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  #1  
Old 01-27-2017, 09:33 PM
Jeanp007 Jeanp007 is offline
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Free Energy Device Successfully Invented

Hello,

I am pleased to present to this forum a complete description of a device I invented that will successfully generate free energy for scrutiny. How this device functions? You know that energy is defined as the work done by a force. This means that if we could have an unbalanced or loose force that is capable of performing work continuously and indefinitely, we have free energy.

How to obtain that unbalanced force? Let say we have an object on which two equal forces are acting in opposite directions (This object could be a cylindrical container full of a pressurized gas. The gas is applying two equal and opposite forces on the base and on the top of the cylindrical container). If we could reverse one of the two forces acting on the object and have the two forces acting on the object in the same direction, we will have a net force or resultant force acting on the object and this resultant force will cause the object to move, producing energy.

Because the system used to reverse one of the two forces acting on the object need help from gravity, the object on which a force reversal has been performed will be able to move in the vertical direction only (up and down). This object will then have an increased weight compared to an identical object on which the force reversal mechanism has been disabled.

By mounting the two objects on a crankshaft, the object on which the force reversal is active will be heavier and it will be able to lift the object on which the force reversal has been disabled. The crankshaft will then spin, propelled by the difference of weight between the two objects. By alternating the force reversal between the two objects, the crankshaft will continue to spin and the force spinning it can be used to supply mechanical energy to a generator and produce electricity or be used to put in motion an automobile.

The link below lead to a PDF file that contains a detailed description of the process described above and everything is explained, with all the drawings and mathematical calculations provided. As you read that file, what you will need to scrutinize closely is the method used to reverse a force acting on an object, without the need for an outside anchoring object. If you agree with the method used to reverse the force, then we have free energy.

I have to tell you also that I have been able to have the proposed design scrutinized by one expert in physics, and this physicist could not find a reason why the device wouldn’t functions as expected. This physicist used to teach mechanical engineering courses at one famous American university.

You may be asking yourself why I decided to publicly disclose this invention this way: Simply put, I have been unable to find any investor who could accept to fund this invention in order to bring it to the market. I have come to a final design for this device several years ago and all my efforts to find an investor have failed because of the general skepticism about this kind of invention by the scientific community. I hope that with the help of the members of this forum we can be able to find a way to move forward with this breakthrough invention, which has the potential of changing the course of humanity.

I have applied for a provisional patent for this invention in the US.

The following link will leads to a full disclosure of this invention. It is better to open it on a PC or a Tablet because the drawings are too big for a Smartphone.
Free energy is here!!!
https://drive.google.com/file/d/0B87...ew?usp=sharing
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  #2  
Old 01-28-2017, 03:39 AM
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Do you have a prototype?
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Old 01-28-2017, 03:45 AM
Jeanp007 Jeanp007 is offline
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I don't have a prototype

I don't have a prototype because I don't have enough money to build one. I hope to find an investor who can provide enough money to build one. What is your impression after reading that theory?
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Old 01-28-2017, 05:51 AM
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this does not seem like a super pricey device to build
and seems kind of like the divide by zero errors that lead to infinite energy predictions
what part of it seems out of your budget ?
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Old 01-28-2017, 07:32 AM
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  #6  
Old 01-28-2017, 08:02 AM
Dingus Dingus is online now
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Quote:
Originally Posted by Jeanp007 View Post
I don't have a prototype because I don't have enough money to build one. I hope to find an investor who can provide enough money to build one. What is your impression after reading that theory?
My impression is that unless you have a working prototype, your theories aren't worth squat.
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Old 01-28-2017, 03:45 PM
wrtner wrtner is online now
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Originally Posted by Dingus View Post
My impression is that unless you have a working prototype, your theories aren't worth squat.
Not true. Reference: Einstein
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Old 01-28-2017, 05:00 PM
Jeanp007 Jeanp007 is offline
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Quote:
Originally Posted by spacecase0 View Post
this does not seem like a super pricey device to build
and seems kind of like the divide by zero errors that lead to infinite energy predictions
what part of it seems out of your budget ?
All the components of the device have to be custom made and as you begin to design, you realize that the device will have more than 50 different components.
I believe it can take around $40,000 to build a good size working model.
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Old 01-30-2017, 06:01 PM
Jeanp007 Jeanp007 is offline
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Looking for an investor

Hey Guys! Does anyone know an investor who could be interested in this invention? I am looking for an investor who could team with me in order to move forward with this invention.
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Old 01-31-2017, 01:12 AM
ricards ricards is offline
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If it works the way you theorize, It you could find an Investor, If he fund you to build it in Industrial scale, do you think its going to be "FREE"?. I doubt that, 40k US dollar is no joke for a prototype.

