However, what we had was $\\angle LNG$ which was the apparent change along the ecliptic from the point of view of the observer.<\/p>\n As you can tell, they’re close, but not quite the same. However, we could get to what we needed from what we had by subtracting $\\angle LNT$ and $\\angle YNM$.<\/p>\n We found both of these angles to be $0;33\u00ba$ in this post<\/a> and $0;50\u00ba$ in this post<\/a>, respectively.<\/p>\n Unfortunately, both of those values were calculated starting from an incorrect value for $\\angle ZNH$. But, the good news is that applying them anyway will (Ptolemy hopes) still get us closer to the correct value! And then we’ll be able to iterate through again to reduce the discrepancy even further.<\/p>\n So let’s get started.<\/p>\n As a quick note, I’m doing these calculations in a Google Sheet<\/a>. In it, each step references the cells of the previous steps. That way, if I make an error, I can correct it upstream and have the sheet calculate everything downstream, saving myself a lot of work. Because Sheets is saving the full value behind the scenes, it also prevents successive rounding errors. However, I’m only displaying the values to the first sexagesimal place, so sometimes things won’t quite add up as if I were doing each calculation discreetly.<\/p>\n I’ve also color coded the components. Green indicates that the values don’t change in each iteration. Red indicates they do.<\/p>\n We’ll begin by using the same diagram as we did previously:<\/p>\n If you need a reminder on how this diagram is laid out, you can find it in the original post<\/a>.<\/p>\n As with before, $arc \\; BG = 95;28\u00ba$. This hasn’t changed since this was determined from the mean motion tables about this circle over the interval given in this post<\/a>.<\/p>\n Next, we can apply our revisions to $\\angle BDG$.<\/p>\n Previously, we stated that $\\angle BDG = 93;44\u00ba$. But we need to subtract $\\angle LNT$ and add $\\angle YNM$. Thus,<\/p>\n $$\\angle BDG = 93;44\u00ba – 0;33\u00ba – 0;50\u00ba = 92;21\u00ba.$$<\/p>\n Its supplement, $\\angle EDH = 87;39\u00ba$.<\/p>\n We’ll then focus on $\\triangle EDH$. Previously, we created a demi-degrees circle about this to solve it. However, I’ll instead use modern trig to save time:<\/p>\n $$\\overline{EH} = 120 \\cdot sin(87;39\u00ba) = 119;54\u00ba.$$<\/p>\n Again, this is in the context where $\\overline{ED} = 120^p$.<\/p>\n Next, $\\angle BEG = 47;44\u00ba$ as it did previously, which again gives us the second angle in $\\triangle BED$. Thus, we can find the remaining angle,<\/p>\n $$\\angle DBE = 180\u00ba – 47;44\u00ba – 87;39\u00ba = 44;37\u00ba.$$<\/p>\n Focusing next on $\\triangle BEH$, in the context where its hypotenuse $\\overline{BE} = 120^p$,<\/p>\n $$\\overline{EH} = 120 \\cdot sin(44;37\u00ba) = 84;17^p.$$<\/p>\n We’ve now found $\\overline{EH}$ in two contexts, so we can use that to convert $\\overline{BE}$ to the context in which $\\overline{DE} = 120^p$:<\/p>\n $$\\frac{119;54^p}{84;17^p} = \\frac{\\overline{BE}}{120^p}$$<\/p>\n $$\\overline{BE} = 170;43^p.$$<\/p>\n We’ll now turn to $\\angle ADG$ for which we will also need to apply our corrections. Previously, this angle was<\/p>\n $$\\angle ADG = 161;34\u00ba.$$<\/p>\n However, we need to add $\\angle KNS$ (which we found<\/a> to be $0;32\u00ba$) and subtract $\\angle YNM$ (which we found<\/a> to be $0;50\u00ba$). Hence:<\/p>\n $$\\angle ADG = 161;34\u00ba + 0;32\u00ba – 0;50\u00ba = 161;16\u00ba.$$<\/p>\n Next, we can find its supplement, $\\angle ADE$:<\/p>\n $$\\angle ADE = 180\u00ba – \\angle ADG = 180\u00ba – 161;16\u00ba = 18;44\u00ba.$$<\/p>\n Now we can focus on $\\triangle ZDE$. It’s already in the context we want in which the hypotenuse, $\\overline{ED} = 120^p$.<\/p>\n We can then find that<\/p>\n $$\\overline EZ = 120 \\cdot sin(18;44\u00ba) = 38;32^p.$$<\/p>\n We then look at $arc \\; AG = 177;12\u00ba$ indicating $\\angle AEG = 88;36\u00ba$.<\/p>\n So, in $\\triangle ADE$, we now know $\\angle ADE = 18;44\u00ba$ and $\\angle AED = 88;36\u00ba$ which means the remaining angle, $\\angle DAE$ is $72;40\u00ba$.<\/p>\n Next, we’ll look at $\\triangle AEZ$. In it, hypotenuse $\\overline{AE} = 120^p$.<\/p>\n Therefore,<\/p>\n $$\\overline{EZ} = 120 \\cdot sin(72;40\u00ba) = 114;33^p.$$<\/p>\n This allows us to convert $\\overline{AE}$ to our context in which $\\overline{ED} = 120^p$.<\/p>\n $$\\frac{38;32^p}{114;33^p} = \\frac{\\overline{AE}}{120^p}$$<\/p>\n $$\\overline{AE} = 40;22^p.$$<\/p>\n We’ll put that aside and look at $arc \\; AB = 81;44\u00ba$ which was the mean motion that occurred between the first and second observation. Therefore, $\\angle AEB = 40;52\u00ba$ as it is the angle that subtends the arc.<\/p>\n Focusing next on $\\triangle AE \\Theta$, $\\angle AE \\Theta = 40;52\u00ba$ and the hypotenuse $\\overline {AE} = 120^p$. Thus,<\/p>\n $$\\overline{A \\Theta} = 120 \\cdot sin(40;52\u00ba) = 78;31^p.$$<\/p>\n We’ll again context switch this into the context where $\\overline{ED} = 120^p$:<\/p>\n $$\\frac{40;22^p}{120^p} = \\frac{\\overline{A \\Theta}}{78;31^p}$$<\/p>\n $$\\overline{A \\Theta} = 26;25^p.$$<\/p>\n And we can use the Pythagorean theorem to find $\\overline{E \\Theta}$ in this context too which gives me $\\overline{E \\Theta} = 30;32^p$.<\/p>\n We’ll now subtract $\\overline{E \\Theta}$ from $\\overline{BE}$:<\/p>\n $$\\overline{B \\Theta} = 170;43^p – 30;32^p = 140;11^p.$$<\/p>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2024\/03\/AlmagestFig-10.7-3.jpg?resize=229%2C300&ssl=1\")
<\/a><\/p>\n![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2024\/03\/AlmagestFig-10.8v2.jpg?resize=300%2C296&ssl=1\")
<\/a><\/p>\n