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Author Topic: Free energy from gravitation using Newtonian Physic  (Read 76212 times)

Offline pequaide

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Free energy from gravitation using Newtonian Physic
« on: February 17, 2007, 07:39:49 PM »
The cylinder and spheres machines

Tethered spheres imbedded (at 180?) in the surface of a hollow cylinder can stop the spinning cylinder when the spheres are allowed to feed out on the end of the tethers. The only quantities of motion that I see might be conserved are kinetic energy (1/2mv?) or Newton?s linear momentum (mv); but not both, for the same v can not satisfy both equations.
I would like to get a few people to make and video tape these cylinder and spheres machines, I would very much like to know which type of motion is conserved. My data shows that it is Newtonian motion. I make my models out of PVC pipe and couplings; fishing string; and one inch spheres from scientific supply houses.
I am not looking for people to tell me what Laws I have broken, or what concept problems I have. I build these machines and they work; I am looking for people to repeat the experiment. I get output energy of about 350% the input energy.
I have video tapes of several models, and the mechanism used to spin them, on DVD. If you think the DVD would be helpful in building your own machines; just post your address and I will send you one free. There are pictures on yahoo/groups/mad-scientist
« Last Edit: February 18, 2007, 01:14:59 AM by pequaide »

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Offline slncspkr

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Re: Free energy from gravitation using Newtonian Physic
« Reply #1 on: February 19, 2007, 07:24:35 AM »
 Hello pequaide.
can you post a small video to see it please?
thank you.
slncspkr

Offline Flit

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Re: Free energy from gravitation using Newtonian Physic
« Reply #2 on: February 19, 2007, 04:03:22 PM »
I get output energy of about 350% the input energy.

What method do you use to measure the output energy and input energy?

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Re: Free energy from gravitation using Newtonian Physic
« Reply #2 on: February 19, 2007, 04:03:22 PM »
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Offline Rosphere

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Re: Free energy from gravitation using Newtonian Physic
« Reply #3 on: February 19, 2007, 05:41:10 PM »
How it is that, "the same v can not satisfy both 'equations;'" are we setting linear momentum equal to kinetic energy?

It sounds like you have a complex system of twirling masses with tethers causing discontinuities.  Which v are we talking about?

var1  initial rotational velocity, (should be ωar1,) assembly at sphere 1 release.

var2  initial rotational velocity, (should be ωar2,) assembly at sphere 2 release.

va1  initial rotational velocity, (should be ωa1,) assembly at assembly release.
(It is unclear without seeing the video if the tube is released from the hand before of after the spheres are released from the tube.)

vs1  initial linear velocity of sphere 1.

vs2  initial linear velocity of sphere 2.

There will be more velocity changes but the speculation grows exponentially without seeing the video.

My guess is that the spheres go around the tube in some cool pattern due to the tethers and the rotational mass of the cylinder.  Energy is transferred from rotational to linear and back again.  An overall gyroscopic effect may prevent the tube from toppling over, maybe throughout the entire event.  Then the whole thing finally comes to rest with the spheres just clearing the table-top?

Or, if what you are saying is true, you might want to warn people to take safety precautions.  At 350% out/in, the acceleration must stress the tethers to failure sending the spheres out as projectiles?  :o
« Last Edit: February 19, 2007, 06:33:54 PM by Rosphere »

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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #4 on: February 20, 2007, 02:33:16 AM »
Measurements were made by video taping the models that have slits in them. The slits come back from the small hole that allows the string of the bolas to go through a diameter of the cylinder. The portion of the bolas (string with two spheres on each end) that protrudes (the tether) through the cylinder can be varied, and the mass of the cylinder can be varied. These variations allow the experimenter to arrange for the cylinder to stop when the tether is at 90? to a tangent of the cylinder.

Before the tether is at 90? (to the tangent) it will counter rotate the cylinder, after 90? it will pull the cylinder in the original direction. By having a slit at the entrance hole of the tether it allows the video camera to take a few frames while the cylinder is stopped. The movement of the string along the slit allows you to calculate the spheres velocity at the end of the bolas when the cylinder is stopped.  The original velocity is determined by a photo tachometer (3.25 rps). Some strobe light photography was also used, with the strobe flashing at 100 flashes per second, which yielded the same results, about 350% output from the original input energy. 

