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2002-05-06 - 11:49 a.m.

Assume a spherical woodchuck
Yeah, the woodchuck is easy. But how do you integrate a hedgehog? Kymber asks from across the room Saturday night.
Stokes Theorem? I say.
Hmmm. That could work.
Lo many years ago Nia made the mistake of signing an email to me with the quote How much wood would a woodchuck chuck if a woodchuck could chuck wood? (which in Latin is Quantum materiae materietur marmota monax si marmota monax materiam possit materiari?)
So I wrote her a proof because Nia hates it when I do stuff like this. But, she did ask, after all.
Assume a spherical woodchuck with a diameter equal to the average length of a woodchuck.
This can be used to calculate the maximum value of STOMACH, a variable defining the amount of wood in the woodchuck's capacity, storage capacity, where the maximum value is STOMACH = 4/3*pi*(diameter/2)^2.
Assume a known rate of ingress, represented by the variable EAT.
Assume a know rate of egress, represented by the variable, POOP.
Assume the woodchuck begins with any empty STOMACH.
Assume the woodchuck can eat the entire time it is awake, a value known from wildlife observations by people like Giselle to be DAY. Giselle is introducted in another long story about a pink duck and a gay man wearing only a towel.
Then, the amount of a wood a woodchuck could chuck would be the integral of integrand (EAT-POOP), integrated from DAWN (time zero when his stomach is empty) to DAY (length of time he is awake), with the limitation, should the integral value exceed Max(STOMACH), the wood chuck will explode.
Watch out they spatter.
I'm sure a bit of web searching would yield the field observations data necessary to calculate an actual value should it be that important to you.
You did ask.
You can imagine the reply from Nia.
Crown trip was wonderous and much will have to be written later.
Yawn. Much later.
Must concentrate on staying awake through work.
Damn Um-may.

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