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### Author Topic: Sum of torque  (Read 121901 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #180 on: June 02, 2015, 01:16:48 AM »
I prefer a liquid it must be more efficiency. Let the door open and like the pressure moves all the time this will give a torque on the device, like that ? Yes, but I'm not sure of:

1/ water will move because it is compressible, even a small volume of water, this amount can't be ignored, but the energy seems very low
2/ an object rotates => no torque, so this is the delay of pressure to be stabilized that can give a torque, here with the door always open, the pressure change all the time. Look at positions: it's necessary to close door and reopened because the torque is counterclockwise and after clockwise, are you agree ?

I drawn the device, and I think there is one wave inside the upper circle and another wave in the lower circle. In the upper the pressure increases, in the lower the pressure decreases, a part of time, and this must give forces F1 to F4. Length of arm can be higher. The material must absord the wave for prevent round trips of waves.

have you tested on a software ? Maybe if you draw something for understand why you want to recover energy from this device I could understand.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #180 on: June 02, 2015, 01:16:48 AM »

#### lumen

• Hero Member
• Posts: 1372
##### Re: Sum of torque
« Reply #181 on: June 02, 2015, 02:31:28 AM »
Ok, you are saying that the release of the pressure into the second chamber causes movement and then the centrifugal force would cause the next compression cycle and the release would be delayed until a time when the release would again act to increase the speed. The delay of the gas may cause a shift in weight that works against the rotation like shifting your weight on a swing to slow yourself down.

Would this be easier if the centrifugal force was used instead to rotate a generator and some of the power was used to increase the speed of the disk.
At some point there would be excess power that could be directly used for other purposes and the design would be achieving the same goal using the same forces but would be easier to design and could operate at lower speeds.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #182 on: June 02, 2015, 09:48:08 AM »
There is no gravity, so no weight. For recover energy, maybe turn the device at the max angular velocity possible to do, and recover energy at constant velocity is better. With a constant velocity the device must be more stable.

Maybe it's possible to use a piezoelectric vibrator for create the waves and test with only one circle. Step2: the valve at left, move out the valve quickly. Step3: return of 180° the device around itself, now the valve is at right. Step4: move in the valve. Step1: return the device around itself of 180°, now the valve is at left. Repeat the cycle.

And you, how do you want to recover energy from this device ? I don't understood your Roberval system
« Last Edit: June 02, 2015, 01:04:28 PM by EOW »

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #182 on: June 02, 2015, 09:48:08 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #183 on: July 19, 2015, 05:00:57 PM »
I don't use gravity here. All volumes are constant. I use only the red shape, I drawn black circle for show the center where I put the springs.

The pressure come from small balls inside the red 'S' shape. The balls are attrack to the center or repuls from it. This gives a pressure p, 2p, 3p on the side where there are the forces F1 and F2. The balls are vey small.

I give the force F1, this need the work W1 and I recover the work W2 from F2. The move like a snake. But for have a continious mouvement, it's necessary to add another part of circle and give the pressure. Put a pressure inside a shape don't need energy in theory.

« Last Edit: July 19, 2015, 09:15:00 PM by EOW »

#### sm0ky2

• Hero Member
• Posts: 3339
##### Re: Sum of torque
« Reply #184 on: July 20, 2015, 03:12:07 AM »
the balls do not behave exactly as a fluid, but much slower. pressure does not have to be constant through the length of the tube.
When pressure varies in one part from the springs, more balls move out of the "higher pressure" area quickly.
this lowers the pressure by a proportionate amount in the part of the tube just before the springs.
meaning, to maintain the original pressure in the rest of the tube, you have to increase the input of balls
to bring the pressure back up by an amount of balls squeezed out by the springs.
the torque comes at a cost.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #184 on: July 20, 2015, 03:12:07 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #185 on: July 20, 2015, 07:41:39 AM »
Right the number of balls of 3p must increase compared to the number of balls of p. Springs are used only for balls that are inside the curved shapes, not in the straight red shape. So the pressure adjust itself in the part where there is no spring: red straigth shape and especially when the curve shape becomes straight, or straight becomes curved. When a space increases between balls of 3p why spring necessary gives the work ? Why it's not a closer ball that can move ?

