Question

A 2WD tractor has to develop 30 kN gross tractive force at the ground. Rolling radius of the rear-wheels is 0.73 m.
Each rear-wheel is driven by a simple planetary final-drive in which sun gear (26 teeth) is input, ring gear (78 teeth) is fixed, and the carrier is bolted to the rear-wheel hub.
The sun-planet external mesh efficiency is 98.5%, while the planet-ring internal mesh efficiency is 99%. Losses upstream of the final-drive are neglected.
The required sun-shaft input torque per rear-wheel (in kN.m) is ________. (Rounded off to two decimal places)

Hint:

Find the wheel torque from the tractive force, get the planetary speed reduction ratio, then divide by both mesh efficiencies.

Answer 1

Correct Answer: 2.81

Step 1: Get the tooth-count based radii and the velocity relation.
Sun teeth $Z_s = 26$, ring teeth $Z_r = 78$, so the planet has $(Z_r - Z_s)/2 = 26$ teeth, and pitch radii scale with tooth count.
For the carrier arm, the planet centre moves with the arm, and this gives the classic result $\omega_{sun}/\omega_{arm} = (Z_s+Z_r)/Z_s = 104/26 = 4$.

Step 2: Work out the output torque needed per wheel.
Half of the 30 kN pull is carried by each rear wheel: $F = 15$ kN.
With rolling radius $0.73$ m, the carrier (wheel hub) torque is $T_{carrier} = 15 \times 0.73 = 10.95$ kN.m.

Step 3: Use power balance across the gear train.
Ideal (loss-free) power balance gives $T_{sun,ideal} \times \omega_{sun} = T_{carrier} \times \omega_{carrier}$, so $T_{sun,ideal} = T_{carrier}/4 = 2.7375$ kN.m.
Real power lost at the two meshes means actual input torque must be larger: $T_{carrier}\omega_{carrier} = \eta\, T_{sun}\omega_{sun}$.

Step 4: Bring in the two mesh efficiencies.
$\eta = \eta_{sun-planet} \times \eta_{planet-ring} = 0.985 \times 0.99 = 0.97515$.
\[ T_{sun} = \frac{T_{sun,ideal}}{\eta} = \frac{2.7375}{0.97515} = 2.807\ kN.m \]

Final Answer:
Each rear wheel's sun shaft must supply \[ \boxed{2.81\ kN.m} \]