So far, what we know about Ptolemy’s second anomaly is that it doesn’t have an effect at conjunction or opposition. Its at its maximum at quadrature, which is to say, a $\frac{1}{4}$ and $\frac{3}{4}$ of the way through each synodic month1. Its effect is to re-enforce whatever anomaly was present from the first anomaly. Ptolemy laid out a conceptual model in Chapter 2, but to determine the parameters of the model, we’ll need to first explore how much this second anomaly impacts things.
To determine this, Ptolemy used observations that fit the following three criteria:
1. The moon’s speed was about at the mean (for that is when the equation of anomaly is maximum)
2. The mean elongation of the moon from the sun was about a quadrant (for then the epicycle was near the perigee of the eccentre)
3. The moon had no longitudinal parallax.
The first of these is important because we’re ultimately going to be looking at the contribution of this second anomaly to the total anomaly. Which means we’re going to need to subtract out the contribution of the first anomaly. Having this near its maximum reduces the proportion that could be due to error. In addition, larger angles are easier to measure than small ones.
For the second, the moon being $90º$ from the sun could hypothetically be at apogee too, so my assumption is that Ptolemy is referring only to the one when it is at at perigee, thus making the apparent angle larger and thus, easier to measure.
The third of these is ultimately the most important. With this, Ptolemy is considering parallax because, at this point, we don’t know just how much of an impact this eccentric addition will have on bringing the moon closer and further from the observer at perigee and apogee. When considering parallax, the effect is magnified when the object gets closer. We didn’t have to consider this with the first anomaly because we didn’t need to observe the ecliptic longitude of the moon to derive that since we used the calculated solar position as a baseline. However, in this case, we do not have that luxury. So how do we assure ourselves that there is no effect from parallax?
Let’s start with a drawing:
Here, I’ve depicted the celestial sphere as the large circle. The circle near the center2 is the celestial equator, and the tilted one the ecliptic, with the winter solstice towards us, but slightly to one side3. The Earth is at the center, with the moon drawn in at two positions4: once closer to the observer at perigee, and once at the mean distance which can represent any distance further than perigee.
I’ve set this up for Alexandria, so let’s take a look at where each of these lunar positions would appear to fall on the ecliptic. To help us, I’ll zoom in a bit:
Here, it’s clear that they both appear to be at the same ecliptic longitude.
Now, let’s wait for the Earth to rotate some and draw in the same lines of sight to the moons in the same positions, again from Alexandria:
In this case, the moon appears at a different place along the ecliptic due to parallax. This is precisely what Ptolemy is wanting to avoid. He wants to ensure that when he takes the measurement, parallax isn’t throwing things off. The question is then, how to know when the appropriate time to do so is?
To understand, let’s add to our drawing.
Here, I’ve added two new circles. The upper, horizontal one is a small circle that represents the daily path of the zenith of Alexandria around the celestial sphere.
The second, vertical one, is an altitude circle5. It’s been a long time since we’ve discussed altitude circles. We introduced them in II.12, and II.13 is a table that gives this angle, but we haven’t really touched them since then. So to review quickly, an altitude circle is a great circle going through the zenith and the object in question, or at least its apparent position on the celestial sphere. Hence, the altitude circle I’ve drawn in has one point on the circle for the zenith and a second intersection on the ecliptic at the moon’s apparent ecliptic longitude6.
As you can clearly see, the altitude circle appears to make a right angle with the ecliptic, or very close to it. We’re not being exact about the math here. But now let’s consider the other case where parallax did occur:
Here we have the same point on the ecliptic, but the zenith has moved as the Earth rotated. This gives a very steep angle of the altitude circle with the ecliptic.
So the takeaway here is that when the zenith has the same ecliptic longitude as the object in question, it’s a good time to observe. Or we can use that angle between the altitude circle and the ecliptic. When it’s $90º$, there will be no effect due to the parallax. How do we know when that is? Well, it’s one of those things from the table in II.13, either the East or West angle depending on which side of the meridian the moon happens to be falling on.
With these requirements out of the way, Ptolemy can start determining the maximum equation of anomaly the moon actually displays, which can then be compared to the first equation of anomaly we worked out in the last book to determine how much of an impact this second anomaly has.
Instead of jumping straight into the proof, Ptolemy immediately gives us the answer stating,
the greatest equation of anomaly is about $7 \frac{2}{3}º$ with respect to the mean position (or $2 \frac{2}{3}º$ different from [the corresponding equation of] the first anomaly).
That’s surprisingly large. Recall that the width of the moon is roughly $\frac{1}{2}º$ so this means that this is over $5x$ as wide as the full moon. Next, Ptolemy lays out his proof starting with the observations.
