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<\/a><\/p>\nIn this setup, the epicycle, centered on $Z$ and which rotates counter-clockwise about $G$, has rotated $120\u00ba$ from apogee at $A$. Similarly, the second epicycle, which rotates around point $B$ with its true center at $H$, has also rotated $120\u00ba$ from apogee, but in the clockwise direction.<\/p>\n
We draw lines of sight to the tangents from the observer, on Earth at $D$, creating points $\\Theta$ and $K$.<\/p>\n
Lastly, we’ve dropped a perpendicular from $D$ onto $\\overline{HZ}$ at point $L$.<\/p>\n
Immediately, we can see that $\\angle GBH = \\angle DGL = 60\u00ba$ because they are the supplements to angles $\\angle ABH$ and $\\angle AGL$, both of which are $120\u00ba$.<\/p>\n
Next, looking at $\\triangle BGH$, this is an equilateral triangle. We know this because $\\angle GBH = 60\u00ba$ as does $\\angle BGH$ as it’s a vertical angle to $\\angle DGL$. Thus, the third angle must be $60\u00ba$ as well making the triangle equilateral.<\/p>\n
This all means that points $H$, $G$, and $Z$ all lie on a straight line.<\/p>\n
Next, we’ll return to the context in which the radius of the eccentre which rotates about $B$ is $60^p$. In other words, $\\overline{HZ} = 60^p$. This was also the one in which all the eccentricities, $\\overline{BH}$, $\\overline{BG}$, and $\\overline{GD}$ are all $3^p$.<\/p>\n
Therefore, we can subtract $\\overline{GH}$ from $\\overline{ZH}$ to determine $\\overline{ZG} = 57^p$.<\/p>\n
Next, let’s take a look at $\\triangle DGL$. Within this triangle, $\\angle DGL = 60\u00ba$. In the context in which the hypotenuse, $\\overline{GD} = 120^p$, then the chord of $\\angle DGL$, $\\overline{DL} = 103;55,23^p$ which Ptolemy rounds off to $103;55^p$.<\/p>\n
We can also determine the remaining side of this triangle, $\\overline{GL} = 60^p$.<\/p>\n
Next, we’ll convert back to our context in which $\\overline{DG} = 3^p$:<\/p>\n
$$\\frac{\\overline{DL}}{103;55^p} = \\frac{3^p}{120^p}$$<\/p>\n
$$\\overline{DL} = 2;36^p$$<\/p>\n
We can also convert contexts for $\\overline{GL}$ or use the Pythagorean theorem to determine $\\overline{GL} = 1;30^p$.<\/p>\n
Next, we can subtract $\\overline{GL}$ from $\\overline{GZ}$ to determine $\\overline{LZ} = 55;30^p$.<\/p>\n
Then, using the Pythagorean theorem on $\\triangle LGZ$ to determine $\\overline{DZ} = 55;34^p$.<\/p>\n
Toomer correctly notes here that Ptolemy is suggesting that this is the closest distance to Earth that the center of Mercury’s epicycle gets. However, note that this is when the center has rotated $120\u00ba$ from apogee. As we showed in this post<\/a>, the perigee is actually at $120;30\u00ba$ from apogee. Toomer also notes that, should this be taken into account, the distance should be $55;33,38^p$, which still rounds to Ptolemy’s value. Thus, Toomer notes that the difference cause is negligible.<\/p>\n Regardless, we can now determine $\\angle \\Theta DZ$ which is half of $\\angle \\Theta DK$, the angle we’re looking for. Ptolemy only shows half of this calculation, converting the triangle to a context for a demi-degrees circle, but not actually stating the angle. Rather, he skips straight to the final answer of $\\angle \\Theta DK = 47;46\u00ba$.<\/p>\n And that’s it for this chapter. In the next one, Ptolemy will discuss corrections he made to the periodic motions of Mercury.<\/p>\n
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