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #186 on: July 22, 2015, 09:53:38 AM »
The idea is to increase the potential energy of this device. I increase the angular velocity of the torus more and more like that I can keep the relative position of the Object O always. I gave the position after 45°. Inside the Object O there is a lot of small balls, the pressure come from spring attached directly on the Object O. I create pressure on 2 sides of the Object O. Outside there is no pressure  (or near 0). The volume of the Object O is constant, so I don't lost the potential energy of the springs. The torus don't receive a torque from the Object O because it is composed of circle lines. And the sum of forces F1+F2 (in vector) at the center C2 is zero BECAUSE the pressure inside the Object O is adjust for have F1=F2 in value. The center C2 is fixed to the torus. The center C1 is fixed on the ground.

It is very important to have partial circle for 2 ends of Object O, like that there is only the torque from F1 and F2.

It is very important too, to have |F1|=|F2| in value, for that the pressure inside the object must be adjust.

I give the energy for rotate more and more the torus and the Object O on it but the Object O has more and more kinetic energy because it turns more and more too around the center C2. Give for receive

Like it's very important to have F1=F2, it's possible to use gravity but the device will be more complex. Pressure from balls are easier to do. The balls are smaller in reality.

Anf if possible to have F1>F2 like that there is a clockwise torque on the Object O and on the torus too.

The Object O keep the same angle relatively to the torus, so if I'm on the torus I see always the same angle from the torus and the Object O.
« Last Edit: July 22, 2015, 09:52:37 PM by EOW »

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #186 on: July 22, 2015, 09:53:38 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #187 on: July 22, 2015, 10:15:36 PM »
I give an image for show 8 positions of the device. It's possible to show the angle between the Oject O and the torus. Noted that the angular velocity of the torus increases like the angular velocity of the Oject O in the same time. It's very important to keep these two angular velocities exactly the same. The angular velocities I gave are in the laboratory reference (not in the torus for example).

I drawn big balls but they are really small in reality, it's important for prevent another torque on the Object O.

I drawn the device with balls above the torus, but it's a 2D drawing so I added another image with the torus inside the Object O, the torus pass throught the Object O. (S4.png)

I shows on s4.png the difference of the distances d1 and d2. Like the distances are not the same I need to adjust the pressure for have F1=F2 with balls.

I don't drawn the springs that attrack the balls but the springs are fixed on the Object O.

The energy is recover after. This device increases the potential energy. The Object O turns more and more quickly for free around its center of rotation C2. But I need to give an energy for rotates the torus, this energy can be recover later. If F1 can be higher than F2, in this case the torus has a clockwise torque too and it's possible to recover the energy in the same time. Noted that the angular velocity MUST increase more and more.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #188 on: July 23, 2015, 10:57:13 AM »
Pressure with balls is like water and gravity. The pressure is more and more higher at bottom, here the pressure at point p1 is greater than at point p0, so even d2>d1 it's possible to have F1=F2.
Balls are very small compared to the shape, in the contrary another small torque appear on the Object O. If the Object is 1 meter of size balls can be like 1cm. I drawn springs but it can be another technologie like magnets.

Note that even I can have F1>F2 the energy from the torus can't be recover in the same time, it's necessary to increase more and more the angular velocity. And at a time, recover all the energy from the torus and from the rotation of the Object O around its center of rotation C2.
« Last Edit: July 23, 2015, 07:58:55 PM by EOW »

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #188 on: July 23, 2015, 10:57:13 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #189 on: July 24, 2015, 07:47:08 AM »
Or for have F1=F2 in value I can use centrifugal force. I put a liquid inside the Object O. Like the Object O keep its relative position the centrifugal forces are always the same. Or use 2 liquids with different density for have F1=F2 and no other force on the center C2.

Or in one part of the Object 1 put a gas and in other part small balls (or a liquid). The goal is to have F1=F2 or F1>F2.

