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<\/a><\/p>\nIn this diagram we have<\/p>\n
the eccentre carrying the epicycle center be $LM$ on center $D$, and the eccentre of the planets mean motion be $NX$ on center $Z$, equal [in diameter] to $LM$.<\/p><\/blockquote>\n
We’ll then draw the line of apsides through these centers, creating $\\overline{NM}$ with $L$ at the intersection with the circle of mean motion and $E$ as the observer on Earth.<\/p>\n
We’ll place the center of the epicycle the first opposition be $A$, connecting that to points $D$ and $E$, as well as extending $\\overline{ZA}$ in both directions, producing point $X$. In the other direction, it is extended to $\\Theta$ where it meets a perpendicular dropped from $E$. Similarly, a perpendicular is dropped onto $\\overline{X \\Theta}$ from $D$ producing point $H$.<\/p>\n
In the last post, we concluded that $\\angle NZX = 79;30\u00ba$ as this was the motion along the circle of mean motion between apogee and the first opposition.<\/p>\n
Thus, its vertical angle, $\\angle DZH$ is as well.<\/p>\n
So, we’ll create a demi-degrees context about $\\triangle DZH$ in which the hypotenuse, $\\overline{DZ} = 120^p$.<\/p>\n
We can then state that the arc opposite $\\angle DZH$, $arc \\; DH = 159;00\u00ba$. Similarly, the supplement, $arc \\; ZH = 21\u00ba$.<\/p>\n
This allows us to find the corresponding chords. Doing so, I find, $\\overline{DH} = 117;59^p$ and $\\overline{ZH} = 21;52^p$.<\/p>\n
Next, let’s refer back to the eccentricity. In the last post, we determined that the distance between the observer on Earth and the center of mean motion was $5;23^p$. In this diagram, that’s $\\overline{EZ}$. So half that is $\\overline{DZ} = 2;42^p$.<\/p>\n
This gives us $\\overline{DZ}$ in the context in which the diameter of the eccentre is $120^p$, which we also know in the demi-degrees context we were just working with, so we can use it to convert the other sides:<\/p>\n
$$\\frac{2;42^p}{120^p} = \\frac{\\overline{DH}}{117;59^p}$$<\/p>\n
$$\\overline{DH} = 2;39^p$$<\/p>\n
and<\/p>\n
$$\\frac{2;42^p}{120^p} = \\frac{\\overline{ZH}}{21;52^p}$$<\/p>\n
$$\\overline{ZH} = 0;29^p$$<\/p>\n
which Ptolemy rounds to $0;30^p$.<\/p>\n
Next, let’s focus on $\\triangle DAH$. In it, we know that $\\overline{DA} = 60^p$ since it’s a radius and we just found $\\overline{DH}$. Thus we can use the Pythagorean theorem to find:<\/p>\n
$$\\overline{HA} = \\sqrt{60^2 – 2;39^2}$$<\/p>\n
$$\\overline{HA} = 59;56^p.$$<\/p>\n
As when we did this for Mars, $\\overline{ZH} = \\overline{H \\Theta}$ and $\\overline{E \\Theta} = 2 \\cdot \\overline{DH}$.<\/p>\n
This allows us to determine<\/p>\n
$\\overline{A \\Theta} = 59;56^p + 0;29^p = 60;25^p.$<\/p>\n
However, Ptolemy is evidently calculating with higher precision but still only showing the first sexagesimal place as he comes up with $60;26^p$.<\/p>\n
Additionally, $\\overline{E \\Theta} = 5;18^p$.<\/p>\n
That’s two sides of $\\triangle AE \\Theta$, so we can use the Pythagorean theorem to determine the remaining side, $\\overline{AE}$:<\/p>\n
$$\\overline{AE} = \\sqrt{60;25^2 + 5;18^2}$$<\/p>\n
$$\\overline{AE} = 60;39^p.$$<\/p>\n
And again Ptolemy has been doing some rounding as he comes up with $60;40^p$<\/p>\n
Now, let’s enter a demi-degrees context about this triangle. In it, the hypotenuse $\\overline{AE} = 120^p$ which we can use as our conversion piece:<\/p>\n
$$\\frac{120^p}{60;40^p} = \\frac{\\overline{E \\Theta}}{5;18^p}$$<\/p>\n
$$\\overline{E \\Theta} = 10;29^p.$$<\/p>\n
We can then find the corresponding arc, $arc \\; E \\Theta = 10;01\u00ba.$<\/p>\n
This means that the angle opposite in the circle, $\\angle EA \\Theta = 5;00,30\u00ba.$ Ptolemy has rounded this to $5;01\u00ba.$<\/p>\n
We’ll set that aside and look at $\\overline{X \\Theta}$. This is $\\overline{XZ}$ (which is $60^p$ since it’s the radius) plus $\\overline{Z \\Theta}$ which is $1^p$. Thus, the total length of $\\overline{X \\Theta} = 61;00^p$.<\/p>\n
That gives us two sides of $\\triangle E \\Theta X$, so we can again use the Pythagorean theorem to find the remaining side:<\/p>\n
$$\\overline{EX} = \\sqrt{61;00^2 + 5;18^2}$$<\/p>\n
$$\\overline{EX} = 61;14^p.$$<\/p>\n
We’ll now create a demi-degrees context about $\\triangle EX \\Theta$ in which, the hypotenuse, $\\overline{EX} = 120^p$. We know that piece in two contexts, so we can use it to convert $\\overline{E \\Theta}$<\/p>\n
$$\\frac{120^p}{61;14^p} = \\frac{\\overline{E \\Theta}}{5;18^p}$$<\/p>\n
$$\\overline{E \\Theta} = 10;23^p.$$<\/p>\n
We can then look up the corresponding chord. I find $arc \\; E \\Theta = 9;56\u00ba$ but Ptolemy finds it to be $9;55\u00ba$. Taking his value, the of the angle opposite it in this circle, $\\angle EX \\Theta = 4;58\u00ba$.<\/p>\n
Thus, we can now find $\\angle AEX$ which is the difference between $\\angle EA \\Theta – \\angle EX \\Theta$:<\/p>\n
$$\\angle AEX = 5;01\u00ba – 4;58\u00ba = 3;00\u00ba.$$<\/p>\n
But at the first opposition, the planet, viewed along $\\overline{EA}$, had an apparent longitude of $23;11\u00ba$ into Scorpio. Thus, it is clear that, if the epicycle center were carried, not on eccentre [of mean distance] $LM$, but on [eccentre of mean speed] $NX$, it would have been at point $X$ on that eccentre, and the planet would have appear along $\\overline{EX}$, differing by $0;03\u00ba$ [from the actual position], and thus, would have had a longitude of $23;14\u00ba$ into Scorpio.<\/p>\n
So there’s our first correction. In the next post, we’ll do the same for the second opposition.<\/p>\n
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