{"id":4222,"date":"2023-09-25T20:03:53","date_gmt":"2023-09-26T01:03:53","guid":{"rendered":"https:\/\/jonvoisey.net\/blog\/?p=4222"},"modified":"2023-10-21T10:36:05","modified_gmt":"2023-10-21T15:36:05","slug":"almagest-book-ix-exploring-mercurys-double-perigee","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2023\/09\/almagest-book-ix-exploring-mercurys-double-perigee\/","title":{"rendered":"Almagest Book IX: Exploring Mercury\u2019s Double Perigee"},"content":{"rendered":"

As a general rule, I try to stay away from using too much modern math as I work through the\u00a0Almagest<\/em>. The goal of this project is to try to understand how astronomers worked in a historical context – not simply examining their work through a modern lens.<\/p>\n

However, Ptolemy’s discussion around Mercury has been greatly frustrating me. There’s several reasons for this. A large one is certainly that the language Ptolemy used is clunky which is challenging for a model that is so complex.<\/p>\n

Therefore, I want to dig deeper into what’s happening with the double perigee and make sure I fully understand it. In particular, I previously showed a diagram from Pedersen which looked at the path the center of the epicycle would trace out for various eccentricities. This graphically showed the distance from earth over time, but I wanted something more quantitative, so in this post, we’ll derive an equation to determine the distance between the center of Mercury’s epicycle and earth as a function of\u00a0 the angle from apogee.<\/p>\n

This method comes straight from Pedersen but I’ll be doing it in the context of Toomer’s translation<\/span>1<\/sup><\/a><\/span>.<\/p>\n

As usual, let’s start with a diagram:<\/p>\n

 <\/p>\n

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

By and large, this image is what we’ve seen before. $\\overline{AG}$ is the line of apsides. Points $Z$, $D$, and $E$ are the same as they were when we set up Mercury’s model<\/a>. In other words, $Z$ is the point around which the eccentre that carries the epicycle rotates clockwise, $H$ is its geometric center, $D$ is the point about which Mercury itself rotates counter-clockwise, and $E$ is the observer on earth.<\/p>\n

As we discussed when we set up Mercury’s model, $\\overline{ED} = \\overline{DZ} = \\overline{ZH}$. Instead, of typing these out every time, we’ll just use a shorthand and call that length $e$ as it is a constant in this model.<\/p>\n

Similarly, we’ll also call $\\overline{KH}$ (the radius of the eccentre the epicycle rides on) $R$ as this is also a constant.<\/p>\n

Lastly, $\\angle AZH$ is always equal and opposite $\\angle ADK$. We’ll call these angles $\\Theta$.<\/p>\n

We’ve connected points $H$ to $K$ and $K$ to $E$. We’ve also extended $\\overline{KD}$ to point $B$ where we’ve dropped a perpendicular on it from $H$.<\/p>\n

So, let’s get started on the derivation.<\/p>\n

First, we’ll concentrate on $\\triangle ZDH$. In this triangle, $\\angle DZH = 180\u00ba – \\Theta$ as it’s the supplemental angle to $\\Theta$.<\/p>\n

This is also an isosceles triangle since two of its sides $\\overline{DZ}$ and $\\overline{ZH}$ are equal. This means that, within this triangle, $\\angle ZDH = \\angle ZHD$.<\/p>\n

We can solve for these angles since the angles in a triangle add up to $180\u00ba$ and we know what the third angle is.<\/p>\n

Thus:<\/p>\n

$$\\angle ZDH + \\angle ZHD = 180\u00ba – (180\u00ba – \\Theta)$$<\/p>\n

$$\\angle ZDH + \\angle ZHD = \\Theta$$<\/p>\n

Since these angles are equal,<\/p>\n

$$\\angle ZDH = \\angle ZHD = \\frac{\\Theta}{2}$$<\/p>\n

Next, we’ll start looking at the angles about $D$. We already stated that $\\angle ZDK = \\Theta$ as this is part of the hypothesis\/model. We also just solved for $\\angle ZDH$. So this means the only piece left on $\\overline{KB}$ is $\\angle{HDB}$ and the sum of these three angles must be $180\u00ba$ since they add up to the straight line.<\/p>\n

Thus, we can write:<\/p>\n

$$\\angle DZK + \\angle ZDH + \\angle HDB = 180\u00ba$$<\/p>\n

And substituting in:<\/p>\n

$$\\Theta + \\frac{\\Theta}{2} + \\angle HDB = 180\u00ba$$<\/p>\n

Solving for $\\angle HDB$ and simplifying:<\/p>\n

$$\\angle HDB = 180 – \\frac{3 \\Theta}{2}$$<\/p>\n

Next, we’ll want to start working on $\\triangle DHB$, but we’ll start by returning to $\\triangle ZDH$ and find the shared side, $\\overline{ZH}$.<\/p>\n

Again, this triangle is isosceles, so we’ll divide it in half about $Z$. This allows us to state:<\/p>\n

$$cos(\\frac{\\theta}{2}) = \\frac{\\frac{1}{2} \\overline{DH}}{e}$$<\/p>\n

Solving:<\/p>\n

$$\\overline{DH} = 2e \\cdot cos(\\frac{\\theta}{2})$$<\/p>\n

Now, we know both an angle and the hypotenuse in $\\triangle DHB$, so we can go ahead and solve the other two sides. I’ll start with $\\overline{BH}$:<\/p>\n

$$sin(180 – \\frac{3 \\Theta}{2}) = \\frac{\\overline{BH}}{\\overline{DH}}$$<\/p>\n

$$\\overline{BH} = \\overline{DH} \\cdot sin(180 – \\frac{3 \\Theta}{2})$$<\/p>\n

