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### Author Topic: Sum of torque  (Read 124810 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #195 on: August 02, 2015, 04:11:13 PM »
The goal is always the same: have a torque on the red object and no torque on the support. The support turns around C1. C1 is fixed to the ground. The red object turns around C2. C2 is fixed to the support. I attrack balls from C2, all balls are attracked from C2. I want to have F1=F2 and the sum of forces of attraction from part1 =  the sum of forces of attraction from part2. For that I use the left part with a higher radius than the right part (ratio 9/7). But I need to have F1=F2, so it's important to change the pressure in the same time. When the radius at left increases, the volume of balls increases too, so I can cancel the forces of attraction left/right. If I take the good value of the pressure I can cancel in the same time F1+F2 on C2. One end of each spring is attached on C2, the other end of the spring is attached on a ball. There is one spring for each ball.

Like that the red object has a torque on it. It's unstable like before, but I accelerate more and more the support in the same time. I win the energy of the rotation of the red object. The center of gravity of the red object is on C2.
« Last Edit: August 03, 2015, 01:00:00 AM by EOW »

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

##### Re: Sum of torque
« Reply #195 on: August 02, 2015, 04:11:13 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #196 on: August 03, 2015, 09:53:59 AM »
I changed the radius of the left part for have a counterclockwise torque on the support.

With the radius at 5.5 at left and the radius 3.5 at right I have :

The vector at left :
x=-6.27
y=1.49

The vector at right:
x=5.21
y=-1.9

The sum is :
x=-1.06
y=-0.41

So the force give a counterclockwise torque on the support.

The force on C2 from F1 and F2 is 0 because I take the pressure at 3.5 for the right and 2.22 for the left.

« Last Edit: August 03, 2015, 07:39:19 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #197 on: August 04, 2015, 01:16:46 AM »
If I attrack balls at outer circle I can have F1=F2 and a counterclockwise torque on the support, no ?

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

##### Re: Sum of torque
« Reply #197 on: August 04, 2015, 01:16:46 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #198 on: August 04, 2015, 06:21:04 PM »
Here I have a counterclockwise torque on the support and each red object receives a counterclockwise torque too.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #199 on: August 11, 2015, 02:13:53 PM »
Black lines = to the center C1. Red line = to the center C2.
Attraction: height = pressure.

Torque1=2.66
Torque2=-4
Torque3=-1
Torque4=+1
Torque5=+0.166
Torque6=-0.833

Sum of torques=-2

Force3+Force4=4
Force5+Force6=2

Sum of F3+F4+F5+F6 (vectors)=sqrt(4²+2²)=4.47

equation of the line of the sum of forces F3+F4+F5+F6 on C2: y=-2x+1; Equation of perpendicular: y=0.5x; Intersection: x=0.4,y=0.2; Distance =sqrt(0.4²+0.2²)=0.447

Torque from forces on C2 from F3+F4+F5+F6 = 4.47 * 0.447 =2

« Last Edit: August 11, 2015, 07:12:22 PM by EOW »

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

##### Re: Sum of torque
« Reply #199 on: August 11, 2015, 02:13:53 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #200 on: August 12, 2015, 11:52:58 AM »
Hello,

A wheel is turning and is moving in translation exactly like a wheel of a bike on the road. I put a gas under pressure P inside the black shape. The black shape is moving in translation (no rotation). Outside the black shape there is no pressure (a theoretical problem to simplify the problem). The angle $$\alpha$$ is very small, it can be at 0.00000001 rd for example.

I drawn several positions of the wheel+black_shape. A point "w" fixed on the wheel turns and moves in translation:

All the volumes are constant. I don't drawn the gaskets between the wheel and the black shape but there are (no gas escapes). I suppose the friction at 0 to study the sum of energy.

I don't need an energy to move in translation the black shape. So, the sum of energy must be at 0 too for the wheel, but:

If I'm looking at points A and B, one is moving more at right than one another. I calculated this with a big precision, like that it's possible to decrease the angle $$\alpha$$:

The radius of the wheel is at 1.

