A 2WD tractor has nominal rear-wheel radius .5 m. Rolling radius is 6% smaller. In 20 wheel revolutions the actual travel is 56 m. Find wheel slip in percent, to two decimals.
Rolling radius is $.5(1-.06)=.47$ m. No-slip travel for 20 revolutions is $D_t=20(2\pi)(.47)=59.06194$ m. Slip is the travel deficit relative to this theoretical distance: $s=100(D_t-56)/D_t=\mathbf{5.18\%}$. Using nominal rather than rolling radius would overestimate slip.
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