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Author Topic: Free energy from gravitation using Newtonian Physic  (Read 78950 times)

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #135 on: February 10, 2008, 11:23:57 PM »
Hello pequaide,

in the Atwood-Simulation you have to switch on the friction-Button and enter 5 for mass m3. The text explaines that only if you use the friction-on-simulation that you will see the influence of mass m3.

The term friction used here is misleading. What they mean is the mass-spin-inertia-resistance of m3 which consumes part of the energy generated be free-fall of m1.

For the case m3 = 5 and m1 = 1 and friction "on" you will get an accelleration of 2.8028572.
After 1 second of free fall mass m1 has momentum P = 2.8028572 kg m/s. Circumference of m3 will deliver the angular-velocity via rot/ s = 2.8028572 / 2 x pi x r .

The spin-momentum then depends on the radius of mass m3.

 In order to calculate the spin-momentum of mass m3 you have to use the relevant formulas for :

Spin-Inertia Is = m3 x r exp^2
Spin-Momentum = Is x omega ( angular-velocity )

Now the spi-inertia and the angular-velocity as well can only be calculated if you have the radius of the pulley-mass m3.

It^s not so easy to calculate this. I have done it by the assumption of r = 0.1 meter and have to look over this calculation again because I am not sure if I made a mistake.

I think that this Atwood-Example is no good for any explanations of what you found.

Regards

Kator

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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #136 on: February 13, 2008, 01:53:41 AM »
In reference to:   www.msu.edu/user/brechtjo/physics/atwood/atwood.html

Indeed the word friction in the friction button is confusing, the word adhesion may be more appropriate. In physics we almost always view friction as something lost. From their own calculations they are not viewing this friction as something lost.

If:  m1 is 2 kg, and m2 is 3 kg, and the friction button is not selected, you get an acceleration of 1.962 m/sec/sec.

If m1 is 2 kg, and m2 is 3 kg, and m3 is zero, and the friction button is selected you get an acceleration of 1.962 m/sec/sec.

The acceleration is the same with the friction button on and the pulley moving, as it is with the button off and the pulley not moving. Their calculations assume no loss for this quantity they call friction.  And certainly we can build very high quality bearings where this is almost true. But it is incorrect to think that there needs to be a frictional loss on the surface of the pulley.

The strings on the ascending and descending sides of the pulley can be independently wrapped and fastened to the pulley; you really don?t need friction to make a pulley work. If the circumference of the pulley was coated with dry ice, you could still use the pulley by wrapping the string from both sides a few times around the pulley in the appropriate direction, and attaching the other end of the string to the pulley. You would get the same results.

Suppose we use a long steel pipe mounted horizontally on dry ice for the pulley. The acceleration of the pipe?s mass, as it spins, is roughly equal to the acceleration of the masses m1 and m2. It makes a difference what the mass of the pipe is. But if we use a larger radius pipe with the same mass; you will get the same acceleration.

I don?t think it was an oversight that the MSU Atwood site makes no reference to the radius of the pulley. It does not matter what size wheel you use. Therefore F = ma is not in reference to angular momentum changes. F = ma makes linear momentum.

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 (radius) from the point of rotation. This uniformity of mass allows the average momentum to be half the distance from the center to the circumference (1/2R). So adding 4 kg to m3 (the spokes have a mass of 6 kg instead of 2 kg for example) would be the same as adding 1 kg each to m1 and m2.


The following is an E-mail to MSU

Your MSU site on Atwood?s machine has been used as a verification of an experiment that is circulating in Europe and the U.S.

Your Atwood calculations guarantee that momentum can be placed and stored in a spinning wheel. The momentum is placed in according to the relationship of F = ma. If it is placed in according to F = ma then most assuredly it must be taken out according to the same relationship. 

Your Atwood site is unusual and useful because it makes prediction concerning the mass of the pulley. Many do not deal with the mass of the pulley and assume it to be without mass.

On your site: if mass 1 and mass 2 are not equal, the unequal portion of the mass accelerates the total mass (m1 + m2) according to the relationship of F = ma.  But the unequal mass also accelerates the pulley or wheel, as you have noted.

According to your site calculations the Atwood accelerates the pulley as if the average mass of the pulley is half way between the center axis and the circumference. In a spoke wheel this would be correct. So the mass of the pulley is accelerated twice as easily as the same mass equally distributed between m1 and m2. This is because the average mass of the spoke pulley is moving half as fast as the circumference velocity, and m1 and m2 are moving at the same velocity as the circumference.

For example: #1 let?s start with 2 kg in m1, and 3 kg in m2, and 0 kg in m3.  Then adding 1 kg to each of m1 and m2 would be equal to adding 4 kg to m3. Since the hanging masses accelerate according to F = ma then the wheel must be accelerating according to F = ma as well. 

When m1 = 2kg; m2 = 3 kg; m3 = 0 kg then acceleration equals 1.962 m/sec.

When m1 = 2kg; m2 = 3 kg; m3 = 4 kg then acceleration equals 1.4014 m/sec.
 
