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Author Topic: problem with helicoids and pressure  (Read 8946 times)

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #25 on: 27/03/2012 12:48:10 »
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Are you saying that fluid pressure contained by the green line will make the screws turn?
No, I'm not saying that.

I would like to know if for you the unique thread on reply #20 (drawing) turn and give torque ?
 

Offline Geezer

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Re: problem with helicoids and pressure
« Reply #26 on: 27/03/2012 20:08:45 »
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Are you saying that fluid pressure contained by the green line will make the screws turn?
No, I'm not saying that.

I would like to know if for you the unique thread on reply #20 (drawing) turn and give torque ?

Do you mean #21?
 
The answer is, generally, no. If it was yes, screws would be not be of any use because they would fall out all the time.
 
If you eliminate the rotational friction, and the friction between the point of contact and the thread, the screw will rotate. If you use a screw with a much greater pitch it will rotate much more easily.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #27 on: 27/03/2012 20:33:08 »
it's really the reply #20, but forget this reply I reformulate: for find my error I'm going step by step: I'm thinking with only one screw, 2 bars everywhere except at small red surface (only one side) where pressure is 1 bar. The screw can only turn not move in translation. Like this the screw turn and give a torque ? In this case, for me, the sum of energy is 0 because we need to move the small red surface in the same time. Are you agree with that ?
 
 

Offline Geezer

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Re: problem with helicoids and pressure
« Reply #28 on: 27/03/2012 23:23:45 »
The algebraic sum of the energies is always zero, but if some of the work energy produces friction, heat energy will be created and the work energy output will be less than the work energy input.
 
If there was no frcition at all in the system you could get as much work out as you put in, but no more. If there is any friction in the system (and there always is) the system converts work into heat and the system is less than 100% efficient.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #29 on: 28/03/2012 04:47:07 »
ok, the screw turn and give torque, the sum of energy is 0 and there is friction.

Second step: now, imagine you have 2 screws:

1/ one in your left hand where outside there is 1 bar except at small red surface where pressure is 2 bars (red surface at down), this screw turn and give torque, the sum of energy is 0 and there is friction. The screw turn in a direction.
2/ one in your right hand where outside there is 3 bars except at small red surface where pressure is 2 bars (red surface at up), this screw turn and give torque, the sum of energy is 0 and there is friction. The screw turn in the same direction than other screw.

Now, move closed 2 screws for put small red surfaces in front of (one is down, other up). The first drawing show a lot of volume where pressure is 2 bars but we can have only one. The volume where pressure is 2 bars is composed by 2 parts of screws down and up and 2 longitudinals surfaces (gaskets). For me, screws turn and give torque, there is friction (no more before) but for the energy the volume where there are 2 small red surfaces don't lost energy (for me) because 2 surfaces are longitudinal axis (gaskets) and 2 surfaces (red surfaces) are screws and screws don't move in translation they turn only. The small volume where pressure is 2 bars move in the longitudinal axis. The volume where pressure is 2 bars is constant. With one screw, the screw give energy and we lost the same energy by move the red surface. Here, we don't lost energy by move red surface because it's the other screw that diseapear the other side of the red surface. Finally, we have the system like first drawing show.

Sure there are 3 pressures, but for one screw we can say everwhere outside the system there is 1 bar, for the red volume (interface) the gaskets can controlled pressure. For other screw, the pressure can be controlled by thin film of pressure and thin film of dynamic wall, this need system for controlled walls but in theory this don't need energy (if friction is not 0 we lost energy but why 100% of energy we received from screws ?). See second drawing for that, it's a part of black screw for see details.
« Last Edit: 28/03/2012 05:48:48 by fgt55 »
 

Offline Geezer

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Re: problem with helicoids and pressure
« Reply #30 on: 28/03/2012 06:44:51 »

ok, the screw turn and give torque, the sum of energy is 0 and there is friction.


There is no need to go any further. You have just proved it cannot produce free energy.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #31 on: 28/03/2012 15:25:49 »
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There is no need to go any further.
Ok, I understand for one screw (for one screw I knew sum of energy is 0) but not for two, it's the problem for me and if I need to explain I can't say with 2 screws it's the same problem... Just for understand the all system ;) Please :)
« Last Edit: 28/03/2012 15:35:54 by fgt55 »
 

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Re: problem with helicoids and pressure
« Reply #32 on: 28/03/2012 18:30:09 »
But I don't understand the system. I don't even know what you mean by "pressure". Is this a fluid pressure, or are you really referring to a force?
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #33 on: 28/03/2012 19:03:14 »
sorry, I'm not fluently in english and I'm not a specialist in physics :(

"pressure": it's air pressure (for example, use anybody fluid) at 1 bar, 2 bars or 3 bars

ask me if you don't understand others things in my message
 

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Re: problem with helicoids and pressure
« Reply #34 on: 28/03/2012 20:26:14 »
OK - So what you have is two screws that mesh together and form a gas-tight seal where they touch (no leaks) and the seal encloses a volume of high pressure gas. The two screws are rotated by some external gearing system.
 
