Having completed an example calculation for converting linear digits to area digits in the previous post, we’ll now do the same calculation for a lunar eclipse. The good news is the setup is the same. While I don’t strictly need to redraw the diagram, I’m going to anyway because the earth’s shadow is so much larger than the moon and drawing it as such helps me visualize things mentally although the respective position of the points doesn’t change at all.
We will now take the disc of the moon to be circle $ABGD$, centered on $E$ while the earth’s shadow will be $AZGH$ at mean distance, centered on $\Theta$. As with the previous post, we’ll let $\frac{1}{4}$ of the moon be eclipsed meaning we have $3$ linear digits covered. This means that $\overline{BD}$, the diameter of the moon, is 12$ linear digits while the amount covered, $\overline{ZD} = 3$ linear digits.
As far as the ratio of the lunar diameter to earth’s shadow at the lunar mean distance, Ptolemy adopts a ratio of $2;36 : 1$. This isn’t something we’ve directly calculated previously. Rather, the closest we’ve come previously was in this post where we calculated the ratio of the earth’s shadow at lunar apogee (instead of mean distance) and found it to be $2;40 : 1$. Presumably, Ptolemy calculated this either at the mean distance or at perigee and then averaged, but did not include the calculation in the Almagest. Regardless, Ptolemy’s value certainly sounds reasonable, so we’ll adopt it without further commentary.
Using it, we can then state that $\overline{ZH} = 31;12$ linear digits.
In addition, we can determine that the distance between the two centers, $\overline{E \Theta} = 18;36$ linear digits using the same logic as we did in the previous post.
Determining the circumference of each, we find that the moon’s circumference is $37;42$ linear digits. For the moon’s shadow it is $98;01$ linear digits.
Next, Ptolemy calculates the area for each. For the moon it comes out to be the same as we calculated for the sun in the previous post1 which was $113;06$ square linear digits. For the earth’s shadow we get $764;32$ square linear digits.
Now we’ll focus on solving the triangles created about point $K$ again. In this, $\overline{AE} = \overline{EG} = 6$ linear digits since they’re both radii of the moon. Similarly, $\overline{A \Theta} = \overline{\Theta G} = 15;36$ linear digits as they’re both radii of the shadow.
Repeating the logic from the previous post, we then determine
$$\overline{K \Theta} – \overline{EK} = \frac{\overline{A \Theta}^2 – \overline{AE}^2}{\overline{E \Theta}} $$
This works out to $11;08$ linear digits implying that $\overline{K \Theta}$ is greater than $\overline{EK}$ by this amount. Thus, knowing $\overline{E \Theta} = 18;36$ linear digits, we can work out that $\overline{EK} = 3;44$ linear digits and $\overline{K \Theta} = 14;52$ linear digits.
Then, using the Pythagorean theorem, we can determine $\overline{AK}$ which is equal to $\overline{KG}$ and is $4;42$ linear digits. Thus, the total length of $\overline{AG} = 9;24$ linear digits.
We can now use this to determine the area of $\triangle{AEK} = 8;46$ square linear digits. Doubling that, we find $\triangle{AEG} = 17;33$ square linear digits. Repeating the same calculation for the shadow, we determine $\triangle{A \Theta G} = 69;52$ square linear digits.
Now we’ll convert to parts, first for the moon, such that $\overline{BD} = 120^p$ which means $\overline{AG} = 94^p$ in that context. Similarly, if we let the diameter of the shadow be $\overline{ZG} = 120^p$, then $\overline{AG} = 36;09^p$.
Doing so allows us to look up the corresponding arc in the Table of Chords. There, we find that $arc \; ADG = 103;08º$ and $arc \; AZG = 35;04º$.
Next, we determine the area of the wedges. For the lunar disc, we take $arc \; ADG$ and divide it by $360º$ to determine the proportion of the circumference the arc represents and then multiply it by the area of the moon we determined above:
$$\frac{103;08º}{360º} \cdot 113;06 = 32;24$$
This result is in square linear digits.
Doing the same for the earth’s shadow using $arc \; AZG$:
$$\frac{35;04º}{360º} \cdot 764;32 = 74;28$$
Now, from each of these, we subtract the corresponding triangles. First for the moon (shape $AKGD$):
$$32;24 – 17;33 = 14;51$$
And for the earth’s shadow (shape $AZGD$):
$$74;28 – 65;52 = 4;36$$
We next add these together to determine the area of the overlap, shape $AZGD$ to be $19;27$ square linear digits. However, this was for when the area of the moon was $113;06$ linear square digits. Thus, we’ll need to convert these to area digits so that the total area of the moon is $12$ area digits.
$$19;27 \cdot \frac{12}{113;06} = 2;04$$
Thus, when the earth’s shadow covers $3$ digits of the moon taken linearly, it covers $\frac{2;04}{12}$ of the area of the moon which Ptolemy expresses as $2 \frac{1}{15}$. This will again get entered into our table which I will display in the next post.
