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[math-fun] Re: Simplest Ovals
by Henry Baker 27 Feb '07

27 Feb '07

After doing some plotting with Maxima/gnuplot on some ovals produced from parabolae, I concluded that the ovals produced by parabolae from isosceles triangles were a little bumpy for my taste. The parabolic top lobe looks pretty good, but the two bottom lobes are too pointy. There appear to be no degrees of freedom to adjust things. Rules: 1. The apex of each of the three parabolae goes through one of the vertices. 2. At the points of intersection of the parabolae (in pairs), the slopes also match. This is kind of cool, as it appears that each triangle has a unique circumparabolic path. If one were to make an amusement park ride based on this path, and trace it appropriately, one would successively undergo uniform acceleration in three different directions, with a discontinuous change in acceleration in between. I wonder if there are rides in amusement parks with this sort of characteristic. I did find after a Google search that mechanical cams are sometimes shaped to impart uniform acceleration at various portions of their turning. At 08:52 PM 2/25/2007, Henry Baker wrote: >I did a little more research on this topic & found that the answer appears to be very simple. > >Consider a moving particle going around the outside of a triangle as fast as it can. It obviously has to change direction in order to complete a circuit of the triangle -- the simplest way to do this is obviously the circumcircle. However, going around the circumcircle may take longer than utilizing this same potential for acceleration & deceleration to navigate closer to the triangle. For example, if the triangle is very long and thin -- approximating a line segment -- then the fastest way to circumnavigate it would be to utilize all of the acceleration to gain as much speed as possible up to the midpoint, and then to utilize all of the acceleration to slow down towards the end. The reason is to avoid the spike in acceleration that would be required to turn the 180 degree corner at the end (we have limited the maximum magnitude of acceleration for the entire round trip). > >It would appear that the fastest circuits around triangles utilize pieces of _parabolas_ to get around the triangle faster -- three pieces to be exact. During each of the three parabolic pieces, the entire ability to accelerate is used to create a uniform acceleration field much like that studied in high school physics by the acceleration due to gravity, except that the uniform acceleration is in the plane of the triangle, and used to slow down the particle just enough to clear each vertex before accelerating to the next. > >One way to think about this curve is to follow Hamilton and consider its "hodograph" -- take all of the _velocity_ vectors and bunch them at the origin. Then the hodograph for the fastest curve can be constructed by taking each side of the triangle and considering it as a vector in the direction that the particle will "trace" it (i.e., stay "close" to it); place each of these three vectors at the origin. Then the acceleration vectors for the fastest curve will trace out the perimeter of the triangle formed by these three vectors (suitably scaled). Since the _magnitude_ of the acceleration is at all times the same, the perimeter will be traced out at constant speed (we aren't charged for the change in direction of the acceleration at the corners of this derived triangle). Around the original triangle, however, the particle will speed up & slow down as it goes around the parabolic path pieces. > >(The intuitive idea is that the longer sides of the triangle give the particle more distance to speed up and more distance to slow down, so the particle can achieve a higher velocity "along" those sides.) > >If the original triangle is isosceles, then the overall shape of the 3 parabolic pieces will look egg-shaped. > >I haven't yet worked out what triangle shape will produce a pleasing egg shape. > >If someone already has a particle simulation applet, then perhaps these data could be put into it. > >At 08:09 AM 2/22/2007, Henry Baker wrote: >>In bicycle/car races, the vehicles have a maximum acceleration potential. For simplicity's sake, let's assume that the max acceleration in any direction is capped at a certain constant. When going around corners, these vehicles "go wide", so that they don't go beyond their maximum acceleration when turning the corner. On the other hand, their ability to "go wide" is limited by the width of the road/track. >> >>So, one way to produce an egg-shape is to consider a racing track of a certain width, whose overall shape is in the form of an isosceles triangle. If the triangle is equilateral, and the width is wide enough, then the curve is circular. However, when we start reducing the width, we get a curve which starts getting pinched in 3 places. If we then lengthen the altitude of the triangle so that it is no longer equilateral, then the curve starts looking egg-shaped. >> >>I assume that curves such as these have been studied. Does anyone know what their equations are? Can anyone point me to any references?

