[LB7 (OP)]

In the geometry:


tad.png (190.95 kB . 1490x762 - viewed 9596 times)

It is not only the sum of forces that is not at 0 but also the sum of torques:

With the dimensions I gave, the surface of the blue area is : 18000 usi watch http://villemin.gerard.free.fr/GeomLAV/Cercle/aaaAIRE/Rectang2.htm

The sum of torques on the all spheres is 18000

The sum of torques from the segment is integrate(x*((2pi)-2*acos(75/113)*(sqrt(75²+x²)-75)/(113-75)) dx from 0 to L = 19676

The sum of torques is not 0

I don't take the sphere packing and I think it is not necessary. Maybe I need to correct the result of the sum of torque from each sphere because the forces are not exactly at the full diameter but I think if I decrease the diameter of the spheres near 0 (in theory) I can have the full diameter, the result is not cumulative.

For the sum of forces, I have 2 radial gradients of pressure, one because the outer circle don't start like the inner but I have a second gradient of pressure because the forces from the friction are not parallel, so I need to correct the sum of forces but I don't think the sum can reach 0 because the straight segment breaks the propagation of these radial gradient of pressure and the sum of forces on each sphere is 0 inside the media.

[LB7 (OP)]

if I consider the sum of torques at 0 I cannot have the sum of forces at 0. If I want the sum of torques at 0 I need to reduce the pressure but if I do that I increase the sum of forces. Nobody knows how to calculate theses pressures ?


https://isamax7.blogspot.com/2021/09/strategie-des-russes.html


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[LB7 (OP)]

With a square (empty shape), there is a torque on the square (around itself):


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[LB7 (OP)]

With the white shape like that it is easier without any calculations:


sfch.png (45.51 kB . 437x489 - viewed 8353 times)


IsaMaX-22 Déc 2021.pdf (599.88 kB - downloaded 1644 times)


x.pdf (698.93 kB - downloaded 529 times)

[LB7 (OP)]

nothing moves except the spheres, and it doesn't change the others torques...

what else ? Maxwell

[LB7 (OP)]

With that simplified shape to reduce the calculations. Like that I can assume the horizontal pressure is the same at the inner circle than the outer circle (because the sum of forces is 0, maybe...). So the sum of torques ? on the sphere it is 6778 (calculated with the area). But how I can have that value with the horizontal surfaces ? If I take piF, the max pressure I have 10562.

The logic (Inner R=75, outer R=113):

1/ The sum of forces I give to the terminal spheres is vertical
2/ The horizontal walls can only give a vertical force
3/ hyp1: I consider the sum of forces equal to 0, so for a given horizontal the pressure is the same (it is possible to think with different diameters inner/outer) because the horizontal forces is at 0 (hyp1), it is very strange to say the pressure is the same for a given horizontal...
4/ The area of the spheres is pi/2*(113²-75²)-AT = 11221-2225*2 = 6771, AT=A1+A2 (link above), A2=0.5*(113²-75²)*(113-75), AD=sqrt(2*113*(11375)), h=sqrt(113²-AD²/4), A1=113²*acos(h/113)-h*sqrt(113²-h²)
5/ hyp2: the sum of torques is 0, so the difference of pressure between top/bottom is area/(113*cos(pi/2-acos(75/113)))²/2 = 1.89 (that result is very strange in comparison with the lateral pressure...)
6/ The vertical bottom force from the horizontal surfaces is 1.89*84.522 = 160.23 and the sum of forces from the terminal forces I need to give from A0 is 2*integrate(cos(atan(x/75)) from 0 to 113*cos(pi/2-acos(75/113) dx = 72.62*2 = 145.25, the vertical forces from the curved surface is 67.6 because it is 75*integrate(1.89*(x-pi/2)/pi) dx from pi/2 to -pi/2 less 113*integrate(1.89(x-a)/(2a)) dx from a to -a with a=pi/2-acos(75/113)=0.72. The sum of vertical forces is 160.23-145.25-67.6 not 0

The hyp1 is false... but I think it is the hyp2 it is easier to think the sum of torques is not conserved, the sum of forces ? maybe but in that case I was wrong about my theory about gravitation, sniff


IsaMaX-6-janvier 2022.pdf (773.13 kB - downloaded 562 times)

[LB7 (OP)]

The logic (Inner R=75, outer R=113):

