From: foltinek@math.utexas.edu (Kevin Foltinek) Subject: Re: REGARDING PEELING AN ORANGE AND OTHER TOPOLOGICAL MATTERS Date: 17 Jan 2000 12:21:51 -0600 Newsgroups: alt.math,alt.math.undergrad,sci.math Summary: [missing] In article Deinst@world.std.com (David M Einstein) writes: > Yes, this is shortest if you allow spinning the ball. However, what > if you do not allow spinning? As I understand it, the problem is the following. Given: a sphere (ball) on a (flat) plane, with the sphere constrained to move by "rolling without slipping", and further constrained so that the sphere is not allowed to spin around the vertical (perpendicular to the plane) axis. Find: a path to follow such that the sphere is in a specified orientation at a specified location, and minimize the path length. The first part of this by itself is interesting: describe the motion of the sphere. I'll be vague and probably make a mistake or two, because I'm doing this off the top of my head... and I don't know how much of this will be at an undergrad level... The configuration space is SO(3)\times R^2. The phase space is SO(3)\times so(3)\times R^2\times R^2, and this is where the constraints show up (which is why this is interesting). Given (R,A,x,v) in phase space, the constraints are (or can be written as): A = [0 0 -a2; 0 0 a1; a2 -a1 0] v = [a1*r; a2*r] (where r is the radius of the sphere). That is, the velocity constraints form a two-dimensional distribution, spanned by A1 and A2. Then, look at the commutator [A1,A2]: its projection onto the so(3) component is [0 a3 0; -a3 0 0; 0 0 0]. Thus, by the single commutator, we can span all of so(3). By commuting [A1,A2] with A1 and A2, we finally span all of velocity space so(3)\times R^2. Therefore, by Chow's theorem, the entire configuration space is reachable; that is, we can reach any orientation/position we desire. Intuitively, what I just said is the following: First, by moving the sphere around a very small square in the plane, we can (almost) rotate the ball around the vertical axis (though, of course, we will slightly change the point of contact). Second, by combining the first with short straight lines in the plane, we can (almost) move the sphere without having to rotate it at all; or, equivalently, we can (almost) rotate the sphere (around any angle) without changing its position. (The fact that these are only approximately true is not important, because the error is "second-order", i.e., if you want to spin by a small angle \theta, the change in the point of contact will be something like r*\theta^2 which is much smaller.) Remark: three constraints on a five-dimensional velocity space means that what we really have here is a special case of E. Cartan's famous five-variables problem. Finding a path, and minimizing its length, are the problems which (in some generality) are dealt with in optimal control theory and sub-Riemannian geometry. For one approach, look up Pontryagin's maximum principle; for another, read Griffiths' "Exterior Differential Systems and the Calculus of Variations". I suspect (though haven't verified) that this problem reduces to a fairly simple (or at least well-known) optimization problem. Kevin. ============================================================================== From: Douglas Zare Subject: Re: REGARDING PEELING AN ORANGE AND OTHER TOPOLOGICAL MATTERS Date: Mon, 17 Jan 2000 17:25:59 -0500 Newsgroups: alt.math,alt.math.undergrad,sci.math Douglas Zare wrote: > David M Einstein wrote: > > > Vaguely beside the point, but if one places a sphere on a plane, and > > then chooses some point, call it A, on the sphere. What is the minimum > > distance that one needs to roll the sphere to have point A resting on > > the point on the plane where the sphere touched the plane originally, > > and what does the > > path look like. > > That is an excellent problem, particularly if you care about the > orientation of the sphere rather than just the point of contact. An > easier one is to describe short nontrivial paths which go from one > configuration of sphere and plane back to itself. A harder one is to > find paths which work simultaneously for several different spheres > (with different radii). I'll post partial solutions to these later. One way to get a short nontrivial path which does not change the configuration is by rolling the ball along a path close to a figure 8. To first order, the rotation about the vertical axis from rolling a ball along a small path is proportionate to the signed area enclosed by the path. Rolling the ball along a small circle will approximately result in a small spinning motion. A figure 8 can enclose 0 signed area. How is this useful when one considers the second problem? If one has a path on the plane, then replacing part of this path A with another segment B has the effect of multiplying the ending configuration of the sphere by a conjugate of the rotation of the sphere produced by rolling ball along the loop (backwards A)B. It helps to know when this actually has no effect on the ending configuration, because these are the alterations which we can use to shorten a path's length. By the isoperimetric inequality, given a path in the plane which does not have constant curvature, there exists two points on the path so that the segment A between them together with a shorter arc of a circle, B, enclose 0 area. These will look like rough, distorted figure 8's for the simplest variations of curvature. There are some technical details that one has to be careful about, but after this one gets that the geodesics in the natural geometry on the total configuration space (or its 2-fold cover) are the results of rolling the ball along arcs of circles in the