{"id":933,"date":"2018-09-02T14:01:36","date_gmt":"2018-09-02T19:01:36","guid":{"rendered":"http:\/\/jonvoisey.net\/blog\/?p=933"},"modified":"2018-10-13T11:29:29","modified_gmt":"2018-10-13T16:29:29","slug":"almagest-book-ii-symmetry-of-rising-times","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2018\/09\/almagest-book-ii-symmetry-of-rising-times\/","title":{"rendered":"Almagest Book II: Symmetry of Rising Times – Arcs of the Ecliptic Equidistant from the Same Equinox"},"content":{"rendered":"

In this next chapter, Ptolemy’s goal is to<\/p>\n

show how to calculate, for each latitude, the arcs of the equator… which rise together with [given] arcs of the ecliptic.<\/p><\/blockquote>\n

To do this, we’ll do a bit of convenient math, breaking the full ecliptic into its traditional 12 parts. However, since these signs are not of equal size, Ptolemy takes an even 30\u00ba for each sign, beginning with Aries, then Taurus, etc…<\/p>\n

The first goal will be to prove that<\/p>\n

arcs of the ecliptic which are equidistant from the same equinox always rise with equal arcs of the equator.<\/p><\/blockquote>\n

<\/p>\n

Unfortunately, the diagram in the translation I’m using is extremely difficult to parse as is Ptolemy’s explanation of it<\/span>1<\/sup><\/a><\/span>. Here’s Ptolemy’s explanation:<\/p>\n

Let ABGD be a meridian, BED the semi-circle of the horizon, AEG the semi-circle of the equator, ZH and\u00a0\u0398K two arcs of the ecliptic such that points Z and\u00a0\u0398 are each supposed to be the spring equinox, and equal arcs have been cut off on opposite sides [of that equinox]: these are arcs AH and\u00a0\u0398K, which are rising at points K and H [respectively]… Let points L and M represent poles of the equator.<\/p><\/blockquote>\n

Some of this should be familiar. The meridian, horizon, and equator are all things we’ve been seeing in the setup for the last several problems. However, we add a complication into this problem when we add the ecliptic which we’d previously been ignoring. But before adding that, let’s first start with just the meridian, horizon, equator, and poles. However, this diagram is going to get quite busy, so instead of tweaking the previous diagrams I’ve been using, we’re going to start completely over with some color coding, so I apologize for the slow buildup on this one. In addition, I’m going to rotate things to hopefully make them make more sense.<\/p>\n

First, let’s just add the horizon (green), celestial equator (blue), meridian (black) and the poles (L & M).<\/p>\n

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

Here, I’ve colored the arcs on the side facing us.\u00a0As noted above, we’ll need the ecliptic which I’ll color orange:<\/p>\n

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

This gives us a few points of intersection that are important. One between the ecliptic and equator (Z) which is the vernal equinox, and another between the ecliptic and horizon (H) which is where the ecliptic is rising. While the ecliptic is touching the meridian, we’re not going to be using those points, so I’ve left them unlabeled.<\/p>\n

Now let’s play some connect-the-dots. In particular, we’ll want to connect, L and M via a great circle passing through E. Additionally, let’s connect Z and H to the pole M via great circles.<\/p>\n

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

That’s pretty cluttered, but we’re still only half done. But before moving on, let’s review what we’ve done and then tidy things up a tiny bit. First, what we’ve done is drawn a figure where the vernal equinox (Z) is above<\/em> the horizon, and noted the point (H) where the ecliptic is then meeting on the horizon. In the next step, we’ll reverse that: We want the vernal equinox\u00a0below<\/em> the horizon, and we’ll again consider the other point of the ecliptic on the horizon.<\/p>\n

But before that, let’s clean things up a tiny bit. First, I’m going to remove the color for the ecliptic because we’re about to reuse it and I don’t want to get us confused<\/span>2<\/sup><\/a><\/span>. In addition, the arcs from Z and H to the meridian aren’t used, so let’s go ahead and truncate them to just leave $arc \\; ZH$.<\/p>\n

