{"id":2139,"date":"2020-08-09T22:42:13","date_gmt":"2020-08-10T03:42:13","guid":{"rendered":"http:\/\/jonvoisey.net\/blog\/?p=2139"},"modified":"2020-08-09T22:42:13","modified_gmt":"2020-08-10T03:42:13","slug":"almagest-book-iv-alexandrian-eclipse-triple-radius-of-the-epicycle","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2020\/08\/almagest-book-iv-alexandrian-eclipse-triple-radius-of-the-epicycle\/","title":{"rendered":"Almagest Book IV: Alexandrian Eclipse Triple \u2013 Radius of the Epicycle"},"content":{"rendered":"

Continuing on with Ptolemy’s check on the radius of the epicycle, we’ll produce a new diagram based on the positions of the Alexandrian eclipses. However, instead of doing it piece-by-piece as I did when we explored the Babylonian eclipses, I’ll drop everything into a single diagram since we already have some experience and the configuration for this triple is a bit more for forgiving on the spacing:<\/p>\n

\"\"<\/a><\/p>\n

<\/p>\n

As with before, the eclipses in order are at A, B, and G.<\/p>\n

From the previous post, we determined that $arc \\; AB = 110;21\u00ba$ of the epicycle and makes a corresponding angle from the center of the deferent ($\\angle{ADB}$) of $7;42\u00ba$. Meanwhile, $arc \\; BG = 81;36\u00ba$ on the ecliptic and produces $\\angle{BGZ} = 1;21\u00ba$.<\/p>\n

Again, we can ask why the diagram was set up this way. Ptolemy explains:<\/p>\n

It is clear that the apogee must lie on $arc \\; AB$, since it can lie neither on $arc \\; BG$ nor on $arc \\; GA$, both of which produce an additive effect and are less than a semi-circle.<\/p><\/blockquote>\n

Again looking at the last post<\/a>, we noted that the motion from A to B produced a rearwards effect along the ecliptic. Hence, B must be to the right of A. Similarly, B to G had a forwards effect, indicating that G must be to the left of B. Lastly, we could determine the angle between G and A by subtracting $\\angle{BDG} – \\angle{GDA}$ and we would find that from A to G should also produce a rearwards effect indicating G should be to the right of A.<\/p>\n

The only way we can make these statements true is if apogee lies on $arc \\; AB$. If we rotated it such that A was to the right of apogee, this would force G to be to the left of A. Similarly, if we rotated things so that B was to the left of apogee, then G would be to the right of B.<\/p>\n

So with those three points in their relative positions, we again draw lines from D to each, producing point E where $\\overline{AD}$ intersects the epicycle. We then connect point E to points B and G. Perpendiculars are dropped from point E onto $\\overline{BD}$ and $\\overline{GD}$ producing points Z and H respectively and making right triangles $\\triangle{BEZ}$, $\\triangle{EZD}$, and $\\triangle{EHD}$. Lastly a perpendicular is dropped from point G onto $\\overline{BE}$ producing point \u0398 and forming right $\\triangle{EG \\Theta}$.<\/p>\n

With all of that out of the way, back to the math!<\/p>\n

As with when we did this for the Babylonian eclipses, we’ll be putting everything in the context of a circle drawn about E, Z, and D. And as we saw last time, point H is also on that circle, so let’s just call it $circle \\; EHZD$ from the outset this time.<\/p>\n

In this circle, we have $\\angle{EDZ} = 7;42\u00ba$. Thus, in this circle where D is on the circumference, the arc opposite it, $arc \\; EZ = 15;24\u00ba$. The corresponding chord, $\\overline{EZ} = 16;4,42^p$.<\/p>\n

Turning our attention to the epicycle, $arc \\; AB = 110;21\u00ba$. So the angle it subtends on the circumference of the epicycle, $\\angle{AEB}= 55;11\u00ba$<\/span>1<\/sup><\/a><\/span>. We can then take the supplementary angle along $\\overline{AD}$ to determine<\/p>\n

$$\\angle{DEB} = 180\u00ba – 55;11\u00ba = 124;49\u00ba$$<\/p>\n

Now looking at $\\triangle{DEB}$ we now know two of the angles ($\\angle{DEB}$ and $\\angle{EDB}$) so we can subtract those from 180\u00ba to say<\/p>\n

$$\\angle{EBD} = 180\u00ba – 124;49\u00ba – 7;42\u00ba = 47;29\u00ba$$<\/p>\n

Next, we’ll create a circle around B, E, and Z to form $circle \\; BEZ$. On this circle, we have $\\angle{EBZ}$ which we just found, so the arc opposite it is twice of that which is to say, $arc \\; EZ = 94;58\u00ba$. Looking up the chord, $\\overline{EZ} = 88;26,17^p$, again in the context of $circle \\; BEZ$.<\/p>\n

But again, we already determined $\\overline{EZ}$ in the context of our circle we’ll convert everything to, $circle \\; EHZD$. This allows us to set up the conversion ratios, to solve for $\\overline{BE}$, which was the hypotenuse in $circle \\; BEZ$, in the context of this circle:<\/p>\n

$$\\frac{\\overline{BE}}{120^p} = \\frac{16;4,42^p}{88;26,17^p}$$<\/p>\n

$$\\overline{BE} = 120^p \\cdot \\frac{16;4,42^p}{88;26,17^p} = 21;48,59^p$$<\/p>\n