And substituting in for $\\overline{DH}$:<\/p>\n

$$\\overline{BH} = 2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(180 – \\frac{3 \\Theta}{2})$$<\/p>\n

Next, we’ll solve for $\\overline{DB}$ in the same triangle:<\/p>\n

$$cos(180 – \\frac{3 \\Theta}{2}) = \\frac{\\overline{DB}}{\\overline{DH}}$$<\/p>\n

$$\\overline{DB} = \\overline{DH} \\cdot cos(180 – \\frac{3 \\Theta}{2})$$<\/p>\n

$$\\overline{DB} = 2e \\cdot cos(\\frac{\\theta}{2}) \\cdot cos(180 – \\frac{3 \\Theta}{2})$$<\/p>\n

Next, we’ll turn our attention to $\\triangle KBH$.<\/p>\n

First, $\\overline{KB} = \\overline{KD} + \\overline{DB}$. We don’t know $\\overline{KD}$ currently, so let’s work on that.<\/p>\n

This triangle is a right triangle, so applying the Pythagoren theorem we have:<\/p>\n

$$\\overline{KH}^2 = R^2 – \\overline{BH}$$<\/p>\n

Substituting in some:<\/p>\n

$$(\\overline{KD} + \\overline{DB})^2 = R^2 – [2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(180 – \\frac{3 \\Theta}{2})]^2$$<\/p>\n

$$\\overline{KD} + \\overline{DB} = \\sqrt{R^2 – [2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(180 – \\frac{3 \\Theta}{2})]^2}$$<\/p>\n

And solving for $\\overline{KD}$:<\/p>\n

$$\\overline{KD} = \\sqrt{R^2 – [2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(180 – \\frac{3 \\Theta}{2})]^2} – \\overline{DB}$$<\/p>\n

And substituting in for $\\overline{DB}$:<\/p>\n

$$\\overline{KD} = \\sqrt{R^2 – [2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(180 – \\frac{3 \\Theta}{2})]^2} – 2e \\cdot cos(\\frac{\\theta}{2}) \\cdot cos(180 – \\frac{3 \\Theta}{2})$$<\/p>\n

We can do a bit of simplification using the identities $cos(180\u00ba – \\Theta) = – cos (\\Theta)$ and $sin(180\u00ba – \\Theta) = sin(\\Theta)$.<\/p>\n

$$\\overline{KD} = \\sqrt{R^2 – [2e \\cdot cos(\\frac{\\theta}{2}) \\cdot sin(\\frac{3 \\Theta}{2})]^2} + 2e \\cdot cos(\\frac{\\theta}{2}) \\cdot cos(\\frac{3 \\Theta}{2})$$<\/p>\n

Finally, we can turn to $\\triangle KDE$. In it, we know that $\\angle KDE = 180\u00ba – \\Theta$ as it’s the supplement to $\\angle ADK$. We also just solved for one side, $\\overline{KD}$.<\/p>\n

It’s not a right triangle, so we’ll need to use the law of cosines to solve for $\\overline{KE}$ which is what we’re really after here as it’s the distance from earth at any given value of $\\Theta$.<\/p>\n

$$\\overline{KE} = \\sqrt{\\overline{KD}^2 + e^2 – 2 \\cdot \\overline{KD} \\cdot e \\cdot cos(180 – \\Theta)}$$<\/p>\n

I could substitute in for $\\overline{KD}$ but this is already messy enough<\/span>2<\/sup><\/a><\/span> . So I’ll just simplify the final $cos$ term again.<\/p>\n

$$\\overline{KE} = \\sqrt{\\overline{KD}^2 + e^2 + 2 \\cdot \\overline{KD} \\cdot e \\cdot cos(\\Theta)}$$<\/p>\n

I’ve turned this into a Google sheet<\/a> to see how it behaves<\/span>3<\/sup><\/a><\/span>. There’s an obvious question of what values to use for $e$ and $R$, so I’ll jump ahead and tell you now that Ptolemy will get a value of $e = 3^p$ when $R = 60^p$.<\/p>\n

We can plot that, and in doing so, get the following:<\/p>\n

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

While there is clearly a double minimum here (indicating a double perigee), we can’t tell directly from the graph where it is. However, since the points I calculated were $0.1\u00ba$ apart, we can look at the data set to get a pretty good understanding.<\/p>\n

Doing so, I find the minimum values at $120.5\u00ba$ and $239.5\u00ba$ (from apogee). This means that, if apogee is $10\u00ba$ into Libra, which places the nearest points in Gemini and Aquarius and is consistent with what Ptolemy reported in the last post<\/a>.<\/p>\n

\n

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

As a general rule, I try to stay away from using too much modern math as I work through the\u00a0Almagest. The goal of this project is to try to understand how astronomers worked in a historical context – not simply examining their work through a modern lens. However, Ptolemy’s discussion around Mercury has been greatly … <\/p>\n

Continue reading “Almagest Book IX: Exploring Mercury\u2019s Double Perigee”<\/span><\/a><\/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_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},"jetpack_post_was_ever_published":false},"categories":[24],"tags":[25,68,14],"class_list":["post-4222","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-almagest","tag-mercury","tag-ptolemy"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-166","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4222","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=4222"}],"version-history":[{"count":16,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4222\/revisions"}],"predecessor-version":[{"id":4241,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/4222\/revisions\/4241"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=4222"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=4222"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=4222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}