KE of system / different reference frames question

[Humber]

Come on now, stay on topic and answer the question!

Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

Speaking of contributions, do you have pictures of your carts? RP said you made a couple of new ones to go with your cow-cart.

No; trolling would be to evade answering questions, even simple ones, to keep a discussion from reaching its inevitable (to most people at least) conclusion.

Need I pay homage to you, the master, before you provide a couple of pictures of your cart? And it was you who brought up the topic of carts.

Come on now, stay on topic and answer the question!

Back to the topic then: answer Dalespam's question.

The black pot speaketh!

How many times do you need to hear that the source of energy is frame dependent before you'll accept reality?

You are embarrassing yourself.

 

[Subductionzon]

Humber, you know better than that. A simple Galilean transformation shows you that the Brennan torpedo being powered by a steam driven cable is the same as a Brennan torpedo going downstream faster than the stream. Again it is all about relative velocities.

 

[Humber]

Humber, you know better than that. A simple Galilean transformation shows you that the Brennan torpedo being powered by a steam driven cable is the same as a Brennan torpedo going downstream faster than the stream. Again it is all about relative velocities.

And about cables. They are an engineering problem. The advantage of Brennan's system, was that it only needed two cables. Rudder control and drive on the same pair. That demanded a differential drive for the prop, and so two counter-rotating props. A complex arrangement that paid off in the distance the torpedo could travel, within the limits of cable capacity.

And the topic is KERS and ground energy.

 

[mender]

And the topic is KERS and ground energy.

How many times do you need to hear that the source of energy is frame dependent before you'll accept reality?

 

[my_wan]

That is the point. They should be the same, but aren't.

Ok, looking over your response I see a more reasonable attempt to respond.

Consider this analogy. We are outside and separate from the space shuttle and get in a fight. We knock each other back and start flying away from the space shuttle at 5 m/s. But wait, we are flying away from each other at 10 m/s. In the car/earth analogy the one of us is the 5 m/s velocity, and associated k_e, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a k_e of the car and the k_p of the battery. Both exactly half just like our fight resulted in half the velocity for each of us relative to the space shuttle.

You are attempting to conserve k_e in your debate with DaleSpam without accounting for where the k_p that went into the battery came from.

 

[Humber]

Ok, looking over your response I see a more reasonable attempt to respond.

Consider this analogy. We are outside and separate from the space shuttle and get in a fight. We knock each other back and start flying away from the space shuttle at 5 m/s. But wait, we are flying away from each other at 10 m/s. In the car/earth analogy the one of us is the 5 m/s velocity, and associated k_e, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a k_e of the car and the k_p of the battery. Both exactly half just like our fight resulted in half the velocity for each of us relative to the space shuttle.

You are attempting to conserve k_e in your debate with DaleSpam without accounting for where the k_p that went into the battery came from.

That has nothing at all to do with the matter at hand, and there is clearly a problem with the relative velocities in each case, as I quite clearly showed you. They are not my numbers. One wonders why you raised the matter, but now want to drop it, in favour of another.

 

[rcgldr]

DDWFTTW like devices rely on the difference in speed between two media. Which media that KE is extracted from depends on the frame of reference.

Is that the topic of this thread?

Yes, from the first post:

This post appeared on a ddwfttw forum: "For those who actually care, relative to any frame other than that of the ground, the ground does have energy. ... KERS ... initial energy from fuel

It doesn't matter what the initial energy source was.

Asume an inertial frame of reference at some initial fixed speed and direction wrt ground. In that frame, the ground (earth) has energy. If a vehicle accelerates or decelerates by applying a force to the ground, then the energy of the ground (earth) is changed wrt the frame of reference. In the cases where the speed of the ground (earth) is increased wrt frame of reference, then energy is added to the ground (earth). In the cases where the speed of the ground (earth) is decreased wrt frame of reference, then energy is extracted from the ground (earth). I don't understand why there is an issue with this concept.

 

[Humber]

This is true, due to Newton's 3rd law.

