Acceleration of a car and jet plane problem

[cjl]

And this is why those of us familiar with the argument get frustrated.

Your land rover has no hope against a sports car off the line if the sports car launches well (either with a high stall torque converter, launch control, or just a careful modulation of the clutch). In addition, at all other speeds, the sports car will still accelerate harder because it has more horsepower. Horsepower is the relevant factor when figuring out how fast you can add kinetic energy to the vehicle, and a higher horsepower will always be able to accelerate faster.

 

[russ_watters]

I suppose it depends on what language and framework you are working under. At the atomic level you need the energy and collisions from atoms in order to generate a force. Therefore, if you have energy from the atoms bumping you have power producing the force.

I'm sorry, but this really isn't generally true. Only glasses have collisions between molecules and even then what you say suggests energy expenditure, which doesn't happen.

 

[cjl]

Good point. But in all cases the wheel torque is the mechanical quantity most directly related to the net force and therefore to acceleration.

Not really though, since you still have to account for tire size to convert from wheel torque to tractive force. This is why power is the easier way to approach it - you can bypass all the details of gear ratios, wheel sizes, etc, and you can directly figure out acceleration based only on knowing vehicle speed and current power production.

 

[russ_watters]

Not really though, since you still have to account for tire size to convert from wheel torque to tractive force. This is why power is the easier way to approach it - you can bypass all the details of gear ratios, wheel sizes, etc, and you can directly figure out acceleration based only on knowing vehicle speed and current power production.

I like this approach. Torque is a component of power, but not the only component. And wheel torque at any instant (along with mass and a fixed wheel size) tells you acceleration, but when people say "acceleration" for a car they are usually talking about average acceleration presented as a time to a speed (0-60, 50-70, etc), not instantaneous acceleration.

By knowing what people want, including seeing through lack of specificity and how "car guy" terminology differs from the standard physics terminology you can bridge the gap and make sure we're answering the question that was intended to be asked.

 

[vysqn]

I never ever thought, that this discussion will develop so much I asked super stupid question showing that my knowledge of physics is misunderstood by myself. But I analized all post and I drew conclusions which i wrote at post #61

 

[Dale]

Not really though, since you still have to account for tire size to convert from wheel torque to tractive force. This is why power is the easier way to approach it - you can bypass all the details of gear ratios, wheel sizes, etc, and you can directly figure out acceleration based only on knowing vehicle speed and current power production.

As you accelerate wheel size remains constant while speed does not. So the relationship with power is certainly not more direct.

When wheel torque peaks acceleration peaks. When power peaks acceleration does not peak. Acceleration can be non zero while power is zero, and at that time it has the same fixed relationship with torque as always. Acceleration will decrease if power is constant, and at that same time it has the same fixed relationship with torque as always.

Sorry, but your argument does not seem consistent with the facts to me. I maintain that in all cases the wheel torque is the mechanical quantity most directly related to the net force and therefore to acceleration.

 

[jim hardy]

Your land rover has no hope against a sports car off the line if the sports car launches well (either with a high stall torque converter, launch control, or just a careful modulation of the clutch). In addition,...
... Horsepower is the relevant factor when figuring out how fast you can add kinetic energy to the vehicle, and a higher horsepower will always be able to accelerate faster.

i don't think that's quite so.

Her point is that with higher ratio gearing
her 136 horsepower off road truck can
at low vehicle speed
deliver more torque to the drive wheels than a sports car equipped with more horsepower and lower ratio(nearer 1::1) gearing .
That's how mechanical advantage works.
Selecting "Low Range" in a 4wd increases mechanical advantage typically between two and four fold.

A 'modulated' clutch transmits torque without loss, ie torque in equals torque out
but it does NOT transmit power without loss
because the clutch disk itself in that modulated clutch,
sandwiched as it is between the flywheel and pressure plate metal surfaces,
rotates at lower RPM than those two surfaces (and the engine that's driving them.)
There's slip between those metal driving surfaces and the composite clutch disk's driven surface.
So a goodly chunk of the engine's power is dissipated as heat at those sliding surfaces so never reaches the driving wheels..
Power lost in the clutch is equal to product torque X (engine rpm - clutch disk rpm) and goes to heat not kinetic energy .

