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<\/a><\/p>\nSince we’ve been through this process so may times, I don’t think this image needs much explanation. But if you need a refresher, refer to our post where we did this for Mars<\/a> or Jupiter<\/a>.<\/p>\n We’ll begin by noting that $\\angle NZX = 55;52\u00ba$ as this was the angle between the first position and apogee as we calculated in the last post<\/a>. This means that $\\angle DZH$ is as well since this is the vertical angle.<\/p>\n So, we’ll focus on $\\triangle DZH$ which contains this angle, creating a demi-degrees context about it in which the hypotenuse, $\\overline{DZ} = 120^p$. In this demi-degrees circle, $arc \\; DH = 111;44\u00ba$ and its supplement, $arc \\; HZ = 68;16\u00ba.$ We can then find the corresponding chords: $\\overline{DH} = 99;20^p$ and $\\overline{HZ} = 67;20^p.$<\/p>\n Next, since we know the distance between the eccentres in the context in which the diameter of the eccentres is $120^p$, we can use that to convert to that context<\/span>1<\/sup><\/a><\/span>:<\/p>\n $$\\frac{3;34^p}{120^p} = \\frac{\\overline{DH}}{99;20^p}$$<\/p>\n $$\\overline{DH} = 2;57^p.$$<\/p>\n and<\/p>\n $$\\frac{3;34^p}{120^p} = \\frac{\\overline{HZ}}{67;20^p}$$<\/p>\n $$\\overline{HZ} = 2;00^p.$$<\/p>\n Now, since $\\overline{DA}$ is a radius with a measure of $60^p$, we can use the Pythagorean theorem on $\\triangle DAH$ to determine $\\overline{AH}$:<\/p>\n $$\\overline{AH} = \\sqrt{60^2 – 2;57^2} = 59;56^p.$$<\/p>\n And as we’ve seen several times, $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{\\Theta E} = 2 \\cdot \\overline{HD}$ which allows us to determine that $\\overline{A \\Theta} = 61;56^p.$<\/p>\n We now know two sides of $\\triangle AE \\Theta$ so we can use the Pythagorean theorem to determine the remaining side:<\/p>\n $$\\overline{AE} = \\sqrt{61;56^2 + 5;54^2} = 62;13^p.$$<\/p>\n We’ll then consider a demi-degrees circle about $\\triangle AE\\Theta$ in which the hypotenuse, $\\overline{AE} = 120^p.$ We’ll need to convert into that context:<\/p>\n $$\\frac{120^p}{62;13^p} = \\frac{\\overline{E \\Theta}}{5;54^p}$$<\/p>\n $$\\overline{E \\Theta} = 11;23^p.$$<\/p>\n Ptolemy comes up with $11;21^p$. Toomer agrees with my result, but we’ll take Ptolemy’s for consistency.<\/p>\n Next, we’ll find the corresponding arc, for which I find $arc \\; E \\Theta = 10;51\u00ba.$ This means that the angle which this arc subtends on the opposite side of the demi-degrees circle, $\\angle EA \\Theta = 5;25,30\u00ba.$<\/p>\n We’ll jump out of the demi-degrees context and back into the one in which the diameters of the eccentres is $120^p.$ In that context, $\\overline{ZX}$ is a radius, so has a measure of $60^p$ to which we can add $\\overline{Z \\Theta}$ to find $\\overline{Z \\Theta} = 64^p.$ That gives us two sides of $\\triangle E \\Theta X$ so we can use the Pythagorean theorem to find the remaining side:<\/p>\n $$\\overline{EX} = \\sqrt{64^2 + 5;54^2} = 64;16^p.$$<\/p>\n We can now examine $\\triangle EX \\Theta$, creating a demi-degrees context about it in which the hypotenuse, $\\overline{EX} = 120^p.$ We then convert into that context:<\/p>\n $$\\frac{120^p}{64;16^p} = \\frac{\\overline{E \\Theta}}{5;54^p}$$<\/p>\n $$\\overline{E \\Theta} = 11;01^p$$<\/p>\n for which Ptolemy finds $11;02^p.$<\/p>\n If we consider the corresponding arc, we find $arc \\; E \\Theta = 10;33\u00ba$ and the angle this arc subtends on the opposite side of the demi-degrees circle, $\\angle EX \\Theta = 5;16,30\u00ba.$<\/p>\n As with before, we can then subtract to find:<\/p>\n $$\\angle AEX = \\angle EA \\Theta – \\angle EX \\Theta = 5;25,30\u00ba – 5;16,30\u00ba = 0;09\u00ba.$$<\/p>\n This is our correction for the first opposition.