Free energy from gravitation using Newtonian Physic

[Kator01]

Hello pequaide,,

it seems you have a different understanding of the term momentum.
Please define what you understnad by the term "momentum"

According to kinematics and SI-Units  :
Momentum = Impuls = force x time = mass x velocity exp2  [kg m sec exp2]

This would give all three masses at speed 1.8   momentum of 9.79 ( see formula 3) and not 5.4

m3 alone with speed 1.8 ( spinning off the system ) will have 1/3 of this = 1kg x ( 1,8 * 1. = 3.24
leaving the rest of the momentum = 6.55 to the balanced masse m1 and m2 at speed 1.8.

There is no more momentum available  than 6.55 which you can transfer to m3 which will end up by then having the momentum 9.79

Simply repeating wrong data or calculations does not have any effect on physical facts.

Regards

Kator

[pequaide]

Last I knew (mv) 3 kg * 1.808 m/sec = 5.425 units of momentum.  It is not three times the acceleration it is three times the velocity. Or how do you come up with 9.79? Try this next example.

A 4 kg mass that is dropped from .051 m will have a velocity of 1.00 m/sec, if it is dropped as a pendulum bob it will be moving 1.00 m/sec horizontally. If it is caught into a horizontal circle it will continue this 1.00 m/sec velocity around the circumference of the circle. If you then release the mass from that circle and catch it in a pendulum it will rise (as a pendulum bob).051m.

A four kilogram mass moving 1m/sec in a circle can give all of its motion to one of its four kilograms, as demonstrated by the cylinder and spheres experiment. A good scientist should ask themselves, what must be the velocity of that one kilogram when it has all the motion? This transfer of motion will leave 3 kilograms at rest, and Newton?s Three Laws of Motion requires that the momentum remain constant. Some complain that the experiment is moving in a circle; but that is a ridiculous complaint because ballistic pendulums move in circles and they are used to establish Newtonian Physics. Others sight direction of motion as a problem, but Atwood?s machines are used to prove F=ma and nearly all the mass in motion of an Atwood is moving in opposite directions (almost half the mass is going up and a little over half is going down).

When a one kilogram mass moving 4 m/sec strikes a 3 kilogram mass at rest the new velocity for the 4 kilogram combination must be 1 m/sec to comply with Newtonian Physics.

Shouldn?t we be able to return to the point from which we came? Which was 4 kilogram moving 1.00 m/sec. When the four kilograms gives all its motion to the one kilogram and then the one kilogram is directed to slide back into the 3 kilograms at rest, shouldn?t that return us to the original motion. The only velocity that could do that is 4 m/sec for the one kilogram mass when it has all the motion.

Mathematically we could state it like this:  4 kg * 1.00 m/sec = 1 kg * 4 m/sec = 4 kg * 1.00 m/sec.

Only 4 m/sec velocity for the one kilogram will bring you back to the point of origin, which was 4 kg moving 1.00 m/sec.

Four kilograms moving 1 m/sec has 2 joules of energy. 1/2mv?

One kilogram moving 4 m/sec has 8 joules of energy. The one kilogram moving 4 m/sec can rise to .8155 meters, four times higher than the total height of the four kilograms at .051 meters. .051 m * 4 = .204 m

Note that I have not even mentioned radius, I don?t need to, velocity is independent of radius. Ballistic pendulums always conserve linear momentum no mater what the length of the pendulum.  Galileo proved that velocity and radius are independent of each other.

Formulas are valueless when they require a quantity that we don?t even need to know. The angular momentum formula requires radius and we don?t even need to know radius.

Energy can be made in the laboratory for $25, which is the cost of a cylinder and spheres machine.

[pequaide]

We have a pendulum where I work that swings for the better part of an hour. The energy transfer in the cylinder and spheres experiment occurs in less than a half second. Put these two facts together and bearing friction percentage becomes minimal. We are looking at about 300% increases in energy in a 3 or 4 second cycle.