...you can directly figure out acceleration based only on knowing vehicle speed and current power production.

add to that knowing what fraction of current power production goes to acceleration.

Either approach can be made to work and anyone's preference is just that - a preference.

old jim

 

[Randy Beikmann]

To simplify this discussion, I would assert that Newton's laws and the laws of thermodynamics/energy are equally important. There's no reason to argue about which is more so here. To accelerate a car, there must be a force F at the tire patch to cause it, so that F=ma (neglecting drag). But, except for the trivial case where v=0, creating this force requires the power P=Fv, where v is the vehicle speed. Using the two equations, you see that P=mav. So as speed increases, it takes more power to produce the same acceleration.

So accelerating the car requires a force, and creating that force requires power, except for the infinitesimal time period when v=0. Why make this complicated?

Note that traction force and engine power each have limits, and either can be the limiting factor to your acceleration. At low speeds the power (P=mav) required to accelerate is small, so acceleration is limited by the maximum available traction force Fmax: then a=Fmax/m. Beyond a certain speed (v>Pmax/Fmax, where Pmax is the maximum power), the engine can't produce enough power to fully utilize the tire traction, so a=Pmax/(mv). (I've neglected drag and driveline friction here, which don't change the main conclusions.)

 

[cjl]

i don't think that's quite so.

Her point is that with higher ratio gearing
her 136 horsepower off road truck can
at low vehicle speed
deliver more torque to the drive wheels than a sports car equipped with more horsepower and lower ratio(nearer 1::1) gearing .
That's how mechanical advantage works.

Selecting "Low Range" in a 4wd increases mechanical advantage typically between two and four fold.

But in reality, that's very unlikely, for a number of reasons. First of all, in low range, a truck or SUV cannot deliver anywhere close to full horsepower to the wheels. The reason for this is that trucks and SUVs tend to have relatively heavy flywheels, and in extremely low gears (such as a low range provides), most of the horsepower will actually be spent just accelerating the engine. The inertia of the engine and driveline prevents you from efficiently delivering that power to the wheels. Also, sports cars are generally geared to provide high acceleration based on their engine properties and the design of the car, so their first gear is generally more than low enough to get the car off the line effectively.

A 'modulated' clutch transmits torque without loss, ie torque in equals torque out
but it does NOT transmit power without loss
because the clutch disk itself in that modulated clutch,
sandwiched as it is between the flywheel and pressure plate metal surfaces,
rotates at lower RPM than those two surfaces (and the engine that's driving them.)
There's slip between those metal driving surfaces and the composite clutch disk's driven surface.
So a goodly chunk of the engine's power is dissipated as heat at those sliding surfaces so never reaches the driving wheels..
Power lost in the clutch is equal to product torque X (engine rpm - clutch disk rpm) and goes to heat not kinetic energy .

True, but a clutch eliminates that drivetrain inertia problem mentioned above because the engine is already at speed, so the torque transmitted to the wheels is quite a bit better. In addition, you can actually get well above the engine's rated torque during a launch in a manual car since you can take advantage of the engine's inertia to help accelerate the car. Using a proper launch, I'd bet on nearly any sports car to beat an older, 136hp Land Rover from 0-20 every time.

add to that knowing what fraction of current power production goes to acceleration.

In most gears, at most speeds, you can reasonably assume that fraction to be about 85%

Either approach can be made to work and anyone's preference is just that - a preference.

old jim

Except that a lot of people use the "torque" arguments to draw erroneous conclusions, which is the source of this whole discussion in the first place.

 

[Dale]

Except that a lot of people use the "torque" arguments to draw erroneous conclusions, which is the source of this whole discussion in the first place.

And some people (such as the OP) use the “power” arguments to draw erroneous conclusions.

 

[jim hardy]

In most gears, at most speeds, you can reasonably assume that fraction to be about 85%

not when you're riding the clutch.

and i reject the 'heavy flywheel' hypothesis as straw-grasping to defend a position.

As i said a preference is only a preference.