The first was an observation made
in the $2nd$ year of Antonius, Phamenoth [VII], $25$ in the Egyptian calendar [139, Feb. 9], after sunrise, and $5 \frac{1}{2}$ hours before noon. The sun was sighted7 $18 \frac{5}{6}º$ into Aquarius and Sagittarius $4º$ was culminating. The apparent position of the moon was $9 \frac{2}{3}º$ into Scorpio, and that was its true position too, since when it is near the beginning of Scorpius, about $1 \frac{1}{2}$ hours to the west of the meridian at Alexandria, it had no noticeable parallax in longitude.
Let’s pause here a moment to verify that there is no parallax as required by the 3rd rule above. To do so, let’s carefully consider some values. First off, the moon was $9;40º$ into Sagittarius. Culminating that that time was “Sagittarius $4º$”. So what exactly does that mean?
I’ll sketch it out.
This diagram is rather oversimplified but should get the point across. Along the bottom we have the ecliptic with the relevant constellations (and Capricornus between them) laid out as it would appear to an observer within the celestial sphere. We can see that the moon is $9;40º$ into Scorpio, the point $4º$ into Sagittarius is on the meridian, aka culminating, and the Sun is $18;50º$ into Aquarius.
As I’ve drawn it here, the altitude circle is such that it looks like a right angle with the ecliptic, but this is a 2D drawing of a curved surface so we can’t necessarily trust it. Thus, we’ll need to do a better job of determining whether or not it really is. To do so, we’ll need to refer to Table II.13. We’ve never really used this table before, so let’s walk through it.
First off, which table to use? Ptolemy said the observation was taken in Alexandria which has a latitude of $31º$. There’s no table that has this value, but the Lower Egypt one is closest at $30;22º$. It’s not perfect, but we can use it.
Next, the angle we’re interested in is the one between the altitude circle and the ecliptic with the moon as its vertex. Since this is to the west8 of the Meridian, we will be using the column marked “west angle”. Additionally, the point is in Scorpius, so that’s the sub table we need to look at.
However, this table works using hours before or after the meridian. What we have is in degrees. So what we’ll need to do is determine how many degrees the moon is from the meridian and convert to hours.
We know that the moon is $9;40º$ into Scorpius, which means there’s $20;20º$ left in Scorpius plus the $4º$ of Sagittarius west of the meridian for a total of $24;20º$.
Recall that the sky rotates $15º$ per hour which means the moon crossed the meridian about $1;37$ hours previously9. That’s what we need to look up in the table. However, we only have $1$ hour and $2$ hours in the table, which means we should do some extrapolation.
The difference between the two is:
$$92;28º – 77;22º = 15;6º$$
And we need $\frac{37}{60}$ of that which is $9;18º$. So we need to subtract that from the former to get the angle to be approximately $83;10º$. It’s not a right angle, but keep in mind, the table we were using wasn’t quite right since the latitude was a bit off. Just taking a quick peek at the one for Rhodes at $36º$ latitude, we can see that the angle increases. We could repeat the above calculation for Rhodes and again interpolate between the two to get a better result. Or we could walk through the procedure to calculate the exact angle as we did in II.12. I’ll spare us, because Toomer notes that doing so would give a value of $83;45º$. Still not a right angle, but close enough Ptolemy is deems it suitable for use.
Ptolemy then backs up slightly, giving a bit of information about where he came up with the value of $18 \frac{5}{6}º$ for the sun: He calculated this from the solar model with $885$ Egyptian years, $203$ days, and $18 \frac{3}{4}$ equinoctial hours having elapsed since the beginning of the epoch. We’ve seen this plenty of times now, so I won’t go through the calculation, and neither does Ptolemy.
This same period can also be used to determine the mean position of the moon. Again, Ptolemy doesn’t show the calculations and we’ve done it enough I won’t either, but Ptolemy determines that the mean moon should have been located $17;20º$ into Scorpius. Doing the math, that’s $91;30º$ away from the sun, so slightly more than a quadrant, but again, close enough Ptolemy declares it good enough to include10.
So now we could subtract the difference in position of the mean moon and observed moon to get the total equation of anomaly:
$$17;20º – 9;40º = 7;40º$$
But what we’re really after is the amount of it that’s in excess of the first anomaly which means we’ll need to determine the first equation of anomaly and subtract that out. Ptolemy again skips the calculation noting only that the moon is $87;19º$ from apogee which we get by applying that interval from epoch with the lunar mean motions table. This tells us there on the epicycle the moon is. That can then be used in the lunar table of anomaly to determine the moon was close to its maximum anomaly. Interpolating, would give a value of $4;57º$. Ptolemy rounds this off to $5º$.
I’ll pause here for a moment to note that the angle from apogee came from the first column which means it should be negative. In other words, the moon would appear at a lower ecliptic longitude than the mean moon, i.e., to the right when observing it from Earth. Since this is the same direction as the observed position from the above observation, we can simply take the difference11 to get the contribution of the second anomaly to be $2;40º$.
