{"id":398,"date":"2018-06-09T19:04:57","date_gmt":"2018-06-09T19:04:57","guid":{"rendered":"http:\/\/jonvoisey.net\/blog\/?p=398"},"modified":"2018-06-26T22:08:48","modified_gmt":"2018-06-26T22:08:48","slug":"almagest-book-i-triangular-lemmas-for-spherical-trigonometry","status":"publish","type":"post","link":"https:\/\/jonvoisey.net\/blog\/2018\/06\/almagest-book-i-triangular-lemmas-for-spherical-trigonometry\/","title":{"rendered":"Almagest Book I: Triangular Lemmas for Spherical Trigonometry"},"content":{"rendered":"

Our next task is to demonstrate the sizes of the individual arcs cut off between the equator and the ecliptic along a great circle through the poles of the equator. As a preliminary we shall set out some short and useful theorems which will enable us to carry out most demonstrations involving spherical theorems in the simplest and most methodical way possible.<\/p><\/blockquote>\n

In opening the next chapter in Book 1, Ptolemy again gives us a goal to work towards, namely, the length of the chord shown in solid red below<\/span>1<\/sup><\/a><\/span>.<\/p>\n

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

But before we do that, we’re going to have to lay out some lemma<\/span>2<\/sup><\/a><\/span> to get us there. There’s going to be several, but for this post, I’m only going to address the first two which come from triangles, whereas the remaining involve circles and a bit of new notation that I’ll want to introduce before getting into them.<\/p>\n

Although we’re working towards spherical geometry, there’s still plenty of triangles and circles we’ll still be able to draw within the sphere. After all, any plane through a sphere cuts a circle. And if you have a chord across that circle and connect it to a side or center, then you have a normal Cartesian triangle.<\/p>\n

So it’s not surprising that Ptolemy starts with some triangles:<\/p>\n

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

The only thing to note in this diagram is that $\\overline{DG} \\parallel \\overline{EH}$<\/span>3<\/sup><\/a><\/span>.<\/p>\n

According to Ptolemy, because these two lines are parallel,<\/p>\n

$$\\frac{\\overline{AG}}{\\overline{AE}} = \\frac{\\overline{DG}}{\\overline{EH}}$$<\/p>\n

Ptolemy didn’t give a proof for this, but it comes from some similar triangles. In particular $\\triangle{ADG}$ is similar to $\\triangle{AHE}$. So the ratio of their sides will be the same as the ratio of their bases.<\/p>\n

Next Ptolemy does something a little funny and tosses a $\\overline{DZ}$ into the mix.<\/p>\n

$$\\frac{\\overline{AG}}{\\overline{AE}} = \\frac{\\overline{DG}}{\\overline{DZ}} \\cdot \\frac{\\overline{DZ}}{\\overline{EH}}$$<\/p>\n

Obviously the $\\overline{DZ}$’s will cancel so we haven’t really changed anything here, simply altered the look of it so we can do a substitution here in a moment. But before that, we’ll apply the same logic as previously to note that $\\triangle{BDZ}$ and $\\triangle{BHE}$ are similar which means that, again, the ratio of their bases and sides will be similar. This allows us to state:<\/p>\n

$$\\frac{\\overline{DZ}}{\\overline{EH}} = \\frac{\\overline{BZ}}{\\overline{BE}}$$<\/p>\n

And wouldn’t you look at that, we have a $\\frac{\\overline{DZ}}{\\overline{EH}}$ in common to both of our equations allowing for some substitution [13.1]<\/span>4<\/sup><\/a><\/span>.<\/a><\/p>\n

$$\\frac{\\overline{AG}}{\\overline{AE}} = \\frac{\\overline{DG}}{\\overline{DZ}} \\cdot \\frac{\\overline{BZ}}{\\overline{BE}}$$<\/p>\n

This is the first lemma Ptolemy set out to prove.<\/p>\n

For the next part, we begin with nearly the same figure as the previous one, but we’ll tack on a bit more:<\/p>\n

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

Here, we’ve dropped the previous parallel line and extended $\\overline{GD}$ out to point H such that $\\overline{AH} \\parallel \\overline{BE}$.<\/p>\n

This time, instead of looking at the ratio of a piece of a side to the full side, we’ll look at the pieces of the sides individually:<\/p>\n

$$\\frac{\\overline{EG}}{\\overline{AE}} = \\frac{\\overline{GZ}}{\\overline{HZ}}$$<\/p>\n

Again, this comes from similar triangles, but it’s also in Euclid’s Elements in Book VI, proposition 2<\/a>.<\/p>\n

