At 10:36 AM 4/4/97 -0800, you wrote:
>> a 235/70 tire would be 87.81617260034in, and the circumference of a
>> 245/60 tire would be 83.4872063454in. So, for every revolution, the
>
>Jon - You're guilty of false precision. I imagine temperature alone
>makes thousandths of an inch irrelevant. More importantly, every tire
>manufacturer's tires are different dimensions - even with the same
>P-metric rating on the sidewall. Also, you might want to calculate
>the diameter with half the tread worn off when picking your tires.
>That'll take a bit more work, and will also depend alot on who made them.
>
>Finally, the section width - and overall rolling diameter of a tire -
>is also dependent on the rim width.
>
:) Yep, very true.
I'm just used to using all those digits though. I've seen many
people round off every calculation (and I've tried it myself). With
every new calculation, the answer gets more and more impresise... I'm
actually quite surprised at how much those harmless round-offs can
throw off the final answer. Sure, they sometimes counteract each other,
but I'd rather know the exact answer, so I leave 'em in there until the
final calculation which I round off.
Once I know the final, precise answer, I could take into effect stuff
like tread depth, rim width, etc... Of course, given that the "real
world" introduces so many variables, our calculations will probably be
off by about the same amount anyway, whichever way we do them... :)
Hey, I was on a mathematical high. ;)
-Jon-
Jon Steiger - Network Administrator for Academic Information Technology
.- steiger@ait.fredonia.edu -- http://www.cs.fredonia.edu/~stei0302/ -.
| DoD# 1038, EAA# 518210, NMA# 117376, USUA# A46209, KotWitDoDFAQ |
| '96 Dodge Dakota SLT V8, '96 Kolb FireFly 447, '91 Yamaha FZR600R |
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I do not speak for SUNY College at Fredonia; any opinions are my own.
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