Distorted rectangular prism

[BMAD]

Suppose you have a rectangular prism with dimensions 4x4x8. Lay this shape down with the small square facing you.   We are going to distort this rectangular prism.  Take the top left corner of the front square and connect it to the bottom left corner of the back corner, take the bottom left corner of the front and connect it to the bottom right corner of the back, and so on. Keep the short edges 4 and the long edges 8, how do the volumes compare? 

Edited July 25, 2016 by BMAD

[araver]

Trying to visualize it, but as Bonanova pointed out

Spoiler

I can't "find" a volume for a 3D object that does not have enough information about its surface.

I tried drawing the 2 green squares and connected the upper to lower points using red lines as the OP suggested.

The 8 initial points uniquely determine a prism because each of the green square edges from the upper square has a parallel pair in the lower square, and as such determine a rectangular face for the polyhedron.

The figure you speak of is not unambiguously defined as it is not an 8-vertex polyhedron. The 4 red lines, along with the original green lines (4+4=8 green lines) do not determine an object with a surface as you are connecting two green pairs which are NOT in the same plane.

I believe you can give it surface unambiguously if you give a certain algorithm to create such such as rotations/twists of "faces" in 3d.

E.g. as a "twisted candy".

One starts with a face being parallelogram with 2 green opposing sides (4 units in length) and 2 red opposing sides (sqrt(16+64) units in length), connects it in half along the red middle points, and twists it around this center line till its edges fit the arrangement. The surface is of course not the face of a polyhedra, but is uniquely determined by the "twist" function which translates the initial planar face (paralleogram) to the final position. Don't have a 3D editor to show animation properly...

Is this the figure you've been meaning in the OP?

[bonanova]

Are we looking at the back and front 4x4 faces from the same direction? Or, does left and right, as they refer to the back face, assume we have turned the shape around so we're looking at outside of the back 4x4 face?

That is, if we label the vertices ABCD on the front face, clockwise starting from the upper left corner like this

A     B

D     C

Then if the 8" edges connect these to the back face vertices EFGH as A-E, B-F, C-G, D-H would your connections be A to H, D to G, and so on?

It sounds as if we are joining the front and back faces of a prism to form a torus that has a square cross section, only we're twisting the shape by 90 degrees before joining the faces. (Kind of like constructing a Mobius strip from a piece of paper. This new shape would have only three sides, just as a Mobius strip has only one side. Interesting.) I wonder, though, with the length (8) being only twice the side (4) whether this is even possible.

One thought:

Spoiler

If we just did the twist, leaving the front and back faces parallel but not joined, the volume would decrease. The cross-sectional area would remain the same, but I believe the length would decrease. Imagine the four 8" edges as being strings that do not stretch. Twisting one of the faces by 90 degrees would bring the two faces closer, by a distance that could be found by inscribing an 8" radius arc in an 8X4 rectangle.

Possibly it's wrong to say the strings can not stretch. Another way to look at this is to slice the original prism into n equal shorter ones and rotate each of them by (90/n) degrees. And then take the limit where n grows large without bound. In this case, the volume is always the sum of the smaller volumes, which do not change upon rotation. Nor would the 8" length of the prism decrease in this case, so the volume would remain 8x4x4..

Either way, if after twisting we then brought the two faces together, the inner part of the torus would compress and the outer part would have to stretch. This means the material of which the original prism is made is compressible. I don't think there is enough information given to determine how much it would compress. Or, we could allow distortion (permitting the lengths of the 8" edges to change as necessary, by saying the interior of the original prism is filled with a (distortable) but incompressible fluid. In that case, the fluid being incompressible, the initial and final volumes will be equal.

I think the information given is insufficient to create an answer, but I may have missed an important point somewhere.

 

[BMAD]

On 7/26/2016 at 2:31 AM, bonanova said:

Are we looking at the back and front 4x4 faces from the same direction? Or, does left and right, as they refer to the back face, assume we have turned the shape around so we're looking at outside of the back 4x4 face?

That is, if we label the vertices ABCD on the front face, clockwise starting from the upper left corner like this

A     B

D     C

Then if the 8" edges connect these to the back face vertices EFGH as A-E, B-F, C-G, D-H would your connections be A to H, D to G, and so on?

It sounds as if we are joining the front and back faces of a prism to form a torus that has a square cross section, only we're twisting the shape by 90 degrees before joining the faces. (Kind of like constructing a Mobius strip from a piece of paper. This new shape would have only three sides, just as a Mobius strip has only one side. Interesting.) I wonder, though, with the length (8) being only twice the side (4) whether this is even possible.

One thought:

  Hide contents

If we just did the twist, leaving the front and back faces parallel but not joined, the volume would decrease. The cross-sectional area would remain the same, but I believe the length would decrease. Imagine the four 8" edges as being strings that do not stretch. Twisting one of the faces by 90 degrees would bring the two faces closer, by a distance that could be found by inscribing an 8" radius arc in an 8X4 rectangle.

Possibly it's wrong to say the strings can not stretch. Another way to look at this is to slice the original prism into n equal shorter ones and rotate each of them by (90/n) degrees. And then take the limit where n grows large without bound. In this case, the volume is always the sum of the smaller volumes, which do not change upon rotation. Nor would the 8" length of the prism decrease in this case, so the volume would remain 8x4x4..

Either way, if after twisting we then brought the two faces together, the inner part of the torus would compress and the outer part would have to stretch. This means the material of which the original prism is made is compressible. I don't think there is enough information given to determine how much it would compress. Or, we could allow distortion (permitting the lengths of the 8" edges to change as necessary, by saying the interior of the original prism is filled with a (distortable) but incompressible fluid. In that case, the fluid being incompressible, the initial and final volumes will be equal.

I think the information given is insufficient to create an answer, but I may have missed an important point somewhere.

 

The square faces are the front and back. If you think of the congruent squares each having vertices abcd in the same clockwise rotation each resting parallel to the other.  Then connect 8 inch edges in the following manner, front A to back B,  front B to back C,  front C to back D,  and front D to back A. 

How does this volume compare to a standard rectangular prism of 4x4x8?

[bonanova]

It depends on what is constant and what can vary. Do the edges all keep their original lengths? Does the cross section remain everywhere square? By "connecting 8" edges" do you mean deforming the shape and "gluing" the square faces to each other, a to b, b to c, c to d and d to a so that the new shape has no vertices (corners)? Do the 8" edges end up as circular arcs? What keeps us from doing this and then squashing the final shape flat so that its volume is zero? I can't visualize a process that gives a final shape having a definitive volume.

[BMAD]

Let me see if i can ask my question better. 

Suppose you have twelve rods of two different sizes. One size is 4 in.  The other is 8 in. There are 8 - 4in. Rods and  4 -  8 in rods.   Use the 4 in rods to construct two squares. Stand your squares up perpendicular to your table and parallel to each other. Now take one 8 in rod and connect it to the top left of one square (first square) and bottom left of the other. Take your second rod, connect it to the top right of the first square and top left of the other. Third rod,  bottom right of first and top right of second. Connect the last rod to the remaining vertices. 

All rods will remain straight through this process. 

Find the volume and compare it to the volume of 4x4x8 

Edited July 28, 2016 by BMAD