Mixing
What is the maximum volume that can be contained by a sheet of paper?
Clash Royale CLAN TAG#URR8PPP
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204
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I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty non-trivial, I believe) question:
Q: By properly folding a common $210mmtimes 297mm$ sheet of paper, what
is the maximum amount of water such sheet is able to contain?
The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?
Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.
As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
geometry optimization recreational-mathematics volume
edited Jul 20 at 22:27
qwr
6,59532654
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
 |Â
show 16 more comments
up vote
204
down vote
favorite
I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty non-trivial, I believe) question:
Q: By properly folding a common $210mmtimes 297mm$ sheet of paper, what
is the maximum amount of water such sheet is able to contain?
The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?
Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.
As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
geometry optimization recreational-mathematics volume
edited Jul 20 at 22:27
qwr
6,59532654
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
35
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
32
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
27
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
6
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
45
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48
 |Â
show 16 more comments
up vote
204
down vote
favorite
up vote
204
down vote
favorite
I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty non-trivial, I believe) question:
Q: By properly folding a common $210mmtimes 297mm$ sheet of paper, what
is the maximum amount of water such sheet is able to contain?
The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?
Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.
As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
geometry optimization recreational-mathematics volume
edited Jul 20 at 22:27
qwr
6,59532654
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty non-trivial, I believe) question:
Q: By properly folding a common $210mmtimes 297mm$ sheet of paper, what
is the maximum amount of water such sheet is able to contain?
The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?
Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.
As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.
geometry optimization recreational-mathematics volume
edited Jul 20 at 22:27
qwr
6,59532654
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
edited Jul 20 at 22:27
qwr
6,59532654
edited Jul 20 at 22:27
qwr
6,59532654
edited Jul 20 at 22:27
qwr
6,59532654
6,59532654
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
asked Jul 18 at 20:32
Jack D'Aurizio♦
270k31265629
270k31265629
35
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
32
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
27
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
6
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
45
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48
 |Â
show 16 more comments
35
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
32
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
27
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
6
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
45
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48
35
35
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
32
32
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
27
27
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
6
6
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
45
45
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48
 |Â
show 16 more comments
5 Answers
5
active
oldest
votes
up vote
96
down vote
accepted
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$phi:PsubsetmathbbR^2tomathbbR^3,qquad |dphi^Tdphi|_2 leq 1$$
of the piece of paper $P$ into $mathbbR^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $phi$ sends the paper boundary $partial P$ to a curve $gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $gamma$, what is the volume of the volume-maximizing surface $M_gamma$ with $phi(partial P) = gamma$? (II) Can we characterize $gamma$ for which $M_gamma$ has maximum volume?
Let's consider the case where $gamma$ is given. We can partition $M_gamma$ into
1) regions of pure tension, where $dphi^Tdphi = I$; in these regions $M_gamma$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $|dphi^Tdphi|_2 = 1$ but $det dphi^Tdphi < 1$.
We need not consider $|dphi^Tdphi|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $tau$ along which $phi$ is an isometry. Since $M_gamma$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_gamma$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $tau$. In other words, for some stress $sigma$ constant along each $tau$, at all points $tau(s)$ along $tau$ we have
$$hatn = sigma tau''(s)$$
where $hatn$ the surface normal; it follows that (1) the $tau$ follow geodesics on $M_gamma$, (2) each $tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $gamma$, the surface $M_gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $gamma$ and $M_gamma$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $tau$ on the surface:
answered Jul 18 at 23:14
user7530
33.4k558109
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
 |Â
show 14 more comments
up vote
55
down vote
This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.
answered Jul 19 at 17:26
Rahul
32.4k361159
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
add a comment |Â
up vote
36
down vote
This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.
If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.
