# hybrid car idea

Discussion in 'Architecture & Engineering' started by vslayer, Apr 23, 2008.

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1. hey Billy_T...here is an awesome free energy everywhere idea...capturing neutrinos and converting them to energy. lolz. Now that you know it, just invent a good capturing device. =p

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3. ### MetaKronRegistered Senior Member

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The thing is that a person might be able to take a heavy moving object and transfer its momentum in a way that increases the kinetic energy available in the system, but it might still not release more kinetic energy than it takes to accelerate the heavy object in the first place. I have an open mind, not a hole in my head.

The point is that it is really easy to inefficiently transfer momentum, and that means that inefficiencies may exist that can be cured.

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5. ### MetaKronRegistered Senior Member

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The smaller weight does need to be going ten times faster to conserve momentum. M1 * V1 = M2 * V2

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7. ...and so the Bible read: 2nd LAW OF THERMODYNAMICS

100% of the energy can not be transformed to work

&

Entropy can be produced but never destroyed

8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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There appears to be no limit to your exhibiton of stupidity!!

If M1 is the weigh dropped on the see-saw and is ten times more massive than M2 sitting on the other end then you stated (an amazingly still do!) that:

M1 = 10*M2 so (10*M2) * V1 = M2*V2

or dividing out the M2 factor from both sides:

10V1 = V2
or in words that V2, the speed of the smaller weight, is ten times greater that the speed of the more massive weight.

then you applied the KE equation to get the input KE1 as M1 *V1^2 and the output KE2 as M2*V2^2 = (M1/10) * (10*V1^2) = 10*M1*V1^2 = 10*KE1

I am sure that is why you falsely think you have a 10 fold gain in KE, (i.e. KE2= 10*KE1). Also why you are now stating KE need not be conserved (that part is true)
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PS We can continue this if you like having your "nose rubbed in it," But to encourage you to finally admit how dumb you were, I ask you:

How did the end of the see-saw with the little weight sitting on it move upwards ten times faster than the end with the big weight was going down?

By asking, I am assuming your stupidity is not infinite.

Last edited by a moderator: May 31, 2008
9. yea guys...remember that you are forgetting about friction here...and transfer of energy to heat...as well as light sometimes...UV energy...

10. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Yes, and the see-saw is sitting on a curved Earth. Hehehe.

11. ### MetaKronRegistered Senior Member

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It is safe to assume that I know how a lever works to trade force for distance and that I meant a hypothetical seesaw with one leg ten times the length of the other. It is also hypothetically rigid. The smaller weight goes ten times as fast. It weighs ten times less. Add it all up you have ten times the kinetic energy on the lighter side. For the sake of the hypothetical model I have excluded friction and the weight of the seesaw. It's easy to see that a realworld model could have a 1 kg weight on the short side and a 50 g weight on the long side, and that the seesaw lever could be a little bit longer to compensate for weight and friction. The increase in speed depends on the ratio between the lengths of the two sides of the seesaw.

You haven't said why you think it is false that there is a tenfold increase in kinetic energy, even while admitting that formula shows that there is, in the hypothetical idealized model.

The fact that a given amount of energy could be converted to light which is incapable of transferring significant momentum proves that a lot of work can be lost in the conversion process. I am talking about minimizing the amount of work that is lost when using chemical or electrical energy to create momentum that is transfered to the wheels of the car.

12. ### MetaKronRegistered Senior Member

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And the first law, the law of conservation of energy, is false. Conservation of momentum is true.

One of the smart guys here told me about this a few months ago.

13. yea well tell me his/her name so that I can label them as idiots

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14. ### MetaKronRegistered Senior Member

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You would be wrong. Look up the term "momentum" on Wikipedia and Google. Try to find three reliable sources.

15. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Yes, I have. I told you that energy is conserved so you can not increase if by a factor of 10 as you are suggesting.

I am sadistically enjoying your repeated and insistant statement that you can get more energy out of a system than the energy input into the system. It is seldom anyone is so publicly foolish. But as you often post NONSENSE, you deserves this public humiliation more than others who only occasionally post NONSENSE.