plane. If you consider the family of geodesics which connect two configurations with the same location in the plane, I believe (well, I need to sharpen one estimate) the shortest one will correspond to the smallest radius circle so that rolling the ball along this circle once produces the appropriate rotation of the sphere. The other geodesics correspond to other circles which might have to be traversed multiple times. The third problem I mentioned is rather messy, and I don't yet know what the geodesics look like in that geometry. Douglas Zare [reformatted --djr] ============================================================================== From: foltinek@math.utexas.edu (Kevin Foltinek) Subject: Re: REGARDING PEELING AN ORANGE AND OTHER TOPOLOGICAL MATTERS Date: 18 Jan 2000 13:52:47 -0600 Newsgroups: alt.math,alt.math.undergrad,sci.math In article Deinst@world.std.com (David M Einstein) writes: > Douglas Zare (zare@math.columbia.edu) wrote: > : By the isoperimetric inequality, given a path in the plane which does > : not have constant curvature, there exists two points on the path so > : that the segment A between them together with a shorter arc of a > : circle, B, enclose 0 area. These will look like rough, distorted > : figure 8's for the simplest variations of curvature. There are some > : technical details that one has to be careful about, but after this one > : gets that the geodesics in the natural geometry on the total > : configuration space (or its 2-fold cover) are the results of rolling > : the ball along arcs of circles in the plane. If you consider the > : family of geodesics which connect two configurations with the same > : location in the plane, I believe (well, I need to sharpen one > : estimate) the shortest one will correspond to the smallest radius > : circle so that rolling the ball along this circle once produces the > : appropriate rotation of the sphere. The other geodesics correspond to > : other circles which might have to be traversed multiple times. > If I understand you correctly the shortest path that brings a > point into contact with the original resting place will be a circle. There are two paths to look at: one path is the path on the plane; the other is the path on the sphere. The curvature of the path on the plane will be the same as the geodesic curvature of the path on the sphere. A circlular arc (constant curvature) on the plane, therefore, will correspond to a circular arc on the sphere (where "circle on the sphere" means the result of slicing the sphere by a plane; not necessarily a "great circle"). For a closed circle on the plane, with curvature k, \int k ds = 2\pi (where the domain of integration is exactly once around the circle); for the corresponding circular arc on the sphere, since the curvatures are equal, \int k ds = 2\pi; but by Gauss-Bonet, the total geodesic curvature of a simple closed path is 2\pi-1/r*(enclosed area). (The "enclosed area" is the area of the enclosed region on the sphere, not the plane.) Therefore, a simple closed circle on the plane and a simple closed circle on the sphere are *never* the same (the latter will have total curvature <2\pi). The "simple" is important here: it is possible, by travelling only on a circle, to return the same point on the sphere to the same point on the plane, by going around the appropriate circle on the sphere enough times. (1/r*(enclosed area) must be an integer multiple of 2\pi.) > This was my original intuition, and may be true (or nearly true) for > small displacements. For the problem: Start with point A on the sphere touching point B on the plane; move so that point C on the sphere is touching the same point B on the plane; minimize arclength travelled. It's still not obvious to me that circles are the shortest paths. I'm a little worried about the "technical details" which Douglas mentioned - as I understand it, his argument is: to first order, rotation of the sphere around the vertical axis is proportional to the enclosed area (this I agree with); a variable-curvature path can be replaced by a circular arc so that enclosed signed area is zero (this I agree with); doing so will therefore not change the rotation around the vertical axis (this I agree with to first order). It's not obvious to me (in fact, I question it) that (1) higher-order effects are not relevant, and (2) [more importantly] changing the path will not change the final point of contact on either the sphere or the plane. > However, you cannot bring the point that is > initially oposite the plane into contact with the plane by rolling it > in a circle (Dave Seaman's rectangle method will do it in 3/2 the > circumference of the sphere). If I have mistunderstood you, forgive me. This (the initially opposite point) might be a case of a "singular path" in some sense. They (in various flavours) tend to show up in systems which are constrained in velocity-space. > Could you > suggest a book that would put me (Someone who has a fairly solid, but > somewhat rusty undergraduate education) on the path to understanding > Kevin's response? Try Warner's book on differential geometry and manifolds (I've forgotten the exact title, but everybody knows the book), which I seem to recall talks about Chow's theorem (and all of its prerequisites). The relevant parts are the ones about tangent vectors, Lie brackets of vector fields, distributions or Pfaffian systems and the Frobenius theorem, and the basics of Lie groups [though a good book on mechanics would probably be more relevant here]. I don't know of any good references for the control theory approach, because almost everything I know about control theory I learned from talking to people at conferences and seminars. Kevin.