Quote
from mpmath import *

mp.dps=100; mp.pretty=True

da=0.000000001
a=pi/2

x1=a-sin(a)
y1=1-cos(a)

x2=a+da-sin(a+da)
y2=1-cos(a+da)

x3=a+pi-sin(a+pi)
y3=1-cos(a+pi)

x4=a+pi+da-sin(a+pi+da)
y4=1-cos(a+pi+da)

print "dx=", x2-x1-(x4-x3)
print "dy=", y2-y1+(y4-y3)

print "dx1=", x2-x1
print "dx2=", x4-x3
print "dy1=", y2-y1
print "dy2=", y4-y3

The result is that the point B moves more at right than the point A, so with the pressure of the gas this would say the point B gives more and energy than the point A need.

dx= 0.000000000000000001000000000000000124479849582226383362699685857886561868869889953568598575440719416999541559130733522
dy= -7.143671195514218638848647176908380996574437467642491632594536520172703974417113722971550136626075185  e-102

I calculated for A and B like I drawn but it's possible to change the angle of the black shape.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #201 on: August 13, 2015, 10:04:56 PM »
The numerical solution:

from mpmath import *

mp.dps=150; mp.pretty=True

D=0.1
S1=pi/2
S2=pi

w=0.0000000000000001

def rectangles(f,a,b,n) :
h=(b-a)/n/1.0
z=0.0
for i in range(n) :
z=z+f(a+i*h)
return h*z

def sx1(t):
y1= -1/( sin(t) / (1-cos(t)) )
if t<=pi/2.0:
a1=atan(y1)+pi
a2=-t-pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>pi/2.0 and t<=pi:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>pi and t<=3.0*pi/2.0:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))
if t>3.0*pi/2.0 and t<=2.0*pi:
a1=atan(y1)+pi
a2=-t+pi/2.0
a3=abs(abs(a1)-abs(a2))

return sqrt( sin(t)*sin(t)+(1-cos(t))*(1-cos(t)) ) * cos(a3)

m1=rectangles(sx1,S1+w,S1+D+w,40000)
m2=rectangles(sx1,S2+w,S2+D+w,40000)

print m1
print m2
print abs(m1)-abs(m2)

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

##### Re: Sum of torque
« Reply #201 on: August 13, 2015, 10:04:56 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #202 on: August 13, 2015, 10:06:16 PM »
The maths solution

All the volumes are constant. I didn't draw the gaskets between the wheel and the black shape but there are (no gas escapes). I suppose the friction at 0 to study the sum of energy. The radius of the wheel is at 1.

If I'm looking at points A and B, one is moving more at right than one another. I calculated this with a big precision, like that it's possible to decrease the angle $$\alpha$$

I don't need an energy to move in translation the black shape. So, the sum of energy must be at 0 too for the wheel, but for find the work of the circle I must integrate

$$\int_{pi/2}^{pi/2+da}(x-sin(x)) sin(x) dx$$
$$\int_{pi/2}^{pi/2+da} (1-cos(x)) cos(x) dx$$
$$\int_{3pi/2}^{3pi/2+da} (x-sin(x)) sin(x) dx$$
$$\int_{3pi/2}^{3pi/2+da} (1 - cos(x)) cos(x) dx$$

$$\int_{0}^{x}(x-sin(x))sin(x)dx=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))$$
$$\int_{0}^{x}(1-cos(x))cos(x)dx=-x/2.0+sin(x)-1/4.0*sin(2.0*x)$$

The result is 0.5 for an angle of 0.1

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxNUMERICAL SOLUTIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[quote]from mpmath import *

mp.dps=200; mp.pretty=True

da=0.000000001
a=pi/2

x1=a-sin(a)
y1=1-cos(a)

x2=a+da-sin(a+da)
y2=1-cos(a+da)

x3=a+pi-sin(a+pi)
y3=1-cos(a+pi)

x4=a+pi+da-sin(a+pi+da)
y4=1-cos(a+pi+da)

w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1-(y4-y3))*cos(a+da)/2

print w1
print w2

print "dx=", x2-x1-(x4-x3)
print "dy=", y2-y1-(y4-y3)

There is a difference of distance:

dx=1.00000000000000012447984e-18
dy=2.00000000000000012422984e-9

But, I calculated the mean works with the worst case:

w1=(x2-x1-(x4-x3))*sin(a+da)
w2=(y2-y1+(y4-y3))*cos(a+da)/2  ***** I need to calculate the integrale

print w1
print w2

And the results of the works are:

w1= 0.00000000000000000100000000000000012397984
w2=-0.00000000000000000100000000000000012422984

I can reduce the angle there is always a difference in the axis x.

