Question

An engine, running at 1200 rpm, drives a 1.2 m diameter rigid wheel on a non-deformable surface at 5 km/h forward speed. The power is transmitted from the engine to the wheel through a transmission gearbox (gear ratio = 3:1), a differential (gear ratio = 4:1), and a final drive (gear ratio = n:1). Neglecting the deformation of the wheel and wheel slip, the value of \(n\) is __________ (Rounded off to 2 decimal places)

Hint:

To calculate the final drive gear ratio, start by calculating the wheel's speed and then use the total reduction factor to solve for \(n\).

Answer 1

Correct Answer: 4.2

Step 1: Calculate the wheel’s speed in rpm
The forward speed of the wheel is given as 5 km/h, and the diameter of the wheel is 1.2 m. The circumference of the wheel is: \[ C = \pi \times D = 3.14 \times 1.2 = 3.77 \, {m} \] The wheel’s rotational speed (in rpm) can be found by dividing the forward speed by the circumference, then converting from meters per minute to rpm: \[ {Wheel speed (rpm)} = \frac{{Forward speed (m/min)}}{{Circumference (m)}} = \frac{5 \times 1000 / 60}{3.77} = \frac{83.33}{3.77} = 22.1 \, {rpm} \] Step 2: Calculate the engine speed
The engine speed is 1200 rpm, and the gear ratios are:
Gearbox: 3:1 (reduces speed by a factor of 3),
Differential: 4:1 (reduces speed by a factor of 4),
Final drive: \(n:1\) (reduces speed by a factor of \(n\)).
The total speed reduction factor is: \[ \frac{1200}{22.1} = 54.3 \] Thus, the combined reduction ratio from the gearbox, differential, and final drive is: \[ 3 \times 4 \times n = 54.3 \] \[ 12n = 54.3 \] \[ n = \frac{54.3}{12} = 4.20 \] Thus, the value of \(n\) is 4.20.