When m1 = 3kg; m2 = 4 kg; m3 = 0 kg then acceleration equals 1.4014 m/sec.


Example #2    m1 with a mass of 20 kg; and m2 with a mass of 21 kg; and with m3 zero:
 Should be equal to m1 equal to zero; m2 with a mass of 1 kg and m3 with a mass of 80 kg.

When m1 = 20kg; m2 = 21 kg; m3 = 0 kg then acceleration equals .2393 m/sec/sec.

When m1 = 0kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec.


Example #3 when force remains constant the doubling of the quantity of mass that is accelerated should reduce the acceleration to half.

When m1 = 0 kg; m2 = 1 kg; m3 = 80 kg then acceleration equals .2393 m/sec/sec this is 1 kg accelerating 41 kg. 80 kg at half the velocity (1/2 R) would be equal to 40 kilograms on the circumference.

When m1 = 0 kg; m2 = 1 kg; m3 = 162 kg then acceleration equals .2393/2 = .1196 m/sec/sec (from the MSU site). This is 1 kg accelerating 82 kg. 162 kg at half the velocity (1/2 R) would be equal to 81 kilograms on the circumference.

Twice the accelerated mass 82kg / 41kg causes half the acceleration .1196/.2393 because force remained constant ( 9.81N ).

When the cylinder and spheres machine takes the momentum back out of the wheel it can place all the motion into an object with a mass as small as the mass (difference between m1 and m2) that accelerated the wheel in the first place. This is where it gets interesting.

This MSU site confirms 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. When the momentum of the wheel is given to the mass used to accelerate the wheel, the mass can rise much higher than the distance it was dropped.

Offline Kator01

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Re: Free energy from gravitation using Newtonian Physic
« Reply #137 on: February 16, 2008, 12:13:14 AM »
Hello pequaide

I have done some research on basics :

Moment of inertia ( can bee accesses only one or two  times. Then you must register )



[url]http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html][http://www.efunda.com/math/solids/IndexSolid.cfm/url]

[url]http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html



Mechanics->Rotation->   Angular momentum
         Rotational kinetic energy
         Moments of Intertia

Here you find the basic formulas.

I found a german scientific paper about calculating the accelleration of two ( or one ) falling mass ( 1 kg ) in the atwood-example using a pulley ( full cylinder, no rim-mass ) of 5 kg. Your assumtions are not true.

See here the formula right above Paragraph 4) Ein Fallversuch (3)

http://www-ik.fzk.de/~drexlin/Mechanik0607/L8.pdf

Assuming  the radius of the pulley to be 0,1 meter the value of accelleration is in exact accordance with the atwood-simulation :

accelleration a = 2.8 m/s exp^2 and not 9,81. This means then for just one mass of 1 kg  accellerating downwards ( no other upward moving mass used )  that if you let this mass fall for 1 second the velocity will be 2.8 m/s. The distance it has fallen downwards is 1.4 m. ( s = v x t / 2 )

By this you can calculate the Input-Energy ( raise the 1 kg up to 1,4 m ) = m x g x h = 1 kg x 9,81 x 1.4 m = 13.734 Joule.

The Spin-Inertia of the pulley J = m/2 x r exp^2 = 2.5 x 0,01 = 0.025 kg m exp^2

Angular velocity of the pully at v = 2.8 m/s of the falling mass -> Omega = v/r = 2.8 m/s / 0.1 m = 28 1/s

This will give you the Spin-Momentum of the pulley  L = J x omega exp^2 = 0.025 x 28 = 0.7 kg m exp^2 / s

The falling mass m at v = 2.8 m/s has momentum of 2,8 kg m/s

Rotational energy of the pulley Wrot =  J x omega exp^2 / 2 = 0.025 x 28 exp^2 / 2 = 9,8 Joule.

Energy of the falling mass m at v = 2.8 m/s = 1/2 x m x v exp^2 = 1/2 x 1kg x 2.8 exp<^2 = 3.92 Joule

Total energy after 1 second = 9.8 + 3.9 = 13.7 Joule ( see above for Input Energy )

What you have to to is to exactly measure the velocity of the spheres and compare this to the rotational energy of the cylinder-and-spheres system before you release the balls.

As I said the atwood-machine is no good example. Spin-momentum has a different dimension ( kg m exp^2 / s ) than
translatory momentum ( kg m/s ), and I have no idea how this small 0.7 Spin-Momentum will convert to translatorry momentum, increasing energy.

There is one thing one can find out by experiment : The proposal of Alan Cresswell on his homepage here :

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

See Diagramm 2-2 ONCE MORE FOR NEWTONIAN APPLEHEADS

Regards

Kator





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Re: Free energy from gravitation using Newtonian Physic
« Reply #137 on: February 16, 2008, 12:13:14 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #138 on: February 19, 2008, 03:28:26 AM »
How much linear momentum does the wheel and falling mass have after the one kilogram has fallen 1.4 meters?

I think it is about 9.8 units of linear momentum.

