[I]t is declared that in the $45^{th}$ year of the calendar of Dionysius, on Parthenon $10$, that the planet Jupiter occulted the southernmost [of the two] Aselli at dawn. Now, the moment [of the observation] is in the $83^{rd}$ year from the death of Alexander, Epiphi [XI] $17\/18$ in the Egyptian calendar [$240$ BCE Sept. $3\/4$], dawn.<\/p>\n
For that time, we find the longitude of the mean sun as $9;56\u00ba$ into Virgo. But, the star called ‘the southern Asellus’<\/span>1<\/sup><\/a><\/span> among those surrounding the nebula in Cancer had a longitude, at the time of our observation [of it], of $11 \\frac{1}{3}\u00ba$ into Cancer. Hence, obviously, its longitude at the observation in question was $7;33\u00ba$ [into Cancer], since to the $378$ years between the observations corresponds to [a precessional motion of] $3;47\u00ba.$<\/p>\n Therefore, the longitude of Jupiter at that moment (since it had occulted the star) was also $7;33\u00ba$ into Cancer. Similarly, since the apogee was $11\u00ba$ into Virgo in our times, it must have had a longitude of $7;13\u00ba$ into Virgo at the observation. Hence, it is clear that the distance of the apparent planet from the then apogee of the eccentre was $300;20\u00ba,$ while the distance of the mean sun from that same apogee was $2;43\u00ba.$<\/p><\/blockquote>\n Using this information, Ptolemy sketches a new diagram:<\/p>\n In this diagram, we have the line of apsides as $\\overline{AG}$. Points $Z$, $D$, and $E$ remain as the equant, center of mean distance, and observer respectively.<\/p>\n The epicycle is centered on $B$ with Jupiter at $\\Theta$. The mean sun is at $L.$ We’ll create $\\overline{ZB}$ which gets extended to the opposite side of the epicycle at $H$, as well as lines $\\overline{DB}$, and $\\overline{E \\Theta}$.<\/p>\n We then drop perpendiculars from $Z$ onto $\\overline{DB}$ at $K$ and from $D$ onto $\\overline{E \\Theta}$ at $M$.<\/p>\n We’ll also extend a perpendicular from $B$ onto $\\overline{E \\Theta}$ at $N$, extending it as far as $X$ where it meets a line created from $D$ that is parallel to $\\overline{E \\Theta}$, thus forming parallelogram $DMNX.$<\/p>\n To get started, we’ll consider $\\angle AE \\Theta$. This is the apparent distance of the planet before apogee. We stated above that the planet was observed to be $300;20\u00ba$ after the apogee, which means that $\\angle AE \\Theta$ is a full circle minus this amount of $59;40\u00ba.$<\/p>\n Additionally, $\\angle AEL$ is the distance between the apogee and the mean sun from the point of view of the observer. Ptolemy has calculated this to be $2;43\u00ba$ for us.<\/p>\n We can add these two angles together to determine $\\angle LE \\Theta = 62;23\u00ba.$<\/p>\n Now, recall that $\\overline{LE} \\parallel \\overline{\\Theta B}$ because the angle of the planet about the epicycle is always equal to that of the mean sun from the point of view of the observer. This makes $\\angle E \\Theta B$ an alternate angle, giving it the same value as $\\angle LE \\Theta$ or $62;23\u00ba.$<\/p>\n We can then focus on $\\triangle B\\Theta N,$ creating a dem-degrees circle about it. In it, the hypotenuse, $\\overline{B \\Theta} = 120^p$ and $arc \\; BN = 124;46\u00ba.$ The corresponding chord, then, $\\overline{BN} = 106;20^p.$<\/p>\n Since we know the size of this (the radius of the epicycle) in the context in which the diameter of the eccentres is $120^p$, we can use that to convert:<\/p>\n $$\\frac{11;30^p}{120^p} = \\frac{\\overline{BN}}{106;20^p}$$<\/p>\n $$\\overline{BN} = 10;11^p.$$<\/p>\n Ptolemy rounds up to $10;12^p.$<\/p>\n We’ll now look at $\\triangle DEM$. In it, we know that $\\angle DEM = 59;40\u00ba$ which means that the remaining non-right angle, $\\angle MDE = 30;20\u00ba$. We’ll now create a demi-degrees context about this triangle in which the hypotenuse, $\\overline{DE} = 120^p$. Additionally $arc \\; DM = 119;20\u00ba.$ Thus, the corresponding chord, $\\overline{DM} = 103;34^p$.