Ugh. Math. Power versus Acceleration

[niky]

Okay, so I've been having an interesting conversation elsewhere, and I've been faced with the most charming of dilemmas.

Given that Power is Work done over time, ergo:

P = W/t

Which can be translated into Force times Velocity:

P = FV

Which can be translated into Mass times Acceleration times Velocity

P = MAV

Some people claim that as V increases, the Power required to keep constant A also increases.

In other words, if an engine has X Power, acceleration will slow down as velocity increases. Even without wind resistance or any other factors.

-

My counter is that we're considering the output of the engine as force. Not power. So acceleration should be constant given constant mass. So... discuss.

 

[Beeblebrox237]

Well, using the equation P=MAV, then yes, as velocity increases then acceleration will decrease, assuming that P is constant.

And now we wait for Famine...

 

[ITCC_Andrew]

Isn't force somewhat dependent on acceleration in itself?


Like, a car that's stationary doesn't really have much force, until it accelerates.


Anyways, I think, given constant mass, gear ratios and wind resistance, the power output needed to maintain constant acceleration is what changes.


Considering it as a car accelerating, even that equation in itself meets up with increasingly powerful variables (drivetrain resistance, wind resistance, and the "I'm about to run out of tarmac" driver fear variable, ) it would obviously take less effort for a bigger motor to reach the same speed.


Again, power doesn't always even lead to acceleration. We've all seen the massive amounts of power transport trucks get, and how slow they remain. That's just mass, wind resistance, and gear ratios at work.


But, if you spun a motor up to speed, without moving tires, to see which motor revs up faster, that doesn't immediately lead to acceleration either.


I'm confused as to why people are having this discussion, given that it seems to be "apples to oranges."


So, please explain in context.

 

[niky]

The question is: What is P and who put it there? And is it correct to say that P is the motive force moving the object?

Or that P simply describes the outcome of the application of the motive force?

This is where I run into trouble explaining that, yes, given constant output from a rocket motor, acceleration of a rocket is constant, unless the mass changes. And since the mass decreases, a rocket at constant thrust will actually experience a positive change in acceleration over time rather than a negative change.

One countered that as a rocket reaches higher velocities, then the power required to accelerate it increases. Obviously... not. Because what matters is the velocity of the exhaust gas relative to the rocket.

-

In the case of the car, we know that acceleration tapers off the faster you go. My take on it is that this is due to gearing. At low gears, you have more force delivered over less time. In higher gears, you have the same force delivered over a longer time frame, which means lower acceleration. (Ignoring friction and drivetrain losses and wind resistance.) They're countering that it's simply P=MAV. I've found a physics discussion which states something similar... taking into account that by putting power through the wheels, you are accelerating the Earth in the opposite direction. , but it doesn't seem proper to me.

 

[Exorcet]

Constant power is not a constant acceleration.

Over 1 second an engine might produce 500 J of energy. If the car is still initially, the kinetic energy has gone from 0 to 500 in one second. If the mass of the car is one kg the velocity has gone from 0 to 32 m/s in 1 second.

If the car started with 500 J then velocity only goes from 32 to 45 (or a change of 13).

The question is: What is P and who put it there? And is it correct to say that P is the motive force moving the object?

Or that P simply describes the outcome of the application of the motive force?

This is where I run into trouble explaining that, yes, given constant output from a rocket motor, acceleration of a rocket is constant, unless the mass changes. And since the mass decreases, a rocket at constant thrust will actually experience a positive change in acceleration over time rather than a negative change.

A rocket is a force producer. An ICE is a power producer. One produce ~constant thrust (force) the other produce ~constant power (rate of change of energy).

 

[ITCC_Andrew]

The question is: What is P and who put it there? And is it correct to say that P is the motive force moving the object?

Or that P simply describes the outcome of the application of the motive force?

This is where I run into trouble explaining that, yes, given constant output from a rocket motor, acceleration of a rocket is constant, unless the mass changes. And since the mass decreases, a rocket at constant thrust will actually experience a positive change in acceleration over time rather than a negative change.

Yes, and the amount of power it takes to put an object in motion decreases once it's in motion; rolling a ball down a hill. First you have to get the rocket to move, (leave the ground). Then you have a bit of momentum...


Or, another way to say it:

It's a lot harder to put your friend into motion, while they're on the swing, than it is to keep them in motion.

 

[Slash]

Without wind resistance I don't believe that power would have to increase to keep something moving at a certain velocity. Add wind resistance into that factor and I would certainly agree that power must increased to act against the increasing opposing force.

