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Author Topic: I have a proven model I built. I am looking for like minded people who know more  (Read 32021 times)

Offline conradelektro

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@MeGaFarr and all flywheel lovers:

Look at these flywheel energy storage https://en.wikipedia.org/wiki/Flywheel_energy_storage and you will understand better, what can be done with a fly wheel.

Quote
Energy storage efficiency: Flywheel energy storage systems using mechanical bearings can lose 20% to 50% of their energy in two hours. Much of the friction responsible for this energy loss results from the flywheel changing orientation due to the rotation of the earth (an effect similar to that shown by a Foucault pendulum). This change in orientation is resisted by the gyroscopic forces exerted by the flywheel's angular momentum, thus exerting a force against the mechanical bearings. This force increases friction. This can be avoided by aligning the flywheel's axis of rotation parallel to that of the earth's axis of rotation.
Conversely, flywheels with magnetic bearings and high vacuum can maintain 97% mechanical efficiency, and 85% round trip efficiency.

Remark: energy storage efficiency does not take into account the losses in the electric motor (about 60% to 80% efficiency) and the losses in the electric generator (also 60% to 80% efficiency).

One can put energy in a fly wheel by spinning it up e.g. with an electric motor. This takes time T1 and energy E1.

Then one can take out energy from a fly wheel by driving e.g. an electric generator, lets say we take out energy E2 for time T2.

Without losses we get T1 x E1 = T2 X E2.

But because we have losses (friction, losses in the electric motor, losses in the electric generator) we get T1 x E1 > T2 X E2 (less can come out as one has put in).

It is true that it takes little energy to keep the fly wheel running at a constant speed once it has reached that speed. But this is also a loss, because this "maintaining energy" can not be recovered.

In simple words: in the ideal case as much energy as is stored in the fly wheel can be recuperated. In the real case much less can be recuperated (because of losses).

I think that the mental error the fly wheel lovers are doing is the following:

Because it takes little energy to maintain the speed of a fly wheel (with no load on the fly wheel) they think, that this remains the case once a load is put on the fly wheel.

Greetings, Conrad

Free Energy | searching for free energy and discussing free energy


Offline Dog-One

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I think that the mental error the fly wheel lovers are doing is the following:

So far, I don't see that Frank has any intention of taking energy back out.  He just likes spinning wheels.  ah hum.   :D

Offline memoryman

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MeGaFaRR, your flywheels (number and size/mass don't matter) start with zero energy (stopped).
They get energy by being rotated. They don't GAIN energy anywhere, they don't create energy, they don't multiply energy. Why do you think that they have more energy coming out than was input? Where would that happen?

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Offline CANGAS

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kinetic energy is calculated by k.e.=m x V^2. Where did that energy come from?

Quote from: Wicki
Kinetic energy
From Wikipedia, the free encyclopedia
Kinetic energy

Ek = ½mv2

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest.

In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is \begin{smallmatrix} \frac{1}{2}mv^2 \end{smallmatrix} .


You are wery velcome
CANGAS 207

Offline memoryman

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CANGAS, the question was addressed to MeGaFaRR. I wanted HIM to answer.

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Offline CANGAS

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memoryman, MY question was for YOU. You left out the 1/2 that is supposed to be in front of the m. I wanted YOU to tell ME and HIM why it is a crucial component of the kinetic energy formula.


many tanks
CANGAS 208

Offline sm0ky2

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"My machine only proves my concept that "I CAN REDUCE THE ENERGY REQUIRED TO ROTATE A FLYWHEEL TO FULL SPEED OR BEYOUND OF ANY POWER UNIT AND/OR INCREASE THE OUTPUT OF STORED ENERGY IN A SET OF FLYWHEELS" done."
Without measuring, how does it prove anything?

without measuring, it proves nothing at all.
But when we DO measure, what do we find?
I will not charge you to being honest, or accuse you of falsehood.
For these things are irrelevant, and will present themselves soon, irrespective of my personal opinion of you.

"done" ? what exactly did you "do"?
Understand there is a difference in "energy" and torque.
a difference between torque and RPM
a difference between RPM and radius.
what is "energy" when compared to radius, RPM and torque?

here is your "done" with 4 times the efficiency if the machine YOU describe.
STILL not "overunity". But considerably more torque, per radii.

https://www.youtube.com/watch?v=hvz9A0c_HeQ





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Offline sm0ky2

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memoryman, MY question was for YOU. You left out the 1/2 that is supposed to be in front of the m. I wanted YOU to tell ME and HIM why it is a crucial component of the kinetic energy formula.


many tanks
CANGAS 208

reading his post, it may have been an unintentional disclusion.
While this does make a mess of the mathematics, I believe your question was "why" it is a crucial component.

If I may, attempt to answer that by stating that is equally as crucial as the ^2 component it refers to.
You see, there is a function here not fully described  by theory.
Theory says it is the square root of the square, which does not work out in the math.
therefore, to more closely estimate the answer to the actual measurements, one half of the square is taken to be the "value".

What does that do, mathematically speaking ?

2^2 = 4, 1/2 of 4 = 2
2^2 = 4;  (sqrt)4 = 2
ok....