All velocities used were Newtonian linear, of course you have to know rps (rotations per second) and radius. But all motion was in the linear frame, which would be the distance traveled around the circumference of the circle in a unit period of time. The end objective would be to release the spheres, and then they would be traveling in a line. Whenever an object is released from a circular motion it will travel in a straight line equal to the quantity of travel around the circumference.

Keeping it simple the original velocity is about, 3.25 rps * 4.81 in * (.0254 m /in) * 3.14 = 1.25 m/sec, and the final velocities of the spheres are about 4.5 m/sec. I admit it is rough data but it is well over kinetic energy conservation, which is our only other choice. I think momentum is conserved.

As far as safety; yes the spheres will on occasion break the string, these are not toys they are somewhat dangerous. Proceed to build at your own risk, and please don?t let children play with them.

I have videos but I can not put them on my computer from D or E drive, I don?t know what the problem is. I might try loading from the camera itself. I can send about any picture you want. The cylinder in the background in the pictures is one with a slit, numerically marked to calculate distance traveled by the string.       

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Re: Free energy from gravitation using Newtonian Physic
« Reply #4 on: February 20, 2007, 02:33:16 AM »
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Offline Flit

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Re: Free energy from gravitation using Newtonian Physic
« Reply #5 on: February 20, 2007, 07:43:35 AM »
Tethered spheres imbedded (at 180?) in the surface of a hollow cylinder can stop the spinning cylinder when the spheres are allowed to feed out on the end of the tethers.

Isn't this conservation of angular momentum?  Similar to an ice skater extending or retracting their arms to alter the speed of a spin?

I may have missed some important point but I don't quite get the purpose of a device that stops spinning, isn't the goal to get one to keep spinning?

Could you please share a few more details so I can understand this better?

Offline helmut

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Re: Free energy from gravitation using Newtonian Physic
« Reply #6 on: February 20, 2007, 03:18:53 PM »
Hello pequaide
Your development is very impressiv . But to be honest,it is, at least to me, very difficult to emagine,how the Parts move ,and how the Energy is to move out.
It would be very very helpfull to see either a Video,or an animatet gif.
Please give ita push,and spend more Information,so that we can learn from you,how the things work in function.

Regards from Duisburg
Helmut

.....................
All velocities used were Newtonian linear, of course you have to know rps (rotations per second) and radius. But all motion was in the linear frame, which would be the distance traveled around the circumference of the circle in a unit period of time. The end objective would be to release the spheres, and then they would be traveling in a line. Whenever an object is released from a circular motion it will travel in a straight line equal to the quantity of travel around the circumference.

Keeping it simple the original velocity is about, 3.25 rps * 4.81 in * (.0254 m /in) * 3.14 = 1.25 m/sec, and the final velocities of the spheres are about 4.5 m/sec. I admit it is rough data but it is well over kinetic energy conservation, which is our only other choice. I think momentum is conserved............... 


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Re: Free energy from gravitation using Newtonian Physic
« Reply #6 on: February 20, 2007, 03:18:53 PM »
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Offline nwman

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Re: Free energy from gravitation using Newtonian Physic
« Reply #7 on: February 20, 2007, 07:57:44 PM »
I am also having a hard time visualising this setup. So I maybe way off but here is a concepts. When it come to any potential OU device you must break it down to its simplest equations and find all the input and out put energy. To keep it simple like the ice skater spinning with there arms out at a slower rpm and then PULLING there arms in to accelerate the rpm. So the work done to pull there arms in is then stored in the rotational energy of the skater. So, when they open there arms again they are spinning at a higher rpm then originally. I believe the energy is being added from the work done to pull the arms/balls to a closer radios?

Tim

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #8 on: February 20, 2007, 11:34:50 PM »
In order to make a perpetual motion machine there must be a portion of the cycle of the machine that makes energy. The cylinder and spheres machines can quadruple (or more) the quantity of energy in less than a second; this makes them applicable to be the impetus for what is envisioned as the classic perpetual motion machines.

