{"id":1512,"date":"2017-10-05T09:45:29","date_gmt":"2017-10-05T13:45:29","guid":{"rendered":"https:\/\/geekhaus.com\/math103_fall2017\/?p=1512"},"modified":"2017-11-25T16:37:23","modified_gmt":"2017-11-25T21:37:23","slug":"open-project-prince-ruperts-cube","status":"publish","type":"post","link":"https:\/\/geekhaus.com\/math103_fall2017\/2017\/10\/05\/open-project-prince-ruperts-cube\/","title":{"rendered":"Prince Rupert\u2019s Cube"},"content":{"rendered":"<h2>Group Member Roles:<\/h2>\n<p>Designer: Grant Folsom<br \/>\nMath Consultant: Summer Xia<br \/>\nWriter: Taylor Zabinsky<\/p>\n<h2>Project Idea:<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1449 size-full\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/download.jpg\" alt=\"\" width=\"259\" height=\"194\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/download.jpg 259w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/download-80x60.jpg 80w\" sizes=\"(max-width: 259px) 100vw, 259px\" \/><\/p>\n<p>Our project idea is to make a Prince Rupert\u2019s Cube. The Prince\u00a0Rupert problem was proposed by\u00a0Prince Rupert of the Rhine, Prince Rupert&#8217;s cube\u00a0is the largest\u00a0cube\u00a0that can pass through a hole cut through another cube without <a href=\"http:\/\/mathworld.wolfram.com\/PrinceRupertsCube.html\">splitting the\u00a0cube\u00a0<\/a>into two pieces. This proposition was developed through a bet that Prince Rupert made and took around 100 years to solve. This spare capacity is normally used to reinforce the edges of the hexagon hole and a cube. According to\u00a0<a href=\"https:\/\/en.wikipedia.org\/wiki\/Prince_Rupert%27s_cube\">Wikipedia<\/a>, the construction of a physical model of Prince Rupert\u2019s cube is made challenging by the accuracy with which such a model needs to be measured, and the thinness of the connections between the remaining parts of the unit cube after the hole is cut through it.<\/p>\n<h2>First Steps:<\/h2>\n<p>Initially, we worked on our design, figuring out the logistics of how we are going to print it. We had to solve the problem of how one cube fits\u00a0inside another cube. After working on multiple designs, we finally learned that one square has to be tilted at a 45-degree angle in order to fit into the other. We then created a hole on one cube from the shape of the other cube. Lastly, we added small curves connecting one plane to the other to ensure that it has support for printing.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1721 aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/first-model.png\" alt=\"\" width=\"500\" height=\"341\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/first-model.png 475w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/first-model-300x205.png 300w\" sizes=\"(max-width: 500px) 100vw, 500px\" \/><\/p>\n<h2>Design Process:<\/h2>\n<p>The initial design was to have the second cube tilted at a 22.5-degree angle. It was then changed to 45 degrees in order to allow the second cube to pass through the first. When the platform was first created, the main concern was keeping the platform stable. Rings were added to keep the platform (cube one) together. When the first iteration underwent a print, it fell apart during the printing process. The rings were redone, made flat instead of round, for better stability and printing process. The second print also had supports which ensured a solid structure and allowed the print to come out better. The second iteration ended up working seamlessly; the cube could fit inside of the platform very nicely.<\/p>\n<h2><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1537 \" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/IMG_4148-e1507239634534-768x1024.jpg\" alt=\"\" width=\"553\" height=\"738\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/IMG_4148-e1507239634534-768x1024.jpg 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/IMG_4148-e1507239634534-225x300.jpg 225w\" sizes=\"(max-width: 553px) 100vw, 553px\" \/><br \/>\nFinal Print of Prince Rupert&#8217;s Cube:<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1729 size-large\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-1024x768.jpg\" alt=\"\" width=\"1024\" height=\"768\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-1024x768.jpg 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-300x225.jpg 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-768x576.jpg 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-678x509.jpg 678w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-326x245.jpg 326w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Final-Print-80x60.jpg 80w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/p>\n<h2>The Mathematics of Prince Rupert&#8217;s Cube:<\/h2>\n<p>Before we start, think about any objects or shapes fitting inside of one another. The first thing that might come to mind is a child&#8217;s toy, where one has to try to fit each shape in the corresponding hole. What if it was possible to fit one single object through 3 different ones? Well, this is actually possible. A round-square-triangle is a single object that is able to fit through a 2D square, circle, and triangle.