LD1
Epic Contributor
So let's try this again.
Imagine the tire is a boloney skin for simplicity. Inflate it to 10 psi. That means every square inch of tire has 10 pounds pushing on it.
Put on the tractor, and set the machine on the ground. Every square inch of tire still has 10 pounds pushing on it. That includes the tire that is in contact with the ground.
The tire will deform until the tractor is supported and every square inch of tire has 10 pounds pushing on it. The total weight per tire will be proportional to the weight of the tractor and the weight distribution front to rear, side to side. The pounds per square inch of contact remains the same.
You can't go to zero because the tire is no longer supporting the tractor and you get a divide by zero error
The tread pattern is going to affect behaviour which is why the R1's sink in until the casing hits the ground. The R3 is the closest to a boloney skin and causes the least amount of damage.
R1's cause more soil compaction on the larger tractor because the bar sinks in with higher pressure until the contact patch levels out and the pressure is distributed.
Still not true. A tire is belted and has sidewalls. If a tire were like a balloon, then yes, I would agree with your logic. But it is not. Even your turf tires, and even if you are on concrete, doubling the load will NOT double the contact patch (which it would have to in order for your "theory" to hold up). Likewise, cutting the air pressure in half does NOT double it either.
Changing the load or air pressure does INDEED change the contact patch. BUT they are NOT linear.