<?xml version="1.0" encoding="UTF-8" standalone="yes"?><oembed><version><![CDATA[1.0]]></version><provider_name><![CDATA[Azimuth]]></provider_name><provider_url><![CDATA[https://johncarlosbaez.wordpress.com]]></provider_url><author_name><![CDATA[John Baez]]></author_name><author_url><![CDATA[https://johncarlosbaez.wordpress.com/author/johncarlosbaez/]]></author_url><title><![CDATA[Rolling Circles and Balls (Part&nbsp;3)]]></title><type><![CDATA[link]]></type><html><![CDATA[<p>In <a href="https://johncarlosbaez.wordpress.com/2012/08/31/rolling-circles-and-balls-part-1/">Part 1</a> and <a href="https://johncarlosbaez.wordpress.com/2012/09/03/rolling-balls-and-circles-part-2/">Part 2</a> we looked at the delightful curves you get by rolling one circle on another.    Now let&#8217;s see what happens when you roll one circle <i>inside</i> another!</p>
<h3>  Four times as big </h3>
<p>If you roll a circle inside a circle that&#8217;s 4 times as big, we get an <a href="http://en.wikipedia.org/wiki/Astroid"><b>astroid</b></a>:</p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Astroid"><img width="400" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/astroid_animation.gif" /></a></div>
<p><b>Puzzle 1.</b> How many times does the rolling circle turn as it rolls all the way around?</p>
<p>By the way: don&#8217;t confuse an astroid with an asteroid.  They both got their names because someone thought they looked like stars, but that&#8217;s where resemblance ends!</p>
<p>You can get an astroid using this funny parody of the equation for a circle:</p>
<p><img src='https://s0.wp.com/latex.php?latex=x%5E%7B2%2F3%7D+%2B+y%5E%7B2%2F3%7D+%3D+1&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='x^{2/3} + y^{2/3} = 1' title='x^{2/3} + y^{2/3} = 1' class='latex' /></p>
<p>Or, if you don&#8217;t like equations, you can get a quarter of an astroid by letting a ladder slide down a wall and taking a time-lapse photo!  </p>
<div align="center"><a href="http://mathforum.org/mathimages/index.php/Envelope"><img src="https://i1.wp.com/math.ucr.edu/home/baez/mathematical/astroid_as_envelope_animation.gif" /></a></div>
<p>In other words, you get a whole astroid by taking the envelope of all line segments of length 1 going from some point on the x axis to some point on the y axis!</p>
<div align="center"><a href=""><img src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/astroid_as_envelope.png" /></a></div>
<h3> Three times as big </h3>
<p>If the fixed circle is just 3 times as big as the one rolling inside it, we get an <a href="http://en.wikipedia.org/wiki/Deltoid"><b>deltoid</b></a>:</p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Deltoid"><img width="400" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/deltoid_animation.gif" /></a></div>
<p><b>Puzzle 2.</b> Now how many times does the rolling circle turn as it rolls all the way around?</p>
<p>By the way: it looks like we&#8217;re back to naming curves after body parts&#8230; but we&#8217;re not: both this curve and the muscle called a deltoid got their names because they look like the Greek letter delta:</p>
<div align="center">
<img src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/delta.gif" /></div>
<p><b>Puzzle 3.</b>  Did the Greek letter delta get that name because it was shaped like a river delta, or was it the other way around?</p>
<p>As you might almost expect by now, if you&#8217;ve been reading this whole series, there are weird relations between the deltoid and the astroid.  </p>
<p>For example: take a deltoid and shine parallel rays of light at it from <i>any direction</i>.  Then the envelope of these rays is an astroid!</p>
<div align="center"><a href="http://xahlee.info/SpecialPlaneCurves_dir/Deltoid_dir/deltoid.html"><br />
<img src="https://i1.wp.com/math.ucr.edu/home/baez/mathematical/astroid_as_caustic.png" /></a></div>
<p>We summarize this by saying that the astroid is a <a href="http://mathworld.wolfram.com/Catacaustic.html"><b>catacaustic</b></a> of the deltoid.  This picture is by <a href="http://xahlee.info/SpecialPlaneCurves_dir/Deltoid_dir/deltoid.html">Xah Lee</a>, who has also made a nice movie of what happens as you rotate the light source:</p>
<p>&bull; Xah Lee, <a href="http://xahlee.info/SpecialPlaneCurves_dir/Deltoid_dir/deltoidCaustic.mov">Deltoid catacaustic movie</a>.</p>
<p>I don&#8217;t completelly understand the rays going <i>through</i> the deltoid, in either the picture or the movie.  It looks like those rays are getting refracted, but that would be a <b>diacaustic</b>, not a catacaustic.  ,I think they&#8217;re formed by continuing reflected rays to straight lines that go through the deltoid.  If you didn&#8217;t do that you wouldn&#8217;t get a whole astroid, just part of one.</p>