Let $x=210textmm$ be the width and $y=297textmm$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $phi$ of the long sides and the angle of inclination $xi$ of the short sides. Then at height $alpha h$, with $0lealphale1$, a rectangular cross-section of the prism of height $hmathrm dalpha$ has volume
$$
left(x-2frac hcosphi+alphacdot2htanphiright)left(y-2frac hcosxi+alphacdot2htanxiright)hmathrm dalpha;,
$$
and integrating over $alpha$ yields the volume
$$
left(left(x-2frac hcosphi+htanphiright)left(y-2frac hcosxi+htanxiright)+frac13h^2tanphitanxiright)h;.
$$
I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result
$$
happrox47.62textmm;,\
phiapprox0.55112;,\
xiapprox0.56838
$$
and resulting volume
$$
Vapprox1.315679370667,l;.
$$
Here's a rough attempt at building this:
P.S.:
This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)
answered Jul 18 at 22:12
joriki
164k10180328
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
add a comment |Â
up vote
12
down vote
This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)
edited Jul 19 at 10:48
Eric Duminil
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
 |Â
show 6 more comments
up vote
-2
down vote
It is unbounded.
Water cannot pass through a small enough hole. The idea is to create a surface from your paper that is made mostly of such small holes.
For example, you can perform many close parallel cuts, so that the paper can be stretched to have a larger surface, while the holes are small enough that the water cannot pass. By making the cuts closer and closer, you can make the surface as large as you want, and it will hold an arbitrarily large volume of water.
answered Jul 31 at 9:10
mnhrdt
1
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
96
down vote
accepted
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$phi:PsubsetmathbbR^2tomathbbR^3,qquad |dphi^Tdphi|_2 leq 1$$
of the piece of paper $P$ into $mathbbR^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $phi$ sends the paper boundary $partial P$ to a curve $gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $gamma$, what is the volume of the volume-maximizing surface $M_gamma$ with $phi(partial P) = gamma$? (II) Can we characterize $gamma$ for which $M_gamma$ has maximum volume?
Let's consider the case where $gamma$ is given. We can partition $M_gamma$ into
1) regions of pure tension, where $dphi^Tdphi = I$; in these regions $M_gamma$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $|dphi^Tdphi|_2 = 1$ but $det dphi^Tdphi < 1$.
We need not consider $|dphi^Tdphi|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $tau$ along which $phi$ is an isometry. Since $M_gamma$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_gamma$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $tau$. In other words, for some stress $sigma$ constant along each $tau$, at all points $tau(s)$ along $tau$ we have
$$hatn = sigma tau''(s)$$
where $hatn$ the surface normal; it follows that (1) the $tau$ follow geodesics on $M_gamma$, (2) each $tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $gamma$, the surface $M_gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $gamma$ and $M_gamma$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $tau$ on the surface:
answered Jul 18 at 23:14
user7530
33.4k558109
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
 |Â
show 14 more comments
up vote
96
down vote
accepted
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$phi:PsubsetmathbbR^2tomathbbR^3,qquad |dphi^Tdphi|_2 leq 1$$
of the piece of paper $P$ into $mathbbR^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $phi$ sends the paper boundary $partial P$ to a curve $gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $gamma$, what is the volume of the volume-maximizing surface $M_gamma$ with $phi(partial P) = gamma$? (II) Can we characterize $gamma$ for which $M_gamma$ has maximum volume?
Let's consider the case where $gamma$ is given. We can partition $M_gamma$ into
1) regions of pure tension, where $dphi^Tdphi = I$; in these regions $M_gamma$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $|dphi^Tdphi|_2 = 1$ but $det dphi^Tdphi < 1$.
We need not consider $|dphi^Tdphi|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $tau$ along which $phi$ is an isometry. Since $M_gamma$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_gamma$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $tau$. In other words, for some stress $sigma$ constant along each $tau$, at all points $tau(s)$ along $tau$ we have
$$hatn = sigma tau''(s)$$
where $hatn$ the surface normal; it follows that (1) the $tau$ follow geodesics on $M_gamma$, (2) each $tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $gamma$, the surface $M_gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $gamma$ and $M_gamma$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $tau$ on the surface:
answered Jul 18 at 23:14
user7530
33.4k558109
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
 |Â
show 14 more comments
up vote
96
down vote
accepted
up vote
96
down vote
accepted
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$phi:PsubsetmathbbR^2tomathbbR^3,qquad |dphi^Tdphi|_2 leq 1$$
of the piece of paper $P$ into $mathbbR^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $phi$ sends the paper boundary $partial P$ to a curve $gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $gamma$, what is the volume of the volume-maximizing surface $M_gamma$ with $phi(partial P) = gamma$? (II) Can we characterize $gamma$ for which $M_gamma$ has maximum volume?
