August 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Rolling a ball (without slipping) along a smooth curve
by Eugene Salamin 10 Aug '08

10 Aug '08

----- Original Message ---- From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Thursday, August 7, 2008 5:05:05 PM Subject: [math-fun] Rolling a ball (without slipping) along a smooth curve The problem of rolling a ball (without slipping) along a smooth curve seems quite interesting in itself. PUZZLE: Suppose a unit sphere rolls on the plane, once around the unit circle about the origin. What net spatial rotation (as a 3x3 matrix) does this impart to the sphere? --Dan --------------------------------------------------------------------------------- Let a ball of unit radius roll on a circle in the xy-plane of radius r in the positive (CCW) direction starting from the point (r, 0, 0) on the positive x-axis. Approximate the circle by N segments, and let t = 2 pi/N. When rolling on the k-th segment (0 <= k <= N-1), the ball rotates about the axis (cos kt, sin kt, 0) by angle rt in the negative sense. Along the 0-th segment, the rotation matrix is R[0] = Rx(-rt), a rotation about the x-axis. To determine the rotation matrix for the k-th segment, rotate the system by -kt about the z-axis, then apply R[0], and finally rotate back by kt about z. With rotation matrices applied right-to-left, we have R[k] = Rz(kt) Rx(-rt) Rz(-kt). The overall rotation for the complete roll around the circle is R = R[N-1] R[N-2] ... R[1] R[0] = Rz[(N-1)t] Rx(-rt) Rz[-(N-1)t] Rz[(N-2)t] Rx(-rt) Rz[-(N-2)t] ... Rz(t) Rx(-rt) Rz(-t) Rz(0) Rx(-rt) Rz(-0) = (Rz(-t) Rx(-rt))^N, using (N-1)t = -t mod 2 pi. For a unit vector n and any vector v, let Kn be the matrix such that Kn v = n X v (the cross product). Then the rotation by angle p about n is given by the matrix Rn(p) = exp(p Kn). We have R = (Rz(-2 pi/N) Rx(-2 pi r/N))^N = ((1 - (2 pi/N) Kz + O(N^-2)) (1 - (2 pi r/N) Kx +O(N^-2)))^N = (1 - (2 pi/N) (Kz + r Kx) + O(N^-2))^N ---> exp(- 2 pi (Kz + r Kx)). Thus the rotation angle is 2 pi sqrt(1 + r^2) in the negative sense about the axis (1, 0, r). When r --> 0, so that the ball is much larger than the circle, the ball rocks a little bit going around the circle, and the net rotation is nearly the identity. When r = 1, the ball and circle having equal radii, the rotation angle is 2 sqrt(2) pi. Suppose the ball is rolled, not along the complete circle, but only along the arc from a to b. Then, it is clear from the above derivation, that we have R(a, b) = Rz(b) exp(- (b - a) (Kz + r Kx)) Rz(-a). If we write this as R(a, b) = Rz(b - a) Rz(a) exp(- (b - a) (Kz + r Kx)) Rz(-a) = Rz(b - a) exp(- (b - a) (Kz + r Kx')), where x' is x rotated by angle a, it is clear that the x-axis plays no special role. If the rotation instead begins at (0, r, 0) on the y-axis, then the formula is the same as that obtained by replacing Kx in the original formula by Ky. Also note that for r --> 0, R(a,b) --> Rz(b - a) Rz(- (b - a)) = 1, and so, despite the appearance of 2 pi in the formula above, the rotation never deviates much from the identity. Before ending, I will mention a nice fact that is important in the theory of rigid bodies. Every rotation about a space fixed axis commutes with every rotation about a body fixed axis. Try it for yourself to see that this is true. It is a simple puzzle to reconcile this observation with the noncommutativity of the rotation group. Gene

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[math-fun] Puzzle to find a function D^2 -> [0,1]
by Dan Asimov 09 Aug '08

09 Aug '08

PUZZLE: Can you find a smooth (C^oo) function f : D -> [0,1] (where D is the closed unit disk in R^2) such that 1) f is onto. 2) The unit circle bd(D) is finv(1), and 3) For 0 < c <= 1, finv(c) is a smooth curve with length = 2pi. where finv(c) denotes the inverse image of c. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Rolling a ball (without slipping) along a smooth curve
by Dan Asimov 07 Aug '08

