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Author Topic: Sum of torque  (Read 120360 times)

Offline EOW

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Re: Sum of torque
« Reply #255 on: October 29, 2015, 10:23:15 AM »
At start, I want the same angular velocity of the spiral1 and the spiral2, but like the diameter is not the same I need to increase the diameter of the tube. Increase the diameter cost nothing in theory. Like that the arm has a torque on it. The angular velocity of each spiral at start is -w.cos(a). The Spiral1 has a counterclockwise torque on it, so it will accelerate in the laboratory reference counterclockwise. The Spiral2 has a clocckwise torque on it so it will accelerate in the lab reference clockwise. In the arm reference, the Spiral1 accelerates from -w.cos(a) to X and the Spiral2 deccelerates from -w.cos(a) to Y, but radius of each spiral change in the same time and it's possible to adjust the thickness of the tube when it goes from the Spiral1 to the Spiral2.

This device must works with the angle a=0: axis of spiral are vertical.

I can use a full material for the tube with a low mass, it's easier to see the volume can be constant from Spiral1 to Spiral2.
« Last Edit: October 29, 2015, 09:24:15 PM by EOW »

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Re: Sum of torque
« Reply #255 on: October 29, 2015, 10:23:15 AM »

Offline EOW

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Re: Sum of torque
« Reply #256 on: October 31, 2015, 09:52:40 PM »
Take the example in the image. There is no arm, just 2 pulleys and one tube. I want to have the same angular velocity for the Pulley1 and Pulley2 even the radius of the Pulley1 is twice than the Pulley2. The pulleys turn counterclockwise. There is pressure from a gas outside at P=100000Pa and inside the closed tube there is p=1000 Pa. The Pulley1 receives a counterclockse torque from F1, the Pulley2 receives a clockwise torque from the force F2. Like I want the same angular velocity if the section (surface) of the tube when it is on the Pulley1 is 2mm * 2 mm. I need to keep constant the surface of the material, it is 4 surfaces by 2*2 = 16 mm². Like the Pulley1 has a radius twice than the Pulley1 I need to move 16 in surface but the length of the arc is 2 times lower it's not 2 mm but 1 mm, so if I change only 2 surfaces, I need to have 2 surfaces at 2*1 and 2 surfaces at 6*1 = 16. So the front surface from F1 is 2*2=4 and from F2 it's 1*6=6. So the Force F2=1.5*F1. Or if I change 4 surfaces I will have 4 surfaces at 4mm*4mm.

The sum of the energy:

1/ The Pulley1 has a counterclockwise torque from F1 the energy is 2*R*F1*wt=8(P-p)Rwt, where (P-p) is the difference of the pressure
2/ The Pulley2 has a clockwise torque from F2 the energy is -R*F2*wt=-16(P-p)Rwt
3/ The volume of the tube increases like 8(P-p)Rwt
Note: move the wall don't need any energy because the force from pressure is perpendicular to the movement except where there are F1 and F2.

Even the sum of energy is constant do that:

Now place all this device on an arm that turns at w. The Pulley1 and the Pulley2 turn at -w in the reference arm. But look at the torque on the arm, it is positive. And the Pulley2 receives a clockwise torque but it turns at -w in the arm reference not in the lab reference ! so it decelerates in the arm reference (accelerates in the lab reference).

The sum can't be constant.

Offline EOW

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Re: Sum of torque
« Reply #257 on: November 02, 2015, 10:12:32 PM »
In the last example, the volume of the tube increase but it's possible to let it constant, for that I need to enter the tube inside itself. Like that repeat the cycle is easier.

I place the center of rotation of the device in the red center like that the force from pressure due to the conic section can't give a torque.

The pulley1 can support the tube in the upper side of the magenta line (third image) and the pulley2 can support the lower side of the magenta line. Like that there is a torque on the arm.

