# When Vehicle Power Dictates Acceleration

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One interesting problem when dealing with a vehicle of a certain mass is to determine what is required in order to get the maximum acceleration while going from one velocity to another.

## Statement

If a moving vehicle has an energy source that has a variable power output, the energy source must be set to its maximum power – during the entire velocity range – to ensure that the vehicle will get its maximum possible acceleration throughout that velocity range.

At any given velocity:

• The force applied to the vehicle dictates the acceleration it gets;
• The power applied to the vehicle dictates the force it gets;
• Therefore, the maximum possible acceleration of the vehicle depends solely on the maximum power available for the vehicle.

When it comes to accelerating a moving vehicle, only power tells the whole story.

## Explanations

### Force requirement

The first basic requirement is given by Newton’s second law: The force F required is equal to the mass m of the vehicle times the desired acceleration a of the vehicle. In simple terms: F = ma.

### Power requirement

But since there is a force in motion, work is done, so there is a second requirement: The power P required is equal to the force F applied to the vehicle times the velocity v of the vehicle. In simple terms: P = Fv.

### Putting it all together

If the two equations are combined together, we get P = mav. This means that as long as there is a mass m and velocity v (i.e. not equal to zero), the power P required is proportional to the desired acceleration a. At this point, we can ignore Newton’s second law because it is indirectly implied in this new equation, i.e if the power requirement is fulfilled, the force requirement is also necessarily fulfilled.

We have been talking about “desired acceleration” and “required power” until now but, in the real world, we are often given a power rating from an energy source and we take whatever acceleration we can get from it. In this case the equation can be rewritten as a = P/(mv).

With this new equation, assuming power and mass are constants, we can see that the acceleration is a function of velocity. Particularly, as the velocity increases, the acceleration will decrease.

Since the mass m is a constraint given by the initial problem, it cannot be modified. The velocity v is also a constraint given by the initial problem, that is, it must be within the desired velocity range. So if one wants to increase the acceleration throughout the velocity range, one has no other choices but to increase the power available to the vehicle. If the power P is doubled, the acceleration a throughout the velocity range will also be doubled (remembering that the acceleration will still decrease as the velocity increases).

### Power is power

Because of the law of conservation of energy, the power available to the vehicle is equal to the power given by the energy source powering the vehicle (not considering losses). The energy source can make its power with:

• a rotational system (P = torque times angular velocity);
• fluid power (P = pressure times volumetric flow rate);
• electricity (P = potential difference times current);
• combustion (P = fuel mass flow rate times fuel heating value);

or any other way one can think of, it does not matter.

Although, in any case, note that there may be some inefficiencies that will lead to some losses due to transformations between the energy source and the point of application on the vehicle. Obviously, only the power available at the point of application on the vehicle is relevant.

### A common mistake

When considering the special case where a vehicle is powered by wheels of radius r, some people like to state they can link the acceleration directly to the wheel torque T, by using the relation F = T/r instead of the power equation we used. Combining this equation with Newton’s second law, they get T/r = ma and claim that it is a more direct way because the wheel radius r is constant (unlike the velocity v).

But where does that radius comes from? Are we allowed to choose any value? The equation F = T/r is subjected to the law of conservation of energy which extends to power, namely, Pin = Pout. With a rotating object, Pin = Tω (where ω is the object angular velocity) and Pout = Fv. This means that T/F = v/ω. So if T/F = r, then v/ω = r as well. The radius r implies a transformation where power is kept constant, and that cannot be ignored. Replacing r with the velocity ratio in the misleading equation will give Tω/v = ma. Thus we get back to our original equation: P = mav.

The introduction of the wheel radius does not simplify the process, it just hides the important notion of conservation of energy. Even with this special case , there are no ways around it, one way or another, power will have to be considered because velocity must be considered when accelerating a moving vehicle.

 It is a special case because the force F doesn’t have to be the result of a rotational system, meaning there may not be any torque involved in powering the vehicle. (For example, when a horse is pulling a buggy.)

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33 replies
1. S
sandy stone says:

As much as I hate to step into an emotional internet discussion, I wonder if there is just a communication issue going on here. Are we talking about instantaneous acceleration at a particular engine's torque peak, or are we talking about accumulated (integrated) acceleration, like trap speed at a drag strip?

2. R
Randy Beikmann says:
jack action

The problem right now is that the question is not about what are all the possible limits of acceleration, it is about a single one: Is power or torque a better indicator of acceleration? Google will send all the users typing this question to the insight I wrote and other similar threads because they have the terms ¨power¨, ¨torque¨ and ¨acceleration¨ in it.If the question is whether power or torque is "a better indicator of acceleration", then I'm afraid there is no general answer. It depends on the full question/problem definition – there are always "if's" that will determine the answer. It's like when someone asks you, "What's the best car to buy?" The answer must be qualified by your purposes, budget, etc.