3^2 = 9, 1/2 of 9 = 4.5
3^2 = 9; (sqrt)9 = 3
see the difference?

the real question is not the importance of the 1/2.
but "why" it differs from the square root of the square.
[1/2(x^2)] vs [(X^2)^(1/2)]

I do not like to make assumptions when it comes to other peoples perspectives.
However, I envision, that by the way your question was asked,
that you do not see this from the perspective which is required to receive your answer.

The answer lies in a fudge factor, that pertains to gravity.
not "every" gravity, but that which applies specifically to an average value encountered here on earth.

It is, simply put, the difference between E = MC^2
                                                    and E = 1/2(MC^2)
and pertains to a fudge factor, of unknown derrivitave, that differentiates a mass from its' momentum.
This is a quantity of earth's gravity, and varies slightly from the constant derived in gravitational fields of lesser or greater value.
such as that of our moon, or the planet Jupiter. [moon; Ke= 1/3(Mv^2) and Jupiter Ke=159(Mv^2)]

mathematically, it is equivalent, per unit mass, to the 'missing' mass of a photon of known energy, as derived from its' momentum.
or that of the difference between the mass of a gluon, and the "expected value" of the mass of its' associated hadron.

This is an incompleteness in our physics, which should someday be described as E=(Mc^2 - some yet unexplained value) instead of the "1/2" currently assumed.

And although your question was probably intended to point out someones mistake, or mistyping,
It is more relevant to our field of research than you probably understand.





Offline verpies

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A frictional clutch is only 50% energy efficient while engaging and transferring energy from one flywheel to another.

To transfer the angular momentum from one flywheel to another, with minimal losses, a torsional spring must be used

Free Energy | searching for free energy and discussing free energy


Offline MeGaFaRR

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without measuring, it proves nothing at all.
But when we DO measure, what do we find?
I will not charge you to being honest, or accuse you of falsehood.
For these things are irrelevant, and will present themselves soon, irrespective of my personal opinion of you.

"done" ? what exactly did you "do"?
Understand there is a difference in "energy" and torque.
a difference between torque and RPM
a difference between RPM and radius.
what is "energy" when compared to radius, RPM and torque?

here is your "done" with 4 times the efficiency if the machine YOU describe.
STILL not "overunity". But considerably more torque, per radii.

https://www.youtube.com/watch?v=hvz9A0c_HeQ

Hp=torque X rpm's
Does a bigger flywheel at same rpm's have more torque?

Offline MeGaFaRR

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Hp=torque X rpm's
Does a bigger flywheel at same rpm's have more torque?

Thanks verpies
If I could afford it I would have used a lot more efficient and useful equipment to prove my concept but I couldn't. I understand the helpful hint and appreciate it but I'm not in the position to do that at the moment. Now I have an associate in China that has read my paper and is not wearing any kind of physics blinders, that is in the process of using my theory to build a machine like I have. But I do believe he has way more resources than I do and he is following this thread so any useful hints he can use will be greatly appreciated.
Thanks
Frank

Free Energy | searching for free energy and discussing free energy


Offline AlienGrey

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Well didn't they invent the rocket, perhaps they will make it into some thing called a cuckoo clock mechanism rocket powered type device and include gravity like the Ausy guy did ? who knows how far it will go ;) however joking apart their does appear to be some sort of magic attached to the spinning fly wheel.
« Last Edit: February 24, 2016, 12:59:07 AM by AlienGrey »

Offline ARMCORTEX

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A frictional clutch is only 50% energy efficient while engaging and transferring energy from one flywheel to another.

To transfer the angular momentum from one flywheel to another, with minimal losses, a torsional spring must be used

Well said verpies, I believe Pejic is using one of those "torsional springs", a new word to learn and a path to study.

https://www.youtube.com/watch?v=44O3crS-B9k

Finally a bit of progress, maybe someday we will unravel the secrets of these flywheel devices.

Very enlightening thread, thanks to MegaFarr

Offline MeGaFaRR

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Well said verpies, I believe Pejic is using one of those "torsional springs", a new word to learn and a path to study.

https://www.youtube.com/watch?v=44O3crS-B9k

Finally a bit of progress, maybe someday we will unravel the secrets of these flywheel devices.

Very enlightening thread, thanks to MegaFarr


I wish I could understand what they are saying

Offline MeGaFaRR

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 Sent: Thursday, February 11, 2016 7:44 PM
To: the_physicist@askthephysicist.com
Subject: New Ask the Physicist question
Email: Megafarr@hotmail.com question-

If a 5 ton flywheel was rotating at 5000 rpm,
in a vacuum with magnetic bearings, would a
1 horsepower motor be adequate to keep the
momentum going?


 On Feb 12, 2016, at 11:22 AM, physicist <the_physicist@askthephysicist.com> wrote:

There is always some friction, no matter how hard you try to eliminate it.
It would seem that for vacuum and magnetic suspension that the friction would be small and 1 hp motor might very well be fine. To do better than guess, I would need a number for frictional torque.

 Does everyone agree with this or is this physicist wrong?
Please do comment as to your thoughts
Frank

 

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