Free energy from gravitation using Newtonian Physic

[pequaide]

I posted a strobe light photograph on mad_scientist. It is a cylinder and spheres device being dropped out of mechanical arms. The center bolt of the arms can be seen (at the top) as the cylinder drops.

Two spheres can be seen; one in front of the cylinder and one behind. Just to the left of lower center there is a line of black holes that are photos of a hole in the cylinder wall. This hole is where the sphere was seated before release. These holes appear to still be moving forward slightly, and it also appears that the mechanical arms throw the cylinder a little off to the left. The releasing of the spheres and the dropping of the cylinder and spheres occurs at the same time.

It can be easily seen that the cylinder is stopping. I have photographs of the hole in the cylinder moving forward and then backward and then forward again.

So the questions a scientist must ask are: where did the motion go, and what is the quantity of that motion.

The photo is fuzzy because I resized it to VGA for my slow computer to send, but portions of the 4M copy are very clear.   

[supersam]

@ all,

as pequaides picture points out, it is not a two dimensional problem!!!!  you must think at least in three dimensions if a a gravity wheel is ever to be achieved that actually works at over 100% efficiency.  the main problem that i see with this idea is that everyone wants to continue to explain why this can't work as a basic wheel, in two dimensions.  when it is basically pr oven by mathematics for years that it can't.  however i believe that if you expand the mathematics and the experiments into at least three dimensions you will realise that it is entirely possible to have a gravity machine that is at the very least highly efficient.

Lol
SAM

[Kator01]

Hello all,

I am refering to post #66 of pequaide.

Assume we have two masses m1 and m2 balanced on a vertical wheel. This system has a spin-inertia

Is = m x r exp2 and is at rest at the beginning. An additional Mass m3 is put on this wheel at lets say 2 o clock-position in an height H = 0.5 meter above bottom dead point ( bdp at 6 pm ) and let go.

Each mass m1 , m2 m3 is 1 kg.
Now potential energy of mass m3 = m x g x h  is the only energy available which can be converted to kinetic energy at dbp. Ekin = 1 x 9,81 x 0.5 [kg x m/sec exp2 x m ) = 4.9 [kg m exp2 / sec exp2].

At bdp this value 4.9 true for the whole mass-system ( balance masses will be accelared ),

Now the formula Ekin = m x v exp2 / 2 = 4.9 is dissovled by V = Ekin = Square-root of {4.9 / 3 x 1kg}
= 1.2 m/sec ( please note that all 3 masses have to enter the formula, since part of the potential energyof mass m3 is used up to accelerate the spin-intertia of the balanced masses m1, m2 )

The resulting velocity of the whole system is much slower than the mass m3 alone falling free this height H = 0.5 meter at bdp ( without beeing attached to the balanced mass-system)

Momentum does not create energy. It just represent the torque = Force x r ( radius).

The assumed velocity in post #66 is calculated wrong.

Sorry, but I can not see any possibility here to have a gain in energy at the end of the process described.

Regards

Kator

[pequaide]

Kator01 An energy of 4.9 comes from a velocity of 1.808 m/sec (1/2 * 3 *1.808 *1.808, 1/2mv?); three kilograms moving 1.808 m/sec is a momentum of 5.424 units. One kilogram moving 5.424 m/sec is also 5.424 units of momentum and the one kilogram moving 5.424 m/sec can rise 1.5 meters. You only dropped it .5 meters.

[Kator01]

Hello pequaide,

sorry, it was very late in the mornig ( about 2 oclock I guess ). I forgot to multiply 4.9 by 2 and then devide by 3.

Now we have some hard work ahead to think it over or better do sound calculation :

System1 consiting just of one mass m3 = 1 kg dropping vertically down from height 0.5 meter :

Energy of mass m3 at bdp ( bottom dead point or 6 pm) will be :

m x g x h = 1kg x 9.81 x 0.5 m = 4.9 [kg m exp2/ sec exp2]   ( 1)

therefore :
velocity at bdp = square-root of g x 2 / m3 = of 3.13 meter/sec. (2)

Momentum or better impulse  of mass m3 at v = 3.13 m/sec

impulse = f x t = mass x velocity exp2 = 1 kg x 3.13 exp2 = 9.79 [kg m sec exp2] (3)

System2 consisting of mass m3 attached to a vertical balanced mass-system m1-m2 :

Energy = 4.9 ( 1)  available for System2 results in

velocity = square-root of g x 2 / m1+m2+m3 = 1.807 m/sec at bdp (4)

Momentum or better impulse  of mass m1+m2+m3 at v = 1.807 m/sec :

impulse = f x t = mass x velocity exp2 = 3 kg x 1.807 exp2 = 9.79 [kg m sec exp2] ( 5 )

Not only is the energy the same but also the  impuls or as you say momentum of system2 equals momentum of system1 ( 3 ) = ( 5 )

Now, I stay with my  previous analysis : 

If you split mass m3 of system2 at bdp to form a new system3 consiting of one vertical  balanced masses m1+m1and one horizontal wheel with one mass m3 and radius 0.5 m ( both of them at v = 1.807 at the moment of seperation) there will be no way to increase the velocity of mass m3 beyond 3.13 m/sec by letting the mass m3 fly off ( for example on a flexible lever from 0.5 to 1 meter )

Indeed you would  have measured a higher velocity ( 3.13 m/sec) of mass m3 spinning of system3 ( after transferring  of energy  of vertical system  m1+m2 to m3-horizontal system if you compare it with resulting  system2 - velocity :

3.13 : 1.8

which is correct but not true in the sense of an energy-gain of 174 % as this 3.13 m/sec is the same velocity mass m3 can reach alone if it drops down 0.5 meter without sharing its energy with m1+m2-system.

Sorry to say, no  way of gaining energy with this system

Kator