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

That’s a bit better, so let’s get started on the next portion.<\/p>\n

In particular, we want to consider the next vernal equinox (below the horizon), but the new arc along the ecliptic needs to be equal to the previous arc along the ecliptic ($arc \\; ZH$)<\/span>3<\/sup><\/a><\/span>.\u00a0So let’s get that drawn in.<\/p>\n

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

Here, I’ve done basically the same thing as we did previously except that the vernal equinox (\u0398) is now the same distance below the horizon and K, the point on the ecliptic, is on the horizon, such that $arc \\; ZH = arc \\; K\\Theta$. I’ve also connected these points to their nearest celestial pole, L.<\/p>\n

One last cleanup to remove the color and make all things equal before getting to the math:<\/p>\n

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

While that’s pretty complicated, it’s also got a sort of beauty to it. A symmetry. And if you’ve noticed the post title, proving symmetry is what we’re after. In particular, Ptolemy wants to demonstrate that $arc \\; EZ = arc \\; E\\Theta$. So let’s get started.<\/p>\n

First off, we already stated that\u00a0$arc \\; ZH = arc \\; \\Theta K$.<\/p>\n

Fortunately, this helps us arrive at a few more equalities. First, $arc \\; MH = arc \\; LK$. This is because they both start on the celestial equator, which means they start equidistant from their respective poles. Since they both end on the horizon which has a constant angle with respect to the celestial equator, the points H and K must also be equidistant from their respective poles.<\/p>\n

In addition, $arc \\; EK = arc EH$. That was covered in this post<\/a><\/span>4<\/sup><\/a><\/span>.<\/p>\n

Also, the two large spherical triangles we’ve drawn, LK\u0398 and MHZ are also equal, as are spherical triangles LEK and MEH<\/span>5<\/sup><\/a><\/span>.<\/p>\n

While the large triangles and the small ones don’t quite fit one within the other, they still do share some vertexes: L and M. And because both the big triangles and the small ones are congruent, $\\angle{KLE} = \\angle{HME}$ and\u00a0$\\angle{KL \\Theta} = \\angle{HMZ}$.<\/p>\n

This means that we can subtract the small angles out of the big ones and end up with the same results. In other words:<\/p>\n

$$\\angle{KL\\Theta} – \\angle{KLE} = \\angle{EL \\Theta}$$<\/p>\n

and<\/p>\n

$$\\angle{HMZ} – \\angle{HME} = \\angle{EMZ}$$<\/p>\n

And since KL\u0398 and HMZ were equal, and we subtracted equal angles from them, that means $\\angle{EL \\Theta} =\u00a0\\angle{EMZ}$.<\/p>\n

These are angles in more spherical triangles, EL\u0398 and EMZ. And since their other sides were equal<\/span>6<\/sup><\/a><\/span>, that means we have a side-angle-side relationship and thus, spherical triangle EL\u0398 = spherical triangle EMZ. If the triangles are equal, then so too must their bases be.<\/p>\n

Therefore,\u00a0E\u0398 and EZ.<\/p>\n

This is only the first of several proofs in this chapter, but because it got quite involved, I think I’ll leave off there for now.<\/p>\n

\n","protected":false},"excerpt":{"rendered":"

In this next chapter, Ptolemy’s goal is to show how to calculate, for each latitude, the arcs of the equator… which rise together with [given] arcs of the ecliptic. To do this, we’ll do a bit of convenient math, breaking the full ecliptic into its traditional 12 parts. However, since these signs are not of … <\/p>\n

Continue reading “Almagest Book II: Symmetry of Rising Times – Arcs of the Ecliptic Equidistant from the Same Equinox”<\/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":[31,25,36,37,35,39,28,26,14],"class_list":["post-933","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-algebra","tag-almagest","tag-celestial-equator","tag-celestial-sphere","tag-ecliptic","tag-equinox","tag-geometry","tag-mathematics","tag-ptolemy"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-f3","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/933","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=933"}],"version-history":[{"count":11,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/933\/revisions"}],"predecessor-version":[{"id":1031,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/933\/revisions\/1031"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=933"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=933"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}