If the problem allows any transmission to be used, including a CVT, then the answer is (assuming no losses) that max engine power is more important than engine torque, since you can hold the engine at its peak power speed at "any" vehicle speed (that is actually stretching it, since even a CVT can't produce an infinite range of ratios), whether you can utilize it or not. But a properly matched fast-shifting DCT could keep the engine speed near enough to the peak power speed so that it's not that much different than the CVT. Now, if the question stipulates low vehicle speeds, all that matters is that the engine can produce enough power to produce the force at the tread that fully utilizes the tires' traction, and the maximum amount of power available (Pmax) is not important.

If prepping a drag racer, launch is the most important aspect – if you get a slow start, it takes a huge amount of power later in the run to try to catch up. I would first concentrate on traction (force capability), because the launch is traction-limited. Then I would set the transmission and final drive gear ratios to multiply the engine torque and produce enough axle torque to use that traction. But it does not take that much power at launch, since P=Fv, assuming no losses. A lower-power car with correct gearing and tires can easily beat a high-power car that is incorrectly set up.

As to how to guide the discussions and correct the wrong conclusions by gearheads out there, my solution was to write a book that (IMHO) clearly and correctly explains all this in detail. :wink: Smart gearheads do use physics.

3. jack action says:
sandy stone

Are we talking about instantaneous acceleration at a particular engine's torque peak, or are we talking about accumulated (integrated) acceleration, like trap speed at a drag strip?We are talking about an acceleration at a given velocity (other than zero). More precisely, over a speed range.

Randy Beikmann

Now, if the question stipulates low vehicle speeds, all that matters is that the engine can produce enough power to produce the force at the tread that fully utilizes the tires' traction, and the maximum amount of power available (Pmax) is not important.Pmax might not be important, but power still is. If the maximum friction force (or any other limiting factor you can think of) allows for, say, 10 000 N at a speed of 1 m/s, knowing the torque produced by the engine alone won´t be enough information, no matter the gearing or tire radius. You have to make sure the engine produces enough power, i.e. the exact same amount needed at the wheel. It doesn´t even matter how much torque it produces and at what rpm. It doesn´t even matter if it is a combustion engine or not. All that matters is how much power it produces.

The acceleration needed at the wheel from my example will be 10 000 N divided by the mass of the vehicle (ignoring rotational inertia to keep it simple). What do I need from the engine? 10 000 W (= 10 000 N X 1 m/s). That´s all you need to know. That is the only thing to remember from the insight I wrote. For more detailed information, I wrote 8 pages about it on my website to explain the acceleration simulator I build (see my signature); It is too much info to fit in a single insight which is meant to answer one simple question.

4. S
sandy stone says:
jack action

The acceleration needed at the wheel from my example will be 10 000 N divided by the mass of the vehicle (ignoring rotational inertia to keep it simple). What do I need from the engine? 10 000 W (= 10 000 N X 1 m/s).I think this is getting to the heart of the matter. As soon as you specify a particular speed, then torque and power are essentially two sides of the same coin, speed being the conversion factor. This whole discussion is similar to arguing whether a pressure differential in a pipeline causes a flow, or whether a flow causes a pressure differential. I would say neither, or both, if you like. They just happen together. In the same way, 10 000 N at 1 m/sec IS 10 000 W, and it's a matter of convenience (or personal opinion) which is considered the primary quantity and which is the derived quantity. Again, only because the speed is specified.

5. jack action says:
sandy stone

This whole discussion is similar to arguing whether a pressure differential in a pipeline causes a flow, or whether a flow causes a pressure differential.Funny that you mention pressure differential and a flow. If you had a hydraulic motor powering my vehicle, I don´t know what would be the its pressure differential or volumetric flow rate, but the combination of both would have to produce 10 000 W. This is the only characteristic required to be sure I can produce the desired force (or acceleration) at the given velocity. It would be a mistake to think that only the pressure is relevant because that is what will cause the wheel torque. Any pressure can produce that torque, as long as it has the appropriate volumetric flow rate.

6. D
Dale says:
jack action

We are talking about an acceleration at a given velocity (other than zero).I am certainly not excluding zero velocity. You would certainly like to exclude it since it makes your claim obviously wrong, but I do not consent to the limitation.

jack action

Is power or torque a better indicator of acceleration?Let’s examine it for the simple lossless case:

Wheel torque is always proportional to acceleration with a simple constant of proportionality.

Power is related to acceleration by a proportionality which is not constant.

At v=0 power becomes infinite, but acceleration and wheel torque maintain their standard proportionality.

At small but nonzero speeds power becomes arbitrararily large but acceleration and wheel torque do not.

Wheel torque and power are not independent but are closely related to each other by speed.

The combination of power and speed that is related to acceleration is itself proportional to wheel torque with the above simple constant of proportionality.

Wheel torque is proportional to acceleration from rest, power is not.