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[math-fun] More oval pix up
by Steve Witham 26 Feb '07

26 Feb '07

Fred supplied pictures of ovals of Greg Fee's and Bill Gosper's, which I've added to http://www.tiac.net/~sw/2007/02/Steve_Gray_oval --Steve

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[math-fun] Re: Simplest Ovals
by Henry Baker 25 Feb '07

25 Feb '07

I did a little more research on this topic & found that the answer appears to be very simple. Consider a moving particle going around the outside of a triangle as fast as it can. It obviously has to change direction in order to complete a circuit of the triangle -- the simplest way to do this is obviously the circumcircle. However, going around the circumcircle may take longer than utilizing this same potential for acceleration & deceleration to navigate closer to the triangle. For example, if the triangle is very long and thin -- approximating a line segment -- then the fastest way to circumnavigate it would be to utilize all of the acceleration to gain as much speed as possible up to the midpoint, and then to utilize all of the acceleration to slow down towards the end. The reason is to avoid the spike in acceleration that would be required to turn the 180 degree corner at the end (we have limited the maximum magnitude of acceleration for the entire round trip). It would appear that the fastest circuits around triangles utilize pieces of _parabolas_ to get around the triangle faster -- three pieces to be exact. During each of the three parabolic pieces, the entire ability to accelerate is used to create a uniform acceleration field much like that studied in high school physics by the acceleration due to gravity, except that the uniform acceleration is in the plane of the triangle, and used to slow down the particle just enough to clear each vertex before accelerating to the next. One way to think about this curve is to follow Hamilton and consider its "hodograph" -- take all of the _velocity_ vectors and bunch them at the origin. Then the hodograph for the fastest curve can be constructed by taking each side of the triangle and considering it as a vector in the direction that the particle will "trace" it (i.e., stay "close" to it); place each of these three vectors at the origin. Then the acceleration vectors for the fastest curve will trace out the perimeter of the triangle formed by these three vectors (suitably scaled). Since the _magnitude_ of the acceleration is at all times the same, the perimeter will be traced out at constant speed (we aren't charged for the change in direction of the acceleration at the corners of this derived triangle). Around the original triangle, however, the particle will speed up & slow down as it goes around the parabolic path pieces. (The intuitive idea is that the longer sides of the triangle give the particle more distance to speed up and more distance to slow down, so the particle can achieve a higher velocity "along" those sides.) If the original triangle is isosceles, then the overall shape of the 3 parabolic pieces will look egg-shaped. I haven't yet worked out what triangle shape will produce a pleasing egg shape. If someone already has a particle simulation applet, then perhaps these data could be put into it. At 08:09 AM 2/22/2007, Henry Baker wrote: >In bicycle/car races, the vehicles have a maximum acceleration potential. For simplicity's sake, let's assume that the max acceleration in any direction is capped at a certain constant. When going around corners, these vehicles "go wide", so that they don't go beyond their maximum acceleration when turning the corner. On the other hand, their ability to "go wide" is limited by the width of the road/track. > >So, one way to produce an egg-shape is to consider a racing track of a certain width, whose overall shape is in the form of an isosceles triangle. If the triangle is equilateral, and the width is wide enough, then the curve is circular. However, when we start reducing the width, we get a curve which starts getting pinched in 3 places. If we then lengthen the altitude of the triangle so that it is no longer equilateral, then the curve starts looking egg-shaped. > >I assume that curves such as these have been studied. Does anyone know what their equations are? Can anyone point me to any references?

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[math-fun] Taylor series "inverter"?
by Steve Witham 25 Feb '07

25 Feb '07

Is there a machine out there, analogous to Plouffe's Inverter ( http://pi.lacim.uqam.ca/eng/ ), or the sequences database ( http://www.research.att.com/~njas/sequences/ ), that can take the first n coefficients of a Taylor series and find a function close to that Taylor series? Better yet (of course) take *any* expression for a function and return the simplest expression of the function based on well- known functions. --Steve p.s. I am behind in my egg-collecting.

