Pressure and torque in water without external gravity

[Gh778]

I would like to calculate the torque on the air object under water. The pressure is only give from water (water attrack itself), there is no other gravity (like Earth for example) or other object around. Without air, in the center the gravity is 0 I think. With asymetric air volume, I think the pressure at left is lower than at right like it's impossible I would like to calculate this torque. How can I do ?

 

[Hassan2]

Hi Gh778

Now this geometry is simple and no 3D imagination is required!
But without gravity, pressure is the same everywhere through the water unless the water is acceleration. Furthermore what keeps the air region in that geometry? there must be a rigid container containing the air. Your problem becomes the simple case of an arbitrary rigid object withing a uniform pressure.

 

[Gh778]

For example, take a planet (1000 km of diameter) of water with big air object inside. Like that we can calculate the torque because the gravity is not small.

 

[mikeph]

The air object cannot experience a torque as it isn't rigid. It will deform and rupture under gravity, or if there is no gravity, it would form a sphere due to surface tension. As it does form a sphere I guess you would get boundary layers of air moving with the water as the interface evolves, but there can't be torque because the air has no resistance to such a shear stress.

 

[Hassan2]

I think the water and the object would separate and the water form a perfec sphere after pushing out the air/object.

 

[Gh778]

My question is to know if there is a torque or not, at start the object was small but some people don't see gravity effect. You can imagine an object big enough for have some gravity effect and small enough to be attach on Earth, 1km for example. A sphere with water inside, the air object seems to have a torque.

My program, the torque is big compared to water force:

#include <stdio.h>
#include <math.h>

//tab for contain all point, a point = a molecule of water for example
double t[1000][1000];

int main(void)
{

int i,j,x,y; // used for loops

double
d=0,
Fx=0, // local force x
Fy=0, // local force y
angle=0,
R=0, // radius
xx=0, // local x force
yy=0, // local y force
id=0, // simulate the gravity formula: 1/d²

// forces on water
TFx1=0, // Torque on left vertical line from direct force
TFy1=0, // Torque on left vertical line from pressure
TFx2=0, // Torque on right vertical line from direct force

// forces on air object
TFy2=0, // Torque on right vertical line from pressure
TFx3=0, // Torque on slope line from direct force
TFy3=0; // Torque on slope line from pressure

printf("\nStart\n"); // put 1 in tab t if there is a molecule of water
for(i=0;i<1000;i++)
{
for(j=0;j<1000;j++)
{
//in the circle put 1
if(((i-500)*(i-500)+(j-500)*(j-500))<=250000)
t[j]=1;
else // outside the circle put 0
t[j]=0;
// put 0 for the air block part of circle
if(((i-500)*(i-500)+(j-500)*(j-500))>=78400 && ((i-500)*(i-500)+(j-500)*(j-500))<=90000 && i>500 && i<750 && j>500)
{
t[j]=0;
}

}
}

// put 0 in the vertical volume
for(i=480; i<520; i++)
for(j=500; j<800;j++)
t[j]=0;
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Caculate force and torque for x=480, y from 500 to 800
// this is the left vertical line
x=480;
for(y=500;y<800;y++)
{
Fx=0;Fy=0;
for(i=0; i<1000; i++)
{
for(j=999; j>=0;j--)
{
d=sqrt(pow(fabs(i-x),2.0)+pow(fabs(j-y),2.0));
if(d!=0)
{
id=(double)(t[j])/d/d;
angle=fabs(atan((fabs(j-y)/fabs(i-x))));
if(i>=x && j>=y)
{
xx=id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<x && j>=y)
{
xx=-id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<=x && j<y)
{
xx=-id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
}
else if(i>x && j<y)
{
xx=id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
} Fx+=xx;
Fy+=yy;
}
//printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
//system("Pause");
}
// printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
// system("Pause");
}
R=((double)y-500.0)/10000.0;
TFx1+=Fx*R;
TFy1=TFy1+(TFy1+Fy)*R;
//printf("\nFy=%f, R=%f, TFy=%f", Fy, R, TFy);
//system("Pause");
}

printf("\nTFx=%f, TFy=%f", TFx1, -TFy1);

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Calculate forces and torque for x=520, y from 500 to 780
// This is the right vertical line
x=520;
for(y=500;y<780;y++)
{
Fx=0;Fy=0;
for(i=0; i<1000; i++)
{
for(j=999; j>=0;j--)
{
d=sqrt(pow(fabs(i-x),2.0)+pow(fabs(j-y),2.0));
if(d!=0)
{
id=(double)(t[j])/d/d;
angle=fabs(atan((fabs(j-y)/fabs(i-x))));
if(i>=x && j>=y)
{
xx=id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<x && j>=y)
{
xx=-id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<=x && j<y)
{
xx=-id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
}
else if(i>x && j<y)
{
xx=id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
} Fx+=xx;
Fy+=yy;
}
//printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
//system("Pause");
}
// printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
// system("Pause");
}
R=((double)y-500.0)/10000.0;
TFx2+=Fx*R;
TFy2=TFy2+(TFy2+Fy)*R;
// printf("\ny=%i, Fx=%f, Fy=%f", y, Fx, Fy);
// system("Pause");
}
printf("\nTFx=%f, TFy=%f", TFx2, TFy2);
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// x=750, y from 126 to 165
// This is theslope right line of air block
x=750;
for(y=626;y<665;y++)
{
Fx=0;Fy=0;
for(i=0; i<1000; i++)
{
for(j=999; j>=0;j--)
{
d=sqrt(pow(fabs(i-x),2.0)+pow(fabs(j-y),2.0));
if(d!=0)
{
id=(double)(t[j])/d/d;
angle=fabs(atan((fabs(j-y)/fabs(i-x))));
if(i>=x && j>=y)
{
xx=id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<x && j>=y)
{
xx=-id*fabs(cos(angle));
yy=id*fabs(sin(angle));
}
else if(i<=x && j<y)
{
xx=-id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
}
else if(i>x && j<y)
{
xx=id*fabs(cos(angle));
yy=-id*fabs(sin(angle));
} Fx+=xx;
Fy+=yy;
}
//printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
//system("Pause");
}
// printf("\nx=%i, y=%i, i=%i, j=%i, d=%f, id=%f, s=%f, xx=%f, yy=%f, angle=%f, Fx=%f, Fy=%f", x, y, i, j, d, id, s, xx, yy, angle, Fx, Fy);
// system("Pause");
}

R=((double)y-500.0)/10000.0;
TFx3+=Fx*R;

TFy3=TFy3+(TFy3+Fy)*R;

// printf("\nTFx=%f, TFy=%f, T=%f", TFx, TFy, T);
// system("Pause");

}

printf("\nTFx=%f, TFy=%f", TFx3, TFy3);
system("Pause");
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
return 0;
}