Originally Posted by
Brocklee
After thinking about it, I should have gone into tire traction more. In my last post I said that less weight on a tire equals more traction, well that's partially true. A better way to put it is it increases tire efficiency, for example (these are pulled from the chassis engineering book I talked about) if we have a tire with a 1000lbs load it will have 1000lbs traction available theoretically, so that tire can produce 1g of cornering force or it's 100% efficient. Now let's take the exact same tire and change the vertical load to 500lbs it will have 700lbs of traction available, well we can see that the lighter tire has less traction, 300lbs less, however we can see that the tire with the 500lbs vertical load will now produce 1.4g's or it is 140% efficient.
Sorry to disagree, but that's not so straight forward. I assume the example here was for a defined set of parameters and most likely not for off road?
The bigger issue in determining the coefficient of friction of rubber on a rough surface hinges highly on 2 things: the tires ability to maintain contact with the ground (the suspension) and the contact patch. The contact patch is very straight forward in on-road scenarios, but on a surface that isn't level means there needs to be enough force behind the tire to conform it to the surface of the track. Tires are designed to deform the right amount for a given weight. Any more and you run into the situation you presented where the mass of the car is increasing faster than the frictional force generated, but any lighter and the coefficient of friction starts falling off faster than the mass of the car. The result is a bell curve-esque graph with an ideal mass.
Now, keeping the tires in contact with the ground is the second part. There is a big reason why most people say a heavier car is better on a rougher track. It's because the suspension can act more efficiently with a heavier car. Let's say 'X' is the mass of the car (just the sprung mass) and look at what happens as X approaches 0 and as X approaches infinity:
As X approaches 0, you are eliminating the mass that the shocks 'push against' to keep the tires in contact with the ground. If you hit a bump and the tires move up the bump, there is nothing for the shocks to push against once the tire is over the bump. The tire will continue upwards in a state of free fall. Picture rolling a tire over a garden hose at a decent speed. The tire would bounce off the hose a couple feet in the air before returning to the ground.
Now, as X approaches infinity you are providing an immovable object for the shocks to push against in their effort to keep the tires on the ground. You could go for an infinitely stiff spring to match and the tires would follow the grounds imperfections perfectly. What it comes down to is the ratio of unsprung mass to sprung mass. The higher the ratio, the better the suspension will be able to keep the tires on the ground.
As an example, 8th scale buggy tires are designed with a 7 to 7.5 lb car in mind, as this is a typical weight for nitro buggies in this current crop of cars. When electric 8th scale first hit the scene, they were tipping the scales at 8 or more lbs. On rougher tracks, many people found their electric buggies to handle better as their suspensions worked more efficiently through naturally whooped out sections. However this extra mass is beyond what the tires were designed for, so losing some weight would benefit the handling of the car on smoother tracks. Some people took it to the extreme, getting their 8th scale buggies down to 6.5 lbs and have found their cars more skittish as the tires can't properly conform to the track surface. It's all a balancing act.
Disclaimer: I've left out the possibility of redesigning tires as the average consumer can't do such a thing. Plus, minimum weights on each class mean that most tires are designed to work their best close to that weight.