Ptolemy could easily stop there, but to drive home the point, Ptolemy shows that he can arrive at a similar answer using Hipparchus’ observations, which would also indicate that the value is consistent over longer intervals. So Ptolemy refers to an eclipse that was observed
in the fifty-first year of the Third Kallippic Cycle, Ephiphi [XI] $16$ in the Egyptian calendar [$-127$ Aug $5$], when $\frac{2}{3}$ of the first hour had passed… [Hipparchus] says, ‘and while the sun was sighted in Leo $8 \frac{7}{12}º$ the apparent position of the moon was Taurus $12 \frac{1}{3}º$, and its true position was approximately the same’. So the true observed distance between the moon and sun was $86;15º$. But when the sun is near the beginning of Leo, at Rhodes (where the observation was made), $1$ hour of the day is $17 \frac{1}{3}$ time-degrees. So the $5 \frac{1}{3}$ seasonal hours (which make up the interval the [the following] noon) produce $6 \frac{1}{6}$ equinoctial hours. Therefore the observation occurred $6 \frac{1}{6}$ equinoctial hours before noon on the sixteenth, while Taurus $9º$ was culminating.
Let’s stop here and unpack by doing another sketch like the one above.
Here, Ptolemy has already calculated the distance between the moon and sun, and again, it’s relatively close to $90º$. This time, the moon is to the east of the meridian, by $3;20º$. So to validate that the angle between the altitude circle and ecliptic is $~90º$, we’ll need to look that up on the east column in the table. In addition, Ptolemy noted this observation was made in Rhodes, which does have a table for it. But before we can use the table, we’ll need to convert the angle to hours, which would be about $13$ minutes after the hour.
Again, doing some extrapolation, I only come up with an angle of about $75º$ which is a bit far off of the right angle we’re expecting12. Ptolemy ends up letting this slide, not even mentioning it or even claiming that the moon didn’t have a parallax which is a bit odd given he does take a moment to affirm that the first two requirements13.
Regardless, Ptolemy gives the period between epoch and the observation as $619$ Egyptian years, $314$ days and $17 \frac{3}{4}$ equinoctial hours14. This would put the sun’s mean position $10;27º$ into Taurus, and the moon’s mean position as $4;25$ into Taurus. The angle of the moon around its epicycle from apogee would then be $257;47º$ which, again looking at our lunar table of anomaly, we see is quite near the maximum value.
So we can again take the difference between the observed position of the moon of $12;20º$, subtract out the mean position of $4;25º$, which would be $7;55º$ which Ptolemy somehow gives as $7 \frac{2}{3}º$15. Next we subtract out the portion of that which was due to the first anomaly which Ptolemy again rounded off as $5º$ leaving $2;55º$ as the amount contributed by the second anomaly, which he again calls $2 \frac{2}{3}º$ finding it consistent with the previous observation and assuring us that calculations resulting from other such observations similarly returned a consistent maximum effect of the second anomaly.
- Which would put it at first or third quarter moon.
- Which appears very thin due to perspective.
- We’ll be drawing in another circle here shortly that will make this helpful.
- They’re pretty small, so you may want to click the picture to enlarge.
- This is why I drew the ecliptic slightly off center. Otherwise this would be an odd looking vertical line.
- We’re again ignoring the lunar altitude.
- As stated in the previous post, Ptolemy is actually giving the calculated position here, then lining the sun up, assuming it to be at this position.
- If you’re not familiar with the night sky you may be a bit confused about how we knew if west was left or right. We know this because the observation was taken above the tropic of cancer, so the moon must always be in the southern half of the sky. Thus, the meridian marks due south on our sketch and if you imagine yourself facing south, west is to your right.
- I should clarify a bit here because the moon didn’t actually cross the moon $1;37$ hours previously. Rather, the current ecliptic longitude of the moon did. The moon would have moved about a degree in that time to arrive at the ecliptic longitude Ptolemy described.
- You may be wondering, given that the moon had only crossed the meridian 2 hours prior, the moon still should have been visible for a few more hours before setting. This implies that Ptolemy could have waited another hour or two to take the observation to get a better angle between the sun. However, this would mean that the angle between the ecliptic and meridian would have been further from a right angle which is the more important of the two. Thus, if anything, Ptolemy should have taken the measurement slightly earlier. That is, if the intent of the measurement was for use in this sort of calculation. It’s also entirely likely that Ptolemy simply selected observations that were already existing for use and this is simply what he had to work with.
- After Ptolemy’s rounding.
- However, this is based on a simple linear extrapolation of a value that’s decidedly not linear. Looking at the table, you can notice that the increment between the 1 and 2 hour rows is much lower than between the noon and 1 hour one we just used. As such, the value between noon and 1 should have risen quite quickly and then leveled off some. This suggests that a more rigorous calculation would improve this, perhaps considerably.
- That the angle between the sun and moon be near $90º$ and that the equation of the first anomaly be near maximum.
- Neugebauer notes that this is off by about 11 minutes.
- Do you too get the feeling Ptolemy was just not feeling it when he wrote this section as there’s been a number of questionable flubs and omissions?