Front steering-pivot points are 1.2 m apart. During a turn on flat concrete, the inner and outer front-wheel steering angles are 50° and 35°. Find the wheelbase required for no front-wheel skid, in metres, to two decimals.
For ideal Ackermann steering,$\cot\delta_o-\cot\delta_i=b/L$,where b is steering-pivot separation and L wheelbase. Therefore $L=1.2/(\cot35^\circ-\cot50^\circ)=2.03718$ m, rounded **2.04 m**. The inner wheel has the larger angle because it follows a smaller turn radius.
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