Calculating wheel torque from engine torque

[xxChrisxx]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

That intersection point is also not the upper limit of rate of acceleration in 2nd gear. The upper limit of acceleration in 2nd gear is around 25 KPH, Fmax in 2nd gear.
Can we agree on that?

You are both again talking at cross purposes, but it feels like a bit of progress. Noone has ever actually disagreed with your assertions about engines and gears. It's more your lack of appreciation of power as a tool for calculating performance and tuning.

Everyone agrees that 2nd gear the peak acceleration in gear is at the peak torque. If you'll look back Jack pointed out the peak force in gear at the peak of torque output with a big orange arrow.

Would it help if we rename the 'Fmax' line as 'power limited acceleration'?

The value of this line is that it is the boundary of performance for a given power output. Think of it this way; any gap between the ratio lines curves and this line is lost performance.

The goal of gear ratio selection is to minimise the area between the 'ratio curves' and 'power limit' curve. There will always be a compromise and some lost performance due to the nature of fixed gears and IC engines.

 

[OldYat47]

No, I don't want to start renaming things as long as I understand precisely what is meant by terms like Fmax. Jack Action is using Fmax as a function of v and the value of P at that v. My argument remains that all the information you need is present in the torque curves. I haven't seen any reason to say that the power curve is of any use in the context of vehicle performance. My secondary argument is that there is no way to calculate acceleration from power. In math terms, for any non-zero value of power there are infinite valid values of force (and by extension, acceleration).

I do feel like we're making progress, which is encouraging.

 

[jack action]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

OK.

So from now on, I will refer to «the highest value of F for that ratio» as Fmax.

From now on, I will refer to «the value of F under maximum engine power» as F@pmax.

Let's suppose we are trying to accelerate. You shift from 1st to 2nd so you wind up at Fmax in 2nd gear.

I'm stopping you right there. I'm not sure if you are talking about Fmax or F@pmax (I think it is Fmax), but in either case, nobody talked about shifting to «wind up at Fmax». I thought we already agreed that we shift when the wheel torque in first gear is the same as the wheel torque in 2nd gear (where 1st gear curve crosses 2nd gear curve)?

From that speed until the intersection of the 2nd gear ratio and the constant power line the vehicle is accelerating more quickly than it is at that intersection point (torque values higher, A=F/M, more F = more A).

Agreed.

For that reason I make the statement that that intersection point is not the point of maximum acceleration. It is at 40 KPH in 2nd gear, but not overall in 2nd gear.

Agreed.

That intersection point is also not the upper limit of rate of acceleration in 2nd gear.

Agreed.

The upper limit of acceleration in 2nd gear is around 25 KPH, Fmax in 2nd gear.

Agreed.

My turn.

Can we agree that when you are at Fmax in 3rd gear at 40 km/h, at that same car speed, you can also be at F@pmax in 2nd gear?

Can we agree that F@pmax in 2nd gear is greater than Fmax in 3rd gear and therefore will produce a larger acceleration (torque values higher, A=F/M, more F = more A)?

For that reason I make the statement that the intersection point is the point of maximum acceleration at 40 km/h, no matter what gear you choose (but you have to be in 2nd gear to reach that point).

Can we agree that this same intersection point is the upper limit of acceleration at 40 km/h? Meaning that if the car is set in any other gear (1st, 3rd, top) at 40 km/h, it will not have a higher acceleration than in 2nd gear.

Can we agree that when you are at Fmax in 2nd gear at 25 km/h, at that same car speed, you can also be at F@pmax in 1st gear?

Can we agree that F@pmax in 1st gear is greater than Fmax in 2nd gear and therefore will produce a larger acceleration (torque values higher, A=F/M, more F = more A)?

For that reason I make the statement that the intersection point is the point of maximum acceleration at 25 km/h, no matter what gear you choose (but you have to be in 1st gear to reach that point).

Can we agree that the intersection point at 25 km/h is the upper limit of acceleration at 25 km/h? Meaning that if the car is set in any other gear (2nd, 3rd, top) at 25 km/h, it will not have a higher acceleration than in 1st gear.

 

[xxChrisxx]

My secondary argument is that there is no way to calculate acceleration from power. In math terms, for any non-zero value of power there are infinite valid values of force (and by extension, acceleration)

You could make the exact same argument about the general case of torque:
Traction force = Engine torque * gear ratio * FD ratio / Rolling radius
For any non zero value of engine torque, there are infinite valid values of gear ratio, FD and wheel size to give a traction force. and by extension.

What happens is you then define the variables to give a specific case.

F = P/v
You have a power @ an engine speed, and the engine speed relation to road speed, is the overall ratio. Ie the exact same thing you have to define above to give the specific case.

It's all one and the same.

 

[cjl]

OK, so part of it is semantics. I don't like using Fmax unless it is the highest value of F for that ratio. But I'm not going to argue semantics. I'll agree to use your terminology.

But let's say for a second that we don't care about the ratio at all. Let's say we can pick the ratio arbitrarily - we have a CVT, for example. If we're going 25mph and want to maximize our force, what gear ratio do we pick? Do we pick a ratio such that the engine is at peak torque? No, because that won't actually give us the maximum tractive force available at 25mph. To obtain the maximum possible tractive force at that speed, we need to pick a ratio such that the engine is at peak power, hence why jack is calling it fmax - given that engine, it is the highest possible tractive force the vehicle could generate at that speed (no matter what gear ratio you pick).

 

[OldYat47]

Jack Action, we are talking the same language finally and I agree with all your points. Whoopee! It was all semantics. And yes, I meant Fmax, not F@Pmax.

You could make the exact same argument about the general case of torque:
Traction force = Engine torque * gear ratio * FD ratio / Rolling radius
For any non zero value of engine torque, there are infinite valid values of gear ratio, FD and wheel size to give a traction force. and by extension.

Not the same thing at all. The bottom line is if you know the drive wheel torque you can directly calculate acceleration.

Ignoring friction, power at the drive wheels is always the same as engine power regardless of gearing. But acceleration does change with gearing. The values of power remain constant as the values for acceleration change depending on the reduction ratio. Knowing the drive wheel power, you cannot calculate acceleration. You have to know velocity, which brings you back to A=[(F*v)/v]/M, or A=F/M.

cjl, that's not correct. Look at a any torque and power vs. RPM set of curves. You can choose to program your CVT so as to maintain any engine speed. Your goal is maximum acceleration. If the torque is lower at (maximum power RPM) than at (maximum torque RPM) then acceleration will be less at maximum power than at maximum torque.

Just in time. I'm "unretiring" for a few days (or more).

