Sum of torque

[EOW]

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 [tex]\alpha[/tex] 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 [tex]\alpha[/tex]:

The radius of the wheel is at 1.

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]

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)

[EOW]

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 [tex]\alpha[/tex]

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

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

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

The result is 0.5 for an angle of 0.1


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Code Select Expand

[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:

Code Select Expand

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.

Code Select Expand

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

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Here all math integrations:

Code Select Expand

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 [tex]\alpha[/tex] at 0.1 rd the result is 0.5

I change the math integrals and I obtain this :

Code Select Expand

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]

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.

[EOW]

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