# Calculating Acceleration from Power: A Deeper Look

• nolaman
In summary: But the equations that claim to do this usually have the conversion from power to force buried in constants, so they fail by definition. In summary, this topic is very old and has been discussed many times. Most equations that claim to do the calculation of acceleration do not work, but those that do have the conversion buried in constants. Knowing power, force, and speed can help calculate the third variable, acceleration.
nolaman
This is a very old topic that I wanted to "publicize". My contention is that you cannot use power to calculate acceleration without first converting power to force. It is possible to calculate an acceleration by assuming some period of time to calculate a change in energy, but this method does not yield instantaneous acceleration. It seems trivial, but take a look at this spreadsheet.

I've seen many, many equations that claim to do this calculation, but not one of them worked, and the ones that worked had the conversion from power to force buried in constants. So before posting any formulas please try them out with the spreadsheet to make sure they work.

You are correct: acceleration is a function of torque at the wheels, so you need engine rpm and gear ratio (or a horsepower and a wheel rpm). Or, put another way, acceleration of a car is a function of the engine horsepower and what gear the car is, not engine horsepower alone.

So how is this useful? If you want more acceleration, stay in lower gears longer.

You also need to know the mass of the vehicle being accelerated.

The last graph on that link shows an idealized case with a CVT (continously variable transmission); power output is constant, torque diminishes as speed increases. Here the inputs would be power and mass of vehicle. (The force at each instant = power / speed).

In fact, knowing the rear wheel horsepower and the gear does not yield a workable solution. As is pointed out in the spreadsheet, at a given engine horsepower the rear wheel horsepower is identically the engine horsepower in all six gears.

Useful, nope. But it is interesting. The formula that takes you from torque to HP must be backed out, that is, you must convert from power back to force to do any acceleration equations. It's a one-way, dead end street.

The weight data is in the spreadsheet.

nolaman said:
In fact, knowing the rear wheel horsepower and the gear does not yield a workable solution.
But knowing the rear wheel horsepower and speed does work, since power = force x speed, then force = power / speed.

nolaman said:
This is a very old topic that I wanted to "publicize". My contention is that you cannot use power to calculate acceleration without first converting power to force. It is possible to calculate an acceleration by assuming some period of time to calculate a change in energy, but this method does not yield instantaneous acceleration. It seems trivial, but take a look at this spreadsheet.

I've seen many, many equations that claim to do this calculation, but not one of them worked, and the ones that worked had the conversion from power to force buried in constants. So before posting any formulas please try them out with the spreadsheet to make sure they work.

Could you provide these 'many many equations that claim to do this calculation?

Cyrus, they all fall apart. Most commonly the conversion from power back to force is imbedded in them, so they fail by definition. The others, most commonly based on energy change, do work after a fashion, but the results are ambiguous: The same amount of energy changenproduces different acceleration rates depending on initial (or final) velocity.

Jeff, that's converting back to force. That was a restriction in my original statement.

nolaman said:
Cyrus, they all fall apart. Most commonly the conversion from power back to force is imbedded in them, so they fail by definition. The others, most commonly based on energy change, do work after a fashion, but the results are ambiguous: The same amount of energy changenproduces different acceleration rates depending on initial (or final) velocity.

Jeff, that's converting back to force. That was a restriction in my original statement.

What is 'they'? Can you please provide said equations. I still don't know what you're talking about.

Jeff Reid said:
But knowing the rear wheel horsepower and speed does work, since power = force x speed, then force = power / speed.

nolaman said:
That's converting back to force.
It's part of the definition of power, a rate of work, which equals force times speed.

For example, 1 horsepower = 550 ft lbf / sec = 550 lbf (ft/sec) = force x speed.

So there is an equation relating these three variables, power, force, and speed. If you know any two of these, you can calculate the third.

The original question is, knowing power in that example can you calculate acceleration without first converting back to force? The answer is no.

You need the total reduction of the driveline (including the drive wheel radius) and the RPM. Fair enough, you need that to calculate acceleration from torque. So now you have enough information to calculate acceleration using force. But you still don't have enough information calculate acceleration from power.

So with RPM, gear reduction, drive wheel radius (those three used only to calculate initial speed), mass, power and initial speed can you calculate acceleration? Once you've got all the details worked out and know the acceleration, what happens when the RPM changes?

It's like the old joke about a fellow who asks a farmer for directions. The farmer replies, "You can't get there from here. You've got to go someplace else first".

I will admit that my original statement is flawed in that you could say that power and acceleration are indirectly related. I must confess that was on purpose.

Cyrus, I didn't collect them. They didn't work so I had no reason to.

Jeff Reid said:
It's part of the definition of power, a rate of work, which equals force times speed.

For example, 1 horsepower = 550 ft lbf / sec = 550 lbf (ft/sec) = force x speed.