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Re: Sum of torque
« Reply #257 on: November 02, 2015, 10:12:32 PM »
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Offline EOW

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Re: Sum of torque
« Reply #258 on: November 03, 2015, 07:20:18 PM »
I change the direction of the force, like that the arm receives a clockwise torque. The volume of the tube is constant so I don't lost any potential energy. Each pulley will accelerate in the lab reference (decelerate in the arm reference). Like that the difference is the bigger pulley lost more energy more quickly (in the arm reference) but I don't gave any energy in fact :)

Example:

Radius of Pulley1 = 1000 mm
Radius of Pulley2 = 500 mm

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 7.4641 mm * 0.535898 mm = 4 mm² (resolv 2x+2y=4 and xy=4)

The tube don't lost any potential energy. The arm don't receive a torque.

Example2:

Radius of Pulley1 = 1000 mm
Radius of Pulley2 = 1000 mm (same radius)

Surface of tube in the Pulley1 = 2 mm * 2 mm = 4 mm ²
Surface of tube in the Pulley2 = 0.472136 mm * 16.9492 mm = 8 mm² (resolv 2x+2y=8 and xy=8)

Here the volume of the tube increases and the arm receives a clockwise torque.
« Last Edit: November 03, 2015, 11:21:55 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #259 on: November 04, 2015, 10:56:38 AM »
At start the squares don't turn around the blue center. The arm turns at w. The blue walls are supported by the squares so each square accelerates counterclockwise. The red center supports the red walls, in one turn of the squares, the torque could be at 0 on the arm, but not because the squares increase their angular velocities and there are less and less time for the torque. So the arm accelerates clockwise and the squares accelerate counterclockwise. Imagine with the thickness T near 0.

Outside the device there is a small pressure.
Inside the blue/red shape there is a bigger pressure from a gas for example.

The pressure at 100000Pa is only in the shape composed by 2 blue walls and 2 red walls (small volume because T is near 0). This shape gives the forces F1 to F4.

I give the Python code to test. With k=0 (no acceleration) there is no torque on the arm. But with k=0.2 for example there is a torque of 58 in one turn of the squares.

The squares turn together with the same angular velocity, they keep constant their relative position from each other.
« Last Edit: November 04, 2015, 09:30:05 PM by EOW »

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Re: Sum of torque
« Reply #259 on: November 04, 2015, 10:56:38 AM »
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Offline EOW

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Re: Sum of torque
« Reply #260 on: November 04, 2015, 11:07:14 PM »
Another position,the arm turned of 45° the squares too

Offline EOW

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Re: Sum of torque
« Reply #261 on: November 05, 2015, 04:55:25 PM »
I change the gas every 90°.
« Last Edit: November 05, 2015, 07:58:22 PM by EOW »

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Re: Sum of torque
« Reply #261 on: November 05, 2015, 04:55:25 PM »
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Offline EOW

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Re: Sum of torque
« Reply #262 on: November 06, 2015, 07:01:47 PM »
This device don't create energy, just use the energy from the temperature of a gas (pressure). The arm turns at w clockwise. The disk don't turn around itself. The pneumatic cylinder has a spring inside and has the pressure P everywhere except between the cylinder and the disk. The disk decreases its angular velocity in the arm reference and the cylinder compress the spring. The temperature of the gas decreases. I drawn the device at the time the pneumatic cylinder must work, exactly at this time, it's a transcient action.

Second image: all the device turns clockwise around the red center. The square can turn around the blue center. The device don't receive a torque but the square receives one. The square receives the forces Fa and Fb and Fc, these forces give F2 on the blue center. Inside the red circle there is no pressure.

