When we built the first lunar model, it was done using observations only at opposition, which is to say, during eclipses which only happen during the full phase. In the last few chapters, we looked at quadrature, which is to say, during first and third quarter moon and derived a second anomaly. But what happens if we consider the moon when it’s somewhere between those phases?
Ptolemy gives the answer:
[W]e find that the moon has a peculiar characteristic associated with the direction in which the epicycle points.
So what does that mean?
Let’s tackle the last part of that first, regarding “the direction in which the epicycle points.”
Every epicycle must, in general, possess a single, unchanging point defining the position of return of revolution on that epicycle. We call this point the ‘mean apogee’, and establish it as the beginning from which we count motion on the epicycle. Thus, point $Z$ on the previous figure is such a point. It is defined, for the position of the epicycle at apogee or perigee of its eccentre, by the straight line drawn through all the centres [of the ecliptic, eccentre, and epicycle] ($\overline{DEG}$ here).
The image being referred to was in the last post, so I’ll post it here again:
What Ptolemy is saying here is that, when describing the position of the moon on the epicycle, we have always done it from apogee, which in this diagram is at point $Z$. This seems like no big deal, until we really think about how to define $Z$. In this drawing it’s easy, because no matter which “center” you take ($D$, the center of the eccentre, or $E$, the center of the ecliptic) it’s still the point on the epicycle that’s furthest from.
But as soon as you move away from the moment where it’s on this nice line, things become more confusing:
Here, you can easily see that depending on which center you take, it creates two different potential points for the apogee of the epicycle. So, based on where the moon should be at apogee from the models, which one is the “correct” one? And this does matter because, again, this is the point will be the one we measure the progression around the epicycle from. So if we choose incorrectly, it will have impact to the anomaly. So which to use?
Turns out, it’s neither. It’s a completely different point which is referred to as the “mean apogee”.
We do indeed find that the direction in which [$\overline{ZG}$] points is a single, unchanging point on diameter $\overline{AG}$1, but that point is neither $E$, the center of the ecliptic, nor $D$, the centre of the eccentre, but a point removed from E towards the perigee of the eccentre by an amount equal to $DE$.
So let’s draw in the point he’s referring to:
Here, I’ve drawn in the point as $I$. It is equidistant from the observer/center of the ecliptic at $E$ as $D$, but in the opposite direction. We know that it’s always opposite $D$ because Ptolemy states2 that the point is towards perigee. Since perigee must be directly opposite $D$, so too must $I$3.
So how to prove this?
As usual, Ptolemy begins with observations:
Hipparchus records that he observed the sun and moon with his instruments in Rhodes in the $197^{th}$ year from the death of Alexander, Pharmouthi [VIII] $11$ in the Egyptian calendar [$-126$ May $2$], at the beginning of the second hour. He says that, while the sun was sighted $7 \frac{3}{4}º$ into Taurus, the apparent position of the center of the moon was $21 \frac{2}{3}º$ into Pisces, and its true position was $21 \frac{1}{3} + \frac{1}{8}º$ [$27;27 \frac{1}{2}º$4] into Pisces. Therefore at the moment in question, the distance of the true moon from the true sun was about $313;42º$ towards the rear.
This gives us the positional data, but as usual, we need to determine the time from the beginning of epoch, which Ptolemy determines to be $620$ Egyptian years, $219$ days, and $18$ equinoctial hours.
At that moment, from our previous model, the mean sun would be $6;41º$ into Taurus and the true sun $7;45º$ into Taurus. Meanwhile, the mean moon would be $22;13º$ into Pisces and $185;30º$ from the apogee on the epicycle.
So the predicted position of the angular separation between the sun and moon should be $314;28º$. Let’s start drawing things out:
Here5, I’ve drawn things out as absolutely simply as possible with just the ecliptic, centered on the Earth at $E$. The position of the mean moon and mean Sun are both highlighted.
We’ll obviously need to draw in the moon’s eccentre, which will be centered on a point on the small circle near the center which I’ve included as a dashed line. The question is, where to put that center?
To answer, let’s think back to when we first added the motion of the center of the eccentre around the observer. Specifically, let’s reconsider the discussion regarding this picture:
To get here, we’d started off with the sun and moon at the vernal equinox. At that time, Z would have been along that same line of sight. The image I’ve reposted above is what would have happened a quarter of a sidereal month later. What we see is that the moon (at $H$) has moved a larger angular distance ($~97;20º$) than $D$, the projection of $Z$ onto the eccentre, did ($~82;41º$). However, at this moment, they were diametrically opposite each other.
But this drawing was done with respect to the sidereal period wherein we measure from a fixed point: The vernal equinox in this case. What happens if we consider a synodic period, i.e., with respect to the sun?
