If you have a 20 inch wheel and fill it completely with weight with no spaces between the weights, using 10 one oz weights attached on the left of the rim and 10 one oz weights on the right side of the rim that are out an inch further from center than the weights on the left, you get leverage. Lets use 10 gm of leverage for this case. "this measure being done by putting the weights that are further out at the 12-6 O'clock positions and measuring the leverage at the 3 O'clock position."

If you go to a 40 inch wheel and fill it completely with weight with no spaces between the weights, using 20 one oz weights on the left, then 20 one oz on the right that are all still 1 inch further from center than the weights on the left and measure the leverage, will the leverage go up or remain the same?

I have tested and shown that if you have only two weights, one on the left, then one on the right that's 1 inch further out, whatever leverage you get will stay the same if you move to a bigger wheel but still keep the weight on the right one inch further out than the one on the left.

I would like to understand this with a completely filled wheel, two different sizes.

Please explain in laymen's terms, I'm not a physics major.

Thank you

As I tried to describe in my response to your other post on this topic (see https://www.askmehelpdesk.com/physics/leveraging-question-586768.html ), the "leveraging force" is really a torque. If you have two equal weights on either side of the fulcrum but one is 1" further out than the other the torque that the wheel feels is equal to the weight times the difference in moment arm lengths. So in this case if you have two one ounce weights the torque applied to the whel is 1 ounce-inch. You could have one example where the two weights are at 20" a and 21", and another where they are at 40" and 41", and in both cases the torque is the same. However, please note that this does not mean that both wheels would start turning at the same rate. The amount of torque that it takes to turn a 40" wheel is greater than the torque required to turn a 20" wheel; so while the torques in the two examples are the same, all else being equal the smaller diameter wheel will start to rotate faster than the larger one. This is because the smaller wheel has less moment of inertia, and so is easier to get turning.