This is false, in general.

If some object gains momentum Δp then p_f=p_i+\Delta p

Dividing both sides by the mass m we get v_f=v_i+\Delta v

Substituting that into the expression for the object's final KE we obtain
KE_f=\frac{m}{2} v_f^2 = \frac{m}{2} (v_i+\Delta v)^2 = \frac{m}{2} (v_i^2 + 2 v_i \Delta v +\Delta v^2) = KE_i + v_i \Delta p + \Delta p^2/2m
\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m

This is different from what you wrote because of the v_i \Delta p term. Your neglecting that term is actually your key conceptual error, so I do hope you spend some time examining the brief derivation. If the object is particularly massive, like the earth, then the \Delta p^2/2m term drops out, but the v_i \Delta p term does not.

Momentum is conserved between the car and the Earth.
The car gains -p, and the Earth gains p, so in the above case, that is Δp. It is p_i, which drops out.
In your first post you obtained a result based upon that fact, which is why you also derived
1e-18J, which is the same result as obtained from KE = p²/2m_earth.

Momentum is always conserved, and is frame independent, so that is a unique and frame independent result.

 

[Humber]

Yes, from the first post:

It doesn't matter what the initial energy source was.

Asume a frame of reference at some initial fixed speed and direction wrt ground. In that frame, the ground (earth) has energy. If a vehicle accelerates or decelerates by applying a force to the ground, then the energy of the ground (earth) is changed wrt the frame of reference. In the cases where the speed of the ground (earth) is increased wrt frame of reference, then energy is added to the ground (earth). In the cases where the speed of the ground (earth) is decreased wrt frame of reference, then energy is extracted from the ground (earth). I don't understand why there is an issue with this concept.

Then there is a futher inconsistency. In the first post, there is an equal change of power
of 100W, or 100J/s, but Dalespam's solution, has twice that change for the ground energy.
You need ground energy, to support the cart, and that will cause you to contradict yourselves, time and time again.
Momentum is conserved, and that says that there is no ground energy, but for a negligible amount. The energy transferred to the ground, due to surface deformation, will swamp that.

 

[my_wan]

That has nothing at all to do with the matter at hand, and there is clearly a problem with the relative velocities in each case. One wonders why you raised the matter, but now want to drop that, in favour of another.

This is merely a reframing of the exact same problem. Perhaps not useful, and you are free to choose another but certainly not dropping the original matter. So let's go back to the original.

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

Nope, it doesn't. Think of it this way then. Wouldn't slowing down fro 100 m/s be the same as slowing down by 50 m/s twice? Yet m50^2 + m50^2 \neq m100^2. Slowing down to half your speed take a lot more energy than half your energy. Look up the "Merton mean speed theorem" if you want to know why it is half the total acceleration to stop the car.

 

[Humber]

Frame independence for the result can be shown by the introduction of an observer, moving relative to the Earth and Car at velocity u. That can also be the centre of mass of the car and Earth.

m = mass of car

ΔKE(car)

= (−m u − p)²/2m − (−m u)²/2m

= (m² u² −2m u·p + p²)/2m − (m² u²)/2m

= −u·p + p²/2m


M = mass of Earth.

ΔKE(earth)

=(−M u + p)²/2M − (−M u)²/2M

= (M² u² + 2M u·p + p²)/2M - (M² u²)/2M

= u.p + p²/2M

The term u.p, cancels out, and is independent of the mass of either the car or the Earth, and has no physical meaning. The total energy is once again

p²/2m + p²/2M, where the second term is negligible.

 

[Humber]

This is merely a reframing of the exact same problem. Perhaps not useful, and you are free to choose another but certainly not dropping the original matter. So let's go back to the original.

But, not ignore;
1) That there can be no difference of the sum velocities from any frame.
2) That momentum is conserved, so there can be no significant energy transfer because of the relative masses of car and Earth
3) That the solution provides an energy change that is independent of Earth mass
4) That the total energy change is greater than the KE of the car.
5) That at least 3 different solutions have been offered.

