Torque on a square screw full with water

[Gh778]

Here the code, I have verified with circular thread (cut if in the program), it's ok, with calculator : torque = pi*2*pi*(2*pi/10)*100000=1 240 251 Nm, and with program I found 1 240 412 Nm 5 digits for accuracy. Torque=v*(uLimit-u)/10.0*(sin(a)/cos(a))*pas*pas;

With square thread:

d=0rd/s => Torque=221 544 Nm
d=-0.5rd/s => Torque=223 990 Nm

There is a difference.

I think if we make uLimit half, the torque becomes half too.

If you divided by 2 proof, you divided by 4 the torque because surface is half and pressure is half. With square thread, not exactly divided by 4 because the thread is square (depend of the start and end).

There always a torque with the square thread if the program is good, especially the torque.

The code:

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

int main(void)
{
double x=0.0,y=0.0,z=0.0; // each point of the circular thread, I use circular thread and after select only points which are on the square thread
double u=0.0,v=0.0; // u=altitude, v=radius
double a=0.0; // angle
double torque=0.0; // local torque
double torqueT=0.0; // total torque
double pas=0.001; // step for integration
double pi=3.1415927;
double vLimit=2.0*pi, uLimit=2.0*pi; // limit of integration, in meters
double c1=3.0, c2=4.0; // limit of the square thread in meters
double d=-0.5; // angle start for the helicoid

do
{
do
{
//calculation of each point of the helicoid
x=v*cos(u-d);
y=v*sin(u-d);
z=u;

// select only point for have the square thread
if( (x>c1 && x<c2 && y<c2 && y>-c2) || (x<-c1 && x>-c2 && y<c2 && y>-c2) || (y>c1 && y<c2 && x<c2 && x>-c2) || (y<-c1 && y>-c2 && x<c2 && x>-c2) )
{
a=atan(uLimit/2.0/pi/v);
torque=v*(uLimit-u)/10.0*(sin(a)/cos(a))*pas*pas; // Here the pressure is (uLimit-u)/10.0 in bar //// pas*pas = surface
torqueT+=torque;
/*
// print part result
printf("\nu=%f, v=%f", u,v);
printf("\nx=%f, y=%f, z=%f",x,y,z);
printf("\na=%f, couple=%f",a, torque);
printf("\nT=%f", torqueT);
system("pause");
*/
}
v+=pas; // increase step of integration
}while(v<vLimit);
v=0.0;
u+=pas; // increase step for integration
}while(u<uLimit);

printf("The total torque is: %f",torqueT*100000.0);
}

 

[Hassan2]

Right. divided by 4. I have got the same result. let me finalize my result and then share it with you.

 

[Gh778]

With step (pas) = 0.0001:

Circular thread give 1 240 254 Nm compared with 1 240 251 Nm the accuracy is on 6 digits

Square thread:

d=0 rd => Torque=221516 Nm
d=-0.5 rd => Torque=223958 Nm

I think the accuracy is big enough for see there is a torque (if the program is good).

 

[Hassan2]

First, In you code, in general, a=atan(1/v) . But for 2pi span, the result is fine.

The equation for local torque is then simplified to Torque=(uLimit-u)/10*pas*pas.

This is due to the upper surface. How about the lower surface. Also remember that the gasket is elastic. If the torque or force from one side causes deformation it, it applies a force on the surface.

 

[Gh778]

How about the lower surface.

The upper surface is for example d=0 rd and the lower surface is with d=-0.5 rd. This is what I said there is a torque (I take upper torque less lower torque).

First, In you code, in general, a=atan(1/v) . But for 2pi span, the result is fine. The equation for local torque is then simplified to Torque=(uLimit-u)/10*pas*pas.

For you the result in my program is not fine ? You suppress sin and cos ?

Also remember that the gasket is elastic. If the torque or force from one side causes deformation it, it applies a force on the surface.

For me the force on the gasket can be near zero. Except the problem of capillary action we can take the film very low as possible. And the square thread fixed the gasket on it so the deformation can be limited. It's only a technical problem.

I don't see where is another torque or another force which cancel this torque.

 

[Hassan2]

You had sin(θ)/cos(θ) which is equal to tan(θ) which is equal to 1/v . I just plugged this in your equation.

Just imagine that there is no water in the lower gap but the gasket is there. The upper force or torque deform the gasket till the thread comes to equilibrium. The force of the gasket would be as large as the hydraulic force.

 

[Gh778]

Just imagine that there is no water in the lower gap but the gasket is there. The upper force or torque deform the gasket till the thread comes to equilibrium. The force of the gasket would be as large as the hydraulic force

But there is a vertical force if we put only water at up surface. Vertical force is cancel with up and down surface only. The study is with up and down surfaces with water. The vertical force don't care about radius. If up and down surface have water, the up/down forces on gasket is 0.

The force on the gasket:

thickness of the gasket : 1µm
at bottom: 0.628 bar
on 1 meter of the gasket we have only : 1e-6*1*0.628*100000 = 0.0628 N

 

[Hassan2]

1. Even with the upper water film, you have a torque. Cancellation of the vertical force is obvious to you but how about the torque when only the upper gap is filled? You just calculated a net force due to the upper water. You need to show that this is canceled even when there is no water in the lower gap.

