A question for the topologically inclined

[Quinn]

There's a topological problem I'm trying to work out, and I'm hoping someone here could help me with it. It involves mapping a 3-D shape onto a 2-D surface. And there is a practical application for it: it's for a Jerry Andrus-like "impossible object" that I've designed and am attempting to build in real life. So anyone who can help me with it may get to see the fruits of your efforts at a future TAM, woo-hoo!

A brief disclaimer: my formal math education ended with advanced high school math, so please forgive me for any terminology and notational conventions that I may have invented myself on the fly.

Okay, here goes. Imagine a rectangular box like this (the top and bottom are squares, and all angles are 90°):

http://www.jqpaxton.com/jref/Picture1.png

Now picture the box with the front and right surfaces removed:

http://www.jqpaxton.com/jref/Picture2.png

Now picture it with two diagonal lines across the faces we just removed: one from the top, right, front corner, to the bottom, right, rear corner; and the other from the top, left, front corner, to the bottom, right, front corner:

http://www.jqpaxton.com/jref/Picture3.png

What I'm interested in is the curved, four-edged shape framed by those two lines, the top front edge, and the bottom right edge:

http://www.jqpaxton.com/jref/Picture4.png

What I'm trying to figure out specifically is what that shape would look like if it were flattened out; in other words, what shape one would need to cut a flat piece of paper into, in order to twist it into that curved shape.

From building a couple rough models, I have a basic idea of what the shape looks like -- something close to this:

http://www.jqpaxton.com/jref/Picture5.png

However, my goal is to figure out how to calculate the exact dimensions, based on the dimensions of the box (so that for any given size box, I can create the correct flat shape that will twist into that curved shape).

So, here are the relevant points that I've figured out so far -- or at least, that I think I've figured out; if I'm wrong about any of them, please let me know:

• Obviously, the height is the same as the height of the box, and the width of the straight edges is the same as the width of the box.

• The width of the shape across the middle (at its thinnest point) = [half the width of the short edges] x [the square root of two].

• Angles: At the top right and bottom left corners, the angle formed by the straight edge and the tangent to the curved edge is 90 degrees. At the other two corners, the angle formed is the same as that formed by the bottom (or top) edge of the box, and the diagonal to the opposite corner of the same face. (The picture below isn't drawn to scale, so the angles don't look the same, but in real life they do.)

http://www.jqpaxton.com/jref/Picture6.png

Assuming all those things are right, here's what I still need to figure out. When I fist started dealing with this, I thought the top and bottom edges of the shape would line up vertically, like this:

http://www.jqpaxton.com/jref/Picture7.png

Then once I thought about it for a moment, I realized they wouldn't, and instead I expected the top right corner to line up directly over the bottom left corner, like this:

http://www.jqpaxton.com/jref/Picture8.png

But the reality is somewhere in between those two. So I need to figure out just how far that displacement is, in relation to the height and width of the box. If I know that figure, then I can combine it with the above info about the angles and the width in the middle (assuming all that was correct) to work out the exact shape.

So if anyone could help me out with this, and/or suggest an easier way of going about it that I hadn't realized, I would appreciate it greatly.

[RecoveringYuppy]

Not sure I'm understanding what you're going for but if I think of doing this with a cube than it seems that all four sides of the face you are constructing would be analogous and would make me wonder why are four edges of your last drawing aren't all curved.

Is it strictly necessary that the piece you cut this from be flat and that there be no stretching as opposed to mere twisting?

ETA: Could you also specify what you want the surface of this new face to look like? Preferably in terms of "if I draw a line across the face from here to here it will be straight (or may be curved, or whatever)"

[Ziggurat]

There are a few difficulties involved here. One of them is that you haven't actually defined your object. You've given boundary conditions for it, but that alone doesn't sufficiently specify it. For example, given those boundaries (which I take to be all straight lines), you could construct a surface such that horizontal lines between the two tilted side boundaries always lie on the surface. Conversely, you could construct the surface such that tilted lines between the top and bottom boundaries lie on the surface. These two methods will not be the same surface.

