Free Energy Device Successfully Invented

[citfta]

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

[Jeanp007]

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

[citfta]

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

[Jeanp007]

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.

[citfta]

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

[Jeanp007]

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

[citfta]

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

[Jeanp007]

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

[bistander]

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

[Jeanp007]

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

[padova]

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.