November 2014 - mathfun@mailman.xmission.com

Re: [math-fun] breaking spaghetti
by Bill Gosper 26 Nov '14

26 Nov '14

MStay> I wrote Destin at "Smarter Every Day" about the problem and he responded that he was already working on an episode about it; he has completed filming and will publish it in December. He has previously used a camera fast enough to catch the shattering of a prince rupert's drop, so he's up to the challenge. (https://www.youtube.com/watch?v=xe-f4gokRBs) ------ These were demonstrated in MIT freshman Chem, along with their larger and more impressive brethren, which I misrecall as Prince Andrew's drops (dele). Thick-walled and hollow. You whacked them with hammer-- nothing. You wrapped them in a towel and dropped in a grain of carborundum--bam! What's their damn name? --rwg

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Re: [math-fun] breaking spaghetti
by Hugh Everett 26 Nov '14

26 Nov '14

It occurs to me that the speed of bending may result is varying numbers of breaks. I imagine the breaking stress being nearly uniform over the noodle. When you continuously increase the stress it could occur that the duration of failure is long enough for other points to also fail before enough stress is released to stabilize the remaining noodle sections.

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Re: [math-fun] breaking spaghetti
by Bill Gosper 25 Nov '14

25 Nov '14

ACW>If I were researching this seriously, I would advance on two fronts. First, I would build a robotic spaghetti-bender so that the forces were more reproducible between trials. There are a good fistful of variables in the way you put stress on the noodle: each hand exerts a translational force on the piece of noodle it's gripping, as well as a torque; the robotic jig ought to provide controls for at least some of those variables. Second, it is clear to me that a thousand frames per second is at least an order of magnitude too slow. Everything interesting to this problem happens between two adjacent frames: in one frame, the noodle is in one piece, and in the next, three breaks have occurred and two pieces are spinning away wildly into the air. We need to resolve the breaks into distinct frames to see what order they happen in; it seems likely that rebound from early breaks causes the late ones. ----------- Exactly. I withdraw my suggestion that it's a recoil effect. I think Nick Horn's MIT paper just covers varying the clamped endpoint angles and their relative positions. Presumably, one combination of these (for a given arc length) produces Warren's circular arc. But pages I've found seem to be saying (unclearly) that simply shoving the unclamped endpoints together (as with the wafer or Warren's strip) produces, for the strip, deflection described by a polynomial cubic in the absolute distance from the middle abscissa. --rwg I believe the Lehigh incident in my brother's inadvertent CC refers to our 1961 (pre-college) campus tour. Lehigh's pride and joy was a huge 5000000lbf destructive testing machine for, e.g., life-size bridge supports. For our amusement, it squashed lengthwise a wooden column that looked a foot square in cross-section.

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[math-fun] Yumi, yumi, yumi, I've got spaghettos in my tumi
by Henry Baker 24 Nov '14

24 Nov '14

Japanese warriors have used asymmetrical bows ("yumi") since perhaps the 3rd C. AD. These asymmetrical bows were typically 1:2, because these bows wanted to be very long (e.g., Robin Hood's band of merry men) , but Japanese warriors were too short (the armor I've seen seems to indicate 5'0" warriors). "the grip (nigiri) has to be positioned at about two thirds of the distance from the upper tip" http://en.wikipedia.org/wiki/Yumi So perhaps the ancient Japanese also discovered the trick of breaking spaghettos into 3 parts... If you're in the L.A. area anytime in the next few months, be sure to see the terrific showing of Japanese armor (including yumi bows) showing at LACMA -- The Los Angeles County Museum of Art.

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[math-fun] "Imitation Game" slanders Turing
by Henry Baker 24 Nov '14

24 Nov '14

FYI -- I hope this post isn't too political. "nobody Â until film-makers came along Â accused [Turing] of being a traitor" The blatantly false story line in this film really is a gross insult to Turing's memory & legacy; given the strict libel laws in England, I'm surprised that this film can even be shown there without getting sued. http://www.theguardian.com/film/2014/nov/20/the-imitation-game-invents-new-…

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[math-fun] Power series
by David Wilson 23 Nov '14

23 Nov '14

Let f(0) = a f'(0) = b (x^4-1)*f(x^4) + (x+1)*f(x^2) - x*f(x) = 0 Find the power series for f.

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Re: [math-fun] Isn't it odd that when you flex a circular wafer,
by drgoose007 22 Nov '14

