# 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 [1], 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.

[1] 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. D
Dale says:

Of your initial three points, only the first one is correct.
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.At v=0 the second statement is completely false for any vehicle, and it is a very poor approximation for low speeds for automobiles. And the third is never correct, there are always other factors involved.

You do make a good point that not all vehicles produce torque as part of their operating mechanism.

For vehicles such as automobiles whose operating mechanism produces wheel torque, most of your arguments are specious. You could simply replace the word “power” with the word “torque” (or better yet “force”) and come up with the same argument for torque (or more generally for force) as being the primary quantity.

The one exception is the conservation of energy, which is a strong and general principle that should be understood.

The “anti torque” argument of your concluding section is also specious. Yes, you can rewrite wheel torque in terms of power, but for vehicles like automobiles you can also write power in terms wheel torque. The torque (or more generally the force) winds up being more closely related to acceleration.

Power is conserved, so it should be used to understand other conserved quantities. Acceleration is not conserved so it should not be too surprising that it is more directly related to non conserved quantities like torque or force.

2. C
cjl says:

Dale, you continue to miss the entire point here.

The question here is, given a set of parameters for your vehicle, which one do you care about for maximizing acceleration. Nobody, *nobody*, specifies a vehicle by tractive effort unless you're talking about a freight train. Rather, you have engine torque, horsepower, weight, etc. For the purpose of maximizing vehicle acceleration, the parameter you should care about is horsepower (more accurately, horsepower to weight). For acceleration at non-zero speeds, you should always be running the engine as close to peak power as possible, and to improve vehicle acceleration, you should always try to improve the power to weight ratio.

EDIT: Oh, and as far as the "conserved quantities" argument goes, you could easily reframe acceleration as being the rate of addition of kinetic energy. This makes it very clear that to maximize the rate of increase of kinetic energy, you need the greatest power possible.

3. D
Dale says:
cjl

The question here is, given a set of parameters for your vehicle, which one do you care about for maximizing acceleration. Nobody, *nobody*, specifies a vehicle by tractive effortI agree with this statement, and if the statement were specified that way then it would be correct.

But that isn’t what he said and what he said is incorrect.

4. R
Randy Beikmann says:

This is something I've thought through quite a bit over the years. What I realized is that you have to look at it from the standpoint of what limits the acceleration, and it depends on conditions. There are three common limitations: 1) Traction, 2) Power, and 3) Wheelstand.

Here's the key: The vehicle can accelerate no quicker than the least of these three acceleration rates. Note that the second depends strongly on speed. So how large is each possible acceleration level? (I'm not going to worry about wheelstand here, by the way.)

Traction-Limited Acceleration Level: Here, all you need to know is the maximum available traction force, and the vehicle mass (barring aero drag, road grade, etc.). So the traction-limited acceleration level is straight from Newton's second law: a_TL=Fmax/m.

Power-Limited Acceleration Level: Here you are assuming that all the available power is going into acceleration (ignoring for now whether there is enough traction to do so). Since F=ma, and Pmax=Fv, then Pmax=mav. The power-limited acceleration level would be just a_PL=Pmax/(mv).

Now let's take an example. Let's say the car's mass m is 1500 kg (3300 lb), the maximum traction Fmax is 8000 N (1800 lbf), and the maximum power Pmax is 180 kW (241 HP) at the tire/pavement interface.

If you evaluated acceleration at 10 m/s (32.8 ft/s), the power-limited acceleration would be about 1.223 G’s, which there isn’t enough traction to accomplish (the tires spin first). The available traction would limit you to a_TL=0.544 G’s. So the maximum possible rate at 10 m/s is 0.544 G’s.

If you evaluated acceleration at 30 m/s (98.4 ft/s), the power-limited acceleration would be about 0.408 G’s, which is lower than the traction limited acceleration, still 0.544 G’s. So the maximum possible at 30 m/s is 0.408 G’s.

There is a critical speed in between those, where the traction-limited and power-limited acceleration are equal. This is where Pmax=Fmax*v_Crit, producing v_Crit=Pmax/Fmax. In our example, this critical speed is 22.5 m/s (73.815 ft/s).