In that case, the sun, which started at the vernal equinox, rotates on the ecliptic counter-clockwise:
In that interval of time, the mean sun would have moved from the vernal equinox to $S$ which is about $7;17º$. Take a look back at those numbers I just put above describing the points $H$ and $D$. Although there’s some rounding errors, you can quickly see that $7;17º$ is very close to the amount over $90º$ that point $H$ ended up at, while it’s very close to the amount under $90º$ that point $D$ ended at. In short, $\angle{SEH} = -\angle{SED}$.
Turns out, this is always the case; the angle between the mean sun and mean moon is always equal and opposite the angle made between the sun an the center of the eccentre. This is known as the double elongation.
So going back to our current discussion, that means that if we know the difference in ecliptic longitude between the sun and the mean moon, we can double that to find the angle from the sun for the center of the eccentre which points to apogee.
Fortunately, that not hard at all to do because we already stated the calculated position of the mean sun and mean moon. So we can quickly determine the difference is $44;28º$.
Before jumping too much into the math, let’s add to our diagram for this chapter:
Here, I’ve drawn in the position of apogee of the eccentre, $A$, which defines the position of point $D$, the center of the eccentre, and gone ahead and put in the mean moon at $B$. I’ve also drawn in lines of sight from the center of the eccentre, $D$, and the center of the ecliptic, $E$, to the mean moon at $B$. Along the line of sight from the observer, I’ve also drawn in the intersections at $\Theta$ and $Z$ where it meets the epicycle.
So far, we’ve determined that $\angle{AEB} = 88;56º$.
Next, we’ll extend a line from point $D$ onto $\overline{EB}$ such that it’s perpendicular. Unfortunately, it’s such a small triangle the new point at that intersection, $K$ is indistinguishable at this scale from $E$. Still, let’s draw it in.
Now, Ptolemy jumps into the demi-degrees method around this new triangle, $\triangle{DEK}$. Within this triangle, $\angle{DEK}$ is on the circumference and still has the same measure as $\angle{AEB}$ which is to say $88;56º$. Thus, the arc opposite that angle would be twice that measure or $177;52º$.
Since $\overline{DE}$ is the hypotenuse, it divides the circle in half and consequently, $arc \; EK = 2;8º$ as it would be the supplement to $arc \; DK$.
From this, we can find the length of the corresponding chords in the context of this circle:
$$\overline{DK} = 119;59^p$$
$$\overline{EK} = 2;14^p$$
We can then use the Pythagorean theorem to determine $\overline{DE}$. However, doing so returns that $\overline{DE} = 120^p$ to the precision we’ve been using.
Next, we need to convert back to the context of the overall circle. In that context, $\overline{DE} = 10;19^p$, which we determined in the last chapter. So, in to that same level of precision, $\overline{DK} = 10;19^p$ as well.
We can similarly determine that $\overline{EK} = 0;12^p$6.
Now let’s focus on the other right triangle we just created, $\triangle{DKB}$. In it, we know $\overline{DB} = 49;41$, which we determined in the last chapter as well. And we just said $\overline{DK} = 10;19$, which gives us two sides. So we can again use the Pythagorean theorem to determine $\overline{BK} = 48;36^p$ to which we can add $\overline{EK}$ to determine $\overline{BE} = 48;48^p$.
We’ll set that aside for now and start considering the anomaly. Here, we’ll use the position of the true sun as our reference point, being $7;45º$ into Taurus. The distance (measuring counter-clockwise) to the mean moon is $314;28º$. Similarly, the distance from the true sun to the true moon is $313;42º$. That’s a difference of $0;48º$. So let’s add that into our drawing now:
Again, this angle is quite small at this scale, but we can easily see there’s a few new points I’ve added7: $H$ which is the true position of the moon, and $L$, which is the intersection of a line dropped from point $B$ to be perpendicular on $\overline{EH}$. Here, the anomaly is $\angle{BEH} = \angle{BEL} = 0;46º$.
Now, we’ll again use the demi-degrees method around this triangle, $\triangle{BEL}$, to do some more figuring.
In that triangle, $arc \; BL$ is double the measure of the angle it subtends so it is $1;32º$ which gives a corresponding chord length of $1;36^p$.
In addition, we know that $\overline{BE} = 120^p$ in the context of this circle since it’s the hypotenuse. However, we previously determined that $\overline{BE} = 48;48^p$ which allows us to context switch back to the larger context wherein $\overline{BL}$ would be $0;39^p$.
Now, Ptolemy drops in another line from $B$ to $H$ forming a new right triangle, $\triangle{BHL}$. I’ll skip drawing it in since we’re getting cluttered in this tiny space, but in it, we know $\overline{BH} = 5;15º$ because it’s the radius of the epicycle which we determined in the last book8.
We can now context switch into this new triangle, and in it, determine that $\overline{BL} = 14;52^p$. From that, we can determine the corresponding arc, $arc \; BL = 14;14º$ as does the angle it subtends. Thus, the angle opposite it on the perimeter, $\angle{BHL}$ is half that or $7;7º$.