 

[jack action]

@OldYat47, I'm glad we agree. Now let me share a thought experiment with you.

Let's say that we are cruising at the constant speed of 35 km/h in 3rd gear and we happen to have the throttle wide open. It would be the yellow dot on the red line in the figure below:

• The green dots represent where Fmax is for each gear;
• The black dots represent where F@pmax is for each gear;
• The blue dots represent the traction force at 35 km/h in each gear.

We now want to accelerate to 36 km/h as fast as we can. What do we do?

1. Any point in top gear provides a smaller traction force than the one we already have, so they're eliminated;
2. Fmax in 3rd gear offers a larger traction force, but we need to be at 37-38 km/h to get it. We can't instantaneously be at 38 km/h to get the traction force to help us reach 36 km/h; That makes no sense;
3. We could downshift in 2nd gear and get the largest traction force possible at 35 km/h (even larger than in 1st gear at 35 km/h);
4. But there is an even greater traction force at Fmax in 2nd gear (or even the largest traction force possible, i.e. Fmax in 1st gear). The problem is that we need to decelerate to 25 km/h (or even 20 km/h in 1st gear) to get it. That is not really smart, especially knowing that once we'll be back again at 35 km/h (remember our goal is to reach 36 km/h), we'll have the traction force that we would have had by staying at 35 km/h and simply downshift into 2nd gear (point #3).

Can we agree that point #3 is the smartest move one can do to get the greatest acceleration?

But can I get a greater traction force at 35 km/h, just by modifying the gear ratio of my 2nd gear? Let's see.

Let's set a 2nd gear with a lower gear ratio. Let's call it gear 2a (orange in the next figure, which is an enlargement of the previous figure):

Hurray! We have more traction force at 35 km/h! Note also that F@pmax moved to the left (to a lower car speed) on the «Constant power» curve.

Let's try an even lower gear ratio. Let's called it gear 2b:

It provides an even larger traction force at 35 km/h! Note that the traction force at 35 km/h in gear 2b is now exactly F@pmax.

It goes so well, let's try a gear ratio even lower. Let's call it gear 2c:

Oh no! Now we got less traction force than with gear 2b. Actually, we have the same amount of force than with gear 2a (because we have the same engine power in both cases, i.e. slightly less than the maximum engine power). Note that F@pmax is now at an even lower car speed, still moving left on the «Constant power» curve.

So, can we agree that when you are at a given car speed - say 35 km/h - to get the maximum traction force (thus, the maximum acceleration), the engine rpm must be where it produces its maximum power?

Can we agree that this will be true at any speed?

Therefore, can we agree that the maximum acceleration at any given speed depends on the engine maximum power?

 

[OldYat47]

Yes, #3 is the best shift option.

And, you can manipulate gear ratios to make maximum power and maximum torque coincide (sort of like horsepower and pound feet of torque are always equal at 5,252 RPM). That may or may not be the best option for optimum vehicle acceleration. It's an artifact. You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

 

[cjl]

cjl, that's not correct. Look at a any torque and power vs. RPM set of curves. You can choose to program your CVT so as to maintain any engine speed. Your goal is maximum acceleration. If the torque is lower at (maximum power RPM) than at (maximum torque RPM) then acceleration will be less at maximum power than at maximum torque.

This is not true. Wheel torque will be highest if the CVT is programmed to maintain engine speed at max power, not at max (engine) torque. Thus, for maximum acceleration, you always want to maintain the engine speed for maximum power, even though the engine torque is not maximized at this value. This is because max power RPM will always be above max torque RPM, which allows you to use a lower gear ratio for better mechanical advantage, providing higher wheel torque despite the lower engine torque.

 

[cjl]

Yes, #3 is the best shift option.

And, you can manipulate gear ratios to make maximum power and maximum torque coincide (sort of like horsepower and pound feet of torque are always equal at 5,252 RPM). That may or may not be the best option for optimum vehicle acceleration. It's an artifact. You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

He's not manipulating gear ratios to make max power and max torque coincide. He's showing you that maximum wheel torque at a given speed always coincides with the gear ratio that allows the engine to be at max power.

 

[jack action]

Yes, #3 is the best shift option.

I'm glad we agree.

You could also set the 2nd gear ratio such that maximum torque occurred at 35 KPH. That case would give the best acceleration from 35 KPH to 36 KPH after a 3rd to 2nd downshift.

I like that. You are so tenacious! Let's see what happens when we redo the same thought experiment as I did in my previous post, but by changing the 2nd gear ratio to a higher gear instead of a lower gear. Let's call it gear 2d:

Can we agree that by setting a higher gear ratio, the whole traction force curve will go down and cover a greater speed range?

Can we agree that Fmax in gear 2d will be smaller and at a faster car speed than in 2nd gear?

Let's go to an even higher gear ratio. Let's call it gear 2e:

Look at that, Fmax is now precisely at 35 km/h! And it is a lot less than the traction force in 2nd gear. Note also that if I just set it to a slightly higher gear ratio, I will be in 3rd gear.

Now you should also be able to appreciate that Fmax in 2nd gear is not the greatest force you can achieve at 25 km/h; F@pmax in 1st gear is much greater at that same speed.

No matter what gear ratio you're in when at Fmax, there is a lower gear ratio that will create a greater force at the same speed with F@pmax.

If you want to move a point from left to right, it will also have to move down. That is because engine power is always the same, which is equal to the wheel power, and wheel power is F X v, so if v increases and power is the same, then F must go down.

Can we agree that Fmax will never offer the greatest possible acceleration at a given speed?

Can we agree that the maximum acceleration at any given speed depends on the engine maximum power?

You may now appreciate more my answer to one of your previous posts:

For example, suppose you have an engine that generates 250 units torque at 7,000 RPM and 233.3 units of torque at 7,500 RPM. The power in both cases is the same (1.75X10^6 units of power). Suppose maximum power occurs at 7,250 RPM. Then it's easy to generate a power curve that would be maximum at that point and yields 242.8 units of torque (1.76X10^6 units of power).

And I answered (I added comments in red to relate to examples from my previous post):

But the car doesn't go at the same speed at 7000, 7250 or 7500 rpm.

Let's say I want the wheel to turn at 1000 rpm. The gear ratio needed for having the engine at 7000 rpm will be 7:1. For 7250 rpm, you will need a 7.25:1 gear ratio and a 7.5:1 for 7500 rpm. What are the wheel torque then, knowing the wheel rpm is the same in all cases?