In some portion of the cycle of the machine momentum must be increased as well, but this is already proven technology; which is the Atwood?s Machine. This is force times time being used to make momentum so this is not a violation of the Law of Conservation of Momentum.

A 9 kg Atwood, with one kg of imbalance, can produce 14.007 units of momentum from what costs 4.429 units of momentum to achieve.  One kg in a 1 meter freefall has a final velocity of 4.429 m/sec, but after it has fallen the same distance on the machine it has a velocity of 1.4007 m/sec (* 10 kg) for 14.007 units of momentum. It only takes 4.429 units of momentum to return the one kilogram mass to the top, to start over. 

The Law of Conservation of Momentum is one of the most respected Laws in physics. All three of Newton?s Laws of Motion are contained in this Law.  Newton?s Laws of Motion are not restricted to objects that only travel and interact in straight lines, the Laws would be worthless if this were true. Newton used the word oblique to describe the interactions of objects at angles; he also dealt with gravity and planetary motion so he was familiar with forces working at angles. 

If the cylinder and spheres do not conserve momentum I think it is the first and only violation of Newtonian Physics.

When the spheres are seated in the surface of the cylinder there mass is moving at roughly the same speed as the cylinder. Let?s say that the three kg cylinder with the imbedded 1 kg spheres is moving at one meter per second around the circumference (center of rotational mass actually) of the cylinder, that would give them 4 units of momentum and 2 units of kinetic energy. To maintain the 4 units of momentum the spheres would have to be moving 4 meters per second when the cylinder is stopped, and that would give them 8 units of kinetic energy. 8 is four times bigger than 2.    Formulas used were 1/2mv? and mv.

Now this kinetic energy is real, for the 4 kg can only rise .050968m * 4 kg =  one kg at .2038m and the spheres (one kilogram) can rise .8155m. .8155m is four times as high as .2038m. Put this rise into an Atwood and we have a perpetual motion machine.   Formulas used were d = 1/2at?    or      d = ? v?/a

Experiments have been conducted on a frictionless plane using a disk and pucks, with the same results.

So the perpetual motion machine would consist of a horizontally mounted wheel driven by an Atwood machine or an overbalanced wheel. The horizontal wheel would act as a cylinder and spheres machine and would flip the driving mass back up to the top.

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Re: Free energy from gravitation using Newtonian Physic
« Reply #8 on: February 20, 2007, 11:34:50 PM »
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Offline nwman

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Re: Free energy from gravitation using Newtonian Physic
« Reply #9 on: February 21, 2007, 12:18:27 AM »
It might just be me but I can't understand the workings of the device. You are throwing numbers around but I don't see the mechanical design in which you are speaking. Could you post a video or schematics please?

Tim

Offline helmut

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Re: Free energy from gravitation using Newtonian Physic
« Reply #10 on: February 21, 2007, 10:10:30 AM »
Hello Pequaide
I am not really not shure to got the Point.So i discribe it as i see it in my vision as a pendulum.

Therefor i imagin a swinging pendulum.
The axes in the middel is surroundet be two counterrotating masses(weights but imbalanced).The counterrotating Masses are also part of a Pendulum,wich is mountet on the axes from the mainpendulum.Both Pendulums are forced to move in the same direktion.
The speed of the counterrotating weights is according to the speed of the Amplitude,that the Pendulum is acting.As there can place resonanz.

But this is just imagency.

I would like to take part at the Offer ,that you have made,to send Pics and or,a DVD.

Therefore see here the Adress  info@satundedv.de
The email adress ist working.
Please writre details for Kontakt
the Page is deactivatet because of reconstruktion.

so thanks in advance
Helmut



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Re: Free energy from gravitation using Newtonian Physic
« Reply #10 on: February 21, 2007, 10:10:30 AM »
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Offline pequaide

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Re: Free energy from gravitation using Newtonian Physic
« Reply #11 on: February 22, 2007, 12:03:10 AM »
I can understand your inquiry for drawings of a functional perpetual motion machine.  We are all a little curious as to what the first perpetual motion machine will look like.  I think it will be an overbalanced wheel or Atwood?s machine that drives a smaller cylinder or disk that flips masses back to the top like the cylinder and spheres machine.