<p>Anyway, at the very least you get part of an astroid, which you can complete to a whole one.  And then, as you rotate the light source, the astroid you get rolls around the deltoid in a pleasant manner!  This is nicely illustrated here:</p>
<p>&bull; <a href="http://mathworld.wolfram.com/DeltoidCatacaustic.html">Deltoid catacaustic</a>, Wolfram Mathworld.</p>
<p>You can also get a deltoid from a deltoid!  Draw all the <a href="http://en.wikipedia.org/wiki/Osculating_circle">osculating circles</a> of the deltoid&#8212;that is, circles that  match the deltoid&#8217;s curvature as well as its slope at the points they touch. The centers of these circles lie on another, larger deltoid:  </p>
<div align="center"><a href="http://xahlee.info/SpecialPlaneCurves_dir/Deltoid_dir/deltoid.html"><img src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/deltoid_as_evolute.png" /></a></div>
<p>We summarize this by saying that the <a href="http://en.wikipedia.org/wiki/Evolute"><b>evolute</b></a> of a deltoid is another deltoid.</p>
<p>There are also fancier ways to get deltoids if you know more math.  For example, the set of <a href="http://en.wikipedia.org/wiki/Trace_%28linear_algebra%29">traces</a> of matrices lying in the group <a href="http://en.wikipedia.org/wiki/Special_unitary_group">SU(3)</a> forms a filled-in deltoid in the complex plane!</p>
<p>This raises a question which unfortunately I&#8217;m too lazy to answer myself.  So, I&#8217;ll just pose it as a puzzle:</p>
<p><b>Puzzle 4.</b>  Is the set of traces of matrices lying in SU(4) a filled-in astroid?  In simpler terms, consider the values of <img src='https://s0.wp.com/latex.php?latex=a+%2B+b+%2B+c+%2B+d&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='a + b + c + d' title='a + b + c + d' class='latex' /> that we can get from complex numbers <img src='https://s0.wp.com/latex.php?latex=a%2Cb%2Cc%2Cd&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='a,b,c,d' title='a,b,c,d' class='latex' /> with <img src='https://s0.wp.com/latex.php?latex=%7Ca%7C+%3D+%7Cb%7C+%3D+%7Cc%7C+%3D+%7Cd%7C+%3D+1&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='|a| = |b| = |c| = |d| = 1' title='|a| = |b| = |c| = |d| = 1' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=abcd+%3D+1.+&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='abcd = 1. ' title='abcd = 1. ' class='latex' />  Do these values form a filled-in astroid?  </p>
<h3> Twice as big </h3>
<p>But now comes the climax of today&#8217;s story: what happens when we let a circle roll inside a circle that&#8217;s exactly <i>twice</i> as big?   </p>
<div align="center">
<img src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_animation.gif" /></div>
<p>Now you can see the rolling circle turn around <i>once</i> as it rolls all the way around the big one.  </p>
<p>More excitingly, if we track a point on the rolling circle, <i>it traces out a straight line!</i>  Can you see why?  Later I&#8217;ll give a few proofs.</p>
<p>This gadget is called a <a href="http://en.wikipedia.org/wiki/Tusi-couple"><b>Tusi couple</b></a>.  You could use it to convert a rolling motion into a vibrating one using some gears.   Greg Egan made a nice animation showing how:</p>
<div align="center">
<img src="https://i1.wp.com/math.ucr.edu/home/baez/mathematical/tusi_gears_animation.gif" /></div>
<p>The Tusi couple is named after the Persian astronomer and mathematician <a href="http://en.wikipedia.org/wiki/Nasir_al-Din_al-Tusi">Nasir al-Din al-Tusi</a>, who discovered it around 1247, when he wrote a commentary on Ptomely&#8217;s <i>Almagest</i>, an important astronomical text:</p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Nasir_al-Din_al-Tusi"><img src="https://i1.wp.com/math.ucr.edu/home/baez/mathematical/tusi_book.jpg" /></a></div>
<p>He wrote:</p>
<blockquote><p>
If two coplanar circles, the diameter of one of which is equal to half the diameter of the other, are taken to be internally tangent at a point, and if a point is taken on the smaller circle&#8212;and let it be at the point of tangency&#8212;and if the two circles move with simple motions in opposite direction in such a way that the motion of the smaller is twice that of the larger so the smaller completes two rotations for each rotation of the larger, then that point will be seen to move on the diameter of the larger circle that initially passes through the point of tangency, oscillating between the endpoints.
