the second year of Antoninus, Mesore [XII], $26\/27$ in the Egyptian calendar [$139$ CE, July $10\/11$], before sunrise, i.e., about $5$ equinoctial hours after midnight (for the mean longitude of the sun was $16;11\u00ba$ into Aries, and the second degree of Aries [i.e., $1-2\u00ba$ of Aries] was culminating according to the astrolabe). At that moment, Jupiter, when sighted with respect to the bright star in the Hyades<\/span>1<\/sup><\/a><\/span>, was seen to have a longitude of $15 \\frac{3}{4}\u00ba$ into Gemini.<\/p><\/blockquote>\n That gives the apparent position with respect to \u03b1 Tau.<\/p>\n Next, Ptolemy gives us the calculated position of the moon because, he says, Jupiter<\/p>\n had the same apparent longitude as the center of the moon, which lay to the south of it.<\/p>\n For that moment<\/span>2<\/sup><\/a><\/span> we find, by means of the calculations [previously] explained:<\/p>\n Moon’s mean longitude: $9;00\u00ba$ into Gemini Hence its true position: $14;50\u00ba$ into Gemini Thus, from these considerations too, Jupiter’s longitude was $15 \\frac{3}{4}\u00ba$ into Gemini.<\/p>\n Ptolemy then gives the time interval from the third opposition to the above observation as $1$ Egyptian year and $276$ days.<\/p><\/blockquote>\n This allows him to calculate the change in the longitude and anomaly as:<\/p>\n in longitude: $53;17\u00ba$ My calculations from the planetary mean motions table agrees with this.<\/p>\n We then add these increases over this interval to the positions stated in the last post<\/a>. Thus, we find the positions at the time of this observation were:<\/p>\n in longitude: $263;53\u00ba$ from the apogee We now produce a diagram of this configuration:<\/p>\n Quickly explaining this diagram, we have $\\overline{AG}$ as the line of apsides. Then points $Z$, $D$, and $E$ are the center of mean motion, mean distance, and the observer, respectively.<\/p>\n All three of these are then connected to $B$, the center of the epicycle with $\\overline{ZB}$ being extended to the far side of the epicycle at $H$, and $\\overline{EB}$ extended to the far side at $\\Theta$.<\/p>\n Jupiter will be at point $K$ and we’ll create $\\overline{EK}$ onto which we’ll drop a perpendicular from $B$ at $N$.<\/p>\n Perpendiculars are also dropped from $D$ and $E$ onto $\\overline{ZH}$ at $M$ and $L$ respectively.<\/p>\n Then, since the mean position in longitude from the apogee of the eccentre is $263;53\u00ba$, $\\angle GBZ = 83;53\u00ba$.<\/p><\/blockquote>\n We can then look at $\\triangle DZM$ which contains this angle and create a demi-degrees context about it in which the hypotenuse, $\\overline{DZ} = 150^p.$ In that context, $arc \\; DM = 167;46\u00ba$ and its supplement, $arc \\; ZM = 12;14\u00ba$.<\/p>\n We can then find the corresponding chords:\u00a0$\\overline{DM} = 119;19^p$ and $\\overline{ZM} = 12;47^p.$<\/p>\n This, then, gets converted to the context in which the diameters of the eccentres is $120^p$:<\/p>\n $$\\frac{2;45^p}{120^p} = \\frac{\\overline{DM}}{119;19^p}$$<\/p>\n $$\\overline{DM} = 2;44^p$$<\/p>\n and<\/p>\n $$\\frac{2;45^p}{120^p} = \\frac{\\overline{ZM}}{12;47^p}$$<\/p>\n $$\\overline{ZM} = 0;18^p.$$<\/p>\n We’ll now focus on $\\triangle DMB$ in which we know two sides. Thus, we can find the third side using the Pythagorean theorem:<\/p>\n $$\\overline{MB} = \\sqrt{60^2 – 2;55^2} = 59;56^p.$$<\/p>\n Then, we can subtract off $\\overline{ML}$ which has the same length as $\\overline{ZM}$ to determine that $\\overline{LB} = 59;38^p.$<\/p>\n As we’ve seen, $\\overline{EL} = 2 \\cdot \\overline{DM} = 5;28^p.