![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2024\/04\/AlmagestFig-10.12.jpg?ssl=1\")
<\/a><\/p>\nWe’ll again reuse the diagram from the last time around.<\/p>\n
This time, $arc \\; HP = 45;28\u00ba$ as we’d found it in this post<\/a> in which we’d called it $arc \\; GM$.<\/p>\n This means that $\\angle H \\Theta P$ has the same measure.<\/p>\n In this iteration, we also showed that the eccentricity, $\\overline{D \\Theta} = \\overline{DN} = 5;55^p$.<\/p>\n Then, using a bit of trig we can state:<\/p>\n $$\\overline{DF} = 5;55^p \\cdot sin(45;28)$$<\/p>\n $$\\overline{DF} = 4;13^p.$$<\/p>\n Additionally,<\/p>\n $$\\overline{F \\Theta} = 5;55^p \\cdot cos(45;28)$$<\/p>\n $$\\overline{F \\Theta} = 4;09^p.$$<\/p>\n Recalling that $\\overline{DG} = 60^p$ since it’s a radius, we then have two sides in $\\triangle FGD$ allowing us to use the Pythagorean theorem to state:<\/p>\n $$\\overline{FG} = \\sqrt{60^2 – 4;13^2}$$<\/p>\n $$\\overline{FG} = 59;51^p.$$<\/p>\n As with before, $\\overline{QF} = \\overline{F \\Theta} = 4;09^p$.<\/p>\n We can then write:<\/p>\n $$\\overline{GQ} = \\overline{FG} – \\overline{QF}$$<\/p>\n $$\\overline{GQ} = 59;51^p-4;09^p = 55;42^p.$$<\/p>\n Additionally, $\\overline{QN} = 8;26^p$ since it’s twice $\\overline{DF}$.<\/p>\n That gives us two sides of $\\triangle NGQ$, so we can use a bit more trig to determine,<\/p>\n $$\\angle NGQ = tan^{-1}(\\frac{\\overline{QN}}{\\overline{GQ}})$$<\/p>\n $$\\angle NGQ = tan^{-1}(\\frac{8;26^p}{55;42^p})$$<\/p>\n $$\\angle NGQ = 8;37\u00ba.$$<\/p>\n Next, we’ll recall that $\\overline{H \\Theta} = 60^p$, again being a radius. We can subtract off $\\overline{Q \\Theta}$ to find that,<\/p>\n $$\\overline{QH} = 51;42^p.$$<\/p>\n That now gives us two of the sides in $\\triangle NHQ$ allowing us to write:<\/p>\n $$\\angle NHQ = tan^{-1}(\\frac{\\overline{QN}}{\\overline{QH}})$$<\/p>\n $$\\angle NHQ = tan^{-1}(\\frac{8;26}{51;42^p})$$<\/p>\n $$\\angle NHQ = 9;16\u00ba.$$<\/p>\n Lastly, we can subtract to find $\\angle GNH$:<\/p>\n $$\\angle GNH = \\angle NHQ – \\angle NGQ$$<\/p>\n $$\\angle GNH = 9;16\u00ba – 8;37\u00ba$$<\/p>\n $$\\angle GNH = 0;39\u00ba.$$<\/p>\n This is also $arc \\; MY$ which is our correction.<\/p>\n Ptolemy comes up with a slightly higher value of $0;40\u00ba$ which is in agreement with the value Toomer gives of $0;39;31\u00ba$. If I carry my calculation out to the second sexagesimal place I find it to be $0;39,22\u00ba$. So again, slightly different but within reason.<\/p>\n That’s the end of the corrections.<\/p>\n We now have everything we need to go through our third iteration which we’ll do in the next couple posts.<\/p>\n
\n