Nope, it doesn't. Think of it this way then. Wouldn't slowing down fro 100 m/s be the same as slowing down by 50 m/s twice?

No, because the first case has 4 times the KE of the second KE = 1/2mv^2.

Yet m50^2 + m50^2 /neq m100^2. Slowing down to half your speed take a lot more energy than half your energy. Look up the "Merton mean speed theorem" if you want to know why it is half the total acceleration to stop the car.

That is not relevant, when the applied force is constant. And the change in momentum is then Δp = F.Δt

The primary problem, is one of misbegotten notions of relative motion; that one may say that a car moving relative to the Earth at v, is the same as the Eatrh moving at -v relative to the car.
That is trivially true, but will not work for non-inertial frames, and there must be acceleration for KE to change. The acceleration of the car, may not be equated with acceleration of the Earth.

The method employed, also results in energy that is independent of the Earth's mass, and so the same as the u.p of my above post. It has no physical meaning.

KERS can be used to show this, and that the above errors of relative motion, are the basis of the Treadmill.

An F1 car, can store energy, that is directly related to its translational KE. That is not the case on the TM

F1 cars have a minimum weight of 640kg, and a maximum wheel diameter of 0.66m.
At a modest 100km/hr (44m/s) can gain ~ 620kJ of translational KE.

The TM, with a belt at 44m/s, is said to be a Galilean Transform of a real car on the road.
The car is on the belt, and is driven as usual, by a roller and motor.

Make that roller a 0.66m cylinder, and say, a reasonable 50kg. The belt speed will be the same as the roller's surface speed.

On the belt at 100km/hr, ( 44m/s), both car's wheel and cylinder will have the angular velocity of;
w =(44m/s)/0.33m
= 133radians/sec.

The moment, I, of the cylinder is approximately
I = m*r^2
= 50*0.33^2
= 5.445

The angular KE of that cylinder is then;
KE = 1/2*I*w^2
= 48kJ.

The most a KERS could recover, is 48kJ of the cylinder, but the car on the road has 620kJ of translational KE due to linear motion. The belt does not imply that motion.

The reason being, that the car has no translational KE, because the TM's transform is false, and based on the same erroneous reasoning that lays claim to ground power.

 

[RCP]

The final recursion?

@my (obi) wan: Your space shuttle analogy was quite interesting. I do hope humber addresses it in full.

You were the one who started this cyber-strand at JREF way back in '08. Bet you never dreamed it would still be going on in 2012, having accrued well over 100k replies combining all the venues where it has appeared. And just when I sense the denouement is on the horizon, here you are once again. Awesome.

@mender: The pics of humber's 3 carts were posted months and hundreds of pages apart at TR, so they won't be easy to find but I assure you they are there and look well constructed. If I do come across them or can get humber to post all 3 again there, I will email you. As humber says, it would be OT itt, and the issue of ground energy seems to lie at the heart of this epic saga. I find myself anxiously awaiting each new reply and it's inevitable rebuttal, as I did when he was discussing it with uncool. After 3 long years I think we are finally getting down to the nitty-gritty.

 

[Humber]

Ok, looking over your response I see a more reasonable attempt to respond.
Consider this analogy. We are outside and separate from the space shuttle and get in a fight. We knock each other back and start flying away from the space shuttle at 5 m/s. But wait, we are flying away from each other at 10 m/s.

In which case the energy to do that will have come from one or both of them.
Each will gain KE = p^2/2m, and must be the same mass to do so. An astronaut inside the space shuttle applying the same impulse, will transfer only KE = 2p^2/2mShuttle.

In the car/earth analogy the one of us is the 5 m/s velocity, and associated k_e, momentum, etc.

When the car is at rest, both the Earth and car at the same speed, whatever that may be.
The only factor that matters, is how much momentum is transferred, to or from the ground. That is not a "relative" quantity.

For the given example; m_car*10m/s = 1000kg.10m/s.
To follow through with the analogy, at 5kph, the KE would be 1/2mv^2 = 12500J
25% of what is present at 10m/s.