07 Aug '08

The problem of rolling a ball (without slipping) along a smooth curve seems quite interesting in itself. PUZZLE: Suppose a unit sphere rolls on the plane, once around the unit circle about the origin. What net spatial rotation (as a 3x3 matrix) does this impart to the sphere? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] rolling a ball along a fractal
by James Propp 07 Aug '08

07 Aug '08

Rich's comment >... >The sum of the edge lengths goes to >infinity (a not-absolutely-convergent series), while the sum of >the edge vectors is finite (in effect, a convergent series). >... is quite right. Moreover, note that if you've got a fractal curve on a sphere of radius 1, no matter how wild it is, there's a unique way to "contact-transfer" the fractal to another sphere of radius 1 by rolling the second sphere along the first sphere so that the point of contact follows the fractal. The way to see this is to change to a coordinate system in which the point of contact and the plane of tangency between the two spheres stays fixed, and both spheres just rotate in place, with each doing the opposite of what the other one does. Both spheres will execute fractal jiggling, but when the jiggling ends, the second sphere has a mirror image of the fractal on the first sphere. So I believe that hoops are a red herring, and my remark >"How can you roll a ball >along a Koch curve when you can't even roll a *hoop* along a Koch >curve?" was a non sequitur (or rather, the implicit argument --- "A hoop is a subset of a ball, so if you can't roll the hoop you can't roll the ball" --- is a non sequitur), sort of on a par with "Since the volume of revolution obtained from revolving y = 1/x (with x greater than or equal to 1) about the x-axis has finite volume, you can fill it with paint, so you can paint it, so it has finite surface area". The reason it's a non sequitur is that the way you roll a hoop is different from the way you roll a ball. This is easiest to see if you roll a hoop and a ball along a polygonal curve. When you come to a corner, you rotate the hoop by some positive angle, but you don't rotate the ball at all --- you just switch to rolling it along a different great circle (that is, along a different hoop). >Rolling a short distance multiplies this by another matrix I+A, >where (the entries in) A are bounded by D, the distance rolled. >The difference between (I+A)(I+B) and (I+B)(I+A) is AB-BA, which >is bounded by twice the product of the two distances. Any change >in contact point, or orientation, is also bounded by the product. >For the snowflake, going down a level produces 4x as many edges >of 1/3 the length. The difference between rolling along an >edge, versus rolling along 4 smaller edges, is proportional to >the square of the edge length. The effect of successive levels >is multiplied by roughly 4/9 for each level, giving a convergent >series. This now looks right to me. Is everyone else as convinced as I am that you can contact-transfer the Koch curve to a sphere of any radius? Jim

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Re: [math-fun] rolling a ball along a fractal
by Mike Stay 06 Aug '08

06 Aug '08

On Wed, Aug 6, 2008 at 8:21 AM, Steve Witham <sw(a)tiac.net> wrote: >>Mike Stay wrote: >> >> > Well, Erin Pearse wrote all over *about* the question: >>> http://arxiv.org/abs/math-ph/0412029 >>> http://arxiv.org/abs/math.DS/0605527 >> > http://www.math.cornell.edu/~erin/dissertation/dissertation.pdf > > Okay, beginner questions. > > Looking at the dissertation, I was confused about Pearse's meaning of > "convex hull", which I thought was an outline. It is. All the links I gave are probably irrelevant, as I completely misunderstood the real question. -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com

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Re: [math-fun] floating point precision and hyperbolic functions
by Henry Baker 06 Aug '08