When you can make both approaches agree you nave mastered the subject.

old jim

 

[cjl]

Of course not when you're riding the clutch, but the clutch is only a factor in the first portion of the acceleration. As for the "heavy flywheel" argument, you'd be surprised how much of the engine's power is sucked up just accelerating the drivetrain (mainly the flywheel) in low gears. It makes a very significant difference.

 

[Dale]

my thoughts on this are that I would rather consider tractive effort (force) than torque. It more directly relates to what is happening to the vehicle as a whole.

I think this is a good suggestion. It avoids the entrenched positions in this weird debate and is accurate.

 

[jack action]

I realize that I am late to the party, but my thoughts on this are that I would rather consider tractive effort (force) than torque. It more directly relates to what is happening to the vehicle as a whole.

Of course it is better to use the tractive effort. It further proves how much futile torque is. Imagine the force is the result of a jet engine: Where is the torque in that vehicle? Only power matters.

My proof is to pitch my Land Rover against their sports car, select low first and floor the accelerator. It almost does not matter how much power they have (except perhaps in the extreme) against my miserable 101 kW (136 BHP) they cannot match my initial acceleration.

Assuming both vehicles have a transmission designed to extract the most from their engines (which they probably have), a Land Rover gets more acceleration at low RPM because it has more power at low RPM than a sports car has at low RPM. That is the only way.

The «extreme» ones that do beat the Land Rover - guess what? - they have more low RPM power than the Land Rover can produce. That is the only way.

OK, so they catch up and pass me - eventually.

How can a vehicle «catch up» without producing more acceleration than the vehicle it overtakes? The reason the sports car produces more acceleration at one point is because it enters its RPM range where it produces more power than the Land Rover (which also enters its most powerful RPM range); and they both stay there by shifting gears.

Yes, the source of acceleration is the torque. But the source of the torque is the power produced.

At $v=0^+$, the amount of torque is dependent on the amount of energy released by the engine $dE$ during a displacement $d\theta$: $T = \frac{dE}{d\theta} = \frac{Pdt}{d\theta} = \frac{P}{\omega}$. Theoretically, for the nanosecond that $v=0$ while power is applied, the wheel torque is infinite and the tire friction force reaches its maximum, leading to slipping. A mere nanosecond. After that, traction force can be derived directly from power (as long as it is not limited by the tire maximum friction force).

 

[Dale]

Only power matters.

Again, no

By the way, you did not respond to my questions to you above.

Assuming both vehicles have a transmission designed to extract the most from their engines (which they probably have), a Land Rover gets more acceleration at low RPM because it has more power at low RPM than a sports car has at low RPM. That is the only way.

The «extreme» ones that do beat the Land Rover - guess what? - they have more low RPM power than the Land Rover can produce. That is the only way.

In all of these arguments you could simply replace the word “power” with “wheel torque” and have an equally valid argument. Your arguments are neutral evidence in scientific terms, they do nothing to distinguish between the two hypothesis.

If you wish to have some actual evidence then you need to consider ways in which wheel torque and power differ. In all those ways acceleration follows torque, not power. But it seems that you are so blinded by the overarching argument that you are more interested in chanting “power” than in actually looking at sound science.

 

[jim hardy]

Of course it is better to use the tractive effort.

which is wheel torque divided by wheel radius.

 

[jack action]

I guess you are talking about these questions:

So then, without changing the configuration, just looking at that plot as is, where does the peak acceleration occur? Is it at the peak torque or the peak power?

Configuration can be changed: they are different gear ratios. With a CVT, you have an infinite number of gear ratios. For any given speed, the peak acceleration will occur at peak power. Maximum acceleration is a function speed.

If it is at peak power then which of the four peak power points has the highest acceleration and why?

They all are. Maximum acceleration is a function of speed. Maximum available power is usually constant with respect to speed.

Do you believe that in all engines the peak power occurs at the same point as the peak wheel torque?

Yes. At any given speed that is.

Yes, that is the one with the largest potential acceleration. And which one has the largest actual acceleration? It is the one with the largest wheel torque.