\u00a0The original design was based off an internet image where the holes were constructed on different planes like a cube figure. Unfortunately, the print failed to maintain the shape for the piece to fit through. The \u201chole\u201d was remade to a flat design, becoming the final print. The piece itself, designed to fit through the three holes, is designed with the three-dimensional shapes. (Cube, cylinder, and a pyramid).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1933 \" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1024x768.jpeg\" alt=\"\" width=\"715\" height=\"537\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1024x768.jpeg 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-300x225.jpeg 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-768x576.jpeg 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-678x509.jpeg 678w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-326x245.jpeg 326w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-80x60.jpeg 80w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1.jpeg 1334w\" sizes=\"(max-width: 715px) 100vw, 715px\" \/><\/p>\n<p>So what is the problem? What is the largest cube that can pass through a cube of a given size?<\/p>\n<p>Let&#8217;s start simple, by fitting two equivalent sized squares inside of each other. As shown in the diagram below, one can see that the height of the square is distinctly shorter than the length of the diagonal.\u00a0Using the <a href=\"https:\/\/mikesmathpage.wordpress.com\/2014\/02\/22\/the-prince-rupert-problem\/\">Pythagorean theorem<\/a>, one can figure out that the length of the diagonal for this square is actually \u221a2.<b>\u00a0<\/b>This means that two squares of equal size can pass through one another on a diagonal, with extra room to spare, as displayed in the diagram below.<\/p>\n<p>*all of the remaining images below were created by Taylor Zabinsky in Word*<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1685 aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1-1024x700.png\" alt=\"\" width=\"666\" height=\"455\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1-1024x700.png 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1-300x205.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1-768x525.png 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/1-1.png 1076w\" sizes=\"(max-width: 666px) 100vw, 666px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1745 \" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-1.png\" alt=\"\" width=\"428\" height=\"237\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-1.png 696w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-1-300x166.png 300w\" sizes=\"(max-width: 428px) 100vw, 428px\" \/><\/p>\n<p>Now let&#8217;s move on to the cubes. Again, by using the <a href=\"https:\/\/mikesmathpage.wordpress.com\/2014\/02\/22\/the-prince-rupert-problem\/\">Pythagorean theorem<\/a> we know that the length of the diagonal for the cube is \u221a3, which is even larger than the diagonal of the square.\u00a0This means that two cubes of equal size can definitely pass through one another on a diagonal, with extra room to spare, meaning that an even larger cube would be able to pass through.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1712 aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/0-1024x679.png\" alt=\"\" width=\"641\" height=\"425\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/0-1024x679.png 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/0-300x199.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/0-768x509.png 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/0.png 1032w\" sizes=\"(max-width: 641px) 100vw, 641px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1746 \" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-2.png\" alt=\"\" width=\"517\" height=\"276\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-2.png 914w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-2-300x160.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/pyth-2-768x410.png 768w\" sizes=\"(max-width: 517px) 100vw, 517px\" \/><\/p>\n<p>Is it possible for a larger cube to fit inside the first cube? If so, how much larger? Well if the original\u00a0cube has a side length of 1, you know that the other cube can be slightly larger than one, a cube that passes through by barely touching the edges of the diagonal. In other words, a cube that has a side length of the diagonal of the original cube would be the largest cube that can pass through. Below I created a diagram showing the math of how it works. Again, using the Pythagorean theorem, you can solve for C or the length of the largest cube that can pass through.