What is suppsed to happen when the screws are rotated?
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #35 on: 28/03/2012 20:58:22 »
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What is suppsed to happen when the screws are rotated?
With 2 screws, each screw is a motor it give energy to a mechanical receptor (if friction is low). Screws are not receptors they are motors. It's not a screw which give a torque to another. This is the differential of pressure of fluid that give a motor torque to each screw. Like screws give energy to receptors we need lost energy somewhere, where ?
« Last Edit: 28/03/2012 21:04:46 by fgt55 »
 

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Re: problem with helicoids and pressure
« Reply #36 on: 28/03/2012 21:36:59 »
Ah, right.
So the thing is a motor that is driven by the P3 pressure?
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #37 on: 28/03/2012 21:45:54 »
"the thing" ?

Black screw has everywhere around it 3 bars except at interface (small volumes between screws) where pressure is 2 bars => the screw turn and give a torque

Red screw has everywhere around it 1 bar except at interface (small volumes between screws) where pressure is 2 bars => the screw turn and give a torque

Like screws turns and can't move in translation, interfaces (small volumes between screws) move up/down, but each interface is composed by 2 surfaces from screws and 2 gaskets's surfaces. Like 2 gaskets's surfaces can be in longitudinal axis we don't lost energy when the interface move.
 

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Re: problem with helicoids and pressure
« Reply #38 on: 28/03/2012 22:34:00 »
"the thing" = the device.
 
OK, so why would the higher pressure make the screws rotate?
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #39 on: 29/03/2012 06:24:23 »
It's differential of pressure of fluid that rotate each screw. Each screw can only turn (bearing). We can thinking with only one screw, the black for example, it has all around 3 bars except at small surface where pressure is 2 bars. This rotate the screw I think ? See drawing please.
« Last Edit: 29/03/2012 06:39:25 by fgt55 »
 

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Re: problem with helicoids and pressure
« Reply #40 on: 29/03/2012 08:31:33 »
But differential pressure will not produce rotation. There is differential pressure between the inside and outside of your bottle of Fanta, but nothing actually happens until you take the cap off the bottle.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #41 on: 29/03/2012 08:51:28 »
but the red surface move it's not a static surface, follow the screw, move in translation.

And the red surface is only on one side of the screw, not 2 sides. See drawing please.
« Last Edit: 29/03/2012 16:45:31 by fgt55 »
 

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Re: problem with helicoids and pressure
« Reply #42 on: 29/03/2012 18:04:33 »
You are missing a very important point.
 
The screws enclose a constant volume of gas even when they are rotated by an external force. It's just the same as the gas inside a sealed bottle. Differential pressure won't do anything uless you allow the gas to expand.
 
Engines that use gas to do work, like turbines, pistons etc. are frequently called "expanders" for that reason.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #43 on: 29/03/2012 19:33:24 »
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The screws enclose a constant volume of gas even when they are rotated by an external force. It's just the same as the gas inside a sealed bottle. Differential pressure won't do anything uless you allow the gas to expand.
You're explain for 2 screws or one ? If it's for 2 screws, I'm agree with your explanation but I'm trying to understand this device with sum of forces (torques too). Because my son can only understand this, sorry :(

It's not the sum of forces that rotate screws ?

For you one screw turn like that (for me one srew turn, it's only a problem with sum of forces, sure with one screw I can explain sum of energy is 0, the difference of pressure rotate with torque the screw but I need the same energy for move red surface) ? If yes, why two screws don't rotate (explanation with forces) ? My son come back tomorrow and I would like to explain this wk, thanks for your help :)
« Last Edit: 29/03/2012 20:06:38 by fgt55 »
 

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Re: problem with helicoids and pressure
« Reply #44 on: 29/03/2012 21:33:03 »
Just because there is a force, it does not mean there is a transfer of energy.
 
The mechanism you describe prevents the "3P" volume from ever changing. If you think there is an energy transfer (which is the basis of all machines that do work) you need to explain how the volume changes.
 
If you look at it very carefully, you will discover that the volume never changes, so it might as well be a bottle of Fanta.
 
It's cleverly designed to deceive, but if you want it to do anything useful, one of the spirals has to be reversed as it is in supercharger link I posted.
 
Do you think the volume changes?
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #45 on: 29/03/2012 21:47:58 »
No, for me the volume don't change, the device is like that. I understand your point of view and it's ok for the sum of energy when I think with volume/pressure. But when I think with force, I see a force that rotate each screw in the same direction, I don't understand why there is no torque ? The torque come from force, like force is not 0, torque is not 0. And like screws can turn mechanically they can give energy to receptors. For me there is a contradiction here.
 

Offline Geezer

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Re: problem with helicoids and pressure
« Reply #46 on: 30/03/2012 00:31:52 »
I think the problem is that there really is no differential pressure, even though the drawaing suggests there should be. I don't understand how the seals (gaskets) described are supposed to be arranged in a real system.
 
If you can draw a practical arrangement, I think you will find the forces all sum to zero.
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #47 on: 30/03/2012 07:10:47 »
This afternoon, I'll build 2 helicoids in cardboard for my son, like that we can see the exact volume and gaskets's surfaces. For now, a 3d software give the interface like the drawing show, for me it's like I see with 2 real screws I have in front of me. Second drawing is top view for see gaskets's surfaces are in longitudinal axis.
« Last Edit: 30/03/2012 07:16:33 by fgt55 »
 

Offline fgt55

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Re: problem with helicoids and pressure
« Reply #48 on: 02/04/2012 14:22:19 »
All is fine with sum of energy. In fact it was the separation (sides) of fixed pressure, walls can't stay fixed, they need to move and this lost energy.
 

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Re: problem with helicoids and pressure
« Reply #48 on: 02/04/2012 14:22:19 »

 

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