I made drawing of such a machine that I estimated would have the output of Glen Canyon Dam, on the Colorado. I sent the drawing to the Patent Office: boy was I naive. They not only had a good laugh they also keep my money.

I have determined that to discuss any such machines is a waste of time (which I am doing now). If I can?t get you to repeat a simple $25 experiment, what good is discussion of one that will cost $1,000.00 or more.   There is not much sense in posting on sites where all they wanted to do was talk (and make junior high jokes). To my knowledge none would actually build the $25 machines, but they were sure willing to berate them.  Plus: who would move on to the $1,000.00 machines before confirming the $25 versions?

The cylinder and spheres machine is in and of itself a huge challenge to the field of physics. If Newtonian physics is true the Law of Conservation of Energy is false. If the cylinder and spheres machine works as proposed the machine itself is one of the most important inventions in history. If it does not make energy it will have to have lost Newtonian momentum, and this too would be a significant event.

Linear momentum can be caught and released from circular motion, with no change in linear momentum. If the rearrangement of motion as in the cylinder and spheres does not conserve circumference momentum then it does not conserve linear momentum. And this too would be a first in the word of physics. So you see that the cylinder and spheres experiment places the world of physics between a rock and a hard place, either the Law of Conservation of Momentum is false or the Law of Conservation of Energy is false.

Offline helmut

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Re: Free energy from gravitation using Newtonian Physic
« Reply #12 on: February 22, 2007, 02:08:35 AM »
Hello Pequaide
First i want to thank you,that you are kontinuisly passing by and keep
the conversation alive.
I think,if we follow the law of Newton in all aspekts,then we put our Minds in chains.
Thats wy i prefer,to try a search after new knowledges.
Stefan has made a diskription to show us ,were we do a further search.
Analog to the energy, of flowing water,which we take out by a wheel,we have to find a prossedure to guide out the energy from aether.
By electric energy:
I think,that the key is hidden in either a electric field,like it is present in a capacitor,to capture electrons between a plate and a dielektrikum.Or electrons are Captured in a coil
as a result of a collaborated magnetic field,that was caused of a travel from a moving field and masses throu a existing field.
Perhaps we have to build a stator,which has both of it.Once the Coil or Magnet and/or
a flat capacitor. And the Rotor might also be a sort of multicapturabe construktion.

And the Power from Gravitation need also a prossedure to be captured and to be placed on another point to feed a secondary aktion.Without a need of palarisation.
Just to create a differenz in energy level and keep this situation as long as there is need for energy.

I am so shure,that the plans for free energy are all ready in our Minds.But to find the key to open our Minds and disclosure the fakts,is our greatest job.

Helmut

 

Offline pequaide

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Re: Free energy from gravitation using Newtonian Physics
« Reply #13 on: April 21, 2007, 02:29:51 PM »
The following is a description of a simple experiment that proves that the Law of Conservation of Energy is false. It is composed of an upper and lower section. The top is a 208.6g cylinder (3 in. PVC pipe coupler with a 3.5 in. I.D.; both the pipe and the coupler have a ? inch wall) with two 66g spheres seated in the coupler?s surface at 180?. It has a 3 in. I.D. pipe inside that has a mass of 22.2g to help seat the spheres. The top portion of the cylinder (see pictures under: ZPEnergy; Free-Energy; files; pequaide) and spheres experiment can be connected to different lengths and diameters of pipe by using the bottom opening of the coupler.

The spheres stop the top cylinder from spinning, as the spheres swing out on the end of tethers, before they reach 90? to tangent if there is no additional mass added to the bottom. Only by adding mass to the cylinder can you force the spheres to stop the cylinder when the spheres are at 90? 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 string (that tethers the spheres) to allow you to evaluate the motion of the cylinder when the spheres reach 90?; the cylinder will maintain the same rate of motion for about 1/10 of a second as the string (tethering the spheres) moves across the slit. One and only one length (mass) of pipe makes the spheres stop the cylinder at 90?, less than adequate mass and the cylinder will stop before 90? and the cylinder is being force backward while the string, holding the spheres, 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.

For this particular cylinder arrangement and tether length the added mass that stops the top (and bottom) cylinder at 90? is about 364.9g of a 3 in. I.D. PVC pipe.