06 Aug '08

I did a complete analysis of the exponent range & precision issues for a Common Lisp math library, which has to deal with IEEE single precision, IEEE double precision & complex numbers. I'll try to dig up some of the code. Basically, in order to achieve the maximum exponent range, you can't just blindly apply A&S formulae because of the exponentials & powers. Unfortunately, most math libraries, including the C math libraries on which many higher-level languages depend, don't achieve the maximum exponent range. At 08:55 AM 8/6/2008, Michael Speciner wrote: >I forgot that math-fun strips out attachments. So below is what I tried to attach in my last email, possibly with some mailer-inserted line breaks. >log1p and expm1 are ln1+ and exp-1 below. (In PostScript natural log is ln and exp takes two args--a base and exponent, adding an extra source of error since e can't be expressed exactly; the PostScript trig functions work in degrees.) >Notes: >log1p(x) = log(1+x) + (x-((1+x)-1))/(1+x) >For |x| <=1, we use >asinh(x) = (log(1+x/sqrt(x^2+1)) - log(1-x/sqrt(x^2+1)))/2, >which is based on >atanh(x) = (log(1+x) - log(1-x))/2; >while for larger x, we use >asinh(x) = sgn(x) log(|x| + sqrt(x^2+1)) >PostScript uses single precision floating point; any McClaurin series are truncated accordingly. > >/sgn {dup 0 ne {dup dup eq {0 gt {1}{-1} ifelse} if} if} bind def >/e 2.718281828459045 def >/ln2 .6931471805599453 def >/log2 {dup ln //ln2 div dup sgn exch abs floor mul % x roundtozero(log2?(x)) > exch 2 2 index exp div ln //ln2 div add >} bind readonly def >/exp-1 {dup abs //ln2 le > {dup 9. div 1. add > 1 index mul 8. div 1. add > 1 index mul 7. div 1. add > 1 index mul 6. div 1. add > 1 index mul 5. div 1. add > 1 index mul 4. div 1. add > 1 index mul 3. div 1. add > 1 index mul 2. div 1. add > mul} > {//e exch exp 1 sub} > ifelse} bind readonly def >/ln1+ {dup 1. add dup ln 3 1 roll dup 3 1 roll 1. sub sub exch div add} bind def > >/Infinity 2 1000 exp def >/-Infinity Infinity neg def >/NaN Infinity dup sub def > >/pi 3.141592653589793 def >/asin {dup dup mul 1. exch sub sqrt atan} bind readonly def >/acos {dup dup mul 1. exch sub sqrt exch atan} bind readonly def >/tan {dup sin exch cos div} bind readonly def >/ctn {dup cos exch sin div} bind readonly def >/sec {1. exch cos div} bind readonly def >/csc {1. exch sin div} bind readonly def >/cbrt {dup 0. ne {dup abs .333333333333333333 exp 1 index 0. lt {neg} if > dup dup dup add 4 1 roll mul div add 3 div} if} bind readonly def > >/sinh { dup abs //ln2 le % use power series for better precision > { dup dup mul dup 72. div 1. add > 1 index mul 42. div 1. add > 1 index mul 20. div 1. add > mul 6. div 1. add > mul } > { //e 1 index exp exch //e exch neg exp sub .5 mul } > ifelse} bind readonly def >/cosh {//e 1 index exp exch e exch neg exp add .5 mul} bind readonly def >/cosh-1 {dup abs //ln2 le % use power series for better precision > { dup mul dup 56. div 1. add > 1 index mul 30. div 1. add > 1 index mul 12. div 1. add > mul 2. div } > {//e 1 index exp exch e exch neg exp add .5 mul 1. sub} > ifelse} bind readonly def >/tanh {dup sinh exch cosh dup //Infinity lt {div} {pop sgn} ifelse} bind readonly def >/asinh { dup dup mul dup 1. le > { 1. add sqrt div dup ln1+ exch neg ln1+ sub .5 mul } > { dup //Infinity lt > { 1. add sqrt 1 index abs add ln } % s^2 finite > { pop dup abs ln //ln2 add } > ifelse exch 0. lt {neg} if } > ifelse } bind readonly def >/acosh { dup 1. ge { dup dup mul dup //Infinity lt > { 1. sub sqrt add ln} > { pop ln //ln2 add } > ifelse } > {/acosh cvx errordict /rangecheck get exec} > ifelse } bind readonly def >/acosh1+ { dup 0. ge { dup dup 2. add mul dup //Infinity lt > { sqrt add ln1+ } > { pop ln1+ //ln2 add } > ifelse } > {/acosh1+ cvx errordict /rangecheck get exec} > ifelse} bind readonly def >/atanh { dup abs dup 1. lt > { pop dup ln1+ exch neg ln1+ sub .5 mul } > { 1. eq > { //Infinity mul } % +-infinity > { /atanh cvx errordict /rangecheck get exec } % error > ifelse } > ifelse} bind readonly def

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Re: [math-fun] floating point precision and hyperbolic functions
by Michael Speciner 06 Aug '08

