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Author Topic: Free energy from gravitation using Newtonian Physic  (Read 79189 times)

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #120 on: January 19, 2008, 01:49:59 AM »
The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal. In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?

The momentum had to be transferred to the spheres, but their mass is constant.

 The spheres? velocity had to change.

When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.

Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?

It all makes sense; the momentum has been transferred to the spheres.

If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast.   ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same).  The energy of the system quadrupled.

What you can not see in the pictures is that there is a hole in the center of the black plastic strip in the middle of the cylinder. The strings going up to the top of the cylinder are the ends of the strings that come from the spheres. After being fed through the hole the strings are secured with the small screws. This is a quick and efficient way to change a broken string. In the test the string comes away from the cylinder with nearly no friction, and while in the slit the strings don?t even touch the slit. But when the bottom of the cylinder strikes the bed or padded floor the interaction is violent.

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Offline Homer S.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #121 on: January 19, 2008, 01:06:08 PM »
@Kator

So you state a differential gear will produce additonal energy? - Very interesting! I always thought a gear will transfer a part of energy into heat by friction loss? :D

...and by the way I'm an mechanical engineer as well (Konstruktion allgemeiner Maschinenbau).  ;)

@pequaide
Quote
The velocities are roughly equal (actually the spheres are going slightly faster because their center of mass is further from the point of rotation) at release, but they immediately become unequal.
Please can you explain why a mass will going faster if its center of mass is further from the point of rotation? That is not true! The velocity of a rotating mass is always constant independent from its rotation radius. Only the angular rate (or rotary speed) will change according the value of the rotation radius! I believe that fact is the main reason for your misunderstandings.

Quote
In an eighth rotation (1/10 sec) or so the horizontal motion of the cylinder is gone. The momentum of the cylinder has disappeared, and there is scant little friction in the system. So where did the momentum go?
The momentum was used to stretch the strings and to transfer the spheres from translation direction into rotation direction.

Quote
The spheres? velocity had to change.
Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.

Quote
When the string is touching the end of the slit the radius of rotation of the spheres is the length of the string to the slit, when the cylinder is stopped. I often make that length equal to the radius of the cylinder. The initial and final radii are equal.

Look at the strobe light pictures. How much does the horizontal distance between the pictures of the sphere increase? Before release the sphere?s pictures were about a half overlapping each other. Now they are how much apart?

It all makes sense; the momentum has been transferred to the spheres.
You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.

Quote
If the cylinder has a mass three times greater than the spheres, the spheres at max are going four times as fast.   ? * 4 kg * 1 m/sec * 1 m/sec is less than ? * 1 kg * 4 m/sec * 4 m/sec (but the momentum is the same).  The energy of the system quadrupled.

Nope, that's only your assumption founded in misunderstood physics...

Quote
This is a quick and efficient way to change a broken string.
Very interesting! Guess why a string may brake? And of course you think there is no need of energy to break a string made of steel???

Regards,
Homer


Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #122 on: January 19, 2008, 03:48:56 PM »
The strings are made of Berkley Fireline, I think this is a fluorocarbon line that has a very low stretch modulus. It is very strong but after the cylinder hits the bed the line drags across the outer edge of the slit. You can see the line slowly being cut by the slit after a few dozen tests. The string won?t be cut until over one hundred tests so I am not too bothered by it.

Homer; You cannot definitely assume that the velocity of the spheres has increased, the cylinder's velocity can slow down as well.

Pequaide; The strobe light photo shows the sphere pictures becoming further apart, what has that got to do with what the cylinder is doing. As the sphere pictures separate the cylinder is shown stopping.

Homer quote; Yes the velocity is smaller than directly after releasing from the cylinder because the spheres were slowed down by the strings fixed at the cylinder.

Pequaide; How can both the cylinder and the spheres slow down, I have pictures where the cylinder is stopped and the spheres are a blur of motion.

Homer quote: The velocity of a rotating mass is always constant independent from its rotation radius.

Pequaide; This is true only if there is no unbalanced force being applied to the internal parts of the system. The cylinder shows that there is unbalanced force by stopping. The force also affects the spheres; it will speed them up There is no outside unbalanced force being applied to the system but the internal parts of the system experience unbalanced force.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #122 on: January 19, 2008, 03:48:56 PM »
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Offline Homer S.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #123 on: January 20, 2008, 01:10:15 PM »
I'll give up because that's wasted time and time is running too fast. Make your trials and setups and become happy!

Bye
Homer

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #124 on: February 01, 2008, 02:03:58 AM »

Iacob alex has a link to an Atwood simulation in his Pulsatory gravitational avalanche post. The simulation shows that the momentum of the pulley has to be incorporated in the F = ma equation to determine acceleration. They do not however state the mass distribution of the pulley: which is necessary to determine if their calculations are correct. Their calculations give the average mass, in motion, of the pulley to be at half the velocity of the circumference.