<\/p>\n And since we know that $\\overline{DE} = 2;45^p$ in the context in which the diameters of the eccentres is $120^p$, we can convert to that context:<\/p>\n $$\\frac{2;45^p}{120^p} = \\frac{\\overline{DM}}{103;34^p}$$<\/p>\n $$\\overline{DM} = 2;22^p.$$<\/p>\n Ptolemy finds it to be $2;23^p.$<\/p>\n Since $DMNX$ is a parallelogram, this is the same as $\\overline{NX}$ which we can add to $\\overline{BN}$ to find that $\\overline{BX} = 12;35^p.$<\/p>\n Next, we can focus on $\\triangle BDX,$ creating a demi-degrees context about it wherein the hypotenuse $\\overline{BD} = 120^p.$ We can use this to convert into the demi-degrees context:<\/p>\n $$\\frac{120^p}{60^p} = \\frac{\\overline{BX}}{12;35^p}$$<\/p>\n $$\\overline{BX} = 25;10^p.$$<\/p>\n We can the find the corresponding arc, $arc \\; BX = 24;13\u00ba$ although Ptolemy finds it to be $24;14\u00ba.$ In either case, this means that the angle this arc subtends on the opposite side of the demi-degrees circle $\\angle BDX = 12;07\u00ba.$<\/p>\n Next, take a look at $\\angle XDB$. This is a right angle. Thus, we can subtract $\\angle BDX$ out of it which gives us $\\angle BDM = 77;53\u00ba.$<\/p>\n We can then add $\\angle MDE$ to find $\\angle BDE = 108;13\u00ba$.<\/p>\n We’ll then find the supplementary angle to this along $\\overline{AG}$ to find $\\angle ADB = 71;47\u00ba.$<\/p>\n Now we’ll focus on that skinny triangle, $\\triangle ZDK,$ creating a demi-degrees circle about it in which the hypotenuse, $\\overline{ZD} = 120^p.$<\/p>\n In that, $arc \\; ZK = 143;34\u00ba$ and its supplement, $arc \\; DK = 36;26\u00ba.$ We can then find the corresponding chords, $\\overline{ZK} = 113;59^p$ and $\\overline{DK} = 37;31^p.$<\/p>\n This can be converted into the context in which the diameter of the eccentres is $120^p$:<\/p>\n $$\\frac{2;45^p}{120^p} = \\frac{\\overline{ZK}}{113;59^p}$$<\/p>\n $$\\overline{ZK} = 2;37^p$$<\/p>\n and<\/p>\n $$\\frac{2;45^p}{120^p} = \\frac{\\overline{DK}}{37;31^p}$$<\/p>\n $$\\overline{DK} = 0;52^p.$$<\/p>\n We’ll then subtract $\\overline{DK}$ off of $\\overline{DB}$ to find that $\\overline{KB} = 59;08^p.$<\/p>\n We can then focus on $\\triangle ZBK$ in which we know two of the sides. Thus, we can find the remaining side:<\/p>\n $$\\overline{ZB} = \\sqrt{59;08^2 + 2;37^2} = 59;11^p$$<\/p>\n which Ptolemy rounds up to $59;12^p.$<\/p>\n We’ll now create a demi-degrees context about this triangle in which\u00a0 the hypotenuse, $\\overline{ZB} = 120^p.$ Converting into that context:<\/p>\n $$\\frac{120^p}{59;12^p} = \\frac{\\overline{ZK}}{2;37^p}$$<\/p>\n $$\\overline{ZK} = 5;18^p.$$<\/p>\n We can then find the corresponding arc $arc \\; ZK = 5;04\u00ba.$ Thus, the angle that this arc subtends on the opposite side of the circle, $\\angle ZBD = 2;32\u00ba.$<\/p>\n If we add this to $\\angle ADB$ we find<\/p>\n $$\\angle AZB = 71;47\u00ba + 2;32 = 74;19\u00ba.$$<\/p>\n This next part is a bit hard to see, so let play with our diagram a bit.<\/p>\n In this diagram, I’ve exaggerated the angles of $\\overline{LE}$ and since that’s parallel to $\\overline{\\Theta B},$ that as well.<\/p>\n I’ve also extended $\\overline{HZ}$ until it hits $\\overline{LE}$, creating a new point at $T$ and I’ve created $\\overline{ES}$ which is parallel to $\\overline{ZH}$.<\/p>\n Then, I’ve marked out the angles that are equal:<\/p>\n $$\\angle LTZ = \\angle LES = \\angle \\Theta BH$$<\/p>\n and<\/p>\n $$\\angle AZH = \\angle AES.$$<\/p>\n Now, let’s look at $\\angle LES$. This is quite clearly $\\angle LEA + \\angle AES$.<\/p>\n However, due to the angles we just stated were equal, we can do some substituting to write:<\/p>\n $$\\angle \\Theta BH = \\angle LEA + \\angle AZH.$$<\/p>\n We know $\\angle LEA = 2;43\u00ba$ as this was the angle of the mean sun past apogee. We just found $\\angle AZH = 74;19\u00ba.$<\/p>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2024\/08\/AlmagestFig-11.11.jpg?resize=300%2C282&ssl=1\")
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