{"id":883,"date":"2018-08-13T13:23:37","date_gmt":"2018-08-13T18:23:37","guid":{"rendered":"http:\/\/jonvoisey.net\/blog\/?p=883"},"modified":"2018-08-13T13:26:46","modified_gmt":"2018-08-13T18:26:46","slug":"almagest-book-ii-symmetries-of-arcs-and-day-night-lengths","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2018\/08\/almagest-book-ii-symmetries-of-arcs-and-day-night-lengths\/","title":{"rendered":"Almagest Book II: Symmetries of Arcs and Day\/Night Lengths"},"content":{"rendered":"

For the past few\u00a0Almagest<\/em> posts, we’ve been working on the following diagram:<\/p>\n

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

As a refresher, AEG is the celestial equator and BED is the horizon of an observer. Z is south. H is the point of the winter solstice as it crosses the horizon (or rises).<\/p>\n

In the first post<\/a> we determined $arc \\; EH$. In the second post<\/a>, we used that to determine $arc \\; BZ$. In the third<\/a>, we’ve determined $arc \\; E \\Theta$ and then came full circle and showed another way to get\u00a0$arc \\; EH$.<\/p>\n

Now, Ptolemy wants to generalize.<\/p>\n

Instead of requiring that H be the winter solstice, we can let it be\u00a0any<\/em> point along the ecliptic<\/span>1<\/sup><\/a><\/span>.<\/p>\n

But instead of diving into equations, Ptolemy discusses this very generally as there is no reason to dive into all the equations again. In essence, we’ve just walked through them. There’s nothing fundamentally special about H having been the winter solstice. So by reassigning it, we can walk through all the same steps we did in the previous posts. But what will change is $arc \\; H\\Theta$. This in turn will change $arc \\; EH$ and $arc \\; E\\Theta$, but as we just noted above, those are things we can calculate.<\/p>\n

Previously we used the winter solstice because it’s an extreme: It’s the point where the sun is furthest from the celestial equator at what Ptolemy has taken to be $\\frac{1}{2} \\frac{11}{83}$ of 360\u00ba. That was a handy number to use. But diving a bit more specifically into what $arc \\; H\\Theta$\u00a0really\u00a0is<\/em>, it’s the arc between the celestial equator and ecliptic. If you’ve been following along, that should sound familiar as we’ve derived an entire table of these values<\/a>.<\/p>\n

So since we’ve generalized this a bit, let’s update our figure. First, let’s recall this diagram I showed a few posts ago<\/a>:<\/p>\n

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

Here, I’d drawn in a dashed line to represent the path the winter solstice (or now any other point on the ecliptic) would take over the course of a single night. I’d only put that in temporarily to illustrate the point, but let’s now solidify that a bit more.<\/p>\n

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

Here, all I’ve done is draw that line as solid instead of dashed and given it a label of L. So a point on the horizon will rise along $arc \\; HL$, first being on the horizon at H, rising to its maximum altitude, L, and then setting on the other side of the celestial sphere.<\/p>\n

But let’s consider another point on the ecliptic opposite E from H, but equidistant.<\/p>\n

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

Here I’ve drawn that in as point K. If we jump back and consider again H as the winter solstice, this would mean that K would be the summer solstice.<\/p>\n

In addition, N would be the north celestial pole. M is the point at which the sun (on the summer solstice) is at its lowest point\u00a0below<\/em> the horizon.<\/p>\n

What should quickly become apparent is that there’s a symmetry going on: Since we said that K was equidistant from E as H, that can also be stated as $arc \\; EH = arc \\; EK$. They’re simply mirrored and flipped.<\/p>\n

Ptolemy sketches out the general proof that this holds true for the rest of the pieces as well:\u00a0$arc \\; E\\Theta = arc \\; EX$,\u00a0$arc \\; H\\Theta = arc \\; KX$, and\u00a0$arc \\; HL = arc \\; KM$.<\/p>\n

But let’s look at those last two in a bit more detail because that’s a big part of what Ptolemy is really wanting to get at with this section. In particular, consider\u00a0$arc \\; HL$ and $arc \\; KM$. We said that $arc \\; HL$ was (half of) the path of the sun\u00a0above<\/em> the horizon, whereas\u00a0$arc \\; KM$ was (half of) the path of the sun\u00a0below<\/em> the horizon. Thus, their complements of their respective small circles (not drawn), must also be the same.<\/p>\n

Putting that into a physical meaning, this would imply that the length of the\u00a0day<\/em> for a point on the ecliptic rising at point H, is the\u00a0same<\/em> as the length of the\u00a0night<\/em> for a point on the ecliptic\u00a0rising at point K and vice versa.<\/p>\n

That’s a very specific pronunciation, but in a more generalized context, this is the explanation of why days are longer in the summer than winter. It’s always fun when math tells us what we already know. It’s generally a sign we’re doing something right.<\/p>\n

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

For the past few\u00a0Almagest posts, we’ve been working on the following diagram: As a refresher, AEG is the celestial equator and BED is the horizon of an observer. Z is south. H is the point of the winter solstice as it crosses the horizon (or rises). In the first post we determined $arc \\; EH$. … <\/p>\n

Continue reading “Almagest Book II: Symmetries of Arcs and Day\/Night Lengths”<\/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":[25,36,37,35,28,14],"class_list":["post-883","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-almagest","tag-celestial-equator","tag-celestial-sphere","tag-ecliptic","tag-geometry","tag-ptolemy"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-ef","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/883","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=883"}],"version-history":[{"count":6,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/883\/revisions"}],"predecessor-version":[{"id":895,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/883\/revisions\/895"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=883"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=883"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}