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### Author Topic: Free energy from gravitation using Newtonian Physic  (Read 85506 times)

• Full Member
• Posts: 117
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #45 on: November 15, 2007, 05:07:10 AM »
d = 1/2v?/a; if velocity is 1 m/sec rise (d) is .051 m, if velocity is 2 m/sec rise is .2038 m. (a) is acceleration 9.81 m/sec  d = 1/2v?/a   or   Ã¢Ë†Å¡2ad = v

If by any means you can transfer the momentum of a massive object to a smaller portion of itself you have made energy. The cylinder and spheres is one means and the rolling ring is probably another.

Yes I see your theory now. The problem is that momentum is a property of the moving mass. Whereever the mass goes, the momentum must go with it. A moving body cannot simply lose mass. The best a body can do is split up into two masses in which case each will have its own momentum based on the velocity at the time of split and the mass of each body.

Therefore the presumption that the velocity of a body will double if it loses half its mass is false.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #45 on: November 15, 2007, 05:07:10 AM »

#### DarkLight

• Newbie
• Posts: 23
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #46 on: November 15, 2007, 08:30:55 AM »
You don't need to change mass when you want to change velocity.

It's enough to convert a part of linear momentum into angular or angular to linear .

You can convert linear motion into 2 circular with opposite directions, and so you have no change in angular momentum of the whole system, but you have change in linear momentum, because part of the energy of the moving body is converted in energy of angular motion.

• Full Member
• Posts: 117
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #47 on: November 15, 2007, 12:16:02 PM »
Sorry Roy,

even if they sit on a rigid bar welded to the arms it will still work, a little uncomfortable perhaps, but it will still work.

Hans von Lieven
Hi Hans,

I know this is trivial but thought you may be interested. I sent two of my junior staff to the park this afternoon armed with two nice big sturdy G Clamps. Their instructions were to clamp the seats ( face to face type) to the two rigid supports so that the seats could not pivot. And then to try and get a swing going.

The result....total failure! No swing possible. I suspect not even with a good jazz player in the background if I had one.

As I said, trivial....but interesting.

ERS

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #47 on: November 15, 2007, 12:16:02 PM »

#### hansvonlieven

• elite_member
• Hero Member
• Posts: 2558
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #48 on: November 15, 2007, 09:57:05 PM »
@ P-motion

Where in the sphere would you mount the fulcrum and how would you keep it swinging perpendicular to the ground for gravity to act upon?

Hans von Lieven

#### hansvonlieven

• elite_member
• Hero Member
• Posts: 2558
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #49 on: November 17, 2007, 05:26:02 AM »
G'day P-motion,

I am having difficulty picturing what you are telling me. Aassuming you mount a number of pendula inside a sphere where would they be and what would they do when at "rest".

Please draw me a diagram, I am at a loss.

Hans von Lieven

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #49 on: November 17, 2007, 05:26:02 AM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #50 on: November 20, 2007, 01:17:17 PM »
Let?s review the steps to a free energy machine.

A 1 kg pendulum bob that swings down .051 m (vertical drop) will have a velocity of 1 meter per and a momentum of 1.

Ten 1 kg pendulum bobs that swing down .051 m  will have a velocity of 1 meter per and a momentum of 10.

A pendulum bob can swing in any direction; north, south, east, or west.

Ten pendulum bobs can swing down into the same circle and all could be moving in a clockwise direction in that circle. By releasing each bob at the bottom of the swing the ten bobs could be moving at one meter per second clockwise in the same circle; one at 360?, 324?, 288?, 252?, 216?, 180?, 144?, 108 ?, 72?, and 36?. The total momentum of the circle will still be ten because each bob could be released at the 360? point (in the rotation of the circle) and then the ten would be moving in a straight line. We will not release them; but we will transfer their motion to only two of them.

Using the cylinder and spheres experiment concept we will transfer all their motion to the masses at 360? and 180?. Two kilograms now have all the momentum (10). They must be moving 5 m/sec.

At 5 m/sec the two 1 kg masses will rise (d = ? v?/a) 1.27 m.   2 * 1.27 = 2.548

The system started with ten kg at .051 m.   10 * .051 = .51

2.548 /.5097 = 5   the energy of the system has increased by 500%.

