Question about gear differential ratio and wheel's torque
I know that the differential gearbox is providing a torque to the wheels; as the relation in the figure indicates : Tw (wheel torque) = Td (differential torque) × (ηd×nd). Where nd:is the differential gear ratio and η:is the effeciency of the differential gearbox. Now this torque at wheels (Tw), which is provided by the differential , is also = Rw×Fx (torque from road friction ) . My question is : Won't the torque from (Fx) on the wheels accelerate (increase) the wheel's rotational speed(ωw) ? If this is the case;now ωw is increasing. And since (ωd=nd×ωw); then ωd(differential or driveshaft rotational speed) should also increase to maintain the gear ratio (nd) constant. So how would this differential/drive shaft rotational speed (ωd) increase to maintain this gear ratio (nd) constant ?
1 Answer
Think of the differential as a 'black box', with a fixed gear ratio (there are certainly more complicated scenarios, but this is typical for most automobiles). Ignore efficiency, for the sake of simplicity. The speed of the driveshaft will always be a multiple (or inverse multiple) of the wheel speed; as is the torque. The multiplication factor is the gear ratio of the differential. It makes no difference whether or not the engine is accelerating the vehicle, or if the car is being driven externally (gravity, being pushed, etc...), the driveshaft speed will always be ratio the wheel speed , and the gear ratio of the differential determines that ratio. Thus the engine rpm, as it drives the driveshaft, is always turning at some ratio of the wheel speed (unless it is disconnected from the drive shaft, i.e. in neutral) . The scenario presented is really simplified, and, typically, there is a transmission in order to offer a greater range of gear ratios to allow performance over a greater range of speeds and loads.
