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### Author Topic: Sum of torque  (Read 120330 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #225 on: September 20, 2015, 02:01:40 PM »
Maybe I don't need the arms, just the sphere and its mass in the center. With the container + balls + springs. The main axis of rotation is the blue axis.
« Last Edit: September 21, 2015, 12:05:50 AM by EOW »

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

##### Re: Sum of torque
« Reply #225 on: September 20, 2015, 02:01:40 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #226 on: September 21, 2015, 09:12:07 AM »
Like that with a gyroscope. Or without a gyrsocope it must work.
« Last Edit: September 21, 2015, 05:48:46 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #227 on: September 22, 2015, 07:29:06 PM »
At start the disks don't turn around themselves. I launch the arms at w. I can put inside the orange disk a gyroscope because in the laboratory reference the orange disk don't turn.

2 springs: one attracts the other repuls.

There is one gyroscope inside the orange disk. The blue disk don't have a gyroscope. The blue arm receives a positive torque. The orange arm don't receive any torque because the gyroscope take the torque.

For 180° Spring1 repuls, the Spring2 attracks. For the next 180° Spring1 attracks, the spring2 repuls.
.

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

##### Re: Sum of torque
« Reply #227 on: September 22, 2015, 07:29:06 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #228 on: September 24, 2015, 09:17:49 PM »
Details of the disks

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #229 on: September 25, 2015, 03:55:53 PM »
The orange disk is not necessary. Onle the blue disk and the gyrsocope. The blue arm receives a torque. And the gyroscope receive the negative torque but it does a precession. I drawn several position of the gyroscope and the blue disk to show the electrostatic charge (or it can magnetism or springs or hydraulic device). The arms turn clockwise but the blue disk turn counterclockwise in the arm reference (the disk like the gyroscope don't turn in the laboratory reference).

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

##### Re: Sum of torque
« Reply #229 on: September 25, 2015, 03:55:53 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #230 on: September 26, 2015, 10:42:24 AM »
No gravity !

The gyroscope must have only one torque on it to precession correctly. So, the angular velocity of the arms must be constant. I add an external device that recover the energy from the arm like that w is constant. The gyroscope don't turn around itself in the laboratory reference without a torque on it. But it turns like that :

I add a torque from springs (or something else) so the gyroscope does a precession, this is only its angular velocity. The torque from the blue arm is not constant in one turn so the device that recovers the energy must be precise. The blue disk must turn at the precession.

When I commute the torque (each 180° of the blue disk in the arm reference) on the gyroscope, the gyroscope will do a nutation. The angular velocity of the precesion is small and it's good. In the arm reference, the angular velocity of the gyroscope will be at -w.cos(a)-d or -w.cos(a)+d with 'd' the angular velocity of the precession. I can choose a gyroscope with a spinning wheel very high, so high that the nutation and the precession will be low if the inertia Is of the gyroscope is very high. Sure, I need to choose a torque on the gyroscope in relation with Is.
« Last Edit: September 26, 2015, 09:40:17 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #231 on: September 27, 2015, 04:30:13 PM »
For start the device I use 2 gyroscope with their wheel turning in the opposite direction. After, I use 2 separated gyroscopes.

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

##### Re: Sum of torque
« Reply #231 on: September 27, 2015, 04:30:13 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #232 on: September 28, 2015, 11:22:52 AM »
w is the angular velocity of the arm in the lab reference
d is the angular velocity of the precession of the gyroscope when it has a torque.

I'm sure that the gyrsocope don't turn in the lab reference when the gyroscope turns at -w.cos(a) in the arm reference and when the arm turns at w. If I give a torque on the gyroscope, the gyroscope will do a precession small but it exists. And maybe when the gyroscope is not exactly at -w.cos(a) in the arm reference and the arm is at w, the gyroscope don't turn in one direction but in two directions: I'm not sure about that ! So, I will change the angular velocity of the arm in the same time I will add 2 torques on the gyroscope so I can't.

Maybe like d << w and like I change the sign of the torque each 180° so each pi/w second, this is not a problem, the gyroscope can have a mean position good enough to work like I want. The modification on the angular velocity from d to w is small.

Example, with w=10 rd/s and d=0.01 rd/s with an angle of 45°, the angular velocity of the gyroscope in the arm is -7+0.01 and 7-0.01 the precession don't change the position of the gyroscope. In 1.4 turns of the arm, the gyroscope must return to its initial position.