It is quite possible that the circle with the bobs could have been a ring dropped as a torsion pendulum.

Or the circle could have been started as an Atwood?s machine using one kilogram accelerating 10 other kilograms. Or start by dropping one kilogram over a frictionless pulley accelerating a 10 kilogram block (ring) on a frictionless plane. These methods would also yield 500% the original energy.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #51 on: November 27, 2007, 11:23:14 PM »
P-Motion: I am encouraged to see so many still trying to make free energy with the overbalanced wheel, but others are making a fundamental mistake. The mass that causes the overbalance (or imbalance) must be given the momentum of the more massive wheel and released, preferably, at the high point of the energy production. This is because overbalanced wheels make momentum not energy; you must engineer a momentum transfer that is then an energy increase. You can produce energy with an overbalanced wheel but you have to arrange a momentum transfer.

Here is what I mean. Construct a 9 kg rim mass wheel and place a 1 kg (overbalanced or imbalance) mass on the edge. Let the mass accelerate (drop vertically) .509 meters, and the wheel and overbalanced mass will have a momentum of 10 and an energy content of 5.

Drop the same 1 kg mass in freefall .509 m and you will get 3.16 units of momentum and 5 units of energy.  There is a momentum increase between freefall and the overbalance wheel but not an energy increase.    Formulas used 1/2mv?: mv: d = 1/2v?/a or Ã¢Ë†Å¡(2*a*d) = v : (a) in the wheel is .981 m/sec (a) in the freefall is 9.81 m/sec

If the overbalanced mass (1kg) had 10 units of momentum it would have 50 units of energy.

So if you find a way to transfer all the momentum of the overbalanced wheel to the smaller overbalance mass you would have an energy producing machine.

I have transferred the momentum using the same principle in three different arrangements; the cylinder and spheres device, a large center disk stopped by two air table pucks, and a heavy puck trailed by a lighter puck that swings out and stops the more massive leading puck.

So I am encouraging people to continue experimenting with overbalanced wheels but you must make the momentum transfer to the smaller imbalanced mass, then let the smaller mass rise (as in a pendulum) and then reconnect the mass to the larger wheel after it has risen.  You will then have more energy than when you started, because the imbalance mass will be higher than when you started.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #51 on: November 27, 2007, 11:23:14 PM »

#### supersam

• Sr. Member
• Posts: 475
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #52 on: November 30, 2007, 05:24:53 AM »
pequaide,

if i am hearing you right, then if we have an axis with two weights on tethers rotating, then all we have to do is shove the weights off of the perpendicular path that the centrifugal force wants to hold them to at the bottom of the rotaion and let the centripital force pull them back into line, for maximum gain in momentum, on the downswing.

in other words if we have two bodies of mass rotateing on an axis and we were able, at the bottom of a swing, to shove the one body aside say thirty degrees, it's actual mass would be decreased significantly, because it will be effectively alot closer to the axis, and therefore alot eaisier to lift, by the other weight that is now in it's fully stretched out maximum radius swing.  therefore generating excess power.

at what rotational speed will the first weight be pulled back into the line of maximum radius for it's downward and aroundward voyage?  i think i can think of a way to shove or skew the weight at the bottom position so that the effective radius will be smaller, and therefore the actual weight.  i am worried about hysterisies in an actual model, but that is something for experimentation.

lol
sam

ps:  if this is possible we better figure out how to stop it or, what to do with the extra power.  and i mean quick!!!

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #53 on: November 30, 2007, 01:07:38 PM »
A ten kilogram mass moving one meter per second in a straight line can be caught on the end of a string and it will then be moving one meter per second around the circumference of the circle. After traveling 360? around the circle it can be release, and it will again be traveling one meter per second in the same original direction. Both momentum (mv) and kinetic energy (1/2mv?) are conserved.

