Kinetic multiplier

[JGarriga]

Hello everybody

I have been a "reader in the shadow" for several years.
I have even participated in some topics long time ago.
Having lost mi nickname and password I reappear today with a new one.
The reason is that I have a design I'm convinced to be a runner but I have not the necessary skills to build it.
Maybe it is impossible to build with the required accuracy.
(just like everyone I guess :) )

In any case I will share it with you.
I know that there are very qualified people in this forum and I would appreciate your remarks.
If someone dares to build it I want to point out that it is a very simple construction although it needs a great accuracy
in some aspects.

Also I will appreciate if someone can do a mechanical simulation to confirm or discard my assumptions.

The copyrights are not relevant because the goal now in our world would be to achieve a device producing clean and constant energy.

So let me to explain it with the help of the attached picture.

First of all it is not intended to be a gravity wheel but a kinetic energy multiplier.
Gravity and centrifugal forces plays just an auxiliary rule.

1.- The wheel A ( it does'nt need to be a wheel. It can be just a rod) is the mass and is turning around a ball bearing in its center.
It is the real provider of the energy.

2.- The wheel B turns also around a ball bearing in its center. It is a kind of lever.

3.- The wheel C is attached to the wheel B without turning in a way that B+C is a pendulum. It is the hammer.

4.- The wheel E is fixed. It does'nt turn. It is the anvil

5.- The pin D is fixed on the wheel A. It is just a stop for C to avoid to work in both directions

6.- The dots circle is just to show that during the swing of the pendulum B+C a collision will take place between C and E

7.- F is the angle in degrees between the line from the center E to the center of B and the line between the center of B and
the point where the collision between E and C takes place.



How it runs ?

When C collides with E there is a kinetic energy stored in the pendulum B+C but being E fixed it must appear a reaction
in the center of the wheel B wich is transmitted to A generating a torque and forcing A to turn.
This torque is enough to close the cycle as long as the angle F is no bigger than 3 degrees.
In the next collision the kinectic energy is bigger because the wheel A is already turning.
Te reaction is bigger and accordingly the torque.

Where is the limit ?
Just in the strenght of the materials.

Nevertheless in that case the limit is fixed by the centrifugal force
Because when the rotational speed of A reaches some point there is no more collisions between C and E.
Then the speed dim until the collision takes place again.

It would be perfect if the center of mass of the assembly remains all the time in the center of A

It can be done in several ways. Forexemple duplicating the assembly B+D+C simmetrically.

Now the troubles:

1.- For a contact angle no bigger than 3 degrees as required, bodies tend to stick together ( Morse effect)
and the movement will stop suddenly.
2.- The ball bearings can't have radial gap between inner and outer ring because it will decrease the angle F at the collision
and the reaction will be dissipated in that movement.
So maybe it would be more suitable with bushing instead of ball bearings.




Thank you very much for your attention.[/img]

[agor95]

Inventors generally do not destroy diagrams etc.

Can you post an image of one of your old designs.

Then we can get your ID re-instated.

Regards

[JGarriga]

Hello agor95.
I'm affraid I can't post wath you say because my old computer crashed.
Anyway my old nick was lap_lap or laplap but having forgotten my password it has been impossible to restore the account.
With this information I imagine it can be easy to find the threads.

[rlortie]

Try member "lap"

http://www.besslerwheel.com/forum/profi ... ile&u=1410

[pequaide]

Kinetic multiplication can be achieved by wrapping a weighted string around a wheel. The multiplication factor is equal to the total mass over the mass of the weight on the end of the string.

A nine kilogram spinning wheel will increase the energy by a factor of ten when it throws a one kilogram mass on the end of a string.

10 kilograms moving one meter per second is 5 joules

1 kilogram moving 10 meters per second is 50 joules.

[JGarriga]

Yes I am this lap rlortie. Thanks for your help.
Apparently I am obsessed with the stationary shaft.
I have tried to build a prototype without success because of my lack of manual skills.


pequaide:

If I have understood imagine the hanging mass of one Kg falling down 1 m
reaching the floor without rebounding, and provided that the string is tied to the wheel then the wheel can wrap again the string in the opposite direction and the mass of 1 Kg will be raised up more than 1 m ?