If you replace a portion of the added pipe with a larger diameter coupler; you can make calculations and construct a system that keeps the spinning momentum of the added mass the same.  But if the rotational 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 changed 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 under these conditions the Law of Conservation of Momentum would be false. You simply can?t pretend that both these two formulas (mv and 1/2mv?) 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?) incompetence.

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.

The straight 3 in. pipe is replaced with a combination of a coupler and a 3 in. connecting pipe. If the connecting pipe for the bottom coupler is cut at 73.2g; the momentum change from the 3 in. (I.D.) 364.9g pipe length would be 12% low. The momentum would be the same for this lower portion if the connecting pipe is cut to a length that gives you 112.8g.

If the connecting pipe for the coupler is cut at 112.8g; the kinetic energy change from the 3 in. (I.D.) 364.9g pipe length would be 11% high. The kinetic energy would be the same for this lower portion if the connecting pipe is cut to a length that gives you 73.2g.

This11% and 12% change for this lower portion doesn?t mean that the spheres motion will be going up or down by 11% or 12% that has to be calculated separately because the spheres are going to absorb all the motion of the connecting pipe whether it has a mass of 73.2g or 112.8g. The 11% only predicts that the 90? stop will be lost if you use the wrong one (73.2g or 112.8g) of the two masses. The spheres themselves would ideally have about 500% of the original energy at the 90? stop.

The 112.8g connecting pipe was used and the 90? stop was maintained, that means that kinetic energy is not what the spheres are responding to. The spheres are responding to momentum, and since there was no change in the rotational momentum there was no change in the 90? stop.

This 11% is well outside of the accuracy of the 90? stop method. I think I could detect a 6g change in the connecting pipe mass, and the 73.2g pipe is 39.6 grams off. If the connecting pipe only had a mass of 73.2g the method would be telling you to add more mass to get it to the stop at 90?, but at 112.8g the 90? stop has been achieved and the method is not telling you to add or subtract any mass. The cylinder and spheres experiment responds to momentum.

Another experiment was conducted using the results from the 3 in. pipe add-on as a base line, and comparing the results collected by connecting a larger diameter cylinder onto the bottom of the top coupler. After being spun the top cylinder is forced into a stop when the spheres swing out to 90?. This is accomplished when a mass of 364.9 g 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 times 364.9, which is 279g. I then placed a 279.1g pipe on the bottom of the top cylinder and began video taping it while spinning and releasing it by hand. Once the appropriate mass is determined the cylinder will stop at 90? at any RPS, being hand held offers no real problem, in fact it proves that the RPS does not matter. And the hand held models are much more readily available for physics class demonstrations. 

At 279.1g the cylinder had a slight backward motion, while the string was in the slit at 90?, that indicates that it is a little light. I added 7.5g and the forward motion was greater than the previous backward motion. That means that the 279.1g 4.0 pipe was about 3g light, 3/279 is about 1% off.  This is the third object (added to the bottom of the top coupler) with different diameters that have the same rotational momentum, and they all cause the same 90? stop. If graphed this would be three points on the same line, a line that indicates that the cylinder and spheres experiment is driven by the Law of Conservation of Momentum.

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 for the cylinder and spheres experiments 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 method of adding different diameters is an entirely different and second method that confirms in my mind that the cylinder and spheres conserves momentum and can increase energy by 300% 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 R.P.S. used if the proper amount of mass at a certain diameter has been added to the bottom of the top coupler. Three proper amounts of mass at a certain diameter have been used to stop the top cylinder when the spheres are at 90? to tangent.

The cylinder will stop with any initial R.P.S. given it by the hand release method. I am going to use 2 rotations per second to get some actual numbers. I think 2 R.P.S. is within the ability of the hand release method.   

A 3.0 inch inside diameter PVC pipe (1/4 in. side wall) with a length just under 10 inches and with a mass of 364.9g 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: 364.9g (3in. I.D. pipe)

Momentum: (mv) 364.9g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .18926

Kinetic energy: (1/2mv?) .5 * 364.9g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .04908


A 3.5 inch inside diameter PVC pipe coupler with a mass of 218.9g combined with a 112.8g connecting 3.0 in. pipe gives the cylinder a nice stop when the spheres are at 90?. The effect rotational mass of the coupler is at about 3.75 in. and the effective rotational mass of the connecting pipe at a 3.25 in. diameter.  Here is the math used for the quantities of motion mentioned.