At any given speed, the largest wheel torque is only dependent on the appropriate gear ratio selected. That appropriate gear ratio is the one that will give the highest power at that speed.

In all of these arguments you could simply replace the word “power” with “wheel torque” and have an equally valid argument.

I'm glad you agree that my argument is as valid as yours.

In all those ways acceleration follows torque, not power.

And wheel torque, as a function of speed, what does it follow?

If someone asks you: «How can I double my car's acceleration?» Are you going to answer: «All you need is a larger wheel torque. Double your gear ratio in your differential and you will double your acceleration. Better yet, just put wheels with half the radius, that will do the job.» You will be right, but be prepared to face a mad customer when he will find out his car goes at half the speed. Then you will probably say: «Oh! you wanted to keep the same speed range? You didn't mention that. If you want to do that, you have no other choices but to double your power if you want to double the acceleration.»

The problem you have right now is that you are trying to fit your definition of acceleration to the problem (i.e. instantaneous) and then say: «See, I'm right.» And, yes, you are. And, yes, it wasn't an error on your part to assume that definition at first.

But the acceleration that the OP talks about is the average acceleration over a given speed range. Again, it is OK you did not get that initially. There are no "right" definitions of acceleration. You went for the pure "physics" sense; car guys go for the more "practical" sense, with the words they know and without necessarily understanding that other definitions might be assumed.

But now that you understand what is at stake, you have enough knowledge to get what the problem is all about. Most car guys without university education don't have all the knowledge you have and cannot get what you are talking about. It is therefore your responsibility to reach to them as they don't get what you are talking about and they get confused and frustrated. They don't care about instantaneous acceleration. Stop talking about it. Stay on the subject at hand. Correct the language if you want, but stay on topic: Average acceleration over a given speed range.

I will cite again what @russ_watters said (emphasis mine):

but when people say "acceleration" for a car they are usually talking about average acceleration presented as a time to a speed (0-60, 50-70, etc), not instantaneous acceleration.

By knowing what people want, including seeing through lack of specificity and how "car guy" terminology differs from the standard physics terminology you can bridge the gap and make sure we're answering the question that was intended to be asked.

It is not about you being right, it about you answering the question asked, even if it requires some decoding on your part.

 

[russ_watters]

It is not about you being right, it about you answering the question asked, even if it requires some decoding on your part.

Agreed, and I will add the caveat that when one knows there is a miscommunication they should endeavor to correct it instead of arguing right/wrong. It is less confrontational to say a person is correctly answering a different question than just to say they are "wrong". Better yet, when in an emphatic argument, one should put effort into looking for the miscommunication.

So for this situation, I would say that the word "acceleration" is being misused, 0-60 time is what is intended and as such power is the critical factor. Further I would say that power and torque are not separate issues, so the entire debate is a misnomer: engine torque is not a relevant parameter on its own, and two otherwise identical cars with different rpm/torque ratios but generating the same horsepower at the same speed must by law of physics be delivering the same torque to the ground.

 

[Baluncore]

Imagine the force is the result of a jet engine: Where is the torque in that vehicle? Only power matters.

The OP referred to Force and Power, not to torque. We know his force is torque for wheeled vehicles, but we do not worry about drive wheel torque while driving because it is hidden from us by the engine-speed to road-speed matching device called a gearbox. For that reason we optimise power = energy conversion, so we can climb hills by increasing PE = m·g·h, or accelerate by increasing KE = ½·m·v².

Torque is unimportant unless the drive wheel radius and RPM is specified. Energy converted by the motor is transmitted as power = torque * RPM. Acceleration will reduce torque as the RPM increases. Vehicle acceleration will be a maximum when the vehicle first starts to move. Since engine torque and power are gear ratio independent, the drive wheel torque can be maximum only in low gear, at low road speed.

It is clear that for a fixed energy flow, torque must fall as speed increases. Since KE = ½·m·v², acceleration will reduce as road-speed increases, even with a perfect gearbox.

 

[Dale]

Maximum acceleration is a function speed

Since you are just outright lying now it is time to close this thread. There is only one speed at which the acceleration is a maximum as is clearly shown in your own figure.

Thread closed.