While we’re looking at the ecliptic, we can notice that<\/p>\n

$$\\angle{ADG} = \\angle{ADB} – \\angle {GDZ} = 7;42\u00ba – 1;21\u00ba = 6;21\u00ba$$<\/p>\n

So in our $circle \\; EGZD$, $arc \\; EH$ is opposite that meaning it will have twice the measure, so $arc \\; EH = 12;42\u00ba$ allowing us to head to the chord table to determine $\\overline{EH} = 13;16,19^p$, again in the context of $circle \\; EGZD$.<\/p>\n

Now back to the epicycle. Here,<\/p>\n

$$arc \\; ABG = arc \\; AB + arc \\; BG = 110;21\u00ba + 81;36\u00ba = 191;57\u00ba$$<\/p>\n

Thus, $\\angle{AEG}$, being on the circumference is half that measure, or $95;59\u00ba$. We can then determine $\\angle{DEG} = 84;01\u00ba$ since it’s the supplement along $\\overline{AD}$. That allows for us to solve $\\triangle{EGD}$ for $\\angle{EGD}$:<\/p>\n

$$\\angle{EGD} = 180\u00ba – 6;21\u00ba – 84;01\u00ba = 89;38\u00ba$$<\/p>\n

Now we’ll create $circle \\; EGH$ since this puts the angle we just found on the circumference. Thus, in the context of this circle, $arc \\; EH$ is twice that angle or $179;16\u00ba$. The arc that subtends this angle would then be $\\overline{EH} = 119;59,50^p$ in the context of this circle.<\/p>\n

But we want to convert back to $circle \\; EHZD$. Fortunately, we’ve around found $\\overline{EH}$ in that context, so we can now set up the conversion ratio to help us find $\\overline{GE}$:<\/p>\n

$$\\frac{\\overline{GE}}{120^p} = \\frac{13;16,19^p}{119;59,50^p}$$<\/p>\n

$$\\overline{GE} = 120^p \\cdot \\frac{13;16,19^p}{119;59,50^p} = 13;16,20^p$$<\/p>\n

Recapping, we’ve now figured out $\\overline{EB}$ and $\\overline{GE}$ in the context of $circle \\; EHZD$.<\/p>\n

Again, we’ll return to the epicycle where $arc \\; BG = 81;36\u00ba$. Thus, $\\angle{BEG} = 40;48\u00ba$.<\/p>\n

Next, we’ll draw $circle \\; EG \\Theta$. In the context of this circle, $arc \\; G \\Theta = 81;36\u00ba$ and the chord, $\\overline{G \\Theta} = 78;24,37^p$.<\/p>\n

We can also determine<\/p>\n

$$\\angle{EG \\Theta} = 180\u00ba – 90\u00ba – 40;48\u00ba = 49;12\u00ba$$<\/p>\n

Therefore the arc opposite this angle, $arc \\;E \\Theta = 98;24\u00ba$ and its chord, $\\overline{E \\Theta} = 90;50,22^p$.<\/p>\n

Now we’ll switch the context from this circle back to $circle \\; EHZD$ by using the chord we’ve found common to both $\\overline{EG}$<\/span>2<\/sup><\/a><\/span>. First let’s do $\\overline{G \\Theta}$:<\/p>\n

$$\\frac{\\overline{G \\Theta}}{78;24,37^p} = \\frac{13;16,20^p}{120^p}$$<\/p>\n

$$\\overline{G \\Theta} = 78;24,37^p \\cdot \\frac{13;16,20^p}{120^p} = 8;40,20^p$$<\/p>\n

Repeating for $\\overline{E \\Theta}$:<\/p>\n

$$\\frac{\\overline{E \\Theta}}{90;50,22^p} = \\frac{13;16,20^p}{120^p}$$<\/p>\n

$$\\overline{E \\Theta} = 90;50,22^p \\cdot \\frac{13;16,20^p}{120^p} = 10;2,49^p$$<\/p>\n

Since everything is now in the context of $circle \\; EHZD$, we can now subtract<\/p>\n

$$\\overline{B \\Theta} = \\overline{BE} – \\overline{E \\Theta} = 21;48,59^p – 10;2,49^p = 11;46,10^p$$<\/p>\n

Now we’ll check out $circle \\; BG \\Theta$. We’ve determined two of the sides of this right triangle ($\\overline{B \\Theta}$ and $\\overline{G \\Theta}$), so we can use the Pythagorean theorem to solve for the third ($\\overline{BG}$)<\/span>3<\/sup><\/a><\/span>:<\/p>\n

$$\\overline{BG} = \\sqrt{{11;46,10^p}^2 + {8;40,20^p}^2} = 14;37,12^p$$<\/p>\n

This is the chord we’ll use to finally switch context to the epicycle. For, in the epicycle, $arc \\; BG = 81;36\u00ba$ and thus its chord, $\\overline{BG} = 78;24,37^p$.<\/p>\n

