Google Sheet<\/a>.<\/p>\nWe’ll start off by noting that $arc \\; XE$ we determined to be $42;40\u00ba$ in the last post (where it was called $arc \\; AL$). The central angle $\\angle E \\Theta X$ has the same measure, as does its vertical angle, $\\angle D \\Theta F$.<\/p>\n
We’ll also quickly note that<\/p>\n
$$\\overline{D \\Theta} = \\overline{DN} = \\frac{1}{2} \\cdot \\overline{\\Theta N} = \\frac{1}{2} \\cdot 11;50^p = 5;55^p.$$<\/p>\n
We can then directly determine $\\overline{DF}$ using some trig:<\/p>\n
$$\\overline{DF} = 5;55^p \\cdot sin(42;40\u00ba) = 4;01^p.$$<\/p>\n
And we can do the same for $\\overline{F \\Theta}$:<\/p>\n
$$\\overline{DF} = 5;55^p \\cdot cos(42;40\u00ba) = 4;21^p.$$<\/p>\n
We now have two sides of $\\triangle FAD$. We just determined $\\overline{DF}$ and $\\overline{DA} = 60^p$ as it’s a radius. Thus, we can use the Pythagorean theorem to solve for $\\overline{FA}$:<\/p>\n
$$\\overline{FA} = \\sqrt{60^2 – 4;21^2} = 59;52^p.$$<\/p>\n
Next, $\\overline{QF}$ is the same length as $\\overline{F \\Theta}$ which is $4;35^p$.<\/p>\n
That can be added to $\\overline{FA}$ to determine $\\overline{QA} = 64;13^p$.<\/p>\n
Similarly, $\\overline{QN} = 2 \\cdot \\overline{FD} = 8;01^p$.<\/p>\n
That gives us two sides in right triangle $\\triangle NAQ$. Thus, we can again use a bit of trig to solve for $\\angle NAQ$:<\/p>\n
$$\\angle NAQ = tan^{-1} (\\frac{8;01^p}{64;13^p}) = 7;07\u00ba.$$<\/p>\n
We’ll quickly note that $\\overline{\\Theta E} = 60^p$ since it’s a radius.<\/p>\n
That allows us determine $\\overline{QE}$ which is:<\/p>\n
$$\\overline{QE} = \\overline{\\Theta E} + \\overline{QF} + \\overline{F \\Theta} = 60^p + 4;35^p + 4;35^p = 68;42^p.$$<\/p>\n
This now gives us two sides in $\\triangle NEQ$: $\\overline{QN}$ and $\\overline{QE}$. So a bit more trig allows us to determine $\\angle NEQ$:<\/p>\n
$$\\angle NEQ = tan^{-1} (\\frac{8;01^p}{68;42^p}) = 6;40\u00ba.$$<\/p>\n
Lastly,<\/p>\n
$$\\angle ANE = \\angle NAQ – \\angle NEQ = 7;07\u00ba – 6;40\u00ba = 0;28\u00ba.$$<\/p>\n
This is the same as $arc \\; KS$ and is what we set about to find as it’s the correction for the first opposition.<\/p>\n
Again, the value derived here agrees with the values found by Ptolemy and Toomer.<\/p>\n
\n <\/p>\n","protected":false},"excerpt":{"rendered":"
Now it’s time to determine the small angles that get applied as corrections. As with before, we’ll start with the first opposition, again using the same diagram.<\/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":[80,17,6],"class_list":["post-4662","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-eccentricity","tag-equant","tag-mars"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-1dc","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4662","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=4662"}],"version-history":[{"count":4,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4662\/revisions"}],"predecessor-version":[{"id":4666,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4662\/revisions\/4666"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=4662"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=4662"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=4662"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/a><\/p>\n