Mass: 218.9 for the coupler and 112.8 for the connecting pipe

Momentum: (mv) 218.9g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .13101

Momentum: (mv) 112.8g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .058506   Total equals .189516

Kinetic energy: (1/2mv?) .5 * 218.9g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .039202

Kinetic energy: (1/2mv?) .5 * 112.8g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .015173           Total = .05437

Momentum change from the 3.0 in. pipe is .189516/.18926 = 0.1%

Kinetic energy change from the 3 in. pipe is .05437/.04908 = 11%

A 4.0 inch inside diameter PVC pipe with a mass of 283g 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: 283g   (4inch added cylinder)

Momentum: (mv) 283g * 1kg/1000g* 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .19195

Kinetic energy: (1/2mv?) .5 * 283g * 1kg/1000g * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 4.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .06510


Momentum changed from the 3.0 pipe was .19195/.18926 = 1%

The kinetic energy change from the 3.0 in. pipe was .06510/.04908 = 33%

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 at 90? 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 fall outside the 2% at 11% and 33%. Kinetic energy is not conserved by the cylinder and spheres experiment.

If the initial momentum of the cylinder and spheres is conserved in the motion of the spheres alone (when the cylinder is stopped) then the systems has over 400% of the original energy.

Momentum: (mv) 208.6g * 1kg/1000g* 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .12484  (top 3.5 in. I.D. coupler with holes for the string and holes to seat the spheres)

Momentum: (mv) 22.2 g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .011515 (small piece of 3 inch I.D. pipe used to seat spheres)

Momentum: (mv) 364.9g * 1kg/1000g* 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .18926  (3 in. PVC pipe about 9.8 inch long)

Momentum: (mv) 132g * 1kg/1000g* 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .084265  (spheres seated on the surface of the coupler, 66g each)

Total initial momentum = .40988

Final momentum will be .40988 for a velocity of 3.1 m/sec   .40988/.132 = 3.1 for the spheres only

Kinetic energy: (1/2mv?) .5 * 208.6g * 1kg/1000g * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.75in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .037356

Kinetic energy: (1/2mv?) .5 * 22.2g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .002986

Kinetic energy: (1/2mv?) .5 * 364.9g * 1kg/1000g * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 3.25in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps = .04908

Kinetic energy: (1/2mv?) .5 * 132g * 1kg/1000g * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps * 4.0in. * 2.54 cm/in *1m/100cm * 3.14159 * 2rps =.026896

Total Initial Kinetic energy = .11632

Final Kinetic energy will be .5 times .132kg times 3.1 m/sec times 3.1 m/sec (1/2mv?) = .63426

This is a kinetic energy change of .63426/.11632       545%   

The change in achievable rise (d =1/2 v?/a) of the initial moving objects (spheres, coupler, pipe) to the final moving objects (spheres) will also be proportional to the change in kinetic energy and is called potential energy.

Offline supersam

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Re: Free energy from gravitation using Newtonian Physic
« Reply #14 on: April 24, 2007, 01:49:20 AM »
pequaide,

please don't take this the wrong way. i am just a dumb ironworker with too much time on my hands.  however much i would like to do your experiment,  it looks fascinating, i am having trouble figuring out, how all these parts are supposed to come together.  i would be more than willing to duplicate your $25 experiment as all it would cost me is time.  i would also like to see your $1,000 experiment because alot of the materials may be available to me from scrap on the construction sites that i work at.  and if i need special machineing of parts, that might also be available from friends, in there spare time. 

but i am having trouble grasping the really intense concept that you have put out so far with my intelectual ability.  can you make it any simpilar for people that have the physical and mechanical skills to get the job done, but don't have the physics and math capabilities, along with the engineering ability that you have?

sorry to be a dumbass
lol
sam

ps.  maybe this is over my head conceptually, but if you can just draw me a print i can build you the world.

 

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