A ten kilogram mass moving one meter per second in a straight line can be caught on the end of a string and it will then be moving one meter per second around the circumference of the circle. The mass can then be subdivided into nine kilograms and one kilogram and all the motion can be given to the one kilogram mass. The one kilogram can be released into the same direct of travel as the original line of motion of the ten kilograms. The one kilogram is now moving in the same original direction and the nine kilograms is stopped; what is the velocity of the one kilogram.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #53 on: November 30, 2007, 01:07:38 PM »

#### supersam

• Sr. Member
• Posts: 475
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #54 on: December 01, 2007, 12:06:41 AM »
pmotion and pequaide,

thanks for your responses.  the idea i have is to take a simple two spoke wheel, around an axis, with the ability to shift the spokes latterally at the bottom of the spin and then bring the spoke back into alighnment at the top of the rotation.

lets say that we use magnets, oriented the same way as weights,    on the end of our spokes.  at the bottom or 6 o'clock position we place another magnet oriented to repel our weight as it swings by, therefore disorting it's rotation laterally, moving it closer to the axis, at the same time the opposing weight is starting to gain momentum from gravity therefore unbalanceing the wheel.

at what speed will the weight that has been shoved laterally be pulled back into line right as the other weight is shoved latterl,y by the magnet?

lol
sam

ps:i hope i am getting the concept across.  i am still trying to get a picture of it in my simple mind.

#### Kator01

• Hero Member
• Posts: 862
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #55 on: December 01, 2007, 02:13:23 AM »

may I draw your attention to the following Website :

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

Scroll down to Diagramm 13 :

INTRODUCTION TO STEEL INERTIA QUANTUM DRIVE

Hope you enjoy this.

Kator

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #55 on: December 01, 2007, 02:13:23 AM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #56 on: December 01, 2007, 02:59:46 AM »
Sam; I think you are making this to complex. A rim mass wheel can be accelerated according to Newton?s Second Law of Motion (F = ma). You could wrap a vertically mounted wheel with a string and hang a mass from the string to accelerate the wheel. Or you could mount the wheel in a horizontal plane and drape a string, with a mass on the end, over a frictionless pulley to accelerate the wheel. Both methods should produce an F = ma relationship.

The cylinder and spheres experiment has given all the motion of from up to 8 units of mass to 1 unit of mass.  A large center disk is also stopped by two unwrapping pucks (on a frictionless plane).

#### Eddy Currentz

• Jr. Member
• Posts: 53
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #57 on: December 03, 2007, 10:12:31 PM »

may I draw your attention to the following Website :

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

Scroll down to Diagramm 13 :

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.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #58 on: December 04, 2007, 12:23:54 AM »
P-Motion: I would make these alterations to your machine. Change length and mass to the metric system, the science world is much more familiar with metric. For example; you rarely see 32 ft/sec/sec as the acceleration of gravity, it is usually 9.81 m/sec/sec.

With this change you have 4.5 kg on each end of a one meter vertically mounted bar (the bar is very light weight) with a high quality center bearing.  Add an extra 4.5 kg to one of the end masses so that the overbalance 4.5 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 O'Clock to 6 O'Clock). The acceleration rate should be one third that of gravitation 9.81* (4.5/13.5) = 3.27 m/sec/sec. This is from F = ma   44.145 newtons / 13.5 kilograms = 3.27 m/sec

At the end of the .5 meter drop (of the extra 4.5 kg) all the 13.5 kg will be moving 1.808 m/sec.  v = Ã¢Ë†Å¡ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec

As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 9.0 balanced kg moving 1.808 m/sec in a vertically mounted wheel and 4.5 kg moving 1.808 m/sec in a horizontally mounted wheel.  Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that it is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel. While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 4.5 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment.

Newtonian Physics predicts that the overbalanced mass will now be moving 1.808 m/sec * 3 = 5.42 m/sec (from the Law of Conservation of Momentum) and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.

Add the extra mass back to the vertical wheel at 90? after you have transfer 2/3 of its energy to another system. You can still start over because you have three times the original energy to work with.

#### Kator01

• Hero Member
• Posts: 862
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #59 on: December 05, 2007, 12:44:08 PM »
Hello all,

I followed this with very much interst and I like to add a bit to this discussion :

I have understood that pequaide has seperated 1/3 of momentum and changed the orientation of
the momentum axis so that both axis are perpendicuar to each other. This is a key-feature in this system.
After this he transfers the momentom of the  bigger-momentum-system to the spun-off-system.

Regards

Kator