{"id":4359,"date":"2023-12-20T10:42:43","date_gmt":"2023-12-20T16:42:43","guid":{"rendered":"https:\/\/jonvoisey.net\/blog\/?p=4359"},"modified":"2023-12-23T14:15:34","modified_gmt":"2023-12-23T20:15:34","slug":"almagest-book-x-the-size-of-venus-epicycle","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2023\/12\/almagest-book-x-the-size-of-venus-epicycle\/","title":{"rendered":"Almagest Book X: The Size of Venus’ Epicycle"},"content":{"rendered":"

Now we’ll turn our attention to finding the size of Venus’ epicycle. Fortunately, the lack of the extra sphere that Mercury had will make this much easier.<\/p>\n

Ptolemy starts with the following diagram:<\/p>\n

\"\"<\/a>In this image, circle $ABG$ is the eccentre carrying the epicycle with its center at $D$ and the observer, on earth, at $E$.<\/p>\n

We’ll take point $A$ as the point $25\u00ba$ into Taurus (apogee) and $G$ as the point $25\u00ba$ into Scorpio (perigee). Points $Z$ and $H$ are tangent to these circles (i.e., at greatest elongation).<\/p>\n

We’ll first create a demi-degrees circle about $\\triangle AEZ$. In this, $\\angle AEZ = 44;48\u00ba$, which is the greatest elongation Ptolemy reported at this position in the previous post<\/a>.<\/p>\n

We can then look at the subtending arc, $arc \\; AZ$ which would then have a measure of twice this, or $89;36\u00ba$. This is then looked up in the table of chords<\/a> to find the corresponding chord, $\\overline{AZ} = 84;33^p$ in this demi-degrees context in which $\\overline{AE} = 120^p$.<\/p>\n

We can then do the same for $\\triangle EHG$ in which $\\angle HEG = 47;20\u00ba$.\u00a0 We then find that $\\overline{GH} = 88;13^p$ in the context where $\\overline{EG} = 120^p$.<\/p>\n

However, these are two different contexts. We’ll temporarily pick one to standardize the other to.<\/p>\n

Ptolemy picks the first one in which $\\overline{AE} = 120^p$ and the radius of the epicycle has a measure of $84;33^p$.<\/p>\n

To convert the other triangle, we’ll again set up a demi-degrees circle about it, and let $\\overline{GH} = 84;33^p$ instead, so we can determine the length of $\\overline{EG}$.<\/p>\n

We’ll then convert the other triangle to this context. Doing so, I find $\\overline{EG} = 114;59^p$. Ptolemy comes up with a slightly different value of $115;01^p$.<\/p>\n

He then adds $\\overline{AE} + \\overline{EG}$ to determine $\\overline{AG} = 235;01^p$ in this context, half of which ($\\overline{AD}$ or $\\overline{DG}$) is $117;30,30^p$.<\/p>\n

We can then subtract to determine $\\overline{DE}$:<\/p>\n

$$\\overline{AE} – \\overline{AD} = \\overline{DE} = 120^p – 117;30,30^p = 2;29,30^p.$$<\/p>\n

Ptolemy rounds this down to $2;29^p$.<\/p>\n

We can then convert this to the larger context we’ve been using for all of the models in which $\\overline{AD} = 120^p$ and $\\overline{DE} \\approx 1 \\frac{1}{4}^p$ in which case we find that the radius of the epicycle, $\\overline{AZ} = 43 \\frac{1}{6}^p$.<\/p>\n

\n

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

\n

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

Now we’ll turn our attention to finding the size of Venus’ epicycle. Fortunately, the lack of the extra sphere that Mercury had will make this much easier.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","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":[14,69],"class_list":["post-4359","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-ptolemy","tag-venus"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-18j","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4359","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=4359"}],"version-history":[{"count":2,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4359\/revisions"}],"predecessor-version":[{"id":4381,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4359\/revisions\/4381"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=4359"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=4359"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=4359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}