On pavement tires don't completely follow tractive force available = COF x weight, this is why wider tires give more traction. I need to dig up the references on this if needed. If I remember correctly on other types of surfaces this "rule" follows closer.
I was going to use this as a first pass at getting the tractive force available in the model.
Yes that is why engineers in the 1960's said dragsters would never exceed 1 G accelerations. Tire traction on pavement is a combination of 3 different types of behavior.
gearing = the tire conforms to the irregularities in the surface creating a geared effect.
adhesion = brief adhesive bonding between the sticky rubber and the pavement at the molecular level.
If you take all those into account some of the super cars built for ultimate performance primarily because of the type of tires they come with, can achieve acceleration in excess of 1 G on pavement but their total acceleration is still limited by the coeff of friction even if it is greater than 1. High performance tires can achieve coef of friction near 1.2 under the proper conditions.
In the case of bonneville, tire traction is in my view mostly limited by the shear strength of salt on salt, and like you mentioned is probably very similar to other loose surfaces like packed dirt roads.
While you and I are in the “radius” school, Hotrod’s approach is to use the diameter. This requires the number to be divided by 2 to produce the TE available. It would probably be safer for him to have simply said divide by 2 rather than the number of drive wheels. That would keep people running motorcycles or dualies from being misinformed.
I do not use the diameter of the tire, (see my original computer code) I am using the rolling radius (lever arm) of the axle to determine the actual force applied at the tire contact patch, on the drive wheels. In the case of all wheel drive you have 4 drive wheels each with independent loading (traction) so the total accelerative force available if the the engine can deliver it is 4 times what a single driven wheel with the same weight.
If you have a motorcycle all your traction comes from a single wheel, and it can only apply the force that the traction of a single tire can deliver. Put two driven wheels on it by converting it to a 3 wheeled motor cycle and you roughly double the possible traction (weight unchanged), put the same engine in an all wheel drive streamliner of the same weight and you have approximately 4 x the possible traction for the engine to use.
If a single tire (1 ft loaded radius) loaded with 1000 lbs on a 0.6 coef of friction salt surface, can only deliver a max of 600 ft-lb of axle torque to the salt as a thrust of 600 pounds. Load two of those same tires with 500 lbs weight and you still have the same traction limit because you not only doubled the number of tires, but you cut the individual loading in half. Increase the load on both tires to 1000 lbs with ballast and now you can deliver 600# of thrust with each tire or 1200 lbs of thrust.
I typo'd in my earlier comment I meant loaded radius but said loaded diameter, we are all on the same page, the only thing that counts is the distance (lever arm length) from the axle to the ground contact patch.