{"id":4672,"date":"2024-07-30T13:50:25","date_gmt":"2024-07-30T18:50:25","guid":{"rendered":"https:\/\/jonvoisey.net\/blog\/?p=4672"},"modified":"2024-07-30T13:50:25","modified_gmt":"2024-07-30T18:50:25","slug":"almagest-book-x-second-iteration-correction-for-equant-second-opposition","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2024\/07\/almagest-book-x-second-iteration-correction-for-equant-second-opposition\/","title":{"rendered":"Almagest Book X: Second Iteration Correction for Equant \u2013 Second Opposition"},"content":{"rendered":"

Next, we’ll work on the correction for the second opposition in our second iteration.<\/p>\n

As with before, we’ll make use of our previous diagram:<\/p>\n

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

In this, $arc \\; XZ = 39;04\u00ba$ as this was the distance we determined the second observation was from the line of apsides.<\/p>\n

This is also equal to $\\angle Z \\Theta X$ and its vertical angle $\\angle D \\Theta F$.<\/p>\n

As a reminder, $\\overline {D \\Theta} = \\overline{DN} = 5;55^p$ in this second iteration.<\/p>\n

With that, we can use some trig to determine $\\overline{DF}$:<\/p>\n

$$\\overline{DF} = \\overline{D \\Theta} \\cdot sin(\\angle D \\Theta F) = 5;55^p \\cdot sin(39;04\u00ba)$$<\/p>\n

$$\\overline{DF} = 3;44^p.$$<\/p>\n

Next, $\\overline{DB} =60^p$ since it’s a radius of one of the eccentres. So we now know two sides in $\\triangle DBF$ allowing us to state via the Pythagorean theorem:<\/p>\n

$$\\overline{BF} = \\sqrt{60^2 – 3;44^2} = 59;53^p.$$<\/p>\n

And using a bit more trig or the Pythagorean theorem, we can determine $\\overline{F \\Theta} = 4;36^p$ which has the same measure as $\\overline{QF}$.<\/p>\n

We can now add $\\overline{BF} + \\overline{FQ}$ to find,<\/p>\n

$$\\overline BQ = 59;53^p + 4;36^ = 46;29^p.$$<\/p>\n

We can also state that<\/p>\n

$$\\overline{NQ} = 2 \\cdot \\overline{DF} = 2 \\cdot 3;44^p = 7;28^p.$$<\/p>\n

That gives us two sides in $\\triangle NBQ$ so we can use a bit of trig to determine:<\/p>\n

$$\\angle NBQ = tan^{-1}(\\frac{\\overline{QN}}{\\overline{QB}})$$<\/p>\n

$$\\angle NBQ = tan^{-1}(\\frac{7;28^p}{64;29^p}) = 6;36\u00ba.$$<\/p>\n

We’ll now add a few pieces together to find $\\overline{QZ}$:<\/p>\n

$$\\overline{QZ} = \\overline{Z \\Theta} + \\overline{\\Theta F} + \\overline{FQ} = 60^p + 4;36^p + 4;36^p = 69;12^p.$$<\/p>\n

That gives us two of the sides in $\\triangle NZQ$ allowing us to use a bit more trig to determine,<\/p>\n

$$\\angle NZQ = tan^{-1}(\\frac{\\overline{QN}}{\\overline{QZ}})$$<\/p>\n

$$\\angle NZQ = tan^{-1}(\\frac{7;28^p}{69;12^p}) = 6;09\u00ba.$$<\/p>\n

Lastly, we can subtract to find<\/p>\n

$$\\angle BNZ = \\angle NBQ – \\angle NZQ = 6;36\u00ba – 6;09\u00ba = 0;27\u00ba.$$<\/p>\n

This is also $arc \\; LT$ which is what we were looking for in this post.<\/p>\n

Ptolemy gives a value of $0;28\u00ba$ whereas Toomer gives a value of $0;26,51\u00ba$ in good agreement with mine<\/span>1<\/sup><\/a><\/span>.<\/p>\n

\n

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

Next, we’ll work on the correction for the second opposition in our second iteration.<\/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":[80,17,81,6],"class_list":["post-4672","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-eccentricity","tag-equant","tag-line-of-apsides","tag-mars"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-1dm","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4672","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=4672"}],"version-history":[{"count":2,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4672\/revisions"}],"predecessor-version":[{"id":4674,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4672\/revisions\/4674"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=4672"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=4672"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=4672"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}