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Author Topic: Sum of torque  (Read 119734 times)

Offline EOW

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Re: Sum of torque
« Reply #315 on: June 27, 2016, 07:23:16 PM »
Thanks Dieter, I will try to simulate it because I don't find the error in this device. I'm not sure the device is well explained...
If someone knows maths I will appreciate comments or questions about the device.

When I wrote the wheel moves and rotates, it is at the time I studied the sum of energy. Like the device can accelerate or decelerate it is not easy to count the energy, so it is easier to think with an external device ED that force the wheel to move to the right and rotate in the same time, that device ED recovers an energy for some forces or moments and needs an energy for others forces/moments. I count these energies.

I corrected some errors in my drawings. I gave the sum of energy with formulas.

Axes:
   The green axis is fixed on the wheel

Fixed bodies:
   The ground
   The device that gives the force Fq

Discreet mobile bodies:
   All the device moves in translation and rotates clockwise like a wheel of a bike

   The red wall
   The outer circle of the wheel and the blue wall (blue color = one body)
   N balls (small like molecules of water, just for simplify the calculations), without mass (simplify)
   N springs, without mass (simplify calculations)
   Gaskets (I don't drawn them)

Connections and constraints:
   The red wall can only turn around the green axis, the red wall is in contact with balls only, the red wall receives the pressure from the balls
   Each ball is attracted by a spring from the center C
   The device like that is unstable but I give the force Fq from an external device fixed on the ground.

Constraints on motion :
   When the wheel moves and rotates, all the device turns and moves, the springs never lost their potential energy because the balls are always in the same relative position

Physical properties of all bodies:

   All bodies are rigid except the springs
   The springs and the balls are without mass
   The wheel has a mass.
   Each ball is attracted with the law 1/d² with 'd' the distance between the center of the wheel and the ball


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Re: Sum of torque
« Reply #315 on: June 27, 2016, 07:23:16 PM »

Offline dieter

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Re: Sum of torque
« Reply #316 on: June 29, 2016, 04:42:19 AM »
I was still waiting for you to answer my personal message, so I can mail you the physics sim files, if you want it.

I'm having some access provider troubles here, so I may be offline for a while.

Offline EOW

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Re: Sum of torque
« Reply #317 on: July 01, 2016, 01:01:07 PM »
Sorry, I was busy these last days. I sent a pm.

I find the error for the last device but I try with a triangle. I changed only the shape and the point of attraction. I sent equations later.

Have a good day :)
« Last Edit: July 01, 2016, 07:18:43 PM by EOW »

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Re: Sum of torque
« Reply #317 on: July 01, 2016, 01:01:07 PM »
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Offline EOW

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Re: Sum of torque
« Reply #318 on: July 02, 2016, 10:07:06 AM »
Maybe like that.

Offline EOW

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Re: Sum of torque
« Reply #319 on: July 02, 2016, 11:51:07 PM »
Maybe directly on the circle (no ground)
« Last Edit: July 03, 2016, 10:03:16 AM by EOW »

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Re: Sum of torque
« Reply #319 on: July 02, 2016, 11:51:07 PM »
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Offline EOW

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Re: Sum of torque
« Reply #320 on: July 03, 2016, 08:07:17 PM »
With the wheel in rotation the energy is :

Energy lost by the moment: √3*R*δ*F = 0.001*√3*1*0.45 = 7.79e-4 J
Energy win by the center: √3*R*δ*F = 7.79e-4 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * 2 *F = 3.84 e-4 J

Offline EOW

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Re: Sum of torque
« Reply #321 on: July 05, 2016, 09:38:15 PM »
With a gas under pressure inside the triangle, the sum of energy is not 0:

L = 1 m =  thickness of the device
P = 1 Pa = pressure of the gas
R = 1 m = radius of the wheel

The force on the red arm is √3 N so F2 = √3 N
The torque on the red arm around the red axis is 1.5 Nm

Energy studied for an angle of δ = 0.001 rd

I apply the force F2 from the ground, I lost an energy because the length increases. The force F2 come from a spring for example. I consider there is no acceleration, I can use a big mass for the wheel of use an external device to keep constant the velocity.

Energy lost by the moment on the red axis: √3*R*δ  = √3e-3 J
Energy win by the center in translation: √3*R*δ = √3e-3 J
Energy needed for the force F2: ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) *2*F = 4.27e-4 * √3 = 7.39 e-4 J
« Last Edit: July 06, 2016, 07:07:47 AM by EOW »

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Re: Sum of torque
« Reply #321 on: July 05, 2016, 09:38:15 PM »
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Offline EOW

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Re: Sum of torque
« Reply #322 on: July 11, 2016, 01:47:58 PM »
I don't need the balls and the springs. I can use the pressure of a gas.