</p></blockquote>
<p>I don&#8217;t quite understand why he was interested in this, but it has something to do with using epicycles to build linear motion out of circular motion.    It also has something to do with the apparent motion of planets between the Earth and the Sun.  </p>
<p>Later Copernicus also studied the Tusi couple.  He proved that the moving point really did trace out a straight line:</p>
<div align="center"><a href="http://www.columbia.edu/~gas1/project/visions/case1/sci.1.html"><img width="400" src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/tusi_copernicus_book.jpg" /></a></div>
<p>However, many suspect that this was not a true rediscovery: al-Tusi had also proved this, and some aspects of Copernicus&#8217; proof seem too similar to al-Tusi&#8217;s to be coincidence:</p>
<p>&bull; George Saaliba, <a href="http://www.columbia.edu/~gas1/project/visions/case1/sci.1.html">Whose science is Arabic science in Renaissance Europe?</a>, Section 2: <a href="http://www.columbia.edu/~gas1/project/visions/case1/sci.2.html">Arabic/Islamic science and the Renaissance science in Italy</a>.</p>
<p>&bull; I. N. Veselovsky, <a href="http://articles.adsabs.harvard.edu/full/seri/JHA../0004//0000129.000.html">Copernicus and Nasir al-Din al-Tusi</a>, <i>Journal for the History of Astronomy</i> <b>4</b> (1973), 128&#8211;130.</p>
<p>In fact, the Tusi couple goes back way before al-Tusi.   It was known to Proclus back around 450 AD!  Apparently he wrote about it in his <i>Commentary on the First Book of Euclid</i>.  Proclus is mainly famous as a philosopher: people think of him as bringing neo-Platonism to its most refined heights.   Given that, it&#8217;s no surprise that he also liked math.  Indeed, he said:</p>
<blockquote><p>
Wherever there is number, there is beauty.
</p></blockquote>
<p>And of course this is what I&#8217;ve been trying to show you throughout this series.</p>
<h3> Why the Tusi couple works </h3>
<p>So, here are three proofs that a Tusi couple really does trace out a straight line.  I posed this as a puzzle on Google+, and here are my favorite three answers.  I like them because they&#8217;re very different.  Since people have thought about Tusi couples since 450 AD, I doubt any of these proofs are original to the people I&#8217;m mentioning here!  Still, they deserve credit.</p>
<p>The first, due to Omar Antol&iacute;n Camarena, is in the style of traditional Euclidean geometry.  </p>
<p>Let O be the center of the big circle.  Let A be the position of the traced point at the instant t when it&#8217;s the point of tangency between the circles, let B be its position at some future time t′ and let C be the point of tangency at that same time t′.   Let X be the center of the small circle at that future time t′.</p>
<div align="center"><img width="280" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_proof.png" /></div>
<p>We want to prove A, B and O lie on a line.  To do this, it suffices to show that the angle AOC equals the angle BOC:</p>
<p>Want: ∠BOC = ∠AOC</p>
<p>The arc AC of the big circle has the same length as the arc BC of the small circle, since they are both the distance rolled between times t and t′.  But the big circle is twice as, so the angle BXC on the little circle must be twice the angle AOC on the big circle:  </p>
<p>Know: ∠BXC = 2 ∠AOC</p>
<p>But it&#8217;s a <a href="http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII20.html">famous fact</a> in Euclidean geometry that the angle BXC is twice the angle BOC:</p>
<p>Know: ∠BXC = 2 ∠BOC</p>
<p>From the two equations we know, the one we want follows!</p>
<p>Here&#8217;s the proof of that &#8216;famous fact&#8217;, in case you forgot it:</p>