 

[ITCC_Andrew]

Without wind resistance I don't believe that power would have to increase to keep something moving at a certain velocity. Add wind resistance into that factor and I would certainly agree that power must increased to act against the increasing opposing force.

Well, I'm also thinking about how the wind resistance and gear ratios kinda work against each other.
If you had 100 gears, (a bit absurd, admittedly,) each of those 100 gears would be small that the wind resistance would have lessened impact.


But, I also see your point. That's entirely true.

 

[Omnis]

Niky, you forgot to multiply by VTEC.

 

[niky]

Damnit. It's supposed to be a two dimensional problem!

A rocket is a force producer. An ICE is a power producer. One produce ~constant thrust (force) the other produce ~constant power (rate of change of energy).

Ah, that is probably a better way to put it, then.

I know that an ICE connected to the wheels loses the ability to accelerate at higher speeds, but I've pegged it down to the amount of force (in terms of rotational energy) falling behind the curve as gear ratios increase the time over which that force is applied... lessening the force at the wheels at higher rotational speeds.

 

[Omnis]

Aren't those figures instantaneous acceleration and velocity? One is a function of the other. At a constant power, it follows that an object should be travelling at x velocity. If it's not at that v, then its acceleration must be greatest at v=0 and zero at v=x.

Please tell me if I'm wrong, and, further, if I'm an idiot.

 

[ITCC_Andrew]

Wheelspin.

Other than that, Omnis, I think you're right.

 

[niky]

Aren't those figures instantaneous acceleration and velocity? One is a function of the other. At a constant power, it follows that an object should be travelling at x velocity. If it's not at that v, then its acceleration must be greatest at v=0 and zero at v=x.

Please tell me if I'm wrong, and, further, if I'm an idiot.

Constant velocity, though, against what force? I'm not arguing that the formula is incorrect, I'm trying to put it into context.

 

[Omnis]

The mass*acceleration. ... More mass held at constant velocity being proportional to power.

It's 2am here though so I'd rather someone else figure it out with more rigor.

 

[niky]

See... here's the interesting part. If we're pushing an object up against gravity, there's your reference for acceleration/velocity. You're maintaining velocity against the pull of gravity. An object moving perpendicular to gravity (a car), what is the reference for velocity/acceleration?

Yes. This makes my head hurt. But it's an argument on another board. I want to WIN, damnit!

 

[Small_Fryz]

Aren't those figures instantaneous acceleration and velocity? One is a function of the other. At a constant power, it follows that an object should be travelling at x velocity. If it's not at that v, then its acceleration must be greatest at v=0 and zero at v=x.

Please tell me if I'm wrong, and, further, if I'm an idiot.

You might not be wrong, but I do think you're an Idiot

 

[ITCC_Andrew]

Popcorn + Beethoven's fifth.

 

[oopssorryy]

Popcorn + Beethoven's fifth.

Are you watching A Clockwork Orange?

 

[CheersMate]

Okay, i try to explain the problem, but it´s gonna be hard to do it in english

You mixed up the momentum with kinetic energy.

A example:

Two cars are driving in the same direction, Car A) is driving with 100mp/h, car B) with 200mp/h.

If you use the principle of linear momentum (p=m*v), Car B) has double the momentum than car A)

Thats right, but now lets brake down to 0 mp/h.

Car B) drives two times the distance of car A) in one second, because the velocity b is twice as high as velocity a.

Kinetic energy is: 1/2*m*v^2

That means car B) has four times the kinetic energy than car A)

Let´s say both cars weight 1kg.

That means for Car A): 1/2*1*27,7^2= 383,645 kg*m^2*s^-2 = J

For Car B) 1/2*1*55,5^2= 1540,125 kg*m^2*s^-2= J

As you can see: The brakes have to "destroy" four times the kinetic energy for Car B) than for Car A) although the velocity is just twice as high.

If you throw a ball two times in the air, one time with velocity a) and the second time with 2*velocity a) the ball will fly four times as high.


Back to the cars:

If your car has 150bhp power, it can deliver 110 KJ (kinetic-) energy per second to your car.

That means, if you need 2 seconds to 100 mp/h, you will need 220 KJ.
But if you want to drive 200 mp/h, you will need four times the energy = 880KJ.
660 KJ are missing and to deliver 660 KJ, your engine needs 6 more seconds.
8 Seconds total to 200 mp/h and not double the time 0-100mp/h.

 

[ITCC_Andrew]

Are you watching A Clockwork Orange?

No. I'm watching Dion. Clearly.

Some day I might actually see A Clockwork Orange, but, I haven't, so I have no idea where you're going with this.