If these 9.8 units of linear momentum is transfer to the one kilogram that has fallen, then it will rise, 4.895 meters. This is an increase to 4.895m/1.4m = 350%.

A rim mass pulley or wheel would make it easier to calculate the momentum of the pulley. And I think you would agree that such a rim mass pulley could be constructed.

I think a ring vertically mounted on dry ice would give you a nearly perfect F = ma, if you dropped an extra mass on a string wrapped around the pulley ring.

Let the ring have a mass of 5 kg with an extra dropped mass of 1 kg. Now the dropped mass and the ring have the same velocity.  Now; do the math again, the system has more momentum than it needs to return to the top. And the cylinder and spheres transfers all the motion of the cylinder and spheres (ring and dropped mass) to the spheres (dropped mass).

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #139 on: February 20, 2008, 01:36:45 AM »
Wrap a string around a 5 kilogram rim mass pulley; suspend a 1 kilogram mass from the string as in a one suspended mass Atwood machine. This is one kilogram accelerating 6 kilograms. Acceleration equals 1/6 * 9.81 m/sec/sec. If you let the mass drop one meter the entire system will be moving 1.808 m/sec.

You can use the kinetic energy formula to prove that the energy of this 5 kilogram rim and suspended one kilogram mass has the same energy as one kilogram that has free fallen one meter. 
 
The square root of (1m * 2* 1/6 * 9.81) = 1.808m/sec,    ? * 6 kg* 1.808 m/sec * 1.808 m/sec = 9.806 joules.  This is the energy of the 6 kilograms after one kilogram has dropped one meter.

The square root of (1 * 2 * 9.81) = 4.43,    ? * 1kg * 4.43 m/sec * 4.43 m/sec = 9.81 joules. This is also the energy of a one kilogram mass that has free fallen one meter. Since energy is Force times distance I guess this should not be surprising. 

They have the same kinetic energy; but they do not have the same momentum. Momentum is Force times time so this should not be surprising either.

The time over which the force acts in the one kilogram free fall for one meter is √ (2 *d / a) = t = .4515 sec
 
The time over which the force acts in the one kilogram dropped one meter over a five kilogram rim mass pulley is √ (2 *d / (a/6)) = t = 1.106 sec


6 kg * 1.808 m/sec = 10.848 units of momentum; 1 kg * 4.43 m/sec = 4.43 units of momentum.  10.848/4.43 = 1.106 sec / .4515 sec

A puck circling on the end of a string, which is wrapping the string around the center pin on a frictionless plane, is not conserving angular momentum. It is conserving linear momentum.

Ballistic pendulums always conserve linear momentum not angular momentum. Even if the incoming and final motion is circular, linear momentum is still (always) the quantity conserved not angular momentum.

How can you trust angular momentum conservation when it can?t pass the simplest of test?

When you reverse the phenomenon of the ballistic pendulum and send off a small projectile from a large circling object, why wouldn?t you expect the same quantity to be conserved as that which was conserved in the forward direction?   

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Re: Free energy from gravitation using Newtonian Physic
« Reply #139 on: February 20, 2008, 01:36:45 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #140 on: March 08, 2008, 05:56:12 PM »
Here are some experiments that would prove that the angular momentum conservation theory in the laboratory is false.

Swing a 1 kg sphere down .4587 meters on the end of a twenty meter pendulum string; it will be moving 3 m/sec.  Have it embed in a 2 kilogram block that is suspended on the end of a one meter pendulum string. The twenty meter string then releases and the 3 kilogram combination will be moving 1 meter per second in a 1 meter pendulum.

Linear Newtonian Momentum is conserved.

Angular momentum is not conserved.

Kinetic energy is not conserved.

Suspend a pendulum bob from a horizontally mounted roller track sled; accelerate both the sled and the bob to one meter per second keeping the string in a vertical position. Have the sled impact a stationary block. The bob should rise .051 meters. Do the same experiment with one, two, and three meter strings, the bob should rise .051 meters in all situations.

Velocity of the bob at impact is maintained in the three different pendulum lengths.

Linear momentum of the bob is maintained in the three different pendulum lengths.

Angular momentum is not maintained in the three different pendulum lengths.

Some use the frame of reference concept to prop up the theory of angular momentum conservation. This frame of reference concept is basically assigning an imaginary axis; usually it is the original axis. One debater equated this assignment of an axis to the frame of references that might be used in linear momentum conservation.

It is true that linear momentum is conserved under different frames of reference. Let?s say a one kilogram mass moving 1 m/sec on a frictionless plane is struck from behind by a 3 kilogram mass moving 3 m/sec.  You can use the frame of reference of any point on; the air table, the one kilogram mass, or the 3 kilogram mass and momentum is always conserved from that frame of reference.

But in changing your axis all axes are using the same point of reference as that used by the air table, it is not a change in reference it is merely a means of pretending that the radius has not changed, which of course has to be avoided because if the radius changes your conservation of angular momentum is lost.

Angular momentum is composed of three quantities, mass, linear velocity, and radius. Radius appears in the formula three times.