If I attrack balls from the torus I can have F1=F2 if I use the mean of the pressure. At bottom I will have the same pressure 10 but I can play with the start of the pressure, in one part I start with 0 to 10 and in other part I start from 8 to 10, like that I can have maybe F1>F2. The sum of forces on C2 from the springs is 0, I change just the mean not the final pressure. Balls must be compressible a littlefor transmit pressure in all directions.
« Last Edit: July 24, 2015, 07:02:06 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #190 on: July 25, 2015, 01:43:34 AM »
Maybe with the attraction from springs like that. I can change the shape for find the good forces: the force on C2 is like the yellow force. Here I have 2 areas with 2 different gravities.

F1, F2, F3 come from the springs, note that like there is a curve (torus) the forces from springs are lower than the forces from pressure so I need to add F7, F8 and F9
F4, F5, F6 come from the forces of pressure
F7, F8, F9 come from the forces of pressure, gravity2 is small compared to gravity1

Dotted arrows are the reported forces on C2 or for construct the sum of forces. The torus turns counterclockwise. Don't forget, the torus accelerates more and more for follow the acceleration of the Object. The angular position between the Object and the torus is ALWAYS the same.
« Last Edit: July 25, 2015, 11:22:09 AM by EOW »

#### Free Energy | searching for free energy and discussing free energy

##### Re: Sum of torque
« Reply #190 on: July 25, 2015, 01:43:34 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #191 on: July 26, 2015, 05:54:02 AM »
The torus and the square object turn counterclockwise. The torus turns around C1. C1 is fixed to the ground. The square object turns around C2. C2 is fixed on the torus. The square object if free to turn on the torus. Consider the mass of the springs and the mass of the balls like zero, like that there is no problem with centrifugal forces, or like the position of each ball don't move in reference of the torus it's possible to cancel with an external arm for each ball the centrifugal force.

First image: the square object without the torus for understand how forces are on it. The are two areas, first Attraction1 where the pressure is higher than the other area Attraction2. The springs attrack more and more and give more and more pressure at "bottom" (bottom of the image). The forces F1, F2, F3 com from the springs. The forces F4, F5, F6 come from the pressure of the first area. The forces F7, F8, F9 come from the pressure of the second area. The sum of forces F1+F2+F3 is lower than the sum F4+F5+F6 because the shape of the torus is curved. I drawn small red forces on springs to look how springs attrack, I drawn only 3 springs but there is one spring for one ball. And the balls are very small.

Second image: it's the same but now with the torus. Look at the sum of forces F1+F2+F3+F4+F5+F6+F7+F8+F9 on the center C2: the yellow force. The yellow force don't give a torque on the torus. So I can accelerate like I want the torus  and after recover all this energy. Even, I can set F7+F8+F9 higher for have a counterclockwise torque on it, I lost a part of a torque on the square object but I win a bigger torque because the distance with C1 is higher.

Third image: I drawn the device for 4 positions. The angle between the square object and the torus is always the same. The square object want to turn counterclockwise, the square object accelerate more and more BUT I accelerate with an external motor the torus like that the square object keep its position. The springs don't move inside the square object so the potential energy of the springs is contant. I noted the letter 'a' for look at the position.

I can't recover the energy in the same time, I need to accelerate the torus and recover the energy after. I recover the potential energy from the square object only, it turns more and more.

The device is unstable, it's necessary to control very well the angular acceleration of the torus. And the center C2 must be with friction as lower as possible.