<\/p>\n [T]he planet at the first opposition, when viewed along $\\overline{AE},$ had an apparent longitude of $1;13\u00ba$ into Libra. Thus, it is clear that if the epicycle centre were carried, not on [circle] $AL$, but on [circle] $NX,$ it would have been at point $X$ [at the first opposition], and the planet would have been seen along $\\overline{EX}, 9’$ in advance of its actual position at $A,$ with a longitude of $1;04\u00ba$ into Libra.<\/p><\/blockquote>\n Moving on to the second correction, we’ll reproduce the diagram for that situation:<\/p>\n In this new diagram, $\\angle NDX = 19;51\u00ba$ as that’s the angle we calculated that Saturn appeared after apogee for the second opposition. Thus, its vertical angle, $\\angle DZH = 19;51\u00ba$ as well.<\/p>\n We then create a demi-degrees circle about $\\triangle ZDH$ in which the hypotenuse, $\\overline{ZD} = 120^p$. In it, $arc \\; DH = 39;42\u00ba$ and its supplement, $arc \\; ZH = 140;18\u00ba.$ The corresponding chords are $\\overline{DH} = 40;45^p$ and $\\overline{ZH} = 112;52^p.$<\/p>\n This gets converted into the context in which the diameter of the eccentres is $120^p$:<\/p>\n $$\\frac{3;34^p}{120^p} = \\frac{\\overline{DH}}{40;45^p}$$<\/p>\n $$\\overline{DH} = 1;13^p$$<\/p>\n and<\/p>\n $$\\frac{3;34^p}{120^p} = \\frac{\\overline{ZH}}{112;52^p}$$<\/p>\n $$\\overline{ZH} = 3;21^p.$$<\/p>\n We can then look at $\\triangle BDH$ in which we know two sides, allowing us to find the third using the Pythagorean theorem:<\/p>\n $$\\overline{BH} = \\sqrt{60^2 – 1;13^2} = 59;59^p.$$<\/p>\n And as we’ve seen several times, $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{\\Theta E} = 2 \\cdot \\overline{HD}.$<\/p>\n Thus, we can add $\\overline{H \\Theta}$ onto $\\overline{BH}$ to determine $\\overline{B \\Theta} = 63;20^p.$ And since we also know $\\overline{\\Theta E} = 2;26^p$ we know two sides of $\\triangle BE \\Theta$ allowing us to find the third using the Pythagorean theorem:<\/p>\n $$\\overline{EB} = \\sqrt{63;20^2 + 2;26^2} = 63;23^p.$$<\/p>\n We then enter a demi-degrees context about this circle in which its hypotenuse, $\\overline{EB} = 120^p$:<\/p>\n $$\\frac{120^p}{63;23^p} = \\frac{\\overline{\\Theta E}}{2;26^p}$$<\/p>\n $$\\overline{\\Theta E} = 4;36^p.$$<\/p>\n The corresponding arc, $arc \\; \\Theta E = 4;24\u00ba$ and the angle this arc subtends in the demi-degrees circle, $\\angle \\Theta BE = 2;12\u00ba.$<\/p>\n We can add $\\overline{Z \\Theta}$\u00a0 onto that to determine $\\overline{X \\Theta} = 66;42^p$ which gives us two of sides of $\\triangle XE \\Theta$. We find the third with the Pythagorean theorem:<\/p>\n $$\\overline{EX} = \\sqrt{66;42^2 + 2;26^2} = 66;45^p.$$<\/p>\n We’ll then create a demi-degrees circle about $\\triangle XE \\Theta$ in which the hypotenuse, $\\overline{EX} = 120^p.$<\/p>\n $$\\frac{120^p}{66;45^p} = \\frac{\\overline{E \\Theta}}{2;26^p}$$<\/p>\n $$\\overline{E \\Theta} = 4;22^p$$<\/p>\n which Ptolemy rounds up to $4;23^p.$<\/p>\n We then find the corresponding arc, $arc \\; E \\Theta = 4;11\u00ba$ for which Ptolemy finds $4;12\u00ba$. Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\\angle EX \\Theta = 2;06\u00ba.$<\/p>\n Finally, we can subtract:<\/p>\n $$\\angle BEX = \\angle \\Theta BE – \\angle EX \\Theta = 0;06\u00ba.$$<\/p>\n Here too, then, it is clear, since the planet, at the second opposition, when viewed along $\\overline{EB},$ had a longitude of $9;40$ into Sagittarius, that if, instead, it were viewed along $\\overline{EX},$ it would have a longitude of $9;46\u00ba$ into Sagittarius. And we showed that, at the first opposition, it would, on the same hypothesis, have had a longitude of $1;04\u00ba$ into Libra. Hence, it is clear that the interval in apparent [longitude] from the first to the second opposition, if it were taken with respect to the eccentre [circle] $NX,$ would be $68;42\u00ba$ of the ecliptic.<\/p><\/blockquote>\n We’ll continue on to the final opposition for Saturn:<\/p>\n<\/a><\/p>\n