There are two routes to describe how the engine's output manifests itself as acceleration of the vehicle; a power centric view, and a torque centric view. They are both absolutely consistent with each other, as, after all, power, speed and torque are inextricably linked together.
These views or approaches differ mainly in how the gearing affects the quantity
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First, the power centric view.
The gearing does not modify power at all. If you had a 100% efficient transmission, and a 100 BHP engine, you would get 100 BHP at the driving wheels. For a typical manual gearbox system, the efficiency is about 95%, so, you would get 95 BHP.
At any road speed and condition, you need to supply a certain amount of power at the wheels to overcome the various sources of drag - rolling resistance, gradient, and aerodynamic drag.
If there is more power being delivered to the wheels than the vehicle's drag power, the vehicle will speed up, and vica versa.
Some maths
Taking Newton's Second law - F=ma
multiply both sides by vehicle velocity
Fv=mav
P=mav
The product Fv is the power at the wheels available for acceleration (Power at the wheels minus drag power). Re-arranging gives
a = P/(mv)
This formula tells us that;
owing to the v on the denominator, for a given power at the wheels, the acceleration you get reduces as the vehicle speeds u[
owing to the P on the numerator, for a given road speed, if you can gear the engine to be turning at a speed where it produces more power, you'll get more power at the wheels, and so, you will accelerate more strongly.
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Now the torque centric point of view
When the engine's torque is passed through a reduction ratio in the gearbox, the output shaft will turn more slowly, but, the torque will be multiplied up by an equivalent ratio.
Again, the transmission efficiency takes it's share. So, if you had 100Nm of torque, and a 10:1 overall gear ratio, you would get 1000 Nm at the wheels with perfectly efficient transmission, but, only 950 Nm if the efficiency were 95%.
Taking the torque at the wheels, and dividing by the rolling radius of the wheel gives the tractive effort - the force pushing the vehicle along.
If we subtract all the sources of drag force from the tractive effort, and are left with a positive value, the vehicle will accelerate, and vica versa.
Again, starting with Newton's Second Law
F=ma
If we consider F to be the net force - tractive effort minus drag force, then, that's it, Newton's Second Law tells all.
We can, however, look a bit further, and split up the left hand side of that equation, and look at the tractive effort and the drag force, fd;
TE-fd=ma (where TE stands for tractive effort)
The tractive effort is the engine's torque multiplied up by the gear ratio, divided by the rolling radius
(t*ng*nf*eta)-fd=ma
where ng is the gearbox ratio, nf is the final drive ratio, and eta is the transmission efficiency
Here's where Tyro's question can be seen most clearly - in lower gears, ng is higher. 1st gear ratio in a typical car's gearbox is somewhere in the region of 3.5 to 4, while top gear may be 0.8 or 0.9. So, changing down increases ng, and providing that the engine's torque doesn't simultaneously reduce by more than the gear ratio step, then, the vehicle will accelerate faster.
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Finally, to repeat a point from earlier - the two approaches give exactly the same answer, they are entirely equivalent.
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