With a CVT power is constant, but both wheel torque and acceleration are not and at all times the wheel torque is proportional to the acceleration.

With a typical powertrain in a fixed gear the peak wheel torque often occurs at lower speeds than the peak power, the peak acceleration occurs at the speed of the peak wheel torque not the peak power.

For constant wheel torque acceleration is constant, while power is not.

So clearly, wheel torque is more directly related to acceleration than power. So what is the value of power: first it is the primary criterion to use when selecting an engine. Second, power is conserved but torque is not, so the relationship between wheel torque and engine torque is not as simple as the relationship between wheel power and engine power. Third, power is more relevant for determining peak speed. I am sure there are many other examples where power is more directly relevant than wheel torque, but acceleration is obviously one of the exceptions where wheel torque is more directly relevant.

jack action

Given those numbers, it must be very clear to anyone that the orange engine can produce way more acceleration than the blue one.Sure, and it does so precisely by converting that engine power into greater wheel torque. As I said above, power is the primary criterion for selecting an engine. The wheel torque is not independent of the engine power, but the wheel torque remains the quantity that is more directly related to acceleration.

7. jack action says:

@Dale , you know I agree with you on every sentence you write because, you said it yourself more than once, we basically say the same thing with different words. The ones I like from your last post are these ones:

Dale

The wheel torque is not independent of the engine power, but the wheel torque remains the quantity that is more directly related to acceleration.I like how you use the double negative. Instead of saying «is dependent of», you say «is not independent of». I prefer the positive version: Acceleration depends on wheel torque and wheel torque depends on wheel power, which is the same as engine power (not considering inefficiencies). And the reason why I can so easily jump to «acceleration depends on power» when considering a vehicle is because the speed range for any vehicle design is fixed. A passenger car has one, a tractor has another one, a race car another one. The speed range is a given, always.

This is why I can relate directly the power curve of an engine to its acceleration as long as the speed range are comparable. Knowing that there are so many people out there who strongly argue that «acceleration depends on wheel torque and wheel torque depends on engine torque», therefore «acceleration depends on engine torque», don't you think my choice of words are better suited to convey the science behind the phenomena? Those people are right with their statement by the way, it just doesn't take into account the speed change associated with the acceleration increase.

As for the difficulty you have about ##v=0##, I will quote another insight written by @anorlunda :
Underlying Assumptions and Limitations

Really there is only one assumption behind Ohm’s law; linearity. Not in the mathematical sense, but rather that a graph of voltage versus current shows an approximately straight line in a given range.

There are always limits to the range even though they may not be explicitly mentioned. For example, at high voltages breakdown and arcing can occur. At high currents, things tend to melt. In the old days, we said that real-world resistance can not be zero. But now we know that superconductors are an exception to that rule. ##R=0## is OK for superconductors.

Students often forget that limits exist. A frequent (and annoying) student question is, “So if ##I=V/R## , what happens when ##R=0##. Ha ha, LOL.” They think that disproves the “law” and thus diminishes the credibility of science in general. Their logic is false.

8. D
Dale says:
jack action

@Dale , you know I agree with you on every sentence you writeEven this one?

Dale

So clearly, wheel torque is more directly related to acceleration than power

9. jack action says:
Dale

Even this one?If so then state it clearly and if not then you need to address my specific justifications for that statement.

Dale

So clearly, wheel torque is more directly related to acceleration than power.If so then state it clearly and if not then you need to address my specific justifications for that statement.In a pure sense, your statement will be true in any case (like when ##v=0##, for example). If you specified some context (like specifying acceleration over a speed range), then you cannot distinguish which of wheel torque or wheel power is more influential on acceleration. My text do specified the limits and context for my statement. Furthermore, I wanted to show that the statement is also true even if there is no wheel torque.

Dale

I am fine with that. Engine power is the start, wheel torque is the end. Why is it so hard for you to admit that the end is more directly related to the result than the beginning? That should be obvious.Like I said above, in the specific case presented, they are equivalent and that is if wheel torque is present. So it ends up to be about which words you want to use to convey the message.

Why do I choose these words? Because I know my audience. I know what they read/heard before and where they get confused. I know the words they need to hear such that they come to the right conclusions. Most importantly, I know what words they don't need to read/hear, because it would only add to their confusion.

When you say «torque» (even if you specify «wheel torque»), they hear «engine torque». Once the torque requirement is fulfilled, most people failed to see the power requirement that must also be fulfilled. Somehow, the force concept is easier to «see» than power for physics newbies. Probably because you can feel a force but not power.

Finally, when the power requirement is fulfilled, the torque requirement is automatically fulfilled (at the desired velocity, that is). It makes things so much easier (no need to know the wheel radius or gear ratios). You should see the nightmare of calculations that some people do when using torque and RPM instead of power. It is very discouraging and it contributes to the hatred of math found in the general population. When I understood how easy it can be, I felt like a kid learning a magic trick!