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Author Topic: Free energy from gravitation using Newtonian Physic  (Read 80770 times)

Offline Janus20

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Re: Free energy from gravitation using Newtonian Physic
« Reply #60 on: December 05, 2007, 12:48:25 PM »
Hello pequade and all,

may I draw your attention to the following Website :

http://www.unifiedtheory.org.uk/

Scroll down to Diagramm 13 :

Try liking some more "think"

INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE

Hope you enjoy this.

Kator


Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor.  ;D

Have some more laughs with me. Be careful not to THINK

Free Energy | searching for free energy and discussing free energy


Offline Eddy Currentz

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Re: Free energy from gravitation using Newtonian Physic
« Reply #61 on: December 05, 2007, 04:44:47 PM »



Thanks for the link! I like the way this guy thinks. I haven't laughed so hard in quite a while, he's got a good sense of humor.  ;D
[/quote]

Have some more laughs with me. Be careful not to THINK
[/quote]Hi Janus. I wasn't laughing derisively at Alan's work, which I think is very good, but rather at his social commentary. I find the dry wit of the British exceedingly amusing. ;D

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #62 on: December 07, 2007, 01:30:10 AM »
Katore: The advantage of having two wheels with perpendicular axes is that while the overbalanced mass is in the vertical wheel gravitational acceleration causes a great deal of momentum, and when the mass that causes the overbalance has been transferred to the horizontal wheel the momentum of the system is constant because the overbalanced mass is no longer under gravitational acceleration.

Another advantage of the horizontal wheel is that it is smaller and close to the range of apparatuses that I have made and evaluated. The disk that I used on a frictionless plane was (if I remember correctly) 7 inches in diameter: a little smaller than the .1 m radius proposed for the horizontal wheel. The PVC pipes used to make cylinder and spheres machines were four and five inches.  I would imagine a large cylinder (say 5 meters) would work but the momentum transfer would occur more slowly because the rate of rotation is slower. In the smaller cylinders the motion has to be slowed by video tape or film photographs using strobe lights in order to see the cylinder stop, because if the string is not released it will restart the cylinder?s motion.

The vertical wheel can have a mass much greater than the 3 (total mass) to 1 (overbalanced mass) proposed. I have PVC pipe models that go up to 8 to 1, and the total momentum transfer is still in the blink of an eye (well maybe two blinks).  I assume there is no upper limit to the mass relationship except that the transfer of momentum to the smaller object will occur more slowly as the small mass unwraps.

I should revise the first paragraph of this entire thread by crossing out the words ?feed out?. I don?t let out the line the string simply unwraps. The string gets longer but it is simply because it is unwrapping. Some have expressed confusion about this and it was a bad choice of words, sorry. The spheres unwrap from the cylinder and absorb all of its motion.

I will try to post a few more pictures. My computer connection speed is so slow I have a hard time doing this.

Free Energy | searching for free energy and discussing free energy

Re: Free energy from gravitation using Newtonian Physic
« Reply #62 on: December 07, 2007, 01:30:10 AM »
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Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #63 on: December 08, 2007, 12:24:07 AM »
Hello pequaide,

is it possible that you please post here the exact link to the mad-scientist(yahoo-goup) ?

I think the construction of your proposed configuration is really not easy if its based on pure mechanical means . I am just trying to figure out an electronic solution for the momentum-transfer. But the main difficulty I have is with imagination of the force-connection between the two systems. Might take some time to ponder all this.

Kator


Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #64 on: December 08, 2007, 04:24:18 AM »
mad_scientist@yahoogroups.com

I have a few pictures posted in files with this group but I have not posted there for quite some time.

Free Energy | searching for free energy and discussing free energy

Re: Free energy from gravitation using Newtonian Physic
« Reply #64 on: December 08, 2007, 04:24:18 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #65 on: December 08, 2007, 05:41:34 PM »
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.   

Offline supersam

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Re: Free energy from gravitation using Newtonian Physic
« Reply #66 on: December 13, 2007, 01:11:54 AM »
@ 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

Free Energy | searching for free energy and discussing free energy

Re: Free energy from gravitation using Newtonian Physic
« Reply #66 on: December 13, 2007, 01:11:54 AM »
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Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #67 on: December 14, 2007, 01:48:04 AM »
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




 




Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #68 on: December 14, 2007, 03:38:45 AM »
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.