250 * 7 = 1750 units torque [--> corresponds to gear 2a]
242.8 * 7.25 = 1760 unit torque [--> corresponds to gear 2b]
233.3 * 7.5 = 1750 units torque [--> corresponds to gear 2c]

Note how the wheel torque is the same when at 7000 or 7500 rpm; That is because the power is the same in both cases. Yet, the wheel torque is greater at 7250 rpm; That is because the power is greater. And the wheel torque increase is directly proportional to the power increase (1.76X10^6 / 1.75X10^6 = 1760 / 1750).

 

[OldYat47]

Tenacious perhaps, but also concerned that I missed the significance of cjl's comment about maximum power and the effects of multiplying torque. He or she is, of course, correct, which means I am missing stuff again. That's distressing. I'll sign off for a while, looks like more time at the neurologist for me. Ya gets old, you gets feeble.

Thank you all for a refresher and a refreshing debate.

 

[Emre]

What about accelerating in first gear from a dead stop?It makes sense that the max achievable acceleration comes when we choose the gear corresponding to max horsepower RPM at a given speed.But what can we say when the speed is 0?I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?Is this the reason why diesel cars accelerate as if they are too quick for their power?

 

[CWatters]

In an ideal engine the power would be high/maximum at all rpm right down to zero. Now since power = torque * angular velocity this would mean the torque should increase to infinity at zero rpm.

Unfortunately we can't make an ideal gas/petrol or diesel engine, they have a maximum torque that limits power at low rpm. However diesel are "more ideal" than gas/petrol. Diesel have greater torque at low rpm so they develop more power at lower rpm.

Electric motors are even "more ideal" than diesel engines. They can develop max torque at very low rpm. Take a look at some of the Teslas vs super car videos. Typically the Tesla wins over short distances but super car has higher top speed..



Some think that one day an electric dragster should be able to take the top fuel drag records..

https://newatlas.com/top-ev-electric-drag-racing-top-fuel/50741/

 

[Randy Beikmann]

What about accelerating in first gear from a dead stop?It makes sense that the max achievable acceleration comes when we choose the gear corresponding to max horsepower RPM at a given speed.But what can we say when the speed is 0?I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?Is this the reason why diesel cars accelerate as if they are too quick for their power?

Remember that power = torque*rotation speed (P=T*omega), so when omega is zero, power is zero. Axle torque might be huge at zero speed (limited by traction), but except for the fact the tires will slip a bit acceleration, you can't use much power right at launch.

 

[jack action]

I mean,why don't we choose the max torque RPM for a dead stop for max acceleration since the wheel torque is engine torque x gear ratio x axle ratio?Am i wrong?

You are not wrong. From a dead stop, you should launch at a rpm where the torque is maximum to get the maximum acceleration (given you have the traction to support it), and that is what racers do. Then you increase the rpm until maximum horsepower and change the gear ratio as often as necessary to keep the rpm constant as much as possible.

 

[miltos]

A better idea is to start launching at max rpm so you can make use of the motor rotational inertia. Using the clutch you should take care to transfer to the wheels the maximum torque, accordind to the max available friction with the road (while you keep accelaration pedal to the floor)

With a low horsepower car, the engine rpm will drop a lot. When they tend to drop below max torque rpm you should idealy stabilize motor rotation speed to that of max torque, by deppresing the clutch pedal.

With a race car, it is likely that the maximum friction will correspond to a motor torgue less than the maximum available. Then, there are more than one way to achieve max accelaration (even without full throttle) and probably other factors are crucial for the starting rpm (eg turbo lag, clutch wear).

 

[Emre]

You are not wrong. From a dead stop, you should launch at a rpm where the torque is maximum to get the maximum acceleration (given you have the traction to support it), and that is what racers do. Then you increase the rpm until maximum horsepower and change the gear ratio as often as necessary to keep the rpm constant as much as possible.

Thanks for the answer.But for many cars when you are close to peak hp RPM at launch,tire spins become too much in comparison to peak torque RPM.Then why do this happen?

 

[miltos]

The answer is at my post above

 

[jack action]

But for many cars when you are close to peak hp RPM at launch,tire spins become too much in comparison to peak torque RPM.Then why do this happen?

RPM plays a role too. If an engine revs at 8000 rpm and is suddenly connected to a drivetrain at 0 rpm, one or more of three things must happen:

• The engine must stall;
• The clutch must slip;
• The tires must slip.

When slipping occurs, the greater the rpm difference, the greater the chance to be dealing with the lower kinetic friction coefficient instead of the higher static friction coefficient (Hence, promoting even more spinning).

If you are at a higher engine torque, and lower rpm, not only the difference in rpm is less and a higher friction coefficient might result, but if the car accelerate fast, that rpm difference will decrease fast. Furthermore, if the torque is higher, the clutch might also slip more, meaning the wheel rpm is even less, thus resulting in an even smaller rpm difference, that will fade even faster.

 

[OldYat47]

Broken record approaching. It's all about the torque available at the driving wheel(s) since acceleration = mass / force. The limiting factor is tire to road static friction. We'll assume any clutch mechanisms work properly, or else the problem becomes indeterminate.
Power is not a factor. For any value for power there are an infinite number of values for acceleration (just change the velocity). Once you "correct" for velocity you are left with force (torque). Rear wheel power always equals net engine power minus drive train losses regardless of the overall gear ratios.
For any set of gear ratios you can plot the torque curve using the overall gear ratios and the engine torque curve. The "best" set of ratios depends on the maximum speed you want to achieve. A set of very low gears may get you to 80 MPH at maximum RPM very quickly but wouldn't be of much use if you had to get to 100 MPH. Shift points are best where the area under the combined torque curves is greatest.
But that's theory. In the real world we have other factors, most significant is wind resistance. So for example, you may want low ratios in the first couple of gears to maximize acceleration when wind resistance is less significant, then higher ratios to reach the speed you want to reach. Tricky, eh? Now you have to consider aerodynamics and create a new set of curves showing the effects of drag on available torque.

 

[OldYat47]

I've been away for a while, partly because of discussions like this one.
Here's a sample problem: You have a vehicle of mass M. Conditions are ideal; no air resistance, infinite tire to road friction, etc. You observe that the driven wheels have radius R. You observe that those wheels are being driven at a rate of power P. Velocity is unknown. Is the vehicle accelerating? If so, what is the rate of acceleration?

 

[jack action]

Here's a sample problem: You have a vehicle of mass M. Conditions are ideal; no air resistance, infinite tire to road friction, etc. You observe that the driven wheels have radius R. You observe that those wheels are being driven at a rate of power P. Velocity is unknown. Is the vehicle accelerating?