The distance traveled around the circumference of the circle in a unit period of time is equal to linear velocity. This linear velocity is then divided by radius to obtain radians per sec. The product of mass and radians per sec is then multiplied by radius two more times. Since radians are linear velocity divided by radius and we multiply by radius two more times this makes two radii drop out of the equation. This leaves us with mass times linear velocity times radius; one radius beyond Newtonian Physics.

This is the funny thing about angular momentum, why not just drop the one extra radius and everything would work just fine. The reason is that angular momentum was developed by Kepler for use with satellites and the extra radius was necessary to compensate for a huge increase in linear velocity caused by gravity. For satellites a small radius means huge linear velocity, and a huge radius mean small linear velocity.

This brings us to yet another proof that angular momentum conservation will not work in the lab. It should be obvious to any mathematician that a formula that works for satellites where you have huge increases in linear velocity caused by gravity can not possibly work in situations where there is no increase in linear velocity caused by gravity.

A thought experiment that should clarify the impossibility of angular momentum conservation in the lab would be to put a pin at the position of the Sun and run a string to the comet at apogee. The comet will continue to rotate around the pin at a distance of apogee.

Place a second pin at a distance from the comet that would have been equal to perigee. Place this second pin between the comet and the first pin so that the string comes in contact with the second pin and the comet then begins orbiting the second pin at a distance of perigee. No linear velocity change has occurred because there is no gravitational acceleration caused by the Sun.

The angular momentum formula is: M ω R? where ω equals radians per sec or linear velocity / r.

M *(linear velocity / r) * r * r = M * linear velocity * r

The two formulas that have to be equal to each other if angular momentum conservation is to be true are:

M * linear velocity * apogee (rotation about pin #1)   =   M * linear velocity * perigee (rotation about pin #2)   Since mass and linear velocity remained constant with the use of pins the equation is false because apogee does not equal perigee. The two equations are equal only if gravity causes a huge increase in linear velocity.

With the Sun back in place; linear velocity at apogee / linear velocity at perigee =   perigee / apogee, and the equation (M * linear velocity at apogee * apogee   =   M * linear velocity at perigee * perigee) is true.   


This two pin experiment can be performed on a frictionless plane; only the scale is changed the principle remains the same. If there is no gravitational acceleration angular momentum conservation does not work. 

Kepler needed no frame of reference, he got it right.

You could release the comet orbiting pin #2 in the direction of pin #1. Have the comet come within a perigee distance of pin #1 and recapture it on the end of a string. Now your point of reference that proponents of angular momentum conservation want to use is the same point around which the comet is circling, except now the radius is diminished. So it is not an issue of position is space it is only that proponents don?t want to use the real radius for their angular momentum calculations. And no wonder they don?t want to use the real radius because that will prove that angular momentum conservation is false.

Proponents could build a cylinder and spheres device and attempt to prove that angular momentum conservation overrides linear momentum conservation, the spheres can have only one velocity in the open position. Only 2 times the original velocity satisfies angular momentum conservation, but it will take 4 times the original velocity to satisfy linear momentum conservation. This is in a 3 to 1 mass relationship between the cylinder and spheres. The experiment should be awarded the Nobel Prize in Physics.

When an object rotating on the end of a string wraps around a thick center post it does not have a change in linear velocity. If a one kilogram object moving one meter per second starts wrapping its one meter string around a stationary post it will be moving at exactly the same speed when the string shortens to .5 meters or .25 meter. This means that angular momentum is diminishing. Also; if the circling object is unwrapping from a post there will be no change in linear velocity, you could release the circling object and direct it to move in the same direction with the same linear momentum at 10 meters radius as well as at 2.5 meters radius.

Proponents of angular momentum conservation will use the frame of reference notion to keep the radius the same. But by doing this they basically ignore the experiment.

Real scientists don?t ignore reality: the radius is really changing and linear velocity is not. The best way to deal with this reality is to accept Newton?s view; (mv) radius has nothing to do with momentum.

I can change the string length on the cylinder and spheres and it will not change the final maximum velocity of the spheres (according to my measurements). Newtonian Physics works: no frame of reference is necessary. 

The reason that this lengthy discussion is necessary is that angular momentum conservation is often used in an attempt to negate the cylinder and spheres experiment, but angular momentum conservation is false in the lab. And there is no such law as the Law of Conservation of Kinetic Energy.

The cylinder and spheres experiment transfers the motion of four units of mass to one unit of mass, by doing this the experiment guarantees that at least one or more of these formulas is false; mv, ? mv?, M ω R?. The spheres can not have two velocities.

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #141 on: May 23, 2008, 12:39:22 AM »
Two spheres can be given the same velocity with different initial quantities of kinetic energy; this proves that the Law of Conservation of Energy is false.  This phenomenon, which is found in the cylinder and spheres experiment, assures us that energy can be made from the gravitational field.

Mechanical arms release the spheres and cylinder at a particular point in the 148 rpm rotation. By adding mass to the cylinder in the form of a 3 inch I.D. PVC pipe the cylinder is stopped by the unwinding spheres just as the tethers of the spheres are at 90? to a tangent line at the surface of the cylinder. If the 3 inch pipe is a few grams too light or a few grams to heavy this 90? stop will not occur.

