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@gmail.com> </div><div>Date:11/21/2014 10:34 PM (GMT-05:00) </div><div>To: Henry Baker <hbaker1@pipeline.com> </div><div>Cc: math-fun <math-fun@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@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

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