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<\/a><\/p>\nIn this figure, the circle $ABGE$, is the “eccentre of mean motion”, i.e., the equant circle<\/span>1<\/sup><\/a><\/span>.<\/p>\n Point $D$ will be the observer at the center of the ecliptic.<\/p>\n As with before, points $A$, $B$, and $G$ are the apparent positions of Mars on the equant circle at the three observations<\/span>2<\/sup><\/a><\/span>. These will be joined to form the line of sight with the observer as $\\overline{DA}$, $\\overline{DB}$, and $\\overline{DG}$.<\/p>\n Line $\\overline{DG}$ is also extended until it meets the side of the circle opposite $G$ at $E$<\/span>3<\/sup><\/a><\/span><\/p>\n We’ll also draw $\\overline{AB}$ and $\\overline{AE}$ and drop perpendiculars from $E$ onto $\\overline{AD}$ at $Z$, from $A$ onto $\\overline{BE}$ at $\\Theta$, and from $E$ onto $\\overline{BD}$ at $H$.<\/p>\n We know $arc \\; BG$<\/span>4<\/sup><\/a><\/span>, is $95;28\u00ba$, as that was the mean motion along the equant circle as derived in the last post. But because we don’t have a point at the center of this circle, we can’t use that to determine any angles right now.<\/p>\n So, this is where we’re going to start having to cheat and use some things we\u00a0do<\/em> know as a first approximation just to get us moving.<\/p>\n At the risk of confusing things<\/span>5<\/sup><\/a><\/span>, let’s pause and look back at the diagram we constructed in the last post.<\/p>\n There, we said that the angle Mars moved from the point of view of the observer about the ecliptic was $\\angle LNM = 93;44\u00ba$.<\/p>\n What Ptolemy is wanting right now (in the context of this diagram from the last post) is $\\angle ZNH$ which, as we said, we don’t have.<\/p>\n But, this angle, $\\angle BNG$, is what we have to work with for now so Ptolemy adopts it.<\/p>\n Jumping back to the context of the diagram we’ve constructed for this post, that means that\u00a0$\\angle BDG = 93;44\u00ba$.<\/p>\n Thus, its supplement, $\\angle EDH = 86;16\u00ba$.<\/p>\n We can now imagine a demi-degrees circle about $\\triangle EDH$. In it, $arc \\; EH = 172;32\u00ba$ as its twice the angle it subtends.<\/p>\n Therefore, in this context, where the hypotenuse, $\\overline{ED} = 120^p$, \u00a0$\\overline{EH} = 119;45^p$.<\/p>\n We’ll hold onto that for a bit, and now focus on $arc \\; BG$ which, as we just mentioned, is $95;28\u00ba$<\/span>6<\/sup><\/a><\/span>. Therefore, we can state that the angle subtending this on the far side of the circle is half that: $\\angle BEG = 47;44\u00ba$. This means we now know the size of two angles in $\\triangle BED$<\/span>7<\/sup><\/a><\/span>. Thus, the remaining angle, $\\angle DBE = 46;00\u00ba$.<\/p>\n Knowing that, we can now focus on $\\triangle BEH$ (which is a right triangle), imagining a demi-degrees circle about it. In it, $arc \\; EH = 92;00\u00ba$ since it is opposite the angle we just found.<\/p>\n Therefore, the chord, $\\overline{EH} = 86;19^p$ in this context where the hypotenuse, $\\overline{BE} = 120^p$.<\/p>\n But, we just found this same segment ($\\overline{EH}$ in the previous context which will allow us to convert $\\overline{BE}$ to that same previous context:<\/p>\n $$\\frac{119;45^p}{86;19^p} = \\frac{\\overline{BE}}{120;00^p}$$<\/p>\n $$\\overline{BE} = 166;29^p$$<\/p>\n in the context where $\\overline{ED} = 120^p$.<\/p>\n We’ll now turn to $\\angle ADG$. This angle is the motion of Mars (from the point of view of the observer) from the first observation to the last. In other words<\/span>8<\/sup><\/a><\/span>,<\/p>\n $$\\angle ADG = 93;44\u00ba + 67;50\u00ba = 161;34\u00ba.