[agor95]

the hanging mass of one Kg falling down 1 m
reaching the floor without rebounding, and provided that the string is tied to the wheel then the wheel can wrap again the string in the opposite direction and the mass of 1 Kg will be raised up more than 1 m ?

Go on; Let it bounce and land on a platform.

Then the wheel will roll up the string without the mass to begin with and then the 1 Kg mass for the rest of the distance.

So distance down will be less than distance up.

Regards

[pequaide]

In this post: I have not dealt with the means by which the wheel started; but we can do that now.

Lets use F = ma to find out how fast the wheel is moving after the one kilogram has dropped one meter.

The (dropping) one kilogram exerts a force of 9.81 newtons on ten units of mass; for an acceleration of .981 m/sec/sec; and a final velocity of 1.4007 m/sec. This is 10 kilogram moving 1.4007 m/sec and is 14.007 units of momentum.

Use the yo-yo despin event to transfer all the 14 units of momentum to one kilogram.

The one kilogram will be moving 14 m/sec.

That one kilogram will rise 10 meters.

You must throw the one kilogram as in the de-spin devices. You can throw one of the nine kilograms in the wheel or you could arrange for the one dropped kilogram to be thrown.

You must throw the one kilogram upward: and then it is a matter of design; and not a matter of physical theory.

[agor95]

The 9.81 newton tangential force is on the wheel's rim of mass 9 Kg.

Is the wheel a disc and what is it's radius?

Naturally the 1 Kg is not in free fall so the final velocity will be less than 1.4007 m/sec.

P.S.
Interesting thread - throw https://en.wikipedia.org/wiki/Trebuchet

[agor95]

JGarriga

Thank for sharing the drawing.

Regards

[pequaide]

The nine kilograms would be a wheel with nearly all the mass in the rim; the radius does not matter.

Free fall would be 4.429 m/sec for a one meter drop; or do you mean the bearing is not perfect?

For an acceleration of .981 m/sec/sec over one meter; the final velocity is 1.4007 m/sec for the suspended mass and the rim. d = ½ v²/a or (1 m * 2 * .981m/sec/sec = v²)

[ME]

-Edited-

I stated earlier: The acceleration of the weight depends on the MoI of the wheel. Hence the radius is important.

I made a mistake :-)
Pequaide is correct - indeed the radius doesn't matter.

[agor95]

Thanks I get it.

Torque and MoI cancel out.

[pequaide]

The small wheel could be at the bottom of the large wheel: and the small wheel could be used to throw a mass to the top of the large wheel.

The thrown masses would be collected on one side of the large wheel and the large wheel would than be used to drive the small wheel.

If the total mass of the large and small wheel were 99 times greater than the thrown mass then the acceleration for the entire system would be .0981 m/sec/sec.

After the mass has dropped one meter the system will be moving .44 m/sec; this is 100 kilogram moving .44 m/sec; for 44 units of momentum. We need about a third of that to throw the mass back up ten meters.

But this is a drop of only one meter in a rise of 10 meters; so we have room for 9 more masses to be placed on the large wheel. This will get the acceleration back up to .981 m/sec/sec: now we need only a tenth of the motion to recycle the system. This is your kinetic multiplication.

[ME]

Torque and MoI cancel out.

Eeh nope, the radii of both moments cancel out.

Torque is still proportional to the radius (t = r x F = I·a).
Just as MoI is still proportional to radius-squared.

The linear acceleration is a factor of the dropping MoI (acting on the wheel) versus the total MoI (the rotation of the system): A reduction in gravitational acceleration (1/10th in the example)
For a ring-shaped flywheel this acceleration also happens to be factor of mass because the radii cancels out.
This factor is the same for the final kinetic energies when the weight hits ground and splits the initial potential energy of the weight.
As the weight moves linearly (p=m·v) and the wheel rotates (L=I·ω, note: proportional to the radius), their "units of momentum" are not compatible.