If the pulley was a ring mounted on dry ice, the calculations would be simple.  The entire mass m1 + m2 + m3 is accelerated by the difference between m1 and m2.      F = ma;     (m2 ? m1) * 9.81 / (m1 + m2 + m3) = a.  This is true because m3 (the rim shaped pulley) is moving at the same velocity as m1 and m2. If m3 is not a ring the calculations becomes more complex, but the moving mass m3 still has to be incorporated in the F = ma equation.

This simulation has value because m3 can be used as a cylinder and spheres experiment.

Make the pulley (mass 3) massive (9 kg), and place nearly all the mass of the pulley in the rim. From their calculations I think they are assuming a more or less uniform (from center to rim as in a solid disk) distribution of mass.

Place the pulley in a horizontal plane. Drape the string to mass 2 over a frictionless pulley. 

Make mass 1 equal to zero and mass 2 equal to 1 kg. With m1 zero there can be no more friction in the string, so you will have to wrap and tie the string to the horizontal pulley. 

This will give you 1 kg accelerating 10 kilograms and an acceleration of .981 m/sec.

When the ten kilograms achieves a velocity of 1 m/sec; mass 2 has dropped .5096 meters.  d = ? v?/a

Transfer the nine units of momentum of the horizontally mounted rim (pulley) into one of those nine kilograms as in the cylinder and spheres device and the one kilogram will rise (as in a pendulum) 4.128 meters. This is an energy increase to 810%

Mechanically arrange the kilogram at 4.128 meters to lift mass 2 back to where it started and you still have 3.618 meters left over.

I was reviewing my video tapes and I came across the ten to one model; the134 grams of spheres stopped 1258 grams of cylinder. In that model the slit was extended 3/16 in. from the sphere?s seat hole in the cylinder; the string is wrapped from one seated sphere nearly half way around the cylinder and entered the cylinder through the slit extended from the seat of the other sphere. The string length to the sphere?s center of mass of this model is about 2.8 times the cylinder?s radius.

 Let?s say the mass of the cylinder with spheres (1392 g) was moving 1 m/sec around the circumference of the circle before release. That would mean that it would take one newton 1.392 seconds to bring the cylinder and seated spheres to a stop. F = ma, a = ∆v/∆t, 1N = 1.392 kg *1 m/sec / time; time = 1.392 kg * 1 m/sec / 1 newton.

After the spheres have all the motion it will still take one newton 1.392 seconds to bring the spheres to a stop. F = m * ∆v/∆t; ∆v = F * ∆t/m; ∆v = 1 newton * 1.392 sec / .134 kg = 10.38 m/sec. The spheres will have to be moving 10.38 m/sec for the one newton to stop them in 1.392 seconds

So Newtonian Physics predicts that the energy change in the system will be; initial energy ? * 1.392 kg * 1 m/sec * 1 m/sec, to final KE, ? * .134 kg * 10.38 m/sec * 10.38 m/sec; which is an increase to 7.22 / .696 = 1038%.

I have scavenged this model to make another model that has slits in the normal positions about half way between the seats, this will change the stopping capacity of the original model.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #124 on: February 01, 2008, 02:03:58 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #125 on: February 02, 2008, 01:38:34 AM »
I think that those that constructed the Atwood simulation assumed that the mass distribution of the pulley was like that of spokes. The spoke arrangement would allow equal mass at every distance from the point of rotation. This uniformity of mass allows momentum (of the concentric hollow cylinders) to be a linear function as the radius is increased from the point of rotation to the circumference. And the average momentum is at half the distance from the center to the circumference. If the mass distribution of the pulley is like that of a solid disk this (average momentum location) is not true, because mass and velocity would increase as the spinning concentric rings (being evaluated) approach the circumference (subdivide the disk into a hundred concentric rings). The average momentum of the solid disk is at about 58% of the distance from the center to the circumference.

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #126 on: February 02, 2008, 02:10:28 PM »
www.msu.edu/user/brechtjo/physics/atwood/atwood.html   I was hoping that typing in the site would post a link to the site.  ???

Put 1000 (average momentum is half way to the circumference, effectively 500 kg) kg in for the mass of the pulley, and 2 kg and 3 kg for the masses of mass 1 and 2. Then calculate the momentum of the pulley after a 1 meter drop. d = 1/2v?/a: a = 1/505 * 9.81

And then calculate the momentum of one kilogram that has dropped one meter. The difference between mass one and mass two is only one kilogram.

And then remember that there is a Law of Conservation of Momentum.