[Kator01]

Hallo pequaide,

I apologize. I was totally on the wrong track. I have done myself what I was suggesting to you and looked in my physics-books.
I had confusion with english physical terms because momentum is impulse in german language.I understand now what you found. I had another concept in the back of my mind ( Alans discovery m*v exp2) and was mixing up the formulas.

Now, the correct way to calculate the final energy of m3 ( after m1+m2 fully stopped) must me done via the momentum-change

F*t = m * Delta V ( change of velocity of mass m3 ) and not by direct energy-calculation since this is what actually happens :

- momentum- ( or impuls- ) transfer.

This formula is the correct link of calculating the final energy of m3. The mistake lies in simply splitting the energy of mass-system m1+m2+m3 into m1+m2 and m3.

Now the momentum of m1+m2 after m3 spun off is 3.6.
If this system is stopped lets say in 1 second the m1+m2 * Delta Velocity is - 3.6 adding as a
+3.6 to mass m3 resulting in v = 5.4 m/sec thus delivering energy = 14.58.

As I said earlier : I think the key lies in changing the direction of momentum-vector of spinning m3 by 90 deg to m1+m2-system. When this is done there cannot be transfered any information from m3-system back to m1+m2-system if
m3 is moving to a bigger radius. The momentum-vectors of both systems are made independant.Now then you can slow down m3-angular-velocity in order to transfer momentum
of m1+m2-system.

The challange lies in finding a practical and simple system. I think of water running down on an pelton-like turbine where
the water leaves at 6 pm entering another horizontal turbine-system and flowing then out radially ( both system impulswise coupled for momentum-transfer )

This is hard engineering-work and it takes some time. Alan is a good person for enginnering-ideas. Write him about this.

If you have systems like a pendulum you mentioned why not show a picture including the principle here ?

Have a look to this pendulum Alan describes  here : http://www.unifiedtheory.org.uk/

( I am sorry the hyperlink-button does not work with my online-watch-system)

at diagramm 16 : A PERPETUALLY RESONANT DAMPED AND FORCED SPRING 

Kator

[pequaide]

 Quote from P-Motion  It would need to be shown how 3 weights falling could increase the potential of an opposing mass above the potential of the falling weights. A math calculation does not show how the increase occurs.
From the math above, the 3 weights could only cause the opposing weight to rise about .5 meters.

All three one kilogram masses will eventually be raised to .5 meters, but they were only dropped .1666 meters

Velocity of an object dropped equals the square root of the product of 2 times acceleration times the distance dropped. In freefall that would be
 
V = sqrt of 2 * 9.81 m/sec * .1666 m  = 1.808 m/sec

This (1.808 m/sec) is the velocity of each of the three one kilogram objects that was dropped (by using a pendulum) .1666 m. The total momentum of the three is (3 kg * 1.808 m/sec) = 5.425 units.

If this momentum is then transferred to one (using the cylinder and spheres device) of these three kilograms the one kilogram will have to have 5.425 m/sec velocity. This will conserve the momentum of the three. This will leave two kilograms at rest. Rest means they have no ability to rise and therefore are at zero elevation. On the other hand the one kilogram is able to rise 1.50 meters.  This places the center of mass of the three objects at .50 meters. d = ? v?/a

The center of mass of the three one kilogram objects started at .1666 meters elevation and ends with the center of mass of all three at .50 m elevation. An energy increase to 300%

You could also use a 2 kg balance wheel (with the mass concentrated on the rim of the wheel) to achieve 5.425 units of momentum. Place 1 kg on the edge of the 2 kg rim. This will give an acceleration of 9.81/3 = 3.27 m/sec. The center of mass of the mounted rim will not change upon acceleration; only the elevation of the overbalanced 1 kg will change.

After dropping the overbalanced 1 kg mass .5 m the three kilograms will have 1.808 m/sec velocity each.  So an elevation of .5 m for all three can be achieved by dropping only one .5 m.

You will have to keep in mind whether you are starting by dropping three kg .1666 m in a pendulum or by dropping one kilogram in a 3 kg (total mass) overbalanced wheel.