re: energy producing experiments

[Delburt Phend]

https://www.bing.com/images/search?view=detailV2&ccid=TMu2hSEA&id=6BB16497F82989ED2B478664F434FFEA6B0CE030&thid=OIP.TMu2hSEAPL8m07F-jJA8ggHaFl&mediaurl=https%3A%


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I don't know where the short names when but here are a few, of what some refer to as Galileo's pendulum.

The energy and therefore speed at the down swing position is the same on both sides; as the upward swing starts. That also means that the magnitude of linear momentum is the same. But what happened to angular momentum conservation.  L = mvr

Angular momentum is not conserved because the length, or radius, does not remain the same.

Kinetic energy is not a conserved quantity; so what is the only Law that explains the restoration of spin in the despin.

[Delburt Phend]

Pasco has a few experiments where they claim angular momentum is conserved. They do this by not changing the radius. The problem is that to get angular momentum you have to multiply linear momentum by the radius. They drop a disk on top of another disk; of the same dimension (Radius).

When you have a before and after experiment where the before and after radius are the same; then you are merely stating that linear momentum has remained the same. You have not searched out whether or not you can change the radius.

In the interrupted pendulum; also known as Galileo's pendulum, the radius is changed.  And angular momentum does not remain the same as the pendulum swings through the down swing position. This experiment is used to prove that energy is a conserved property. Conservation of energy should not be a surprise because nothing (mass and velocity) has change.

Change the mass, as in a ballistic pendulum, and the conservation of energy drops out; with a totally make-believe quantity of heat. The double despin proves that the motion is still there; it has not been lost as heat.

[sm0ky2]

The wheel is roughly the same shape as the gyroscope. And the interesting thing is that the energy transfer to the spheres only takes about one third rotation. Experiments have shown that the stop occurs in the same quantity of rotation no matter what the rate of rotation. The quantity of friction in one third rotation must be very small. So you have a huge quantity of energy produced with very little friction and in very little time.

Ok so, to understand this i want you to consider the energy transfer along the string:


From our perspective we can view this as a 'wave', which has an amplitude and velocity
It's frequency, however, is pre-determined by our analysis.
It is a wave, traversing a string, therefore we know it has a wavelength equal to the free motion of the string. At 1/3 we find the node of maximum change in amplitude, and thus maximum transfer of rotational momentum.


Tension plays a major role in the scalar equation here
This is derived from the force between the two masses which approaches its' peak at the same node
and subpeaks at subsequent nodes. (as does the remainder of momentum transfer; remember that an object that 'stops' and changes direction from our perspective doesn't necessarily have a 0-momentum and to know the true value we would first have to determine our own)


We dont need to apply relativity here, most of that is semantical for our application.
We can now apply a peg at the 1/3 node, such that as the cylinder rotates and the smaller masses apply tension to the string, at 1/3 rotation and at 1/3 the length: the string hits the peg.
Then we can observe the momentum change directions in almost a pure fashion

[Delburt Phend]

A disk with a 1.278 meter diameter and a thickness (height) of one decimeter can have a mass of one thousand kilograms.

To throw this disk up 30 meters will require a velocity of 24.26 m/sec. From d = ½ v²/a

This would be 24.26 m/sec * 1000 kg = 24,260 units of linear Newtonian momentum.

A stack of these disks that is 30 meters high would have a mass of 300,000 kilograms.

When this 300,000 kg stack is dropped 5 cm it will have a velocity of .9904 m/sec. This is 297,136 units of linear Newtonian momentum. You only need 24,261 units of the 297,136 units to reconfigure the stack, and the stack has only dropped half of the available distance.

So here is the procedure; you let the entire stack drop 5 cm. You transfer 24,261 units of momentum to the lower disk, you throw the disk back up to the top and the stack is ready to be dropped again.

The remaining 272,859 units of momentum and the remaining drop of 5 cm can be used to spin electric generators.

The 24.261 m/sec velocity of the one thousand kilograms can be achieved by use of the despin event.

[Delburt Phend]

John Mandlbaur; successfully argues that angular momentum conservation does not work in lab. You may find it interesting. john@baur-research.com