Wheel Weight

Wheel weight is critical for a variety of reasons. First, weight is weight, and any weight anywhere on the car will make it harder to accelerate, harder to turn and harder to stop. Second, when accelerating the engine not only has to accelerate the car, it also has to spin the wheels, so the moment of inertia- the rotational equivalent of weight- of all rotating parts also affects how quickly the car accelerates. Third, the moment of inertia also affects braking ability, especially with ABS, where the ability of the wheels to lock, and then quickly begin rotating again several times a second is critical. Finally, wheels are unsprung weight, meaning that they are not supported by the suspension and there fore must follow every bump and crack in the road. More unsprung weight makes it more difficult to keep the tire in contact with the ground, and more difficult to keep ride comfortable.
The only point that the following discussion addresses, though, is the moment of inertia, and how much effect that has on the acceleration and deceleration of the car. The same torque that accelerates the car also has to spin all four wheels, so the effective weight of the wheel works out to be the actual weight, plus whatever the effect of the rotational inertia is. The easiest way to figure out how much the rotational inertia adds is to figure out the kinetic energy of the wheel. When you accelerate any mass from a standstill to a certain speed, the kinetic energy will be:
kinetic energy= 1/2 M(VV)
M is the object’s mass and V is the velocity. A rotating object has kinetic enery too, and that can be determined with this formula:
kinetic energy from rotation=1/2I(rpm x rpm)
I is the moment of inertia, and rpm is, well, how fast the object is spinning.
What we want to do with these formulas is to back up that old rule of thumb about a pound of rotating weight being equivalent to two, three, or even four pounds of non-rotating weight. The total equivalent would be related to the total kinetic energy. To find out how much rotational inertia would be equivalent to a pound of weight, we just have to look at kinetic energy.
To relate kinetic energy from rotating, we first have to find the relationship between speed and rpm. For a wheel, this relationship is simple. For every revolution, the wheel travels a distance equal to the total circumference of the tire, so the linear kinetic energy can be expressed in terms of rpm, like so:
Kinetic energy= ½ M[(rpm x circumference)(rpm x circumference)]
So, the easiest way to prove our rule of thumb is to answer the following question: If a 1-lb object moving along at a given speed has a certain kinetic energy, how much rotational inertia would an object need to have to obtain the same kinetic energy? The answer comes by simply equating the two forms of kinetic energy like this:
1/2 I(rpm x rpm)= ½ M[(rpm x circumference)(rpm x circumference)]
and solving the equation for M.
Remember in high school when you used to always complain about how this algebra stuff was never going to be useful in real life? Remember how solving for M didn’t seem to have anything to do with making pizza at Dominos every Saturday night? Little did you know solving for M would be critical when you went to buy wheels for your new ride. Don’t you feel stupid now?… You should call your algebra teacher and apologize! What… you still don’t want to do the mat? Fine… I’ll do it for you. The answer is….
M=I/(rolling circumference)squared
There, I did it! Just plug in I and the rolling circumference and you have it all figured out… easy rite?
Oh yeah, how do you determine the rotation inertia of a wheel? Rotational inertia depends on shape as well as on weight, and for a shape as complex as a wheel, you can’t really do it without elaborate and sensitive equipment or a detailed computer model of the wheel, or an embarrassingly simple wheel. That was another part of the grand plan that fell apart with all this math… an elaborate rotational inertia measuring device to have its debut with the next wheel guide.
Before going to all that effort, lets test things out with a simple model of a wheel. Let’s say we have a three-piece wheel where the center is a solid, flat disk and the rim sections form a simple cylindrical shape. These shapes are simple enough that we can calculate their rotational inertia with simple formulas as long as we know what each individual part weighs. All we have to do is use these formulas:
I(cylinder)= M(RR) I(disk)= ½ M(RR)
where M is the mass, and R is the radius of the part. To get a good estimate of the individual component weights to plug into these formulas, I did some research. Now assume that we are using a 17x8-inch wheel. For a 17-inch wheel, the center section is actually only about 15.5 inches in diameter. (r is 7.75 inches), and one of the center sections should weigh bout 8 lbs.
Now, a brief glance at the actual center section of a wheel will show you that most of the mass is centered around the hub, and as the spokes go out, they typically get thinner and thinner.
For the rim section, remember that most of the rim material is at least an inch in from the edge of the wheel, since the wheel’s diameter is measured all the way at the edge of he wheel. Let’s say that the rim section of a 17 inch wheel can be approximated by a cylinder that is more like 16 inches in diameter (R is 8). Including the bolts to hold the wheel together, such a rim would weigh about 10 lbs. Plugging these numbers into the formulas and adding them together, we get a total rotational inertia of 880.25 lb-inches squared.
Now, for the rolling circumference, we can assume that we mounted a 235/40-17 tire on that wheel, and calculate the rolling circumference as 76.65 inches.
Finally, we have enough numbers to plug into our formula and prove the rule of thumb. Remember, our estimate of rotational inertia is a little higher than reality, but it should be a pretty good approximation. So, enough suspense…. What’s the answer?
M= 880.25 lb inches-squared/(76.65.in x 76.65in) = 0.1498 lb
Uh… 0.1498 lbs? That’s less than 2.5 ounces. So, according to this estimate, the effective weight of an 18 lb wheel, not including the tire is about 18 lbs, 2.5 ounces. The rule of thumb would have predicted as much as 64 lbs. So much for the rule of thumb! Basically, what this is telling us is big fat spokes or little skinny spokes don’t really make any difference as long as the wheels weigh the same amount. Lightweight wheels are still vitally important to handling, since maintaining contact with the ground is the cornerstone of cornering, but in a straight line, a pound is a pound.

got it??? sorry so technical, but hey.. learning is learning.



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