With ideal conditions (i.e. no resistances), it will always be accelerating with power $P$.

If so, what is the rate of acceleration?

The acceleration $a$ will always be (if we ignore rotational inertia):
$$a = \frac{P}{Mv}$$
no matter the velocity $v$ (which constantly changes since it is accelerating). Note how the tire radius $R$ is irrelevant. Note also that how the power is delivered is also irrelevant (it could be through the driven wheels, a rocket, wind on a sail, ...).

That is the beauty of using the power figures: All those little details don't matter.

 

[OldYat47]

Look at your equation. Note that P/(M X v) is force. Acceleration cannot be calculated from power since the two are not related. In the equation you quote you can have an infinite number of values for acceleration at any given power. The acceleration in your equation is dependent on the velocity. Let's say you observe a 10 kg mass and you observe that the power driving the mass is a constant 10 watts (there's a handy power consumption dial you can see). How fast is the mass accelerating? The rate of acceleration is indeterminate, it's dependent on the velocity. The higher the velocity the lower the acceleration. Using power you get an infinite number of values for acceleration. that's the problem with using power, and the beauty of sticking with a = M /v. You get one value.

 

[OldYat47]

Now let's look at your statement that a mass will always be accelerating with power P. Suppose that the mass is not in a perfectly frictionless environment. There is a constant 1 Newton drag on the object. As the velocity increases the factor P/v eventually equals 1 Newton. At that point the mass no longer accelerates. In the real world the effects of drag and friction affect the net force available to drive any mass. At some point those effects balance the driving force, the net force goes to zero, acceleration goes to zero, but power at that point does not drop to zero. Power would still be the product of the driving force, ignoring the effects of drag and friction, and the velocity.

 

[jack action]

The acceleration in your equation is dependent on the velocity.

The rate of acceleration is indeterminate, it's dependent on the velocity.

It is the whole point of the exercise. It is a fact that the rate of acceleration is dependent on the velocity for a vehicle propelled by driven wheels.

Let's say you observe a 10 kg mass and you observe that the power driving the mass is a constant 10 watts (there's a handy power consumption dial you can see). How fast is the mass accelerating?

Here's another question for you. Let's say you observe a 10 kg mass and you observe that the wheel torque $T$ driving the mass $M$ is a constant 10 N. How fast is the mass accelerating?
$$a = \frac{F}{M} = \frac{T}{Mr}$$
Oups! I don't know the wheel radius $r$! But I know one relationship that is always true though: $v = \omega r$, where $\omega$ is the wheel angular velocity. So:
$$a = \frac{T\omega}{Mv} = \frac{P}{Mv}$$
In your mind, $T$ is an independent characteristic of the vehicle. It is not. If you know $r$ and $T$, then you also know $P$ and $v$. So I can just as well state that acceleration is indeterminate because it is dependent on the wheel torque, which is a function of $v$. An acceleration of 10 m/s² at 10 km/h is not the same thing as an acceleration of 10 m/s² at 300 km/h. The difference? The amount of power (not torque) necessary to achieve the objective.

Consider the following three equations:
$$a = \frac{F}{M}$$
$$a = \frac{T}{Mr}$$
$$a = \frac{P\frac{1}{v}}{M}$$

• The first equation states: «acceleration is proportional to $F$ and $\frac{1}{M}$».
• The second equation states: «acceleration is proportional to $T$, $\frac{1}{r}$ and $\frac{1}{M}$».
• The third equation states: «acceleration is proportional to $P$, $\frac{1}{v}$ and $\frac{1}{M}$».

What the first and second equations hide is that $F$ and $T$ are function of $v$ as well when you consider a limited amount of wheel power. For a given wheel power, the wheel torque $T$ will always be $\frac{P}{\omega}$ and the force $F$ will always be $\frac{P}{v}$, no matter how you design your drivetrain.

How do $F$ and $T$ vary with with $v$? You will find that:
$$F \propto \frac{1}{v}$$
Or:
$$T \propto \frac{1}{v}$$
The fun thing with the equation considering power, $\frac{1}{v}$ is directly in the equation, thus showing how $a \propto \frac{1}{v}$, which is the important fact to understand and that is not clear from $a = \frac{F}{M}$ or $a = \frac{T}{Mr}$. You cannot keep $F$ or $T$ constant across a speed range, unless you change the power accordingly.

At some point those effects balance the driving force, the net force goes to zero, acceleration goes to zero, but power at that point does not drop to zero.

The same can be said about the driving torque: It will not drop to zero when the acceleration will be zero. Considering the sum of resistances $R$, the equations are only modified in this way:
$$a + \frac{R}{M} = \frac{F}{M}$$
$$a + \frac{R}{M} = \frac{T}{Mr}$$
$$a + \frac{R}{M} = \frac{P\frac{1}{v}}{M}$$
Given a constant power $P$ and an increasing velocity, the right hand-side of the equations will decrease, either:

• $F$ will decrease;
• $T$ will decrease;
• or $v$ will increase.

At one point, it will be equal to $\frac{R}{M}$ and $a$ will have to be equal to $0$. Note that $F$, $T$ or $\frac{1}{v}$ will never be zero, even if the acceleration is zero. If $R = 0$, the acceleration will constantly decrease with velocity, but never reach zero, just like $F$, $T$ or $\frac{1}{v}$.

 

[OldYat47]

I said that in "real"conditions (surface of the Earth, for example) the net force drops to zero due to increasing drag and friction (opposing the driving force). I agree with your formulas although I'd write them a = (driving force - drag and friction) / mass. When the net force (driving force - drag and friction) drops to zero there is no acceleration. Power does not drop to zero, it's still the driving force times the velocity.

Look at your equation [a + (R/M)] = [P(1/v)/M]. (P X 1/v) is force, not power, and v must be known as well as P.

Your last set of statements illustrates another difference between using power instead of force. I can restate saying power changes with velocity even if the sum of forces and the rate of acceleration remain constant. So power and acceleration are not related. There are many ways to twist this around. For example, if you know any two of the factors of a = F / M you can calculate the third. But to calculate power you must also know the velocity, because power is a rate (time dependent).

 

[jack action]

But to calculate power you must also know the velocity, because power is a rate (time dependent).

Acceleration is also a rate, which is not only time dependent, but speed dependent. And that is why power is such an important notion when considering acceleration. Looking at the power equation:
$$\frac{dv}{dt} = \frac{P}{Mv}$$
$$Mvdv = Pdt$$
$$\frac{1}{2}Mv^2 = Pt$$
A nice conservation of energy representation.

