Torque on a square screw full with water

[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).

 

[Hassan2]

Yes if we put the thread in water, we the buoyancy force do work for us. that's fine. But if the thread has a weight, we should consider it. If the thread is put in the water and the density is one, we should get the vertical force equal to the weight.

 

[Gh778]

Maybe for understand what I said it's better to continue to have water up/down surfaces not external/internal surfaces. But add water at bottom of the square thread. Yes, with density of 1, the square thread has a weight equal to the up force give from sum of up/down surfaces and bottom surface (the small surface under the square thread). With water at up/down surfaces and water under bottom surface, the weight is cancel with up force from water. The torque is always here and turn down the square thread and give energy. Sure we need to remove water, but water is at the same height and this don't change the potential energy of water I think.

Edit: and another strange thing, it's the surface at bottom of the thread. If we stop the thread at corner or in the middle of a side, the top/bot surface is not the same. The up/down surface torque cancel the up force, ok. But if the surface at bottom change the up force change.

 

[Gh778]

I imagined 3 cases :

1/ Up and down surfaces only with water (the code case), like we calculated T=F for every d

2/ All system in water (fixed threads and square thread) except up and down surfaces with air. The square thread has a density of 1. The weight can’t be cancel by up and down surfaces. The small bottom surface change with d, so the up force change with d. So the torque give by external/internal forces must cancel the difference of up force. So torque T1= f(d)

3/ We put only square thread in water without fixed threads. The square thread can’t move up, it turn only. How T=T1 ? Because T is not function of d but T1 is function of d.

So if the case 3 is T=T1 how the case 2 can be exist ?

Edit1: it's ok for me, all is fine and all sum energy is 0 in all cases.

Edit2: Not really, in fact, the differential torque we calculated is aroudn 2% of total torque (example) but the surface increase more than 50% at bottom...

Edit3: With our code, for d=0 or d=pi/4 the torque is near the same=879805 or 879759. But the bottom surface is near 50% more. The surface at bottom when d=0 is Sd0=(c2-c1)*Δd*Rmoy=1*0.5*3.5=1.75 m². The bottom force is 1.75*6.28/10*100000=109900 N. 50 % more is 50000N more ! I will try to calculate with accuracy the bottom surface.

Edit4: why for d=-pi/8 and d=pi/8 we have not the same torque with our code ?

Edit5: The drawing show a square thread in water. The weight of the square is 0 (for simplified the study). The up/down forces are canceled with up forces like our code show. But it rest the another torque. I don't see where the system loss energy like that ? So I'm not sure the code is good.

 

[Gh778]

I think there is an error in the code because we can use the square thread in two cases:

1/ (first drawing) The square thread is in water all surfaces of the square thread are in contact with water. The thread can turn only around Z axis. The square thread can't move in Z axis or others. Our code said there is a torque Tu with up/down surfaces so the external/internal surfaces must give a torque Te which cancel this Tu torque because the thread can't turn alone in water !

2/ (second drawing) Like I said before, the thread pass through the water. We have only up/down surfaces and external/internal surfaces in contact with water. Our code said the square thread has a up force and the torque , this force cancel the torque. But we have Te=Tu (hypothesis 1/) which want to turn the square thread and this is not possible because the sum of energy must be 0.

I thought the code was good because we have compared it with known values. Have you an idea ?

 

[Hassan2]

For torque calculation. the code is for cases with no water on its external/external surfaces. If you want it for case 1 and 2, you should modify it.

 

[Gh778]

All seems ok, I understand my error(s), thanks

But for our code (with water only up/down surfaces), if the equation is:

x=v*cos(u-d);
y=v*sin(u-d);
z=u/2.0; // change the pitch

Torque = Force, but when we turn 2pi, we move up less than we turn.

Example:
d=0 => T=F=1319612
d=0.5 => T=F=1309722

Difference =9890

Energy torque = 9890 * 2 * pi = 62140 J
Energy force = 9890 * pi = 31070 J

There is a big diffenrece, it's my method ? I have change a lot of thing in the code but never I find Wt=Wf

 

[Gh778]

I understood my error, it's because the code don't take the pitch in parameters, we need to divided torque by 2 !

 

[Hassan2]

For a pitch higher than 1, the normal vector is different. for a pitch of m ( z= m*u), the normal vector becomes:

normal=(-msinu, mcosu , -v)

If we normalize it we have

$\vec{n}=(\frac{-msinu}{\sqrt{m^{2}+v^{2}}},\frac{mcosu}{\sqrt{m^{2}+v^{2}}}, \frac{-v}{\sqrt{m^{2}+v^{2}}})$

In a general method we can write the dT as the cross product of dF and r:

$\vec{dT}=\vec{dF} \times \vec{r}$

where


$\vec{r}=(vcosu, vsinu, 0)$

The direction of dF is the negative of the normal vector and its magnitude is

$\left|dF\right|=PdS=P v du dv /cos\beta =Pvdudv \frac{ \sqrt{m^{2}+v^{2}}}{v}$


and $\vec{dF}=-\left|dF\right| \vec{n}$

and $dF_{z}=-Pvdudv$

For torque we have

$\vec{dT}=-|dF|\vec{n} \times \vec{r} =P dudv( -v^{2}sinu, v^{2}cosu , mv)$

so and $dT_{z}=Pmvdudv$

 

[Gh778]

in this case for d=0 and m=0.5, F=1.3e6 and T=2.6e6 it's not the contrary we must find ?

 

[Hassan2]

in this case for d=0 and m=0.5, F=1.3e6 and T=2.6e6 it's not the contrary we must find ?

That's correct. The force and the torque are not the same but their work are the same.

Added: You are right. let me check the derivation.

 

[Hassan2]

The mistake was in the normal vector. I think its correct now. can you verify the normal with your method? see post 158

 

[Gh778]

all is fine like that.