energy producing experiments

[greendoor]

Yes, that is the best I can do.

M1 = 1000 kg
v1 = +0.313 m/s

M2 = 1001 kg
v2 = -0.313 m/s

P1 = M1 * V1 = 1000 kg * +0.313 m/s = 313 kg-m/s

P2 = M2 * V2 = 1001 kg * -0.313 m/s = -313.313 kg-m/s

System momentum = P1 + P2 = 313 kg-m/s + (-313.313 kg-m/s) = -0.313 kg-m/s

I get a system momentum of -0.313 kg-m/s (not 626 kg-m/s)


Note that a 1 kg mass travelling at -0.313 m/s has a momentum of -0.313 kg-m/s
and the mass difference between M1 and M2 is 1 kg

It seems to me that by this logic, any balanced flywheel (not matter what mass and speed) has zero Momentum ...

[Omnibus]

@Fletcher,

Out of respect for pequaide & greendoor I suggest that we leave this thread to them …

I don’t think nonsense should be respected unless you want to substitute politeness for truth or just enjoy wasting your time. I’ve yet to be convinced that the idea in question isn’t nonsense.

On the contrary indeed ! - energy can not be created out of nothing & gravity acts vertically within the context of a differential field environment i.e. there is no horizontal or sideways component to g-force allowing said "displacement" - displacement sideways can only occur by the intervention of another force - this is not at odds with the foundation of natural sciences - however, treating gravity as having some sideways displacement capability or being anything other than conservative, without experimental proof of such claims, vis-a-vis unsupported, is against the foundation of natural sciences, & is at least quite delusional or at best wishful.

Energy can be created from nothing (what is meant by ‘nothing’ is that it is from no preexisting energy reservoir). In this case this is proved by the experimental fact that the center of mass is always situated sideways with respect to the axis of rotation at every position of the wheel. Producing of energy out of nothing is further confirmed by the fact that the net torque (the sum of all eight torques) is always non-zero at all positions of the wheel. Evidence for that I presented already which you either missed to notice or are choosing to ignore for some reason.

[Omnibus]

@greendoor,

It seems to me that by this logic, any balanced flywheel (not matter what mass and speed) has zero Momentum ...

Add 'excess' to 'Momentum' and you'll be correct. You're claiming production of excess momentum and @Wubbly showed you there isn't such. You wanted numerical proof, you got it.

[Fletcher]

@Fletcher,

Out of respect for pequaide & greendoor I suggest that we leave this thread to them …

I don’t think nonsense should be respected unless you want to substitute politeness for truth or just enjoy wasting your time. I’ve yet to be convinced that the idea in question isn’t nonsense.

On the contrary indeed ! - energy can not be created out of nothing & gravity acts vertically within the context of a differential field environment i.e. there is no horizontal or sideways component to g-force allowing said "displacement" - displacement sideways can only occur by the intervention of another force - this is not at odds with the foundation of natural sciences - however, treating gravity as having some sideways displacement capability or being anything other than conservative, without experimental proof of such claims, vis-a-vis unsupported, is against the foundation of natural sciences, & is at least quite delusional or at best wishful.

Energy can be created from nothing (what is meant by ‘nothing’ is that it is from no preexisting energy reservoir). In this case this is proved by the experimental fact that the center of mass is always situated sideways with respect to the axis of rotation at every position of the wheel. Producing of energy out of nothing is further confirmed by the fact that the net torque (the sum of all eight torques) is always non-zero at all positions of the wheel. Evidence for that I presented already which you either missed to notice or are choosing to ignore for some reason.

Hey omnibus - I'll listen to anybody's opinion, no matter how pathetic it is ;7)

I could hardly miss it omnibus - its over quite a few threads over quite a few forums - it just happens to be wrong - any device that shifts masses to create torque using gravity only can be summarized as a pendulum - pendulums do what pendulums do - once moved [by a force] into a state of imbalance they move under the influence of gravity to find balance, which incidently is their position of least Pe - that's why the system CoG shifts downwards & to beneath the center of rotation [axle usually] - that means when in the balanced position there is zero torque & why systems oscillate between positive & negative torque until there is zero torque aka balanced !

Now I'm done - wubbly is asking the right questions IMO.

[greendoor]

I have heard about an physics teacher who would set up a very large pendulum, and then pull it up to a students face, and then let it swing down and back up again. The idea is for the student to be brave enough not to flinch - knowing for certain that the pendulum will never return to the exact same spot, and therefore can't hurt him/her.

I propose a similar experiment to prove/disprove this theory - which I shall call "Pequaide A".

Set up a very large balanced see-saw. Have 1000 kg masses on each end. Let's imagine the see-saw can fall, say, 2 meters. Lie with your head underneath one of the 1000 kg weights. It's as light as a feather, because it is perfectly balanced by the opposite 1000 kg weight.

OK so far?

Now - for the experiment control, let's take a box of Cornflakes. Let's drop the box of Cornflakes from 2 meters, right onto your face. I think most of us are tough enough to withstand a full frontal assault from a box of Cornflakes dropped flatside onto our face.