To move the Earth by 5m/s;

KE = 1/2mv^2 = .5*6e24kg*5^2 = 7.5e25J

To give you an idea how much energy that is, consider a 100MW power station.
In continuous operation, it produces 100e6J/s, and so 3600*24*365*100e6J ~ 3.2e15J per year.

That is 7.5e25J/3.2e15J ~ 2.3e10 years' output, which is 2.3e10/4.6e9 or 5 times the age of the Solar System. I doubt that a car with 50kJ is going to make an impression.

You are attempting to conserve k_e in your debate with DaleSpam without accounting for where the k_p that went into the battery came from.

I have not conserved energy, but only momentum. The proffered solutions provides more than the 50kJ of the car. F1 cars are festooned with instruments, and they measure everything that can be measured, and they will say no energy from the ground - there are no instruments to measure it. Only one set of results, that can be broadcast worldwide, and to planes and space stations.

KE is relative, and as such can be calculated from velocity, yet have no physical meaning.
That is never the case with momentum, which many, including all cart enthusiasts, prefer to ignore.

 

[my_wan]

@RCP
The "cyber-strand" deserved the attention it got and not at all surprising. It also had some people willing to stick it out more than I was and they deserve more credit. I only found it worthwhile to support the claim and blow a few whistles because the physics was in their favor.

However, I also failed to recognize that I was debating one of the same people here, and on a physically related set of perceptions about Galilean relativity to boot. Probably not worth sticking it out under the circumstances.

 

[Humber]

@RCP
The "cyber-strand" deserved the attention it got and not at all surprising. It also had some people willing to stick it out more than I was and they deserve more credit. I only found it worthwhile to support the claim and blow a few whistles because the physics was in their favor.

However, I also failed to recognize that I was debating one of the same people here, and on a physically related set of perceptions about Galilean relativity to boot. Probably not worth sticking it out under the circumstances.

It is not just about the cart, but those very ideas of Galilean Relativity. Haven't you noticed that every example, is a two object inertial case, or how that is somehow suspended, so that things can accelerate? The most trivial possible. Definitions are stretched to breaking point, and then used as a bludgeon.

Do you have a rebuttal to the problem of the F1 car on the belt? It completely falsifies the treadmill, and the Galilean notions that define it. Or how an ordinary domestic car, can be said to accelerate the massive Earth? That one, defies all experience of the physical world.

 

[my_wan]

I have not conserved energy, but only momentum.

And didn't DaleSpam also point out that energy must also be conserved? Or do you not believe energy is conserved? Yet you have half the energy put in the battery as potential energy, half in removing the cars momentum relative to Earth yet your method wants a third half to account for the second half of the k_e? The 50 kJ you want accounted for is in fact 50 kJ k_e, not momentum. You can't spend 50 kJ to stop the car and put 50 kJ in the battery and still expect another 50 kJ out of that 100 total kJ.

Said quiet nicely here:

Just as the Earth lost 100 kJ (50 kJ to the car's KE and 50 kJ to the battery), as the car went from 0 to 10 m/s the Earth will gain 100 kJ (50 kJ from the car's KE and 50 kJ from the battery) as the car goes from 10 to 0 m/s. There is no free lunch here.

I don't really see much else to add.

 

[rcgldr]

Do you have a rebuttal to the problem of the F1 car on the belt?

I don't see any problem here. If both the belt and the F1 car are initially at rest (say from a ground frame of reference), and then the F1 uses it's engine to convert chemical potential energy from fuel into kinetic energy of the belt and F1 car, and assuming no losses to heat, then all of the energy winds up as an increase in KE of the belt and F1 car, with most of the KE going into the belt because the F1 car has much more mass than the belt.

It completely falsifies the treadmill.

Falsifies what?

 

[Dale]

Momentum is conserved between the car and the Earth.
The car gains -p, and the Earth gains p, so in the above case, that is Δp.

