongation is equal with one with Mercury as a morning-star and the other as an evening-star, then this means that they’re being mirrored along the line of apsides due to the symmetry previously discussed.
To be fair, the anomaly (and therefore the elongation) are more complicated than that since Mercury’s model is not a simple epicycle or eccentre. Thus, the notion that this method should work is flawed unless the extra components of the anomaly are somehow either minimized at these points or cancel out in some manner. If so, it’s not obvious as to why this should be the case.
Regardless, I’ll show a diagram here in a moment that will hopefully make the core idea (mirroring across the line of apsides) more clear, but first let’s cover the observations Ptolemy gives so they can be included in the diagram.
[$1$] In the sixteenth year of Hadrian, Phamenoth [VII] $16/17$ in the Egyptian calendar [$132$ Feb. $2/3$], in the evening, we observed Mercury, by means of the astrolabe [armillary sphere] instrument, at its greatest distance from the mean longitude of the sun. Also, from a sighting with respect to the bright star in the Hyades5, it was seen then to occupy a longitude of $1º$ into Pisces. At the time in question, the sun’s mean longitude was $9 \frac{3}{4}º$ into Aquarius. So the greatest elongation from the mean as an evening star comes out as $21 \frac{1}{4}º$
There’s two parts to this observation. In the first part, Ptolemy attempts to determine Mercury’s position on this date by measuring the angular distance along the ecliptic between it and the sun using an armillary sphere.
This is a pretty challenging observation for several reasons. The most obvious is that this would have to be done as the sun is setting, but before it is set as it would otherwise be impossible to sight the sun. To do that, you’ll need to actually find Mercury in the sky before it’s fully dark. This is possible as its brightness at maximum elongation is $-0.3$ making it among the brightest objects in the sky. The only stars brighter than that are Sirius ($-1.46$), Canopus ($-0.74$), and α Centauri ($-0.27$), none of which are anywhere near the ecliptic meaning that Mercury would be the first object visible in this region of the sky.
Second, is that observing the sun or any other object near the horizon will result in atmospheric refraction throwing off your measurement6.
Although Ptolemy doesn’t address refraction, perhaps his awareness of it is why he attempts to reinforce the accuracy of his position for Mercury by also measuring its distance (again, along the ecliptic) from the bright star α Tau. However, as Toomer notes, Ptolemy is claiming a position of Mercury of $331º$ in ecliptic longitude while α Tau is given a longitude of $42 \frac{2}{3}º$. This is a difference in ecliptic longitude of nearly $72º$ which is a large angle to measure accurately. Toomer questions the truth of this observation.
Regardless, Ptolemy is giving the position of Mercury at this time at $1º$ into Pisces when it had an elongation of $21 \frac{1}{4}º$.
[$2$] An, in the eighteenth year of Hadrian, Epiphi [XI] $18/19$ in the Egyptian calendar [$134$ June $3/4$], at dawn, Mercury [was observed] at greatest elongation, appearing very small and dim; from a sighting with respect to the [same] bright star in the Hyades, it was seen to occupy [the position of] $18 \frac{3}{4}º$ into Taurus. Now, at that time, the mean sun was $10º$ into Gemini. Here too, then, the greatest elongation from the mean [sun] as morning-star was $21 \frac{1}{4}º$, equal [to the elongation in [$1$]].
This time, no observation is made directly against the sun and the ecliptic longitude is only given with respect to α Tau.
This observation, combined with the last provides a pair of observations with equal and opposite7 elongations which can be used for the purpose described above with no further ones. However, Ptolemy gives an additional set of observations to further validate the line of apsides this would imply, so I’ll include those before moving on.
[$3$] Again, in the first year of Antoninus, Epiphi [XI] $20/21$ in the Egyptian calendar [$138$ June $4/5$], in the evening, we observed Mercury by means of the astrolabe at its greatest distance from the sun’s mean longitude. From a sighting at that moment with respect to the star on the heart of Leo8, it was seen to occupy [the position of] $7º$ into Cancer. But at the time in question, the mean sun was $10 \frac{1}{2}º$ in Gemini. Therefore, the greatest elongation [of Mercury] as evening star comes out as $26 \frac{1}{2}º$9.
