[math-fun] Packing problem
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place. Are there good heuristics for avoiding the rattler problem when packing cylinders? For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter? If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler? My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches. At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable. Jim Propp PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality.
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer. On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear. On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
How do you move the tiny one without violating the no-slip condition? In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero. (Maybe this shows that I should rethink my model of friction?) Jim On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Or rethink your model of the ladder (and frame)? WFL On 4/23/19, James Propp <jamespropp@gmail.com> wrote:
How do you move the tiny one without violating the no-slip condition?
In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero.
(Maybe this shows that I should rethink my model of friction?)
Jim
On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I don't mind any particular model, but I think I just don't understand the no-slip condition. Can you rigorize it? On Mon, Apr 22, 2019 at 9:51 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Or rethink your model of the ladder (and frame)? WFL
On 4/23/19, James Propp <jamespropp@gmail.com> wrote:
How do you move the tiny one without violating the no-slip condition?
In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero.
(Maybe this shows that I should rethink my model of friction?)
Jim
On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
No great insight to offer, I'm afraid. It's just that your door-frame should perhaps have nonzero depth? WFL On 4/23/19, Allan Wechsler <acwacw@gmail.com> wrote:
I don't mind any particular model, but I think I just don't understand the no-slip condition. Can you rigorize it?
On Mon, Apr 22, 2019 at 9:51 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Or rethink your model of the ladder (and frame)? WFL
On 4/23/19, James Propp <jamespropp@gmail.com> wrote:
How do you move the tiny one without violating the no-slip condition?
In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero.
(Maybe this shows that I should rethink my model of friction?)
Jim
On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Tue, Apr 23, 2019 at 7:24 AM Allan Wechsler <acwacw@gmail.com> wrote:
I don't mind any particular model, but I think I just don't understand the no-slip condition. Can you rigorize it?
How about this: if two points (on two distinct rolls, or on a roll and on the bungee cord) are in contact at some instant, then their instantaneous relative velocity vector must have its tangential component equal to zero? I like the simplicity of that criterion. Note that under that criterion, you CAN fit an upright six-foot ladder through a six-foot doorway as long as your instantaneous forward speed is zero at the moment of contact. Likewise, under this proposed criterion, the 1,esilon,1 shishkebab is unstable, because the small roll can slip out from between the two larger rolls, and then the two larger rolls can move closer to each other, allowing the cord to shorten. And that accords with my physical intuition. I see a slight lurking complication. If points on the bungee cord have distinct identities (as they must if one is going to impose my proposed version of a no-slip condition), then what does it mean for the cord to shorten? Do we want to require that it shortens uniformly? If not, then do we allow some parts of the cord to lengthen while others shorten? It’d be nice to have an idealization of the roll-wrapping problem that was simultaneously physically realistic and mathematically sweet, but I don’t feel we’re there yet. Jim Propp
On Mon, Apr 22, 2019 at 9:51 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Or rethink your model of the ladder (and frame)? WFL
On 4/23/19, James Propp <jamespropp@gmail.com> wrote:
How do you move the tiny one without violating the no-slip condition?
In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero.
(Maybe this shows that I should rethink my model of friction?)
Jim
On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
isn't this problem equivalent to packing circles of different diameters in a plane so that each circle touches (at least) three others? Like in http://users.telenet.be/Wouter.Meeussen/FOAM2D.htm Wouter. -----Original Message----- From: James Propp Sent: Tuesday, April 23, 2019 3:42 PM To: math-fun Subject: Re: [math-fun] Packing problem On Tue, Apr 23, 2019 at 7:24 AM Allan Wechsler <acwacw@gmail.com> wrote:
I don't mind any particular model, but I think I just don't understand the no-slip condition. Can you rigorize it?
How about this: if two points (on two distinct rolls, or on a roll and on the bungee cord) are in contact at some instant, then their instantaneous relative velocity vector must have its tangential component equal to zero? I like the simplicity of that criterion. Note that under that criterion, you CAN fit an upright six-foot ladder through a six-foot doorway as long as your instantaneous forward speed is zero at the moment of contact. Likewise, under this proposed criterion, the 1,esilon,1 shishkebab is unstable, because the small roll can slip out from between the two larger rolls, and then the two larger rolls can move closer to each other, allowing the cord to shorten. And that accords with my physical intuition. I see a slight lurking complication. If points on the bungee cord have distinct identities (as they must if one is going to impose my proposed version of a no-slip condition), then what does it mean for the cord to shorten? Do we want to require that it shortens uniformly? If not, then do we allow some parts of the cord to lengthen while others shorten? It’d be nice to have an idealization of the roll-wrapping problem that was simultaneously physically realistic and mathematically sweet, but I don’t feel we’re there yet. Jim Propp
On Mon, Apr 22, 2019 at 9:51 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Or rethink your model of the ladder (and frame)? WFL
On 4/23/19, James Propp <jamespropp@gmail.com> wrote:
How do you move the tiny one without violating the no-slip condition?
In my model of slippage, you can’t fit a six-foot-tall ladder upright through a six-foot-high doorframe because it will get stuck, even though the moment of contact (and friction) has duration zero.
(Maybe this shows that I should rethink my model of friction?)
Jim
On Mon, Apr 22, 2019 at 6:23 PM Allan Wechsler <acwacw@gmail.com> wrote:
I think I'm not understanding what you mean by "shishkebab". If you put the tiny one between the other two, so that the centers are collinear, then any little perturbation of the tiny one reduces the energy by allowing the two big ones to get a little closer.
On the other hand, it seems to me that 1, 1, epsilon is stable as long as the centers are reasonably close to collinear.
On Mon, Apr 22, 2019 at 6:16 PM James Propp <jamespropp@gmail.com> wrote:
In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place.
Are there good heuristics for avoiding the rattler problem when packing cylinders?
For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter?
If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler?
My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches.
At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable.
Jim Propp
PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hi Jim, To have a realistic model of friction, you need static and kinetic coefficients along with an idea of the normal forces at all points of contact. Then rate of change for the velocity vectors depends on weighing of the frictional forces against the gravitational forces, by Newton's Law. Say that you hold two rolls adjacent with longitudinal axes parallel to Earth's surface. Then place a third roll into the crevice between the two so that it rests in stable equilibrium. Tilting the whole configuration toward vertical, you should find a critical angle where the third roll begins to slip down the fixed two. This is the old problem of the block and the inclined plane, where inevitability of slipping is explained by the fact that increasing the angle toward vertical takes a limit toward zero normal force from the ramp on the block. How do normal forces get distributed in yr cording problem? Draw a graph whose vertices are the rolls, and whose edges indicate contact points. Then label all the edges with values for the force, by requiring zero sums on each vertex (static equilibrium). Don't forget that the cord is an elastic medium that stretches uniformly, and puts on an external force to any of the tubes it touches. At the detail level, this picture looks complicated, but maybe the details aren't necessary. The edge set divides the region interior to the cord into disjoint tiles. We can go from stable configuration to unstable by adding a small-enough tube somewhere on a region interior to any one tile. What is "small enough"? Given a set of radii on the vertices of the tiles, this can be calculated explicitly. However, I think we already know that the rattlers have to be pretty small relative to their boundary. I would think that the best strategy is to start off going for a smoothly varying gradient of radii. If this doesn't work at first, it seems plausible that a recursive procedure of letting the rattlers fall out and then putting them back into the configuration on the boundary should, after so many iterations, reach a stable configuration. --Brad
participants (5)
-
Allan Wechsler -
Brad Klee -
Fred Lunnon -
James Propp -
Wouter Meeussen