06 Aug '08

I forgot that math-fun strips out attachments. So below is what I tried to attach in my last email, possibly with some mailer-inserted line breaks. log1p and expm1 are ln1+ and exp-1 below. (In PostScript natural log is ln and exp takes two args--a base and exponent, adding an extra source of error since e can't be expressed exactly; the PostScript trig functions work in degrees.) Notes: log1p(x) = log(1+x) + (x-((1+x)-1))/(1+x) For |x| <=1, we use asinh(x) = (log(1+x/sqrt(x^2+1)) - log(1-x/sqrt(x^2+1)))/2, which is based on atanh(x) = (log(1+x) - log(1-x))/2; while for larger x, we use asinh(x) = sgn(x) log(|x| + sqrt(x^2+1)) PostScript uses single precision floating point; any McClaurin series are truncated accordingly. /sgn {dup 0 ne {dup dup eq {0 gt {1}{-1} ifelse} if} if} bind def /e 2.718281828459045 def /ln2 .6931471805599453 def /log2 {dup ln //ln2 div dup sgn exch abs floor mul % x roundtozero(log2?(x)) exch 2 2 index exp div ln //ln2 div add } bind readonly def /exp-1 {dup abs //ln2 le {dup 9. div 1. add 1 index mul 8. div 1. add 1 index mul 7. div 1. add 1 index mul 6. div 1. add 1 index mul 5. div 1. add 1 index mul 4. div 1. add 1 index mul 3. div 1. add 1 index mul 2. div 1. add mul} {//e exch exp 1 sub} ifelse} bind readonly def /ln1+ {dup 1. add dup ln 3 1 roll dup 3 1 roll 1. sub sub exch div add} bind def /Infinity 2 1000 exp def /-Infinity Infinity neg def /NaN Infinity dup sub def /pi 3.141592653589793 def /asin {dup dup mul 1. exch sub sqrt atan} bind readonly def /acos {dup dup mul 1. exch sub sqrt exch atan} bind readonly def /tan {dup sin exch cos div} bind readonly def /ctn {dup cos exch sin div} bind readonly def /sec {1. exch cos div} bind readonly def /csc {1. exch sin div} bind readonly def /cbrt {dup 0. ne {dup abs .333333333333333333 exp 1 index 0. lt {neg} if dup dup dup add 4 1 roll mul div add 3 div} if} bind readonly def /sinh { dup abs //ln2 le % use power series for better precision { dup dup mul dup 72. div 1. add 1 index mul 42. div 1. add 1 index mul 20. div 1. add mul 6. div 1. add mul } { //e 1 index exp exch //e exch neg exp sub .5 mul } ifelse} bind readonly def /cosh {//e 1 index exp exch e exch neg exp add .5 mul} bind readonly def /cosh-1 {dup abs //ln2 le % use power series for better precision { dup mul dup 56. div 1. add 1 index mul 30. div 1. add 1 index mul 12. div 1. add mul 2. div } {//e 1 index exp exch e exch neg exp add .5 mul 1. sub} ifelse} bind readonly def /tanh {dup sinh exch cosh dup //Infinity lt {div} {pop sgn} ifelse} bind readonly def /asinh { dup dup mul dup 1. le { 1. add sqrt div dup ln1+ exch neg ln1+ sub .5 mul } { dup //Infinity lt { 1. add sqrt 1 index abs add ln } % s^2 finite { pop dup abs ln //ln2 add } ifelse exch 0. lt {neg} if } ifelse } bind readonly def /acosh { dup 1. ge { dup dup mul dup //Infinity lt { 1. sub sqrt add ln} { pop ln //ln2 add } ifelse } {/acosh cvx errordict /rangecheck get exec} ifelse } bind readonly def /acosh1+ { dup 0. ge { dup dup 2. add mul dup //Infinity lt { sqrt add ln1+ } { pop ln1+ //ln2 add } ifelse } {/acosh1+ cvx errordict /rangecheck get exec} ifelse} bind readonly def /atanh { dup abs dup 1. lt { pop dup ln1+ exch neg ln1+ sub .5 mul } { 1. eq { //Infinity mul } % +-infinity { /atanh cvx errordict /rangecheck get exec } % error ifelse } ifelse} bind readonly def

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Re: [math-fun] rolling a ball along a fractal
by Steve Witham 06 Aug '08