OK?

Now - place your face underneath the 1000 kg mass at one end of the see-saw. Raise the see-saw so this mass is now 2 meters above your face. Place the box of cornflakes on this see-saw.

You will see the 1000 kg weight slowly - very slowly - start to accelerate. It gets faster and faster - you can see it accelerating ...

Do you pull you face away, and run away like a chicken?

Or do you let the 1000 kg weight smack you in the face?

Remember - it's fully balanced, and effectively weightless ...

But it has momentum - a lot of momentum by my calculations ...

Who do you trust? Michael, Broli & Fletcher? Would you leave your face underneath it?

Personally - I wouldn't. I value my head too much.

[Omnibus]

@Fletcher,

any device that shifts masses to create torque using gravity only can be summarized as a pendulum - pendulums do what pendulums do - once moved [by a force] into a state of imbalance they move under the influence of gravity to find balance, which incidently is their position of least Pe - that's why the system CoG shifts downwards & to beneath the center of rotation [axle usually] - that means when in the balanced position there is zero torque & why systems oscillate between positive & negative torque until there is zero torque aka balanced !

That's incorrect when applied to the wheel at hand. I explained it already. You said you've read it but the above shows you haven't understood what you've read. Please go back several pages and read carefully the demonstrated difference between the behavior of a pendulum vs. the behavior of the wheel in question. I've even presented two WM2D files to illustrate that difference.

I know it's very hard to overcome indoctrination but at least give it a try.

[Omnibus]

@greendoor,

You'd better face the simple numerical example @Wubbly gave rather than bump your face into cornflake boxes and you'll find the answer.

[Omnibus]

The only interesting example so far, seemingly similar to the Atwood case, is the "cherry pit" case which explores the difference in a steady-state in absence and presence of balancing forces. Unfortunately, it isn't clear how this effect can be applied to create continuous excess energy.

[Wubbly]

In the Atwoods example I can clearly see one mass moving up in a linear fashion and one mass moving down in a linear fashion.
It looks like a linear problem to me so I used linear equations.

A flywheel is a rotating system and rotating systems have a different set of equations.
If I crack open an elementary physics books, it says

Angular Momentum = Iw

"I" = moment of inertia. (For a uniform disk, the moment of inertia I = MR^2 / 2)
(For a thin ring, the moment of inertia I = MR^2)

"w" is in radians/second.

So Angular Momentum would have units of kg*m*m/s (the radians drop since they are dimensionless)

I'm not sure why there is an extra "m" in the units for angular momentum. I'm still trying to wrap my
brain around rotating systems, but the point is, mass for linear systems is similar to "I" - moment of inertia
of rotating systems, but you can't just substitute mass for moment of inertia. You need to know the system you are rotating
and calculate "I". (e.g. For a thin uniform Rod, I = 1/12 * ML^2).

A flywheel would not have zero momentum because you would apply a different set of equations "Iw" for angular momentum.

[Wubbly]

Looking at the Greendoor Atwoods example after the 10 m drop of M1 and the 10 m rise of M2

M1 = 1001 kg
V1 = -.313 m/s

M2 = 1000 kg
V2 = .313 m/s

If we could somehow stop the system and "capture" the kinetic energy of each mass at this point, we would have

KE1 = M1 * v1 * v1 / 2 = 1001 kg * (-.313 m/s)^2 /2 = 49.03 J

KE2 = M2 * V2 * V2 / 2 = 1000 kg * (0.313 m/s)^2 /2 = 48.98 J

Total kinetic energy = KE1 + KE2 = 98.01 J


And then we ask ourselves what is the potential energy of a 1 kg mass at a height of 10 m?

PE = m * g * h = 1 kg * 9.81 m/s/s * 10 m = 98.1 kg m^2 / s^2 = 98.1 J


The kinetic energy of the atwoods machine, after the 10 meter move, is basically identical to the potential energy of a 1 kg mass at a height of 10 meters.


If we could somehow take 1 kg off of the 1001 kg mass, use the "captured" kinetic energy to push it up 10 meters,
we would have basically reset the system (assuming no friction of course).

I'm sorry, but I don't see the extra energy in this Atwoods machine.

[broli]

Wubbly I don't like to burst your bubble but anyone who understood what was happening so far acknowledged that vectors are meaningless.

In these experiments it's not the direction of the bodies that matters but their quantity of momentum/inertia/motion. In your textbook this quantity alone is meaningless not because it makes sense but because up till now it has had no use as a quantity. The direction was always needed in collision analysis. Not to mention that same textbook will tell you energy cannot be created.

What we are doing is using LOGICAL newtonian physics for practical solutions. Do you think I or the others would care less that the vector sum is 0? I already have shown how conservation of angular momentum can be debunked using conservation of linear QUANTITY momentum. Later on I will post the equation for the radius increase in time. Not because I have anything to prove for the skeptics but because I want to help pequaide out to have some mathematical model he can compare his results to.