Looking at it only on the point of view of force, you get:
$$\frac{dv}{dt} = \frac{F}{M}$$
$$Mdv = Fdt$$
Here I have a conservation of momentum equation that is equally true, but if I have a limited amount of power, I cannot integrate it as easily because $F$ will actually be a function of $t$: As the speed increases (thus as time increases too), the force $F$ will decrease. The exact way this will happen is given by the previous power equation.

Say you have a car going 100 km/h with an acceleration of 5 m/s², and you wish to double that acceleration. Looking at it your way, all you have to say is: « $a=\frac{F}{M}$, let's double the driving force $F$, easy peasy. » So, based on this, you change your driven wheels with ones with a radius half as much as your original wheels: Bam! you have just double the force $F$ ( $F = \frac{T}{r}$, $T$ is constant ). But when you bring the car at 100 km/h, guess what? You still have an acceleration of 5 m/s². The only thing different is that now your car speed range is divided by two: you can get driving force at much lower speeds (if you couldn't before), but you also have cut your top speed in half (if you had such a limit before). But in both cases, at the same speeds, you get the exact same accelerations.

Even if you look at $F = \frac{T}{r}$ and say « Let's keep $r$ constant and double $T$ instead, by changing the gear ratio in the drivetrain. » Then you will get the exact same result.

Looking at it my way, I say: « $a=\frac{P}{Mv}$, let's double the driving power $P$, easy peasy. » So, based on this, I change the engine to one that is twice as powerful as the original engine: I truly doubled the force (and the torque) at 100 km/h (actually, I doubled it at any speed), so I effectively get an acceleration of 10 m/s² at 100 km/h.

How can you say that power is not related to acceleration if when you double the power you actually double the acceleration at any speed? And when you only consider the driving force only, it doesn't work at all (unless you increase the driving force with power only).

 

[OldYat47]

I can say power is not related to acceleration by looking at the physics formulas involved. The best example is one I stated above: With a constant force and a constant mass acceleration is constant, but power increases continually as velocity increases. From that I can say that power and acceleration are not related.
I will agree that at any specific speed acceleration increases if power at that speed increases. That's because when you convert from power to force by doing the (P/v) calculation the force is greater if the power is greater.
Let's look at that decreasing the wheel radius approach. Reducing the radius by half does indeed double the force and double the rate of acceleration. What happens to the power? (Power after the radius reduction at any given velocity) = (power before the reduction at twice that same specific velocity). Since the velocity is halved the power stays the same at half the velocity as before the radius reduction. Same power, twice the acceleration. Another example of how power and acceleration are not related.
I'm out of time for tonight, have things to attend to. The last thing I want to point out is that you say a = P/Mv. Once again, P/v = force, that equation works out to
a=F/M

 

[jack action]

With a constant force and a constant mass acceleration is constant, but power increases continually as velocity increases.

But if you double or halve the power at any speed, you will also double or halve the acceleration at that speed. Conclusion: The acceleration is proportional to power.

Since the velocity is halved the power stays the same at half the velocity as before the radius reduction. Same power, twice the acceleration.

Yes, but the 'doubled' acceleration is not at the same speed anymore. Don't you think it matters?

Here are the examples put in graphics. A 3500 lb, 4-speed car. Two possible engines: One with a 100 hp maximum power, ranging between 4000 & 6500 rpm; One with the power output multiplied by 1.625 at every rpm:

Here are the results for the acceleration vs speed:

You can see that the green line produces exactly 1.625 the acceleration of the 100 hp engine (blue line) at any speed. I also put as dotted lines, the equivalent acceleration of engines of constant 100 hp & 162.5 hp (Of course, they match with the peak output of the engines).

The red dotted line is the result of reducing the drivetrain ratio by 1.625 (either by changing the differential overall gear ratio or reducing the wheel radius). The accelerations are now a perfect match with the ones from the green line, but they happen at speeds 1.625 smaller.

Note how the red dotted line accelerations for the 2nd, 3rd & 4th gears are an exact match for the original accelerations of the 1st, 2nd & 3rd gears. Note how the red dotted line follows the 100 hp line, just like the blue line. All you need to do is add a 5th (lower) gear to the original set-up and it would follow perfectly the 1st gear on the red dotted line (without reducing the top speed with the last gear).

Now imagine you have an engine producing a constant 100 hp from 0 rpm (like an ideal electric motor). Changing the overall gear ratio would change nothing to the constant 100-hp curve on the previous graph, as you would already cover the acceleration from 0 km/h with any overall gear ratio (or wheel radius).

For the sake of discussion, let's assume the maximum acceleration the car can handled because of traction limitation is 0.35 g. Can you see how the car with the red dotted line will have the exact same acceleration at any speed than the original car? Can you see that the one with the more powerful engine will have a much greater acceleration overall and reach 150 km/h a lot faster than any of the other two set-ups?

I ask again: How can you say power is not related to acceleration if when you double the power you actually double the acceleration at any speed? And when you only consider the driving force only, it doesn't work at all (unless you increase the driving force with power only).

 

[OldYat47]

Let's look at the physics formulas.

Acceleration = (force / mass). Force is defined as (mass X acceleration), so acceleration = (mass X acceleration / mass). Masses cancel, result is acceleration.

Power = (force X velocity) so power = (mass X acceleration X distance / time). Solving for acceleration you get acceleration = [time / (mass X distance)] which is clearly wrong since acceleration is (distance / time squared). Case closed, there is no relation between power and acceleration. Can't be simpler than that.

It's not generally true that increasing power increases acceleration. Power must increase at the same rate as velocity to maintain constant acceleration. Whenever the rate of the increasing power is not greater than the rate of increase of velocity the rate of acceleration decreases. Constantly increasing power, constantly decreasing acceleration. And I am sure that in your charts the acceleration calculation involved at some point dividing power by velocity, which is force.

What is that relationship between power, acceleration and velocity? It's P/v which is force.

 

[Randy Beikmann]

My approach is to consider a given vehicle speed V, and examine what limits its acceleration. I'll start by assuming that a) the transmission has gearing that puts the engine at its peak power speed at vehicle speed V, and b) the vehicle won't be limited by wheelstand. For simplicity, I'll also neglect drag of all types (it doesn't change the conclusions, however). Then there are two maximum acceleration possibilities at that vehicle speed:

1) Traction-limited acceleration a_TL.
2) Power-limited acceleration a_PL.

The actual acceleration limit for the vehicle at speed V is the smaller of those two - you must both have the traction force to accelerate at that rate, and the power to produce that traction force.

