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<\/a><\/p>\nAs with before, the only thing that’s really changed in this diagram is that we’ve now placed point $G$ as our third opposition.<\/p>\n
We’ll first recall that we determined $arc \\; NX = 32;51\u00ba$ as a first approximation. Thus, the central angle, $\\angle NZX$ is as well.<\/p>\n
We’ll then enter a demi-degrees context about $\\triangle DZH$ in which the hypotenuse, $\\overline{ZD} = 120^p$. In that circle, $arc \\; DH = 65;42\u00ba$ allowing us to find the corresponding chord $\\overline{DH} = 65;06^p$.<\/p>\n
Similarly, we can find that supplementary $arc \\; ZH = 114;18\u00ba$, so the corresponding chord, $\\overline{ZH} = 100;49^p$.<\/p>\n
We then convert these to the context in which the diameter of the eccentres is $120^p$:<\/p>\n
$$\\frac{2;42^p}{120^p} = \\frac{\\overline{DH}}{65;06^p}$$<\/p>\n
$$\\overline{DH} = 1;28^p$$<\/p>\n
and<\/p>\n
$$\\frac{2;42^p}{120^p} = \\frac{\\overline{ZH}}{100;49^p}$$<\/p>\n
$$\\overline{ZH} = 2;16^p.$$<\/p>\n
We’ll now look at $\\triangle GDH$. In it, we know $\\overline{GD} = 60^p$ since it’s a radius, and we just found $\\overline{DH}$, so we can use the Pythagorean theorem to find $\\overline{GH} = 59;59^p$.<\/p>\n
Again, $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{E \\Theta} = 2 \\cdot \\overline{DH}$.<\/p>\n
Therefore, we can subtract $\\overline{H \\Theta}$ from $\\overline{GH}$ to determine $\\overline{G \\Theta} = 57;43^p$.<\/p>\n
We can then look at $\\triangle E \\Theta G$. In it, we just determined $\\overline{G \\Theta}$ and we know that $\\overline{E \\Theta} = 2;56^p$, so we can use the Pythagorean theorem to determine $\\overline{EG} = 57;47^p$.<\/p>\n
We’ll then convert into a demi-degrees context about that triangle in which the hypotenuse, $\\overline{EG} = 120^p$:<\/p>\n
$$\\frac{120^p}{57;47^p} = \\frac{\\overline{E \\Theta}}{2;56^p}$$<\/p>\n
$$\\overline{E \\Theta} = 6;06^p$$<\/p>\n
although Ptolemy rounds down to $6;05^p$.<\/p>\n
We can then find the corresponding arc $arc \\; E \\Theta = 5;49\u00ba$ although Ptolemy comes up with $5;48\u00ba$.<\/p>\n
Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\\angle EG \\Theta = 2;54\u00ba$.<\/p>\n
We’ll now look at $\\overline{GX}$ which is a radius, so has a measure of $60^p$. We can then subtract off $\\overline{Z \\Theta}$ to find that $\\overline{\\Theta X} = 55;28^p.$<\/p>\n
That gives us two of the sides in $\\triangle E \\Theta X$, so we can find the remaining side, $\\overline{EX} = 55;33^p.$<\/p>\n
We’ll then convert into a demi-degrees context about that triangle:<\/p>\n
$$\\frac{120^p}{55;33^p} = \\frac{\\overline{E \\Theta}}{2;56^p}$$<\/p>\n
$$\\overline{E \\Theta} = 6;20^p.$$<\/p>\n
We can then look up the corresponding arc for which I find, $arc \\; E \\Theta = 6;03\u00ba$ although Ptolemy finds $6;02\u00ba$. Thus, the angle which this arc subtends on the demi-degrees circle, $\\angle EX \\Theta = 3;01\u00ba$.<\/p>\n
We can then subtract:<\/p>\n
$$\\angle GEX = \\angle EX \\Theta – \\angle EG \\Theta$$<\/p>\n
$$\\angle GEX = 3;01\u00ba – 2;54\u00ba = 0;07\u00ba.$$<\/p>\n
Hence, since the planet at the third opposition, when viewed along $\\overline{EG}$, had a longitude of $14;23\u00ba$ into Aries, it is clear that, if it had been on $\\overline{EX}$, it would have had a longitude of $14;30\u00ba$ into Aries.<\/p><\/blockquote>\n
So that takes care of our three corrections.<\/p>\n
In the next post, we’ll take these into consideration and recalculate the eccentricity and line of apsides.<\/p>\n
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