<div align="center"><img width="280" src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_proof_2.png" /></div>
<p>The second proof is due to Greg Egan.  He distilled it down to a moving picture:</p>
<div align="center">
<img src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_animation_egan.gif" /></div>
<p>It&#8217;s a &#8216;proof without words&#8217;, so you may need to think a while to see how it shows that the Tusi couple traces out a straight line.</p>
<p>The third proof is due to Boris Borcic.  It uses complex numbers.   Let the big circle be the unit circle in the complex plane.  Then the point of contact between the rolling circle and the big one is:</p>
<p><img src='https://s0.wp.com/latex.php?latex=e%5E%7Bi+t%7D&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='e^{i t}' title='e^{i t}' class='latex' /></p>
<p>so the center of the rolling circle is:</p>
<p><img src='https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Cfrac%7Be%5E%7Bit%7D%7D%7B2%7D+%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;displaystyle{ &#92;frac{e^{it}}{2} } ' title='&#92;displaystyle{ &#92;frac{e^{it}}{2} } ' class='latex' /></p>
<p>Since the rolling circle turns around once clockwise as it rolls around the big one, the point whose motion we&#8217;re tracking here:</p>
<div align="center">
<img width="150" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_animation.gif" /></div>
<p>is equal to:</p>
<p><img src='https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Cfrac%7Be%5E%7Bit%7D%7D%7B2%7D+%2B+%5Cfrac%7Be%5E%7Bi%28%5Cpi+-+t%29%7D%7D%7B2%7D+%3D++%5Cfrac%7Be%5E%7Bi+t%7D+-+e%5E%7B-i+t%7D%7D%7B2%7D++%3D+i+%5Csin+t+%3D++i+%5C%3B+%5Cmathrm%7BIm%7D%28e%5E%7Bit%7D%29+%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;displaystyle{ &#92;frac{e^{it}}{2} + &#92;frac{e^{i(&#92;pi - t)}}{2} =  &#92;frac{e^{i t} - e^{-i t}}{2}  = i &#92;sin t =  i &#92;; &#92;mathrm{Im}(e^{it}) } ' title='&#92;displaystyle{ &#92;frac{e^{it}}{2} + &#92;frac{e^{i(&#92;pi - t)}}{2} =  &#92;frac{e^{i t} - e^{-i t}}{2}  = i &#92;sin t =  i &#92;; &#92;mathrm{Im}(e^{it}) } ' class='latex' /></p>
<p>So, this point moves up and down along a vertical line, and its height equals the height of the point of contact, <img src='https://s0.wp.com/latex.php?latex=e%5E%7Bit%7D.&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='e^{it}.' title='e^{it}.' class='latex' /></p>
<p>But the same sort of argument shows that if we track the motion of the <i>opposite</i> point on the rolling circle, it<br />
equals:</p>
<p><img src='https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Cfrac%7Be%5E%7Bit%7D%7D%7B2%7D+-+%5Cfrac%7Be%5E%7Bi%28%5Cpi+-+t%29%7D%7D%7B2%7D+%3D++%5Cfrac%7Be%5E%7Bi+t%7D+%2B+e%5E%7B-i+t%7D%7D%7B2%7D++%3D++%5Ccos+t+%3D+++%5Cmathrm%7BRe%7D%28e%5E%7Bit%7D%29+%7D+&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;displaystyle{ &#92;frac{e^{it}}{2} - &#92;frac{e^{i(&#92;pi - t)}}{2} =  &#92;frac{e^{i t} + e^{-i t}}{2}  =  &#92;cos t =   &#92;mathrm{Re}(e^{it}) } ' title='&#92;displaystyle{ &#92;frac{e^{it}}{2} - &#92;frac{e^{i(&#92;pi - t)}}{2} =  &#92;frac{e^{i t} + e^{-i t}}{2}  =  &#92;cos t =   &#92;mathrm{Re}(e^{it}) } ' class='latex' /></p>
<p>So, this opposite point moves back and forth along a <i>horizontal</i> straight line&#8230; and its <i>horizontal</i> coordinate equals that of the point of contact!   </p>
<p>You can see all this clearly in the animation Borcic made:</p>
<div align="center">
<img src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/tusi_couple_animation_borcic.gif" /></div>