I calculated for A and B like I drawn but it's possible to change the angle of the black shape, this could change the sign of the result.

from mpmath import *

mp.dps=200; mp.pretty=True

da=0.000000001
a=pi/2

def rectangles(f,a,b,n) :
h=(b-a)/float(n)
z=0
for i in range(n) :
z=z+f(a+i*h)
return h*z

def sx(x):
return (x-sin(x)) *sin(x)

def cx(x):
return (1-cos(x)) *cos(x)

def sxp(x):
return (x+pi - sin(x+pi)) *sin(x+pi)

def cxp(x):
return (1 - cos(x+pi)) *cos(x+pi)

m1=rectangles(sx,a,a+da,10000)
m2=rectangles(cx,a,a+da,10000)
m3=rectangles(sxp,a,a+da,10000)
m4=rectangles(cxp,a,a+da,10000)

print m1
print m2
print m3
print m4

print m1-m2+m3-m4

The result is:

Sum of works = 0.00000000514

But I'm not sure about my integration

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxMATH INTEGRATIONxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Here all math integrations:

x=pi/2.0
p1=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p2=-x/2+sin(x)-1/4*sin(2*x)
x=pi/2.0+da
p3=1/4*(-2*x+4*sin(x)+sin(2*x)-4*x*cos(x))
p4=-x/2+sin(x)-1/4*sin(2*x)

x=pi/2.0
p5=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p6=1/2*(-x-2*sin(x)-sin(x)*cos(x))
x=pi/2.0+da
p7=-x/2-sin(x)+1/4*sin(2*x)+(x+pi)*cos(x)
p8=1/2*(-x-2*sin(x)-sin(x)*cos(x))

The result is 0.00000000514

The difference is small because the angle is small, if I take $$\alpha$$ at 0.1 rd the result is 0.5

I change the math integrals and I obtain this :

x=pi/2.0
p1=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p2=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=pi/2.0+da
p3=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p4=-x/2.0+sin(x)-1/4.0*sin(2.0*x)

x=3.0*pi/2.0
p5=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p6=-x/2.0+sin(x)-1/4.0*sin(2.0*x)
x=3.0*pi/2.0+da
p7=1/4.0*(-2.0*x+4.0*sin(x)+sin(2.0*x)-4.0*x*cos(x))
p8=-x/2.0+sin(x)-1/4.0*sin(2.0*x)

print p1
print p2
print p3
print p4
print p5
print p6
print p7
print p8

z1=-(p3-p1)+p4-p2
z2=p6-p5+p7-p8
print abs(z1)-abs(z2)

It is the same result. And if I take the angle à pi the sum is 0.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #203 on: August 26, 2015, 07:58:16 AM »
The idea is to use small balls and springs to have a pressure inside a container like a fluid do with the gravity. The black container is full of small balls and springs, the pressure is from 0 to pmax in the top view (not the side view). One spring for one ball, I didn't draw all springs, only three. Balls are compressible or there are in random position (not like I drawn), I need the lateral force from the pressure. I drawn big balls but there are smaller in reality. In the top view I drawn the disk like an ellipse because there is an angle on the side view.

I attrack the disk with the springs like the balls. IT IS THE DIFFERENCE with older ideas. The red disk has 2 axis of rotation, one around the black axis and the other around itself. It is very important to have this axis of rotation around itself too, because the energy come from this axis.

The container is an object. The disk is an object.

Look at the forces on the red disk. I attrack the red disk like the  blue balls. This gives a torque on the red disk, it's logical. Like it's a disk there is no other torque on it.

Look at the forces on the red arm. Like the disk is attracked like the balls it's like an object with the same density in water with gravity, the force on the red arm is F2. F2 is only located on the arm, it is the sum of forces on the disk. F2 can't give a torque on the red arm.

Look at the forces on the container: only F1. There are lot of F1 because they are forces from the pressure but each F1 is parallel to the axis of rotation, so no torque.

Like I attrack the balls AND the red disk the force on bottom (top view) from pressure is cancelled by the force from springs, so there is no force on the side where there are green points.

