Torque on a square screw full with water

[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.