Posted by Chris Case [162.158.57.47] on Wednesday, June 22, 2016 at 09:24:22 :
In Reply to: Power Slide in a Power Wagon? posted by Greg Coffin [172.68.35.20] on Wednesday, June 22, 2016 at 08:35:36 :
It's called the Co-efficient of Friction. Braking or acceleration, rubber tires on dry pavement have the friction of 70% of the weight on the tires. Empty truck, let's use 1,000# per tire. So it takes 700# to break each loose, 1400 for the pair. Engine makes 100lb/ft of torque, times rear end gear ration, zo aboyt 600, divide by the radius of the tire in feet, 1.5, = 400ft'lbs of torque from engine. They can't make two wheel scratch, not enough torque to overcome the CoF.
But can we add an additional 1,000lbs of side thrust? I think you would need a .2 G turn? What speed, what radius? More maths... Hmm, center of gravity at about 3', tread width of about 6, means ummm 45 degrees, or one-to-one. Or 5,000 lbs of centripedal force from cornering before rolling.... I'm at the limit of my logic/math.... but it looks theoretically possible, if the driver can turn the wheel fast enough to make the G force before running out of speed.
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