2. The calculation is not like that because the forces on the gaskets are not only the hydraulic forces ( these forces are negligible as you showed). The force on the thread presses it again the gasket. The thin gasket must take all the force to in the equilibrium. In case of torque cancellation,the force of the gasket must be large enough to cancel the torque of the water.

 

[Gh778]

You just calculated a net force due to the upper water.

Not at all, the upper force is give with square thread when d = 0 rd. The lower force is give with square thread when d=-0.5 rd (it's an example).

Show the 2 drawings below (circular thread for understand easily)

First drawing d=0 rd
Second drawing d= -0.5 rd
Third drawing with 2 threads

 

[Hassan2]

For d=.5 ,I get a big difference even for circular thread. In my code in my code
x=v*cos(u);
y=v*sin(u);
z=u;

I consider the d in the if branch:
if( z>d && (x>c1 && x<c2 && y<c2 && y>-c2) |...

 

[Gh778]

for d=-0.5 I don't find any difference with the circular thread: 1 240 313 with step =0.0005

Don't change on the third digit. Are you sure ?

 

[Hassan2]

OK I corrected my code. Now only for square thread I have a nonzero torque.

 

[Hassan2]

Now calculate the torque around two other axes, both passing through the center ( u= pi) one parallel with x-axis and another parallel with y axis. It must be zero for a circular thread but for a square thread I expect an nonzero one.

 

[Gh778]

Why ? if I put the square thread on an Z axis of rotation, others torque can't do anything, I think. For me the contrary torque must be on Z axis. But where ?

 

[Hassan2]

You are right they can't don anything if we have calculated them by correct forces. Because of the bearings, we don't need the other two torques because they are canceled by bearing not by the gasket.

But according to my calculation, there is a net force in -Z direction. This force must be canceled by the lowers gasket because nothing else can cancel them. The gasket forces are perpendicular to the thread surface. They can't have z component only. Their other component cause a toque around z axis. Can you calculate the total force? Then we can distribute it uniformly on the two cords of the lower gasket and recalculate the torque.

 

[Gh778]

But according to my calculation, there is a net force in -Z direction.

Are you sure ? The water is not connected from upper surface to lower surface. The square thread is cut horizontaly at bottom like at top. For me the up force is equal to the down force. See the drawing.

Edit: And another problem (without capillary effect) I can take a gasket very small like the thickness of water 1µm for example, I don't know how this small surface can cancel the torque on Z ?

 

[Hassan2]

In the code, i changed the torque equation to vertical force by dF=(uLimit-u)*v*pas*pas

Maybe the code is wrong!

Added: In the drawing, since the same heights have the same pressure, it is like connected water.

 

[Gh778]

For me it's logical, it's the same surface, the same height of water, how the Z forces can be differents ?

 

[Hassan2]

Actually you are right. The vertical forces can't be more than the water weight which is negligible.

Added: Something interesting is appearing here. Suppose we have no water in the lower gap. If we calculate the vertical force it becomes high even if the water eight is negligible. So the force has nothing to do with the weight.

 

[Gh778]

Something interesting is appearing here. Suppose we have no water in the lower gap. If we calculate the vertical force it becomes high even if the water eight is negligible. So the force has nothing to do with the weight.

I think it's important not forget the fixed thread for see what weight the water give.

The vertical forces can't be more than the water weight which is negligible.

I don't think like that. I'm seeing all forces on the fixed AND square threads, this must done the weight of water. But the square thread see what ? A force up equal to surface S by the mean of height of water less a force down equal to surface S by the mean of height of water. Surfaces are equals, heights of water are equals so the square thread see 0 force up/down.

But for the torque, I don't see where is my error. The program seems to be good. The vertical force is 0. This is strange for now.

Thanks for your help and your patience Hassan2 ;)

 

[Hassan2]

I improved the code ( program). The results show that the torque and the force for both surfaces are the same , except for small numerical errors: here is the result:

d: 0.000000 The total torque is: 221715.207569 Total Force: 8.803893
d: 0.314159 The total torque is: 221715.207569 Total Force: 8.803893
d: 0.628319 The total torque is: 221714.721935 Total Force: 8.803872
d: 0.942478 The total torque is: 221713.735984 Total Force: 8.803829
d: 1.256637 The total torque is: 221712.480470 Total Force: 8.803783

 

[Gh778]

Can you give your code ?