Traction-limited acceleration a_TL is simply F_TL/m, where F_TL is the total traction-limited thrust at the drive tires. Of course, it is determined by the tires, the weight distribution, and the CG height. It is not inherently a function of speed, if we neglect aerodynamic lift, etc.

Power-limited acceleration is F_PL/m, where F_PL is the power-limited force. With the assumption that the gearing is right, the power all supplied to the tires, so P_Max=F_PL*V. Thus F_PL=P_Max/V, and a_PL = P_Max/(mV). So obviously this is a strong function of speed.

So the maximum acceleration at speed V is the smaller of F_TL/m and P_Max/(mV).

What does this mean? At low speeds, the power-limited acceleration would be huge, but you don't have the traction to achieve it. The acceleration is traction-limited, and nearly constant. If you look at acceleration just after launch, actual vehicles do accelerate this way.

As you accelerate, you eventually reach a speed where the power-limited acceleration equals the traction-limited acceleration, that is, where V=P_Max/F_TL. From that speed on, the acceleration is power-limited, and acceleration (using max power) is reduced, varying as 1/V. You can use this to calculate the maximum speed where you can spin the tires.

To sum up, acceleration starts as a constant level line vs. speed, and then tapers as a hyperbola.

 

[russ_watters]

I can say power is not related to acceleration by looking at the physics formulas involved. The best example is one I stated above: With a constant force and a constant mass acceleration is constant, but power increases continually as velocity increases. From that I can say that power and acceleration are not related.

I'm sorry, but that is akin to saying "if I close my eyes, the world ceases to exist." There is a relationship there and you even half-described it (the relationship is linear).

Let's look at the physics formulas.

Acceleration = (force / mass). Force is defined as (mass X acceleration), so acceleration = (mass X acceleration / mass). Masses cancel, result is acceleration.

All equations are reducible to 1=1 (or A=A). There is no point in doing what you just did.

Power = (force X velocity) so power = (mass X acceleration X distance / time). Solving for acceleration you get acceleration = [time / (mass X distance)] which is clearly wrong since acceleration is (distance / time squared).

You can't solve for acceleration when acceleration isn't in the equation. That's just an error/gibberish.

Case closed, there is no relation between power and acceleration. Can't be simpler than that.

Doing the math wrong doesn't make it cease to exist. It's really simple:

P=FV and A=F/M

Solve the first for F: F= P/V

Plug into the second: A=P/(VM)

 

[OldYat47]

??
Excluding drag and friction the shape of the acceleration curve is dependent on (and follows) the shape of the force curve as long as there is no tire spin. If there is tire spin the acceleration curve will sag relative to the force curve until traction is restored either by speed matching (rotational vs. vehicle forward speed) or traction force matching up.
At any gear ratio there will be some velocity where the engine is at peak torque and another velocity where the engine is at peak power.
P_max / (mV) is a force because you are dividing power by velocity.

 

[Rick stievenart]

Also remember that the driven axle tire's radii will not be constant...

 

[Randy Beikmann]

??
Excluding drag and friction the shape of the acceleration curve is dependent on (and follows) the shape of the force curve as long as there is no tire spin. If there is tire spin the acceleration curve will sag relative to the force curve until traction is restored either by speed matching (rotational vs. vehicle forward speed) or traction force matching up.
At any gear ratio there will be some velocity where the engine is at peak torque and another velocity where the engine is at peak power.
P_max / (mV) is a force because you are dividing power by velocity.

From launch until when the acceleration is power-limited, we can assume that the driver (or a traction control system) is controlling the throttle so that the tire slip is just right to utilize the car's maximum traction, which is almost independent of velocity. So during that time, a=F_TL/m. It's a good assumption borne out by data.

I stated in my assumptions that for any vehicle speed V, the gearing was such that the engine speed was at its peak HP speed. This could be enforced by a CVT, or approximated by a transmission with many gear ratios. That doesn't mean the engine is *producing* peak power, since the throttle doesn't have to be fully depressed when it would produce more force at the tires than traction will allow.

P_Max/(mV) is not a force. P_Max/V is a force I call F_PL. When I divide F_PL by m, I get an acceleration I call a_PL for power-limited acceleration. So P_max/(mV) is power-limited acceleration. You can check the units: it's not a force.

I think you are making this too hard. Don't worry about anything except 1) how much traction the car has, and 2) how much power it has. Any other factors are small ones that will have mild effects on the actual results. Think of how a real car accelerates: it accelerates quickly just off launch, up to a certain speed, and then the acceleration tapers off. The tapering happens at speeds lower than where aerodynamic drag is important (unless your power-to-weight ratio is extremely high).

If you pick up a car magazine and plot speed vs. time from their tests, you'll see exactly what I am describing: an initially linear increase in speed, after which velocity increases more and more slowly.

 

[jack action]

Power = (force X velocity) so power = (mass X acceleration X distance / time). Solving for acceleration you get acceleration = [time / (mass X distance)] which is clearly wrong since acceleration is (distance / time squared).

This is wrong.

To respect the relationships you defined, solving for acceleration:

acceleration = [time / (mass X distance)] X power

acceleration = [time / (mass X distance)] X force X velocity

acceleration = [time / (mass X distance)] X force X [distance / time]

acceleration = [time / (mass X distance)] X [mass X acceleration] X [distance / time]

Time, distance and mass all cancel out and you are left with acceleration = acceleration, just like in your first case.

Please, don't look at me (and others) as attacking you, I'm not. I'm just trying to let you see the world from another point of view that is very helpful. Note that I agree with everything you say (except for this little mishap you just did ).

It's not generally true that increasing power increases acceleration.

100 % true. Here is an example. If you turn on the A/C in a car, the engine power will increase and that power increase will not affect the acceleration of the car.

Power must increase at the same rate as velocity to maintain constant acceleration.

That is another good example: If you increase the velocity of the car, it requires power, just like turning on the A/C. In that case too, the car acceleration will not be affected.

Whenever the rate of the increasing power is not greater than the rate of increase of velocity the rate of acceleration decreases.

Not true. If the speed is doubled and the power is doubled, the acceleration will stay constant. The acceleration will decrease only if the power increase is lower than the velocity increase.

Let me put all of our statements on a force-velocity graph:

Let's consider the orange dot as our starting point where the force = 1 and velocity = 1. The power needed is force X velocity = 1.

You , you focus on the horizontal line, where the force is constant and the velocity increases. On the graph, the velocity is 2 and thus the power has also increase to 2 while the force (acceleration) is kept at 1. That is when you say «See, increasing power doesn't mean increasing acceleration», which is totally true.