From this response I understand that you agree with the proof of post 59 as it applies to a single body and that here you are correctly pointing out that in the KERS scenario I must use the formula twice, once for \Delta KE_e and once for \Delta KE_c, with \Delta p_e=-\Delta p_c in order to conserve momentum.

Is this a correct understanding of your position? If I am misunderstanding then can you please provide a clear yes/no answer to whether or not you agree with the proof of post 59.

 

[Humber]

I don't see any problem here. If both the belt and the F1 car are initially at rest (say from a ground frame of reference), and then the F1 uses it's engine to convert chemical potential energy from fuel into kinetic energy of the belt and F1 car, and assuming no losses to heat, then all of the energy winds up as an increase in KE of the belt and F1 car, with most of the KE going into the belt because the F1 car has much more mass than the belt.

Falsifies what?

The amount of KE recoverable by the KERS is due only to the KE of wheels and the mechanism of the TM, and not related to the translational KE of the car, in any way.

 

[Dale]

ΔKE(car)

= (−m u − p)²/2m − (−m u)²/2m

= (m² u² −2m u·p + p²)/2m − (m² u²)/2m

Note, a slight math error marked in red. A negative times a negative is a positive, so that term should be +2m u·p

ΔKE(earth)
= u.p + p²/2M

So you clearly agree with my formula, \Delta KE = v_i \Delta p + \Delta p^2/2m, in the case where u=v_i and p=\Delta p. Is there any time that you do not agree with it?

Btw, it would really help if you were more careful about distinguishing your quantities. Ie. Is your p the change in momentum or is it the initial momentum? I follow the usual convention where \Delta p is the change in momentum, p_i is the initial momentum, and p_f=p_i+\Delta p is the final momentum.

 

[Humber]

From this response I understand that you agree with the proof of post 59 as it applies to a single body and that here you are correctly pointing out that in the KERS scenario I must use the formula twice, once for \Delta KE_e and once for \Delta KE_c, with \Delta p_e=-\Delta p_c in order to conserve momentum.

Is this a correct understanding of your position? If I am misunderstanding then can you please provide a clear yes/no answer to whether or not you agree with the proof of post 59.

Only the change in the car's momentum is of concern, so, \Delta p_e=-\Delta p_c That is all there is. It is transferred to the Earth.
In the first case, the car's 50kJ, is transferred to the battery. So, the second case, which is a mirror, should see the energy come from the battery to the car. The total remains in each case at 50kJ.

In the example you gave, and in the first case, the energy to the ground is a result of a change of Earth's velocity, 1e-21m/s. That results in 5e-18J

In the other case, the change in Earth's velocity is Vcar*(Mearth - Mcar)/Mearth
Vcar =10m/s
Mearth = 1e25kg
Mcar =1e3kg

so the change in KE is related to (10 - 1e-21)m/s.
That , I think, accounts for the difference of relative velocities and energy.

The energy "transferred back to the car" should be Mearth/Mcar*5e-18J =50kJ

 

[Dale]

Only the change in the car's momentum is of concern, so, \Delta p_e=-\Delta p_c That is all there is. It is transferred to the Earth.
In the first case, the car's 50kJ, is transferred to the battery. So, the second case, which is a mirror, should see the energy come from the battery to the car. The total remains in each case at 50kJ.

In the example you gave, and in the first case, the energy to the ground is a result of a change of Earth's velocity, 1e-21m/s. That results in 5e-18J

In the other case, the change in Earth's velocity is Vcar*(Mearth - Mcar)/Mearth
Vcar =10m/s
Mearth = 1e25kg
Mcar =1e3kg

so the change in KE is related to (10 - 1e-21)m/s.
That , I think, accounts for the difference of relative velocities and energy.

You are still dodging the question. Do you agree with the proof post 59 or not? Please be clear.

 

[Humber]

Note, a slight math error marked in red. A negative times a negative is a positive, so that term should be +2m u·p


So you clearly agree with my formula, \Delta KE = v_i \Delta p + \Delta p^2/2m, in the case where u=v_i and p=\Delta p. Is there any time that you do not agree with it?

