Why don't Perpetual Gravity Wheels Work ?

[primemignonite]

Yes Bill, but as of now matters have not yet advanced
past trying.

[pequaide]

I have transferred the motion of five units of mass to one unit of mass, and I have carefully measured the original and final velocities. Both of these conservation laws (mv and 1/2mv²) can not be true; one must be false. Newtonian momentum wins; it is as simple as that.

You can bring in imaginary friends (other forms of energy) to pretend that energy is still conserved; but it simply is not. Energy increases dramatically. If you double the momentum of an object you quadruple the energy, you just can’t get around it.

So don’t try to get around this fact; use it.

[greendoor]

Yes, momentum is observed to be conserved, be it linear or rotational - I take that as a given or absolute.

It would appear that energy is also conserved but it seems many literally take that to mean kinetic energy - Ke is only one type of energy - it is one side of a many sided polygon - Ke just happens to be the only form we can use for mechanical arrangements where a force is needed to be applied to move something else - sum ALL the energy's & I think they energy equation balances ?!

So in some examples you can change the quotient of Ke [the usable kind] but always at the expense of the others that just haven't had time to play yet.

There I go, over simplifying things again.

Sure - there are losses. In the Newtons cradle there would be some heat & impact noise to account for a small amount of energy loss. This is why the balls don't fly upward quite as high as the original dropped ball. Even a simple pendulum or flywheel has losses.

I'm not afraid of losses. Assuming a free energy input source is found, we can afford some losses. For the sake of calculation, I am happy to write of 50% of the energy input to losses.

Let's say we have 1000 kg moving at 1 m/s.
Momentum = 1000 kg*m/s
Kinetic Energy = 1/2*1000*1^2 = 500

Assuming we transfer this momentum to 1 kg moving at 1000 m/s
Momentum = 1000 kg*m/s
Kinetic Energy = 1/2*1*1000^2 = 500,000 (1000x gain in energy???)

This is an extreme example - so I will allow that the transformation was a bit lossy, and let's say that the velocity ended up at 500 m/s instead of 1000 m/s. (Generous enough?)

In that case - allowing for 50% loss of momentum (even though we can be pretty sure momentum is mainly conserved ...)

1/2*1*500^2 = 125,000 (that's still 250 x starting energy ...)

Or let's say that 75% loss of momentum occurred ...

1/2*1*250^2 = 31,250 (that's still 62.5 x starting energy ...)

OK - let's say 90% of momentum was lost ... leaving 10%

1/2*1*100^2 = 5,000 (that's still 10 x starting energy ...)

It just gets crazy, doesn't it? Are you seriously trying to tell me that BOTH momentum & energy are always conserved???? I don't think so.

This is why I am seriously doubting the theory of the conservation of energy. To the point where it would appear to be a fraud.

What does it mean? To me, this means that calculations involving energy can be misleading when applied incorrectly. I trust the basic Newtonian calculations involving momentum more than I trust the Leibniz theory of Energy.

AFAIK - these strange numbers don't mean that energy is created. To my mind, in means the theory of energy is a bit screwed up, and shouldn't be relied upon. But momentum/motion - that is undeniable.

FWIW - the theory I am working on does not depend on this kind of maths for energy gain. And when I use the word 'energy' - I mean this in the very loose sense of motive power moving a load, because I don't really know what words to choose now.

[Fletcher]

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Look to the index on the right side & scroll down to "ballistic pendulums" - read what they say about momentum & energy conservation, especially kinetic energy.

EDIT 1 : actually I steered you there to look at the experimental setup more than anything - this type of testing is now pretty much redundant but IMO its an excellent way to grasp energy changes because what available force/energy there is from momentum etc is immediately translated into lifted height of the pendulum & so Pe = mgh applies [easy to see just how much energy to do work is actually available & what happens to the rest].

EDIT 2 : put your values into the calculator greendoor, except the large mass [1000 kgs] isn't moving relative to the 1 kg at 1000m/s to gist your scenario.

[broli]

Fletcher I don't think you see what Greendoor has done. This is also why pequide has been promoting on this forum for a very long time but his presentation skills suck. I could/can help with the illustration part if he just talked like a normal human :p.

BUUUUT to get to the point. What they are really saying is. WHAT IF momentum could be fully transferred from one object to another. Then indeed conservation of energy wouldn't hold while conservation of momentum does.

The only way we find the velocities of bodies after collision is using exactly Co energy equation and Co momentum equation to find a solution for the speed. So obviously you can not trick it like this. You have to use another momentum transform system that does not use the COE equation but only the COM equation.

And I believe this is what pequide's toy is all about. Only am I having a hard time understanding any of it.

[pequaide]

In the ballistic pendulum diagram they first tell you that energy is not conserved; that energy is lost to internal energy (imaginary friends). And then they try to tell you energy (PE) is conserved. They are misleading you: PE is conserved after 95% of the PE is lost. The energy of the bullet is not conserved and that is what we were interested in. The gravitational potential energy of the bullet is not conserved. 95% of the gravitational potential energy of the bullet is lost.

Ballistic pendulums with bullets typically lose 95% or more of their energy. That energy is never found or measured. It is assumed to have been conserved, by a mysterious friend.