Let's consider the case where $gamma$ is given. We can partition $M_gamma$ into
1) regions of pure tension, where $dphi^Tdphi = I$; in these regions $M_gamma$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $|dphi^Tdphi|_2 = 1$ but $det dphi^Tdphi < 1$.
We need not consider $|dphi^Tdphi|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $tau$ along which $phi$ is an isometry. Since $M_gamma$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_gamma$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $tau$. In other words, for some stress $sigma$ constant along each $tau$, at all points $tau(s)$ along $tau$ we have
$$hatn = sigma tau''(s)$$
where $hatn$ the surface normal; it follows that (1) the $tau$ follow geodesics on $M_gamma$, (2) each $tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $gamma$, the surface $M_gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $gamma$ and $M_gamma$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $tau$ on the surface:
answered Jul 18 at 23:14
user7530
33.4k558109
This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.
First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$phi:PsubsetmathbbR^2tomathbbR^3,qquad |dphi^Tdphi|_2 leq 1$$
of the piece of paper $P$ into $mathbbR^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $phi$ sends the paper boundary $partial P$ to a curve $gamma$ in the plane.
We can thus partition your problem into two pieces: (I) given a fixed curve $gamma$, what is the volume of the volume-maximizing surface $M_gamma$ with $phi(partial P) = gamma$? (II) Can we characterize $gamma$ for which $M_gamma$ has maximum volume?
Let's consider the case where $gamma$ is given. We can partition $M_gamma$ into
1) regions of pure tension, where $dphi^Tdphi = I$; in these regions $M_gamma$ is, by definition, developable;
2) regions where one direction is in tension and one in compression, $|dphi^Tdphi|_2 = 1$ but $det dphi^Tdphi < 1$.
We need not consider $|dphi^Tdphi|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $phi$ a short map.
Let us look at the regions of type (2). We can trace on these regions a family of curves $tau$ along which $phi$ is an isometry. Since $M_gamma$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_gamma$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $tau$. In other words, for some stress $sigma$ constant along each $tau$, at all points $tau(s)$ along $tau$ we have
$$hatn = sigma tau''(s)$$
where $hatn$ the surface normal; it follows that (1) the $tau$ follow geodesics on $M_gamma$, (2) each $tau$ has constant curvature.
The only thing I can say about problem (II) is that for the optimal $gamma$, the surface $M_gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $approx 1.236$ liters, less than Joriki's solution).
I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $gamma$ and $M_gamma$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper
Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).
EDIT 2: A sketch of the orientation of $tau$ on the surface:
answered Jul 18 at 23:14
user7530
33.4k558109
edited Jul 26 at 19:16
answered Jul 18 at 23:14
user7530
33.4k558109
answered Jul 18 at 23:14
user7530
33.4k558109
answered Jul 18 at 23:14
user7530
33.4k558109
33.4k558109
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
 |Â
show 14 more comments
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
6
6
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
@user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary.
– user7530
Jul 19 at 9:20
1
1
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
@user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins.
– user7530
Jul 19 at 9:24
1
1
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
@user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack.
– user7530
Jul 19 at 9:41
12
12
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
P.S. This is the solution of the paper bag problem cut in half, isn't it?
– Rahul
Jul 19 at 15:23
2
2
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
@user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway!
– Mario Carneiro
Jul 21 at 10:17
 |Â
show 14 more comments
up vote
55
down vote
This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.
answered Jul 19 at 17:26
Rahul
32.4k361159
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
add a comment |Â
up vote
55
down vote
This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.
answered Jul 19 at 17:26
Rahul
32.4k361159
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
add a comment |Â
up vote
55
down vote
up vote
55
down vote
This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.
answered Jul 19 at 17:26
Rahul
32.4k361159
This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.
answered Jul 19 at 17:26
Rahul
32.4k361159
answered Jul 19 at 17:26
Rahul
32.4k361159
answered Jul 19 at 17:26
Rahul
32.4k361159
answered Jul 19 at 17:26
Rahul
32.4k361159
32.4k361159
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
add a comment |Â
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
17
17
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside.
– Jack D'Aurizio♦
Jul 19 at 17:42
10
10
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.)
– Nominal Animal
Jul 19 at 22:50
4
4
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer.
– Rahul
Jul 20 at 5:30
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
@Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks!
