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Author Topic: Sum of torque  (Read 133519 times)

Offline dieter

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Re: Sum of torque
« Reply #165 on: February 12, 2015, 11:52:52 PM »
Looks good, as far as I understand it.


BTW. I'm off for a break


BR

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Re: Sum of torque
« Reply #165 on: February 12, 2015, 11:52:52 PM »

Offline EOW

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Re: Sum of torque
« Reply #166 on: February 13, 2015, 09:21:06 AM »
Ok. I drawn anothers views at different positions. The torque from F7/F8 is not constant it depends of the position, so forth image shows the system lost energy but it's possible to change the position of the friction and have a clockwise torque to the support3 (fifth image). The sum of torque from F3 and F4 to support1 and support2 is 0 like I show before.

with

angular velocities (lab ref):
Support1 = 0
Support2 = 0
Support3 = +2w
Pulley1 = +4w
Pulley2 = +12w


The motor need -12FRwt
The friction gives +2FRwt
The Support3 gives +12FRwt
The Support2 need 0
The axes of pulleys gives +4FRwt

The sum is at +6FRwt
« Last Edit: February 13, 2015, 08:39:55 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #167 on: February 14, 2015, 09:57:46 AM »
With one motor and one brake on a support. The motor drives the brakes with a crossed belt. The motor turns clockwise at +10w around itself. The brake turns counterclockwise around itself at -10w. The support turns clockwise at +w. So, the motor turns at +11w in lab ref and the brake turns at -9w in lab ref. The motor needs -11FRwt, the brake gives +11FRwt (the brake turns at -10w and the support turns at +w, the difference is 11w), torque F1/F2 gives +2FRwt to the support, torque F3/F5 needs -FRwt to the support, the sum is at +FRwt. All angular velocities are constant. I guess no mass for pulleys and belt.
« Last Edit: February 14, 2015, 05:44:05 PM by EOW »

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Re: Sum of torque
« Reply #167 on: February 14, 2015, 09:57:46 AM »
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Offline EOW

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Re: Sum of torque
« Reply #168 on: February 15, 2015, 01:06:12 PM »
I can replace the friction with an electric generator for example. I guess the efficiency of the motor, generator, belt, etc is 1 (no losses). The motor gives F1 and F2 to the Pulley1. The motor gives F3 and F4 to the support. The belt gives the force F5 to the Pulley2. The belt gives the force F6 to the Pulley1. F5 gives F7 to the axis of the Pulley2. F6 gives F8 to the axis of the Pulley1.The generator gives F9 and F10 to the support.

I noted R the radius of the Pulley1 and the Pulley2.

With |2F1|=|2F2|=|2F3|=|2F4|=|F5|=|F6|=|F7|=|F8|=|2F9|=|2F10|=|F|

The motor need to give -10FRwt
The support reveices -FRwt from F2 and F3
The support receives -FRwt from F9 and F10
The support receives +2FRwt from F7 and F8
The generator receives +11FRwt because the difference between the rotor and the stator is 11w

The sum of energy is +FRwt

All angular velocities are constant. The sum of torque on the support is 0, so its angular velocity is constant. The energy lost from the motor is -10FRwt and the generator can recover +11FRwt.
« Last Edit: February 15, 2015, 04:27:32 PM by EOW »

Offline EOW

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Re: Sum of torque
« Reply #169 on: February 16, 2015, 01:59:48 PM »
The stator of the motor is fixed on the support so it turns clockwise at +w. The stator of the generator is fixed on the support so it turns clockwise at +w.

The rotor of the motor turns clockwise at +10w relatively from the stator, so the rotor turns clockwise at +11w (lab ref), the power needed for the motor is -10FRw and the support needs -FRw. The rotor of the generator turns counterclockwise at -11w but the stator turns clockwise at +w so the difference is 12w, but the support receives a torque -FR and need the power FRw.

Forces on axes of the pulleys give a power 2FRw to the support and with -2FR from 2 startors, the sum of torque on the support is 0.