So there is an equation relating these three variables, power, force, and speed. If you know any two of these, you can calculate the third.

This may sound flippant but what if your car has no wheels?

We're now in Zen-land, and the answer is that the car has nowhere to go.

nolaman said:
We're now in Zen-land, and the answer is that the car has nowhere to go.

But if a car loses its wheels to a scouser and ends up on bricks does it still have horsepower and further is it moving? Ahhhhh...

Apologies to scousers but well you know.

I genuinely didn't mean it to be flippant, but I think in principle I agree with the OP, there is a certain removal from the subject.

nolaman said:
The original question is, knowing power in that example can you calculate acceleration without first converting back to force? The answer is no.

You need the total reduction of the driveline (including the drive wheel radius) and the RPM. Fair enough, you need that to calculate acceleration from torque. So now you have enough information to calculate acceleration using force. But you still don't have enough information calculate acceleration from power.

So with RPM, gear reduction, drive wheel radius (those three used only to calculate initial speed), mass, power and initial speed can you calculate acceleration? Once you've got all the details worked out and know the acceleration, what happens when the RPM changes?

It's like the old joke about a fellow who asks a farmer for directions. The farmer replies, "You can't get there from here. You've got to go someplace else first".

I will admit that my original statement is flawed in that you could say that power and acceleration are indirectly related. I must confess that was on purpose.

Cyrus, I didn't collect them. They didn't work so I had no reason to.

I have no idea what you're talking about. I took a course in Vehicle Dynamics a year ago.

Why don't you just buy a book on vehicle dynamics and use it, why is what you are trying to do so hard?

You can't even provide any equations when asked...

I think the problem is you're trying to plug and chug and have no idea about the scope of the use of your equations.

Garbage in = Garbage out

So I should have collected a bunch of half-baked and incorrect formulas?!

It's a simple question. Let me restate it this way: In the example in the spreadsheet, can a single, consistent formula be written that calculates acceleration from power without converting power to force? If not, then power is not mathematically related to acceleration.

This all goes back to an old discussion among gearheads: What's important in vehicle performance, torque or power? There's a split camp, one saying torque, the other power. One of my points is that when you look at the torque curve you are looking at the shape of the acceleration curve. The relationship between the torque curve and the actual acceleration curve is a single scaling number for each gear, consistent across the RPM range.

Looking at the power curve very few people could guess where the acceleration is higher, lower or static. And even then it takes a very educated guess.

This being a physics forum I thought maybe I'd pose the problem here. I don't mean to upset anyone. BTW, I'm a retired M.E., rotating equipment specialist.

Jeff Reid answered your question. You are correct that (assuming no dissipative losses) power alone doesn't give you acceleration: you need momentum too.

P = F x v

F = m x a

From that you find a = F / m = (P/v) / m = P / (m x v)

In other words, power divided by momentum gives you the acceleration.

And it doesn't matter how it is applied, through what gear ratio or whatever, and what's the diameter of the wheel.

Gear ratio serves simply to allow an engine with a given rpm/torque curve to allow to put out its maximum power at the given velocity. If you KNOW that your engine is delivering a certain amount of power, then you do not need to know the gear ratio.

OmCheeto
That formula contains the conversion back to force (and requires calculating the velocity to find momentum). Momentum / velocity = force.

So far I haven't seen a formula that fills the criteria, and I've been looking a long time.

nolaman said:
That formula contains the conversion back to force (and requires calculating the velocity to find momentum). Momentum / velocity = force.

Yes, but you should understand that you do not need specific gearing ratios or whatever. Also, you do not need to know velocity to have momentum (although usually this is the way one calculates momentum). If mass is unknown, but momentum is known, that's good enough.
So far I haven't seen a formula that fills the criteria, and I've been looking a long time.

You can look forever, you won't find it because it doesn't exist. From power alone, you cannot calculate the acceleration, simply because for the same power, you can have different accelerations if your system under study has different momenta.

This is a bit like asking for the formula that gives you the two elements of a division, if you know the result (if you know that the result is 5, then you want to find: 35 and 7, because 35 divided by 7 equals 5) and you look all over the place for it. You won't find such a formula, simply because *different answers yield the same given*. In the same way, if you have 100 W of power, this can lead to an acceleration of 10 m/s^2 or of 0.1 m/s^2, depending on whether the momentum was 10 kg m/s or was 1000 kg m/s.

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I was trying to stay within the example. You have a vehicle with known gear ratios, weight, drive wheel radius. You are provided a horsepower curve. That's what you have to work with. So to match a given point on the power curve with momentum you need the gear ration and drive wheel radius.

Yes, I am certainly convinced that there is no way to do it. Someone referred me to this forum as a possible resource.

Thanks for the reply. I like the "35, 7" problem analogy. I might start using it, if you don't mind.

