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<\/a><\/p>\nAdditionally $X$ is on $\\overline{ZB}$ where it crosses the circle of mean motion and $N$ has been relocated near perigee on the same circle.<\/p>\n
To begin, recall that our first approximation<\/a> for $\\angle XZN$ was $0;35\u00ba$ before the perigee. We’ll first look at $\\triangle DZH$. In it, the hypotenuse, $\\overline{ZD} = 120^p$ and $arc \\; DH = 1;10\u00ba$. We can then look up the corresponding chord for which I find $\\overline{DH} = 1;13^p$.<\/p>\n Similarly, we can take the supplement of $arc \\; DH$ to find $arc \\; ZH = 178;50\u00ba$ and its corresponding chord $\\overline{ZH} = 119;59,37\u00ba$ which Ptolemy just calls $\\approx 120^p$.<\/p>\n We’ll now convert these pieces back to the context in which the diameters of the eccentres is $120^p$:<\/p>\n $$\\frac{2;42^p}{120^p} = \\frac{\\overline{DH}}{1;13^p}$$<\/p>\n $$\\overline{DH} = 0;01,39^p$$<\/p>\n which Ptolemy rounds to $0;02^p$<\/span>1<\/sup><\/a><\/span>.<\/p>\n $$\\frac{2;42^p}{120^p} = \\frac{\\overline{ZH}}{119;59,37^p}$$<\/p>\n $$\\overline{ZH} = 2;41,59^p$$<\/p>\n which Ptolemy rounds to $2;42^p$.<\/p>\n Ptolemy then considers $\\triangle BHD$ to calculate $\\overline{BH}$ within it. However, because $\\overline{DH}$ is so small, he doesn’t show any work and just concludes that it is “negligibly smaller than the hypotenuse, $\\overline{BD}$.”<\/p>\n In short, he takes $\\overline{BH} = 60^p$ as well.<\/p>\n As we’ve seen before, $\\overline{\\Theta H} = \\overline{HZ}$ and $\\overline{E \\Theta} = 2 \\cdot \\overline{DH}$.<\/p>\n