Even if you ever succeed, finding an Investor and building something up into industrial scale, there is still the Oil & Gas to compete to and not to mention other "FREE" Energy like Solar, Wind, Geothermal, Tidal. etc.. which is already up and running.

what I'm trying to say is, Its not enough to be like Einstein and Know it all, you have to be practical as well. your paper is good (probably too good for me), but the world isn't as nice as you.

Our ability to judge people is only based on our experience with other people and our own behavior, If you think people are so nice to fund your expensive project for the sake of "Rocket Science", I think you are a very nice person for just thinking that.

Overall I think your work is worth a look, but just too technical for me.
I apologize for the harsh words but I could find other words to describe it.
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Old 01-31-2017, 05:23 AM
Jeanp007 Jeanp007 is offline
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The proposed device is cheap to manufacture massively

Ricards,

Although I estimated that a good size working model would cost $40,000 to make, I have to mention also that if the same working model was manufactured massively, it would cost less than $200 to make a unit. The cost of making a single working model is so high because machine shop are simply too expensive. For example a single part that you can buy off the shelves for $ 10 can easily cost $500 if custom made by a machine shop.

Also, the proposed device is highly competitive compared to current energy technologies. For example, this device functioning as an automobile engine would cost less than $300 if produced massively. As you can see, an automobile engine that cost only $300 to make and which doesn't need fuel to operate is highly competitive compared to today's engine technologies.

The same is true for a generator that produce electricity. A 15kw generator based on this technology can cost $1000 to manufacture and it can be easily sold to the public for $3500. I don't see how any technology used currently to generate electricity can compete with the new technology I am proposing.

Any one who may know an investor who could be interested in this breakthrough technology?
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Old 01-31-2017, 12:59 PM
Dingus Dingus is online now
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Originally Posted by wrtner View Post
Not true. Reference: Einstein
You're no Einstein. If you had that level of credibility, people would be seeking you out to make your inventions, not the other way around.
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Old 01-31-2017, 03:05 PM
ricards ricards is offline
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40k for custom? then 200$ if manufactured? seems to me now you're just guessing the numbers.

you mentioned machining, are you really that certain that you're theory is right? what if It does not work the way you theorize? you're putting a big risk to your investor, that is not the way to get their attention, Investor do not want to spend their money without the assurance of getting it back bigger. have we not learned about JP Morgan what he did with Tesla? If you do not know of these men I suggest you research on them too.

As what I have advised "Be Practical" try not to perfect things on your first model It would really cost a lot, focus on "making it work first", this way you wont be asking that much, you can even fund yourself just to prove your theory. Heck If you could theorize something like what you presented, I'm pretty sure you can think of a different design using cheap materials and still apply your theory.

most people here are working on a self-sustaining machine and doing that by sustaining themselves (see the relationship?), very few (yourself included) ask for financial help. you should start by proving your theory first and build a working cheap prototype that you can show anyone. Trust me you wont just attract alot Potential Investors, you will also attract unwanted attentions (if you know what I mean ).

If you still want to have it your way, (find an investor) I must say you have come to the wrong place.

peace out.
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Old 01-31-2017, 04:53 PM
T-1000 T-1000 is offline
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Quote:
Originally Posted by Jeanp007 View Post
Hi,

This shared document was mentioned in other forum and I would like to forward quick question which was raised after looking on the document:

The mechanical device in the link looks OK until water runs out in the columns. I quickly checked whole document and do not see anything about water replenishment. Did I miss something? It would consume energy produced when moving water back upwards...

Cheers!
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Old 02-01-2017, 12:25 AM
Jeanp007 Jeanp007 is offline
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There is no movement of water inside the container

The water inside the container does not move at all, because the piston is immobile inside the container. What happen is the upward force applied by the water on the piston is reversed and made to act on the container in downward direction.

This container then become heavier than the other container which has the force reversal disabled. It is the difference of weight between the two containers that produce useful energy; as the heavier container descends, lifting the lighter container.