The sum of torques is not 0, but the device is unstable. If I let the device like that, the springs will move and lost their potential energy. So it is necessary to accelerate more and more to increase the kinetic energy of the device. The red disk has a mass so like there is a torque on it, it will accelerate but I accelerate the black arm (the container) at the same acceleration EXACTLY the same acceleration. Like that the red disk keeps its position inside the black container. I need an energy to accelerate the container but this energy can be recover later. I win the energy from the acceleration of the red disk.

I can put several disks inside the container and no balls between disks like that I don't have torque on the red arm.
« Last Edit: August 26, 2015, 07:51:17 PM by EOW »

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

##### Re: Sum of torque
« Reply #203 on: August 26, 2015, 07:58:16 AM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #204 on: August 26, 2015, 09:54:11 AM »
To look of 2 axis of rotation of the red disk. Like before each spring attrack.

I noted all surfaces Sx:

The torque on the  container:

S1 and S2 cancel themselves the torques on the recipient
S3 and S4 give the forces F1: F1 are parallel to the axis of rotation: no torque
S5: no pressure (top)
S6: the force from pressure is cancelled by the force from springs BECAUSE I attrack the disk too

The torque on the disk around itself S7 to S9:

S7: it's a sum of segments, at final the torque from springs is always there
S8 and S9 give forces parallel to the axis of rotation (red), there is a difference of surface between S8 and S9

The sum of forces on the disk:

There are some forces like F2, but these forces can't give a torque on the red arm because they are parallel on it.

Torque on the red arm:

With severals disks inside the container I can suppress balls between disks. And if I reduce the size of the container I can suppress the difference from S8/S9

« Last Edit: August 26, 2015, 07:58:00 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #205 on: August 26, 2015, 10:29:57 PM »
The problem on the red arm come from the difference of surface S8/S9, so if I put balls inside the part of the red disk (only the part that is inside the container) I can cancel the pressure from S8/S9 (I attrack the balls inside the red disk from springs from the container). The force on the curve part (circle) don't change the torque on the disk. The red disk has always its torque around itself.
« Last Edit: August 27, 2015, 07:51:31 AM by EOW »

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

##### Re: Sum of torque
« Reply #205 on: August 26, 2015, 10:29:57 PM »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #206 on: August 27, 2015, 04:01:38 PM »
First image: if I take this simple device the force F will lost an altitude so a potential energy will be lost. But the disk will turn more and more. The sum of energy is constant or not... ?

Second image: I put balls inside the part of disk that is inside the container. I attrack balls with springs. The disk will have the force F so a torque. But the container don't have any torque. The container has balls +springs.

« Last Edit: August 27, 2015, 06:18:59 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #207 on: August 30, 2015, 10:03:58 AM »
If I take this device:

(http://s5.postimg.org/kzk1nvd8z/a70.jpg)

The support turns at w0 and the purple disk can turn around itself, at start the angular velocity of the purple disk is 0.
I can adjust the angle alpha like I want before start but the angle is constant after start.

If alpha=0° a point A fixed on the purple disk will be at differents position with time:

(http://s5.postimg.org/91n2npyvn/a73.jpg)

If alpha=2 or 3° it's the same but the point moves up/down like that:

(http://s5.postimg.org/rej2kjjyr/a71.jpg)

And like that:

(http://s5.postimg.org/67t41rbgz/a72.jpg)

The angle can be increased more at 20° for example.

So if the point A moves down or up, it's possible to attach a mass, this mass will move up (under gravity) and the purple will turn a little. I win the energy of the rotation of the purple disk and the mass that moves up. THERE IS NO TORQUE ON THE SUPPORT.

« Last Edit: August 30, 2015, 01:27:13 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #208 on: September 03, 2015, 03:17:18 PM »
The video of the device, I simulated with Ansys, 10 seconds, it's possible to look at the rotation of the disk, at start the disk has no rotation around itself, a small black circle is fixed on the disk to look at the rotation :

https://youtu.be/Pjc4dIf1aWI

If with this device the sum of energy is constant. I can use the last device with the balls+springs but WITHOUT MASS, or a mass as lower as possible. Like that I don't have these problems of inertia. I can recover the torque on the red axis from the pressure and accelerate more and more the device.
« Last Edit: September 03, 2015, 10:46:28 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #209 on: September 04, 2015, 08:04:40 PM »
I can place the disks like that:

http://s5.postimg.org/i21birqon/image.png

to increase the angular velocity of each disk