Btw, it would really help if you were more careful about distinguishing your quantities. Ie. Is your p the change in momentum or is it the initial momentum? I follow the usual convention where \Delta p is the change in momentum, p_i is the initial momentum, and p_f=p_i+\Delta p is the final momentum.

Momentum adds linearily, so there is no dependency on p_i
p^2/2mcar is the KE of the car with momentum p
p^2/2mearth is the change in the Earth's KE when p is transferred to it.

 

[Dale]

Momentum adds linearily, so there is no dependency on p_i

This is an ambiguous statement. There is no dependency of what quantity on p_i?

Also, clearly p_i depends on p_i even though momentum adds linearly, so adding linearly is not a justification for a lack of dependency in general. If you want to show that x doesn't depend on y because y adds linearly then you need to take the linearity property, use it in the expression for x, and show that y drops out algebraically.

Btw, I note that you still avoid the question.

p^2/2mcar is the KE of the car with momentum p

This is true in general.

p^2/2mearth is the change in the Earth's KE when p is transferred to it.

This is true only in the case where the initial velocity is 0. Note the word "change" in bold. The change of x is not the same thing as x except when x is initially 0.

 

[Humber]

This is an ambiguous statement. There is no dependency of what quantity on p<sub>i?

Also, clearly pi depends on pi even though momentum adds linearly, so adding linearly is not a justification for a lack of dependency in general. If you want to show that x doesn't depend on y because y adds linearly then you need to take the linearity property, use it in the expression for x, and show that y drops out algebraically.</sub>

_{It does. The existing momentum is pi,e the total after the change is pi,e + pi,c, so the change is just pi,c
The change in KE for the car is pi,c^2/2mc and for the Earth pi,c^2/2me
Works both ways.}

 

[Dale]

It does. The existing momentum is p_i,e the total after the change is p_i,e + p_i,c, so the change is just p_i,c
The change in KE for the car is p_i,c^2/2m_c and for the Earth p_i,c^2/2m_e
Works both ways.

Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.

 

[Humber]

Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.

The point is that it's not. The idea of the car accelerating the Earth to 10m/s is not physically realizable. In the real world, an F1 car is on the track. The KERS is applied. If viewed from the frame of the car, and from the frame of the ground at the same time, there will be no difference. To start with the Earth at 10m/s relative to the car, is a totally different physical situation. In the first case the calculation is done with the car having 50kJ, which after 10s, is transferred to the battery. In the second, it's already there, at t = 0.
Makes no sense.

 

[Humber]

Then it should be no problem to demonstrate it algebraically, this time using clear variable names to distinguish between initial, final, and change quantities.

It's your notation i = initial, c = car, e= earth. There is one transfer of 10000kg.m/s to the momentum, p_i,e. It makes no difference what that is. The resultant KE depends on the Earth's mass.

 

[Ken G]

Momentum is frame independent, otherwise, there could be violations of conservation.

I figured this was your core error, you don't understand the meaning of conservation laws. Conservation laws don't mean that the quantities are fixed, regardless of frame. They mean that once you pick a frame, the total quantities will stay the same in that frame. If you change frames, the quantities change. That's how conservation laws work. Do you get this, or not?

If follows that changes are also frame independent.

Wrong, the changes are frame independent, but that doesn't "follow" from anything, that statement, and only that statement, is the conservation law. Think about this as long as it takes.

Total energy = p^2/2mcar + p^2/2mearth is correct.

No, not if you think that p is a change in momentum, which is how you are using it. Doesn't it concern you that you do not get the correct answers, and you conclude that all the experts are wrong, but when they tell you what you are doing wrong, you just claim you are right? I can tell you right now, you will never learn anything that way. Is it all right with you to never learn anything?

If the energy change of the car were the same as the change in the ground, p^2/2mcar - p^2/2mearth = 0, then obviously the KERS could not store any energy.

Notice where once again you associate p^2/2m with energy changes. Wrong.