22 Nov '14

I vaguely remember looking at Nick's "Materials' Span Book" back in '75--had to know some pretty hairy stuff to pass the 3 day test to qualify for FSGC (Florida State General Contractor). . I thought Blade was a freakin genius, until found out it was an open book test. . but still - - for a non mathemation on 2sixpacks/day admiral feat.. certainly your stuff much more hairy. ¿ Remember Lehigh ? Sent from my Verizon Wireless 4G LTE Smartphone <div>-------- Original message --------</div><div>From: Bill Gosper <billgosper(a)gmail.com> </div><div>Date:11/21/2014 10:34 PM (GMT-05:00) </div><div>To: Henry Baker <hbaker1(a)pipeline.com> </div><div>Cc: math-fun <math-fun(a)mailman.xmission.com> </div><div>Subject: Re: [math-fun] Isn't it odd that when you flex a circular wafer, </div><div> </div> On Fri, Nov 21, 2014 at 5:35 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: > It would seem that you have stiffness competing with a long lever-arm. > I think it's hairier than that. I think Alan Adler told me that stiffness varies with the cube of thickness, while breaking strength varies as the square, and I here suggested that the ubiquity of surveillance cameras might detect sagging tree branches and predict their failure. (Assuming the typical failure mode where the branch first cracks internally lengthwise, simulating two half-thickness beams.) > > It should either break right next to your fingers or right in the middle > (by symmetry). > > With different shapes, you could get it to break next to your fingers > instead of in the middle. > > For example, if you held a diamond (regular rhombus) shape, it should > break right next to your fingers. > What of an elongated rhombus? > > You need to have a shape that spreads the load so that it collapses > somewhere else other than by your fingers. > > But if the shape is too elongated -- e.g., a long ellipse held on the > major axis -- then it won't break next to your fingers, but also won't > break anywhere else -- it'll simply bend into nearly a circular shape. > > So you want a quasi-elliptical shape whose major axis isn't too much > larger than its minor axis. In this way the forces will be spread out in > such a way that it is forced to break in the middle. > > At least for elliptical shapes, it should be possible to compute the > optimal eccentricity using standard undergraduate calculus. > > If you wanted to break it asymmetrically, the shape itself would have to > be asymmetrical. > > This seems like a problem suitable for the calculus of variations. > Whereas Hilarie's spaghetto invites catastrophe theory? I haven't researched it, but suspect the breaks are sequential, the 2nd when the longer remnant snaps forward beyond straightness. --rwg > > At 02:30 PM 11/21/2014, Bill Gosper wrote: > >it fails along a diameter--the widest point? Presumably, sufficient > elongation > >parallel to that diameter will bifurcate the line of probable failure. > E.g., is there a > >shape that half the time breaks 1:2 and half 2:1? Convex? Could it be > an ellipse? > >Is there a shape where the crack might be uniformly anywhere? > >--rwg > > -- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe(a)googlegroups.com. For more options, visit https://groups.google.com/d/optout.

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[math-fun] Isn't it odd that when you flex a circular wafer,
by Bill Gosper 22 Nov '14

22 Nov '14

it fails along a diameter--the widest point? Presumably, sufficient elongation parallel to that diameter will bifurcate the line of probable failure. E.g., is there a shape that half the time breaks 1:2 and half 2:1? Convex? Could it be an ellipse? Is there a shape where the crack might be uniformly anywhere? --rwg

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Re: [math-fun] Isn't it odd that when you flex a circular wafer
by Warren D Smith 22 Nov '14

22 Nov '14

The curve of the bent wafer will be the one which minimizes integral curvature(s)^2 * width(s) ds i.e. total elastic energy. Here s is arclength along curve. This is subject to the constraint that the 2 endpoints of the curve must lie some fixed distance apart. I think an effectively equivalent constraint is that integral curvature(s) ds = fixed constant. You can now write down Euler-Lagrange equation, solve it, etc. to find the shape of the bending curve. The location of breakage will be the point where |curvature| is maximized. The "optimum design" width(s) curve will be one which uniformizes curvature, therefore, and will be simply width=constant, i.e. a strip; then the bending curve will be an arc of a circle. [ http://people.csail.mit.edu/bkph/AIM/AIM-612-OCR.pdf considers a curve with given endpoints and orientations there, and variable arclength. Not the problem we want. ] Re the claim spaghetti always breaks into 3 pieces, this seems inexplicable with above model. So need a better model. I think it must have to do with some kind of "nonlinear and/or time-dependent elasticity" property of spaghetti. i.e. crack formation. (Above model had been linear, i.e. energy is integral of quadratic.) Possibly dynamical effects involved too. Suggest seeking insight via a high speed video camera.

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Re: [math-fun] Isn't it odd that when you flex a circular wafer,
by Henry Baker 22 Nov '14

22 Nov '14

It would seem that you have stiffness competing with a long lever-arm. It should either break right next to your fingers or right in the middle (by symmetry). With different shapes, you could get it to break next to your fingers instead of in the middle. For example, if you held a diamond (regular rhombus) shape, it should break right next to your fingers. You need to have a shape that spreads the load so that it collapses somewhere else other than by your fingers. But if the shape is too elongated -- e.g., a long ellipse held on the major axis -- then it won't break next to your fingers, but also won't break anywhere else -- it'll simply bend into nearly a circular shape. So you want a quasi-elliptical shape whose major axis isn't too much larger than its minor axis. In this way the forces will be spread out in such a way that it is forced to break in the middle. At least for elliptical shapes, it should be possible to compute the optimal eccentricity using standard undergraduate calculus. If you wanted to break it asymmetrically, the shape itself would have to be asymmetrical. This seems like a problem suitable for the calculus of variations. At 02:30 PM 11/21/2014, Bill Gosper wrote: >it fails along a diameter--the widest point? Presumably, sufficient elongation >parallel to that diameter will bifurcate the line of probable failure. E.g., is there a >shape that half the time breaks 1:2 and half 2:1? Convex? Could it be an ellipse? >Is there a shape where the crack might be uniformly anywhere? >--rwg

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