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[math-fun] Re: Interesting ovoids
by dean＠math.ucdavis.edu 24 Feb '07

24 Feb '07

Franklin T. Adams-Watters wrote: > I asked, what happens if the 1 in that equation is replaced by some > other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us > period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, > x/y, and then loops. ... > When you graph a(n) vs. a(n+1), with suitable parameters the result is > an ovoid. ... > The obvious next step is to find a non-trivial function f_c which > satisfies the functional equation f_c(x,y) = f_c(y,(y+c)/x). ... > This is where I got stuck - I wasn't able to find such an f_c. Maybe > somebody with better analytic skills than I have will be able to. Let f_c(x,y) = (c + (c+1)(x+y) + x^2 + y^2 + x^2 y + x y^2)/(x y). I wish I could claim that I used great analytic skills, but all I did was look at some special cases, assume that the curves satisfied polynomial equations of low degree, and use Mathematica to solve for the coefficients. After a few cases, the pattern was obvious, and led to the invariant f_c. I did notice that this recurrence is related to a 7-th order Somos recurrence: Define s(0), ..., s(4) arbitrarily and then define s(n) for n>=5 so that a(n) = s(n) s(n+5) / (s(n+2) s(n+3)). Then s satisfies the Somos recurrence s(n) s(n-7) = s(n-1) s(n-6) + c s(n-3) s(n-4). I'm sure there must be an easy derivation of f_c from known facts about Somos sequences, but I couldn't find one. Both Franklin's recurrence and Somos's have been studied by many people, so this is probably all well-known to someone. David W. Cantrell wrote: > Here's an example. Suppose a(1) = 1 and a(2) = 2 and c = 1/2. The > sequence itself is not periodic. The continuous function, of which the > sequence is a restriction, has a period of roughly 5.248; a graph of > one cycle is shown at > <http://img453.imageshack.us/img453/744/recursioncu0.gif>. > Its maximum and minimum are roughly 2.549 and 0.741, resp. The max and min are roots of 4x^4 - 80x^3 + 228x^2 - 128x + 1 = 0 Numerically, they're about 2.549193378506151504365717868828 and 0.741012926112203188665166488474 Dean Hickerson dean(a)math.ucdavis.edu

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[math-fun] Picture of Steve Gray's oval
by Steve Witham 23 Feb '07

23 Feb '07

Fred Lunnon sent me a picture of an oval based on Steve Gray's suggestion. >What about the locus of points such that d(A)+kd(B) is constant, I pulled out a bunch of quotes from the list to put it in context: http://www.tiac.net/~sw/2007/02/Steve_Gray_oval/index.html --Steve

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Re: [math-fun] Simplest Ovals (WAS: sections of quadratic surfaces)
by Daniel Asimov 23 Feb '07

23 Feb '07

There ought to be a simple criterion under which there exists a unique "best" simple closed curve in the plane. E.g.g., suppose 1) The curve has bilateral symmetry; 2) the curvature attains two local maxima with values k = M and k = m; 3) the interior of the curve has a fixed area A. Then I'd think there might be a unique "optimal" such curve having minimum length (or equally, fix the length and maximize the area), and that the unique optimal curve would automatically be analytic. (If so, then quite possibly condition 1) is redundant.) --Dan

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Re: [math-fun] Simplest Ovals (WAS: sections of quadratic surfaces)
by Daniel Asimov 23 Feb '07