The wheel rotates and turns before I studied the sum of energy.

There are 4 bodies:
A: the wheel + 2 red walls, the center of the wheel receives the black force, the outer circle of the wheel receives the red forces
B: the black wall, can turn around the red axis (i), receives the gray forces
C: the spring, gives F2 to the black arm and -F2 to the ground
D: the ground, it is fixed, receives the green force

Radius of the wheel = 1 m
All forces at sqrt(3) N
P = 1 Pa

F1 come from the difference of pressure of the gas on the black wall
F2 come from the spring to the black arm
-F2 come from the spring to the wall
Fi come from the black arm on the axis
-Fi come from the axis to the black arm
Fo come from Fi to the center of the wheel
-Fo come from the center to the outer circle
F4 come from the difference of pressure of the gas on the red walls

I studied the sum of energy for a small angle δ like  0.0001 rd for example

3 walls (2 red+1black) shape a equilateral triangle, there is a gas under pressure at 2P inside the triangle. The surface of the triangle is always constant, the gas never lost its potential energy. Outside the triangle there is the pressure P.

The wheel moves to the right and rotates clockwise like a wheel of a bike, BUT THERE IS NO FRICTION BETWEEN THE GROUND AND THE WHEEL.

Like I want the surface of the triangle constant, I need the spring and the force F2.

There is a gasket between the black arm and the red wheel but I don't drawn it.

The mass is in the center of the wheel and in the outer circle of the wheel.

The center of the wheel (move to the right) win the energy the moment on the wheel lost (the moment is counterclockwise and the wheel rotates clockwise) : +1.5Rδ-1.5Rδ = 1.5 e-3 -1.5e-3 = 0 J
The spring lost 7.4e-4 J
The force F4 works very few compared to others, it is  √3Rδ²/2 = 8.6e-7 J

The sum of energy is not at 0, in this case the device destroy an energy


« Last Edit: July 11, 2016, 05:45:00 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #323 on: July 12, 2016, 09:41:40 AM »
I drawn forces with less length .

First: destroy an energy
Second: create an energy

I don't drawn them before but there are spokes on the wheel
« Last Edit: July 12, 2016, 08:28:05 PM by EOW »

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Re: Sum of torque
« Reply #323 on: July 12, 2016, 09:41:40 AM »
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Offline EOW

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Re: Sum of torque
« Reply #324 on: July 13, 2016, 02:45:35 PM »
In fact, I don't need pressure nor the black and red walls, just a bike and a spring. I drawn only the wheel not the bike. There is NO friction between the ground and the wheel.

The force Fo gives the same energy than the torque -Fo/Fw on the wheel lost. But the spring lost an energy.

The energy won by the force Fo to the bike is lost by the torque -Fo/Fw on the wheel but the spring lost an energy. The sum of energy is not at 0. The center win FRδ. The wheel lost FRδ. If F=1 N and R=1m, the energy lost/win is δ. For δ=0.001 rd the energy win/lost is 0.001 J.

The length increases: dl = ( √ ( (θ-sinθ)² + (1-cos(θ))²) - √ ( ((θ+δ)-sin(θ+δ))² + (1-cos(θ+δ))²) ) with θ at 300° The energy for the force F2 is at dl*F2 = 4.27e-4 * 1 = 4.27 e-4 J
The spring lost 4.27e-4 J

Maybe I can attach the spring with a wheel bigger (trochoid)  like that the spring can win/lost more.
« Last Edit: July 13, 2016, 09:25:00 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #325 on: July 14, 2016, 10:36:15 PM »
The spring is attached to the blue object and the black arm. The ground receives nothing. The blue device lost the energy that the wheel win in translation but the wheel lost an energy in rotation. The spring lost near nothing.

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Re: Sum of torque
« Reply #325 on: July 14, 2016, 10:36:15 PM »
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Offline EOW

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Re: Sum of torque
« Reply #326 on: July 30, 2016, 04:27:36 PM »
The spring directly on the wheel, no force from the ground or to the ground. The trajectories of the points X and Y are not the same.

Offline Gabriele

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Re: Sum of torque
« Reply #327 on: July 31, 2016, 11:12:12 AM »
No my friend

Offline Gabriele

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Re: Sum of torque
« Reply #328 on: July 31, 2016, 11:14:19 AM »
Better...how do you link the spring to the wheel?

Offline EOW

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Re: Sum of torque
« Reply #329 on: July 31, 2016, 11:39:43 AM »
The springs is attached on the wheel, two ends of the spring is attached on the wheel. Note, there is no axis of rotation like a wheel of a bike, the mass is only at the outer circle.



     

 

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