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1778 size-large\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-17-at-10.33.14-AM-1024x596.png\" alt=\"\" width=\"1024\" height=\"596\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-17-at-10.33.14-AM-1024x596.png 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-17-at-10.33.14-AM-300x175.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-17-at-10.33.14-AM-768x447.png 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-17-at-10.33.14-AM.png 1278w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/p>\n<p>To visually prove this proposition, take a cube and look along the body diagonal (corner to corner), you see the profile of a <a href=\"https:\/\/mikesmathpage.wordpress.com\/2014\/02\/22\/the-prince-rupert-problem\/\">hexagon<\/a>.\u00a0 You can put a square in this hexagon whose side length is greater than the original side length of the cube.\u00a0 This means that you can pass a larger cube completely through a smaller one.<\/p>\n<p>Therefore; the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Prince_Rupert%27s_cube\">solution<\/a> to the problem is a maximally sized inner cube with sides of length 6% or 1.0606&#8243; (3 sqrt(2))\/4) larger than the unit cube with side lengths of 1.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1675 size-large aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/2-1024x640.png\" alt=\"\" width=\"1024\" height=\"640\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/2-1024x640.png 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/2-300x188.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/2-768x480.png 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/2.png 1228w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1679 aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-11-at-1.12.30-PM.png\" alt=\"\" width=\"526\" height=\"354\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-11-at-1.12.30-PM.png 998w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-11-at-1.12.30-PM-300x202.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/Screen-Shot-2017-10-11-at-1.12.30-PM-768x517.png 768w\" sizes=\"(max-width: 526px) 100vw, 526px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1691 size-large aligncenter\" src=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/4-1-1024x674.png\" alt=\"\" width=\"1024\" height=\"674\" srcset=\"https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/4-1-1024x674.png 1024w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/4-1-300x198.png 300w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/4-1-768x506.png 768w, https:\/\/geekhaus.com\/math103_fall2017\/wp-content\/uploads\/2017\/10\/4-1.png 1160w\" sizes=\"(max-width: 1024px) 100vw, 1024px\" \/><\/p>\n<h2>\u00a0Thingiverse &amp; Tinkercad<\/h2>\n<p>You can find our design for Prince Rupert&#8217;s cube on <a href=\"https:\/\/www.thingiverse.com\/thing:2584133\">Thingiverse<\/a> here and <a href=\"https:\/\/www.tinkercad.com\/things\/i8b2b8lVfZZ\">Tinkercad<\/a> here.<br \/>\nYou can find the design for the square-circle-square on <a href=\"https:\/\/www.tinkercad.com\/things\/e6mbgSWp2xR\">Tinkercad<\/a> here and <a href=\"https:\/\/www.tinkercad.com\/things\/j3Hjd3rCxBe\">here<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<div class=\"mh-excerpt\"><p>Our project idea is to make a Prince Rupert\u2019s Cube. The Prince\u00a0Rupert problem was proposed by\u00a0Prince Rupert of the Rhine, Prince Rupert&#8217;s cube\u00a0is the largest\u00a0cube\u00a0that can pass through a hole cut through another cube without splitting the\u00a0cube\u00a0into two pieces. This proposition was <a class=\"mh-excerpt-more\" href=\"https:\/\/geekhaus.com\/math103_fall2017\/2017\/10\/05\/open-project-prince-ruperts-cube\/\" title=\"Prince Rupert\u2019s Cube\">[&#8230;]<\/a><\/p>\n<\/div>","protected":false},"author":28,"featured_media":1729,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[31],"tags":[],"coauthors":[24],"_links":{"self":[{"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/posts\/1512"}],"collection":[{"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/users\/28"}],"replies":[{"embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/comments?post=1512"}],"version-history":[{"count":64,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/posts\/1512\/revisions"}],"predecessor-version":[{"id":2500,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/posts\/1512\/revisions\/2500"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/media\/1729"}],"wp:attachment":[{"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/media?parent=1512"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/categories?post=1512"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/tags?post=1512"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/geekhaus.com\/math103_fall2017\/wp-json\/wp\/v2\/coauthors?post=1512"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}