What I want you to do now is focus on the vertical line, where the velocity is maintained at 1 and the force is multiplied by 2. In this case also, the power is increased to 2. I double the force (acceleration), I double the power needed.

Finally, let's consider the diagonal line. Now, both the velocity and the force are doubled. The resulting power is now 4. I have to multiply by 2 for the velocity increase and multiply by 2 again for the force increase.

You keep saying that «If you increase power, you do not necessarily increase acceleration», which is true. The statement that I want you to see is that «If you want to increase the acceleration, you MUST increase the power.» Look at the graph:

• Anything above and to the right of the orange dot MUST have an increase of power;
• Anything below the orange dot is not increase in acceleration;
• Anything to the left of the orange dot is a decrease in velocity, therefore it cannot be an acceleration, it is a deceleration.

If you consider the vertical line (constant velocity), acceleration is proportional to power: you double the power, you double the acceleration.

If the velocity increases, if you want to double the acceleration, you still have to double the power AND you have to increase it further more to compensate for the velocity increase as well. (And if you turn on the A/C at the same time, you need to add more power too.)

Yes, power can be used for something else, but if you want to double the acceleration, you will have to AT LEAST double the power.

 

[OldYat47]

I said, "Power must increase at the same rate as velocity to maintain constant acceleration.", and, "Whenever the rate of the increasing power is not greater than the rate of increase of velocity the rate of acceleration decreases."

jack action said, "Not true. If the speed is doubled and the power is doubled, the acceleration will stay constant. The acceleration will decrease only if the power increase is lower than the velocity increase." Which is exactly what I said.

Russ Watters, I said
Power = (force X velocity) so power = (mass X acceleration X distance / time). Force is defined as (mass X acceleration).
I left a step out, though. Here it is.
Let's start with power and assume we want to find acceleration. Power = (mass X acceleration X distance / time). To find acceleration from power we have to multiply power by the inverse of [mass X (distance / time)]. So we need a factor whose dimensions are [1 / (mass X velocity)]. Once again power is converted to force and you wind up with force / mass.

What about the issue of an infinite number of values for power for any value of acceleration? If you've got an infinite number of answers for the same question then there is no relation.

 

[jack action]

@OldYat47 :
You are completely ignoring what I'm telling you and I'm not sure why. Maybe my posts are too long. This is the important quote from my previous post I want you to focus on:

You keep saying that «If you increase power, you do not necessarily increase acceleration», which is true. The statement that I want you to see is that «If you want to increase the acceleration, you MUST increase the power.»

Tell me this isn't true.

 

[Randy Beikmann]

What about the issue of an infinite number of values for power for any value of acceleration? If you've got an infinite number of answers for the same question then there is no relation.

The power used at the tires is directly related to acceleration. The power going into acceleration is equal to the traction force F, multiplied by the vehicle speed V, so P=FV. We also know that F=ma. Substituting, this means that P=maV. So power in accelerating is dependent on vehicle mass m, the acceleration rate a, and the speed V.

So there's no problem - if you know the mass, acceleration, and vehicle speed, you can calculate the (single) value for the required power.

 

[russ_watters]

I said, "Power must increase at the same rate as velocity to maintain constant acceleration."

And that's true! That's the linear relationship you are claiming doesn't exist!

Let's start with power and assume we want to find acceleration.

What you are or should be doing is relating acceleration to power. That's the relationship you are saying doesn't exist. Mathematical handwaving that doesn't connect acceleration to power - when you easily could - doesn't say anything useful.

Power = (mass X acceleration X distance / time).

To find acceleration from power

This is just basic algebra:
P=MAD/T
A=PT/(MD) = P/(VM) -- Just like I said.

...we have to multiply power by the inverse of [mass X (distance / time)]. So we need a factor whose dimensions are [1 / (mass X velocity)]. Once again power is converted to force and you wind up with force / mass.

If you want to find if power is related to acceleration, you shouldn't cancel-out the power, you should leave it there and just re-arrange the equation around it. I feel like either you don't understand basic algebra here. All you do to re-arrange an equation is multiply both sides by the factor you want to move from one side to the other. For example:
Y=5X
Y*1/5=5X*1/5
Y/5=X

What about the issue of an infinite number of values for power for any value of acceleration? If you've got an infinite number of answers for the same question then there is no relation.

As you can see from the equation that you don't want to admit exists, there is only one value of acceleration per value of power.

 

[OldYat47]

This is my second to last post on this subject. I'm going all the way back to define what I'm saying, because it seems I'm speaking in a different language.
I say that power and acceleration are not related. I say that because there is no formula that will give a result in acceleration knowing the power and the mass. So far all the formulas in all the replies that claim to do that use velocity to convert power to force. Compare: acceleration = (force / mass) and acceleration = [power / (mass X velocity)]. Note that the second equation does not relate power to acceleration because (power / velocity) = force. So the second equation is identical to the first equation.

I think (I hope) that we can all agree that, given a constant mass and a constant force, acceleration is constant. I think we can all agree that if acceleration is constant power will increase directly proportional to speed. I think we can all agree that, given a constant mass and a constant power, acceleration will decrease directly proportional to velocity. These last two statements say that power and acceleration are not related. Why? Because in the second statement acceleration is constant, power is increasing, and in the third statement power is constant and acceleration is decreasing.

I said this was my second to last post. If anyone posts a formula that directly relates power to acceleration without using velocity to convert power to force I will post "Good job!". So you need a formula that will solve this question:

Some mass is moving. The power involved at some instant is known. Velocity at that instant is unknown. What is the acceleration of the mass at that point in time?

P. S.
For any given mass and any given acceleration there exists an infinite number of values for power.
(acceleration = (power / velocity) / mass. Insert any acceleration and any mass. There exists an infinite number of combinations of (power / velocity) that will make the equation correct. Once you divide power by velocity you get a single value for force.

 

[jack action]

I don't understand why you assume the mass is constant in $a = \frac{F}{M}$, but you can't assume the mass and the velocity are constant in $a = \frac{P}{Mv}$. You know the mass can change while the vehicle is moving, right? You can consider the mass of fuel that is burned along the journey, or even the case where a tanker is emptying its load while moving.

Let's look at the statements you feed us but assuming mass is not constant when using the force equation:

I say that because there is no formula that will give a result in acceleration knowing the power and the mass.

«I say that because there is no formula that will give a result in acceleration knowing only the force.»

I think we can all agree that if acceleration is constant power will increase directly proportional to speed.

«I think we can all agree that if acceleration is constant force will increase directly proportional to mass.»

I think we can all agree that, given a constant mass and a constant power, acceleration will decrease directly proportional to velocity.