This 3 inch pipe (366g) is then replaced with a 4 inch I.D. pipe (282g), when spun the pipes have about equal Newtonian momentum and the 90? stop is maintained.  The kinetic energy difference between the spinning 3 and 4 inch pipe is about 32%.

The 90? stop of the cylinder can only be maintained if the velocity of the spheres remains constant at that point; the stop is sensitive to momentum changes of the cylinder but is indifferent to the cylinder?s kinetic energy.

The only form of energy that the spinning cylinder and spheres has is motion energy, Kinetic energy is not conserved and there is no way to pretend that energy is coming from some other source; consequently the Law of Conservation of Energy is false.

The top of this cylinder and spheres experiment is a 3.5 in. I.D.  208.4g PVC pipe coupler with two 66g spheres seated in its surface. The top has a short slice of a 3 in. I.D. pipe inside the coupler which has a mass of 22.2g; this helps seat the spheres. Both the PVC pipe and the coupler have a ? inch wall.  There is a 7g plastic strip, for the center hole of the tether string, which crosses a diameter of the coupler.  The top can be connected to different lengths and diameters of PVC pipe.

Without added mass the spheres stop the top cylinder before the tether lines reach 90? to a tangent line at the surface of the cylinder. Only by adding mass to the top cylinder can you force the spheres to stop the cylinder when the spheres are at 90? (the tether and spheres are straight out from the cylinder) to tangent. Mass is added by placing a straight pipe in the bottom of the top coupler. A slit is placed in the cylinder behind the entry hole of the tether string to allow you to evaluate the motion of the cylinder at that point; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string move across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90?, less mass and it stops before 90? and the cylinder is being force backward while the string is in the slit at 90?.  Too much added mass and the cylinder will still be moving forward while the string is in the slit, because the spheres are unable to stop the cylinder. 

For this particular tether length and cylinder arrangement the added mass (mass that is added to the bottom of the top cylinder) that stops the cylinder at 90? is about 366g of a 3 inch I.D. PVC pipe.

 You can then replace the added 3 inch PVC pipe with a larger diameter 4 inch I.D. PVC pipe; you can make calculations and construct a system that keeps the spinning momentum of the added mass (4 inch pipe) the same.  But if the momentum is kept the same the kinetic energy of the spinning larger diameter pipe is not the same.

So if the entire system maintains its 90? stop it is reasonable to assume that nothing has changed and that the momentum of the system is being conserved since the input momentum had not change and the apparent output had not changed either.

Now; one could claim that the 90? stop still occurs but that the spheres are moving faster or slower, but the spheres get to the same place, and the spheres are stopping the cylinder at 90?, and they are doing all this in the same period of time. Velocity must be the same.  And if velocity is the same then The Law of Conservation of Momentum would be true. You simply can?t pretend that both momentum and energy are conserved.

This discrepancy between momentum and kinetic energy exists in ballistic pendulums, there kinetic energy is assigned an imaginary friend (heat) to hide the formula?s (1/2 mv?) conservation problem.

In the cylinder and spheres experiment it is totally impossible to have this phantom friend of heat come to the rescue, because kinetic energy is increasing not decreasing.

Here again is the main point.

Two 66g spheres are embedded in the surface of a 4 inch O.D. PVC pipe coupler. The spheres are connected to the cylinder with a tether that can unwrap from the cylinder. As the tether unwraps all the motion of the spinning system is transferred to the spheres and the cylinder stops rotating. The spheres are at 180? around the cylinder.

The original momentum of this model is proportional to 208.4g * 3.75 (top cylinder coupler) + 22.2g * 3.25 (seating pipe) + 366g * 3.25 (mass added 3 in I.D. pipe) + 132 * 4.28 (spheres seated on the surface of the cylinder) + 7g (plastic strip for center hole of the tether string) * 1.5 = 2619. The final momentum would be the same and is held only by the sphere because the cylinder is stopped 2619 / 132 = 19.84.  The final velocity of the spheres has increased 4.62 times

The original kinetic energy of this model is proportional to ? * 208.4g * 3.75 * 3.75 (top cylinder) + ? * 22.2g * 3.25 * 3.25 (seat pipe) + ? * 366g * 3.25 * 3.25 (mass added pipe) + ? * 132 * 4.28 * 4.28 (spheres seated on the surface of the cylinder) + ? * 7 * 2.0 * 2.0 (the center of kinetic energy of the rotating strip is closer to 2.0 in. than 1.5 in.) = 4738.1. The final velocity of the spheres has already been determined by momentum conservation and the increase is proportional to 19.84/4.28. Therefore the final kinetic energy is ? * 132g *19.84 * 19.84 = 25,979 = 548 %      25,979 / 4738.1

After being spun  and released the top cylinder is forced into a stop when the spheres swing out to 90?. This is accomplished when a mass of 366g is added to the top cylinder in the form of a 3.0 in. I.D.(.25 in. side wall) PVC pipe placed in the bottom portion of the coupler.  There is one and only one mass that will cause a perfect 90? (to tangent) stop. 

