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# New Book

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### Author Topic: Sum of torque  (Read 73270 times)

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #75 on: December 19, 2014, 12:35:30 AM »
All this device is on a support not drawn. The support is turning clockwise at w1. A method for cancel the torque of the counterclockwise blue disk. There is no torque on 3 disks. But the motor gives a torque to the support. If the support is turning at w1 very high, the energy can be recover.

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

##### Re: Sum of torque
« Reply #75 on: December 19, 2014, 12:35:30 AM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #76 on: December 19, 2014, 07:10:20 PM »
It's only a motor, 2 disks on a support. The motor give a counterclockwise torque on 2 disks, so the support receives a clockwise torque. There is friction between 2 disks at point A and B. If 'd' is very small all the counterclockwise torque on the support can be canceled. Like the support is turning clockwise at w1, it's possible to recover energy from the support because w1 can be very high, in the contrary the energy needed for give friction is very small (local angular velocity not dependant of w1).

The motor gives: -F1 to the red disk and -F2 to the magenta disk. The motor receives F1 and F2.

The red disk gives to the magenta disk F3, the red disk receives -F3

The magenta disk gives to the red disk F4, the magenta disk receives -F4

Axis of rotation of the magenta disk is on the support. Axis of rotation of the motor is on the support. When all disks are turning they are like the image all the time.
« Last Edit: December 20, 2014, 12:08:44 AM by EOW »

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

##### Re: Sum of torque
« Reply #76 on: December 19, 2014, 07:10:20 PM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #77 on: December 20, 2014, 10:24:25 AM »
Look at the image. The support receives a clockwise torque Fd. Each disk receives a torque equal at 0. The motor give energy to friction but if the support is turning very quickly the energy won is high.
« Last Edit: December 20, 2014, 09:08:14 PM by EOW »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #78 on: December 21, 2014, 01:07:38 AM »
With one motor it could be ok.

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

##### Re: Sum of torque
« Reply #78 on: December 21, 2014, 01:07:38 AM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #79 on: December 22, 2014, 08:53:43 AM »
Maybe like this ?

The support is turning clockwise at w1, it receives a clockwise torque from 3 motors. There is no counterclockwise torque from pulleys or belts on the support. w1 can be very high. The energy gave by motors is lost but the energy gave to the electromagnetic devices can be recover.

I need only special belts that can push only, or push only or push and pull. I drawn 6 pulleys but it can be 100.

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

##### Re: Sum of torque
« Reply #79 on: December 22, 2014, 08:53:43 AM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #80 on: December 22, 2014, 07:51:20 PM »
With an acceleration.

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #81 on: December 24, 2014, 06:22:00 PM »
Look at the images.

The green disk is very small for cancel all torque on the support but it can be attached to a bigger disk with a mass. The green disk I drawn is a pulley attache to a bigger disk. If the diameter of the blue ring is 0.1m and the diamter of the green disk is 0.001m, if w=1 rpm, then the angular velocity of the green disk must be at 100 rpm. If I applie a force F=20 N (a torque of 1Nm), the support receive a torque = 0.1/2*5 - (0.1-0.001)/2*20 = 0.01 Nm and it can be very low. The work on the green disk is 100Cwt=0.001/2*20*100 = 1*t Nm. The work on the support is 0.01*w*t=0.01*t. So it's possible to compare these 2 different energies.
« Last Edit: December 24, 2014, 11:23:43 PM by EOW »

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

##### Re: Sum of torque
« Reply #81 on: December 24, 2014, 06:22:00 PM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #82 on: December 25, 2014, 10:05:11 AM »
At start, all velocities are at 0. I accelerate the support clockwise, I let pulleys (all pulleys have mass) like they want to turn, all the pulleys are free to turn like they want. The green ring want to turn at -w (in the support frame reference) like the blue pulleys. But the difference of radius force the blue pulleys to follow the green ring. The support receives a clockwise torque, value = 1/10Fd. When I turn the support, I give the energy for turn only the support with all pulleys but I don't give the energy for turn around themselves pulleys. So I receive a clockwise torque and in the same time pulleys turn around themselves, I don't care about the value of this angular velocity, positive or negative, because a value is necessary greater than 0.

For find the final angular velocity of the pulley I can write:

1/2J1w1²+3*1/2J2w2²=C, C is calculated with a final value of w
and
w2=Kw1

There is a difference of 1/10 torque.
« Last Edit: December 25, 2014, 09:14:56 PM by EOW »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #83 on: December 25, 2014, 10:48:26 PM »
Pulleys can't have the same angular velocity, one must accelerate, other need to decelerate. Each pulley receives a torque FR, but the support receives the torque -F(R1-R2). And if I set the belt ON when the support is at w, 2 pulleys are turning at -w. So the work won by 2 pulleys is 2FRwt and the support lost F(R1-R2)wt.

Cycle:

1/ Turn the support with the belt OFF at w, in the labo frame reference disks don't turn, if a torque is appied on a disk its kinetic energy is increasing
2/ Set the belt ON
3/ The support receives -F(R1-R2), one disk +FR1 the other disk -FR2. FR1 increases the kinetic energy and FR2 too.  It's like -FR2 must decrease the energy of the smaller disk but not, in reality (labo frame reference) the smaller disk don't turn and the torque increase the kinetic energy of the smaller disk.
« Last Edit: December 26, 2014, 10:18:02 AM by EOW »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #84 on: December 26, 2014, 04:49:32 PM »
No motor ! it's only a transcient analysis. All velocities are in labo frame reference.