[$4$] Similarly, in the fourth year of Antoninus, Phamenoth [VII] $18/19$ in the Egyptian calendar [$141$ Feb $1/2$], at dawn, [Mercury was observed], again, at greatest elongation: from a sighting with respect to the star called Antares it was seen to occupy [a position of] $13 \frac{1}{2}º$ into Capricorn, while the mean sun was $10º$ into Aquarius. Here too, then, the greatest elongation from the mean as morning-star was $26 \frac{1}{2}º$, equal [to the elongation in [$3$]].
This gives us the second pair of observations.
So now let’s sketch all that out.
The first thing to note is that this diagram is from the point of view of the observer. There is no reference to the model for Mercury here. We’re simply laying out the positions along the ecliptic for Mercury at these times as observed from earth.
Second, we can see in this diagram that Ptolemy cleverly chose positions for [$1$] and [$4$] in which the sun was at the same ecliptic longitude as well as for observations [$2$] and [$3$].
For now, we can clearly see that there’s a line we could draw that would bisect the circle and about which, the observations would be a reflection. That line is the line of apsides we’re searching for10.
But we can’t simply eyeball it; we’ll need to be more rigorous than that.
Ptolemy doesn’t show the math, but we can reason it out pretty easily.
At position [$1$], Mercury was observed to be at $331º$ ecliptic longitude and at [$2$] it was at $48;45º$ which is a difference of $77;45º$. The line of apsides should bisect this difference which means it should intersect the ecliptic $\approx 38;53º$ from either of these points. Thus, subtracting this from [$2$], we find it should intersect at $9;53º$ into Aries which Ptolemy rounds off to $10º$.
The line of apsides would then go through the observer and likewise intersect the ecliptic $10º$ into Libra.
We could repeat this calculation also with the solar position or points [$3$] and [$4$], but would get the same result11.
At this point, Ptolemy is not looking into whether the point in Aries is apogee or perigee, but he’ll return to that.
Instead, he next asks whether the positioning of the lines of apsides is consistent or whether it too precesses12?
To determine this, he essentially repeats the procedure adding several new points earlier in time.
[$5$] In the $23^{rd}$ year in Dionysiys’ calendar, Hydron $21$, at dawn, Stilbon13 was $3$ moons14 to the north of the brightest star in the tail of Capricorn15. At that time, the star in question had a position, according to [the coordinate system defined by] our origin, namely that beginning with the solstical or equinoctial points, of $22 \frac{1}{3}º$ into Capricorn. Mercury, obviously, had the same longitude, and the mean sun was at $18 \frac{1}{6}º$ into Aquarius: for that moment was in the $486^{th}$ year from Nabonassar, Choiak [IV] $17/18$ in the Egyptian calendar [$261$ BCE, Feb $11/12$], dawn. Therefore, the greatest elongation from the mean [of Mercury] as morning-star was $25 \frac{5}{6}º$.
There’s an odd phrase in this section in which Ptolemy refers to “[the coordinate system defined by] our origin.” Obviously, the part in brackets is from Toomer attempting to clarify and smooth out Ptolemy’s language, but what is being referred to here is that this is sufficiently far back in time that Ptolemy will need to correct for the precession of the equinoxes. Indeed, this was a period of $398$ years, and so Ptolemy subtracts $4º$16 from the star’s position as given in the star catalog17 to give its position in that time.
Ptolemy then notes that he was unable to find an exact match (i.e., where the elongation was $25 \frac{5}{6}º$ with Venus as an evening-star) from the observations passed down to him from that same era. Instead, he uses two that are reasonably close:
[$6$] [Firstly], in the same $23^{rd}$ year in Dionysius’ calendar, Tauron $4$, in the evening, [Mercury] was $3$ moons behind [i.e., to the rear of] the straight line through the horns of Taurus and it seemed as if it was going to be more than $3$ moons to the south of that [star] common [to Auriga and Taurus18] when it passed by it. Thus, its position according to our coordinates was $23 \frac{2}{3}º$ into Taurus. That moment was again in the $486^{th}$ year from Nabonassar, [Mechir [VI]] $30$/Phamenoth [VII] $1$ in the Egyptian calendar [$261$ BCE, April $25/26$], evening, at which time the longitude of the mean sun was $29 \frac{1}{2}º$ into Aries. So the greatest elongation from the mean as evening star was $24 \frac{1}{6}º$.
