![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2022\/03\/AlmagestFig6-5lunar.jpg?ssl=1\")
<\/a><\/p>\nWe 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\u00a0 $\\overline{BD}$, the diameter of the moon, is 12$ linear digits while the amount covered, $\\overline{ZD} = 3$ linear digits.<\/p>\n
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<\/a> 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\u00a0Almagest<\/em>. Regardless, Ptolemy’s value certainly sounds reasonable, so we’ll adopt it without further commentary.<\/p>\n Using it, we can then state that $\\overline{ZH} = 31;12$ linear digits.<\/p>\n 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<\/a>.<\/p>\n 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.<\/p>\n 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 post<\/span>1<\/sup><\/a><\/span> which was $113;06$ square linear digits. For the earth’s shadow we get $764;32$ square linear digits.<\/p>\n 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.<\/p>\n Repeating the logic from the previous post, we then determine<\/p>\n $$\\overline{K \\Theta} – \\overline{EK} = \\frac{\\overline{A \\Theta}^2 – \\overline{AE}^2}{\\overline{E \\Theta}} $$<\/p>\n 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.<\/p>\n 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.<\/p>\n 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.<\/p>\n 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 $\\overline{AG} = 36;09^p$.<\/p>\n Doing so allows us to look up the corresponding arc in the Table of Chords<\/a>. There, we find that $arc \\; ADG = 103;08\u00ba$ and $arc \\; AZG = 35;04\u00ba$.<\/p>\n Next, we determine the area of the wedges.\u00a0For the lunar disc, we take $arc \\; ADG$ and divide it by $360\u00ba$ to determine the proportion of the circumference the arc represents and then multiply it by the area of the moon we determined above:<\/p>\n $$\\frac{103;08\u00ba}{360\u00ba} \\cdot 113;06 = 32;24$$<\/p>\n This result is in square linear digits.<\/p>\n Doing the same for the earth’s shadow using $arc \\; AZG$:<\/p>\n $$\\frac{35;04\u00ba}{360\u00ba} \\cdot 764;32 = 74;28$$<\/p>\n Now, from each of these, we subtract the corresponding triangles. First for the moon (shape $AKGD$):<\/p>\n $$32;24 – 17;33 = 14;51$$<\/p>\n And for the earth’s shadow (shape $AZGD$):<\/p>\n $$74;28 – 65;52 = 4;36$$<\/p>\n 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.<\/p>\n $$19;27 \\cdot \\frac{12}{113;06} = 2;04$$<\/p>\n 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.<\/p>\n
\n