1/ The sum of forces I give from A0 to the terminal spheres is vertical
2/ The horizontal walls can only give a vertical force
3/ hyp1: I consider the sum of forces equal to 0, so the sum of horizontal forces is 0 and the sum of vertical forces is 0, so for a given horizontal the pressure is the same (it is possible to think with different diameters inner/outer) because the horizontal forces is at 0 (hyp1), it is very strange to say the pressure is the same for a given horizontal...
4/ The area of the spheres is pi/2*(113²-75²)-AT = 11221-2225*2 = 8996, AT=A1+A2 (link above), A2=0.5*(113²-75²)*(113-75), AD=sqrt(2*113*(113-75)), h=sqrt(113²-AD²/4), A1=113²*acos(h/113)-h*sqrt(113²-h²)
5/ hyp2: the sum of torques is 0, so the difference of pressure between top/bottom is area/(113*cos(pi/2-acos(75/113)))²/2 = 2.51 (that result is very strange in comparison with the lateral pressure...)

Maybe the following is false:

6/ The vertical bottom force from the horizontal surfaces is 2.51*84.522 = 212 and the sum of forces from the terminal forces I need to give from A0 is 2*integrate(cos(atan(x/75)) from 0 to 113*cos(pi/2-acos(75/113) dx = 72.62*2 = 145.25, the vertical forces from the curved surface is 67.6 because it is 75*integrate(2.51*(x-pi/2)/pi) dx from pi/2 to -pi/2 less 113*integrate(2.51(x-a)/(2a)) dx from a to -a with a=pi/2-acos(75/113)=0.72. The sum of vertical forces is 212.23-145.25-89.77 not 0

Because nothing can affirm the pressure is linear from top to bottom. But in that case the law of vertical pressure is not progressive, the gradient of pressure must be higher from top than it is at the bottom. It is very strange to have that sort of gradient of pressure from the geometry. I will ameliorate my logic to break that interrogation.

[LB7 (OP)]

The law of the gradient cannot be one time up/down for one example and down/up for the other example.

For the calculations:

Area of the spheres (sphere packing at 1):

A2=0.5*sqrt(R2²-R1²)*(R2-R1)
AD=sqrt(2*R2*(R2-R1))
H=sqrt(R2²-AD²/4)
A1=R2²*acos(H/R2)-H*sqrt(R2²-H²)
AT=A1+A2
AI=pi/2*(R2²-R1²)-AT
L=R2*cos(pi/2-acos(R1/R2))
P=AI/(L²/2)
FD=P*L
FU1=2*R1*hyp(asin(L/R1))
FU2=P*(R1*pi/2-R2*(pi/2-acos(R1/R2))

FV=FD-FU1-FU2

[LB7 (OP)]

For example:

R1=8,R2=10 => FV=-3.8
R1=5,R2=10 => FV=0.86

The sign changed so the gradient must change. So one hypothesis is false the sum of torque = 0 or the sum of forces = 0.

[LB7 (OP)]

Inside the blue container, I can add an empty shape. The torque on the shape 1 is the same than the torque on the shape 2 but the number of the lack of blue spheres is not the same... Why the torque is the same ? because for a given horizontal the pressure is the same (hypothesis) and here I choose the length of the horizontal segment identical.

[LB7 (OP)]

I drew the container and the empty white shape. It proves the sum of torques or the sum of forces (it is not an exclusive OR) cannot be at 0 so the sum of energy too.

[LB7 (OP)]

I can took the empty shapes very small. I can change the position of the white shape inside the blue container, A1 passes passes through the vertical that passes by A0. Like that I'm sure the torque is the same on the 2 white shapes but the lack of blue spheres is not the same !

[LB7 (OP)]

Yes, it's done I broke at least 2 laws of conservation with that logic. The energy is not conserved


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[LB7 (OP)]

Just to inform I posted some information in french here :

https://overunity.com/18897/un-debut-de-resume-de-mon-histoire/msg560027/#msg560027

[LB7 (OP)]

For those who are interesting in the story of what country did what to me and my girlfriend about my invention, I resumed it on the pdf. It is in french but it is easily translated in english I think.


IsaMaX-6 Déc 2021.pdf (284.88 kB - downloaded 600 times)


IsaMaX-20 Déc 2021.pdf (537.39 kB - downloaded 734 times)