[$7$] [Secondly], in the $28^{th}$ year in Dionysius’ calendar, Didymon $7$, in the evening, [Mercury] was practically on a straight line with [the stars in] the heads of Gemini, and lay to the south of the southern one by $\frac{1}{3}$ of a moon less than twice the distance between [the stars in] the heads. This moment is in the $491^{st}$ year from Nabonassar, Pharmouthi [VIII] $5/6$ in the Egyptian calendar [$256$ BCE, May $28/29$], evening, at which time the longitude of the mean sun was $2 \frac{5}{6}º$ into Gemini. Thus, this [greatest] elongation was $26 \frac{1}{2}º$.
So how do these two observations make up for the lack of a match in elongation from [$5$]?
Recall that the elongation from [$5$] was $25 \frac{5}{6}º$. These two observations bracket that with [$6$] being $24 \frac{1}{6}º$ and [$7$] being $26 \frac{1}{2}º$. Thus, Ptolemy interpolates using these two observations:
We derived the location of the mean position for a [greatest] evening elongation of $25 \frac{5}{6}º$ from the difference between the above two observations [[$6$] and [$7$]]: The difference between the mean positions [of the sun] at the two observations is $33 \frac{1}{3}º$, and the difference between the greatest elongations [was] $2 \frac{1}{3}º$. Thus, to $1 \frac{2}{3}º$ (which is the amount by which $25 \frac{5}{6}º$ exceeds $24 \frac{1}{6}º$) correspond to [a position of the mean sun of] approximately $24º$.
I’ll pause here to walk through the math a bit more slowly.
First, the two observations don’t perfectly bracket the desired elongation. The difference between the two, as Ptolemy states, is $2 \frac{1}{3}º$. The desired elongation is $1 \frac{2}{3}º$ higher than [$6$] or $\frac{2}{3}º$ lower than [$7$]. Thus, we need to apply this same fraction ($1 \frac{2}{3}º \div 2 \frac{1}{3}º$) to the difference in the sun’s mean position as well.
The difference in the sun’s mean position was $33 \frac{1}{3}º$ so we’ll do:
$$33 \frac{1}{3}º \cdot 1 \frac{2}{3}º \div 2 \frac{1}{3}º = 23;48$$
Which Ptolemy rounds off to $24º$19.
If we add this amount to $29 \frac{1}{2}º$ in Aries, we shall get the mean position at which the greatest evening elongation is calculated to be equal to the greatest morning elongation of $25 \frac{5}{6}º$.
Doing this math, $29;30º + 24º = 53;30º$ which is $23 \frac{1}{2}º$ into Taurus which is what Ptolemy gets.
Thus, Ptolemy is ready to calculate the half way point between [$5$] and the interpolation of [$6$]/[$7$].
[T]he point halfway between Aquarius $18 \frac{1}{6}º$ and Taurus $23 \frac{1}{2}º$ is at Aries $5 \frac{5}{6}º$.
Thus, the point on the ecliptic in question is $5 \frac{5}{6}º$ into Aries. Although Ptolemy doesn’t address it right now, this is roughly $4º$ lower than the position in Ptolemy’s time which is consistent with his rate of precession of $1º$ per hundred years20.
However, Ptolemy doesn’t stop there. He gives another set of observations from a similar time period to check, yet again, this value.