So, the motor needs -10FRw and the generator gives 12FRw.
« Last Edit: February 16, 2015, 06:41:09 PM by EOW »

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Re: Sum of torque
« Reply #169 on: February 16, 2015, 01:59:48 PM »
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Offline EOW

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Re: Sum of torque
« Reply #170 on: February 19, 2015, 03:21:20 PM »
A closed container has a pipe enrolled on a grey disk. For enter the pipe inside the container I need to give the energy E1. All the device is at a linear velocity V, with V = constant. When the pipe move out the container I recover the energy E1. The container has 2 forces F2 and F4 on it, this forces give an energy E2 = (F2+F4)*V*t. All concentric forces on the pipe cancel themselves because the pipe has a mass and with rotation there is centrifugal forces. P2 > P1, and imagine the device with P1=0 in theory. There is no gravity.

Offline EOW

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Re: Sum of torque
« Reply #171 on: February 20, 2015, 07:40:00 AM »
In fact, all forces from pressure*surfaces cancel themselve => the container has no net force on it. The force come from the centrifugal forces only. The gas can't give a contrary force, I think the gas give a force that add the net force of the centrifugal forces.

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Re: Sum of torque
« Reply #171 on: February 20, 2015, 07:40:00 AM »
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Offline EOW

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Re: Sum of torque
« Reply #172 on: May 31, 2015, 03:01:52 PM »
The idea is to use the device I drawn in the message #40. But here I use pressure. No gravity. No modification of the volumes. I suppose no friction. It's a 2D but it's possible to imagine in 3D replace circles with spheres. Outside the pressure is constant at P (atmospheric pressure for example at 1 bar). A basic device is composed by an arm and a circle. The black arm can turn around the blue axis. The circle is fixed to the arm. The circle don't turn around itself. I drawn 2 basic devices but it possible to have N devices, in the third image I drawn 16 basic devices. Circles are smart, walls between circles can be removed : red in the drawing. I suppress wall like that a part of each circle receives a torque on it. Like all circles don't turn around the same center, the torque is not the same.

First drawing: the device turns from time=0 to time=2.
Second drawing: the device turns from time=0 to time=0.5

It's possible to look at torque on each arm, like the distance d1 is greater than d2, the torque is higher counterclockwise from time=0 to time=1.

At time=1, the pressure inside circles must changed. This don't need energy. It possible to think with a device with N basic devices (arm+circle) and X circles are time=0, X circles are at time=1, X circles are at time=2, etc. For change the pressure inside circle it's easy: take the pressure inside one and replace from another cirche that is in another time (or position). Like olumes are constant, change pressure don't need energy in theory, a little in practise.

The third drawing shows a device with 16 circles, 2 circles at each time, all 16 amrs turn at the same angular velocity. Two circles work together. At the exact position: time=1 time=4 time=6 and time=8 there is no work. At time=1it's necessary to put P/2 inside circles so change with the circles at time=4.

Inside circles the pressure is 2P from time=8 to time=1, from time=4 to time=6.
Inside circles the pressure is P/2 from time=1 to time=4, from time=6 to time=8.

Offline EOW

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Re: Sum of torque
« Reply #173 on: June 01, 2015, 09:08:39 AM »
First image: for look the sum of torque from time=0 to time=1.  There is no torque if radius are the same ! need one small circle and one bigger. Look at third image.

Second image:  look of the rotation of the gaskets. Even from time=0 to time=1, the device have a net torque on it and gaskets can't compensate it.

For change the pressure inside the circles it's possible :

a/ to exchange physically circles with a device with 16 circles (low angular velocity of the device)
b/ to use an external device, this device will need an energy but this energy will be transform in heating of gas

I explain with a gas but it's ok with a liquid too (less losses for change the pressure inside circles).
« Last Edit: June 01, 2015, 01:12:24 PM by EOW »

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Re: Sum of torque
« Reply #173 on: June 01, 2015, 09:08:39 AM »
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Offline EOW

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Re: Sum of torque
« Reply #174 on: June 01, 2015, 04:27:30 PM »
With gas: there is no torque if circles have the same diameter. The torque is not the same with 2 diameters but the angular velocity must be different for one circle. So the sum of energy is conserved.