So we’ll convert $\\overline{DE}$ using another set of conversion ratios:<\/p>\n

$$\\frac{\\overline{DE}}{120^p} = \\frac{78;24,37^p}{14;37,12^p}$$<\/p>\n

$$\\overline{DE} = 120^p \\cdot \\frac{78;24,37^p}{14;37,12^p} = 643;35,11^p$$<\/p>\n

Next, we’ll do the same for $\\overline{GE}$<\/span>4<\/sup><\/a><\/span>:<\/p>\n

$$\\frac{\\overline{GE}}{13;16,20^p} = \\frac{78;24,37^p}{14;37,12^p}$$<\/p>\n

$$\\overline{GE} = 13;16,20^p \\cdot \\frac{78;24,37^p}{14;37,12^p} = 71;11,55^p$$<\/p>\n

From this chord, now in the context of the epicycle, we can determine the corresponding arc, $arc \\; GE = 72;47,10\u00ba$<\/span>5<\/sup><\/a><\/span>.<\/p>\n

We’ve already shown that $arc \\; ABG = 191;57\u00ba$, so the remainder, $arc \\; GA$ (going clockwise) will be $168;3\u00ba$.\u00a0 From this, we can subtract, $arc \\; GE$ to determine $arc \\; EA = 95;15,50\u00ba$. Again converting to the corresponding chord, $\\overline{EA} = 88;39,35^p$.<\/p>\n

As with the last set of eclipses, Ptolemy pauses here to note $arc \\; AE$ wasn’t a semi circle and $\\overline{EA}$ wasn’t $120^p$ so the center of the epicycle isn’t falling on this line, indicating this A and E weren’t at apogee\/perigee. As such, we’ll need to redraw our diagram, now including apogee (L), perigee (M), and the center of the epicycle (K).<\/p>\n

\"\"<\/a><\/p>\n

 <\/p>\n

We’ll again create the equal products based on Euclid\u2019s Elements<\/em> III.36<\/a>, but this time we’ll chose the first eclipse at A instead of point B:<\/p>\n

$$\\overline{AD} \\cdot \\overline{DE} = \\overline{LD} \\cdot \\overline{DM}$$<\/p>\n

Here, we can calculate $\\overline{AD}$:<\/p>\n

$$\\overline{AD} = \\overline{AE} + \\overline{DE} = 88;39,35^p + 643;35,11^p = 732;14,46^p$$<\/p>\n

This gives us everything we need for the left side of this equation. So we can rewrite<\/span>6<\/sup><\/a><\/span>:<\/p>\n

$$732;14,46^p \\cdot\u00a0 643;35,11^p = 471263;37,49^p =\\overline{LD} \\cdot \\overline{DM}$$<\/p>\n

And again based on\u00a0Elements II.6<\/a> we can write:<\/em><\/p>\n

$$\\overline{LD} \\cdot \\overline{DM} + \\overline {KM}^2 = \\overline{DK}^2$$<\/p>\n

So plugging in and rearranging:<\/p>\n

$$\\overline{DK}^2 = 471263;37,49^p + {60^p}^2 = 474863;37,49^p$$<\/p>\n

$$\\overline{DK} = 689;6,12^p$$<\/p>\n

And now we’ll do one last context swap to determine the radius of the epicycle on the context of the deferent:<\/p>\n

$$\\frac{\\overline{KM}}{60^p} = \\frac{60^p}{689;6,12^p}$$<\/p>\n

$$\\overline{KM} = 60^p \\cdot \\frac{60^p}{689;6,12^p} = 5;13,27^p$$<\/p>\n

Ptolemy’s value ends up being higher by just enough that he justifies rounding this up to $5;14^p$, but either way, this is in excellent agreement with the radius we found from the Babylonian eclipse triple<\/a>.<\/p>\n

So I’ll leave things here, and again save the equation of anomaly and position of the mean moon for the next post!<\/p>\n

\n

\"\"<\/a><\/p>\n

I think today may have been a record day for progress. The four posts that went up today covered almost 8 pages of material! The first one was definitely the hardest and I’d been working on it for a few days, but from these four posts, that was 1.11% of the\u00a0Almagest<\/em> right there! However, it was also a marathon day, spending somewhere around 12 hours working today.<\/p>\n

\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Continuing on with Ptolemy’s check on the radius of the epicycle, we’ll produce a new diagram based on the positions of the Alexandrian eclipses. However, instead of doing it piece-by-piece as I did when we explored the Babylonian eclipses, I’ll drop everything into a single diagram since we already have some experience and the configuration … <\/p>\n

Continue reading “Almagest Book IV: Alexandrian Eclipse Triple \u2013 Radius of the Epicycle”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[24],"tags":[31,25,46,28,26,19,14],"class_list":["post-2139","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-algebra","tag-almagest","tag-eclipse","tag-geometry","tag-mathematics","tag-moon","tag-ptolemy"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-yv","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/2139","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/comments?post=2139"}],"version-history":[{"count":7,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/2139\/revisions"}],"predecessor-version":[{"id":2152,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/2139\/revisions\/2152"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=2139"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=2139"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=2139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}