Please read carefully the theory and you will understand better this process.
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Old 02-01-2017, 12:36 PM
citfta citfta is offline
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A mathematician has pointed out on another forum the fundamental flaw in your calculations. On page seven diagram two you show a bent lever and forces applied to that lever and from that you calculate the downward force on the fulcrum. Those calculations are incorrect. The force Fs cannot just be added to the force Wm to get the downward force on the fulcrum. The force Fs is actually adding an upward force to the lever because of the backward bend to the lever. You need to review the formula for calculating vector forces. Sorry but your device cannot work to produce OU. Any physics student can do the math for you so that is probably why you can't find an investor.

Respectfully,
Carroll
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Old 02-01-2017, 08:51 PM
Jeanp007 Jeanp007 is offline
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A brief review of how levers works

Hello Carroll,

Are you saying that the force Fs can never lift the Mass m on Diagram 2? If the force Fs lift the Mass m, What would be the downward force acting at the fulcrum of the lever according to your mathematician? The link below will show you a more elaborate analysis of levers mechanism. Please review it and then let me know what you think.

https://drive.google.com/file/d/0B87...ew?usp=sharing
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Old 02-01-2017, 09:26 PM
citfta citfta is offline
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The lever example I gave is only one of the many mistakes in your PDF. Another mistake is confusing the pressure inside a container with the weight of the container. Changing the pressure will not change the weight.

Let me give you a very simple test you can perform to prove this to yourself. Weigh a couple of quarts of water. Be sure and use a short and wide container and a very accurate scale. Now put that same water into a very tall and narrow container and measure only the added weight of the water.

If I understand your theory the taller container will have a higher pressure at the bottom and will therefore weigh more. I do know from experience the pressure at the bottom of the taller container will be higher. But the weight of the water will be exactly the same.

Respectfully,
Carroll
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Old 02-01-2017, 10:47 PM
Jeanp007 Jeanp007 is offline
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Pressure is force per unit of area

Carroll,

I do not confuse pressure and force. Pressure is defined as force per unit of area. The unit of pressure is normally the PSI which means "pounds per square inch". If you put the same quantity of water in two cylindrical containers that have different are of the base; the two containers will still have of course the same weight. This is because the water level in the container with the bigger area of base will be lower, meaning that the pressure at the base of this container will be lower as well. If you multiply the pressure at the bottom with the area of the base, you will have the weight of that container and this will be the weight seen if you put that container on the scale.

Now, it you put the same quantity of water in the container that has a smaller area of the base, the water level in that container will be higher than the case of the first container. In this case, we will then have a higher pressure of the water on the base of this narrower container. However, if we multiply the pressure at the bottom of this container with the area of the bottom of this container, we will have the same value for the weight of the container as in the first case. This is because although the pressure at the bottom will be higher than in the first case, this pressure will be acting on a smaller area of the base and the product :" Pressure*area of base" will have the same value for both containers.

I have to tell you also that my device function by reversing a force using a mechanism that receives help from gravity, and it doesn't have anything to do with pressure as you mentioned it.
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Old 02-02-2017, 01:45 AM
citfta citfta is offline
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This will be my last attempt to show you why your device will not work like you think it will. The following are only some of the reasons why it will not work. This discussion is referring to figure 2 on page 9.

After you open the door to allow the pressure of the column of water to enter the lower section that pressure will always be there. Closing the door again will not remove the pressure. So at that point your device is now inert.

The masses on each side of the crossbar will always weigh the same no matter where you move them to. You are just changing position not the weight. Shifting the weight to the fulcrum does not change the weight of the total device.

Adding the bearing to the L shaped piece does not change the vector angles. You still have a situation where the total weight on the left side of the crossbar is equal to the total weight on the right side of the crossbar.

Moving the water up and down against the pistons does not change the weight of the total device on either side of the crossbar.

And how is the water supposed to get back above the door so the pistons can go back down?

Carroll
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Old 02-02-2017, 03:11 AM
Jeanp007 Jeanp007 is offline
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Carroll,

What you just said about opening and closing the door would be true if the container was full of a gas. But because water is incompressible, when the door is closed, the water above the door is totally separated from the water below the door. In this case, you can even remove the V-shaped lever and the Mass m and the piston will remain in place and will not move upward.

But if you remove the Mass m while the door is open, the piston will pop-up out of the container and the water will spill from the container.

Now, what you said about the side masses show that you don't understand correctly the physics of levers. Did you read that analysis I made for you about levers mechanism in previous posts?