So for this vehicle, the maximum acceleration below 22.5 m/s is a_TL=0.544 G’s. Above that, the maximum is a_PL=(180000 watts)/((1500 kg)*v), so the maximum possible acceleration decreases hyperbolically with speed.

What’s the takeaway? At low speed, very large traction forces would be needed to convey max power, so acceleration is limited by available traction. At high speeds, accelerating at the traction limit would require very large amounts of power, so acceleration is limited by power. The speed Pmax/Fmax divides the two regions.

5. R
Randy Beikmann says:

Here is a plot of what I was describing. The max power could produce the dashed orange line, and the available traction could produce the dashed blue line. In actuality, acceleration is limited to the lesser of the two (the gray solid line).

View attachment 234264

6. jack action says:

I've been thinking long and hard about how to show you how your point of view is misleading, especially if you are addressing rookies in the field.

You are saying that all you need to create an acceleration at velocity ##v## is wheel torque because only forces are involved with accelerations;

I'm saying that you all you need to create an acceleration at velocity ##v## is wheel power because the power requirement is not the same at different velocities.

Both of these statements are true. But let's take a look at the same vehicle with a slightly different problem: Finding the maximum velocity.

Finding the maximum velocity is a pure «force» problem, just like acceleration. The tractive effort must be equal to the aerodynamic drag. Based on this, one can say:
$$T propto v^2$$
And that is technically correct: The wheel torque ##T## must be proportional to the squared velocity.

But I say ¨Hold on!¨ There is also power involved. The wheel power ##P## must be equal to the drag power. This statement is the same as the previous one, but we multiply both forces by ##v##. This leads to:
$$P propto v^3$$
Both of the above equations are true. But one of them conveys a lot more information than the other. The «torque» equation is misleading. A rookie will think that if he can put a gear ratio large enough, he will somehow get the wheel torque he wants from an energy source with a given torque. That is not true.

But with the «power» equation, he can assume that if he chooses the appropriate gear ratio, he can get the desired wheel torque AND accomplish that at the required velocity, as long as the energy source gives the required power.

The «power» equation implies the «torque» equation AND gives additional – and important – information.

I can assure you that your point of view will confuse more rookie than mine. I´ve been there, so did a lot of my peers at their beginnings. Sadly, even with a lot of experience, some still refuse to see the law of conservation of energy and pretend that only engine torque matters with acceleration, not engine power. And all of this is because they argue that «only wheel torque matters», which is directly linked to engine torque via the gearbox.

I guess what I´m saying is: Don´t feed the trolls.

7. D
Dale says:
jack action

You are saying that all you need to create an acceleration at velocity v is wheel torqueI would not say “all you need” because power and wheel torque are closely related and saying “all you need” is an oversimplification that misses a lot of important physics.

jack action

I'm saying that you all you need to create an acceleration at velocity v is wheel powerAnd this is clearly wrong. It is not “all you need”. As @Randy Beikmann clearly and correctly pointed out it is not even the limiting factor at low speeds.

jack action

Both of these statements are true.I disagree, neither is true. The truth is that either “all you need” statement is myopic and misses the interplay between the various factors involved. The truth is that wheel power and wheel torque are closely related.

However, in the end it is undeniable that the quantity most directly related to acceleration is wheel torque, not power. I say undeniable, except that you did exactly that, you denied it! You are so embroiled in your black and white “all you need” that you make statements which are false and arguments which are specious.

jack action

But let's take a look at the same vehicle with a slightly different problem: Finding the maximum velocity.For that problem I would unambiguously choose the power equation also. This is a different question and hence has a different answer. In that case power is clearly the more appropriate quantity to use and so I would use it.

jack action

I guess what I´m saying is: Don´t feed the trolls.What I am saying is: teach physics!

I am uninterested in choosing sides in the ongoing debate amongst car enthusiasts. But here, on Physics Forums, we should teach correct physics.

Whether you like it or not, torque is an important concept in physics and cannot and should not be neglected when it is the most appropriate concept for the question at hand. This is not a political campaign where we have the torque party and the power party and the other party consists of trolls. This is physics where both torque and power are essential concepts and each have their legitimate place in describing the world. Torque is not always the most relevant quantity, nor is power.