Now, if I put liquid inside circles, with same diameter, centrifugal forces are not the same inside one circle to another because the centers of rotation are not the same. The torque must be different. Maybe the liquid move between 2 circles. The velocity in a fluid change like the speed of sound in the fluid, the pressure can't be faster than this. So if the device turn at high velocity, and like the red wall change all the time with the angular position, the pressure cannot be the same and the sum of torque is not 0.

Offline lumen

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Re: Sum of torque
« Reply #175 on: June 01, 2015, 07:40:57 PM »
I think the design could be made less complicated by reducing the working principal to just two working components.
1: centrifugal force
2: Roberval action
I have done some testing on the principal and have not found any cause for the device to fail so I am now building a test model.
There must be a condition to cause the device to fail but it's simply not apparent.
 

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Re: Sum of torque
« Reply #175 on: June 01, 2015, 07:40:57 PM »
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Offline EOW

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Re: Sum of torque
« Reply #176 on: June 01, 2015, 08:35:25 PM »
What is Roberval action ? The balance ? there is no gravity here.

Yes, sure for a non continous machine it's possible to test with just 2 circles. Better efficiency with several circles. But do you tested with liquid or gas ? It can be tested with a "velocity" higher than the speed of sound. The "velocity" is the velocity of the red line. Where are you discussed about that ? The speed of sound in water is 1500m/s and for test, it's not easy. I drawn lines of equal pressure from another circle when the door is open. The force F from the red line don't give the same torque, it's logical. But, this additionnal torque don't work the same in one turn (of 2 circles) and it works a small time, "waves" (curved lines black, red, magenta and blue) move at 1500 m/s

Offline lumen

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Re: Sum of torque
« Reply #177 on: June 01, 2015, 09:31:21 PM »
Roberval action is simply the mechanics that maintains the orientation as the chambers rotate around the circle.
If they were aligned to north before starting, they would remain pointing north while rotating.
In your drawing 1-1 is pointing the same direction as 2-2 and 3-3 then 4-4 , 5-5, 6-6. They point the same direction as the disk rotates.

This is what forces the gas (or liquid) to move from one chamber to another by centrifugal force.
There is always an apparent gain in energy even at slow speeds.

 

Offline EOW

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Re: Sum of torque
« Reply #178 on: June 01, 2015, 09:40:11 PM »
Have you already talked about this idea somewhere ? Have a link for Roberval ?

Sure, it's work at any velocity but:

1/ for test with sensors, you need to detect something that run at 1500m/s
2/ with 2 circles, the device like I drawn don't work, in one turn the sum of work is 0, the door must be closed some steps and open others steps

With a high velocity, the waves  don't have time to stabilize them, so it's better. For me it's the delay that can give energy and it's logical because the angular velocity of circles can be very high. If the device lost energy when the liquid moves (water is compressible so a few liquid will move), this energy is near constant, but the angular velocity can be 10 times or 100 times higher.

The centrifugal force is mv²/R, so with a big v and a big R, the centrifugal force is not high, but the distance that circle run is high, so it must be ok. Tell me more about your tests, on a software ?



Offline lumen

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Re: Sum of torque
« Reply #179 on: June 02, 2015, 12:15:10 AM »
Your plan to use a Gas and extract energy is not exactly what I was doing but there are many ways to extract energy from the two basic systems combined.

I plan to generate electrical current directly from centrifugal force using a roberval system on a rotating disk.
The tests I have been doing are to find out if the torque from the centrifugal force can in some unknown way apply back against the rotation of the disk generating the centrifugal force.

Torque on a roberval device cannot apply in any way to cause rotation of the disk but centrifugal force is radial and some things change because of the vector of the force has vector differences over the working area involved.

Regardless of these facts there is a way to correct for that problem so again I'm left searching for the cause of failure in such a device.

It will be interesting to know the exact cause of failure for such a device because it appears so elusive.




 

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