I have a proven model I built. I am looking for like minded people who know more

[CANGAS]

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

[sm0ky2]

"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

[sm0ky2]

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.

[verpies]

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

[MeGaFaRR]

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?