[$8$] Again, in the $24^{th}$ year in Dionysius’ calendar, Leonton $28$, in the evening, [Mercury] was a little more than $3º$ in advance of Spica, according to Hipparchus’ reckoning. Thus, at that moment, its longitude, according to our coordinates, was $19 \frac{1}{2}º$ into Virgo. That moment is in the $486^{th}$ year from Nabonassar, Payni [X] $30$ in the Egyptian calendar [$261$ BCE, Aug. $23$], evening, at which time the longitude of the mean sun was Leo $27 \frac{5}{6}º$. Therefore, the greatest elongation from the mean [sun] as evening-star was $21 \frac{2}{3}º$.
Again, Ptolemy is unable to find a corresponding elongation for Mercury as a morning-star, so he again interpolates two observations:
[$9$] In the $75^{th}$ year in the Chaldaen calendar, Dios $14$, at dawn, [Mercury] was half a cubit [ca. $1º$] above [the star on] the southern scale [of Libra]21. Thus, at that time, it was $14 \frac{1}{6}º$ into Libra, according to our coordinates. This moment is in the $512^{th}$ year from Nabonassar, Thoth [I] $9/10$ in the Egyptian calendar [$236$ BCE, Oct. $29/30$], dawn, at which time the longitude of the mean sun was $5 \frac{1}{6}º$ into Scorpio. Therefore, the greatest morning elongation was $21º$.
[$10$] In the $67^{th}$ year in the Chaldaen calendar, Apellaios $5$, at dawn, [Mercury] was half a cubit [ca. $1º$] above the northern [star in the] forehead of Scorpius22. Thus, at that time, it was at $2 \frac{1}{3}º$ into Scorpio, according to our coordinates. This moment is in the $504^{th}$ year from Nabonassar, Thoth [I] $27/28$ in the Egyptian calendar [$244$ BCE, Nov. $18/19$], dawn, at which time the longitude of the mean sun was Scorpio $24 \frac{5}{6}º$. Therefore, this [greatest morning] elongation was $22 \frac{1}{2}º$.
Again, Ptolemy must interpolate between these two to estimate a point at which the elongation was $21 \frac{2}{3}º$.
In these two observations again, then, since the difference between the two mean positions [of the sun] is $19 \frac{2}{3}º$, and the difference between the greatest elongations is $1 \frac{1}{2}º$, it follows that to $\frac{2}{3}º$ (which is the amount by which $21 \frac{2}{3}º$ of the required elongation exceeds the $21º$ of the lesser [of these two]) corresponds to $9º$.
Again, I’ll pause to do the math:
$$19 \frac{2}{3}º \cdot \frac{2}{3}º \div 1 \frac{1}{2}º = 8;44º$$
Ptolemy again rounds this to $9$ despite $8 \frac{3}{4}º$ being a much better approximation.
If we add the latter to Scorpio $5 \frac{1}{6}º$, we get the mean position at which the greatest morning elongation becomes equal to the greatest evening elongation of $21 \frac{2}{3}º$: this point is $14 \frac{1}{6}º$ into Scorpio.
Checking the math: In Scorpio, $5 \frac{1}{6}º + 9º = 14 \frac{1}{6}º$.
And thus we can average this with the elongation from [$8$] to determine the point where the line of apsides intersects the ecliptic:
And the point halfway between Leo $27 \frac{5}{6}º$ and Scorpio $14 \frac{1}{6}º$ is, again, about $6º$ into Libra.
Again, this is about $4º$ lower than in Ptolemy’s time and Ptolemy now notes that both of these are consistent with his rate of precession:
From the above, and also because the phenomena associated with the other planets individually fit [the assumption], we find it consistent [with the facts to assume] that the diameters through the apogees and perigees fo the five planets that shift, has the same speed as that of the sphere of the fixed stars. For the latter move about $1º$ in $100$ years, as we demonstrated; and here too, the interval from the ancient observations, in which the apogee of Mercury was in about the $6^{th}$ degree [of the signs in question], to the time of our observations, during which it has moved about $4º$ (since it [now] occupies the $10^{th}$ degree), is found to comprise approximately $400$ years.
And that’s all for this chapter. In it, we’ve concluded Mercury’s line of apsides ran through the points $10º$ into Aries and Libra in Ptolemy’s time and precesses with the sphere of fixed stars in this model.
