In the third chapter of Book IV, we’re going to do some more work with the lunar motion. We already did a fair amount of that at the beginning of the last chapter in which Ptolemy essentially quoted some numbers for various lunar periods from his predecessors. Now, we’ll extend those.
First, Ptolemy wants to
multiply the mean daily motion of the sun, which we derived, ca. $0;59,8,17,13,12,31$ degrees per day, by the number of days in one [mean synodic month], $29;31,50,8,20$ days, and add to the result the $360º$ of one revolution.
Let’s use a picture to see exactly what’s happening here:
Since we’re dealing with a synodic month (which is a full cycle of the lunar phases), let’s consider a time when the moon is at its full phase at $M_1$. The moon goes around its orbit1 counter-clockwise. It would be full again when it returns to position $M_1$ except during that time, the Sun has moved as well. How much?
Well, since we previously stated the amount of time it took for one synodic month, we know that’s how long the sun will have been moving. And we worked out the sun’s mean motion in the last book, so multiplying those together we can get the distance the sun travelled in that time as $29;6,23,1,24,2,30,57º$ which would be the change in ecliptic longitude or $\angle S_1 E S_2$. This is a vertical angle with $M_1 E M_2$, so it also represents the extra distance the moon had to travel in that time on top of the $360º$ it would have taken to get back to $M_1$ for a total of $389;6,23,1,24,2,30,57º$ the moon travels in ecliptic longitude during 1 synodic month.
If we take that and divide that by the number of days in the synodic month ($29;31,50,8,20$ days), you get that it travels $13;10,34,58,33,30,30º$ per day.
Next, we’ll work out the daily motion in anomaly. First, recall from this post that there’s a period of return in longitude and anomaly every $251$ synodic months and $269$ anomalistic months. Thus, we can set the relationship $251$ synodic months = $269$ anomalistic months.
Since we’re after the day anomalistic motion, we’ll start by multiplying the number of anomalistic months by the $360º$ it travels on the epicycle during that time which tells us in those $269$ anomalistic months, the moon travels a grand total of $96,840º$ around the epicycle. That happens in $251$ synodic months, so if we multiply that by the number of days in a synodic month, we get that there are $7,412;10,44,51,40$ days in $251$ synodic months or $269$ anomalistic months.
So we can then divide the total distance travelled on the epicycle by the number of days to get:
$$\frac{96840º}{7,412;10,44,51,40 days} = 13;3,53,56,29,38,38^{\frac{º}{day}}$$
This is the daily motion in anomaly.
Now lets return to the full Hipparchian period in which Ptolemy previously stated that $5,458$ synodic months = $5,923$ draconitic months. We can do the same thing.
Multiply the $5,923$ draconitic months by the $360º$ it travels to get a total distance of $2,132,280º$. This happens in $5,458$ synodic months which have a total of $161,177;58,58,3,20$ days. So dividing the total distance by the number of days we get the moon having a daily motion in latitude of $13;13,45,39,40,17,19^{\frac{º}{day}}$.
Next, Ptolemy jumps backwards a little bit. In the first of these, we did this based on a synodic month in which the moon travels more than $360º$ in ecliptic longitude. Now, he steps back to consider the daily motion based on the sidereal month in which the moon does travel only $360º$. But as a shortcut, instead of going through all the calculations again, he simply subtracts the sun’s motion in one day from the daily motion in longitude based on the synodic month to get $12;11,26,41,20,17,59^{\frac{º}{day}}$. This motion is also in ecliptic longitude, but to distinguish it from the motion for a synodic month which we derived previously, he refers to this as the “motion in elongation.”
If we think carefully about what we’ve done here, we’ve found the difference between the motion of the sun and the motion of the mean moon. So what this truly measures is the pace of the moon along the ecliptic with respect to the sun instead of the fixed point of the vernal equinox. So if the moon was a $0º$ elongation which is a new moon, one day later it would be $12;11…º$ in advance of the sun along the ecliptic.
Now let’s divide those further, breaking each day up into its 24 hours to figure out the hourly motion.
The mean hourly motions come out as follows:
- in longitude: $0;32,56,27,26,23,46,15$º/hour
- in anomaly: $0;32,39,44,50,44,39,57,30$º/hour
- in latitude: $0;33,4,24,9,32,21,32,30$º/hour
- in elongation: $0;30,28,36,43,20,44,57,30$º/hour
Instead of subdividing the daily motions, we could instead multiply by 30 to get the mean monthly motion. Note, in some cases this will go past 360º, so we’ll subtract 360º out if it does so we can only concentrate on how far the moon will be advancing on a month-to-month basis. For this we get the monthly mean motion to be:
- in longitude: $35;17,29,16,45,15$º/month
- in anomaly: $31;56,58,8,55,59,30$º/month
- in latitude: $36;52,49,54,28,18,30$º/month
- in elongation: $5;43,20,40,8,59,30$º/month
And we can do the same thing to figure out how far the moon will advance on each in a single Egyptian year of 365 days, again subtracting out complete revolutions:
- in longitude: $129;22,46,13,50,32,30$º/year
- in anomaly: $88;43,7,28,41,13,55$º/year
- in latitude: $148;42,47,12,44,25,5$º/year
- in elongation: $129;37,21,28,29,23,55$º/year
And lastly, Ptolemy multiplies these by 18 (again subtracting out full revolutions) “for convenience”:
- in longitude: $168;49,52,9,9,45$º/18 years
- in anomaly: $156;56,14,36,22,10,30$º/18 years
- in latitude: $156;50,9,49,19,31,30$º/18 years
- in elongation: $173;12,26,32,49,10,30$º/18 years
That gets us some great basic numbers to work with, but if you’ve been following along, you should know when Ptolemy starts calculating things like this, he’s about to go crazy with it and make tables of everything in between. So that’s what we’ll be doing in the next post!