The torque on the red circle from F1 is:

C1=(R1+R2/2)*R2*P  (I suppose the depth is 1)

The torque on the red circle from F2 from the blue center is:

C2=R1*R2*P+(R1+(R2²-2R1²)/2/R1)R1+ (R1-(R1+(R2²-2R1²)/2/R1))*R1

The square in the contrary receives a torque of:

C3=(R1+(R2²-2R1²)/2/R1)R2*P

Numerical application:

R1=10
R2=1
C1=10.5
C2=10.50012
C3=0.05

If the red circle accelerates more and more, the square keeps its relative position with the red circle, so the square is always like I drawn. I need to give an energy to turn the red circle (the device) but I win the energy of the rotation of the square around itself.
« Last Edit: November 06, 2015, 10:59:26 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #263 on: November 07, 2015, 09:21:30 PM »
I can use one or 2 half torus. Outside the pressure is near 0 and inside the grey container there is 1bar. The arm turns clockwise at w. At start the half torus don't turn around itself so it turn at -w in the arm reference.

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Re: Sum of torque
« Reply #263 on: November 07, 2015, 09:21:30 PM »
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Offline EOW

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Re: Sum of torque
« Reply #264 on: November 08, 2015, 07:15:45 PM »
Shows the image
« Last Edit: November 08, 2015, 10:59:18 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #265 on: November 09, 2015, 10:30:28 PM »
The torque on the half disk is 2 times the torque on the device.

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Re: Sum of torque
« Reply #265 on: November 09, 2015, 10:30:28 PM »
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Offline EOW

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Re: Sum of torque
« Reply #266 on: November 10, 2015, 07:13:15 PM »
The device is always like I drawn. F1/F1' gives counterclockwise torque. F2/F2' gives a counterclockwise torque. The only problem is springs rotates the torus in the clockwise direction, so I will attract the torus 1 with the container 2 and attract the torus2 with the container1. I can't attract all the part of the torus but the bigger part so the torque is counterclockwise.
« Last Edit: November 11, 2015, 12:17:17 AM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #267 on: November 11, 2015, 08:36:05 AM »
I hatching the area where I exchange the attraction (transparency mode):

Container 1 attract the torus 2
Container 2 attract the torus 1

Before the green line attract the green torus and the red line attract the red torus. But now, the green line attract the red torus where the hatched area is (don't forget there are 2 torus) and the red line attract the green torus where the area is hatched. The green line attract the rest of the green torus (not hatched area), and the red line attract the rest of the red torus like that there is no force on the green line and on the red line. Like that I don't have any force on the red line and on the green line.

I drawn container1+torus1 on one image and the same for the container2+torus2.

With one device the sum of energy is keeped. F1 and F2 want to turn counterclockwise but the attraction from the springs on the torus want to turn clockwise, so the sum is constant.

With 2 devices, I attract from the green line a part of the torus1 and a part of the torus2. The torque on the torus1 is near 0 or maybe it's counterclowise when I see the surface of the hatching compared of the all surface inside the container. It's the same for the torus2.

The device turns around the red center, it is always like I drawn: it's important to keep constant the length of the springs. The springs don't lost any energy.
« Last Edit: November 11, 2015, 07:44:36 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #268 on: November 12, 2015, 02:26:02 PM »
The green line is fixed to the Container1, the red line is fixed to the Container2.
I don't attract all the torus, only the part that is inside the Container1 or the Container2.
The springs keep their length constant, the device turns around the red axis but all turns at the same angular velocity.
« Last Edit: November 12, 2015, 05:49:16 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #269 on: November 13, 2015, 07:56:33 PM »
The device can be limited to one torus and one container. But the walls of the torus are separate of the interior torus. The inner wall and the outer wall of the torus is in contact with pressure, this pressure can't give any torque to the torus but gives the force F1 and F1 gives a counterclockwise torque on the device. Now, when I attract the interior of the torus it will rotates clockwise, but it's not the same energy I lost. With inner radius at 7 and outer radius at 8.72 the torque on the interior torus is 11.8 but the force F1 gives a torque of 17 and the force on the walls of the container gives a counterclockwise torque too.

At start, all the device turns counterclockwise.

 

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