06 Aug '08

Jim Propp originally asked: >Is there a sensible way to say what it means to roll a ball across a plane, >without slipping, so that the point of contact traces out a snowflake curve? Oh, you mean rolling a *ball* *across* a plane!! Like rolling it on a table. Not rolling a circle around the outside of the curve. (Nor rolling a ball over the surface of a sponge.) We are talking about curves rather than dust, right? All the fractal curves I have heard about have details that get smaller as you approximate them more closely. If not, I don't know what you have. >Maybe there's a dichotomy here, and it makes sense to roll a ball along a >fractal curve as long as its fractal dimension is sufficiently low, but not >if the fractal dimension exceeds some cutoff? So the Q is whether rolling the ball along the curve approaches (for small details) stamping it like silly putty on newsprint. Think of rolling the contact point of the ball around a tiny circle over and over. The ball probably slowly precesses. Maybe the precession could "unwind" the snakings of a non-rectifiable curve so that although it converged on the plane it would diverge on the sphere? Or maybe just the rotation of the sphere could diverge even if the distance it rolled would converge. Or maybe they both have to converge no matter what (finite) dimension of the curve. Example: Start with a regular clockwise polygon and replace the sides with clockwise epicycles nearly as big as the original. But, to continue talking about the question Jim *didn't* ask, >Mike Stay wrote: > > > Well, Erin Pearse wrote all over *about* the question: >> http://arxiv.org/abs/math-ph/0412029 >> http://arxiv.org/abs/math.DS/0605527 > > http://www.math.cornell.edu/~erin/dissertation/dissertation.pdf Okay, beginner questions. Looking at the dissertation, I was confused about Pearse's meaning of "convex hull", which I thought was an outline. The outlines of all his examples were simple polygons. For instance, the Koch curve's outline is a triangle, the Koch snoflake's is a hexagon, etc. To me, those are the hulls and there's nothing fractal about those hulls. There are definitely fractals whose outlines are also fractals, he just doesn't use any of them as examples. --Steve

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Re: [math-fun] floating point precision and hyperbolic functions
by Michael Speciner 06 Aug '08

06 Aug '08

Certainly, calculating asinh that way loses precision near 0, because the arg to ln is near 1 and thus loses most of the bits off the right end. In fact, trying to implement a lot of the hyperbolic functions in a straightforward manner has similar problems--there's neither a log1p or expm1 function in PostScript. It turns out that log1p can be straightforwardly implemented from log, using the fact that log(1+x) ~ x for small x. I don't know a similar trick for expm1, but the McClaurin series converges quite fast. I've attached some PS code that defines expm1, log1p, and a set of trig and hyperbolic functions. --ms On Wed, Aug 6, 2008 at 8:21 AM, Steve Witham wrote: > > ...PostScript.... > > The problem is with [lack of] precision. PS starts out with only single > > precision floating point, and then a lot more precision gets thrown away for > values near zero. > > Can you explain the 2nd stmt? I thought PS used pretty > standard single-precision. Do you just mean that calculating > asinh( x ) = ln( x + sqrt( x^2 + 1 ) ) > loses precision? > > Btw, I have a page on fixing asinh(x) for x << -1, > http://www.tiac.net/~sw/2007/02/asinh_perspective > but never figured out how to fix it for around zero. > > --Steve

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Re: [math-fun] rolling a ball along a fractal
by James Propp 05 Aug '08

05 Aug '08

>Here's an easier one: can you roll a hoop with r>=1/6 (so that it can >cross the largest gap) over the Cantor set? With that set, you >necessarily have only single points of contact for the hoop; I assume >that the edge of the hoop "sticks" to the point and the diameter >traces out a circular arc until the hoop hits the next point. No >matter what radius you choose, near the beginning of the trip, the >edge of the hoop is very nearly flat compared to the distance d >between points, and so the hoop will rotate by an amount very close to >2pi*r/d radians. The missing rotation as compared to rolling along a >line (which we can define as the limit as d->0 of "sticky" rolling >over a lattice with spacing d) clearly converges. Let me make sure I understand Mike's idea for the level-one approximation to the Cantor set: you roll the hoop so that the point of contact goes from (0,0) to (1/3,0), then swing the hoop down pivoting on the point (1/3,0) until the hoop touches the Cantor set at both (1/3,0) and (2/3,0) (so that the bottom point of the hoop is at (1/2,s) for some s < 0), then swing the hoop up pivoting on the point (2/3,0) until the hoop's lowest point is (2/3,0), and then roll the hoop so that the point of contact goes from (2/3,0) to (1,0). Have I got that right? It's a cute idea, but I don't see how it applies to the Koch curve problem. >If r is big enough then I claim that rolling the hoop over the Koch >curve is identical to rolling it over the Cantor set (once if you go >across the bottom, twice if you go across the top), since the hoop >can't reach the concave parts of the fractal. Can you elaborate? Jim Propp

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