STATIC TIRE BALANCING
Tire out-of-balance is usually in the tread area (A); see illustration, since the tread is the heaviest part of the tire and the part farthest away from the center of the axle. In the illustration, (A) represents the heavy spot; in the example let's say it is 5 ounces. (B) is the area of the rim where weights are usually applied. (C) is the center of the axle, (D) is the diameter of the rim, (E) is the diameter of the tire and here we are using a 34-inch diameter tire for the example. Area (A) is 17 inches away from the center of the axle (a 34 inch diameter tire divided by 2 equals 17 inches) and area (B) is 7 1/2 inches away from the center of the axle (a 15 inch rim divided by 2 equals 7 1/2 inches).
In this illustration the 5 ounces of imbalance is far out from the center of the axle (17 inches) and an attempt to balance the assembly will require putting weights on the edge of the rim which is much closer to the center of the axle (7 1/2 inches). To balance the tire in this example by placing weights at point (B) to correct the 5 ounces of out-of-balance at point (A) will require considerably more than five ounces of weight to bring the tire into balance. These figures will change depending on the number of ounces of weight a tire is out of balance and the diameter of the tire in relation to the diameter of the rim it is mounted on. In this particular example the amount of weight required to bring this 5-ounce out of balance into balance with the tire mounted on a 15 inch diameter rim would be 10.6 ounces. The 'taller' a tire is and the 'shorter' the rim is the more weight it takes to bring the assembly into balance. Even more weight is required when using the stick-on type of weights applied to the drop center part of the wheel because the weight is being applied even closer to the axle. This is sort of like a 200 pound kid trying to seesaw with a 100 pound kid...it just does not work unless the heavier kid moves closer to the center of the board.