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Topic: Seeking help from the topologically inclined 


I'm working on building a Jerry Andruslike "impossible object" that I've designed, and I've got it nailed down except for one piece that's giving me problems. It's essentially a long strip that twists 90 degrees along its length. What I'm trying to figure out is [i]what shape I will need to cut a flat piece of wood/cardboard/whatever into, so that when I twist it, it will form the shape I want.[/i] Describing it in more detail than that is all but impossible without pictures. So I had a detailed, illustrated post all written and ready to go, and then realized that the pictures weren't coming up in the post because of some discrepancy in the forum software. So instead, I've put up an entire separate page with the whole thing spelled out and illustrated in detail. It can be found at: http://www.mentallyimpossible.com/box.htm If any of the mathematically inclined folks here would be kind enough to take a look and give me any feedback (ranging from a solution to reasons why what I've said is wrong and/or insufficient), I would be most grateful. Thanks in advance. 


I don't think that shape is possible to make from a flat object if the material doesn't have stretch in it. Especially after seeing that what was earlier touching the long edge along the height of the box turned into a curve when you flattened it. It might be possible to assume some stretching properties of the material to make the calculation of the flat shape possible, but it of course won't be exact. /Tomas 


Tomas, thanks for your input. But I was able to make a "flat" version of it with the overlapping strips of tape, and the tape had no stretch. One of the pictures (the one just below the two cardboard box pictures) is a photograph of it. I enhanced the color in Photoshop, but it is an actual photo of the piece I sliced out. 


I realize now that a curve on an unstretchable material still can form a straight line. It was a bit unintuitive to me at first. Good call. So it _might_ be possible. :) Thanks for a very interesting problem. /Tomas A few observations: If you consider a box double the width and double the depth of your box, you'll realize that the surface you so far made with tape will be unaltered and part of the bigger surface made for the bigger box. Once you realize that, you'll also realize that it must hold true that if you instead halve the depth and width of your box you simply have to cut the new surface out from your old one. This means that the curve on the lower part in step 7 is a constant when only width and depth of the box are changed. That means that the gap mentioned in step 7 is a constant also. So what happens when the width and depth of the box are less than this constant gap on the lower side in step 7? Either that's a paradox (saying that it actually can't be a flat surface) or the smiling curve on the upper side eventually turns into a frown as the width and depth of the box are diminished. If tape trials show that the gap on the lower side is not a constant when different widths and depths are tried or if the upper curve never turns into a frown, it tells us that this is actually not a mathematically flattenable (is that a word?) surface, even though it can be close. /Tomas Also worth noting is that no doublecurved surface can ever be flattened. It's a classic construction problem for sheet metal workers. Only problem here is that I have no idea if the 3D shape discussed here is considered doublecurved or not. Anyone? /Tomas 


I'm giving it some thought, and I'll let you know if I come up with anything. I have put more minds on it, however. I posted your page to StumbleUpon, with the keywords "Mathematics", "Design", "Topology", and "Impossible Object", so you'll have plenty of eyes and minds pondering over the problem. Just one thought, concerning the thinnest point. If, as you state on the page, this is equal to the distance from the center of one of the end squares to the corner, then it's equal to half the diagonal of the square. If we were to cut this square in half along that diagonal, we'd have a triangle, and a right triangle at that. With that triangle, your midpoint would be half the length of the hypotenuse. Now it's just a matter of using the Pythagorean theorem. If we replace [i]a[/i] with the depth (D), and [i]b[/i] with the width (W), we get a hypotenuse of D^2 + W^2. However, your midpoint, as you say, wouldn't be the full length of the hypotenuse, but rather half that distance. So, we take the formula and divide by 2. So, you've got one straight end that's D long, and another straight end that's W long. This gives us (assuming you're right about the characteristics of the midpoint) a thin midpoint of (D^2 +W^2)/2. If you'll always be using an equal depth and width, this could also be stated as ((2D)^2)/2 or ((2W)^2)/2. 


Half the diagonal will be D/sqrt(2) if the width and depth are equal. If you still have any tape left I'd suggest making a very short but very wide and deep box and tape the desired surface. Often doing extreme sizes will show if something is possible. If the surface in such a box is possible to flatten I think it's safe to say that this problem has a solution. /Tomas Posted: Sep 10, 2008 12:59am  Last night, when trying to fall asleep, I made another observation, which I think is the key to finding the equation, if there is one: If you draw a normal from the lower curve at any point, the length of it between the two curves is known and a linear function of how far you have traveled from the centre of the curve along the lower curve. Let's call that distance t. Also, _where_ this normal crosses the upper curve is known, as it is the ratio between the diagonal and the height times t. I've not set those equations up yet, but I'm sure they will give the solution. Especially they'll show that the lower curve is dependent on the height of the box only. /Tomas 


Tomas and Scott, thanks much for your help, and sorry for not getting back to the thread for a couple days; I'm on the road with limited net access. Scott, you're right about the thickness at the thinnest point. I had worked that out, but didn't want to clutter up the page explaining it. Also, I'm not familiar with StumbleUpon. Got a link? Tomas, I was following you up until that last post, but that one surpassed the limits of my math knowledge. I'm not familiar with "normals." However, I'm more than willing to put in the work to learn. I'll see if I can brush up on that and get back to you. Oh, and for what it's worth, I need pieces like this for three different sized "boxes": one where the height:width ratio is 4:1, one where it's 8:1, and one where it's 12:1. 