*snip*

Very well done! For a German speaker (Wahr?) that was, so far as I can see, the best English I've seen on these forums.

 

[Exorcet]

Yes, and the amount of power it takes to put an object in motion decreases once it's in motion; rolling a ball down a hill. First you have to get the rocket to move, (leave the ground). Then you have a bit of momentum...


Or, another way to say it:

It's a lot harder to put your friend into motion, while they're on the swing, than it is to keep them in motion.

Well when in motion, 0 power is required to keep it in motion if we ignore drag, gravity, friction, etc. But to keep it accelerating, you need power. Acceleration is a change in velocity which is a change in energy for a constant mass.

If there was no gravity you would have to work very hard to make your friend move the normal way on a swing.

Without wind resistance I don't believe that power would have to increase to keep something moving at a certain velocity.

It has to increase because the amount of energy you put it becomes smaller and smaller compared to the amount of energy already in the object.

Ah, that is probably a better way to put it, then.

I know that an ICE connected to the wheels loses the ability to accelerate at higher speeds, but I've pegged it down to the amount of force (in terms of rotational energy) falling behind the curve as gear ratios increase the time over which that force is applied... lessening the force at the wheels at higher rotational speeds.

It's not the gears, but just the nature of energy which goes with V^2. Power is to energy what force is to momentum, but momentum goes with V.

And while the rotational energy of the engine is related to force, it's not a force itself. Take a rocket engine out of a rocket and light and it will go somewhere. Do the same with an ICE and it will make noise, parts will spin, but it won't move. It will just dump energy into the rotating parts that can be harnessed by something else.

Aren't those figures instantaneous acceleration and velocity? One is a function of the other. At a constant power, it follows that an object should be travelling at x velocity. If it's not at that v, then its acceleration must be greatest at v=0 and zero at v=x.

Please tell me if I'm wrong, and, further, if I'm an idiot.

A constant power implies constant acceleration at a given velocity but not a fixed velocity, unless there is an opposing and equal power (like drag).

In a vacuum (and on a frictionless but not tractionless road and maybe a CVT) a car will accelerate to infinity, but it will take a very long time to get there.

 

[BobK]

Okay, so I've been having an interesting conversation elsewhere, and I've been faced with the most charming of dilemmas.

Given that Power is Work done over time, ergo:

P = W/t

Which can be translated into Force times Velocity:

P = FV

Which can be translated into Mass times Acceleration times Velocity

P = MAV

Some people claim that as V increases, the Power required to keep constant A also increases.

In other words, if an engine has X Power, acceleration will slow down as velocity increases. Even without wind resistance or any other factors.

-

My counter is that we're considering the output of the engine as force. Not power. So acceleration should be constant given constant mass. So... discuss.

Problem is, the engine doesn't deliver constant power, it follows a curve.

Isn't force somewhat dependent on acceleration in itself?

Force is mass times acceleration, often expressed as F=MA. Acceleration in turn is change in velocity per unit time, or V/T. And velocity is distance per unit time. This gives us F=D/T^2

Like, a car that's stationary doesn't really have much force, until it accelerates.

A car doesn't have force, it has a force applied to it.

Anyways, I think, given constant mass, gear ratios and wind resistance, the power output needed to maintain constant acceleration is what changes.

Gear ratios have nothing to do with it. Nor does wind resistance have anything to do with it, as this is a theoretical physics problem primarily.

Considering it as a car accelerating, even that equation in itself meets up with increasingly powerful variables (drivetrain resistance, wind resistance, and the "I'm about to run out of tarmac" driver fear variable, ) it would obviously take less effort for a bigger motor to reach the same speed.

The, err, fear variable?

Again, power doesn't always even lead to acceleration. We've all seen the massive amounts of power transport trucks get, and how slow they remain. That's just mass, wind resistance, and gear ratios at work.

Please take a course in basic physics.

But, if you spun a motor up to speed, without moving tires, to see which motor revs up faster, that doesn't immediately lead to acceleration either.


I'm confused as to why people are having this discussion, given that it seems to be "apples to oranges."


So, please explain in context.

I hope I did, somewhat. If you can track down a copy of Isaac Asimov's Realm of Measure he explains it beautifully.

Yes, and the amount of power it takes to put an object in motion decreases once it's in motion; rolling a ball down a hill. First you have to get the rocket to move, (leave the ground). Then you have a bit of momentum...


Or, another way to say it:

It's a lot harder to put your friend into motion, while they're on the swing, than it is to keep them in motion.

This is pure bull. You obvious lack of knowledge of basic physics is evident.

Damnit. It's supposed to be a two dimensional problem!