– Nominal Animal
Jul 20 at 14:20
add a comment |Â
up vote
36
down vote
This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.
If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.
Let $x=210textmm$ be the width and $y=297textmm$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $phi$ of the long sides and the angle of inclination $xi$ of the short sides. Then at height $alpha h$, with $0lealphale1$, a rectangular cross-section of the prism of height $hmathrm dalpha$ has volume
$$
left(x-2frac hcosphi+alphacdot2htanphiright)left(y-2frac hcosxi+alphacdot2htanxiright)hmathrm dalpha;,
$$
and integrating over $alpha$ yields the volume
$$
left(left(x-2frac hcosphi+htanphiright)left(y-2frac hcosxi+htanxiright)+frac13h^2tanphitanxiright)h;.
$$
I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result
$$
happrox47.62textmm;,\
phiapprox0.55112;,\
xiapprox0.56838
$$
and resulting volume
$$
Vapprox1.315679370667,l;.
$$
Here's a rough attempt at building this:
P.S.:
This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)
answered Jul 18 at 22:12
joriki
164k10180328
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
add a comment |Â
up vote
36
down vote
This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.
If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.
Let $x=210textmm$ be the width and $y=297textmm$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $phi$ of the long sides and the angle of inclination $xi$ of the short sides. Then at height $alpha h$, with $0lealphale1$, a rectangular cross-section of the prism of height $hmathrm dalpha$ has volume
$$
left(x-2frac hcosphi+alphacdot2htanphiright)left(y-2frac hcosxi+alphacdot2htanxiright)hmathrm dalpha;,
$$
and integrating over $alpha$ yields the volume
$$
left(left(x-2frac hcosphi+htanphiright)left(y-2frac hcosxi+htanxiright)+frac13h^2tanphitanxiright)h;.
$$
I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result
$$
happrox47.62textmm;,\
phiapprox0.55112;,\
xiapprox0.56838
$$
and resulting volume
$$
Vapprox1.315679370667,l;.
$$
Here's a rough attempt at building this:
P.S.:
This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)
answered Jul 18 at 22:12
joriki
164k10180328
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
add a comment |Â
up vote
36
down vote
up vote
36
down vote
This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.
If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.
Let $x=210textmm$ be the width and $y=297textmm$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $phi$ of the long sides and the angle of inclination $xi$ of the short sides. Then at height $alpha h$, with $0lealphale1$, a rectangular cross-section of the prism of height $hmathrm dalpha$ has volume
$$
left(x-2frac hcosphi+alphacdot2htanphiright)left(y-2frac hcosxi+alphacdot2htanxiright)hmathrm dalpha;,
$$
and integrating over $alpha$ yields the volume
$$
left(left(x-2frac hcosphi+htanphiright)left(y-2frac hcosxi+htanxiright)+frac13h^2tanphitanxiright)h;.
$$
I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result
$$
happrox47.62textmm;,\
phiapprox0.55112;,\
xiapprox0.56838
$$
and resulting volume
$$
Vapprox1.315679370667,l;.
$$
Here's a rough attempt at building this:
P.S.:
This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)
answered Jul 18 at 22:12
joriki
164k10180328
This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.
If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.
Let $x=210textmm$ be the width and $y=297textmm$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $phi$ of the long sides and the angle of inclination $xi$ of the short sides. Then at height $alpha h$, with $0lealphale1$, a rectangular cross-section of the prism of height $hmathrm dalpha$ has volume
$$
left(x-2frac hcosphi+alphacdot2htanphiright)left(y-2frac hcosxi+alphacdot2htanxiright)hmathrm dalpha;,
$$
and integrating over $alpha$ yields the volume
$$
left(left(x-2frac hcosphi+htanphiright)left(y-2frac hcosxi+htanxiright)+frac13h^2tanphitanxiright)h;.
$$
I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result
$$
happrox47.62textmm;,\
phiapprox0.55112;,\
xiapprox0.56838
$$
and resulting volume
$$
Vapprox1.315679370667,l;.
$$
Here's a rough attempt at building this:
P.S.:
This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)
answered Jul 18 at 22:12
joriki
164k10180328
edited Jul 19 at 10:28
answered Jul 18 at 22:12
joriki
164k10180328
answered Jul 18 at 22:12
joriki
164k10180328
answered Jul 18 at 22:12
joriki
164k10180328
164k10180328
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
add a comment |Â
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
1
1
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
Just a note on the version with glued corners -- those wasted triangles could be opened into cones.