Aside, one thing I keep getting is formulas that don't work. I mentioned this in an earlier thread. Yesterday someone sent me this one:

acceleration = power / (mass * (vf - vi))

Where do they get this stuff??

nolaman said:
I was trying to stay within the example. You have a vehicle with known gear ratios, weight, drive wheel radius. You are provided a horsepower curve. That's what you have to work with. So to match a given point on the power curve with momentum you need the gear ration and drive wheel radius.

Yes, that's correct, but that's simply because you do not give the power, but you have to deduce it from the rpm of the motor. So power is not a given here. If it were, then
acceleration = power / momentum would do.

But you don't know what is the power unless you deduce the rpm.
For a given velocity, you have:
1) velocity + mass allows you to calculate momentum
2) velocity + wheel diameter + gear ratio allows you to calculate rpm
3) rpm + horsepower curve gives you power

So now you have power and momentum, and you can calculate acceleration.

With the given information you can calculate engine horsepower, which is the same a rear wheel horsepower. You can also calculate speed.

However, every time I try this there is an error of +10.5%. Take a look here:

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So I figured out where the error was coming from (sloppy mistake), so the formula does work.

Sadly, it contains the conversion from power back to force. I hadn't noticed it until I started looking for the source of the error.

With power in this case defined as (torque * (rev/sec) * 2 * pi), and momentum defined as mass times velocity, and velocity defined as (mass * (rev/sec)*2*pi) * (drive wheel radius) / (gear ratio), the power term reduces to torque.

Hey guys,

I am trying to simulate/calculate a car's accerelation (and velocity) by using torque and RPM and gear ratios. I use SI metrics (kg, Nm, m/s2 etc...)

I have reviewed spreadsheet and I have some question related with it.

1. What's primary and final gear ratio? And how it is related with gear ratio? Is it some kind of a motorcycle special thing?

2. While calculating over all ratio (at cell D15) why ratio is multiplied by 12 (what is the meaning of this constant)

3. While calculating over all ratio (again) why ratio is divided by wheel radius?

4. During calculation of accerelation (at cell H19), the value is divided by 32.2. Why?

I may not get constant values beacuse of metric system.

Thank you all.

1. You are correct that this is particular to motorcycles. Most motorcycles have a reduction chain and sprocket or gearset between the crankshaft and the transmission input shaft. There is also a reduction in the rear wheel drive between the transmission output shaft and the rear wheel. Cars will have a rear drive reduction (in the differential), most have direct drive from the clutch to the transmission. Particularly some front wheel drive cars have a reduction between the clutch and the transmission input shaft. Check vehicle specifications carefully. If there is no primary reduction (as an example) just plug the value 1 into that cell in the spreadsheet.

2. The overall ratio from engine to ground varies with the rear wheel radius. Since the spreadsheet is in English units (messy) and torque is in foot-pounds, the ratio of 12 to the rear wheel radius in inches is used. So you've got drive train reduction (transmission and differential) and wheel and tire radius correction. In SI units if you're using Newton-meters for torque and centimeters for wheel radius you would use an appropriate correction to maintain dimension agreement [(10 cm/M)/(radius in cm)], or express the tire radius in meters.

3. Refer to #1 and #2. If the wheel and tire combination is 1 foot in radius (12 inches) then it will produce twice the torque reaction on the vehicle of a wheel and tire combination 2 feet (24 inches) in radius. Correction for a tire 2 feet in radius = 12/24 in English units.

4. It's that messy English system again. One pound of weight = (one pound)/(32.2 feet/second^2) mass. Pound weight vs. pound mass. A = F/M.

Hope this helps.

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## 1. How is acceleration calculated from power?

Acceleration can be calculated from power using the equation: a = P / mv, where a is the acceleration, P is the power, m is the mass of the object, and v is the velocity.

## 2. Can acceleration be calculated using any power unit?

No, the power unit used must be consistent with the other units in the equation. For example, if the mass is given in kilograms and the velocity in meters per second, then the power should be given in watts.

## 3. How does increasing power affect acceleration?

In general, increasing power will result in an increase in acceleration. This is because power is directly proportional to the acceleration, as seen in the equation a = P / mv. However, other factors such as friction and air resistance may also affect the acceleration.

## 4. Is acceleration the only factor affected by power?

No, power can also affect other factors such as velocity and force. For example, increasing power will result in an increase in velocity, as seen in the equation v = P / ma. It can also affect the force applied on an object, as power is equal to the product of force and velocity.

## 5. Can power be used to calculate acceleration for all types of motion?

Yes, power can be used to calculate acceleration for all types of motion as long as the necessary variables (mass, velocity, and power) are known. However, other factors such as friction and air resistance may also need to be taken into account for more accurate calculations.

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