23 Feb '07

I plotted this with equal scale using the Mac utility "Grapher", and from the plot it's clear that this closed curve is strictly convex everywhere. It does have arcs (near the maximum of curvature) where the curvature comes very close to 0. NOTE: There is a converse to the four-vertex theorem (which says that the curvature function k(s) (s=arclength) of a C^2 simple closed planar curve must have a least two local maxima and two local minima). The converse says that for any continuous real-valued function on the circle with at least two local maxima and two local minima is the curvature function of some simple closed curve in the plane. (Cf. < http://www.ams.org/notices/200702/fea-gluck.pdf >.) We want our egg to be strictly convex, bilaterally symmetric, and as simple as possible. Hence we'd like the curvature function to be everywhere positive, satisfy k(t) = k(-t) (0 <= t <= 2pi), and have the fewest possible critical points: two local maxes and two local mins. One simple way to obtain such a function is via k(t) = (cos(t) + 1/2)^2 + 1/2 (or more generally, k(t) = (cos(t) + a)^2 + b^2 where 0 < a < 1 and b > 0). I'd like to see simple closed curves with curvature functions like these. --Dan -------------------------------- RWG wrote: << >> Cheap and smooth: r=(5+cos(t))^2+(3+cos(2*t))^2 . WFL>Looks more like a squash or pear to me Whoa, nonconvex even? It looks quite Mossy when I plot it with Macsyma, equalscale:true, but I don't know how to get equalscale in Mma, which gives me various avocados. (Are there any birds with nonconvex eggs?) >--- must be some funny-shaped birds in your part of the world, Bill! Amen. I recently rounded a corner on which calmly browsed an East African crested crane, perhaps in town to see "The Last King of Scotland." Or maybe it just wanted some action with an American whoopee crane. >> P.S. Fred, they're called *whooping* cranes, not whoopee cranes.

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[math-fun] Interesting ovoids
by franktaw＠netscape.net 22 Feb '07

22 Feb '07

This discussion of ovals reminds me of a problem I was working on a while ago, which I gave up since I was not getting anywhere. The problem starts with the remarkable observation that if we take a(n+1) = (a(n)+1)/a(n-1), the resulting sequence is periodic with period five. (I saw this in one of Martin Gardner's columns in Scientific American; I don't know what issue at this point. He puts it into the mouth of the fictional Dr. Matrix, but I think he gave a reference for it later. It probably also appeared in one of his books.) (This excludes cases where a zero appears in the sequence. It is sufficient to take a(1) and a(2) both positive.) I asked, what happens if the 1 in that equation is replaced by some other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, x/y, and then loops. In general, the sequence exhibits quasi-periodic behavior. It goes up and down regularly, with the average distance between peaks being some real number. This quasi-period is between 4 and 5 for c>1, and between 5 and 6 for 0 < c < 1; however in general the quasi-period depends on the initial values as well as on c. (Of course, if the quasi-period is rational, the sequence is actually periodic, with period equal to the numerator of the quasi-period.) Note that such a sequence has a fixed point u_c where u_c = (1+sqrt(4c+1))/2. (So in particular, u_1 is the golden ratio, phi, aka tau.) When you graph a(n) vs. a(n+1), with suitable parameters the result is an ovoid. If a(1), a(2) is close to u_c,u_c you get an oval, approaching a circle as you approach u_c,u_c. When the starting values are far from this point, the graph approximates an isoceles right triangle, with the right angle near the origin. The obvious next step is to find a non-trivial function f_c which satisfies the functional equation f_c(x,y) = f_c(y,(y+c)/x). The function must obviously have a minimum or maximum at u_c,u_c. Then the ovals should be the curves defined by f_c(x,y) = k. This is where I got stuck - I wasn't able to find such an f_c. Maybe somebody with better analytic skills than I have will be able to. Regardless, have a look at some of these curves. Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.

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[math-fun] Re: sections of quadratic surfaces
by James Propp 22 Feb '07

22 Feb '07

Dan Asimov wrote: >If we replace R^3 by its "projective compactification" by adding the P^2 of >the "lines at infinity", the result is P^3, projective 3-space. >... >Then the conic sections become the intersection of one of these conoids with >an "affine projective plane" in P^3. All the nondegenerate intersections are >equivalent to ellipses, but the segmentlike degenerate ellipse at issue becomes >a circle. (At least it's compact.) This is a nice point of view. Thanks, Dan! Jim Propp

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