 

[Hassan2]

double x=0.0,y=0.0,z=0.0; // each point of the circular thread, I use circular thread and after select only points which are on the square thread
double u=0.0,v=0.0; // u=altitude, v=radius
double a=0.0; // angle
double torque=0.0, force=0 ;// local torque
double torqueT=0.0,forceT=0; // total torque
double pi=3.1415927;
double vLimit=2*pi, uLimit=2*pi; // limit of integration, in meters
double c1=3, c2=4; // limit of the square thread in meters
double d=0;
double P=0;
int Nu=50000;
double pasu=uLimit/Nu;
int Nv=5000;

double vMax=1.02*c2*sqrt(2);
double pasv=(vMax-c1)/Nv;
v=c1;

do
{
do
{
x=v*cos(u-d);
y=v*sin(u-d);
z=u;

if((x>c1 && x<c2 && y<c2 && y>-c2) || (x<-c1 && x>-c2 && y<c2 && y>-c2) || (y>c1 && y<c2 && x<c2 && x>-c2) || (y<-c1 && y>-c2 && x<c2 && x>-c2))
{

P=(uLimit-z)/10.0;

force=P*v*pasu*pasv;
torque=P*pasu*pasv;
torqueT+=torque;
forceT+=force;

}
v+=pasv;
}while(v<vMax);
v=c1;
u+=pasu;
}while(u<uLimit);

 

[Gh778]

1/ Are you sure about your results, I have test with d=-0.5 rd and I have the same torque with my code ?

2/ I don't understand Vmax in the code could you explain ?

Edit

3/ What about the energy:

d=0.5 rd, T=219075 Nm, F=859799 N
d=0 rd, T=221516 Nm, F=879656 N

Difference:
T=2441 Nm
F=19857 N

On 1 turn, 6.28 m, 2pi
Wt=T*2*pi=15337 J
Wf=F*6.28=124701 J

Are you sure about the Z force ?

Edit: there's a problem with Z force because F mean is like 220000 N = (4^2-3^2)*pi/10*100000

 

[Hassan2]

Sorry, there was a problem in my code. They don't become the same.

The Z force is not multiplied by 100,000. so the force is about 8.7 N. Yes there is a net force. Perhaps we are making a mistake in integral limits.

Edit. It is multiplied. Yes there is a problem. let's find it.

 

[Hassan2]

The side of squares are 6 and 8, not 3 and 4.

 

[Gh778]

you're right ! so what's the problem with energy ?

 

[Hassan2]

The differences seems to be error because we use differential area of polar coordinates for a square, which is wrong. It gives large errors. both the torque and the force are the same for different d. The reason is that when d is a multiple of pi/2 , the torque and force are the same. in fact d=pi/2. In fact with d= pi/2, the errors are equal because the square is in the same position of when d=0.

 

[Gh778]

It gives large errors.

even if the step is very small ? I don't understand.

You're agree with that: d=0 rd, T=221516 Nm, F=879656 N ?

both the torque and the force are the same for different d

I'm sure the torque is not equal at different position of d. But now, I think the difference of the vertical force must cancel the energy from the differential torque.

Edit: I have an idea for confirm results. When the radius increase of the thread, sure the surface increase for torque and for force too, but the attack angle decrease when the radius increase, this decrease the torque and increase Z force. Maybe it's for that ?

Edit2: With your code, for d=-0.5 rd and Nu=500000 and Nv=50000 we have T=223955 Nm and F=899509 N

 

[Hassan2]

You can test with a constant pressure ( P=1). Still there is an error. When the pressure depends on z, the error is much higher though. Also, it's not that torque decrease or increases with d. The changes are kind of sinusoidally ( if you draw the torque vs. d).

 

[Gh778]

1/ I would like to find where is the error in the code or in the method

2/ If I delete "-z" for have the same pressure in my code, I have the same result for the torque with d=0 or d=-0.5

With step = 0.001:
d=-0.5 => T=443083
d=0 => T=443076
d=0.5 => T=443082

With step = 0.0001:
d=-0.5 => T=443027.75
d=0 => T=443027.61
d=0.5 => T=443027.74

 

[Hassan2]

Can you do the calculation for uLimit = pi/4, for a constant pressure? It must be independent of d too ( I think).
In my opinion. When we integrate over 2*pi or even over pi/2 or pi, the "area" errors cancel one another due to symmetry. For a variable pressure, the " area" errors are not canceled because in fact they are multiplied by different pressures.

Sorry: In the previous message, I forgot to mention that I integrate over pi/4 radians.

 

[Gh778]

I done the calculation, you're right the torque is not the same. But for me it musn't be the same (even the pressure is constant) because the radius is not the same, the angle too, so why the torque must be the same ? It's not a part of circular thread it's a part of square thread. The radius change like I said in the message #87, see below:

θ change with the radius, it is the slope of the thread. The pitch is the same at every radius but the length is bigger when the radius is increasing. With pitch = 1, when we pass from r=1 to r=3 we pass from sin(θ)=0.707 to sin(θ)= 0.316. The torque is not the same at each point of the thread, bigger if the radius is bigger.

I don't understand where you see a problem with the numerical integration, can you explain more please ?

Edit: When we calculate for pi/4, with d=0 the radius is more close to the center. With d=-0.5 for example we are more far from the center.

 

[Hassan2]

See the figure. The code checks the black points being in the region. If yes, it includes the element area in the summation.
The code can be improved but still the error would exist but smaller one.

 

[Hassan2]

But the error must become small when we choose smaller mesh. There must be another problem.

Edit: I was wrong. For pi/4 the force is not independent of d. As the square rotates, the area of thread which is in the first quarter varies.