I should quickly note that, at this point, Ptolemy has not actually told us which point (the one in Aries or the one in Libra) is the apogee and which is the perigee – He has only referred to the line of apsides through these points. I suspect we’ll return to that at some point as it is probably rather important to the theory.
As a final note, there are some criticisms of Ptolemy’s methods here. Toomer notes that there is an “acute critique” of this method. It comes from Fredrick Sawyer and is evidently published in an Appendix to Bernard Goldstein’s Remarks on Ptolemy’s Equant Model in Islamic Astronomy which was published in what I suspect is a journal called either ΠΡΙΣΜΑΤΑ (Prismata) or Festschrift fur Willy Hartner from $1977$ (p $169-181$). However, I can’t seem to find a copy online.
Regardless, Toomer gives a brief synopsis stating:
He shows that mere equality of mean maximum morning and evening elongations is an insufficient criterion for positing symmetry to the apsidal line, although the observations of Ptolemy actually chose are in fact (grosso modo23) symmetric.
Toomer also cites this paper by Curtis Wilson, (p $225$) although Toomer doesn’t call out any specific points and a brief scan of the page doesn’t immediately call anything to my attention either.
I also find a strong critique of this technique in this paper from Noel Swerdlow (see pages $57-59$) which indicate that Ptolemy’s position of the line of apsides was $40º$ off!
- It has an eccentricity of $0.21$
- I.e., the tangent line to its orbit as seen from Earth.
- Ptolemy may well have not been aware of the true theoretical maximum elongation. For example, Pliny the Elder gives Mercury a maximum elongation of $23º$ in Natural History.
- We haven’t really discussed it yet, but since Mercury and Venus are inferior planets with maximum elongations, their motion is going to be tied directly to the sun’s position.
- α Tau aka. Aldeberan.
- Greek astronomers were aware of this phenomenon. This includes Ptolemy who even had an early model of it in his work on Optics. However, Ptolemy almost never addresses this in the Almagest. The only instance in which he does so is in IX.2, in which he states, the ” interval [between the star and planet] appears to the observer as greater near the horizon, and less near mid-heaven”.
- Since Mercury is being observed on opposite sides of the sun.
- α Leo aka. Regulus.
- Toomer has a footnote on this calculation which states that Ptolemy performed it incorrectly. He states that the position given for Regulus in the star catalog is $2 \frac{1}{2}º$ into Leo and concludes that this should have made the difference $34 \frac{1}{2}º$. However, it’s not clear what he means. Presumably he means difference between the position of Regulus and Mercury, which Ptolemy doesn’t ever give and instead only gives the result of that difference placing Mercury $7º$ into Cancer. Thus, I cannot find any merit to this comment.
- Recall that the observer, at the center of the ecliptic is on the line of apsides in Ptolemy’s model.
- That Ptoelmy gets such immaculately similar values for the pairs of elongations suggests that these numbers were fudged to illustrate the theory.
- Ptolemy actually already answered this question when developing the model that “the whole plane [revolves] about centre $E$, [shifting] the apogee towards the rear by the same amount as for other planets”. But now he’ll attempt to prove it.
- Another name for Mercury in Hellenistic times.
- The angular diameter of the moon is about $\frac{1}{2}º$.
- δ Cap aka. Deneb.
- As is consistent with his rate of precession of $1º$ per century.
- Which was given as ♑ $26;20$.
- α Tau aka. Elnath.
- Toomer criticizes this rounding in the footnote as this would better be expressed as $23;50$ if rounding needs to occur, but ultimately admits that this linear interpolation is only a crude approximation anyway.
- Again, we should be questioning these numbers as the rate of $1º$ per century is incorrect. That Ptolemy seems to conveniently keep arriving at it is an indication that he’s fudging the numbers. But as I’ve noted elsewhere, this seems to be consistent with the mindset of astronomers of the time in which an elegant and well constructed model was given more consideration than the observations.
- α Libra aka. Zubenelgenubi.
- β Sco aka. Acrab.
- Lit. “Roughly Speaking”.