Re: DYNO RESULTS -- STOCK vs AFTERMARKET WHEELS(May hurt your brain..)

["Jason" <jason87vw@nac.net>]

Uuuuhhhh! Yeah, what he said...

-----Original Message-----
From: Steve H <kracker@UDel.Edu>
To: scirocco-l@scirocco.org <scirocco-l@scirocco.org>; 16v Jason
<jason@scirocco.org>
Date: Wednesday, April 19, 2000 12:06 PM
Subject: Re: DYNO RESULTS -- STOCK vs AFTERMARKET WHEELS(May hurt your
brain..)


>Well, you wanted to know this, so here it goes...
>
>I may as well give you all of the info for just about anything you may want
>to know 'scientifically speaking about tires/wheels...
>
>To find the amount of force that is being generated in your wheel and or
>tire at a given speed (Tension)
>
>a (static) = [(M(weight(tire)) + M(cylinder(wheel)))/(M(cylinder)](g)
>
>To find out actual speed from any given RPM (Note: you must calculate or
>your gear ratio's...)
>
>v = 2piRN ; where N is the RPM's, and the uncertainty of R is negligent.
>
>To see how much energy you are getting from your wheels (When at speed, and
>you 'glide' in Neutral)
>
>I = Sigma [(Mi) (R^2)i].....Now this presents a problem, because we are
>taking the sum of all of the weight of each metallic particle in the wheel
>(Needed because the outside of the rim spins faster than the inside) along
>with the sum of the rotations...This can be simplified if we just calculate
>for the tire using :
>
>I=.5m((R1)^2+(R2)^2), where R1  the radius from the center of the wheel to
>where the rim meets the tire, and R2 is the radius from the center of the
>wheel to the outside tread.
>
>Now, to answer your question, to find the force needed to accelerate your
>tire is basic : F=ma
>And the Torque required is t=Ft r = (m)(a)(r)...
>Now we can throw them all together to come up with a plausible
>solution...Looking up above, there is the EQ for the rotational Inertia, so
>taking that answer and applying it to t=m [(a)(r)]r, we can get the total
>amount of torque acting upon your tire....
>
>****Notice : Here's the bad part...All of these numbers are theoretical,
and
>most are near impossible to obtain without the proper equipment...The major
>downfall is the great number of variables here...For example, weight of the
>different sides, the tire expands the faster its turned (therefor, R
>changes), the force of friction between the tire and the surface, the
amount
>of power lost (thus created into heat)...ect, ect, ect, ect.....So
>basically, for real basic approach, the F=MA would be your best bet....If
>you use it make sure you are in Kg, m/s^2, and your force answer will be in
>Newton's...
>
>Have Fun,
>  Steve
>
>
>--
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