StumbleUpon is located at: http://www.stumbleupon.com/ This is a site with a special plugin for Firefox browser users, that allows them to discover new sites that meet their interests. User set up their interests in their StumbleUpon (SU) accounts, and then click on new button in their browser. Whenever they click this button, they are taken to a site related to their interests that was discovered by other users (any user can discover a site, and add tags describing the subjects). So, now that I "discovered" your site as far as SU is concerned, and added tags like "Mathematics", "Design", "Topology", and "Impossible Object", anyone who has those terms set up in their account, or any SU user who searches for those terms, will eventually come across your page describing the problem! In short, you have more eyes, so you have a better chance of getting help. 


I'll just write down what I have so far, but have not had time to solve yet, for w = width and h = height: x=x(t), y=y(t), l=l(t) and all differentials are made in regards to t. l^2(y''^2 + x''^2) + 2l(x'y''x''y') = 4(sqrt(2)  1)r^2 where r = w / h and l = (2(sqrt(2)1)t / h + 1)w / sqrt(2) So much about that differential equation looks familiar to me that I'm sure it's solvable. It was a hell of an expression before I boiled it down, and I can post my reasoning so far if anyone is interested. The solution might be to get rid of t and also make it into a separable differential equation. A little help to go from t to x and y might be t = [integral 0:x](sqrt(1 + (dy/dx)^2)dx) I tried doing it all in nonparametric form, y=y(x), but the expression gets 1024 times nastier so I'm really hoping the above is solvable, but I'll keep attacking it from both ways. If they deliver the same solution I'll be surer it's correct. Worth noting is that x'^2 + y'^2 = 1 which helped me a lot during this. /Tomas 


It seems that this surface is most certainly nondevelopable. If it were constrained by only three of the lines it would be a hyperboloic paraboloid with a non zero Gaussian curvature. Constraining by the fourth side further complicates the surface, and I doubt it would bring the Gaussian curvature to 0, which is required in order to make the surface developable into a flat plane. I could be wrong, and will tinker a bit with solutions. Though I think an approximation to the surface is probably the best you will get. 


Wow, what a subject. Quinn, you deserve much credit for getting as far as you did with your self professed "limitations". It is no small problem. I think TomasB has it on the run through the Diffy Que routine. Mr. Coyle, I think, has beaten us all if only by using the proper vocabulary. A great display I've seen in a museum has a stick centered on the edge of a turning wheel but at a slant. The wheel turns slowly letting the stick trace out a shape. If the stick was striaght up and down the shape traced out would be a cylinder, but the stick is slanted. A flat piece of board surrounds the wheel and the stick, and on it two Curved holes, shaped like back to back parens )( allow the straight stick to pass around and around through the curved holes. I think this is called a Riemann surface. By way of experimentation you might make smaller models and dip them into soap and study the resulting bubble. With the right size you could get the bubble cast in latex rubber. Once dried the latex could be cut off and tested to see if it could lie flat. I'm sure you could glue a latex sheet to your outline but the stretchability question, I think, can only be answered by flattening the shape coming from the model. You could also glue up a layer of tissue onto the surface of the model. Any stretching suspected by Mr. Coyle would be evident in tears to the surface. The matching distances from part 7 remind me of the Boomerang or bannana illusion, where the two curved pieces of wood look different in length depending on which one is inside or outside. The term Normal refers to a line perpendicular to another line or surface. The spike of a nail is normal to the head till you take a few whacks at it. In the case of your curves a normal is perpendicular to a tangent on the curve at that point, like a nail you've poked through the surface going in or going out. Fascinating problem. Thank you for bringing it up. Considering things today I'd much rather be kept awake thinking of this than the other things going on. Elliot 


[quote]On 20090205 23:45, plungerman wrote: A great display I've seen in a museum has a stick centered on the edge of a turning wheel but at a slant. The wheel turns slowly letting the stick trace out a shape. If the stick was striaght up and down the shape traced out would be a cylinder, but the stick is slanted. A flat piece of board surrounds the wheel and the stick, and on it two Curved holes, shaped like back to back parens )( allow the straight stick to pass around and around through the curved holes. I think this is called a Riemann surface.[/quote] There used to be such a display at the [b]Museum of Science and Industry[/b] in Exposition Park, Los Angeles; I remember seeing it a few times when I was a kid. The surface the stick traces out is a circular hyperboloid of one sheet; the curved holes are the two branches of an hyperbola. For a description of [i]Riemann Surfaces[/i] (not a particularly useful one for nonmathematicians, though it has pretty pictures), look [url=http://mathworld.wolfram.com/RiemannSurface.html]here[/url]. 