---------------------------------------------------------------------------------------------------------------------------------------------------------

Second method: I friction the gyroscope and the orange disk. The orange disk increases its angular velocity in the lab reference. The gyroscope precession.
« Last Edit: September 28, 2015, 08:51:56 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #233 on: September 29, 2015, 02:02:06 PM »
The arm moves the center of gravity of the gyroscope and the torque from the friction change the orientation of the gyroscope.

There is an energy from friction, an energy from the rotation of the orange torus. The arm don't receive a torque.

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

##### Re: Sum of torque
« Reply #233 on: September 29, 2015, 02:02:06 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #234 on: September 29, 2015, 03:27:05 PM »
And without a gyroscope. There are vertical disks (green). All others disks are with a slope (like before). All the device turns at w around the black center C.

I drawn ellipses for disks because it's a top view. I drawn one blue disk alone, one orange disk alone and one green disk alone to show the side view.
« Last Edit: September 29, 2015, 09:26:09 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #235 on: September 30, 2015, 01:29:03 PM »
And this idea with the balls and springs to create a pressure like a fluid under gravity. There is a torque on the red center. Use a law in 1/d^2 for the attraction. For example at 1 meter the pressure is 1 and at 2 meters the pressure is 4 not 2. Better with d^3 and don't start at p=0 at the top of the half disk => the disk in at 6 meter at proof for example.
« Last Edit: September 30, 2015, 06:33:27 PM by EOW »

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

##### Re: Sum of torque
« Reply #235 on: September 30, 2015, 01:29:03 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #236 on: October 01, 2015, 10:38:42 PM »
With the device and with gravity.  :

I place at the top of the orange disk a small mass $$m$$ after the device is launched. With an angle of 60°. This mass will move down of $$2R.sin(60°)$$ so the energy lost is $$2R.sin(60°)mg$$. With $$g$$ the gravity. The disk is turning in the arm reference at $$-0.5w$$ and the arm turns at $$w$$. The mean torque on the red arm is $$\frac{2}{\pi}Rmg.sin(60°)$$ so the erergy is a the torque by the angle of the arm, if the disk rotates of $$\pi$$ the arm rotates of $$2\pi$$ so the energy won is $$\frac{2}{\pi}Rmg.sin(60°).2.\pi$$ it's twice than the energy lost by the mass when it moves down.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #237 on: October 03, 2015, 08:00:58 PM »
There is gravity. A small mass 'm' is put at the point 'A' at top of the disk with the linear velocity V+wRcos(a) with 'w' the angular velocity of the arm, 'a' is the angle from the vertical, 'R' is the radius of the disk.  The mass 'm' is ejected at the point 'B' with the linear velocity V+wRcos(a). 'V' depends of the length of the arm. The mass will move down and lost an energy, the arm accelerates and will win an energy. In the image or in the video, the angle 'a' is at 60°.

The mass 'm' lost the energy 2mgRsin(a), with 'g' the gravity.

The mean torque on the arm is 2/pi*mg*cos(a). Like a=60°, when the disk turn of pi (from the top to the bottom, the disk must turn of pi), the arm turns of 2pi. So the energy win by the arm is 2pi*2/pi*mgR*cos(a)=2mgR

There is a difference of energy of 2mgR(1-sin(a)). And at start the disk don't turn around itself, when the mass 'm' is at the point 'B' the disk turns a little.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #238 on: October 05, 2015, 02:38:23 PM »
I changed the direction of the gravity like that the mass don't lost any energy. The arm receives a torque when it rotates. Gravity is always like I drawn even the arm rotates.

Or I can use 2 springs if F1>F2, the goal is to have F1+F2 (in vector) like the slope of the disk. F1=1.73N and F2=1N for example when a=60°. The arm receives a negative torque that cancel the energy 2RFsin(a). One spring win the energy RF another spring win 1.73*1.73RF+RF=3RF
« Last Edit: October 05, 2015, 10:26:15 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #239 on: October 06, 2015, 10:54:32 PM »
If I use a spring, the arm will receive a negative torque but the spring will receive a positive torque. But if I turn the arm with 180° not 360° the point A turns around the center C of little less than 180°, it's maybe something like 175° in the example. The difference of the energy come from the fact the arm turns of 180° but not the point A, so the spring will receive a difference.