Now that we’ve worked out the distance to the moon at the time of the observation, we can put this information back into our lunar model diagram to work out the true scale. We’ll begin with a drawing of our lunar model at the time depicted:
Although putting things in the context of the ecliptic isn’t necessary, I like seeing what we’re working with, so I’ve used the values we determined in this post to lay things out1. Since we’ve seen this sort of diagram several times, I don’t think it necessary to go over all the details.
Our goal in this post is going to be to find $\overline{EL}$ in the context we’ve been working in for most of this book, as $\overline{EL}$ is the distance from the earth to the true moon at the moment of observation. We can then compare that to our last post in which we determined that distance in earth radii in order to express all other aspects of the model in earth radii.
To get started, the double elongation, $\angle{AEB} = 156;26$ as we saw in the post where we calculated the parallax. Therefore, its supplement, $\angle{ZEB} = 23;34º$.
Now let’s construct some additional right triangles to help us along.
Here, we’ve extended $\overline{BE}$ such that it meets a perpendicular dropped from $D$ at $M$. In addition, we’ve dropped a perpendicular from $Z$ onto $\overline{BE}$. This means that $\angle{DEM} = 23;34º$ as well since it’s a vertical angle.
As we’ve demonstrated before, $\triangle{DEM}$ is congruent to $\triangle{ENZ}$ so anything we work out for one is true of the respective sides and angles in the other. So if we consider demi-degrees circles around them, we get that
$$arc \; DM = arc \; ZN = 47;08º$$
And their supplements:
$$arc \; EM = arc \; EN = 132;52º$$
Thus, the corresponding chords:
$$\overline{DM} = \overline{ZN} = 47;59^p$$
$$\overline{EM} = \overline{EN} = 110;00^p$$
in the context of these small circles where their hypotenuses, $\overline{DE}$ and $\overline{EZ}$ are $120^p$. But in the context of the overall diagram, we know that $\overline{DE} = \overline{EZ} = 10;19^p$. This allows us to context switch back to the big picture and get:
$$\overline{DM} = \overline{ZN} = 4;08^p$$
$$\overline{EM} = \overline{EN} = 9;27^p$$
In addition, we also know that the radius of the eccentre, $\overline{BD} = 49;41^p$.
If we consider $\triangle{BDM}$ we now know two of the sides ($\overline{BD}$ and $\overline{DM}$) which means we can use the Pythagorean theorem to find $\overline{BM}$
$$49;41^2 = 4;08^2 + \overline{BM}^2$$
$$\overline{BM} = 49;31^p$$
From that, we can subtract $\overline{EM}$ to get $\overline{BE}$ to be $40;04^p$. And we can then subtract $\overline{EN}$ to determine that $\overline{BN} = 30;37^p$.
Now consider $\triangle{BZN}$. Again, we now know two sides: $\overline{BN}$ and $\overline{NZ}$. So we can again use the Pythagorean theorem to find $\overline{BZ} = 30;54^p$.
Next, we’ll do a demi-degrees circle around this triangle. In it, $\overline{BZ}$ is now the hypotenuse with a measure of $120^p$ in this context. Then also in that context, $\overline{ZN} = 16;2^p$ and the arc subtending it, $arc \; ZN = 15;21º$. Therefore the angle opposite it on the circumference, $\angle{ZBN} = 7;40º$.
Considering now the epicycle itself, this means that $arc \; \Theta K$, which subtends that angle, must also be $7;40º$.
Looking back at the post where we determined the parameters for this observation, we stated that the angular distance around the epicycle from mean apogee, $H$, was $arc \; HKL = 262;20º$. Since $\overline{HK}$ bisects the epicycle, that means that $arc \; H \Theta K = 180º$ which we can subtract out of $arc \; HKL$ to determine $arc \; KL = 82;20º$.
To that, we can add $arc \; \Theta K$:
$$7;40º + 82;20º = 90;00º$$
So, coincidentally, $\angle{\Theta BL}$ is a right triangle. That means we can use the Pythagorean theorem on $\triangle{EBL}$ because we again know two sides: $\overline{EB} = 40;04^p$ as we showed above, and $\overline{BL} = 5;15^p$ as it’s the radius of the epicycle. Therefore, $\overline{EL} = 40;25^p$. This is the distance to the moon in the context is which $\overline{BL} = 5;15^p$, $\overline{EA} = 60^p$, and $\overline{EG} = 39;22^p$.
We can then compare that to our result from the last post where we determined, in a context where $1^p$ is equal to the radius of the earth, that the distance from the observer on earth, $E$, to the moon $L$ was $39;45$ earth radii.
We can then use that restate those previous contexts I just mentioned. When doing so, we get:
$\overline{EA} = 59;00 \; earth \; radii$
$\overline{EG} = 38;43 \; earth \; radii$
$\overline{BH} = 5;10 \; earth \; radii$
Thus, we have restated our lunar model in a far more grounded set of units, which is what we set off to do.
But before leaving off, let’s leave Ptolemy’s world for a bit and do a bit of a reality check on this. The actual average distance to the moon in earth radii is about $60$. That aligns pretty well with the value we determined for $\overline{EA}$. And this is actually no accident. Because $\overline{EA}$ is also the average distance at syzygy (i.e., during solar eclipses). As we’ve seen, Ptolemy was particularly concerned with eclipses so the belief is that Ptolemy chose values that would make this work. In addition, Neugebauer dives deeper into the numbers and finds a number of other errors that happen to cancel eachother out pretty well2.
However, the values at all other points are grossly wrong as when the mean moon is at $G$ and perigee on the epicycle, this would imply a distance of $33;33 \; earth \; radii$ when in reality the closest approach is only about $57 \; earth \; radii$. As we’ve pointed out before, this would be stunningly obvious since the moon would appear significantly larger in the sky.
Regardless, when we began this quest to discover lunar distances, we stated that Hipparchus attempted to determine lunar distances from solar distances. So obviously there is a path between the two. And in the next chapter, we’ll go the other way around, using the lunar distances to attempt to determine the solar distance.
- I’ve also added the positions of the mean and true sun over in Libra. While we didn’t calculate these, Ptolemy listed them along with the lunar calculations he skipped the math on. I didn’t include them in the post because they aren’t important there, or really here, but again, I like seeing the big picture.
- See page $103$ of History of Ancient Mathematical Astronomy.