Transfer the momentum of the pulley to the one kilogram and you have a huge amount of energy.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #126 on: February 02, 2008, 02:10:28 PM »
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Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #127 on: February 03, 2008, 12:12:01 AM »
Hello pequaide,

thank your very much for your effort in continoiusly explaining more and more of some important details.
Looks I am the only one left at the time as a discussion-partner.

I plan to build a simple version with a disc and spheres. The spheres in my model are attached to a ball-bearing fastened to a center-axis.
This is different mechanically in comparison to your experimental setups in the sense that
the center-point of the tethers can rotate freely and do not wind up on the axis.

My question is this : What would you expect to happen according to your experience ?
 I will read this attwood-page and come back to this thread if I have new ideas/insights/ etc.

Regards

Kator

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #128 on: February 03, 2008, 04:05:40 AM »
A disc is fine; I use one for the air table model. But you must wrap the tether of the spheres around the circumference of the disc at least one disc radius. I have no model much shorter than one cylinder radius, but they might work depending on the difference in the spheres' mass to cylinder mass.

As the tether unwraps from the disc the force in the tether forces the disc to accelerate (decelerate; reduce the rate of spin) in the opposite direction, the disc will slow down and stop and then begin rotating in the opposite direction.  If you release the tethered spheres when the disc is stopped the disc will of course remain stopped, and the spheres would be at maximum velocity.

I am going to insert a picture of an air table model, since a photograph is worth a thousand words.

The screwdriver goes through a hole that captures a loop on the end of each string. In operation there is a pin in this hole that is tied to a string that is looped around the operator?s finger.

The pucks on the end of the strings must unwrap from the disc about one radius. The loose strings are held up against the plastic wall, by the fingers, and they hold the pucks in place as the disc is accelerated.  When the disc is released the strings are released and the pucks will unwrap from the disc. If you pull the pin when the disc is stopped the disc will not be restarted, in the opposite direction. The system works very well.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #128 on: February 03, 2008, 04:05:40 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #129 on: February 05, 2008, 03:02:39 AM »
Let me review something in this model. If the pucks stop the disc before the string is at 90? (to a line tangent to the circle at the point where the string enters the circle) then the disc will be restarted in the opposite direction (unless of course the string is released).

By adding mass to the disc the disc can be stopped just as the string becomes straight. In this model the string leaves the disc at 90?.  The 30 pound test line was used for photographic reasons. 10 lb test would work for the air table.

Kator01 Possibly a better answer to your question about whether your design will work or not would be this; how are you going to stop the momentum of the spinning disc? How are you going to apply force to the disc in the opposite direction of the spin?   

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #130 on: February 07, 2008, 12:37:19 AM »
Hello pequaide,

assume the same setup as you have tested and presented here - but the only difference is that the tethers of both pucs are fastened to a ball-bearing on a center-axis. The disc also can rotate around the
center-axis with another ball-bearing. So the difference wound be that the tethers do not wind up on the axis thus becoming shorter at each turn but can rotate free araound the axis and keep their length.
I have no idea of the effects of this additional degree of freedom. There are some key-variations to be tested to fully understand the principle which make the full-stop of the disc or cylinder possible.

Anyway your last description ( disc rotates backwards at 90 degree-position of the pucs = bouncing back )  indicates to the elastic-impact-case  if the mass of the pucs is too big and the half-elastic-case if the masses are in the right proportion.

I have to build it. There is no other way to find out.

Regards
Kator

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Re: Free energy from gravitation using Newtonian Physic
« Reply #130 on: February 07, 2008, 12:37:19 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #131 on: February 07, 2008, 03:43:37 AM »
I will send a diagram.

Point D is where the string enters the cylinder or drapes over a pin on the disc.

Line C-A is tangent to the disc or cylinder at point D.

Line D-B is perpendicular to line C-A

If the disc was moving counterclockwise, a puck on the end of a string in quadrant ADB would force the disc to accelerate clockwise. A puck on the end of a string in quadrant CDB would force the disc to accelerate counterclockwise.

So you might see the puck in quadrant ADB slow the counterclockwise rotation of the disc; stop it; and restart it in a clockwise direction. As long as the puck is in quadrant ADB all acceleration is in the clockwise direction. If the disc is too massive the puck will not stop the counterclockwise rotation of the disc. I have stopped cylinders nine times heavier than the spheres.

While the puck and string are on line D-B the puck places no rotational force upon the disc.

While the puck, on the end of a string, is in quadrant CDB (and still attached to the disc) it will accelerate the disc counterclockwise. If the disc is stopped the puck will accelerate it counterclockwise. If the disc is still moving counterclockwise it will accelerate counterclockwise. If the disc is moving clockwise it will accelerate counterclockwise. If the disc is to light there will be two stops of the disc; the motion will be counterclockwise (original motion); clockwise (stopped in ADB and restarted); counterclockwise (puck in CDB), I have seen it many many times.

