KE of system / different reference frames question

In summary, the conversation discusses the concept of using the ground as a source of energy for a moving object, such as in the case of KERS. It explains how the energy in a given frame of reference can be repartitioned and how this can be misunderstood if one is thinking too literally. The conversation also addresses the issue of where the energy for KERS comes from, with one person arguing that it comes from the car's fuel and another suggesting it comes from the kinetic energy of the car itself. Overall, the conversation highlights the importance of considering reference frames when discussing energy and its sources.
  • #71
DaleSpam said:
It is correct. Do you agree or not?

It sounds like you can't actually see anything wrong with it but you don't like the conclusion, so you illogically reject it.
You're pretty good at this; some people take quite a few more posts to see that.:smile:
 
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  • #72
Humber said:
When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.
This is known as an assumed close.:wink:
 
  • #73
mender said:
You're pretty good at this; some people take quite a few more posts to see that.:smile:

He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.
 
  • #74
mender said:
This is known as an assumed close.:wink:

And you are known as a "troll", I believe?
 
  • #75
Humber said:
vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.
How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.
 
  • #76
Humber said:
And you are known as a "troll", I believe?
The black pot speaketh!
 
  • #77
Humber said:
He has contradicted his own result, and made several other errors in the process.
He will need to correct those, not I.
Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?
 
  • #78
Humber said:
You have now made 4 unforced errors, and contradicted your own result.

When that happens, you will need to find out where your error is. I am quite willing to let you live with it, if you choose to do so.
None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?
 
  • #79
mender said:
Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.
 
  • #80
my_wan said:
How can a relative velocity not be the same in each case? If we are on separate meteors then whatever velocity I say you have is exactly the same velocity you say I have. There can be no difference. "Difference" by definition means to subtract, not add. "And" is not explicitly an addition or subtraction operator, it merely associates two terms for which an operator must be provided.

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

Case 1
[itex]v_{i,c}=p_{f,c}/m= 10 \ m/s[/itex] Initial velocity of car
[itex]v_{f,c}=p_{f,c}/m= 0 \ m/s[/itex] Final velocity of car

[itex]v_{i,e}=p_{f,e}/M= 0 \ m/s[/itex] Initial velocity of Earth
[itex]v_{f,e}=p_{f,e}/M= 10^{-21} \ m/s[/itex] Final velocity of Earth

Obviously, when the car is at rest, the final velocities of car and Earth should be the same,
and -10e-21m/s, if the Car has accelerated the Earth in the opposite direction.
Total change for the car = 10m/s + 10e-21m/sCase 2;
[itex]v_{i,c}=p_{f,c}/m= 0\ m/s[/itex] Initial velocity of car
[itex]v_{f,c}=p_{f,c}/m= -10 \ m/s[/itex] Final velocity of car

[itex]v_{i,e}=p_{f,e}/M= -10m/s \ m/s[/itex] Initial velocity of Earth
[itex]v_{f,e}=p_{f,e}/M= -9.999999999999999999999 \ m/s[/itex] Final velocity of Earth

This makes no sense at all.
 
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  • #81
my_wan said:
Dale is correct that you either must provide the basis of your disagreement or chalk up such statements above to diversionary tactics. Dale attempted to establish a baseline with his post. You can either work with it or choose another baseline that helps to actually make your point clearer. Diversions, such as above, which the only apparent purpose is to obscure the baseline of the discussion do not fit anywhere.

I don't need to do that. The error is quite clear.
The change is 1e-18J in the first case and agrees with KE =p^2/2m, but not the revised calculation. The error there, is rather obvious.
 
  • #82
mender said:
Sounds like it may be a good time to establish a baseline.

Humber, what's 2 + 2?

You are trolling, and have nothing to contribute, but simply want to avoid the consequences for your ddw cart, when it is proven that there is no ground energy.
I have done that by two different means, and now you want to spread your bluff denial, and emulation of Spork, to here.
 
  • #83
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.
 
  • #84
Humber said:
there is no ground energy.
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.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.
 
  • #85
mender said:
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.

Is that the topic of the OP? Again, you are trolling, and exposing yourself as a zealot.
 
  • #86
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.
 
  • #87
rcgldr said:
DDWFTTW like devices rely on the diffrence in speed between two media. Which media that KE is extracted from depends on the frame of reference.

In the case of a Brennan_torpedo, from a water based frame of reference, the retracted wires are the source of KE. Change the frame of reference so that the wires are attached to "non-moving" posts, with water flowing downstream. Using the posts as a frame of reference, then the water is the source of the KE.