$$<\/p>\n Next, we can find $\\angle ADE$. This angle is:<\/p>\n $$\\angle ADE = 180\u00ba – \\angle ADG = 180\u00ba – 161;34\u00ba = 18;26\u00ba.$$<\/p>\n Now we can focus on $\\triangle ZDE$, imagining a demi-degrees circle about it.<\/p>\n In that, $arc \\; EZ = 36;52\u00ba$ as it’s twice $\\angle ADE$. Thus, the corresponding chord, $\\overline{EZ} = 37;57^p$, again in the context where $\\overline{ED} = 120^p$.<\/p>\n Now, let’s actually consider $arc \\; AG$ on the circle we actually have. As a reminder, this circle is the circle about the equant or the circle about which the mean motion takes place. Thus, $arc \\; AG$ is the sum of the mean motion over this interval or:<\/p>\n $$81;44\u00ba + 95;28\u00ba = 177;12\u00ba.$$<\/p>\n This can be related to $\\angle AEG$ which would have half that measure or $88;36\u00ba$.<\/p>\n That means we now know two of the angles in $\\triangle ADE$ ($\\angle AED$ and $\\angle ADE$). Thus, we can determine the remaining angle:<\/p>\n $$\\angle DAE = 180\u00ba – 88;36\u00ba – 18;26\u00ba = 72;58\u00ba$$<\/p>\n And knowing that, we can turn to $\\triangle AEZ$, again imagining a demi-degrees circle about it.<\/p>\n In it, the arc opposite the angle just described $arc \\; ZE$ is twice the angle or $145;56\u00ba$, the corresponding chord of which, $\\overline{EZ} = 114;44^p$, in this context where the hypotenuse $\\overline{AE} = 120^p$.<\/p>\n We’ve now established $\\overline{EZ}$ in two contexts, so we can use that previous one to convert everything to that context:<\/p>\n $$\\frac{37;57^p}{114;44^p} = \\frac{\\overline{AE}}{120^p}$$<\/p>\n $$\\overline{AE} = 39;42^p.$$<\/p>\n As a reminder, the context we just converted to was the one in which $\\overline{DE} = 120^p$.<\/p>\n Putting that aside from now, let’s take a look at $arc \\; AB$. This, again, is the mean motion that occurred between the first and second observation which we determined to be $81;44\u00ba$ in the last post. Thus, $\\angle AEB = 40;52\u00ba$ as it is the angle that subtends this arc.<\/p>\n We can then focus on $\\triangle AE \\Theta$, again imagining a demi-degrees circle about it.\u00a0In it, $\\angle AE \\Theta = 40;52\u00ba$. The hypotenuse, $\\overline{AE} = 120^p$.<\/p>\n In it, $arc \\; A \\Theta = 81;44\u00ba$ as well and the corresponding chord, $\\overline{A \\Theta} = 78;31^p$.<\/p>\n Continuing on, we can then use the Pythagorean theorem on the remaining side to determine $\\overline{E \\Theta} = 90;45^p$.<\/p>\n Again, we’ll convert this to the context in which $\\overline{DE} = 120^p$ using $\\overline{AE}$ as the common side in these contexts:<\/p>\n $$\\frac{39;42^p}{120;00^p} = \\frac{\\overline{A \\Theta}}{78;31^p}.$$<\/p>\n I come up with $\\overline{A \\Theta} = 25;59,33^p$ which should round to 26;00^p, but take Ptolemy’s value here for consistency which is to say, $\\overline{A \\Theta} = 25;58^p.$<\/p>\n Ptolemy uses the demi-degrees conversion to also convert $\\overline{E \\Theta}$ but we can also convert using the Pythagorean theorem which gives me $\\overline{E \\Theta} = 30;02^p$ in that context.<\/p>\n Now, recall that we previously showed that the entire length of $\\overline{BE} = 166;29^p$. We can subtract $\\overline{\\Theta E}$ which we just determined off of this to determine $\\overline{B \\Theta}$:<\/p>\n $$\\overline{B \\Theta} = 166;29^p – 30;02^p = 136;27^p.$$<\/p>\n Turning now to $\\triangle AB \\Theta$, we then know two sides of it, so can use the Pythagorean theorem to determine $\\overline{AB} = 138;53^p$<\/span>9<\/sup><\/a><\/span> in the context in which $\\overline{ED} = 120^p$.