# When Vehicle Power Dictates Acceleration

[Total: 7    Average: 3.3/5]

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.)

Tags:
33 replies
1. T
Tom.G says:

At the risk of intervening in an apparent theological war, does it help to state that Power is a derived unit from Force (Torque), Speed, and Time?
That might even make it easier to explain to non-physicists.

2. D
Dale says:
jack action

In a pure sense, your statement will be true in any caseExcellent. I am glad you agree.

jack action

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.In the many cases where they are equivalent then it is indeed merely a choice of words. However, I carefully chose examples where they were not equivalent. Since they are not always equivalent, it is not always just a stylistic choice.

3. jack action says:
Dale

However, I carefully chose examples where they were not equivalent. Since they are not always equivalent, it is not always just a stylistic choice.Agreed.

Let me show you how people do the calculations when they focus on torque (I've been there). First they go for your simple equation:
$$ma = frac{T_w}{r}$$
What is my wheel torque at velocity ##v##? Let's find out the engine rpm ##omega_e##:
$$omega_e = G_rfrac{v}{r}$$
Where ##G_r## is the gear ratio. Then, let's find the engine torque at the given engine rpm by examining the torque-rpm curve. Once you know the engine torque, we find the wheel torque:
$$T_w = G_r T_e$$
Finally we go back to our original equation and determine the acceleration. But is it the maximum acceleration we can get? The only way to be sure is to repeat the process with other gear ratios, build a table and compare acceleration values. Then I can choose which gear ratio is better suited to get maximum acceleration. But as I increase my speed the acceleration is always lower than at lower speeds. How do I know if this lower speed acceleration is the maximum I can get? Let's try another gear ratio, then. In the end, it is just how patient you are when it comes to try an infinity of possibilities, gear ratio-wise.

Yes, they are still people doing it this way. Essentially, they are solving the following equation, one set of brackets at a time:
$$ma = frac{left(G_r T_eright)}{left(frac{omega_e}{G_r}rright)}left(frac{omega_e}{G_r}right)$$
It is extremely messy and it shows that you are not understanding something about physics. What I want them to see is:
$$ma = frac{P_w}{v}$$
and because:
$$P_w = P_e$$
Then you can simply say:
$$ma = frac{P_e}{v}$$
No need to convert velocity to angular velocity. Just convert your power-rpm curve to a power-velocity curve, and where you have the most power, you know you have the most acceleration possible at that speed. Once you're satisfied with this, then find out what gear ratios and wheel radius combination you need.

Here is a real life question I had from one of my website reader (yes, even after he read my website where I don't focus on gear ratio or tire radius):
I would like to know what factors you take into account when selecting gear and diff ratios for performance applications.

I only have minimal manufacture specifications (weight, weight distribution, frontal area, rolling resistance) and a simple power curve to go off. I also have a speed trace of the track it would be going around.

The current list of factors I am using to derive and tune ratios are:

-Acceleration force per gear. (accounting for resistive forces I.e. drag)

-Tire grip *Wheel Spin* (accounting for vertical forces on the axel)

-RPM drop and where in the powerband it drops to

-Short shifting

I was wondering if there were any major factors I am overlooking and should look into while selecting ratios.

I have attached my main tool if you have time to look at it. I appreciate that you’re probably quite busy and would completely understand if you didn’t have the time. But even a few pointers would be world of help.This came with an Excel spreadsheet filled with the tables mentioned above, and more. He is obviously drowning in numbers. This was my answer:
The basic concept to remember is that, for any speed, the acceleration is proportional to wheel power.

Since power varies with RPM, you want to coincide the RPM where you get the maximum power with the speed where you spend the most time accelerating. Say you know you spend a lot of time accelerating between 150 and 200 km/h, then you should have one gear ratio that gives you peak power RPM when you are at 175 km/h. If you spend most of your time in that speed range, you may also want to set the lower and higher gear ratios closer to that ideal ratio. This will increase your average power in that speed range (closer to peak power).

This is what is shown in the next figures. It is actually based on the numbers from your attached file.

Wherever the lines crosses, these are your ideal shift points. You can see how the average power (between shift points) in the 150-200 km/h range is greater in the second figure, compared to the first figure. Of course, this is only an example, and the power lost at 130 and 200 km/h might be unacceptable. In such case, you either have to set a smaller overall speed range by bringing closer the 1st & 5th gear ratios (i.e. less power at low speed and a lower top speed possible) or you can add gear ratios to fill the voids (i.e. a 7-speed gearbox is required). Note that as you increase the number of gear ratios within a given speed range, you set the average power closer to the peak power (but shifting may cause some losses and/or get more complicated).

