re: energy producing experiments

[Kator01]

oh, I have to apologize .
"earn" is the wrong word.. i fell back into my german thinkig. Of course I mean gaining energy.

As I said the topic was fully discussed in the above mentioned thread, you can go there and reread all post there.

My, telecom and smOky2 questions are related to the practical technical means of energy extraction that is what concept of a machine could be developed for this principle.
More than 8 years have passed and still no solution found

Are you serious in telling the community here that throwing this set-up in the air is a practical way to extract energy ?

I even posted a solution in reply #109 here:

http://overunity.com/1995/free-energy-from-gravitation-using-newtonian-physic/105/#.WKEGX1IeCuI

The problem of energy-extraction was recognized back then and obviously could not be solved ....after 8 years



Mike

[sm0ky2]

For me there is no "question", more so I was attempting to
describe the situation from a physics standpoint.
But apparently physics and mathematics fall to the wayside
When it comes to digitized frame rate of an unscaled distance
Judged by a visual interpretation of a recording.


The cylinders and spheres is an elementary physics lab
Generally covered in the first two years of engineering
Bachelors degrees. This is a simple experiment, and all
Energies and momentii can be accounted for, both physically
And mathematically.


But hey who am I?
This guy says he "created energy"
And his camera frame rate proves it!!


Maybe he can figure out how to get the energy out.....


All I saw from the video was Mr hand putting way more momentum into
The cylinder that the balls were able to overcome in the time it was airborn
You can clearly see the momentum transfer to the cylinder when the balls
Impact the side at 90-degrees.
If you don't think this occurs......
Hold the strings in your hands and spin
Then stop spinnning


Make a fist so it's 'round' like the cylinder
Or hold your hands down at your side so your
Hips act like the cylinder


Tell me what happens when the balls run out of string
But are still moving.


Remember TeatherBall?
When the balls gets wrapped it smacks the pole!
Make a big noise and the pole vibrates.
The same thing happens to the pvc, except it is free
to move from the collision.
A steel cylinder will behave slightly different than the pvc


The inertia from the collision is what determines how much
energy is transferred to the cylinder and the how much reflects
With the ball.
Steel vs pvc is not fully elastic, nor is it a fully inelastic collision
But contains properties of both in different degrees, based on the
Materials properties.
Pvc is fairly rigid, but it flexes, bends and heats up more so than
Two pieces of stainless colliding.
Steel pretty much just bounces off like the balls on a pool table


Two masses of same weight steel ball and pvc ball
You can compare to steel vs steel and visibly see this


The two steel balls will behave just as the pool que and a ball.
The pvc collision will absorb some of the energy and both balls
will roll differently


But changing your materials will not change your experiment
The balls and cylinder have two momentums
First being thrown in the direction of travel, this is a linear momentum
Second is the rotational momentum from the twisting action
These two fight each other through the whole experiment
Because their vectors are different.
The linear momentum, being of greater magnitude, is dominant.

[sm0ky2]

The momentum transferred to the cylinder, mostly goes back into the balls
Some is transferred. But each ball cancels out the other ball in this regards
In fact, if you fix the lengths of your strings so they are equal-----
Such that both balls hit at the same time.
None of the momentum will be transferred to the cylinder
(Except the flexing factor of the materials)
And both balls will repel 180-degrees from the collision
With almost all of the momentum they had when they ran
Out of string.
But again this is in a vector opposing the forward momentum
Of the entire device.

[sm0ky2]

If you want to really perform this experiment in the way
You propose:
Perhaps a turntable or stationary rotatable accommodation
Could get rid of the forward toss that prohibits a full examination
Of the forces the balls impose to the cylinder
Of course the cylinder should not be attached to the turntable
So that it may freely respond to changes in velocity in almost every
Available vector.

[Delburt Phend]

You seem to be focusing on the last few frames of the experiment. Here the cylinder is moving 20 mm in four frames and we can leave it at that. I record maximum speed for the cylinder of 4 frames per 20 mm right at what you think is a glance. If the sphere and cylinder touch they are moving at the same speed. But the experiment is over at this point.

Maximum energy was achieved about .066 sec before the glance you see. And this is the second time maximum energy is achieved.

The cylinder is stopped 16/240 seconds after release: here the spheres have all the motion. This is where the energy of the system is highest. Most cylinder and spheres experiments do not proceed past this point.

The end frames confirm that all the motion is maintained throughout the experiment. And only linear momentum can be maintained throughout. Because linear momentum is the only thing a ballistic pendulum conserves.

The fist and tether ball are not similar experiments. The fist and steel post are not at liberty to rotate. When the sphere comes close to the cylinder at the end of the experiment both the cylinder and the spheres are moving at the same speed; just like they were at the beginning of the experiment.

All you need to do to extract the energy is point the motion up and cut the string.

Lets make a small machine to see what it looks like.

A one kilogram mass will proceed up 25 meters if it has a velocity of 22.15 m/sec. D = ½ v²/a

This is 22.15 units of momentum.

Make a 25 meter string with 1 kilogram at each meter of length: except there is no mass on one end. Place the string vertically with a one kilogram mass at the top. That would leave you with no mass on the bottom of the string. When the string is lowered one meter that would leave you with one kilogram on the bottom; and a kilogram mass at each meter length in between and no mass at the top.

To restore the one kilogram to the top would require that you accelerate the bottom kilogram to 22.15 m/sec and point the motion up and cut the string.

Place this 25 kilogram string on an Atwood with a balanced mass of 97.67 kilograms; That would be 25 kilograms accelerating 122.669 kilograms for an acceleration rate of 25/122.669 * 9.81 m/sec/sec = 2 m/sec/sec.

After a drop of one meter the entire mass of 122.669 kg would be moving 2 m/sec. From d = ½ v²/a

That is 122.669 kg * 2 m/sec = 245.338 units of momentum.

Place all this momentum in the bottom one kilogram using the cylinder and spheres and it will rise 3067.8 meters, actually air resistance would prevent that much rise but you have vastly more than enough motion to restore the one kilogram at the bottom of the string back up to the top of the string.