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<\/a><\/p>\nAs we demonstrated in the last post<\/a>, $\\angle AZL = 57;05\u00ba$ which means that its vertical angle $\\angle DZH$ is as well.<\/p>\n We immediately create a demi-degrees circle about $\\triangle DZH$ in which its hypotenuse, $\\overline{DZ} = 120^p.$ In it, $arc \\; DH$ will be $114;10\u00ba$ and its supplement $arc \\; ZH = 65;50\u00ba.$ The corresponding chords are $\\overline{DH} = 100;44^p$ and $\\overline{ZH} = 65;13^p.$<\/p>\n We then convert this into the context in which the diameter of the eccentres is $120^p$ and the distance between the centers is $3;25^p.$<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{DH}}{100;44^p}$$<\/p>\n $$\\overline{DH} = 2;52^p$$<\/p>\n and<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{ZH}}{65;13^p}$$<\/p>\n $$\\overline{ZH} = 1;51^p.$$<\/p>\n We can then look at $\\triangle ADH$ in which we know two sides and determine the remaining side using the Pythagorean theorem:<\/p>\n $$\\overline{AH} = \\sqrt{60^2 – 2;52^2} = 59;56^p.$$<\/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 $\\overline{A \\Theta} = 61;47^p$ and $\\overline{E \\Theta} = 5;44^p.$<\/p>\n That gives us two sides of $\\triangle AE \\Theta$ so we can again use the Pythagorean theorem to determine<\/p>\n $$\\overline{AE} = \\sqrt{61;47^2 + 5;44^2} = 62;03^p.$$<\/p>\n We can then convert this into a demi-degrees context about this triangle in which the hypotenuse, $\\overline{EA} = 120^p$<\/p>\n $$\\frac{120^p}{62;03^p} = \\frac{\\overline{E \\Theta}}{5;44^p}$$<\/p>\n $$\\overline{E \\Theta} = 11;05^p.$$<\/p>\n We can then find the corresponding arc, $arc \\; E \\Theta = 10;36\u00ba.$ This means the angle this arc subtends on the opposite side of the demi-degrees circle $\\angle EA \\Theta = 5;18\u00ba.$<\/p>\n Jumping back to our starting point, we said that $\\angle AZL = 57;05\u00ba.$ We can subtract $\\angle EA \\Theta$ from this to find $\\angle AEL = 51;47\u00ba.$<\/p>\n That [$51;47\u00ba$], then was the amount by which the planet was in advance of the apogee at the first opposition.<\/p><\/blockquote>\n We still need the position at the second opposition to be able to determine the interval so we’ll do that now.<\/p>\n From our last post<\/a>, we determined that $\\angle BZL = 18;38\u00ba.$ Again, the vertical angle $\\angle DZH$ is the same, allowing us to create a demi-degrees circle about $\\triangle DZH$ in which the hypotenuse, $\\overline{DZ} = 120^p.$<\/p>\n In that context, $arc \\; DH = 37;16\u00ba$ and its supplement, $arc \\; ZH = 142;44\u00ba.$<\/p>\n The corresponding chords are, $\\overline{DH} = 38;20^p$ and $\\overline{ZH} = 113;43^p.$<\/p>\n We’ll convert these back into the context in which the diameters of the eccentres is $120^p$:<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{DH}}{38;20^p}$$<\/p>\n $$\\overline{DH} = 1;05^p$$<\/p>\n and<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{ZH}}{113;43^p}$$<\/p>\n $$\\overline{ZH} = 3;14^p.$$<\/p>\n We can then look at $\\triangle DBH$ in which we now know two sides. So, we find the remaining side using the Pythagorean theorem:<\/p>\n $$\\overline{BH} = \\sqrt{60^2 – 1;05^2} = 59;59^p.$$<\/p>\n Next, we’ll use the fact that $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{\\Theta E} = 2 \\cdot \\overline{HD}$ to determine $\\overline{B \\Theta} = 63;13^p$ and $\\overline{E \\Theta} = 2;10^p.$ This gives us two sides of $\\triangle EB \\Theta$ so we can use the Pythagorean theorem to determine:<\/p>\n $$\\overline{EB} = 63;15^p.$$<\/p>\n We then create a demi-degrees circle about this triangle in which the hypotenuse, $\\overline{EB} = 120^p$:<\/p>\n $$\\frac{120^p}{63;15^p} = \\frac{\\overline{E \\Theta}}{2;10^p}$$<\/p>\n $$\\overline{E \\Theta} = 4;07^p.