This is great, thank you so much! You gave me a lot of insight into the topic and I hadn't thought of it that way.Nobody thinks of it that way when they are new at this. Too many even refuse to do otherwise as they get experience. Why? Because they focus on «acceleration means torque».

4. R
Randy Beikmann says:

Jack, the example you showed here gave valuable insight to the user of your website, especially since it was applied to a specific situation. You used the concept of "power flow" through the driveline to simplify the problem of choosing gear ratios, using a systematic way to do it instead of trial and error. The plots showed how changing the grouping of gear ratios could more closely approximate a "constant power" relationship across a certain range of vehicle speeds.

I don't believe the article in "Insights" did justice to the way you explained things here. Now I better understand what you were after.

I do believe, though, that comparing alternate methods is very valuable in gaining insight. If things had started off with you teaching him how to do it by following the power through the driveline (rather than torque and force), he never would have appreciated the way it made things easier for this purpose. Initial frustration is a good motivator.

For some other purposes, I do still like plotting the acceleration capability vs. speed, including the case where only power limits it. That makes it evident that, even ignoring aerodynamic drag, the capability to accelerate diminishes with increasing speed. Also, if you need the performance envelope of a vehicle on a road course to do simulations, you can express it in terms of longitudinal and lateral accelerations (or forces), but not power since it does not enter into lateral acceleration.

The performance envelope might come in handy for your user's problem of matching gearing to a road course. Besides "smoothing" the power delivery in a specific speed range, he might also want to make sure he doesn't have to shift while accelerating out of a critical corner.

5. jack action says:
Randy Beikmann

I do still like plotting the acceleration capability vs. speed, including the case where only power limits it. That makes it evident that, even ignoring aerodynamic drag, the capability to accelerate diminishes with increasing speed.That is why I prefer ##ma = frac{P}{v}## rather than ##ma = frac{T}{r}##. With the torque equation, it leaves the impression that you can get any level of acceleration, at any speed. With the power equation, you clearly see that the acceleration will decrease as the speed increases.

6. andrewkirk says:

Thank you @jack action for your Insight article and also @Randy Beikmann and @Dale for your contributions to this discussion. I have long had a confusion about why people talk about both power and torque, and the relationship between the two, and had never found time to go into it enough to understand the role they play in the context of a motorised vehicle. This discussion finally cleared it up for me, especially Randy's graph on page 1!

7. J
jim hardy says:

(What ? no Latex in Insights comments ?)

Well of course we are free to choose tire radius r, when we build the vehicle.
We could select gears and wheel size to make torque and thrust numerically equal just to simplify the algebra.
That’d make acceleration = Torque/Mass

Radians per second X r is velocity
and Torque X radians per second is power
Power, velocity and Torque form a triad
so if you know any two you you know the third.
So to argue that using one is “better ” than using the other is pointless.

“”The radius r implies a transformation where power is kept constant,”
How ?
No more than “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” implies constant power.
Both those assumptions would require an infinitely variable transmission, and one of those is not assumed.” implies it.

For any real engine torque and power are both functions of RPM(velocity)

so the choice of approaching it from power or torque is a personal preference, nothing more.

Acceleration is the key to drag racing
So
Let’s assume distance s and use the familiar equation of motion s=(1/2) at^2 to figure time
where
distance = s
acceleration = a
time = t
velocity = v
power = P

We could set distance = 1320 feet but it’s cleaner to just use the symbol “s”..

s= (1/2) at^2
time t = sqrt(2s/a)
obviously to minimize t you maximize a
since i chose gears and wheel radius to make torque equal thrust,
a= Torque/m
which makes time t = sqrt(2s X m/Torque)
and i can lump constants to get
time t = K/sqrt(Torque)
but Torque is not a constant, it’s a function of engine rpm(velocity), so a proper solution involves integration.

Since P = mav
a = P/mv
substituting that into s= (1/2) at^2

s = (1/2) (P/mv) X t^2
t = sqrt(2s X mv/P)
again i can lump constants
t = K X sqrt (v/P)
Neither v nor P is constant (neither was Torque)

so you have a choice of which function to integrate, Torque or the function v/P

Torque and Power are readily available from the engine’s performance curves, but v will have to be written as a function of torque and time .

I just don’t understand your anti-Torque stance.

What ? no Preview in Insights Comments ?

old jim

8. J
jim hardy says:

???? All those previous comments showed only after i posted mine.
This INSIGHTS board needs some serious software work.

9. Greg Bernhardt says:
jim hardy

(What ? no Latex in Insights comments ?)Should be fixed in a month or two. Comment in the forums threads.

10. J
jim hardy says:

I’ve not read the prior comments , but certainly would have read them before commenting if they’d shown,

so if my remarks are repeat of somebody else’s please excuse me.. I’ll find them after a while.

No response necessary – i’ll catch up…

old jim