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Post by schlager7 on Jun 20, 2005 12:00:01 GMT -6
Okay, I'll admit that I saw this in a Fencing Net thread and just HAD to share. Granted, it is out of context... but it was out of context there, too.
We may now present competing equations to determine this.
mine:
N*(rC/(v^2)) + 2Sb/K = 42
where N is # of bouts, r is the radius of the earth, v is the airspeed of an unladen swallow, S is the average curvature of a penguin, b is the number of people in attendance, and K is the air temperature on the kelvin scale.
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Post by Geezer on Jun 22, 2005 14:36:00 GMT -6
Was this the number of transit changes while hitch-hiking through the galaxy?
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Post by schlager7 on Jun 24, 2005 11:58:32 GMT -6
I wonder what the "C" in the equation stands for.
Isn't that supposed to represent the speed of light?
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Post by Geezer on Jun 24, 2005 12:34:39 GMT -6
Ah, Ha! Perhaps it is the Universal Factor for Superfluous Comments. As in: C..."I told you so!"
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Post by fox on Jul 6, 2005 14:56:21 GMT -6
How about
BB=v(-n/G)+W(C-2s)
BB is the probability a fencing blade will break
v is velocity of the fencer wielding the blade
n is the number of hits the fencer still must land on his opponent to win
G is how close his girlfriend is standing to watch measured in centimeters
W is the wobble of the earth's rotation
C is the number of feet his coach will make him cover in footwork drills the next evening if he loses
s represents the number of functional spare weapons the fencer has at the tournament)
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