The challenge is to add just enough mass to the disc so that the disc stops while the string and puck are on line D-B. If the string is then released from the disc the disc will remain stopped. Or if the string enters a slit in the cylinder it will remain stopped until the string reaches the other end of the slit and accelerates it counterclockwise.

Fireline is extremely inelastic; especially with very light pucks on 10 lb test. I envision no elastic activity. 

Okay; lets say your two bearing (one axis) design is placed on an air table. The bearing for the string is just above the bearing for the disc. The bearing for the disc is below the surface of the disc. The string of the puck is fastened to its bearing just above the disc and the string proceeds out to the circumference of the disc; the string then drapes (horizontally) over a pin and proceeds along the circumference clockwise about one disc radius.

You will probably need to hold the pucks in place until just at release. How are you going to accelerate the disc?

It should work. If the disc is about three or four time the mass of the puck it should get you in range of adding just enough mass to the disc to make the disc stop while the puck is on line D-B. I can?t give you exact mass numbers because I don?t know the length of the string that wraps around the disc. If the string of the puck is clear of the disc (just above) it should work great.

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #132 on: February 08, 2008, 02:40:54 AM »
www.msu.edu/user/brechtjo/physics/atwood/atwood.html

On the Atwood site the Green S in the upper left corner of the page identifies the site as originating from Michigan State University, I would imagine the site was made by students but checked over by their professor before release. Therefore the concepts presented by the site represent those of the University. It should not be surprising that one of the United States? major universities thinks F = ma.

This gives us a source of extra momentum. The spinning wheel (pulley) has a much larger amount of momentum than the momentum developed from the freefall for the same distance of the same mass that gave the momentum to the wheel.

A five kilogram rim mass pulley accelerated by dropping a one kilogram mass one meter (as in an only one suspended mass Atwood) has 9.04 units of momentum; and the one kilogram that could have freefell the same distance has only 4.43 units. When the 9.04 units of momentum are given to one kilogram it will rise 4.167 meters, 4.167 times higher than the dropped kilogram.

We know from ballistic pendulums that linear motion and circular motion are completely interchangeable and experience no loss of motion in either direction. So the rotating five kg rim (above) has 9.04 units of momentum. v = Sqrt (1 m * 2 * 1kg/6kg * 9.81m/sec; mv = 1.808 m/sec * 5 kg = 9.04 units

We know that a one kilogram mass moving 5 m/sec will combine with a four kilogram mass at rest and the combined 5 kilograms will be moving 1 m/sec., for momentum conservation. Can going from 5 kg moving 1 m/sec to one kilogram in motion do anything but also conserve momentum?

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #133 on: February 10, 2008, 12:46:50 AM »
Hello pequaide,

thank your for your detailled explanations, I think I have all information necessary for building the setup I have in mind.
I will use spheres held in place by two electromagnets powered by lithium-polymer-batteries. The have enough amperage ( 1.5 Ampere/h ) to feet the electromagnets. Electromagents and batteries are mounted on the bottom-side of the disc and thus rotate with it. There are 2 vertical holes in the disc.  Two  iron pins in these holes are magentically powered from the electromagnet below thus holding the spheres in place on top at the rim of the disc. Via remote-control ( high-frequency-sender ) and the receiver-board ( also rotating on the disc ) the electromagnets are shut off at definite speed thus releasing the spheres.

By chance I lately was getting a ceiling-mounted horizontal-working fan. The rotating disc ( with the fan-blades ) is powered by an electromotor ( AC) and has quite a good amount of mass. Motor and the shaft ( central axis)  are mounted together. So the shaft is fixed on the ceiling. Motor and shaft do not move, just the disc with the blade is rotated by an inner gear. This is exactly what is needed for this experimental setup I have in mind.

You  also can use such a fan for your experiments. The only thing to change here is a ball-bearing to be mounted on the upper shaft with the tethers and spheres attached to it.
I will post a picture of this fan. You will understand it better then.

Regards

Kator



Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #134 on: February 10, 2008, 02:52:46 AM »
Hello pequaide,,

concerning the atwood-calculations can you please give more details as to the basic assumptions of the rim-mass-pulley ?

The mass-inertia of a rim-mass depends of the mass-distribution  and the radius.

How did you calculate the spin-momentum of the pulley ?

Using 1 kg for mass m1 and 5 kg for m3 ( pulley ) I get 2.8028572 m/sec exp^2 in the simulation.

Now you have to assume a radius first before you can calculate the angular momentum.

Thank you

Kator

 

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