Question Bank • QB_FP_097 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses

Question Statement:

A tractor transmission has an engine speed of $1800\text{ rpm}$. The gear ratios are: primary gearbox reduction $3.2 : 1$, bevel differential pinion-crown reduction $4.0 : 1$, and planetary final drive reduction $3.5 : 1$. If the drive wheel rolling radius is $0.70\text{ m}$, calculate the forward speed of the tractor in $\text{km/h}$.

Answer Key:

10.55

Step-by-Step Solution & Explanation

Total overall reduction ratio: $G = 3.2 \times 4.0 \times 3.5 = 44.8$ Drive axle rotational speed: $N_{\text{axle}} = \frac{N_{\text{engine}}}{G} = \frac{1800}{44.8} = 40.1786\text{ rpm}$ Forward speed $v$: $v = \frac{2 \pi N_{\text{axle}} r}{60} = \frac{2 \pi \times 40.1786 \times 0.70}{60} = 2.945\text{ m/s}$ In $\text{km/h}$: $S = 2.945 \times 3.6 = 10.60\text{ km/h}$
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