– amI
Jul 19 at 20:40
3
3
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
@aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface.
– Henning Makholm
Jul 20 at 0:50
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
can we obtain a similar result by "pleating" the border, like in aluminium trays (?)
– G Cab
Jul 21 at 13:21
add a comment |Â
up vote
12
down vote
This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)
edited Jul 19 at 10:48
Eric Duminil
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
 |Â
show 6 more comments
up vote
12
down vote
This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)
edited Jul 19 at 10:48
Eric Duminil
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
 |Â
show 6 more comments
up vote
12
down vote
up vote
12
down vote
This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)
edited Jul 19 at 10:48
Eric Duminil
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)
edited Jul 19 at 10:48
Eric Duminil
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
edited Jul 19 at 10:48
Eric Duminil
1,6561516
edited Jul 19 at 10:48
Eric Duminil
1,6561516
edited Jul 19 at 10:48
Eric Duminil
1,6561516
1,6561516
answered Jul 18 at 20:57
AppoopanThaadi
1298
answered Jul 18 at 20:57
AppoopanThaadi
1298
answered Jul 18 at 20:57
AppoopanThaadi
1298
1298
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
 |Â
show 6 more comments
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
3
3
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
I believe that something more spherical-shaped like a boat is able to contain $geq 1.2l$.
– Jack D'Aurizio♦
Jul 18 at 21:05
2
2
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise.
– Pi_die_die
Jul 18 at 21:14
4
4
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
@Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently.
– Jack D'Aurizio♦
Jul 18 at 21:23
3
3
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
@Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres.
– Jack D'Aurizio♦
Jul 18 at 21:37
1
1
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion.
– Pi_die_die
Jul 18 at 22:29
 |Â
show 6 more comments
up vote
-2
down vote
It is unbounded.
Water cannot pass through a small enough hole. The idea is to create a surface from your paper that is made mostly of such small holes.
For example, you can perform many close parallel cuts, so that the paper can be stretched to have a larger surface, while the holes are small enough that the water cannot pass. By making the cuts closer and closer, you can make the surface as large as you want, and it will hold an arbitrarily large volume of water.
answered Jul 31 at 9:10
mnhrdt
1
add a comment |Â
up vote
-2
down vote
It is unbounded.
Water cannot pass through a small enough hole. The idea is to create a surface from your paper that is made mostly of such small holes.
For example, you can perform many close parallel cuts, so that the paper can be stretched to have a larger surface, while the holes are small enough that the water cannot pass. By making the cuts closer and closer, you can make the surface as large as you want, and it will hold an arbitrarily large volume of water.
answered Jul 31 at 9:10
mnhrdt
1
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
It is unbounded.
Water cannot pass through a small enough hole. The idea is to create a surface from your paper that is made mostly of such small holes.
For example, you can perform many close parallel cuts, so that the paper can be stretched to have a larger surface, while the holes are small enough that the water cannot pass. By making the cuts closer and closer, you can make the surface as large as you want, and it will hold an arbitrarily large volume of water.
answered Jul 31 at 9:10
mnhrdt
1
It is unbounded.
Water cannot pass through a small enough hole. The idea is to create a surface from your paper that is made mostly of such small holes.
For example, you can perform many close parallel cuts, so that the paper can be stretched to have a larger surface, while the holes are small enough that the water cannot pass. By making the cuts closer and closer, you can make the surface as large as you want, and it will hold an arbitrarily large volume of water.
answered Jul 31 at 9:10
mnhrdt
1
answered Jul 31 at 9:10
mnhrdt
1
answered Jul 31 at 9:10
mnhrdt
1
answered Jul 31 at 9:10
mnhrdt
1
1
add a comment |Â
add a comment |Â
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35
This is a very nice problem!
– Dr. Sonnhard Graubner
Jul 18 at 20:36
32
It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty!
– gimusi
Jul 18 at 20:38
27
Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it?
– joriki
Jul 18 at 21:18
6
Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos.
– Rahul
Jul 19 at 5:43
45
It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!"
– Ister
Jul 19 at 11:48