It's been suggested that a lighter car will have less rolling resistance because the weight doesn't squash the tires down as much.
For a long time I accepted that as being intuitively true. It seemed right.
Today I finally crunched some numbers and figured it out. The answer came to me when I realized that the size of a tires contact patch is a function of the weight it carries and the tire pressure, and that rolling resistance is determined by how much the contact patch deforms with load. The size of the contact patch indicates how much load and deformation it has. Right?
Here comes the math:
I'm using my car as an example. It weighs 2650 pounds and has a 61/39% weight distribution. Tire pressure is 50psi.
2650 pounds x 61% = 1616.5 pounds on the front wheels.
1616.5/2 = 808.25 pounds on each front tire.
808.25/50psi tire pressure = 16.165 square inches of contact patch.
Lets say that I remove 50 pounds from the car.
2600 pounds x 61% = 1586 pounds on the front wheels.
1586/2 = 793 pounds on each front tire.
793/50psi = 15.86 square inches of contact patch.
What tire pressure do I need to run in order to get the same size contact patch as a 50 pound lighter car?
808.25 (weight carried by one front tire of the stock weight car) divided by 15.86 square inch (contact patch size of a 50 pound lighter car) = 50.96psi tire pressure.
Or if my car was 50 pounds lighter, how much could I reduce pressure and have the same size contact patch as the heavier car?
793/16.165 = 49.06psi.
Not a big difference! Not yet anyway....
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