Quote:
Originally Posted by InFlight
Axial Bearing Load = (Torque / radius avg) tan (20°) sin(45°)
= (Torque/0.463) (.364) (.707)
Axial Gear Force = 172 (Ounce*In) * (.364)(.707) / 0.463 in
= 95.6 oz or 5.9 lbs
With a Swerve drive gear reduction, this is still a substantial force on a small FR6ZZ
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Much smaller than the thrust loads applied to the bearings in all traction wheeled skid steers.