Also, you talked about moving the water up and down against the piston and the water having to go back above the door: This show me that you did not carefully read my theory. I stated very well in my theory that the water does not move at all inside the container. What I do is reversing one force that was applying a lifting effect on the container and have it acting on the container in a downward direction;increasing this way the weight of the container. what you need to analyze is whether the force reversal has been achieved successfully.

Finally, I have to tell you that I have been able to have the proposed design scrutinized by a physicist who used to teach mechanical engineering courses at one prestigious American university; and this physicist could not find a reason why the device wouldn't function as expected.

Cordially,

Jeanp007
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Old 02-02-2017, 12:40 PM
citfta citfta is offline
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You are still ignoring a couple of things. Here is a copy and paste from the math guy to better explain it than I could.

Quote:
If the arm y has been bent though an angle theta (in your diagram 2 that is an angle of about 120 degrees) then your force Fs has both a horizontal and a vertical component. Its downward component is Fs*cos(theta). Hence the downward force at the fulcrum is Wm+Fs*cos(theta). If theta was 90 degrees Fs has no downward component, and the downward force at the fulcrum would simply be Wm. However there is a sideways force Fs*sin(theta) and if the fulcrum point is not a fixed bearing (i.e. its is like the knife edge illustrated) then the lever will tend to slide sideways. When Fs*cos(theta)=Wm there is zero force on the fulcrum leaving only the sideways force. Since as illustrated your x is greater than y, and the Force Fs acts at a distance from the fulcrum point slightly less that y, to achieve balance Fs is greater than Wm. At an angle where Fs*cos(theta)>Wm the lever will move off the knife edge and be propelled into space by the applied force. It becomes quite clear if theta is 180 degrees (the lever is bent right back on itself) then Fs acts vertically upwards and the lever takes off in an upward direction, or if the fulcrum is a fixed bearing, there is now an upward force on that bearing, and to prevent take off the bearing needs to connect to ground.

You have not summed your forces vectorally, and that is your mistake. You have just assumed that the balance situation for your bent lever system will apply the same weight at the fulcrum as the case for a straight lever, and that is wrong. End of Quote.

You are also neglecting the fact that as the water pushes against the piston to apply pressure to the fulcrum the fulcrum is also applying an opposite force back to the piston and the water. Equal and opposite reaction.

Carroll
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Old 02-02-2017, 04:33 PM
Jeanp007 Jeanp007 is offline
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A mathematician and not a physicist.

Hello Carroll,

I am sorry to tell you this but your mathematician is indeed a mathematician and is not a physicist nor an engineer. All the analysis he made about that lever is totally wrong and I don't have anywhere to start explaining his errors.

So, I am going to propose you this: Try to find someone who is a physicist or a mechanical engineer and show him the analysis I made about levers physics and this analysis made by this mathematician and then ask him who does he think is right.

I think that this is the only way you can know the truth about this.

Cordially,

Jeanp007
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Old 02-02-2017, 05:06 PM
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Am I missing something?

Hello Jean,

You're telling us that if we put your apparatus (say the assembly on the left side of the beam) with a remote control for the pink door (valve) in a sealed box and put it on a scale that you can change the weight by simply actuating the valve.

I don't think so.

Regards,

bi
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Old 02-02-2017, 06:22 PM
Jeanp007 Jeanp007 is offline
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Yes, if you put on container assembly on a balance or a scale, you will see that the container has an increased weight when the door is open than when the door is closed.

When the door is open, the container will weight Wc= F+Fa+ 2Wm and when the door is closed the container will weight Wc'= F-Fa+2 Wm
with F= Pi*H*R^2 and Fa= Pi*h*(R^2-r^2) The difference of weight will then be Wc-Wc'= 2Fa

It is this force 2Fa that will do work and produce energy. You will have to read carefully the theory, without overlooking anything, in order to realize that this is the case however.

As you read the theory, you must pay closer attention to the way I use to reverse the force Fa because this the key to this device.

The link below will show you an elaborated analysis of lever physics and this should be helpful for you to understand better my theory. Remember that the V-shaped lever is the key to this design.

https://drive.google.com/file/d/0B87...ew?usp=sharing
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Old 02-06-2017, 09:04 AM
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Jeanp007
Don't pay attention to certain individuals here in this forum.
They are notorious big mouth, and members of same brotherhood.

Beside that, even though your Idea looks impossible, if you could make some small prototype as proof of the concept,or something like that.
That would go far.
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