An engine at 1200 rpm drives a rigid 1.2 m diameter wheel at 5 km/hour without slip or deformation. Gear reductions are 3:1 in the gearbox,4:1 in the differential and n:1 in the final drive. Find n, to two decimals.
Forward speed is $5000/60=83.3333$ m/min. Wheel speed is $N_w=v/(\pi D)=83.3333/(\pi1.2)=22.10485$ rpm. Overall reduction is $1200/N_w=54.28672$,which equals $3(4)n$. Thus $n=54.28672/12=\mathbf{4.52}$. All ratios here represent speed reductions.
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