Now that we’ve revised our lunar model to include the position of the “mean apogee” from which we’ll measure motion around the epicycle, we need to discuss how we can use this to determine the true position of the moon.
As a general statement, we know how to do this: Take the position of the mean moon, determined by adding the motion since the beginning of the epoch, and add or subtract the equation of anomaly. The problem is that our revisions in this book mean the table for the lunar equation of anomaly we built in Book IV is no longer correct.
Instead, to determine the equation of anomaly, we’ll start with the motion around the epicycle1 and need to factor in the double elongation of the moon from the sun.
To see how to do so, let’s get started on a new diagram:
If this looks familiar, it should as it’s very close to the diagram we used in the last post. However, there’s been a few minor changes. First, I’ve obviously removed the ecliptic as we’re not working with specific observations. Next, instead of dropping $\overline{EX}$ onto $\overline{MN}$ such that it would land perpendicular to the latter, we extended the line $\overline{NX}$ from $N$ onto $\overline{EB}$ to that they are perpendicular. Similarly, we’ve moved point $L$ such that it now extends from $H$ to create $\overline{HL}$ which falls perpendicularly on $\overline{EZ}$. Everything else remains exactly the same.
As such, this implies we’re taking effectively the same setup in the last post, where $arc \; MH = 333;12º$ going clockwise and the double elongation, which is based on the position of the sun which I’ve not drawn in here, is $90;30º$. We’ll be taking these as knowns, the first because we can look it up in our table from our lunar tables of anomaly, and the second because it could be taken from observation.
Now let’s review some of the constants. In the last two posts, we showed that $\overline{DK} = \overline{NX} \approx 10;19^p$. Next, the radius of the eccentre, $\overline{DG} = \overline{AD} = 49;41^p$. We also know the radius of the epicycle, $\overline{BH} = 5;15^p$.
In addition to these values, we can also use a few other values which aren’t constant, but we determined in the last post and since we’re using the same configuration, we’ll simply refer to that post instead of recalculating. First off, $\overline{EK} = 0;05^p$. From that, we can quickly state that $\overline{EX} = 0;5^p$ as well. As a quick proof, let’s pull out those angles and triangles we made:
Here, we know that $\angle{NXE} = \angle{DKE} = 90º$ as $\overline{NX}$ and $\overline{DK}$ were both dropped so they would be. In addition, we know that $\angle{NEX} = \angle{DEK}$ because they’re vertical angles. Lastly, we know $\overline{DE} = \overline {EN}$ which we’ve shown previously. This means these triangles are congruent due to Angle-Angle-Side congruence.
We also showed in the last post, that $\overline{BK} = 48;36^p$ from which we’d subtracted $\overline{EK}$ to get $\overline{BE} = 48;31^p$. And since we just showed that $\overline{EK} = \overline{EX}$ in our new setup, we can subtract again to state $\overline{BX} = 48;26^p$.
Lastly, we’d shown that $\overline{DK} \approx 10;19º$ and since $\overline{DK} = \overline{NX}$ due to the aforementioned congruence between $\triangle{DEK}$ and $\triangle{NXE}$, that gives us the length of two sides in $\triangle{BNX}$. As such, we can use the Pythagorean theorem to determine the hypotenuse:
$$48;26^2 + 10;19^2 = \overline{BN}^2$$
$$\overline{BN} = 49;31^p$$
We can use that to context switch into a demi-degrees circle around $\triangle{BNX}$ wherein $\overline{BN}$ would equal $120^p$ since it’s the hypotenuse. In it, we would have $\overline{NX} = 25;00^p$, from which we can determine the corresponding arc, $arc \; NX = 24;03^p$. Looking at the angle subtended by that arc, $\angle{NBX}$ it will have half that measure or $12;01º$, regardless of whether we’re in the demi-degrees context or not. And that’s a vertical angle with $\angle{MBZ}$ so we can state that $\angle{MBZ} = 12;01º$ as well as is its corresponding arc, $arc \; ZM$.
Now let’s think some more about the epicycle. We previously stated that $arc \; MZ = 333;12º$ which means that $arc \; HM = 26;48º$. From that, we can subtract out $arc \; ZM$ to determine $arc \; HZ = 14;47º$ as would the angle it subtends, $\angle{HBZ}$.
Now we can jump into another demi-degrees circle around $\triangle{HLB}$. In it, we just determined $\angle{HBL}$ so we can state the arc subtended by it, $arc \; HL$ is twice the measure or $29;34º$. Since the hypotenuse, $\overline{HB}$ divides the circle in half, we can determine the supplement of $arc \; HL$, $arc \; LB = 150;26º$.
Knowing these arcs, we can look up the corresponding chords to state $\overline{HL} = 30;37^p$ and $\overline{LB} = 116;02^p$ in the context of this demi-degrees circle.
We’ll now context switch out of the demi-degrees circle to the overall drawing using the knowledge that $\overline{BH} = 120^p$ in the context of that circle, but is $5;15^p$ in the larger context since it’s the radius of the epicycle.
This gives us $\overline{BL} = 5;05^p$ and $\overline{LH} = 1;20^p$.
Now we can add $\overline{BL}$ onto $\overline{BE}$ to get $\overline{LE} = 53;36^p$ which, along with $\overline{LH}$ is two of the sides in right triangle $\triangle{LEH}$. So we can again use the Pythagorean theorem to determine the hypotenuse, $\overline{EH} = 53;37^p$.
We’ll switch into a final demi-degrees circle about this triangle. In it, $\overline{HL} = 2;59^p$ and its corresponding arc, $arc \; HL = 2;52º$. Thus, the angle it subtends, $\angle{HEL}$, which is the angle that defined the equation of anomaly, is half that, or $1;26º$.
Thus, we have taken two inputs, the double elongation and the motion around the epicycle from the mean apogee, and been able to calculate the equation of anomaly which is what we set out to do in this post.
