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### Author Topic: Free energy from gravitation using Newtonian Physic  (Read 88142 times)

#### Kator01

• Hero Member
• Posts: 866
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #180 on: January 12, 2009, 12:04:21 AM »
Hello Pequaide,

well, I would love to do it myself but I have  no technical means ( mechanic ) to produce the whole setup.
One thing I had in mind was to use a ceiling-ventilator-motor ( low rot/min) but I have to adapt the mechanics and have to find a way to figure out the inertia-moment of the rotor.
Any way - using electromagnets would make it more complicated because of the wiring and the external trigger-electronics for switching the electromagnets including the power-supply ( batteries ) which must all be mounted on the rotor-plane.
I think using permanent magnet would be the simplest solution - although the preliminary break-away-force-test is more time-consuming.

Then I was thinking about security.
Question : How much energy is left when the spheres break away ? Is it dangerous ?
My quess is that if you will hit the exact force there will be not much energy left as it is all consumed up in the break-away-process.

I even think that you can calculate the force before and than use ( by the prelimninary static test ) the corresponding washer-magnet-configuration for the real test.

I think that even in you present disc-setup on the table you can use this idea. There are so many forms of magnets awailable you can find one form which fits your gray disk.

http://www.monstermagnete.de/catalog/index.php?language=en&osCsid=31a1972c8fc03bc8ae5efdcdac512f8f

http://www.supermagnete.de/eng/index.php?switch_lang=1

Or look at these half-shell-magnets on the long bar ( upper-middle )

http://www.kundelmagnetics.com/pages/Kundel_Motor_Parts.html

Unfortunately I will not be able to do it myself.

Regards

Kator01

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #181 on: January 14, 2009, 01:29:53 AM »
I switched sides, now the fishing line (the white nylon cord was on the left) is on the right side of the gray puck, and of course the white disk spins the opposite direction.

I changed to a narrower ribbon.

I drilled a hole in the white disk and placed the fishing line through it, this holds the tether to a length of 4 3/32 inches to the center of mass of the gray puck.

I also added 700g at 7.75 inches from the center of the 9 inch radius drive wheel. This is roughly equal to 600 additional grams moving at the same speed as the ribbon. The white disk still appeared to stop; this I think puts us at 7.8 times the mass of the gray puck. This means that the gray puck is capable of finishing with 7.8 times the original energy. Final 1/2 * 456g * 7.8m/sec * 7.8m/sec; original Â½ * 3556g *1m/sec * 1 m/sec

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #182 on: January 14, 2009, 02:15:20 AM »
I replaced the drive wheel with a sled of approximately equal inertia.  The sled and the white disk were stopped as the gray puck swung out on the end of the tether (fishing line). You just quickly see the stop; it was not quite as distinct as the wheel. I found out why when I measured the mass of the sled. The sled had a mass of around 3.4 kg, much greater than the estimated inertial of the wheel (2.8kg). Video tapes will reveal the true nature of the observed stops. The wheel stop may actually be reversing, or the sled stop may not be quite coming to a complete stop (but it is close).

This should be a fairly good proof that linear motion and circular motion is one and the same thing. How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #183 on: January 14, 2009, 05:07:34 AM »
"How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?"

Using standard vector calculus and their fingers?

It's a bit of a more complicated problem than is usually found in freshman engineering dynamics textbooks, true. But it reminds me very much of that damn sophomore class I took, once long ago, in a galaxy far far away...

Once all the variables are made constant, and all the constants set to zero, the problem reduces to one already solved.
We just have to figure out which one.

Seriously, momentum conservation calculations can be difficult and misleading, particularly when there is impact between objects of widely differing velocities and/or masses. Throw in rotating elements and it's easy to get misled down the wrong garden trail. If one recalls that momentum IS conserved, at least in this universe, the analysis may become a little easier, because at least you'll know when you made a mistake somewhere.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #184 on: January 15, 2009, 12:06:55 AM »
Good mathematics allows you to make predictions, so knowing the original motion you should be able to predict the final motion.

So tell me the original angular momentum of 3 kg moving 1m/sec in a straight line.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #185 on: January 16, 2009, 06:16:41 AM »
I don't know how you are planning to maneuver me into following your garden path, so I'm not even going to start. I can't see how your device moves, so I don't know how or where straight paths become curved ones, nor do I know the degree of curvature, nor a multitude of other things that are needed for a correct analysis. I'm sure we could arrive at one, with some work. But kinematic analyses are not really my forte, so maybe I shouldn't even have started replying here. But it sounds like you are trying to claim a violation of conservation of momentum, and that's even less likely than a buoyancy drive or a gravity wheel, just on first principles, that much I do know.

