**FIRST EMAIL**/FRC Game Hint #2

[dlavery]

Originally Posted by billbo911

I see your point and must acknowledge, in fact, my equation was in error.

I had made an assumption, for which I must apologize, that the weather conditions on that October day in 1961 (class of 1962) would have been quite warm. You see, where I live on the west coast, October afternoons can easily be in the mid 80's to upper 90's. I chose to use the "Smoot = (1/(364.4-ear))*(Harvard Bridge length)" equation assuming there would have been quite a bit of expansion of the bridge due to the temperature. Not knowing along which axis the ear was measured, I assumed,(there I go again), it's length. Approximating a 5' 7" tall man's ear to be 2.25", I figured this would be close enough to account for the assumed expansion. I then subtracted this amount, incorrectly I might add, to account for the assumed expansion. When in fact, instead of subtracting it from the divisor, it should have been subtracted from the product to yield a length that had been corrected for temperature. In other words, if my assumption was correct, the equation should have been Smoot = ((1/364.4)*(Harvard Bridge length)) - ear.

Now, based on your need for accuracy and tolerance, the equation would be:
Smoot = (1/(364.4 +/- ear))*(Harvard Bridge length), correct?

Ahhh, I think we may have identified the source of the confusion. You had assumed that the length of "an ear" in the original measurement was a 2.25 inches, the size of a five-foot-seven-inch-tall man's ear. However, in this particular case, the original lore reveals that the "ear" is allegedly the width of the ear hole in the football helmet that Smoot was while lying prone upon the bridge. So, an approximation of 1.0-1.5 inches may be more correct. This interpretation may be further supported when considering the effects of the local weather at the time the measurement was made. A football helmet would be a suitable form of protection against inclement weather during such an excursion, and makes for a plausible rationale for the difference in the tolerance coefficient.

There, have I smeared enough mud on this to cover my error???

I dunno about mud, but we have both smeared a lot of SOMETHING on this topic!

-dave


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