I will tell you now the nature of the two errors:

(1) You do not understand the system
(2) Even in the analysis of the poorly understood system, you make a false assumption to over simplified it.
NO, I did NOT admit "that formula shows that there is," {Energy gain} In my post 85, I told what I guessed was YOUR false reasoning that lead you to the false conclusion of a ten fold energy gain. In post 85, I did not bother to point out exactly where your mistake is. I do that in my next post, to keep this one shorter.

16. ### MetaKronRegistered Senior Member

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Not a factor of ten, a multiple of ten.

Didn't I just say that even knowing how much kinetic energy is in my hypothetical heavy object, I don't know how much energy it takes to get that much kinetic energy into it? And no, kinetic energy is not conserved so you can't use that rule. The way that kinetic energy and momentum work, you can conserve one or the other but not both.

m1*v1 = m2*v2 cannot coexist with the false equation 1/2m1v1^2 = 1/2m2v2^2

17. ### MetaKronRegistered Senior Member

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Billy, you're the one who told me that any high school physics student knows the law of conservation of momentum. It is nearly trivial to prove, as we have already in this thread, that this contradicts the law of conservation if we apply such a law to kinetic energy.

18. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I fail to see any difference between "a factor of" and "a multiple of" ten, but I will not argue symantics with you.

Instead, as you continue to stubbornly display your stupidity, I will now explicity show where the second of your two errors is. Because you have a false assumption, an "error of analysis," you computed 10, for the energy gain ("multiple") instead of 2.5, but that is still impossible and due to your lack of understanding of the system. (Which is your first, and more serious error. - I may explain that to you later.)

OK, Below I do the analysis of your FALSELY UNDERSTOOD system correctly for you. I have waited for you to correct your mistake, but doing this analysis correctly seems to be beyond your capabilities.

Like you, I will assume an idea see-saw, i.e. it has no mass and yet is completely rigid and does not absorb any energy.* My notation is easy to remember: All small letters refer to the small mass, m, even though its speed after contact, s, is ten times greater than the big mass’s speed, S, due to fact mass m is sitting initially on the see-saw's lever arm that is 10 times longer. I.e. s =10S and also by your assumption, M =10m gives the relationship between the two masses. Before contact the speed of m is zero, i.e. v = 0, and speed of the big mass is called V.

For notational convenience, I measure energy in “half Joule” units. (This avoids the usual 0.5 factor.)

Thus, T1, the Total energy input to the system is MV^2; and after M's contact with the see-saw:
The total energy, T2 = MS^2 + ms^2 = MS^2 + (M/10)(10S)^2 = 10MS^2.

You think that the energy gain, G = (T2 / T1) = 10(S / V)^2 is 10, because you falsely assume that S = V. This S = V error is but ONE of your errors. You cannot even analyze the wrong system correctly! S < V because the small mass has acquired kinetic energy. Conservation of energy implies that the velocity of the big mass must become less. But in post 89 you foolishly state:
They are both true.

I will now do your false system's analysis correctly; i. e. without your false S = V assumption, but keep in mind the results are still NONSENSE, as you do not understand the system. That is your major error, but I may (or may not) explain it to you later. For now, I only provide a hint in the footnote. I.e. I am giving you time to correct your major conceptual error by yourself.

To find the relationship between S and V, you should have applied “conservation of momentum” correctly instead of assuming S = V (but still will get the wrong result due to your other, more important, conceptual error) .

The pre-contact total momentum, M1 = MV. The after contact the total momentum, M2, is:

M2 = MS + ms = MS +10mS = MS +10(M/10)S = MS + MS so: M2 = 2MS.
{I show each step as I realize your abilities are limited.}

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Now you apply (incorrectly) “conservation of momentum” i.e. M1 = M2 or using the two bold equations just above, MV = 2MS and,
then dividing by 2M, conclude S = V/2, which is very different from your false assumption.

Then from the above, G = 10(S / V)^2 = 10(1/2)^2 = 10/4 = 2.5, not 10.

I.e. the impossible energy gain is not your 10, but a still NON-SENSICAL, 2.5.

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*This implies an infinitely large force at the moment of contact, but as F = ma was not used in the analysis, that did not cause mathematical problems. This does provide a hint for understanding the problem correctly. (I.e. is a hint for seeing your major error.)