If it's not spinning as it translates, it has zero angular momentum, and its linear momentum is mv, or 3 kg-meters/sec. As you well know.
But as soon as it is forced into a curved path, by a centripetal acceleration, it acquires angular momentum. I think.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #186 on: January 16, 2009, 12:33:53 PM »
Newton said F = ma. Newtonâ€™s momentum conservation was linear. You absolutely can not conserve both linear and angular and Newton did enough experiments to know this.

Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time.  This will be linear Newtonian momentum; it is not angular momentum being produced.  The angular momentums of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.

#### Kator01

• Hero Member
• Posts: 866
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #187 on: January 19, 2009, 12:44:14 AM »
Hello TinselKoala,

I also had a difficult time to visualize this process.
For better understanding have a look at the pics way back at the pages 9, 12 and 17. Especially the three phases of the spheres-movement until they are fully swung out ( 90 degrees to tangent) are very well shown on page 17.
The only difference is that here instead of the puc steel-balls are used and a cylinder instead of the white disc.
The prime mover for the spinnig ot the complete setup is not shown there as this can be done in very different ways.

Regards

Kator01

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #188 on: January 19, 2009, 05:58:24 PM »

From Wikipedia:

"Definition

Angular momentum of a particle about a given origin is defined as:

\mathbf{L}=\mathbf{r}\times\mathbf{p}

where:

\mathbf{L} is the angular momentum of the particle,
\mathbf{r} is the position vector of the particle relative to the origin,
\mathbf{p} is the linear momentum of the particle, and
\times\, is the vector cross product.

As seen from the definition, the derived SI units of angular momentum are newton metre seconds (NÂ·mÂ·s or kgÂ·m2s-1 or joule seconds). Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

Well, the math symbols didn't translate, but you can see from the definition on Wikipedia
http://en.wikipedia.org/wiki/Angular_momentum
that angular momentum is defined as the vector cross product of the particle's (or object's) position vector and its LINEAR MOMENTUM.

Linear momentum is conserved; angular momentum is conserved; and the two are related by vector mechanics, just as I have said.

An excellent basic text on these matters is
Statics and Dynamics
Vector Mechanics for Engineers
By Johnston and Beers
(now in its at least 12th edition)
Wherein you will find your problem and many many other similar ones analyzed with excruciating thoroughness.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #189 on: January 19, 2009, 06:05:02 PM »
Newton said F = ma. Newtonâ€™s momentum conservation was linear. You absolutely can not conserve both linear and angular and Newton did enough experiments to know this.

Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time.  This will be linear Newtonian momentum; it is not angular momentum being produced.  The angular momentums of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.

Sorry, you are wrong about your assertion re the wheels and masses. Please don't make me construct an apparatus to prove it.
You have said nothing about moment arms in your thought experiment. If one wheel's mass is concentrated at the rim, and the other's is at the hub, there will be differences in the circumferential velocity as the linear momentum of the falling mass is coupled to the angular momentum of the wheels. If the (PE-KE) of the mass as it leaves the wheels is the same, then the angular momentum of the wheels will be the same, but if the mass distributions are different the circumferential velocities will be different--hence the velocity of the weight when it drops off the wheel will be different--hence the (PE-KE) of the weight will be different--
you can see how the complexities multiply.
But you may rest assured, momentum, whether angular or linear or some combination, is conserved. There are ample experiments, all around you, that confirm this fact.
Ever ride in an automobile, or an aircraft? Then you can be glad of CofM.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #190 on: January 19, 2009, 06:05:56 PM »
Hello TinselKoala,

I also had a difficult time to visualize this process.
For better understanding have a look at the pics way back at the pages 9, 12 and 17. Especially the three phases of the spheres-movement until they are fully swung out ( 90 degrees to tangent) are very well shown on page 17.
The only difference is that here instead of the puc steel-balls are used and a cylinder instead of the white disc.
The prime mover for the spinnig ot the complete setup is not shown there as this can be done in very different ways.

Regards

Kator01

If one picture is worth a thousand words, then one video is worth a thousand still pictures.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #191 on: January 20, 2009, 02:14:07 AM »
TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (Nâ€¢mâ€¢s or kgâ€¢m2s-1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

Keplerâ€™s formula worked without all this hocus pocus. A joule is a unit of energy. A â€œpseudovectorâ€ they have got to be joking. Are you sure this isnâ€™t pseudoscience.

Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer.  After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829

Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414

I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.

I am sure you could find similar experiments on the internet if you knew what the experimenter called it.