Indeed. Makes my head hurt, too.

No. I'm watching Dion. Clearly.

Some day I might actually see A Clockwork Orange, but, I haven't, so I have no idea where you're going with this.

And this is somehow relevant to the question? You've managed to go completely off-topic.

 

[niky]

And while the rotational energy of the engine is related to force, it's not a force itself. Take a rocket engine out of a rocket and light and it will go somewhere. Do the same with an ICE and it will make noise, parts will spin, but it won't move. It will just dump energy into the rotating parts that can be harnessed by something else.

Actually, I'm starting to see where I've gone off tangent. I was looking at the loss of acceleration due to gearing changes and how they affect force or torque, but I forgot that a CVT with theoretically infinite gears would give you a classically downsloping acceleration curve... going down as velocity goes up. Because as the velocity at the wheels increases, the acceleration provided by the same power decreases.

 

[Danoff]

Actually, I'm starting to see where I've gone off tangent. I was looking at the loss of acceleration due to gearing changes and how they affect force or torque, but I forgot that a CVT with theoretically infinite gears would give you a classically downsloping acceleration curve... going down as velocity goes up. Because as the velocity at the wheels increases, the acceleration provided by the same power decreases.

I'm not entirely sure I followed that but I think it's wrong.

The power, force, velocity, acceleration, of the engine can remain exactly constant regardless of the speed of the car. The force, distance over which that force is applied, acceleration all correspond to the crank and not the vehicle. Gearing is what allows you to divorce the rotational force imparted by engine from the rotational speed of the wheels.

 

[Exorcet]

I'm not entirely sure I followed that but I think it's wrong.

The power, force, velocity, acceleration, of the engine can remain exactly constant regardless of the speed of the car. The force, distance over which that force is applied, acceleration all correspond to the crank and not the vehicle. Gearing is what allows you to divorce the rotational force imparted by engine from the rotational speed of the wheels.

I think his post sounds OK. Yes, you can have constant force from a car, but that requires increasing power. This might be the case for some cars, but only within a single gear usually (unless you shift in such a way to avoid this, but I don't think that case pertains to the topic).

Specifically, the bit above about the CVT car is true if the engine is being held to peak power, which would be for max acceleration. The car would be producing constant power but also an increasingly lower acceleration as speed went up.

 

[Slash]

Well, I'm also thinking about how the wind resistance and gear ratios kinda work against each other.
If you had 100 gears, (a bit absurd, admittedly,) each of those 100 gears would be small that the wind resistance would have lessened impact.


But, I also see your point. That's entirely true.

Basically I summed it up so that gear ratios wouldn't come into play, in this case. That said I agree with you.

 

[Danoff]

I think his post sounds OK. Yes, you can have constant force from a car, but that requires increasing power.

For constant power, it is defined as work over time. Work is force times distance.

So constant power is force x distance/time.

There is nothing about that that suggests getting a constant force requires increasing power. If your engine is operating at the same RPM, distance per time is constant (the velocity of the car doesn't matter). So for constant RPM, you have constant power, constant force, and constant distance over a constant unit of time. All you need is gearing.

 

[Exorcet]

OK, I wasn't thinking of it like that. But the question seems to be about a car's ability to accelerate vs speed. The gearing is going to try to keep the power as constant as possible. The car will have some amount energy already at the start (may be zero) and it will increase due to the power of the engine. From the power and KE = .5mV^2, the acceleration can't be constant (taking power to be constant or nearly constant).

 

[Danoff]

OK, I wasn't thinking of it like that. But the question seems to be about a car's ability to accelerate vs speed. The gearing is going to try to keep the power as constant as possible. The car will have some amount energy already at the start (may be zero) and it will increase due to the power of the engine. From the power and KE = .5mV^2, the acceleration can't be constant (taking power to be constant or nearly constant).

I'm not following. I'm not talking about gearing for the purpose of keeping the engine's RPM at a particular place in the power band, I'm talking about gearing to provide mechanical advantage to rotational force exerted by the engine.

 

[Exorcet]

Yes but to maximize that force, you want to maximize power. So if the driver is trying to accelerate as fast as possible, he'll keep the engine as close to peak power RPM as possible. He's basically trying to keep the power output constant.

The power is going to the KE of the car (ignoring losses and rotation) and the relative change in KE decreases as KE gets bigger.

Gearing doesn't change this, CVT or not. For a car with discrete gears every time you hit peak power within a gear while accelerating, it corresponds to hitting peak power for a specific gear ratio with a CVT. Your gear ratio is constantly decreasing and so is your rotational force at the wheels.