Let’s pull these two triangles out and exaggerate the angles to be able to see what we’re dealing with here:
First, we can find $\angle{BHE} = 172;53º$ as it’s the supplementary angle to $\angle{BHL}$ which we just determined to be $7;7º$. We’d also previously determined $\angle{BEL} = 0;46º$, so that’s two of the angles in $\triangle{BEH}$ known which means we can determine $\angle{EBH} = 6;21º$. Since this angle is subtended by $arc \; \Theta H$, we can state that this arc has the same measure. In addition, we should note that $\Theta$ is the position of the true perigee of the epicycle. What this implies is that $H$, the true position of the moon, is only $173;39º$ from the true apogee. But we previously stated that the true moon should be $185;30º$ from the apogee of the epicycle. What this means is that $\Theta$, the true apogee, is clockwise from $H$ and that the mean perigee is somewhere counter-clockwise.
So now that we’ve got some figures to play with regarding the distances from apogees and perigees, we’ll draw in a new point, $M$ which we’ll let be the mean perigee. Because $H$ should be $185;30º$ from it, measuring clockwise since that’s the direction the epicycle moves about its center, that means it should be some distance counter-clockwise from $H$.
We’ll extend a line from $B$, through $M$ until it falls on $\overline{AG}$ at point $N$. Here, I’ve drawn it such that it falls such that it’s equidistant from $E$ ad $D$. At this point, we don’t know that for a fact, as it’s what we’re out to prove, but for the purposes of this drawing we’ll assume it for now. In addition, we’ve dropped $\overline{EX}$ onto $\overline{BN}$ such that they’re perpendicular9.
As previously stated, $arc \; \Theta H = 6;21º$. In addition, we stated that the true moon should be $185;30º$ from apogee, which means it should be $5;30º$ from perigee. Thus, we can add these two together to determine $arc \; \Theta M = 11;51º$ as is the angle it subtends, $\angle{EBX}$.
Now we’ll create another demi-degrees circle around $\triangle{EBX}$. In it, $arc \; EX$ is opposite the angle we just found, so has twice the measure, or $23;42º$. Converting that to the corresponding chord, we get that $\overline{EX} = 24;39^p$ in the context of this triangle. Also in this triangle, we know that $\overline{BE} = 120^p$, again because it’s the hypotenuse.
We can use that to convert back to the context of the larger circles where we previously determined $\overline{BE} = 48;48^p$ which tells us that, in this larger context, $\overline{EX} = 10;2^p$.
We’ll pause a moment to pull out a few angles:
Reviewing, we know from the very first angle we determined in this post that $\angle{AEB} = 88;56º$ which tells us that $\angle{BEN} = 91;4º$ as it’s the supplement. Similarly, we determined $\angle{EBN} = 11;51º$. So we can subtract those two from $180º$ to determine $\angle{ENB} = 77;05º$.
This happens to be the same angle as $\angle{ENX}$ so we’ll do one last demi-degrees triangle around $\triangle{ENX}$. In it, $arc \; EX = 154;10º$ since it’s opposite $\angle{ENX}$ and the corresponding chord, $\overline{EX} = 116;58^p$. In addition, we know $\overline{EN} = 120^p$ since it’s the hypotenuse.
Since we know that in the larger context $\overline{EX} = 10;2^p$ we can use that to context switch back to the overall diagram and determine that $\overline{EN} = 10;18^p$. This is very nearly the same as $\overline{DE}$ which we determined previously to be $10;19^p$. So, while this isn’t a rigorous proof, for at least the case we’ve considered here, we’ve demonstrated that there exists a point, $N$, that, within rounding, is equidistant from $E$, but opposite it from $D$ which is what the original proposition was.
To demonstrate more thoroughly, Ptolemy will
show that we get the same result at the opposite sides of the eccentre and epicycle.
But we’ll save that for the next post.
- I’ve removed $A$ in the above diagram since there would also potentially be two of them. So refer to the first diagram.
- And will prove momentarily.
- Pedersen, in Survey of the Almagest, refers to this as the “third” lunar model. In addition, both he and Neugebauer in History of Ancient Mathematical Astronomy, take a rather literal translation, calling this the “inclination” or “prosneusis” (which I believe to be the literal Greek word). However, Toomer notes that the same word is used in an entirely unrelated context later on. As such, I’m following his lead and not giving this any particular name.
- Toomer notes that this calculation here is an adjustment to account for parallax but it is not explained in the text.
- I’ve been thinking for awhile now that drawing a larger ecliptic circle to mark the signs would be helpful so I’ve finally gone and created one. I hope you all like it too!
- I’m not showing the math here as we’ve walked through this process of context switching between circles plenty of times.
- I’ve also dropped point $K$ since we’re finished with it now.
- Again, Ptolemy has done some rounding here as the values we found from the Babylonian and Alexandrian eclipses were slightly lower.
- Again, things are quite squished and they’re nearly indistinguishable.