Getting back to a DDWFTTW cart, from an air (true wind) based frame of reference, the ground (earth) is the source of KE.

Is that the topic of this thread? And the torpedo, as the reference states, was directly powered by on-shore steam engines. That has nothing at all to do with relative motion.
 
  • #88
mender said:
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.

mender;3692808
 
  • #89
Back to the topic then: answer Dalespam's question.
DaleSpam said:
None of this answers the post you quoted; you are merely attempting to dodge the question.

Do you agree or disagree with the proof of post 59?
 
  • #90
Humber said:
You can continue to show that you have no control at all of the zeal that forces you to make a fool of yourself.
Come on now, stay on topic and answer the question!
 
  • #91
mender said:
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.
 
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  • #92
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.
 
  • #93
Subductionzon said:
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.
 
  • #94
Humber said:
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?
 
  • #95
Humber said:
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 ke, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a ke of the car and the kp 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 ke in your debate with DaleSpam without accounting for where the kp that went into the battery came from.
 
  • #96
my_wan said:
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 ke, momentum, etc. The other one of us is the energy that went into the battery. So the total energy is accounted for with a ke of the car and the kp 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 ke in your debate with DaleSpam without accounting for where the kp 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.
 
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  • #97
rcgldr said:
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.

Humber said:
Is that the topic of this thread?
Yes, from the first post:

Humber said:
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.
 
  • #98
DaleSpam said:
This is true, due to Newton's 3rd law.

This is false, in general.

If some object gains momentum Δp then [itex]p_f=p_i+\Delta p[/itex]

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

Substituting that into the expression for the object's final KE we obtain
[itex]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 [/itex]
[itex]\Delta KE = KE_f - KE_i = v_i \Delta p + \Delta p^2/2m [/itex]

This is different from what you wrote because of the [itex]v_i \Delta p[/itex] 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 [itex] \Delta p^2/2m [/itex] term drops out, but the [itex]v_i \Delta p[/itex] 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 pi, 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 = p2/2mearth.

Momentum is always conserved, and is frame independent, so that is a unique and frame independent result.
 
  • #99
rcgldr said:
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.
 
  • #100
Humber said:
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.

Humber said:
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 [itex]m50^2 + m50^2 \neq m100^2[/itex]. 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.
 
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  • #101
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.
 
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  • #102
my_wan said:
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.

my_wan said:
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.

my_wan said:
Yet [itex]m50^2 + m50^2 /neq m100^2[/itex]. 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.
 
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  • #103
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. :cool:

@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.
mys_solved2.gif
 
  • #104
my_wan said:
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.

my_wan said:
In the car/earth analogy the one of us is the 5 m/s velocity, and associated ke, 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.

my_wan said:
You are attempting to conserve ke in your debate with DaleSpam without accounting for where the kp 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.
 
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  • #105


@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.
 
<h2>1. What is kinetic energy of a system?</h2><p>Kinetic energy is the energy an object possesses due to its motion. It is a scalar quantity that depends on the mass and velocity of the object.</p><h2>2. How is kinetic energy of a system calculated?</h2><p>The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.</p><h2>3. Does the kinetic energy of a system change in different reference frames?</h2><p>Yes, the kinetic energy of a system can change in different reference frames. This is because the velocity of the object may be different in different frames of reference, resulting in a different value for kinetic energy.</p><h2>4. Can kinetic energy be negative?</h2><p>No, kinetic energy cannot be negative. It is always a positive quantity, as it represents the energy of an object in motion.</p><h2>5. How does the kinetic energy of a system affect its motion?</h2><p>The kinetic energy of a system is directly proportional to its velocity. As the kinetic energy increases, so does the velocity of the system. This means that a system with a higher kinetic energy will have a greater tendency to continue moving and will require more force to stop it.</p>

1. What is kinetic energy of a system?

Kinetic energy is the energy an object possesses due to its motion. It is a scalar quantity that depends on the mass and velocity of the object.

2. How is kinetic energy of a system calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. Does the kinetic energy of a system change in different reference frames?

Yes, the kinetic energy of a system can change in different reference frames. This is because the velocity of the object may be different in different frames of reference, resulting in a different value for kinetic energy.

4. Can kinetic energy be negative?

No, kinetic energy cannot be negative. It is always a positive quantity, as it represents the energy of an object in motion.

5. How does the kinetic energy of a system affect its motion?

The kinetic energy of a system is directly proportional to its velocity. As the kinetic energy increases, so does the velocity of the system. This means that a system with a higher kinetic energy will have a greater tendency to continue moving and will require more force to stop it.

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