I then replaced the 3.0 in. pipe with a 4.0 I.D. PVC pipe. I calculated its relative velocity (at the same RPS) to be about 4.25/3.25 that of the 3.0 pipe. Therefore to maintain the same momentum its mass will be 3.25/4.25 time 366g, which is 280g. I then placed a 282g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it from the mechanical arms. The cylinder stops very nicely just as the spheres achieve 90? to tangent and the string is then entering the slit.   

To conserve kinetic energy the 4.0 in. pipe would have to have a mass of 213g, about 30% off. I see no way that kinetic energy is being conserved in this experiment. That makes the Law of Conservation of Energy false.

My original estimations of a 350% energy increase were based upon attempts to actually measure the velocity of the cylinder with embedded spheres (using a mechanical release) and then measure the spheres alone at the 90? stop.  I used strobe light photograph, video tapes, and photo gates. This different diameters of added mass method is an entirely different and second method that confirms that the cylinder and spheres conserves momentum and can increase energy by 400% or more. This will be the driving force of perpetual motion machines. 

These 90? stop experiments prove that the quantity of motion lost by the cylinder is momentum. Newton?s Three Laws of Motion require that the momentum lost by the cylinder must be gained by the spheres, because the force in the tether string must be equal in both directions. 

The cylinder stops with the spheres at 90? with any RPS used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Two proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90? to tangent.

Once the appropriate mass is determined the cylinder will stop with any initial RPS given, the mechanical release spins at 2.5 RPS.

A 3.0 inch inside diameter PVC pipe with a length just under 10 inches and with a mass of 366g gives the cylinder a nice stop when the spheres are at 90?. The effective rotational mass of the cylinder is at about a 3.25 in. diameter.  Here is the math used for the quantities of motion mentioned.

Mass: 366g

Momentum: (mv) 366g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2373

Kinetic energy: (1/2mv?) .5 * 366g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .07692


A 4.0 inch inside diameter PVC pipe with a mass of 282g gives the cylinder a nice stop when the spheres are at 90?. The effect rotational mass of the cylinder is at about 4.25 in. Here is the math used for the quantities of motion mentioned.

Mass: 282g

Momentum: (mv) 282g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2391

Kinetic energy: (1/2mv?) .5 * 282g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =   .10135


Momentum changed from the 3.0 pipe was .2373/.2391 = 1%

The kinetic energy change from the 3.0 in. pipe was .10135/ .07692 = 32%

When the tether achieves 90? it enters a slit, if the cylinder is stopped it remains stopped as the tether string crosses the slit. If the spheres stop the cylinder before 90? they will have started moving the cylinder backward and the cylinder will be moving backward while the string is in the slit. If the spheres have not yet stopped the cylinder before they reach 90? then the cylinder will be moving forward while the string is in the slit.

The cylinder will be stopped only if it has a certain mass, add 6g and the cylinder will still be moving forward in the slit, subtract 6g from the proper mass and the cylinder will be moving backward. It is accurate within 2% of the added mass. Momentum conservation falls within this 2%, kinetic energy conservation falls outside the 2% at 32%. Kinetic energy is not conserved by the cylinder and spheres experiment.


If the momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has above a 400% increase in energy.

Momentum: (mv) 208.4g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .1559

Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .01439

Momentum: (mv) 366g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .2373

Momentum: (mv) 132g * 1kg/1000g* 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .1127

Momentum: (mv) 7g * 1kg/1000g* 1.5in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps = .002095


Total initial momentum = .522385

Final momentum will be .522385 for a velocity of 3.9575 m/sec:  .522385/ 132g spheres because the spheres have all the motion.

Kinetic energy: (1/2mv?) .5 * 208.4g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =   .05831

Kinetic energy: (1/2mv?) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =   .0046659

Kinetic energy: (1/2mv?) .5 * 366g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =  .07692

Kinetic energy: (1/2mv?) .5 * 132g * 1kg/1000g * 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 4.28in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =   .048114717

Kinetic energy: (1/2mv?) .5 * 7g * 1kg/1000g * 2.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps * 2.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2.5rps =   .0005572

Total Initial Kinetic energy =   .1886 

Final Kinetic energy will be 132g times .5 * 3.9575 * 3.9575 = 1.03376 / .1886 = an increase to 548%

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Re: Free energy from gravitation using Newtonian Physic
« Reply #141 on: May 23, 2008, 12:39:22 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #142 on: June 03, 2008, 03:41:49 AM »
I replaced the 366 gram 3.0 inch inside diameter PVC pipe (1/4 inch side wall) with a 4.5 inch inside diameter slice of a PVC pipe coupler (with a 9/32 inch side wall) that had a mass of 249.2 grams.

After being spun and released (as described in previous posts) the top cylinder behaved in exactly the same manner with the 249 gram coupler as when the 366 gram pipe was in position. The cylinder made a beautiful stop just as the spheres reached full extension and the string was entering the slot in the top cylinder.