$$<\/p>\n Next, we find the corresponding arc, $arc \\; E \\Theta = 3;56\u00ba.$ Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\\angle EB \\Theta = 1;58\u00ba.$<\/p>\n Finally, we can subtract this from $\\angle BZL$ to determine that $\\angle BEL = 16;40\u00ba.$<\/p>\n Therefore, at the second opposition, the apparent position of the planet was $16;40\u00ba$ to the rear of the apogee. And we showed that, at the first opposition, it was $51;47\u00ba$ in advance of the same apogee. Therefore, the interval in apparent [longitude] from the first opposition to the second is computed as the sum of the above amount, $68;27\u00ba$, in agreement with the distance found from the observations<\/a>.<\/p><\/blockquote>\n We still need to find the second interval and thus, the third opposition.<\/p>\n Again, in our last post<\/a>, we determined that $\\angle GZL = 56;30\u00ba$ which is the same as its vertical angle $\\angle DZH$.<\/p>\n We enter the demi-degrees circle about $\\triangle DZH$ in which the hypotenuse, $\\overline{DZ} = 120^p.$ In it, $arc \\; DH = 113;00\u00ba$ and the\u00a0 supplement $arc \\; ZH = 67;00\u00ba.$ The corresponding chords are $\\overline{DH} = 100;04^p$ and $\\overline{ZH} = 66;14^p.$<\/p>\n We now convert back to the context in which the diameters of the eccentres is $120^p$:<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{DH}}{100;04^p}$$<\/p>\n $$\\overline{DH} = 2;51^p$$<\/p>\n and<\/p>\n $$\\frac{3;25^p}{120^p} = \\frac{\\overline{ZH}}{66;14^p}$$<\/p>\n $$\\overline{ZH} = 1;53^p.$$<\/p>\n We then find $\\overline{GH}$ using the Pythagorean theorem:<\/p>\n $$\\overline{GH} = \\sqrt{60^2 – 2;51^2} = 59;56^p.$$<\/p>\n And then we can $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{\\Theta E} = 2 \\cdot \\overline{HD}$ to determine $\\overline{G \\Theta} = 61;49^p$ and $\\overline{E \\Theta} = 5;42^p.$<\/p>\n That’s two of the sides of $\\triangle GE \\Theta$ so we’ll again solve for the remaining side via the Pythagorean theorem:<\/p>\n $$\\overline{GE} = \\sqrt{61;49^2 + 5;42^2} = 62;05^p.$$<\/p>\n We’ll then create a demi-degrees circle about this triangle in which the hypotenuse, $\\overline{GE} = 120^p.$ We then convert into that context:<\/p>\n $$\\frac{120^p}{62;05^p} = \\frac{\\overline{E \\Theta}}{5;42^p}$$<\/p>\n $$\\overline{E \\Theta} = 11;01^p.$$<\/p>\n Next, we look up the corresponding arc, $arc \\; E \\Theta = 10;32\u00ba$ which means that the angle this arc subtends on the opposite side of the demi-degrees circle, $\\angle EG \\Theta = 5;16\u00ba.$<\/p>\n We subtract this from $\\angle GZL$ to determine $\\angle GEL = 51;14\u00ba.$<\/p>\n That [$51;14\u00ba$], then, is the amount by which the planet was to the rear of the apogee at the third opposition. And we showed that at the second opposition, it was $16;40\u00ba$ to the rear of the same apogee. So, the distance in apparent [longitude] from the second opposition to the third, is computed as the difference, $34;34\u00ba$, which is, again, in agreement with that derived from the observations<\/a>.<\/p><\/blockquote>\n So the model checks out.<\/p>\n But still remaining in this chapter is to determine the position of Saturn about its epicycle at one of the oppositions.<\/p>\n It is immediately clear, since the planet at the third opposition had a longitude of $14;14\u00ba$ into Capricorn, and was shown to be $51;14\u00ba$ to the rear of apogee, that the apogee of its eccentre had, at that moment, a longitude of $23;00\u00ba$ into Scorpio while its perigee was diametrically opposite at $23;00\u00ba$ into Taurus.<\/p><\/blockquote>\n We’ll now draw a new diagram:<\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n