Using $\\overline{DZ}$ again just to break things up, we can say:<\/p>\n

$$\\frac{\\overline{GZ}}{\\overline{HZ}} = \\frac{\\overline{GZ}}{\\overline{DZ}} \\cdot \\frac{\\overline{DZ}}{\\overline{HZ}}$$<\/p>\n

This next step is a little tricky, but $\\triangle{ADH}$ is similar to $\\triangle{BDZ}$. The reason is that two of their sides are common ($\\overline{HD}$ extends right into $\\overline{DZ}$ as does $\\overline{BD}$ into $\\overline{DA}$). In addition, $\\angle{DBZ} = \\angle{HAD}$ because these are alternate interior angles. That’s enough to justify those two triangles being similar, and in fact, you can see it if you imagine flipping $\\triangle{BDZ}$ up around point D so it rests inside $\\triangle{ADH}$. Anyway, similar triangles again allows us to state:<\/p>\n

$$\\frac{\\overline{DZ}}{\\overline{BD}} = \\frac{\\overline{DH}}{\\overline{AD}}$$<\/p>\n

Doing a bit of algebraic rearranging we can state:<\/p>\n

$$\\frac{\\overline{DH}}{\\overline{DZ}} = \\frac{\\overline{AD}}{\\overline{BD}}$$<\/p>\n

Now we’ll return to the componendo theorem<\/span>5<\/sup><\/a><\/span> we saw previously since we have the handy $\\frac{a}{b} = \\frac{c}{d}$ form where $a = \\overline{DH}$, $b = \\overline{DZ}$, $c = \\overline{AD}$, and $d = \\overline{BD}$. Thus, applying the theorem we get<\/p>\n

$$\\frac{\\overline{DH} + \\overline{DZ}}{\\overline{DZ}} = \\frac{\\overline{AD} + \\overline{BD}}{\\overline{BD}}$$<\/p>\n

But if we look at what the numerator in both of these physically represent in the diagram, we see that $\\overline{DH} + \\overline{DZ} = \\overline{HZ}$ and $\\overline{AD} + \\overline{BD} = \\overline{AB}$. Substituting these in:<\/p>\n

$$\\frac{\\overline{HZ}}{\\overline{DZ}} = \\frac{\\overline{AB}}{\\overline{BD}}$$<\/p>\n

Let’s flip both of those fractions over:<\/p>\n

$$\\frac{\\overline{DZ}}{\\overline{HZ}} = \\frac{\\overline{BD}}{\\overline{AB}}$$<\/p>\n

Which is something we had before we went on this most recent detour of similar triangles and componendo. So we can substitute that in:<\/p>\n

$$\\frac{\\overline{GZ}}{\\overline{HZ}} = \\frac{\\overline{GZ}}{\\overline{DZ}} \\cdot \\frac{\\overline{BD}}{\\overline{AB}}$$<\/p>\n

One more substitution will finish this one off, so let’s notice that $\\triangle{GZE}$ and $\\triangle{GHA}$ are similar (or apply the theorem from Euclid again<\/a>). As with above, that means the ratio of their similar sides will be equal as well, allowing us to state:<\/p>\n

$$\\frac{\\overline{GZ}}{\\overline{HZ}} = \\frac{\\overline{EG}}{\\overline{AE}}$$<\/p>\n

That substitutes right into the above giving us: <\/a><\/p>\n

$$\\frac{\\overline{EG}}{\\overline{AE}} = \\frac{\\overline{GZ}}{\\overline{DZ}} \\cdot \\frac{\\overline{BD}}{\\overline{AB}}$$<\/p>\n

And that’s the second lemma Ptolemy wants to introduce [13.2].<\/p>\n

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

Our next task is to demonstrate the sizes of the individual arcs cut off between the equator and the ecliptic along a great circle through the poles of the equator. As a preliminary we shall set out some short and useful theorems which will enable us to carry out most demonstrations involving spherical theorems in … <\/p>\n

Continue reading “Almagest Book I: Triangular Lemmas for Spherical Trigonometry”<\/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":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[24],"tags":[31,25,28,26,14],"class_list":["post-398","post","type-post","status-publish","format-standard","hentry","category-almagest","tag-algebra","tag-almagest","tag-geometry","tag-mathematics","tag-ptolemy"],"acf":[],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p9ZpvC-6q","_links":{"self":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/398","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=398"}],"version-history":[{"count":35,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/398\/revisions"}],"predecessor-version":[{"id":570,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/posts\/398\/revisions\/570"}],"wp:attachment":[{"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/media?parent=398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/categories?post=398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/jonvoisey.net\/blog\/wp-json\/wp\/v2\/tags?post=398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}