Last edited by a moderator: Jun 1, 2008
19. ### MetaKronRegistered Senior Member

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So, Billy, you calculate a different amount of gain in kinetic energy but simply refuse to believe it. Perhaps you distracted yourself with all of the digs that you were taking at me.

Also, you're such a mathematician and you don't know the difference between a factor of ten and a multiple of ten?

20. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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My value is the correct one for your, conceptually-wrong, misunderstanding of the system. You need to think more cleary about the physics of the system to understand fully the system. That failure to understand the problem is your more important error, although your second error was bad - i.e. making a false, unjustified assumption.

Perhaps you most important error is your unfounded arrogance, which allows you to repeatedly contradict at least one of the two best founded laws of physics. Every intelligent high school student of physic accepts that BOTH energy and momentum (of any closed system) are constant or "conserved" under all possible processes.

I will give you a second hint now:

I have called the original (just before the big mass contacts the see-saw) momentum M1 and after its contact, M2 is the total momentum of the system.

As stated earlier, M1 = MV, and that is correct for both your misunderstood system and the correct system.

Howerer, only for your misunderstood model of the system is M2 = MS + ms, again with the symbols as defined in my prior post. When you correctly understand the system, you will understand that although momentum is conserved; i.e. M1 = M2, M2 < MS + ms, not equal to MS + ms.

It is this failure to understand the system and the false equating of M2 to the sum of MS + ms, which leads you to the obvious error (an energy gain of 2.5times). You should know that you NEVER can get more energy out of the system than you put into it.

Only you unfounded arrogance allows to accept this "free lunch" result and repeatedly contradict all physicists since Newton. If you were more humble, and you certainly should be, then you would not be stating that every physicist that has ever lived since Newton is wrong and only you are correct about the conservation of both momentum and energy.

I now have given you two hints about the nature of your conceptual error. If you still need more I will give you another one. I would prefer you understand what is your serious conceptual error in understanding the system, instead of me explaining it to you. Try to think about what is the system or ask me for a third hint.

21. ### MetaKronRegistered Senior Member

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I'm not interested. Too many personal attacks.

22. ### MetaKronRegistered Senior Member

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The truth is, Billy, that you are just a demented, sick, sad old man. You do not look better when you make all those putdowns, you just look sad. You couldn't make a positive contribution to the world if you tried.

23. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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OK, but for the benefit of other readers who do wish to learn, here is the third hint I would have given you - it essentially points directly at your mistake:

Imagine that the see-saw is not on earth but strapped to a tiny asteroid whose mass (for simplicity) is equal to M +m. Then consider what that "infinite force" of the first hint (the footnote in post 95) does in terms of the action / reaction force pair that slows the big mass down and flips the little mass up. Then you would understand that this is a three-body, not a two-body system.

I will call the momentum of the asteroid, after contact, Ma. Then the correct equation for M2, the after contact total momentum is:

M2 = MS + ms + Ma.

That is why I told you earlier, as a hint, that M2 < MS + ms.

The Earth is so massive that you do not accelerate it much, even with the infinite, but zero duration, force which the idealized (rigid, mass less etc.) see-saw requires exists. (More properly called, it is an “impulse,” not a “force.) But that infinite force impulse does make Ma very significant to the correct solution of the problem, even when acting on the massive Earth.

In the asteroid case I suggested, Ma = MS + ms. I.e. The momentum you neglected is half the original momentum and MS +ms is the other half. (This may be true even with the Earth, I would need to work the details to see if that is the case, but suspect it may be.)

When you do the problem correctly, you will find that the energy in the little mass plus the energy still in the big mass “is not even equal to the energy that was in the big mass before contact with the see-saw.

Certainly energy given to the little mass is not your gain of 10 nor even 2.5 times multiplied. Even non-physicists know “there is no free lunch.”

Too bad you did not want to get my last "give-it-away" hint and get the correct answer by yourself, but at least you may have learned to be a little more humble and not so arrogant as to tell all physicists since Newton that they are all wrong about conservation of both momentum and energy. It is now quite clear to all who have been reading along that you have a great deal to be humble about as displayed in your many NONSENSE posts.

I have not keep a record, but surely you are now the leader in making nonsense posts.