« Last Edit: July 26, 2015, 12:34:58 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #192 on: July 28, 2015, 06:58:19 AM »
I try to explain like I drawn the forces:

Image1: Water with gravity, there are force from pressure at left and at right and forces at bottom (the weight). I did not drawn all forces, just the first, the middle and the last, because there are a lot of forces from pressure.
Image2: I replace gravity+water with springs+balls, like that I have the same forces from pressure at right and at left, sure there is no "weight" F1+F2+F3 cancelled by -F1-F2-F3. I use compressible balls or I change the arrangment like that I have a pressure on the side walls. Springs attrack exactly like gravity or with another law. I did not drawn all springs, but there is one spring for each ball.
Image3: The forces on the square object only (not forces on the torus). There are 2 differents areas where springs don't attrack with the same force, like 2 different gravities. I have forces from pressure at left and at right. I have only the up forces from the springs. I don't have the down forces because there are on the torus.
Image4: Now I drawn the forces on the torus too. There only a point of connection between the torus and the square object, it is the center C2. So, the sum of forces on the square object will be on the center C2. The sum of forces is drawn by the yellow force. The goal is to cancel the torque on the torus, and have the yellow force like I drawn. F4+F5+F6>F1+F2+F3 because there is a curve so it's necessary to have the forces F7+F8+F9.
Image5: Details of forces
Image6: The device
Image7: The device at 4 positions, look at the point 'a' it rotates like the torus. The angular velocity of the torus is exactly the same than the square object.

There is a torque on the square object so it will accelerate more and more, if you let the square object alone on C2 the device will turn a little and springs will lost their potential energy. But if the square object is place on the torus, and it the torus is accelerate more and more EXACTLY like the square object in this case the angle between the square object and the torus is always the same. There is no torque on the torus, look at the yellow force so I can accelerate it without lost an energy, I can recover the energy I give to the torus later.

Cycle: just accelerate the torus for have the same angular velocity of the square object. It necessary to accelerate more and more the torus and never stop until you can.
Recover energy from the rotation of the square object.

I think it's possible to have a better efficiency with a torque on the torus from F7+F8+F9 because when F7+F8+F9 increase it don't decreases the forces F1+F2+F3 but only F4+F5+F6, I lost a part of torque on the square object but I win the torque on the torus and like the radius is higher the efficiency is higher. Or reduce the distance between C1 and C2 like that it could be easier to turn the torus.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #193 on: July 30, 2015, 01:56:45 PM »
I noted:

'c' the lenght of the square
'R' the external radius of the torus
'd' the distance c1c2 x or y axis
Values: c=3 and R=6.5, d=4.34

I consider the pressure like the height of the balls like I can choose with springs.

**************************************************************************CALCULATION*********************************************************************

******************************************Springs**************************************
Start integration x:
s1=d-c/2=2.84

Middle integration x:
sm=d=4.34

End integration x:
s2=d+c/2=5.84

Height:
H=d+c/2=5.84

Integration1:
$$\frac{1}{c/2}\int_{s1}^{sm} (H-\sqrt(R^2-x^2))(d-x) dx$$

Integration2:
$$\frac{1}{c/2}\int_{sm}^{s2} (H-\sqrt(R^2-x^2))(d-x) dx$$

Values: -0.20+1.63

Sum of forces:
$$\frac{1}{c}\int_s1^s2 (H-\sqrt(R^2-x^2)) dx$$

Value: 1.15

******************************************Right side**************************************

$$\frac{1}{c/2}\int_{0}^{c/2} x(-x+\frac{c}{2}) dx$$

$$\frac{1}{c/2}\int_{c/2}^{c} x(x-\frac{c}{2}) dx$$

Values: -0.37+1.87

Sum of forces:
Value: c/2=1.5

******************************************Left side**************************************

I want cancel sum of forces on C2, so I need Fx: 1.5-1.15=0.35

Torque:

Value: -0.35*4.34=-1.5

*******************************************Results**************************************

Torque on the object: -0.20+1.63-0.37+1.87-1.5=1.42

Torque on the support: 0
« Last Edit: July 30, 2015, 11:19:45 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #194 on: August 02, 2015, 12:50:40 AM »
The red part is full with small compressible balls. All springs are attached on C2 and each spring attrack a ball. The force F1 come from the springs. The support (black) turns around C1, C1 is fixed to the ground. The red object turns around C2, C2is fixed on the support.

The support and the object turn at the same angular velocity.

The force F1 don't give a torque to the support. The force F2 don't give a torque to the support. The force F2 give a torque to the object.