The Ford Expedition has a 65.5-in. track and a 20-ft turning radius. The P265170R17 tires have...

The Ford Expedition has a 65.5-in. track and a 20-ft turning radius. The P265170R17 tires have a nominal radius of 15 in. If the vehicle was designed with a solid rear axle (i.e., no differential), calculate the amount of wind-up in the rear axle when the vehicle performs a 180" turn at minimum radius. (Answer: 2.18 revolutions.) (with explanation and details please)

Expert Solution

Turning radius R = 20 ft.

Track t = 65.5 in = 5.458333 ft.

Turning radius of outer wheel = R+t/2 = 22.72916667 ft

Turning radius of inner wheel = R-t/2 = 17.2708333 ft

Radius of tire r = 15 in = 1.25 ft

Legth of the path of outer wheel L_o = (180/360)*2*(R+t/2)

Number of revolutions of outer wheel n_o = L_o/(2r )= (180/360)*(R+t/2)/r = 9.09167 revolutions.

Legth of the path of inner wheel L_i = (180/360)*2*(R-t/2)

Number of revolutions of inner wheel n_i = L_i/(2r )= (180/360)*(R-t/2)/r = 6.9083 revolutions.

Wind-up in the rear axle = number of revolutions of outer wheel - number of revolutions of inner wheel = n_o - n_i = 9.09167 - 6.9083 = 2.183 revoltuions

samet mamat answered 2 years ago

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