There are a few clues that I haven’t been able to figure out. Particularly, in the problem set up, Ptolemy states that the additional arcs we’re drawing in have a pole at the angle we’re concerned with and “radius the side of the [inscribed] square.” I haven’t been able to make heads or tails of this statement.<\/p>\n
Another is that the translation states that the first of the proofs we’ll go through is “derivable from Theodosius Sphaerica<\/em> II 9.” As we’ll see, about the only step that really happens in that proof is equating the angle to the arc, I imagine this must be what it’s referring to, but there doesn’t seem to be any translations of\u00a0Sphaerica<\/em> that I’ve been able to track down.<\/p>\n UPDATE (8\/27\/2022): Big thanks to reader Eric Piphany who found a document<\/a> that contains many of the theorems from\u00a0Sphaerica<\/em>. In particular, the one to which Ptolemy is referring to here states:<\/p>\n In a sphere, if two circles cut off arcs of each other and a great circle is drawn through their poles, then it bisects the arcs cut off.<\/p><\/blockquote>\n In this case, it doesn’t seem that Ptolemy’s statement is “derivable” from this one, but rather, follows directly from it.<\/p>\n My best guess is that it only works when the auxiliary great circle we’ll be drawing in for each of these, has the vertex of the angle as one of its poles as that seems to be done in all of the derivations. Regardless,\u00a0we’ll just have to take Ptolemy’s word for when this works for now.<\/p>\n Also, a quick note before we get started, each time we do another derivation, Ptolemy changes what symbols are representing, so pay careful attention.<\/p>\n Solstices<\/em><\/p>\n So to get started, first Ptolemy wants to look at the angle between the ecliptic and meridian at the solstices. As such, we’ll need a diagram with a meridian (ABGD) and the ecliptic (AEG) such that A is the winter solstice.\u00a0He then draws in BED such that it’s perpendicular to both making A and G its poles as well as AEG having poles B and D.<\/p>\n This immediately completes the problem because the angle we’re looking for, is $\\sphericalangle DAE$, which subtends $arc \\; DE$. and since E is on AEG and D is its pole, it is, by definition, 90\u00ba as is the angle.<\/p>\n Since we said that point A was the winter solstice, this means that it’s the first point in Capricorn. And by the rule that we derived in the last post this, plus the first point in Cancer (which is the summer solstice) must add up to 180\u00ba. Therefore, the angle at the first point in Cancer (at the summer solstice) is also 90\u00ba.<\/p>\n Equinoxes<\/em><\/p>\n Next, we’ll go for the equinoxes. To define the equinox, we’ll need the celestial equator (AEG)\u00a0and<\/em> ecliptic (AZG) in addition the the meridian (ABGD). Now, we’ll position the now autumnal equinox on the meridian and draw in arc BZED such that A is one of its poles. This time, our objective will be to find $\\sphericalangle DAZ$ which subtends $arc \\; DZ$.<\/p>\n As with before, $ard \\; DE = 90\u00ba$ so we just need to add on $arc \\; EZ$. Similarly $arc \\; AZ = 90\u00ba$ which means we can look up $arc \\; EZ$ in our table of inclinations between the celestial equator and ecliptic<\/a>\u00a0for when the arc of the ecliptic is 90\u00ba. Thus, $arc \\; EZ = 23;51,20\u00ba$ and $arc \\; DZ = 113;51,20\u00ba$ which Ptolemy just rounds off to 113;51\u00ba. Thus, $\\sphericalangle DAZ = 113;51\u00ba$.<\/p>\n Again, A was the autumnal equinox which is the first point in Libra. Thus, by our the proof in our last post, the vernal equinox, the first point in Aries, must be $180\u00ba – 113;51\u00ba = 66.9\u00ba$.<\/p>\n Intermediate Points<\/em><\/p>\n So now we’ve figured out four of the points around the ecliptic. Next we’ll need to get the ones between the solstices and equinoxes.<\/p>\n As with before, we’ll have our meridian circle as ABGD. We’ll include the celestial equator (AEG) and ecliptic (BZD) such that Z is the autumnal equinox<\/span>1<\/sup><\/a><\/span>. And again we’ll have an extra great circle in there, HEK, that has B at its pole.<\/p>\n This time, the objective is to find $\\sphericalangle KB \\Theta$.<\/p>\n To start, we’ll do the first point in Virgo, which means that this point will need to be at B, therefore making\u00a0$arc \\; BZ = 30\u00ba$ as that’s the arc from the first point in Virgo to the autumnal equinox, Z (i.e., it’s the full sign of Virgo).<\/p>\n There’s also a few other things that fall out of this setup. First off, because B is a pole of HEK, this makes $arc \\; BH$ and $arc \\; B \\Theta$ 90\u00ba<\/span>2<\/sup><\/a><\/span>. Although it’s not explicitly stated in the problem set up, AEG has its pole on the meridian as well (at some point to the left of K). This is important because it means that point E is on two great circles (AEG and HEK) that both have their poles on another (the meridian) which can only happen if it’s the pole of that second. Thus, $arc \\; EH = 90\u00ba$ is\u00a0as well.<\/p>\n As so often happens, we’ve now got a Menelaus configuration. This allows us to state:<\/p>\n $$\\frac{Crd \\; arc \\; 2BA}{Crd \\; arc 2AH} =\u00a0\\frac{Crd \\; arc \\; 2BZ}{Crd \\; arc \\; 2Z \\Theta} \\cdot\u00a0\\frac{Crd \\; arc \\; 2E \\Theta}{Crd \\; arc 2EH}$$<\/p>\n As usual, let’s take stock of what we know.<\/p>\n Since we stated above $arc \\; EH = 90\u00ba$, this means that $Crd \\; arc \\; 2EH = 120$.<\/p>\n Next up, let’s look at $arc \\; BA$. This is the arc between the celestial equator and ecliptic when the arc of the ecliptic is 30\u00ba. For this, we can refer to our table for such values<\/a>\u00a0to determine $arc \\; BA =\u00a011;39,59$. From there we can state\u00a0$Crd \\; arc \\; 2BA = 24;15,57$\u00a0which Ptolemy rounds off to 24;16 in this section.\u00a0We can subtract $arc \\; BA$ from\u00a0 $arc \\; BH$ which we above said was\u00a090\u00ba to determine that $arc \\; AH = 78;20\u00ba$ and thus, $Crd \\; arc \\; 2AH = 117;31$.<\/p>\n Since $arc \\; BZ$ is the whole of Virgo, or 30\u00ba, we can say $Crd \\; arc \\; 2BZ = 60$.<\/p>\n And because $arc \\; BZ = 30\u00ba$ we can subtract that from $arc \\; B\\Theta$ which we again said above was 90\u00ba to determine $arc \\; Z\\Theta = 60\u00ba$ and therefore $Crd \\; arc \\; 2Z \\Theta = 103;55,23$<\/p>\n That gives us everything we need to solve for $arc \\; E\\Theta$.<\/p>\n $$\\frac{24;16}{117;31} =\u00a0\\frac{60}{ 103;55,23} \\cdot\u00a0\\frac{Crd \\; arc \\; 2 E\\Theta }{120}$$<\/p>\n Solving for $Crd \\; arc \\; 2 E\\Theta$:<\/p>\n $$Crd \\; arc \\; 2 E\\Theta = 120 \\cdot \\frac{24;16}{117;31} \\cdot \\frac{ 103;55,23}{60} $$<\/p>\n
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