Free energy from gravitation using Newtonian Physic

[DrWhat]

I know it's asking a lot and you have written a lot of detail already, but would you be able to provide even the shortest video clip to show us what is happening.

Thanks

DrWhat

[pequaide]

Lab. notes and comments

I was trying to build what I thought was a duplication of my last 4 inch O.D. coupler, which had a 90? stop with a 364.9g 3 in. I.D. lower pipe, added. To my amazement it took a 390g pipe to make the new cylinder stop at 90?. This is about a 4% mass increase (233.6g cylinder + 364.9g added mass; and 234.2g + 390g). Then I checked the tether length and I had inadvertently increased it by about 6%. So: the longer the tether the more mass the spheres are capable of stopping, even when the spheres mass remains constant. I had known his but had not put numbers on it before. The longer tether length apparently allows a longer period of time for the force to act. 

The top coupler (with seats for the embedded spheres at 180?) has a mass of about 234g; when it has no pipe added to its bottom opening, and the spheres masses are not added. The spheres have a mass of 132.8g. With this mass ratio of 132.8g to 234g the spheres stop the cylinder within about 4 cm from the seat. This means that the tether tightens immediately; as soon as the spheres leave the seat, and all of the force before the stop is tangent to the cylinder, because I don?t think the string (tether) has come away from the cylinder this early in the swing. The cylinder is moving strongly backwards while the string is in the slit (behind the point where the string enters the cylinder).

A 3 kilogram cylinder with two 1/2 kilogram spheres moving one meter per second has 4 units of momentum; before the spheres are released.

After the cylinder is stopped the spheres have 4 units on momentum and a velocity of 4 m/sec. The spheres can rise .8155 meters.  d = ? v?/a

If one kg (of spheres) is placed in a ten kg Attwood?s machine (with 9 kilograms balanced and the one kg of imbalance) and the spheres? mass is allowed to fall .8155 m, it will give you 12.65 units of momentum.  The spheres are now back to their original level and ready to be reloaded into the cylinder and spheres machine, but now the system has 316% the original momentum.

So you see; it doesn?t matter if the kinetic energy formula is defective or not, this system can make energy with or without the kinetic energy formula.

By using the distance formula: we know that the potential rise of 4 kg moving 1 m/sec is only .051, and the rise of 1 kg moving 4 m/sec is .8155, this is 4 times the rise as is indicated by the kinetic energy formula which predicts 4 times the energy.

Whatever the usefulness of the kinetic energy formula; the energy is still there, it is only a matter of utilizing it.

Videos: people can't get them off the DVDs I make. Can anyone help?

[hartiberlin]

Hi,
for  converting your DVD files to MPEGs,
use VOB2MPEG, see:

http://www.videohelp.com/tools/VOB2MPG

After this you can use

virtualdubmod

http://sourceforge.net/project/showfiles.php?group_id=65889

to convert MPEGs to AVIs, e.g.with DIVX.com codec.


Or you can use also DVD2AVI for all in one in ONE step.

http://www.protectedsoft.com/download_dvd2avi.php

Regards, Stefan.

[pequaide]

Let me introduce this concept for your constructive comments.

Divide a ring (thin walled hollow cylinder) into 360 equal masses distributed evenly around the circumference of the ring. The circumference of the ring is a smooth rigid circle; let?s say it is made of light modern fibers.

Let?s roll the ring down a smooth rigid incline so that it is moving across a smooth horizontal surface at one meter per second. That means the center of mass is moving one meter per second. Lets mark all the masses from 1 to 360; which is of course the degrees of the circle, but the marks are now rolling with the circumference of the circle.

When the mass marked 360 is on the bottom it is moving zero meters per second (in relation to the surface).  At the moment when 360 is on the bottom 180 is on the top moving 2 meters per second.  Suppose at this moment we release both the 180 mass and the 360 mass. The mass marked 360 will be at rest on the surface. The mass marked 180 can be caught on the end of a pendulum string. The mass marked 180 is now a pendulum bob and it will rise (d = ? at? or d = ? v?/a) .2039 meters. Now the center of mass of the masses marked 360 and 180 is at .1019 meters above the center of mass of the rim (180 can raise 360 .1019 meters off the surface and 180 is still .1019 meters higher than the top of the rim).

Now let?s do the same with the masses at 359 and 179, and 358 and 178, and repeat the process around the circle. This leaves the entire mass .1019 meters above the center of mass of the rolling ring; that now exists of only light modern fibers.

The question arises: How far did the ring roll down the incline to attain a velocity of one meter per second?

An object in freefall needs to only drop .051 meters to attain one meter per second velocity.

A puck on a frictionless plane need only drop .051 meters to attain one meter per second velocity.

A pendulum bob need only drop .051 meters to attain one meter per second velocity.

I would guess it is the same for a cart.

Is the ring different; or is this a free energy source?

[Mr.Entropy]

Is the ring different; or is this a free energy source?

The ring is different, because it rolls down the incline and ends up spinning at the bottom.  Some of the work done by gravity as it descends must be spent to spin it, in addition to the portion spent to send it forward.

You have shown above that you pretty much know how to calculate how much further the ring must fall in order to end up rolling at 1m/s.

Cheers,

Mr. Entropy