[Fletcher]

@Fletcher,

any device that shifts masses to create torque using gravity only can be summarized as a pendulum - pendulums do what pendulums do - once moved [by a force] into a state of imbalance they move under the influence of gravity to find balance, which incidently is their position of least Pe - that's why the system CoG shifts downwards & to beneath the center of rotation [axle usually] - that means when in the balanced position there is zero torque & why systems oscillate between positive & negative torque until there is zero torque aka balanced !

That's incorrect when applied to the wheel at hand. I explained it already. You said you've read it but the above shows you haven't understood what you've read. Please go back several pages and read carefully the demonstrated difference between the behavior of a pendulum vs. the behavior of the wheel in question. I've even presented two WM2D files to illustrate that difference.

I know it's very hard to overcome indoctrination but at least give it a try.

Two things you fail to understand omnibus - first WM has a choice of different calculation methods & accuracy - depending on which one you choose will result in small cumulative errors either rounding up or down - if you have results that might appear suspicious to some then it would be logical to run the sim using all the available calc methods & average the results & see if the thing spits out the exact same results - then you would have reason to be more confident of your ability to use the program correctly - I would add that since Bessler's wheels could overcome ordinary system losses & do not unsubstantial amounts of work then anything that was dependant on an almost frictionless [loadless] environment [like your model] would give me the more than enough reason to be deeply suspicious about its validity.

Secondly - I'll say it again for your benefit - gravity does not have any lateral component, only vertical - any lateral movement of masses can only be initiated by reaction with another force e.g. friction on a ramp etc - in your rendition of the Abling wheels the ramps [whatever shape you use] cannot add energy to reset the device - your 'squeeze the cheery pip' analogy is also flawed for the same reason - wheel momentum must combine with ramp friction to accelerate & raise up the weight etc - if using gravity alone nothing physical you can do will change the most basic tenant that gravity only has a vertical component & that it does not matter what path your mass takes, either falling or rising, nothing will overcome or mitigate that most basic premiss.

P.S. I haven't gone back & read the files as I followed it thru the first time on overunity.com.

[pequaide]

Wubbly F = ma; v = at or a = v/t, substituting v/t for a in F = ma we get F = mv /t. Solving for mv we get Ft = mv. For a certain quantity of force applied for a certain quantity of time there is a certain quantity of linear Newtonian momentum produced.

The Atwood’s is a machine that was used to prove Ft = mv.

You could use a 100 kg block on dry ice and accelerate it with a one kilogram mass on a string draped over a pulley, and you would prove Ft = mv.

You could use a rim mass wheel and you will prove Ft =mv.

I know, because I have done all these experiments.

You could place all the mass of the MSU Atwood’s in the pulley and you would still prove Ft = mv. (the radius of gyration is .5)

The rule or Law (Ft = mv) will not be tricked or violated by the improper use of vectors or the improper use of angular momentum. It (Ft = mv) is a cornerstone concept in physics.

[greendoor]

Looking at the Greendoor Atwoods example after the 10 m drop of M1 and the 10 m rise of M2

M1 = 1001 kg
V1 = -.313 m/s

M2 = 1000 kg
V2 = .313 m/s

If we could somehow stop the system and "capture" the kinetic energy of each mass at this point, we would have

KE1 = M1 * v1 * v1 / 2 = 1001 kg * (-.313 m/s)^2 /2 = 49.03 J

KE2 = M2 * V2 * V2 / 2 = 1000 kg * (0.313 m/s)^2 /2 = 48.98 J

Total kinetic energy = KE1 + KE2 = 98.01 J


And then we ask ourselves what is the potential energy of a 1 kg mass at a height of 10 m?

PE = m * g * h = 1 kg * 9.81 m/s/s * 10 m = 98.1 kg m^2 / s^2 = 98.1 J


The kinetic energy of the atwoods machine, after the 10 meter move, is basically identical to the potential energy of a 1 kg mass at a height of 10 meters.


If we could somehow take 1 kg off of the 1001 kg mass, use the "captured" kinetic energy to push it up 10 meters,
we would have basically reset the system (assuming no friction of course).

I'm sorry, but I don't see the extra energy in this Atwoods machine.

Wubbly - you are failing to see that there are two essential parts to pequaide's system. Part A is the 'Atwoods' part.

You are quite correct. The Energy calculations are biased in favor of Velocity because Velocity is Squared in these calculations.

Part B of peqaide's system is that all the momentum of the heavy masses (2000 kg) is then transfered to the 1 kg mass.

From the Law of Conservation of Momentum, we know that if we achieve this transfer, the Velocity of the 1 kg will be greatly multiplied. It is at this point that the energy calculations will show the massive increase in energy.

You have to remember that these kinematic energy equations were based on falling masses that constantly accelerate under the influence of gravity. They are a 'book-keeping' or 'accounting' system for comparing quantities of motion. They are mathematical abstractions that don't really model what is actually happening in the moving masses.

But whatever. Words.

[Omnibus]

@greendoor,

It is at this point that the energy calculations will show the massive increase in energy.

No, they will not. Read carefully what @Wubbly wrote and try to understand it.