Lynnmor wrote:
Jframpey wrote:
Yeah! Air softBB's why didn't I think of that?
OK, that's the way I'll go. I have 15" tires - I'll use 3 oz. I have an electronic balance so I'm set.
WooHoo - Thaks
I wrote this 7/13/16, in another thread, but never got an answer:
I have had brake drums that took 5.5 oz. to bring them into balance. When mounting a 4 oz. out of balance tire to that drum, it would require 1.5 oz. to 9.5 oz. to bring the assembly into balance, depending how they are oriented. How many oz. of beads would I add?
I find the heavy spot of the drum and the heavy spot of the wheel/tire and then mount them opposite, then balance the assembly.
Ignoring a dissertation in physics I can say that you will require less than 9.5oz as the beads run on the inside of the tire and have a greater effect due to the distance..
basically.. the turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. the extra distance is of course the distance from the rim edge to the inside of the tire..
typical lead balance weights are heavier than necessary due to the fact that they are attached to the rim on a smaller diameter and therefore have a smaller moment..
consider a tire of 30 inches diameter with a rim of 15 inches diameter and a 4 oz balance weight.
the weight is 4 oz at 7.5 inches radius so the moment is 4*7.5 = 30
the same moment could be obtained with a weight of 30 / 15 = 2 oz if it were possible to mount a weight on the circumference of the tire.