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### Author Topic: Sum of torque  (Read 145652 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #60 on: December 07, 2014, 02:52:20 PM »
No friction, no gravity, no external motor. The black arm is turning clockwise at w1. The gyroscope is turning at w3=0 like that it keeps its position in the labo frame reference. Springs give a clockwise torque on the black arm, the kinetic energy of the black is increasing. The gyroscope don't lost energy because it turns in a perpendiculary plan.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #61 on: December 07, 2014, 06:27:36 PM »
No gravity. No heating dissipation. The black arm is turning clockwise at w1. A motor accelerate more and more a slave disk (inertial disk). I guess w1 constant because I recover the energy from the black arm for keep constant w1. The belt don't receive a torque, all forces Fa, Fb, Fc, and Fd cancel themselves. The motor receives -F but this force is in the axis, so no torque from -F. The slave receives F, and the black arm receives a clockwise torque +FR . The energy won is +FRw1t. The energy gives by the motor is in the kinetic energy of the slave, so this energy is not lost, it can be recover later. The belt gives only forces where I drawn them (special belt), look at second image.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #62 on: December 09, 2014, 09:35:00 AM »
It's difficult to have forces like I drawn with a belt. With an hydraulic system it would be easier. An external hydraulic system (not drawn) gives a pressure in the hydraulic pipes. This pressure accelerates 2 inertial disks. The sum of forces on each hydraulic pipe is 0, so there is no torque. The black arm is turning at w1 clockwise. The disk1 is turning clockwise at w2, with w2=0 at start. The disk2 is turning counterclockwise at -w3, with w3=0 at start.

The black arm receives two forces F and -F, this gives a clockwise torque FR on the arm. The energy recovered is FRtw1. An external system (not drawn) recover the energy from the black arm and keep constant w1. If w1 is very high the work from FRtw1 is very high too.

The hydraulic system must give an energy E, but all this energy is in the kinetic energy of 2 disks. So this energy E can be recover later. At final, the energy gave by the system is FRtw1.

All the system is turning like the second image.

It's possible to compare with the system in the third image. The energy is conserved (I added a ring on the disk1, the mass is in te grey inner part). The only difference if the system in the first image

I think the hydraulic pump must turn at w1 like that all the device is turning at w1 except the disk1 and the disk2.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #63 on: December 09, 2014, 06:26:18 PM »
A big green plate has N small grey disks. Disks can turn around themselves. The plate turns at w clockwise like disks at start. I have on the plate 2*N devices (not drawn) that can ejects small black masses. These devices don't give a force or a torque on the green plate because masses are the same and I eject them with the same velocity. These masses shock grey disks and increase the rotationnal velocity of disks. Sure at final I have a mass, alone, that MUST give a torque FR (if F is the force that a mass give) during a time 't'. But I can change the radius of the mass because I have time, the energy for change the radius is 0 because the force is always perpendiculary to the trajectory, but if the radius decreases the rotationnal velocity is increasing because kinetic energy is constant=1/2mr²w². So I can shock the plate with a clockwise torque. I need the same energy for accelerate disks when w=0 or w=100, but in the case where w=100, the kinetics energy of each disk is increasing more than when w=0:

Second image, I can use springs.

Thrid image: use electromagnetic system for increase rotationnal velocity of each grey disk. Prepare a lot of masses M (black) for have a continous movement.

Don't forget, masses rotate at w so their velocity is V clockwise even the force reduce the velocity, the direction can give a clockwise torque.

The  5° image is a basic system, with only a spring for understand the main idea !
« Last Edit: December 09, 2014, 08:59:26 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #64 on: December 11, 2014, 09:21:43 AM »
No gravity. No friction. Only one motor. An external device not drawn recover energy from the black arm for have w1 constant.

First image: I guess the energy is conserved in this case. It is the basic system that I replicate. At start, all the device is turning clockwise at w1. The motor need energy E1 for turn the inertial disk. There is a torque FR on the black arm but I guess this is the motor that give this energy.

Second image: I use 2 basic devices. The torque on the black arm is 2FR and the energy from the motor is the same than in the case 1/. The inertial disk has te same kinetic energy than in the case 1.

Third image: I can use N basic devices. The torque is NFR.

Fifth image: the small disk is turning at R/rw but I want the big is turning at w, so I can use magnet and like there is a sliding so the external ring can turn more slowly than the small ring
« Last Edit: December 11, 2014, 06:52:30 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #65 on: December 13, 2014, 04:25:24 PM »
Look at images please. The motor accelerates more and more the disk with N intermediate devices. All the device is turning at w1. The motor gives a counterclockwise torque to the main axis (red color) -300 J, the disk win +100 J and with 3 intermediates devices I recover 3*200 J, the sum is 400 J in my example.

The first image shows a basic device motor+belt+disk, I compare the energy when w1=0 and w1=100 at start. I can find the conservation of the energy.

The goal is to accelerate more and more a disk with a motor and have a lot of devices that give the torque from (F5,F6) * (R2-R1) on the main support. An external device recover energy from the support for keep constant w1.

Like I drawn the reductor, it is divided by 3.3, but the rapport motor/disk is 1.45, so this must say with N=100, the disk accelerate very few, so I can guess the energy won by te disk is 0. The motor lost 300 J in my example but I can set N like I want, and each intermediate device win 200 J.
« Last Edit: December 13, 2014, 10:03:22 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #66 on: December 14, 2014, 10:35:18 AM »
w1 is constant, an external device recover energy from the red axis. I guees no friction with pulleys, belts and gears. I guess a mass as lower as possible with pulleys, belts and gears. The motor gives only energy for the brake. The support receives a clockwise torque from (N-1)F(R1-R2) with N the number of intermediate devices. I think I can take w=2w1. w and w1 are constant.
« Last Edit: December 14, 2014, 05:41:36 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #67 on: December 15, 2014, 08:24:33 PM »
I resumed the device in the image.
« Last Edit: December 15, 2014, 10:43:38 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #68 on: December 18, 2014, 11:42:35 AM »
The mass of the belt give Fc1 and Fc2, with Fc1 > Fc2, like that the support has a net torque on it. An external device recover energy from the main axis, like that w1 is constant. A motor on the support drive the Pulley1 but with a mass = 0 for the belt the torque is 0 on the support. But if the mass is not 0, there is a net torque on the support.

(http://imagizer.imageshack.us/v2/xq90/673/sM8cws.png)

(http://imagizer.imageshack.us/v2/xq90/540/0D83oF.png)
« Last Edit: December 18, 2014, 11:07:08 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #69 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.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #70 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 »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #71 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: 383
##### Re: Sum of torque
« Reply #72 on: December 21, 2014, 01:07:38 AM »
With one motor it could be ok.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #73 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.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #74 on: December 22, 2014, 07:51:20 PM »
With an acceleration.