Secondly, how do you want to "flatten" your surface? Are you going to simply project it onto a 2D surface? Or are you going to try to "squash" it, distorting it in a manner which minimizes some measure of "stretch"? Because both of the surfaces I suggested above have intrinsic curvature, and cannot be made flat without doing one or the other (analogy: you can "flatten" a cylinder by unrolling it, but you can't flatten a sphere without squishing it). It sounds like you're more interested in the latter, but if you are, keep in mind that even if you find the shape, you likely won't be able to stretch a material like paper to match it.

[Molinaro]

Take a rectangular piece of paper who's small diameter is equal to the width of the top (and bottom) attachment points. Make it about twice the length needed.

Attach it at the top pair of points and bottom pair, with plenty of extra running off each end.

Trim to desired fit.

You could then remove it to see the required shape.

[Thabiguy]

What I'm interested in is the curved, four-edged shape framed by those two lines, the top front edge, and the bottom right edge:

http://www.jqpaxton.com/jref/Picture4.png


There is no single curved, four-edged shape framed by the four lines that you specified. There is an infinite number of curved shapes having those four edges.

I assume you mean the one which satisfies the following condition: each plane parallel to the top or bottom of the box intersects the shape in a straight line connecting the two points in which the plane intersects the "diagonal" edges.

If that is the case, then there is no flat sheet of paper that you could twist into this shape. This is because the shape that you have described is not flat.

Proof, by contradiction:
Let's assume that the shape is flat and can be created by twisting a flat sheet of paper.

First, keep this in mind: Lines and curves drawn on a paper do not change their length when the paper is twisted or flattened. (I leave the proof as an exercise to the reader.) This is used extensively in the proof.

Next, let's consider your shape ABCD, where A is the "top left front" point, B is the "top right front" point, C is the "bottom right front" point and D is the "bottom right back" point.

Then consider the line AC. It is the shortest curve connecting the points AC, and it lies entirely on the paper when it is twisted.

Now let's flatten out the paper: points A and C will become A' and C'. And the line AC becomes A'C' and it will be a straight line. (Proof by contradiction: let's assume that the curve connecting A' and C' created by flattening the line AC is not a straight line. Then there is a straight line A'C' that is shorter than this curve. When the paper is re-twisted, the curve that is created by twisting the straight line A'C' will become a curve that 1) connects A and C, 2) is shorter that the line AC. But this is a contradiction, because there is no curve connecting A and C that would be shorter than the line AC.)

Similarly, we can prove that the straight line AB (which lies entirely on the twisted paper) will become, after the paper is flattened, a straight line A'B'.

Now, the angle B'A'C' will be the same as the angle BAC. (This is a little bit more complicated to prove, and I will leave it as an interesting exercise to the reader. It can be proven in a "boring" way using limits as one approaches the top of the shape, but maybe someone can come up with a more elegant proof.)

From |AB| = |A'B'| and |AC| = |A'C'| and angle BAC = angle B'A'C', it follows that |B'C'| = |BC|.

Now let's flatten the paper, draw the straight line B'C', and re-twist the paper. There now must be a curve connecting B and C that is 1) equally long as the line B'C', i.e. equally long as the line BC, and 2) lies entirely on the twisted paper. Let's call this curve X.

There is only one curve connecting B and C that is as long as the line BC: the line BC. This line, however, does not lie on the twisted paper. This means that any curve that does lie on the twisted paper is longer than the line BC - including our curve X. This is a contradiction, because curve X should be as long as the line BC.

And that concludes the proof.

ETA: I now see that Ziggurat has said the same thing with many less words, while giving even more additional information in his answer. Oh well. That's what happens when one spends too much time writing a post without checking the forum. :o

[RecoveringYuppy]

Obviously, the height is the same as the height of the box, and the width of the straight edges is the same as the width of the box.

BTW I got a feeling this is definitely wrong on two counts given what you appear to be intending.