Free Energy | searching for free energy and discussing free energy

Re: Free energy from gravitation using Newtonian Physic
« Reply #68 on: December 14, 2007, 03:38:45 AM »
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Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #69 on: December 16, 2007, 10:52:33 PM »
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

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #70 on: December 17, 2007, 12:46:17 PM »
Three 1 kg masses moving  1.808 m/sec have 4.9 joules of energy. 1/2mv?

A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum.

1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after being dropped .5 meters. The kilogram moving 5.424 m/sec will rise 1.5 m.

Free Energy | searching for free energy and discussing free energy

Re: Free energy from gravitation using Newtonian Physic
« Reply #70 on: December 17, 2007, 12:46:17 PM »
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Offline AB Hammer

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Re: Free energy from gravitation using Newtonian Physic
« Reply #71 on: December 17, 2007, 06:12:27 PM »
Greetings

 You will find, when it happens, that math will not have solved it but a dupication of a natural efect in nature. Then and only then will the math be figured out, and then math will be able to improve it.

This is my prediction ;)

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #72 on: December 17, 2007, 06:50:02 PM »
Hallo pequaide,

I had an error in calculation ( 2 ). I wrongly  typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly

your  statement :

"A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"

is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.

and therefore  energy is :

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

and not 14,71.

I would suggest you go back and study the relevant kinematic formulas in your science-books.
This is a basic condition for any further serious discussion. Otherwise we waste our time

I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second  falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.

I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.

Velocity is calculated by the formula :

velocity at bdp = square-root of Epot x 2 / m3 = of 3.13 meter/sec. (2)
                         = square-root of 4.9 x 2 / 1 kg = 3.13 m/s


Regards

Kator

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #73 on: December 17, 2007, 06:51:44 PM »
Hallo pequaide,

I had an error in calculation ( 2 ). I wrongly  typed in the letter g instead of Epot = 4.9. But calculation was done with the value 4.9 correctly

your  statement :

"A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy., but the same momentum"

is wrong because 1 kg mass falling from height 0.5 has a calculated velocity = 3.13 m/sec and
not 3 times 1.8 m/sec.

and therefore  energy is :

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

and not 14,71.

I would suggest you go back and study the relevant kinematic formulas in your science-books.
This is a basic condition for any further serious discussion. Otherwise we waste our time

I will stop here with any further explanations. Please come back with a clear cut explanation why you calculate velocity of mass m3 = 3 times 1.8 meter / second  falling from height 0,5 meter.
Explain in detail how you want to achieve this 5.424 m/s of mass m3.

I have given the scientific formulas to do the correct calculation of velocity of a mass m3 falling down 0.5 m.

Velocity is calculated by the formula :

velocity at bdp = square-root of Epot x 2 / m3 = of 3.13 meter/sec. (2)
                         = square-root of 4.9 x 2 / 1 kg = 3.13 m/s


Regards

Kator

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #74 on: December 18, 2007, 12:56:08 AM »
1 kg of overbalance placed on a two kilogram balanced wheel will develop a velocity of 1.808 m/sec for all three kilograms after only the one kilogram of imbalance has been dropped .5 meters. You have three kilograms moving 1.808 m/sec. That is 5.424 units of momentum. Thats 3 * 1.808 m/sec. This is after only one kilogram has dropped .5 meters in a three to 1 over balanced wheel .

I have a machine that transfers the motion of several kg (three kg) into one kilogram (1 kg). If the motion of three kilograms moving 1.808 m/sec (3 * 1.808 = 5.424) is transfered to one kilogram the one kilogram must be moving 5.424 m/sec if it conserves momentum.  1 * 5.424 = 5.424 m/sec.

A one  1kg mass moving (3*1.808) 5.424 m/sec has 14.71 joules of energy. 1/2mv?

If you don't like the overbalance wheel concept, let each 1 kg mass drop on its own. Each 1 kg will need to drop .1666 meters to be moving 1.808 m/sec. That is a momentum of (3 * 1.808) 5.424. Give that momentum to 1 kg and it will rise 1.5 meters. That is three times as high as .1666 m* 3.   

 

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