Question

While turning a 4-wheel tractor equipped with the Ackermann steering mechanism the axis of the inner front wheel makes an angle of 45°, whereas the angle made by the axis of the outer front wheel is 30° with horizontal line parallel to the front axle. The kingpins of the axle are 1300 mm apart. The spindle genths are essentially zero. What should be the wheel base of the tractor if there is no slippage of any of the wheels?

(a) 1886.78 mm

(b) 1775.83 mm

(c) 1336.59 mm

(d) 2000.01 mm

The question was asked during an internship interview.

This intriguing question comes from Numerical Problems on Calculation of Different Tractor Parameters topic in portion Lubrication System and Tractor’s Classification of Farm Machinery

Answer 1

Correct option is (b) 1775.83 mm

For explanation I would say: Cot θ – Cot Φ = \(\frac{c}{b} \)

Cot30° – Cot45° = \(\frac{c}{b} \)

b = \(\frac{1300}{Cot30° – Cot45°} \)

b = 1775.83mm.