8. R
Randy Beikmann says:
jack action

Sadly, even with a lot of experience, some still refuse to see the law of conservation of energy and pretend that only engine torque matters with acceleration, not engine power.That energy is conserved is indisputable, but where it goes depends on the situation. At low vehicle speeds (say 5 mph), your engine will not be able to run at maximum power Pmax, because if you put your right foot to the floor, the driving tires will spin until your engine goes to redline. At no engine speed would you be able to reach Pmax=T_Engine*omega, because the traction will limit the torque that can be developed. Plus, some of the energy the engine creates will go into the vehicle's kinetic energy, but some will be converted to heat at the tread as the tires slide against the pavement. There simply isn't enough traction to handle the force you would calculate at the tire/road interface in assuming Pmax there.

A good way to illustrate the difference is to calculate the time to accelerate from zero to a given speed, using constant power vs. Newton's laws. Suppose I want to calculate the time for the car I used above to go from 0-45 mph (66 ft/s). If I did so setting power = Pmax, and set final kinetic energy (0.5*m*v^2) equal to work done (Pmax*t), I would get t=1.69 seconds. If I used Newton's laws, then the final velocity equals the traction-limited acceleration multiplied by the time, or v = a_TL*t. Solving, this produces t=v/a_TL, which in this case is 2.51 seconds, and is the actual answer. Why is the actual answer so much slower than the power-based calculation? Because applying the entirety of Pmax to accelerating the vehicle is impossible at low speeds, as noted above.

I would never advise anyone to always favor the energy approach over Newton's laws, or vice versa – in fact you must consider both factors to ensure a correct solution. Both should be considered, and the result should be the same. If they're not, you're missing something. In this case it would be that "power being conserved" is only applicable if there are no factors involved that dissipate some of it as other forms of energy, in this case heat.

I would never advise anyone to always favor the energy approach over Newton's laws, because there is no reason to – in fact you must consider both factors to ensure a correct solution. Both should be considered, and the result should be the same. If they're not, you're missing something. In this case it would be that "power being conserved" is only applicable if there are no factors involved that dissipate some of it as other forms of energy, in this case heat.

9. D
Dale says:
Randy Beikmann

I would never advise anyone to always favor the energy approach over Newton's laws, or vice versaWell said!

10. jack action says:
Randy Beikmann

because the traction will limit the torque that can be developedWhat if there is no traction limit? Well the passengers inside the vehicle will die if the acceleration is too great. That is another limit.

What if there are no passengers inside? Well the mechanical components will break apart under a certain acceleration level. That is another limit.

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.

Let me show you the problem at hand. If you discuss with a «torque» person and ask him which of 2 engines you have can produce the largest acceleration knowing that you have the exact same transmission in both cases, you can only change the differential gear ratio. He will say ¨Acceleration depends on wheel torque, so show me the torque of the engines¨:

After seeing that, he will suggest the blue engine. Not asking you anymore questions. But you say ¨Some people look only at the engine power, what do think about that?¨:

He will answer back ¨It doesn´t matter, only torque creates acceleration¨. So you tell him ¨But they insist, they compare the power curves over the RPM/RPM[SUB]max[/SUB] ratio¨:

He will answer ¨It doesn´t prove anything. See the torque curves with that same condition¨:

¨There, those scientists don´t know anything about real life. It´s only theory.¨ And he will add ¨Besides, go on PF and you´ll see that scientists don´t all agree on that: some say that only torque matters, just like me.¨

This is the implication of not choosing carefully your words about the subject. If you think I am creating something out of nothing, see this thread where the OP cannot stop opposing engine builders with scientists. Even his nickname is Moretorque for pete´s sake. You can find hundreds of discussions like this over the net and these discussions were going on before the net existed.

I don´t care about any other limits (traction, human tolerance, material resistance, etc.) can lower the threshold of maximum acceleration. Given those numbers, it must be very clear to anyone that the orange engine can produce way more acceleration than the blue one. There are no ´if´ or ´but´, or any kind of discussion possible about it.

Otherwise, one could also argue that acceleration is not proportional to the force as well, since ##F = vfrac{dm}{dt} + ma##. So even if there is a force, the acceleration could be zero, therefore you cannot assume proportionality. Even if it´s true, it is a really dishonest – and not helpful at all – argument.

Don´t feed the trolls.