«I think we can all agree that, given a constant force, acceleration will decrease directly proportional to mass.»

If anyone posts a formula that directly relates power to acceleration without using velocity to convert power to force I will post "Good job!".

«If anyone posts a formula that directly relates force to acceleration without using mass to convert force to acceleration I will post "Good job!".»

Some mass is moving. The power involved at some instant is known. Velocity at that instant is unknown. What is the acceleration of the mass at that point in time?

«Some mass is moving. The force involved at some instant is known. Mass at that instant is unknown. What is the acceleration of the mass at that point in time?»

For any given mass and any given acceleration there exists an infinite number of values for power.

«For any given acceleration there exists an infinite number of values for force.»

Why are you allowed to assume that mass is constant and we are not allowed to assume that velocity is constant too?

Again, I challenge you to show us a mathematical equation (or real life example) where two vehicles are accelerated from $v=0$ to $v= v_f$, one reaching $v_f$ faster than the other, and that the fastest one is not using more power than the other one. Even more precise, try to show me an example where the increase in instantaneous power is not directly proportional to the increase in instantaneous acceleration when comparing both vehicles.

I showed you my math, but you never comment on it. Do you understand what I do? Ask questions if you don't understand some of it. Again, you are not wrong, it is just that you are setting arbitrary limits without justification (mass is constant, velocity cannot be constant) .

We cannot be all wrong and you being the only one who's right.

 

[jack action]

@OldYat47 :

OK, I got another image for you, to help you show where you err.

The area $A$ of a rectangle is the length $l$ times the width $w$, or $A = lw$.

The volume $V$ of a box is the area $A$ times the height $h$, or $V = Ah$.

But we can also say that $V = lwh$, right?

Is the volume $V$ related to width $w$? If I increase $w$, does that translate into more volume? I think everyone agree with the fact that it does.

Of course, one can say: «but if you decrease $h$ at the same time, the volume will stay the same.» That is true, still, everyone say that there is a direct relationship with the width of a box and the volume of the box. It is not because the volume of the box is also related to another dimension (which can increase or decrease independently of $w$), that the relationship between $w$ and $V$ disappears.

Assume now that $V$ is the power, $A$ is the force, $l$ is the mass, $w$ is the acceleration and $h$ is the velocity:
$$V=lwh$$
$$P=Mav$$
Can we not say the same thing?

Is the power $P$ related to acceleration $a$? If I increase $a$, does that translate into more power? I think everyone should also agree that it will.

Of course, one can say: «but if you decrease $v$ at the same time, the power will stay the same.» That is true, still, everyone say that there is a direct relationship with the acceleration and the power. It is not because the power is also related to another dimension (which can increase or decrease independently of $a$), that the relationship between $a$ and $P$ disappears.

Conclusion: If you have an increase in acceleration, you must have an increase in power too.

It is simple algebra.

 

[russ_watters]

This is my second to last post on this subject. I'm going all the way back to define what I'm saying, because it seems I'm speaking in a different language.
I say that power and acceleration are not related. I say that because there is no formula that will give a result in acceleration knowing the power and the mass.

[Moderator hat]
It does feel like we are speaking different languages and the language problem is with math: you seem genuinely unable to perform basic algebraic manipulation of an equation. This is a knowledge gap and one we can fix, but only if you make an effort to learn.

This discussion feels like a debate, but it cannot be: this is easy/settled physics and is not debatable. So if you do choose to continue posting in this thread, you will need to change your approach from trying to debate to trying to learn. PF rules prohibit arguments against established science. [/Modhat]

 

[OldYat47]

OK, now I have to respond. I'm not debating, I'm pointing out the simple physics and algebra. I'll run through some of it again.
My basic claim is that power and acceleration are not related, that there is no algebraic function that allows calculation of acceleration directly from power. To calculate acceleration using power you must also know velocity and use velocity to convert power to force. The result is rate of acceleration at one velocity only.
I also stated that
[power / (mass X velocity)]
is not that equation since
(power / velocity) = force,
so the equation is identical to [acceleration = (force / mass)].
I also pointed out that using
acceleration = [power / (mass X velocity)]
for any given rate of acceleration and mass there is an infinite number of values for power and make the equation true. This is mathematically true, just select the velocity that makes it all work out.
I could expand and rephrase that statement: Any value of power can accelerate any mass at any rate of acceleration.
Example question: Can 0.1 (watt) accelerate a 100 (kilogram) mass at a rate of 100 (meters / second^2)?
Answer: Yes, when the velocity is 0.00001 (meters / second).

So my premise remains that power and acceleration are not mathematically related. You can't calculate acceleration directly from power, you must convert power to force first. If you can calculate acceleration directly from power please reply with the equation.

 

[jack action]

I'll try again before they close this thread (I fought for you @OldYat47 ).

You stated that $a = \frac{F}{M}$ and that $M$ is constant, so $a$ is related to $F$. True.

But $F = \frac{P}{v}$ and we can assume that $v$ is constant (you can do it with $M$, why not with $v$?), so $F$ is related to $P$. Also true.

If $a$ is related to $F$ and $F$ is related to $P$, therefore $a$ has to be related to $P$. Otherwise it doesn't make sense.

Sorry, but the equation you are looking for is $a = P$ and that doesn't exist, just like $a = F$ doesn't exist either. But $a \propto F$, $F \propto P$ and $a \propto P$ all exist, and that is what we are trying to explain to you.

 

[jack action]

If you can calculate acceleration directly from power please reply with the equation.

On my website, I have an acceleration simulator where I explain how I find the acceleration, which is found with equation 1b, which is basically $a = \frac{F}{M}$, where $F = \frac{P}{v}$, when there is enough traction. I can calculate the acceleration of any vehicle from power only and it is pretty accurate. You cannot have a better proof than this.

 

[OldYat47]

Please re-read the last paragraph of my last comment. My challenge is for someone to demonstrate going from power to acceleration without first converting power to force by dividing force by velocity. Please review the algebra.
Let's assume we have some constant mass.
Acceleration is proportional to force because with any force there is a single unique value for acceleration. Values for acceleration change directly with changes in values of force.
Force is not proportional to power. We know this because you can have many different values for force at any given power. (velocity dependent)
Acceleration is not proportional to power. We know this because you can have many different values for acceleration at any given power. (velocity dependent)
Dividing power by velocity = force, which we agree on. But that takes power out of the equations and puts force into them.

 

[jack action]

Please re-read the last paragraph of my last comment.

I wish I could, but it seems you are the one who is not reading our comments.

You are simply repeating the same thing over and over again without challenging what is told to you. If you did, you would see where you err.