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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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