![\"\"](\"https:\/\/i0.wp.com\/jonvoisey.net\/blog\/wp-content\/uploads\/2020\/08\/AlmagestFig4-7.jpg?ssl=1\")
<\/a><\/p>\n<\/p>\n
Here, we’ve connected point B to K and from point K, extended a line creating point N such that this new line is perpendicular to $\\overline{BD}$, thus producing right triangle, $\\triangle{BKN}$. The line then continues on to point X which is on the circumference of the epicycle.<\/p>\n
Since we’re looking for the position of the mean moon (which is represented by any of the points along $\\overline{DL}$) this means what we’re really working towards is an understanding of $\\angle{KDN}$ which represents the equation of anomaly. We can then combine that apparent position of the moon<\/span>1<\/sup><\/a><\/span> to determine the position of the mean moon. So let’s get started.<\/p>\n In the last post, we determined that, in the context of the epicycle, $\\overline{DK} = 690;8,42^p$, $\\overline{DE} = 631;13,48^p$, and $\\overline{BE} = 117;37,32^p$.<\/p>\n We can then determine the length of $\\overline{NE}$ as it is half the length of $\\overline{BE}$. This comes from the corollary to III.1 in Euclid’s Elements<\/em><\/a>, which states that if a line ($\\overline{KX}$) cuts a chord ($\\overline{BE}$) into two equal parts, then the center of the circle (K) lies on the cutting line. Taking that in reverse, we can state that, if a line passing through the center of a circle ($\\overline{KX}$), intercepts the chord ($\\overline{BE}$) at right angles, it bisects the chord.<\/p>\n Thus, $\\overline{NE} = \\frac{1}{2} \\overline{BE} = 58;48,46^p$. That can be added to $\\overline{DE}$ to determine $\\overline{DN} = 690;2,34^p$.<\/p>\n Now we’ll context switch to circle about points D, N, and K, which would have $\\overline{DK}$ as the diameter. Setting up our conversion ratios:<\/p>\n $$\\frac{\\overline{DN}}{690;2,34^p} = \\frac{120^p}{690;8,42^p}$$<\/p>\n $$\\overline{DN} = 690;2,34^p \\cdot \\frac{120^p}{690;8,42^p} = 119;58,56^p$$<\/p>\n Converting it to the corresponding arc length we get $arc \\; DN = 178;2\u00ba$. Thus, the angle that is subtended by this arc, $\\angle{DKN}$ is half of that<\/span>2<\/sup><\/a><\/span> or 89;1\u00ba.<\/p>\n This is a central angle on the epicycle, and looking at the arc it subtends on said epicycle, we can therefore state that $arc \\; XEM = 89;1\u00ba$ as well. This can then be subtracted from $arc \\; LM$ which is 180\u00ba, to determine $arc \\; LBX = 90;59\u00ba$.<\/p>\n We’ll leave that hanging temporarily and look at $arc \\; BGE$ which we determined in the last post to be 157;10\u00ba. This is the arc corresponding to $\\overline{BE}$ which we showed above was bisected by $\\overline{KX}$. If the chord is bisected, so too is the arc, which means that $arc \\; BX = \\frac{1}{2} \u00a0\\cdot 157;10\u00ba = 78;35\u00ba$.<\/p>\n We can then subtract that from $arc \\; LBX$ to state<\/p>\n $$arc \\; LB = 90;59\u00ba – 78;35\u00ba = 12;24\u00ba$$<\/p>\n Thus, taken from the center of the epicycle, the moon at point B was 12;24\u00ba from apogee.<\/p>\n But for the observer, what effect would this produce? We actually had the necessary information to answer this question earlier in the post, when considering $\\triangle{DKN}$ in the context of the circle about those points. In it, we determined $\\angle{DKN} = 89.1\u00ba$ and $\\angle{KND}$ is a right angle. Thus, we can subtract those values from 180\u00ba to determine $\\angle{KDN} = 0;59\u00ba$ which is the equation of anomaly for this eclipse.<\/p>\n What does that mean? Well, going back a few posts, we determined the\u00a0 position of the sun for this second eclipse to be 343;45\u00ba ecliptic longitude<\/span>3<\/sup><\/a><\/span> which would mean that the moon would have been at 163;45\u00ba ecliptic longitude which is in 13;45\u00ba into Virgo. But if the mean position of the moon was actually 0;59\u00ba ahead of that, then the mean lunar position was actually 14;44\u00ba into Virgo<\/span>4<\/sup><\/a><\/span> or 164;44\u00ba ecliptic longitude.<\/p>\n While Ptolemy doesn’t appear to do so here, knowing $\\angle{KDN}$ means that we could also find the mean position of the moon during the other two eclipses since we already determined $\\angle{BDA}$ and $\\angle{BDG}$<\/span>5<\/sup><\/a><\/span> and could thus, simply add on this additional angle.<\/p>\n So at this point, we’ve answered all of the questions I put forth at the beginning of my last post. In that post itself, we determined the radius of the epicycle. In this one we determined the position of the mean moon and the equation of anomaly for the second eclipse.<\/p>\n However, this is a marathon chapter and we’re only half way through. In the remainder, Ptolemy will be repeating what we’ve just done in this and the past two posts, but for another set of three eclipses that he observed himself in Alexandria. My suspicion is that he does this as an independent check on Hipparchus’ numbers which he’s just walked us through using the Babylonian eclipses. Fortunately, because we’ve already seen the process, working through this repeat should be much simpler and will hopefully serve to illustrate the method better since this example was pretty tough to follow the first time through!<\/p>\n