Case 1/ At left, the support don't turn, the small disk turns at -6.2rd/s, the big disk turns at -2rd/s. The big disk will drive the smaller, I need a spring for adjust angular velocities. I noted R=R2. The big disk win 6wFRt, the small disk lost 6.2wF3Rt and the spring win the rest. The sum of energy won/lost is well 0.

Case 2/ At right, it's the same case except that the support is turning clockwise at w. The sum of forces are the same. Except one thing the support receive a clockwise torque +2wFRt and like it turns it can produce an energy. Here, the sum of energy won is 2wFRt.

Forces can be like I drawn because in the support frame reference: the bigger disk turns at -w and the smaller at -5.2w, the smaller turns more quickly even the radius is not the same.

I use a spring for take the difference of angular velocities. The spring has no potential energy at start, but it could if necessary.

In the second case the device win energy.
« Last Edit: December 26, 2014, 09:24:47 PM by EOW »

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

##### Re: Sum of torque
« Reply #84 on: December 26, 2014, 04:49:32 PM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #85 on: December 28, 2014, 07:22:19 AM »
The power is transmit like that:

With 2 basic devices (not drawn) :

I can reverse the angular velocity without give a torque on the support if the diameter of P2 and P3 are very small. The motor gives -FRwt to the support but receives 2FRwt from pulleys. The motor needs the energy -2Frwt, P4 gives Frwt. P4 has a mass no others pulleys. The motor accelerates only P4. Forces F1/F2 gives 2FRwt. P2 drives P3 like a gear (not drawn). The sum is well at 0. So, I need to place at least 3 basic devices in serial.

With 4 basic devices (the image):

Motor is P1
P2 is driven from P1 with the belt
P3 is driven from P2 with the gear (each pulley has a gear fixed to it)
P4 is driven from P3 with the belt
P5 is driven from P4 with the gear (each pulley has a gear fixed to it)
P6 is driven from P5 with the belt
P7 is driven from P6 with the gear (each pulley has a gear fixed to it)
P8 is driven from P7 with the belt

And for reverse the velocity of bigger disk, I can use gears with small diameter in the center of each pulley P4 and P5, like that the torque on the support is near 0. The device can win 2FRwt

The power transmit from P4 to P5 is higher if the angular velocity is high: 101 to -99 for example. If I take 3 to -1 I transmit 3FRwt to -FRwt and I think the power is reduced of 1/3. With 101/99 the ration is 99%.
« Last Edit: December 28, 2014, 05:55:00 PM by EOW »

#### dieter

• Hero Member
• Posts: 891
##### Re: Sum of torque
« Reply #86 on: December 29, 2014, 04:36:01 AM »
Most of this is over my head  , but I higly respect the amount of work you put into it. The lack of responses is rather cold, but see it that way: Many people follow the creed "when I can't say anything negative about it then I'd say rather nothing at all.", so in that sense this is good ^^

The most fascinating demonstration of a sum of torque that I have seen is gyroscopical precession, where a rotating wheel that is supported only on one side of the axle does not tip over and fall down, because the torque of the tipping sums up with the rotation and the torque chases the angular momentum as long as the rotation continues. (Or something like that)

Peace

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

##### Re: Sum of torque
« Reply #86 on: December 29, 2014, 04:36:01 AM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #87 on: December 29, 2014, 07:24:12 AM »
I can reduce the angular velocity without give a torque on the support. It's possible if I multiply the number of pulleys. 3 pulleys receive the energy from one with sliding. The small pulley receives 4FRwt, but in the support frame reference it turns at 3w, I can transmit the power at 3 bigger pulleys without lost energy and with an angular velocity of w. The small pulley lost 4FRwt but each bigger pulley win 2FRwt because in the labo frame reference its angular velocity is 2w not w.

For that I need to have a sequential device. I can't give energy from the motor and recover in the last disk all the time. I need to give power 1/3 time to each pulley P3, P4, P5. The rest of the time the pulley is not driven.
« Last Edit: December 29, 2014, 04:00:01 PM by EOW »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #88 on: December 29, 2014, 07:50:56 PM »
Or with a gyroscope like that. The goal is to decelerate (in the support frame reference) all purple disks because they turn at -w in the support frame reference. In the labo frame reference disks don't turn (w=0), so if they decelerate in the support frame reference, they accelerate in the labo frame reference. The gyroscope don't turn, it can cancel near all torque from 2 ending disks, can't cancel all the torque, there are forces fx more and more bigger with the angle. Need to add a lot of gyroscope all around the support with different angles because the gyroscope don't turn.

There is friction between disks and between the arm of the gyroscope and 2 ending disks. Friction can be replace by any system for recover energy like electromagnetic.

The gyroscope receives a torque around the axis z but it turns around the axis x.

I can place the disks in the 'center' like the gyroscope:
« Last Edit: December 29, 2014, 11:13:57 PM by EOW »

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

##### Re: Sum of torque
« Reply #88 on: December 29, 2014, 07:50:56 PM »

#### EOW

• Sr. Member
• Posts: 330
##### Re: Sum of torque
« Reply #89 on: December 30, 2014, 09:33:42 AM »
With this cycle of water. Imagine the size of the container (fluid) very large. The mean density of object to move is 1.1. Density=0 in half part and density=2.2 in the other half part.

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

##### Re: Sum of torque
« Reply #89 on: December 30, 2014, 09:33:42 AM »