The problem is that the debate between torque and power is full of myths (mostly created by magazines for ricers...) making things lot more complex than they are and doing nothing but generate confusion in people not well enough educated on physics basics. The widespread use of Imperial units that, for example, don’t make a clear distinction between torque and work or introduce meaningless “magical numbers” then makes things even worse.
For example the often repeated “power sells cars torque wins races” is misleading. Even if the concept it tries to express is potentially correct it makes it sound like if the two elements, torque and power, were two unrelated things concerning different rpm ranges and, even worse, it suggests that one is more important than the other.
The absolutely correct concept that it tries to explain is that the power (and torque) limited in a short rpm range is useless in real life and it’s better to have maybe lower peak power but a more favourable power (hence torque) curve in the whole rpm range. The problem is the way it’s enounced, calling one “power” and the other “torque”. (why it’s so I don’t know, but it’s something many people do, particularly journalists; look at any car test and you’ll find quite often they refer to “torque” while talking about the “thrust” they feel at low rpm and use the term “power” while talking about the “thrust” at high rpm)
That distinction is particularly stupid, first because the thing “pushing” the car is always the same at any rpm, the force at the wheels; second because at any given rpm, more torque means more power at that same rpm and viceversa.
So while asking for an high torque in the whole rpm range you are asking for more power in that same rpm range.
Consequently the correct phrase should sound something like “an high peak power sells cars, a favourable power distribution on the whole rpm range wins races”. It doesn’t sound as nice and compact as the original but at least it’s correct.
Same applies to another commonly used phrase “in a given gear, maximum acceleration is with engine at peak torque”
That’s false, or, better said, isn’t necessarily true, so to enounce it like it was a general principle is wrong.
In a given gear you have the force from the engine on the wheels that, obviously, follows the torque curve, so you indeed have maximum force at the wheels at maximum torque.
But that doesn’t necessarily correspond to maximum acceleration of the car; in fact if the gear ratio is fixed then engine rpm is strictly proportional to car speed, and the other two main forces acting on the car in acceleration, aero drag and rolling resistance, depend by car speed too, particularly the former increases way more than linearly with speed.
The maximum acceleration is achieved when the difference between force from the engine and (aero drag + rolling resistance) is the largest. Consequently if in a given gear the sum (drag + r.r.) grows more rapidly with speed than the engine torque does with corresponding rpm (as it’s the case in longer gears) the difference will become smaller even if torque increases so you’ll have, in a specific gear, maximum acceleration at rpm way lower than peak torque.
And that lead us to the typical question related with the debate “torque vs power”, the problem of car’s acceleration.
We just saw that all the main forces applied in car’s acceleration depend by speed. And for this reason it makes sense to use speed itself as the independent variable while studying acceleration. If you do it, and apply basic physics, you’ll find that for a given speed, the maximum car’s acceleration is available when the engine is giving maximum power.
Let’s use the method of forces.
F = ma
F = Force from the engine – aero drag – rolling resistance = ma
Force from the engine = Torque at the wheel / wheel radius = Torque at the crank * gear ratio / wheel radius = Torque at the crank * (engine rpm / wheel rpm) / wheel radius = (Torque at the crank * engine rpm) / (wheel rpm * wheel radius) = Engine Power / car speed.
Consequently we have :
Engine Power / car speed – (aero drag + rolling resistance) = ma
If we consider a given car speed we have that (aero drag + rolling resistance) is constant, so the equation says that car’s acceleration is proportional to engine power, hence to maximise acceleration at any car speed you have to maximise engine power.
Obviously you obtain the same result using the power method :
Power of force from engine – power of resisting forces = variation of kinetic energy vs time :
Engine power – (aero drag + r.r.) * car speed = d ( 0.5 * m * car speed ^2) / dt =>
Engine power – (aero drag + r.r.) * car speed = m * car speed * d car speed / dt = m * car speed * a
That is exactly the same equation obtained above and consequently gives exactly the same condition : the maximum force from the engine to the wheels is available when the engine gives maximum power. Then at a given car’s speed (hence with aero drag and rolling resistance constant) that will necessarily correspond to maximum acceleration at that speed.
(notice another important thing from that equation : even if aero drag and rolling resistance were zero, acceleration would still decrease as car’s speed increases)
In the ideal case of a perfect CVT gearbox, then the gear ratio would continuously change with speed, continuously keeping the engine at peak power and obtaining the ideal acceleration. Consequently in that situation a very peaky engine, with just a very high power peak but with low power at any other rpm would be all you need because you are going to use it only at 1 single rpm.
What makes things more complicated is that we have to deal with a real gearbox that usually has a limited number of gear ratios.
This mean that we can’t, at each speed, keep the engine at peak power, rpm will vary in a given range and consequently the engine power will not be constant but will change depending by car speed.
Still the basic requirement stands, to maximise force at the wheels (hence the acceleration at a given speed) you have to maximise power from the engine.
To minimize the time needed to accelerate from speed A to speed B, you have always, at each speed between A and B, to select amongst the available gear ratios the one that puts the engine at the rpm giving the highest power.
In practice this means to upshift only when the power corresponding to the rpm you are in the current gear ratio is lower than the power corresponding to the rpm you’ll be after the upshift, which consequently means to keep a given gear ratio well past peak of power.
If you respect that condition you’ll have the maximum force at the wheels at each speed between A and B.
Obviously the resulting acceleration is going to be worse than the ideal you would achieve with the CVT because the engine will work mostly at an rpm different from max power, hence will mostly give a power lower than the maximum.
And since you are spending very little time at peak power rpm and lot of time elsewhere in the rpm range it’s evident that the engine power at each different rpm in the whole range becomes now more important than the single peak power value.
Consequently what we theoretically need in real life is an engine that, at each rpm of the range used, has as much power as possible, and that obviously corresponds to have a torque at each rpm as high as possible.
Problem is that generally torque (and consequently power) generated by an engine in a wide rpm range is like a short blanket, covering a place you leave uncovered somewhere else, so while “designing” the power curve you have to look for the compromise giving the best end result, and often it becomes convenient to renounce to a bit of power at a given rpm and gain it somewhere else in the used range.
What it’s important to understand though is that you are not renouncing to power to obtain torque, you are renouncing to torque (hence power) at a given rpm to obtain torque (hence power) at another rpm.
Hondanisti wrote:
tq is the force that moves the object from rest or in braking, slows the object
rpm is the time that the force is applied.
their product is work.
DaveKillens wrote:
Horsepower is a measurement of how much work is being done.
The power, product of torque and rpm (number of revolutions per minute), is not the work. The power tells you how much work is done
in the unit of time.
The work is the torque applied times the number of revolutions completed.
that, for a rotating object, rpm times the torque equals power?
