Torque on a square screw full with water

[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.

 

[Gh778]

I would like to understand how you calculate the torque without use radius and why the force is dependent of the radius in your code, could you explain please ? The results seems to be good but I don't understand why. Tan(a)=1/v ?

EDit: ok I understood, from my code: tan(atan(uLimit/pi/2.0/v))=1/v all is fine, I think we need the other plan and this must be v for the force.

 

[Hassan2]

First of all many of my comments during the discussion were wrong but now we are at a point to agree that when the gaps are filled with water, the total vertical force depends on d. This is easier to understand if we choose a rectangular thread rather than a square thread. The lower surface is in fact rotation of the upper surface by d radians. You can show this by your software. Suppose the initial position is so that the tip of the longer side is at u=0; If u increases by say .2 radian, the longer tip moves up and a shorter part replaces it. So, we have less area in the higher pressure now, which means smaller force.

This force causes the square thread to sit on the gaskets and the result is a reaction force normal to the square thread. The force has a component to cancel the torque around Z axis.

 

[Gh778]

I don't see the problem like that but I try to understand your point of view.

1/ Ok, we're agree the Z force is dependant of d like torque, and we know the values of force and torque with great accuracy
2/ We're agree that is only Z move up/down and Z turn only

3/ For me (but it's only an intuition) the energy of the Z force must cancel the energy of the torque. Imagine a system which give energy for torque, this system recover energy of the move up/down of the square thread, so the sum of energy must be to 0. The system can adjust the tickness of the film for make it constant (never touch the fixed thread).

 

[Hassan2]

I don't know what kind of energy you are talking about. We have no moving object yet. When we calculated the torque and the force, in fact we had locked by assumption, so that the gaskets are not stressed. Now That we have the force and the torque, we can unlock thread and let move under force and torque. Immediately they cancel the effect of one another. A small amount of water may over flow from the lower gap and the level may come down a little, in the upper gap. Then their will be no motion to have energy.

 

[Gh778]

I turn the square thread between big fixed threads. Mechanically for me it's possible.

For you:

1/ Is it possible to turn mechanically the square thread with film of water always in the same position ? If not why ?

2/ If the film is constant, the torque and the force are always the same ?

3/ If you turn 2pi you move up 6.28m (the example of code) the energy must be the same I think.

 

[Hassan2]

1.Yes in our calculation, I assumed the water films to covering all the surfaces of the fixed thread and the square thread just sliding on it. Of course the water film must be contained in a very thin plastic bag, otherwise would spill out.

2. If the films were where constant, yes they would be the same. But in my assumption, the film is not fixed on the square thread.

3. We turn it by an external torque? Of course the system can't do work for us. I think we want to show that the net torque due to water is zero and it has become easier now. Its enough to add the gasket force to our calculation.

 

[Gh778]

2/ For me the gaskets are fixed to the square thread, so the film can't move (sure the mechanical system need to control move up/down and the rotation for that).

3/ Yes, we apply external torque for turn the thread.

Its enough to add the gasket force to our calculation.

For me like the gasket is like 1 µm of thickness (the thickness of the film) I don't know how the forces or torque on the gasket is enough for cancel the difference because the surface is so small ?

4/ Are you agree with 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

 

[Hassan2]

1. If you put a cube with 1000kg, on a ring of gasket with 1um diameter, of course the gaskets take all the force and react with the same force. The gasket force becomes equal to the cube weight.

2. Nice calculation! It helped me find a big mistake in the code. dM=P*v*du*dv/cos(θ)*sin(pi/2-θ)=dF exactly, because the angle between F and r is pi-θ

 

[Gh778]

1/ but the system recover energy, imagine a hydraulic cylinder for recover energy, nothing is put on the gasket

2/ So what's change in the code ? For now, we have verified datas with circular thread and with the square thread. For you F=T ? With (8²-6²)*2pi/2/10*100000= 879645N, it's that we found with code.

Edit: ok, all is fine for me now, Wt=Wf. Thanks

 

[Gh778]

I think something is wrong in our code because if we put the square thread directly in water (without fixed threads, only square thread) the thread can turn without move up/down and the sum of energy is not 0, we turn and have the energy of the torque but don't move up/down. Sure there are external/internal surfaces but like I said before they don't give torque (if the external/internal surface give torque, they give energy by create torque in the case we put the square thread between fixed thread, density of square thread=1 and all the system in water, except up/down surface with film of air).