The same results were achieved because the 3.0 inch 366g pipe and the 249.2 gram 4.5 inch pipe have the same Linear Newtonian Momentum while being spun. Linear Newtonian Momentum is what is being conserved.

The kinetic energy of the two spinning pipes is not the same; kinetic energy is about 47% off.

This is also experimental proof that angular momentum conservation in the lab is a false concept.  The angular momentum formula is mr?θ were theta is angular motion in radian per sec.  Both the 3 inch pipe and the 4.5 inch tube have the same theta when being spun by the mechanical arms; their masses are 366g and 249g as mentioned, Their rotational masses are at a radius of 3.25in./2 and 4.78in./2 respectively. The proportional equivalent of linear momentum is (366 * 3.25/2 * θ) / (249 * 4.78/2 * θ). Since theta is equal in both experiments, the equation shows that linear momentum has not changed.

Angular momentum is 249 *4.78/2 *4.78/2 * θ / 366* 3.25/2 * 3.25/2 * θ = 1.47, or 147%, or 47% off.

The cylinder stops nicely when the spheres are fully extended and the string is just entering the slot (described in previous posts) when either the 366g 3 inch cylinder is on the bottom or the 249g 4.5 inch cylinder is on the bottom of the top cylinder, with spheres attached.  The spheres? motion identifies the 3 and 4.5 inch cylinders as being identical.

The only thing identical about the 3 inch and 4.5 inch cylinder is that they have the same linear Newtonian momentum when spun.

The 3 inch and 4.5 inch cylinder do not have the same Kinetic Energy when spun at the same rate. The cylinder and spheres experiment does not conserve Kinetic Energy, and the Law of Conservation of Energy is false.

The 3 inch and 4.5 inch cylinder do not have the same angular momentum when spun at the same rate. The cylinder and spheres experiment does not conserve angular momentum; the concept of angular momentum conservation (in the laboratory) is false.

Offline zerotensor

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Re: Free energy from gravitation using Newtonian Physic
« Reply #143 on: June 03, 2008, 04:18:43 AM »
@pequaide:

It seems to me that you are neglecting the different moments of inertia for a cylinder and a pendulum.  Your linear force equations must be replaced by the angular torque equivalents to properly model the system.  The rotational kinetic energy is what you need to calculate;  not isolated, linear KE components.

Erot = 1/2 I omega2


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Re: Free energy from gravitation using Newtonian Physic
« Reply #143 on: June 03, 2008, 04:18:43 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #144 on: June 04, 2008, 01:31:09 AM »
Why do you think that I am looking for a formula that works? I have a formula that works; mv.

The spheres? motion is very sensitive to added or subtracted mass, you can?t miss calculate the added or subtracted quantity without getting a distinct change in the behavior of the spheres.

 Mass was incrementally added to the 3 inch pipe until the spheres stop the spinning system when the spheres were exactly at full extension. If you add too much mass the spheres can not stop the spinning cylinder, you must have exactly the correct mass.

 I then calculated the correct mass (length) of the 4.5 inch I.D. coupler that would give you the exact same linear Newtonian momentum; and hit the correct mass (cut the coupler the right length) first time, dead on.

I had previously used the same (mv) formula to calculate the correct mass (that mass that would make the cylinder stop just as the spheres were fully extended) of a 4 inch I.D. pipe, and I had cut the pipe a few grams short (where upon I added about 4 grams). The spheres behaved the same way for all three added pipes.

Your formula (Erot = 1/2 I omega2  ) will not work because Newton?s formula does work.   I am sure you won?t claim that (Erot = 1/2 I omega2  ) = mv.

This is a bona fide experiment with a formula that works not a philosophical conjecture.

Offline zerotensor

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Re: Free energy from gravitation using Newtonian Physic
« Reply #145 on: June 04, 2008, 06:25:14 AM »
pequaide:

It is clear that you do not fully grasp the concept of angular momentum.

The formulae and equations of angular momentum can be derived from their linear counterparts in a straightforward way.  The benefit of using the angular formulation is that one does not have to add up a large number of individual linear momenta.

Let's apply this to the example you gave earlier:
Quote
Wrap a string around a 5 kilogram rim mass pulley; suspend a 1 kilogram mass from the string as in a one suspended mass Atwood machine. This is one kilogram accelerating 6 kilograms. Acceleration equals 1/6 * 9.81 m/sec/sec. If you let the mass drop one meter the entire system will be moving 1.808 m/sec.
Only this time, instead of a pulley with all of its mass concentrated on the rim (which is rather unphysical), lets substitute a pulley with a uniform mass distribution.

To proceed as you did in the example, we would have to sum the linear momenta of each infinitessimal mass element, for each ring of mass in the pulley, for the duration of the applied force at the rim.  This can be done, and once the calculus dust settles, you will find that you have obtained the formulation for the behavior of the system in the language of (*horror*) angular momentum.  In fact, that is how we got these equations to begin with.