It is similar to an Atwoodâ€™s machine.

TinselKoala quote; Please don't make me construct an apparatus to prove it.

What is you construct one and it proves you are wrong; would that be worth the effort.

In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.

I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #192 on: January 20, 2009, 03:32:37 AM »
TinselKoala quote; As seen from the definition, the derived SI units of angular momentum are newton metre seconds (Nâ€¢mâ€¢s or kgâ€¢m2s-1 or joule seconds).Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

Keplerâ€™s formula worked without all this hocus pocus. A joule is a unit of energy. A â€œpseudovectorâ€ they have got to be joking. Are you sure this isnâ€™t pseudoscience.

Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer.  After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829

Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/secÂ². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414

I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.

I am sure you could find similar experiments on the internet if you knew what the experimenter called it.

It is similar to an Atwoodâ€™s machine.

TinselKoala quote; Please don't make me construct an apparatus to prove it.

What is you construct one and it proves you are wrong; would that be worth the effort.

In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.

I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).

First, the quote is from Wikipedia, as I clearly state. Second, it's not hocus-pocus, it's recognized vector mechanics. It's how we landed a robot probe on Titan, for one thing. Third, note that the units of angular momentum are kilogram-meters squared-per second,  that is, (kg*m*m)/sec, which is exactly the units you have in your calculations, which would be correct, if only they were, well, correct.  F does indeed = MA, because that's how they are defined. Acceleration is the response of mass to a force. Force is what it takes to accelerate a mass. Mass responds to force by acceleration. Mass is that which resists acceleration by force. And so forth.
In your thought experiment you have made some incorrect assumptions, and without examining the details of your confirmatory "experiment" I cannot evaluate it.
I can imagine your thought experiment as outlined above. But I still cannot figure out how the device in the pictures is supposed to move. You'll just have to show a video, for me to get it, I'm afraid.
And yes, the Joule is a unit of energy, and angular momentum is a means of energy " storage " which is why it can also be expressed in Joule seconds. It's called a "pseudovector" because, by convention, it points out the top axis of whatever is spinning or curving, according to a right-hand rule. By convention. Whereas a "real" vector looks like it points in the direction it's going. Again, by convention.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #193 on: January 21, 2009, 01:02:00 AM »
Pardon me: the formula is correct. It is Keplerâ€™s formula. You have my apologies.

\mathbf{r} is the position vector of the particle relative to the origin,

\mathbf{p} is the linear momentum of the particle,

\times\, is the vector cross product.

This is where I got mixed up; I thought they were multiplying by yet another cross produce. But they are merely saying r * p, which is correct.

Thank you for sticking to your guns and not getting irritated with me.

But I will reiterate; this is a formula that works for satellites because in space radius has an effect upon (p) linear velocity. In the laboratory radius has no effect upon p.

TinselKoala quote: In your thought experiment you have made some incorrect assumptions,

I accelerate the sled and white disk and gray puck by tying a string to the opposite end of the sled. I drape the string over a pulley and suspend a mass from it. That brings the red ribbon up tight against the white disk and the (duck tape) flap that holds the gray puck up against the white disk. I hold one of the upright bolts until every thing is ready. After release and acceleration the suspended mass hits the floor about the same time that the ribbon releases the gray puck.  I assume the puck holds its position up against the white disk until the ribbon feeds past the flap as the system spins. Once the puck begins to swing out on the end of the fishing line (or tether) it absorbs the motion of the white disk and the sled. The puck has a mass of 456g and the system has a total mass of 3514g.

The consolidation of motion in the puck has been used by NASA, what is in question is â€œdoes the puck have the same linear Newtonian momentum as the original systemâ€?

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #194 on: January 21, 2009, 11:35:23 PM »
The MSU Atwoodâ€™s site (page 18 reply 175 this thread) allows the pulley wheel to have mass without throwing off their calculations. So apparently MSU thinks that the pulley itself accelerates in an F = ma relationship.

I searched the web for a while and did not come up with many vertical wheels being accelerated by a string wrapped around the circumference with a mass on the end. Maybe I will have to repeat the experiment and keep the data this time. I never thought I would be required to prove that F = ma.

Does anyone expect the rim to accelerate under some other Law? The block on a frictionless plane, attached to a string draped over a pulley with a mass on the end, accelerates according to F = ma.  How is the rim different?

If the block on the frictionless plane is accelerated linearly and then is captured into a circular path by a string does it lose its linear quantity of motion when it enters the circle? If it does then how does it gain the linear momentum back after the string is cut?

Most texts say angular momentum is conserved in the laboratory and then give the ludicrous experiment of the ice skater as an example.