<\/p>\n Now we’re finally ready to convert into the context in which the diameter of this circle is $120^p$. But, how to do so given that we haven’t even drawn in a diameter of this circle?<\/p>\n Ptolemy makes use of $\\overline{AB}$. However, we haven’t determined the length of this chord in the context in which the diameter of the circle we’re discussing has a diameter of $120^p$. Ptolemy doesn’t really explain so I’ll break from his discussion briefly.<\/p>\n Here, I’ve stripped everything out of the diagram except points $A$ and $B$ and the point between them. I’ve also drawn in the diameter of this circle, its center at $T$, and connected $\\overline{BT}$ and $\\overline{AT}$.<\/p>\n In this, we know $arc \\; AB = 81;44\u00ba$. We can then look at $\\overline{AB}$ which is, conveniently, already inscribed in a triangle with a radius of $60^p$. This means that we can go directly to the chord table to find $\\overline{AB} = 78;31^p$ in this context where the diameter of this circle of mean motion is $120^p$<\/span>10<\/sup><\/a><\/span>.<\/p>\n We now have $\\overline{AB}$ in the previous context as well as the one we’re ultimately looking for, allowing us to convert to the context in which this circle, as well as the ecliptic circle and circle of mean distance, have measures of $120^p$:<\/p>\n $$\\frac{78;31^p}{138;53^p} = \\frac{\\overline{ED}}{120^p}$$<\/p>\n $$\\overline{ED} = 67;50^p.$$<\/p>\n And for $\\overline{AE}$:<\/p>\n $$\\frac{78;31^p}{138;53^p} = \\frac{\\overline{AE}}{39;42^p}$$<\/p>\n I’m forced to pause here because Ptolemy makes a few mistakes in a row.<\/p>\n First, in solving for $\\overline{AE}$ I find it to be $22;27^p$ which Toomer agrees with. Yet, somehow, Ptolemy comes up with $22;44^p$.<\/p>\n Next, Ptolemy converts the chord to the corresponding arc, $arc \\; AE$.<\/p>\n When I do so with Ptolemy’s value, I find it to be $21;50\u00ba$. Ptolemy manages to come up with $21;41\u00ba$. If I use the correct value for the chord, I find the arc to be $21;34\u00ba$, again in agreement with Toomer.<\/p>\n Thus, it’s hard to explain where Ptolemy came up with these values. Toomer has a few notes on what values are found in various manuscripts but concludes that it is most likely that the error is Ptolemy’s and not simply an error in transcription. Please refer to his footnote $40$ for details.<\/p>\n As far as what to do with this error, I’ll adopt Ptolemy’s value of $arc \\; AE = 21;41\u00ba$ for the arc as we’ll now add it to $arc \\; AG$ to get $arc \\; EABG$:<\/p>\n $$177;12\u00ba + 21;41\u00ba = 198;53\u00ba.$$<\/p>\n This means that the remaining $arc \\; GE = 161;07\u00ba$.<\/p>\n The corresponding chord, $\\overline{GE} = 118;22^p$, again in the context where the diameter of this circle is $120^p$.<\/p>\n So what of it?<\/p>\n Now, if $\\overline{GE}$ had been found equal to the diameter of the eccentre, it is obvious that the centre would lie on $\\overline{GE}$, and the ratio of the eccentricity would immediately be apparent. But, since it is not equal [to the diameter], but makes [arc] $EABG$ greater than a semi-circle, it is clear that the centre of the eccentre will fall within the latter.<\/p><\/blockquote>\n What Ptolemy is saying here is that if $\\overline{GE}$ would have equaled $120^p$, then it would have meant it passed through the center of the circle.<\/p>\n Had that been the case, then “the ratio of eccentricity would immediately be apparent.” Ptolemy doesn’t say how, but I suspect we’ll come to that.<\/p>\n However, this point has gotten quite long, we’ll continue in the next post!<\/p>\n
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