$<\/p>\n This means we know two sides of $\\triangle ELB$, so we can find the remaining side:<\/p>\n $$\\overline{EB} = \\sqrt{59;38^2 + 5;28^2} = 59;53^p$$<\/p>\n although Ptolemy gets $59;52^p$ likely due to having computed with greater precision.<\/p>\n We’ll now create a demi-degrees circle about this triangle in which the hypotenuse, $\\overline{EB} = 120^p$ and we’ll convert $\\overline{EL}$ into that context:<\/p>\n $$\\frac{120^p}{59;53^p} = \\frac{\\overline{EL}}{5;28^p}$$<\/p>\n $$\\overline{EL} = 10;57^p.$$<\/p>\n Ptolemy comes up with $10;58^p.$<\/p>\n We then find the corresponding arc, $arc \\; EL = 10;30\u00ba.$<\/p>\n Thus, the angle opposite it on the demi-degrees circle, $\\angle EBL = 5;15\u00ba.$<\/p>\n Next, recall that $\\angle GZB = 83;53\u00ba.$<\/p>\n If we add these two, we get $\\angle BEG = 89;08\u00ba.$<\/p>\n Furthermore, since the approximate longitude of the perigee, $G$, is $11\u00ba$ into Pisces, and the apparent longitude of the planet, as viewed along $\\overline{EK}$, was $15;45\u00ba$ into Gemini, $\\angle KEG = 94;45\u00ba.$<\/p><\/blockquote>\n We can then subtract $\\angle BEG$ from this to determine $\\angle KEB = 5;37\u00ba.$<\/p>\n We’ll now create a demi-degrees circle about $\\triangle BEN$ where the hypotenuse, $\\overline{EB} = 120^p$. In it, $arc \\; BN = 11;14\u00ba.$ As a quick note for later, this also means that the angle this arc subtends, $\\angle BEN = 5;37\u00ba.$<\/p>\n We can then find the corresponding chord, $\\overline{BN} = 11;45^p$ although Ptolemy rounds down to $11;44^p$.<\/p>\n This gets converted back to our context in which the diameter of the eccentre equals $120^p:$<\/p>\n $$\\frac{59;52^p}{120^p} = \\frac{\\overline{BN}}{11;45^p}$$<\/p>\n $$\\overline{BN} = 5;52^p.$$<\/p>\n Ptolemy comes out a bit low at $5;50^p.$<\/p>\n Next, we can state that $arc \\; HK = 41;18\u00ba$ as this is the distance we determined that Jupiter was about its epicycle from the apogee<\/span>3<\/sup><\/a><\/span>. Thus, so is $\\angle HBK$.<\/p>\n We also determined that $\\angle EBZ = 5;15\u00ba$ which we can subtract off of $\\angle HBK$ to determine $\\angle KB \\Theta = 36;03\u00ba.$<\/p>\n Additionally, we showed that $\\angle \\Theta EK = 5;37\u00ba.$<\/p>\n We can then subtract:<\/p>\n $$\\angle BKN = \\angle KB \\Theta – \\angle \\Theta EK = 30;26\u00ba.$$<\/p>\n We’ll then create a demi-degrees context about $\\triangle BNK$ in which the hypotenuse, $\\overline{BK} = 120^p$. We can then state that $arc \\; BN = 60;52\u00ba$ and the corresponding chord, $\\overline{BN} = 60;47^p.$<\/p>\n We can use this to convert $\\overline{BK}$, the radius of the epicycle, back to our context in which the diameter of the eccentres is $120^p$:<\/p>\n $$\\frac{5;50^p}{60;47^p} = \\frac{\\overline{BK}}{120^p}$$<\/p>\n $$\\overline{BK} = 11;31^p.$$<\/p>\n Ptolemy rounds down to $11;30^p$.<\/p>\n You may well note that I’ve called out a number of instances in which Ptolemy did a bit of rounding. Toomer notes that this has caused a moderate amount of accumulated error as a precise value during these calculations would have let to a result of the radius of the epicycle being $11;38^p.$ Toomer speculates that Ptolemy was likely aiming for a nice round number, as seems to be his preference<\/span>4<\/sup><\/a><\/span>.<\/p>\n That concludes Chapter $2$!<\/p>\n
\nMoon’s [mean] anomaly counted from the epicycle apogee: $272;05\u00ba$<\/p>\n
\nand its apparent position at Alexandria: $15;45\u00ba$ into Gemini.<\/p>\n
\nand in anomaly: $218;31\u00ba.$<\/p><\/blockquote>\n
\nand in anomaly: $41;18\u00ba$ from the apogee of the epicycle<\/p><\/blockquote>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2024\/08\/AlmagestFig-11.10.jpg?resize=287%2C300&ssl=1\")
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