Instead of going through a process of re-inventing the wheel every time, we notice that a general rule can be applied to these kind of systems.  It turns out that (in the absence of friction and the like), the total angular momentum is conserved.  This makes calculating the behavior of these systems much easier, provided we know or can calculate the distribution of mass in the system.

The systems you cite as exceptions to the rule appear at first glance to violate the conservation law, but upon closer inspection, they all involve varying moments of inertia, and can be modeled perfectly well using the concept of angular momentum.

While there is nothing wrong with using a linear approach to tackling these kind of problems, it is often far easier to use the angular analogs.

Your assertion that angular momentum is not conserved, while linear momentum is conserved demonstrates a deep misunderstanding of the underlying physics:  The conservation law for angular momentum can be directly derived from the linear equations.

Of course, when we venture outside of the bounds of Newtonian physics, these concepts need not apply -- but I don't think we need to resort to relativity or QM or any other theory to model your machine-- good old fashioned classical mechanics will do just fine.





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Re: Free energy from gravitation using Newtonian Physic
« Reply #145 on: June 04, 2008, 06:25:14 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #146 on: June 05, 2008, 01:21:01 AM »
Well said.

But there is one major problem with angular momentum conservation; it does not work (in the lab). Linear momentum conservation does work, it?s a pragmatic issue.

My text books give me the formula of mr?θ for angular momentum. If I were to calculate the mass (cut to length) of the 4.5 inch coupler from the mass (that is stopped at full extension by the spheres) of the 3.0 inch pipe I would have cut it to: 366g * 3.25in. * 3.25in. * 1 radian/sec = 3865.9;    3865.9 / 4.78 / 4.78 / 1 radian/sec = 169 grams.

At 169 grams for the 4.5 inch I.D. coupler (added to the bottom of the top cylinder) the spheres would not have stopped the combined cylinders exactly at full extension.  The combined cylinder system would have stopped well before full extension of the spheres; and the cylinder would be moving backwards while the string is in the slot.

But the cylinder made a nice clean stop at full extension using the 249 gram length of 4.5 in. I.D. coupler. This is linear Newtonian momentum conservation.

What people assume and what really works are often two different things.

Early physics was founded upon experimentation, now it seems science just assumes everything.

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #147 on: August 08, 2008, 12:46:20 PM »
A pulley can change the direction of a force but it does not change the quantity of force. A 12 newton clockwise tangent force at any position on the circle of the pulley can be balanced with a 12 N counterclockwise tangent force at any position on the circle.  Six 2 newton clockwise tangent forces at any position on the circle can be balanced with a 12 N counterclockwise tangent force at any position on the circle.

A two kilogram mass moving 1 m/sec clockwise in a circle will need two newtons of force to be applied for one second counterclockwise in a tangent direction to get it to stop. The mass can move around the circle as the force is being applied, and the applied force can move around the circle as well because the position of the mass or the position of the applied tangent force is unimportant. Note: that the size of the circle with a center bearing is unimportant, and a balanced wheel could be mounted vertically or horizontally.

This means that a moving wheel can apply its motion to an object in a linear path without loss of motion to the system, and vise versa. 

This means that a massive large diameter moving wheel can apply its motion to a small diameter light wheel without loss of motion to the system. The moving mass of the cylinder in the cylinder and spheres experiment could be contained in the motion of a separate larger wheel; it need not have the same radius as that of the spinning spheres. In fact; linear motion could be used for the moving mass of the cylinder.

The larger wheel or linear motion being used (instead of the same radius that the spheres use) should eliminate the concept of angular momentum conservation. But more importantly it opens a vast array of possible arrangements for the cylinder and spheres experiment, so much so that we might have to call them momentum (linear) consolidation experiments.

So let us review the simplicity of this energy making machine and see if Newtonian Physics has been violated.

A one kilogram mass moving five meters per second collides with a four kilogram mass at rest.  The combination proceeds at 1 meter per second. The combined mass is then caught on the end of a string (of any length) at 90?, the mass will proceed around the circumference of the circle at one meter per second. The circle then acts as a cylinder and spheres experiment and throws all the motion back into one of the five kilograms. If the original motion is to be conserved as Newton would predict the motion must now again be five meter per second.

Momentum is conserved so the event in the proceeding paragraph should sound logical to you, now calculate the energy changes. E = 1/2mv? 

My data confirms Newton?s predictions, and the Law of Conservation of Energy is false.

Offline spyblue

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Re: Free energy from gravitation using Newtonian Physic
« Reply #148 on: August 08, 2008, 04:20:22 PM »
Hello pequaide, I?m very impressed with your experiments and knowledgment.. can you send me your experiment video please ?!
my e-mail is felipe.bit@gmail.com I would like if you can add me on msn messenger.. my contact is felipexz@hotmail.com

Congratulations for your great work
Thank you
Felipe

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #149 on: August 08, 2008, 11:27:14 PM »
I think we have narrowed our inability to place video on the internet to DVD formatting problems. I will probably buy a new DVD recorder next week. And one of my constituent uses hotmail so I will ask him about that. Thanks

 

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