Newtonian momentum however just keep plodding away cranking out win after win.

For example: A 10 g bullet moving 896 m/sec will rise (or could rise under ideal conditions) d =1/2v²/a, 40,918 meters, a joule is a netwon meter so 9.81 N/kg * .010 kg * 40,918.2 m = 4,014 joules. The bullets gravitational potential energy is 4,014 joules. The bullets momentum is 8.96. You can check the quantity of energy with 1/2mv².

After a collision with a 2.5 kilogram block the Newtonian momentum will remain the same with the combination moving 3.57 m/sec, 2.510 kg * 3.57m/sec = 8.96. The gravitation potential energy of the combined bullet and block is d = 1/2v²/a =.64958 meters * 9.81 newtons / kg * 2.510 kg = 16 joules.

So why did they tell you gravitation potential energy is conserved?

The gravitational potential energy of the 16 joule block with bullet is conserved, or they are telling you that an object moving 3.57 m/sec will rise .64958 meters. So?

16 joules / 4,014 joules; over 99% of the original gravitational potential energy is lost.

Fletcher you say “[easy to see just how much energy to do work is actually available & what happens to the rest].

I don’t think it is easy to believe. I think it is ludicrous to believe that we can’t control friction any better than this, 99.6% of the energy is lost, when energy is supposedly something that has to be conserved?

Newtonian momentum however just keep plodding away cranking out win after win

[broli]

pequaide, do you use live messenger or yahoo messenger? If so pm me with your address I would like to have live chats with you something very lacking on this forum. I can help you spread your idea.

[Fletcher]

Fletcher I don't think you see what Greendoor has done.

BUUUUT to get to the point. What they are really saying is. WHAT IF momentum could be fully transferred from one object to another. Then indeed conservation of energy wouldn't hold while conservation of momentum does.

The only way we find the velocities of bodies after collision is using exactly Co energy equation and Co momentum equation to find a solution for the speed. So obviously you can not trick it like this. You have to use another momentum transform system that does not use the COE equation but only the COM equation.

And I believe this is what pequide's toy is all about. Only am I having a hard time understanding any of it.

Thanks for the assist broli - you ask what if momentum could be fully transferred ? - the ratio of transference depends on the elastic properties of one body compared to the other & nothing is 100% inelastic [read 100% efficient] - I believe the ballistic bullet example is a trusty way to show this because one body 'catches' the other & their combined momentum & mass swings upwards to a new gravitational potential energy.

Calculating using CoE & CoM isn't the ONLY WAY as I've just pointed out - experiment & experimental results have more validity than equations - they come after the experiments so if you have something where you question the CoE then you can at least observe the momentum & gravitational potential energy & deduce that a large portion of the kinetic energy before collision has been transformed into some other type, not of immediate use to us in a mechanical sense.

[Fletcher]

The onus is on each & every one of us to experimentally prove their theories - if that can't be done with hard physical evidence then a lower standard of proof can be attempted [I'd place sims in there] but a purely mathematical argument on its own rings bells for me.

Now I know you are attempting to experimentally prove your position pequade & if broli or the likes can assist that standard of proof, then you ought to consider it, IMO ?

[pequaide]

A 2 or 3% lose of momentum can still yield 400, 500, or 600% increases in energy.

[Fletcher]

A 2 or 3% lose of momentum can still yield 400, 500, or 600% increases in energy.

From your equations that would appear the case but the real question is how much of that increase is usable energy & can that be demonstrated without inputting additional energy into the system i.e. a closed system ? - Hell, I'll settle for a self sustaining wheel no matter how it happens, providing no fuel is added ;)

P.S. full credit to anybody trying to tip over Newton's Laws who succeeds & I may seem like a stick in the mud but my intention is not to be deliberately obstructive or antagonistic or parrot what I think I learnt - more power to your elbow but the burden of proof is high.

[broli]

pequaide thanks for that. That's something to think about deeply. Your bullet starts with immense amounts of energy. And suddenly gives a laughable amount back. Where did all the energy go to? I'm sure my professor would yell "heat, sound and bullshit" after which he buries it and moves on with his brainwashing.

[broli]

A 2 or 3% lose of momentum can still yield 400, 500, or 600% increases in energy.

From your equations that would appear the case but the real question is how much of that increase is usable energy & can that be demonstrated without inputting additional energy into the system i.e. a closed system ? - Hell, I'll settle for a self sustaining wheel no matter how it happens, providing no fuel is added ;)

P.S. full credit to anybody trying to tip over Newton's Laws who succeeds & I may seem like a stick in the mud but my intention is not to be deliberately obstructive or antagonistic or parrot what I think I learnt - more power to your elbow but the burden of proof is high.

Actually newton's law are not being tipped here. On the contrary they are made use of to prove the point. The CoE one is the one being tipped over.

[Fletcher]

Ah - I stand corrected broli, CoE, of course - I guess bernoulli & others who based their gas & liquid laws around CoE theorem would take more than a passing interest in these discussions.

[Michael]

You are so very wrong pequaide. In fact I am willing to negotiate a ten thousand dollar bet on the matter. Energy cannot be destroyed.