I just wish the toyos weren't so heavy
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If u got some light weight wheels like the methods I don't see it really posing to much of a problem with the added weight. Compared to other tires. If I remember correctly each 1lb of rotational mass is the equivalent of like only 10 lbs of dry weight. Heck if you really want to do the math here's the formula lol.
"I've heard that 10:1 bit of wisdom before, but I don't think it stands up to scrutiny. It is absolutely true that a pound saved in the wheel or tire is worth more than a pound saved somewhere else, but not a factor of 10 more. I did a "back of the envelope" calculation using basic physics principles, and what I find is that a 1 pound savings in wheel and/or tire weight is equaivalent to about a 1.5 pound savings in the non-rotating weight of the car. What I did was compare the power needed to accelerate a 1 pound mass in a straight line versus the power needed to increase the rotational veocity of a wheel with a 1 pound mass concentrated at the rim an equivalent amount.
Here's the anaysis. To keep the math simple, let's assume the following:
1. The diameter of the wheel is 18", or in other words the radius of the wheel (r) is 9 inches, or 3/4 ft.
2. The radius of the tire (R) is one foot.
3. All the weight savings for a 1 pound reduction in wheel weight is concentrated at the outer edge of the rim - that is, 3/4 ft from the center. This is a bit unrealistic, but is the most favorable assumption with respect to trying to show the advantage of reducing rotating mass.
The instantaneous power needed to linearly accelerate a given mass (m) is given by P(linear) = m*a*v, where a = the acceleration, v is the instantaneous velocity.
The power needed to increase the rotational velocity of a wheel is P(rotation) = I*alpha*omega, where I = rotational moment of inertia of the wheel & tire, alpha is the rotational acceleration (in radians per sec squared), and omega is the instantaneous rotational velocity (in radians/sec). Alpha is related to the linear acceleration of the car by a = alpha*R, and omega is related to the car's velocity by v = omega*R. So the power required for rotational acceleration can be expressed in terms of a and v by: P=Iav/R^3. As for the value of I: for a 1 pound mass concentrated at a distance r from the center of rotation, I = mr^2. Putting it together: P(rotation) = I*alpha*omega = mr^2av/R^3. Since r = 3/4 ft and R = 1 ft, that means you have P(rotation) = 9/16 * mav.
Note that P(rotation) is 9/16 of the amount of P(linear). Now if you can save that 1 pound of mass in the wheel you save on the power needed to both spin the wheel up to speed and to acelerate on down the road, so the total savings in power because of the lighter wheel is (1+9/16)*mav, or just more than one and a half times the power savings if the weight was some place other than the wheel. So while saving a pound in the wheel is definitely more advantageous than saving a pound of weight in the rest of the car, it's no more than about 50% better (not 900% better as claimed).
Of course this says nothing about the other improvements that come about from a ligher unsprung weight - namely faster damping of vibrations and less transfer of energy from the wheel into the car when you hit a bump."
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