Now that Ptolemy has straightened us out regarding which angle we need to be using, Ptolemy sets out
a convention method for making the above kind of correction of the angles and arcs, if anyone wants to make it when the [differences] involved are so small.
Fortunately, we can use the same diagram we ended with last time which was the generic case:
Ptolemy lays his method on us in his familiar brick of text:
[W]e double the amount of the angle [between the altitude circle and ecliptic], and entering with this as argument into the Table of Chords, take the chord corresponding to it, and also the chord corresponding to its supplement. We multiply both of the latter separately by the [moon’s] latitude, in degrees, divide each of the products by $120$, and record the results [separately]. As for the result derived from the first angle, we subtract it from the relevant arc from the zenith [to the ecliptic] when the moon is on the same side [of the ecliptic] as the zenith, but add it when it is on the opposite side [of the ecliptic to the zenith]. We square the result, add that to the result derived from the supplementary angle, also squared, and take the square root of the sum: This will give us the corresponding arc [$ZE$ or $ZD$] which is required.
Next, we take the result which we recorded from the [second,] supplementary angle, multiply it by 120, and divide the result by the arc we found [$ZE$ or $ZD$]. With the resulting [chord] we enter into the [body of the] Table of Chords [I.11], take the corresponding arc [in the column of argument], and halve it. If the corrected arc [$ZE$ or $ZD$] is greater than the original [$ZB$] we add the result to the amount of the original angle, but if [the corrected arc is] less [than the original], we subtract it. The result with be the corrected angle.
Fortunately, Ptolemy actually works out an example for once so we’ll walk through it step by step.
Here, we’ll use the same figure as above and let $arc \; ZB = 45º$, $\angle{ABZ} = 30º$, and the latitude of the moon, $arc \; DB$ and $arc \; BE = 5º$1. Walking through this:
[W]e double the amount of the angle [between the altitude circle and ecliptic]
The angle here is $\angle{ABZ}$ so we’ll double it to $60º$.
and entering with this as argument into the Table of Chords, take the chord corresponding to it
Chord of $60º = 60^p$
also the chord corresponding to its supplement
The supplement of $60º$ is $120º$ for which the chord is $\approx 104^p$.
As a quick note, in Ptolemy’s example, he reverses the order of the next two steps. I’ll follow along with the above text instead of the example’s steps.
We multiply both of the latter separately by the [moon’s] latitude, in degrees
So for the former $60 \cdot 5 = 300$ and $104 \cdot 5 = 520$
divide each of the products by $120$, and record the results [separately]
$$\overline{KB} = \overline{BL} = 300 \ 120 = 2;30º$$
$$\overline{DK} = \overline{EL} = 520 \ 120 = 4;20º$$
As for the result derived from the first angle, we subtract it from the relevant arc from the zenith [to the ecliptic] when the moon is on the same side [of the ecliptic] as the zenith, but add it when it is on the opposite side [of the ecliptic to the zenith].
If we take the case where the moon was at $E$, this was on the same side of the ecliptic so we subtract this from $arc \; ZB$ which would give $arc \; ZL = 42;30º$.
If the moon were at $D$, then we would add instead for $arc \; ZK = 47;30º$
We square the result, add that to the result derived from the supplementary angle, also squared, and take the square root of the sum: This will give us the corresponding arc [$ZE$ or $ZD$] which is required.
This is essentially us using the Pythagorean theorem on triangles $\triangle{ZEL}$ and $\triangle{ZDK}$. We just determined the vertical sides in our drawing ($\overline{ZL}$ and $\overline{ZK}$) and we previously determined the other sides ($\overline{EL}$ and $\overline{DK}$), so we can use the Pythagorean theorem to determine their hypotenuses:
$$ZE = \sqrt{{42;30}^2 + {4;20}^2} \approx 42;46º$$
$$ZD = \sqrt{{47;30}^2 + {4;20}^2} \approx 47;44º$$
Next, we take the result which we recorded from the [second,] supplementary angle, multiply it by 120
So,
$$4;20 \cdot 120 = 520$$
and divide the result by the arc we found [$ZE$ or $ZD$]
These steps are the ones involved in a demi-degrees method about the triangles we just solved. So this step is just context switching us into that.
$$\overline{EL} = 520 / 42;46 \approx 12;08^p$$
$$\overline{DK} = 520 / 47;44 \approx 10;50^p$$
With the resulting [chord] we enter into the [body of the] Table of Chords [I.11], take the corresponding arc [in the column of argument], and halve it.
This is still part of our normal demi-degrees method and it getting the angle opposite each of these chords which would be $\angle{EZL}$ and $\angle{DZK}$, which we noted we need to either add or subtract from $\angle{ZBZ}$ in our last post. So we’re almost there!
Looking those arcs corresponding to those chords we get $arc \; EL \approx 11 \frac{3}{5}º$ and $arc \; DK \approx 10 \frac{1}{3}º$.
Halving each:
$$\angle{EZL} \approx 5 \frac{4}{5}º$$
$$\angle{DZK} \approx 5 \frac{1}{6}º$$
If the corrected arc [$ZE$ or $ZD$] is greater than the original [$ZB$] we add the result to the amount of the original angle, but if [the corrected arc is] less [than the original], we subtract it. The result with be the corrected angle.
So here we need to compare $arc \; ZE$ and $arc \; ZD$ to $arc \; ZB$. It is less, so we need to subtract the angle we just found from $\angle{ABZ}$. If more, we need to add. We find that $arc \; ZE$ is less and $arc\; ZD$ is more. Thus:
$$\angle{ABZ} – \angle{EZL} = \angle{A \ Theta Z} = 30;00 – 5 \frac{4}{5}º = 24 \frac{1}{5}º$$
$$\angle{ABZ} + \angle{DZK} = \angle{AHZ} = 30;00 + 5 \frac{1}{6}º = 35 \frac{1}{6}º$$
That completes the procedure and Book V.
