Tractive effort and speed from gear ratios

Given: $m = 1300$ kg, $T_e = 150$ N m, $G_1 = 3.8$, $G_4 = 0.8$, $G_f = 4.1$, $\eta = 0.9$, $r = 0.30$ m, $f = 0.015$.

(i) Wheel torque in first gear

$$T_w = T_e\, G_1 G_f\, \eta = 150 \times 3.8 \times 4.1 \times 0.9 = 2103.3\ \text{N m}$$

(ii) Tractive force

$$F_t = \frac{T_w}{r} = \frac{2103.3}{0.30} = 7011\ \text{N}$$

(iii) Initial acceleration

Rolling resistance:

$$F_{rr} = f\, m g = 0.015 \times 1300 \times 9.81 = 191.3\ \text{N}$$ $$a = \frac{F_t - F_{rr}}{m} = \frac{7011 - 191.3}{1300} = 5.25\ \text{m/s}^2$$

This is the engine-limited value. The real value is smaller because of air drag and rotating inertia, and it is also limited by tyre grip: with $\mu \approx 0.8$ and about 60 % of the weight on the driven axle, the grip limit is about $0.8 \times 0.6 \times 9.81 \approx 4.7$ m/s², so the driven wheels would spin at full torque.

(iv) Vehicle speed at 3000 rpm

$$v = \frac{2\pi N r}{60\, G\, G_f}$$

First gear:

$$v_1 = \frac{2\pi \times 3000 \times 0.30}{60 \times 3.8 \times 4.1} = 6.05\ \text{m/s} = 21.8\ \text{km/h}$$

Top gear:

$$v_4 = \frac{2\pi \times 3000 \times 0.30}{60 \times 0.8 \times 4.1} = 28.7\ \text{m/s} = 103.4\ \text{km/h}$$

Answer: $T_w = 2103$ N m; $F_t = 7011$ N; $a = 5.25$ m/s²; $v = 21.8$ km/h (first) and $103.4$ km/h (top) at 3000 rpm.