<p>The point of contact, the tip of red arrowhead, is <img src='https://s0.wp.com/latex.php?latex=e%5E%7Bit%7D.&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='e^{it}.' title='e^{it}.' class='latex' />   The two opposite points on the rolling circle are <img src='https://s0.wp.com/latex.php?latex=%5Ccos+t&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='&#92;cos t' title='&#92;cos t' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=i+%5Csin+t.&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='i &#92;sin t.' title='i &#92;sin t.' class='latex' />  So, the rectangle here illustrates the fact that</p>
<p><img src='https://s0.wp.com/latex.php?latex=e%5E%7Bit%7D+%3D+%5Ccos+t+%2B+i+%5Csin+t&#038;bg=ffffff&#038;fg=000&#038;s=0' alt='e^{it} = &#92;cos t + i &#92;sin t' title='e^{it} = &#92;cos t + i &#92;sin t' class='latex' /></p>
<p>It&#8217;s interesting that this famous formula is hiding in the math of the Tusi couple!  But it shouldn&#8217;t be surprising, because the Tusi couple is all about building linear motion out of circular motion&#8230; or conversely, decomposing a circular motion into linear motions.</p>
<h3> Credits </h3>
<p>Few of these pictures were made by me, and none of the animations.  For most, you can see who created them by clicking on them.  The animation of the astroid and deltoid as envelopes come from here:</p>
<p>&bull; <a href="http://mathforum.org/mathimages/index.php/Envelope">Envelope</a>, Math Images Project.</p>
<p>where they were made available under a <a href="http://www.gnu.org/copyleft/fdl.html">GNU Free Documentation License</a>.  Here&#8217;s another animation from there:</p>
<div align="center"><a href="http://mathforum.org/mathimages/index.php/Envelope#2._A_Deltoid_as_the_envelope_of_Wallace-Simson_lines"><img width="455" src="https://i0.wp.com/math.ucr.edu/home/baez/mathematical/deltoid_simson_lines_animation.gif" /></a></div>
<p>This is a way of creating a deltoid as the envelope of some lines called &#8216;Wallace-Simson lines&#8217;.  The definition of these lines is so baroque I didn&#8217;t dare tell it to you it earlier.   But if you&#8217;re so bored you&#8217;re actually reading these credits, you might enjoy this.  </p>
<p>Any triangle can be circumscribed by a circle.   If we take any point on this circle, say M, we can drop perpendicular<br />
lines from it to the triangle&#8217;s three sides, and get three points, say P, Q and R:</p>
<div align="center"><a href="http://mathforum.org/mathimages/index.php/Envelope#2._A_Deltoid_as_the_envelope_of_Wallace-Simson_lines"><img src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/deltoid_simson_line_1.png" /></a></div>
<p>Amazingly, these points lie on a line:</p>
<div align="center"><a href="http://mathforum.org/mathimages/index.php/Envelope#2._A_Deltoid_as_the_envelope_of_Wallace-Simson_lines"><img src="https://i2.wp.com/math.ucr.edu/home/baez/mathematical/deltoid_simson_line_2.png" /></a></div>
<p>This is the <a href="http://mathworld.wolfram.com/SimsonLine.html"><b>Wallace&#8211;Simson line</b></a> of M.  If we move the point M around the circle, we get lots of Wallace&#8211;Simson lines&#8230; and the envelope of these lines is a deltoid!</p>
<p>By the way: the Wallace&#8211;Simson line is named after <a href="http://en.wikipedia.org/wiki/William_Wallace_%28mathematician%29">William Wallace</a>, who wrote about it, and <a href="http://en.wikipedia.org/wiki/Robert_Simson">Robert Simson</a>, who didn&#8217;t.  Don&#8217;t confuse it with the <a href="http://en.wikipedia.org/wiki/Wallace_Line">Wallace line</a>!  That was discovered by <a href="http://en.wikipedia.org/wiki/Alfred_Russel_Wallace">Alfred Russel Wallace</a>.</p>
<div align="center"><a href="http://en.wikipedia.org/wiki/Wallace_Line"><img src="https://i1.wp.com/math.ucr.edu/home/baez/ecological/wallace_line.png" /></a></div>
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