Referring to your last drawing. If the green paper under the black line is the same length as the height of the box then that means you are expecting a straight line to connect those two points. That means the surface you are describing will be folded along that line, rather than the curve you are apparently going for. Since you are going for a curve that traverses the interior of the box rather than following that corner edge, that line needs to be longer than the height of the box.

As I mentioned earlier, I don't think the two edges you've shown as straight will be straight. Do you remember the pythagorean theorem? If so just take a simple example and calculate a few lengths. If you take your object to be four units high, and 3 wide, then the diagonals you've shown as black lines are then 5 units long. But if you calculate the length of the direct line between the midpoint of the top front line and the midpoint of the bottom right line you come up with 4.26. So that implies those lines need some curvature too (how true this actually turns out to be depends on how you answer the questions about what specific curvature you want and what stretching you can tolerate).

[RecoveringYuppy]

ETA: I now see that Ziggurat has said the same thing with many less words, while giving even more additional information in his answer. Oh well. That's what happens when one spends too much time writing a post without checking the forum. :o

But you had fun thinking about it didn't you?

[gdnp]

A question for the math geeks:

I will accept that there are an infinite number of shapes that could fit those boundary conditions. But it seems to me there should be one "best" shape. That would be the shape that would be formed if you created the edges out of wire and then dipped the surface into bubble liquid.

What are the characteristics of that shape? Is it the one with the smallest surface area? Or is it determined by some other factor(s)?

When I was a kid we had some watery plastic resin that dried into hard plastic that we used to make suncatchers out of bent wire. That could be used to produce the shape in question, which could then be cut off the wire and flattened.

[I Ratant]

From building a couple rough models, I have a basic idea of what the shape looks like -- something close to this:
.
Take a piece of foam to your dimensions, and make that cut.
I'm doing that right now.

[GreedyAlgorithm]

A question for the math geeks:

I will accept that there are an infinite number of shapes that could fit those boundary conditions. But it seems to me there should be one "best" shape. That would be the shape that would be formed if you created the edges out of wire and then dipped the surface into bubble liquid.

What are the characteristics of that shape? Is it the one with the smallest surface area? Or is it determined by some other factor(s)?

When I was a kid we had some watery plastic resin that dried into hard plastic that we used to make suncatchers out of bent wire. That could be used to produce the shape in question, which could then be cut off the wire and flattened.
Minimum surface area is certainly a criterion. But bubble liquid doesn't actually find such a surface automatically or else we could solve NP-complete problems in polynomial time. :D

[Thabiguy]

But you had fun thinking about it didn't you?
:eusa_eh: :eusa_think: :eusa_think: :eusa_eh: :eusa_think: :eusa_wall: :eusa_think: :eusa_think: :sour: :boggled: :hypnotize :faint: ... :idea: :clap::yahoo :D

What are the characteristics of that shape? Is it the one with the smallest surface area?

Basically, smallest surface area, though this is more physics than math. There's a whole lot of information here: Wiki on soap bubbles (http://en.wikipedia.org/wiki/Soap_bubble).

[I Ratant]

It "pinches" in the middle..

[RecoveringYuppy]

Where does it pinch top to bottom?

[I Ratant]

where Does It Pinch Top To Bottom?
.
1.35"<1.85"

[jwwalker]

To be picky, this problem isn't topological, it's geometric.

[RecoveringYuppy]

.
1.35"<1.85"
No, I'm asking about the pinch at right angles to that.

[DrRocket]

...

So if anyone could help me out with this, and/or suggest an easier way of going about it that I hadn't realized, I would appreciate it greatly.[/QUOTE]

Ziggurat has put his finger on the problem. Your question is not one of topology but one of geometry. And because the surface that you have described is curved in the geometrical sense of the term, it cannot be flattened out perfectly. That is it cannot be put into one-to-one correspondence with a portion of the plane in a manner that preserves distance and angles. You can do that with a cylinder (just roll it out) but not with a surface with real curvature, such as yours.

[I Ratant]

Couple more images...
Wrapped string around the shape.
The curve of the surface is obvious with an oblique view