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Reactionless device using the principle of Pascal for fluids


esposcar
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Just now, Ghideon said:

Are the white balls required for the setup; what kind of effect are they supposed to have on the outcome? If "all options are good" then why not remove the white balls allow for an easier analysis?

You are absolutly right, but it would require the input pistons to have a different dispacement length to compensate the friction and yes, could work also, of course. But I thought that maybe it would be more descriptive.

12 minutes ago, esposcar said:

You are absolutly right, but it would require the input pistons to have a different dispacement length to compensate the friction and yes, could work also, of course. But I thought that maybe it would be more descriptive.

I will post three frame pic without ball and without maths. Maths are already done, and we will see which piston pusher arrives with more speed to its corresponding top, that would do the picture more clear for analisis.

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26 minutes ago, esposcar said:

You are absolutly right, but it would require the input pistons to have a different dispacement length to compensate the friction and yes, could work also, of course. But I thought that maybe it would be more descriptive.

Ok! But it might be more descriptive if friction is shown as an arrow, labeled "friction force" or similar? To me it isn't obvious why friction is best compensated for by the white balls; they affect the outcome differently.

26 minutes ago, esposcar said:

I will post three frame pic without ball and without maths. 

Ok! Math will be required at some point later. But initially it is good to have a complete and consistent picture of the setup. It is probably necessary to understand* what the math is supposed to model.

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1 hour ago, Ghideon said:

Ok! But it might be more descriptive if friction is shown as an arrow, labeled "friction force" or similar? To me it isn't obvious why friction is best compensated for by the white balls; they affect the outcome differently.

Ok! Math will be required at some point later. But initially it is good to have a complete and consistent picture of the setup. It is probably necessary to understand* what the math is supposed to model.

Yes exactly. I had some problems with the render. I post one clear pic that show what are the intention of the device. There is a new component. I have attached bars perpendicular to the piston-pushers, and as the pushers advance, so the bar will, and they will hit their "wall". But one bar will arrive with more speed than its counterpart and because of that will carry more momentum.

Paso1.png.ea8cf6d22be44da5c3fc286063849159.png

Put a heavy mass over the output piston of the left, and you will have even a bigger difference of speed ;)

Edited by esposcar
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3 hours ago, esposcar said:

If you disagree then prove it wrong empirically, I have done my homework, do yours. And when you can show that what I show is wrong mathematically, you reply me and I will apologize and realize that you show me mathematically that does not work, so stop the bla bla bla and do the homework.

In speculations, the onus is entirely upon the promoter (you in this case).

So I don't have to prove anything.

 

However since my objective is to save you a great deal of fuitile and fruitless work (which is why I have been so plain and blunt).

Fluid Mechanics is quite different from Solid Mechanics.

For instance it is impossible to apply a force of any description to a fluid.

Think what happens if you drop a 100kg cannonball into a tank of water.

Does it apply 1000N to the water?

Of course not.

It just falls through the water.

So let us look at a very simple pipe system.

plugs1.jpg.dcf655fea0769f8f3bb044c3a2ce3d94.jpg

 

A pipe with plugs at A and B containing a fluid.

Starting (as you have done) assuming there is zero fluid friction, what happens if you push with any force whatsoever as shown in the diagram?

Do you think you develop any pressure within the fluid?

The answer is no you don't

You will find, if you try this experiment, that you simply displace plug A sideways until you stop pushing.

You will also find that you cannot apply a specific force F of say 10, 100 or 1000 N.

(in practice all you will need is enough push to overcome the friction holding the plugs in place, but you cannot increase the push beyond that.)

 

Can you now see what is missing from your analysis?

 

There is nothing pushing back at A to overcome.

 

If you are as good at mechancis as you claim, draw a free body diagram of the fluid.

That would be the correct way to analyse the situation.

Edited by studiot
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4 minutes ago, studiot said:

In speculations, the onus is entirely upon the promoter (you in this case).

So I don't have to prove anything.

 

However since my objective is to save you a great deal of fuitile and fruitless work (which is why I have been so plain and blunt).

Fluid Mechanics is quite different from Solid Mechanics.

For instance it is impossible to apply a force of any description to a fluid.

Think what happens if you drop a 100kg cannonball into a tank of water.

Does it apply 1000N to the water?

Of course not.

It just falls through the water.

So let us look at a very simple pipe system.

plugs1.jpg.dcf655fea0769f8f3bb044c3a2ce3d94.jpg

 

A pipe with plugs at A and B containing a fluid.

Starting (as you have done) assuming there is zero fluid friction, what happens if you push with any force whatsoever as shown in the diagram?

Do you think you develop any pressure within the fluid?

The answer is no you don't

You will find, if you try this experiment, that you simply displace plug A sideways until you stop pushing.

You will also find that you cannot apply a specific force F of say 10, 100 or 1000 N.

(in practice all you will need is enough push to overcome the friction holding the plugs in place, but you cannot increase the push beyond that.)

 

Can you now see what is missing from your analysis?

 

There is nothing pushing back at A to overcome.

 

If you are as good at mechancis as you claim, draw a free body diagram of the fluid.

That would be the correct way to analyse the situation.

The first example of the water that you describe like that is not a pascal system. Surf a bit in the net and do some research. At this level I prefer proves than bla bla bla, but anyway, your opinion is taken in consideration. Thank you.

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13 minutes ago, esposcar said:

The first example of the water that you describe like that is not a pascal system. Surf a bit in the net and do some research. At this level I prefer proves than bla bla bla, but anyway, your opinion is taken in consideration. Thank you.

Where exactly did I say it was?

Please do not imply something I did not say, particularly if you are going to be rude about it.

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7 hours ago, esposcar said:

I post one clear pic that show what are the intention of the device.

Thanks! So now we have a complete but simpler picture describing the suggested propulsion mechanism, OK?

Some more questions:
-According to your idea; in which direction is the device supposed to move in the latest picture? 
-When will the device accelerate? The amount of acceleration and is not important yet, I just want to know at what events it accelerates. For instance; Does the device accelerate as soon as the pistons initially affect the fluids? Or does the device start to move when the rods finally hit the walls?  

I 've a few ideas about an (even more) simplified view of the setup, hopefully useful to highlight some issues. But I need the direction of expected movement to draw it correctly.

 

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8 hours ago, Ghideon said:

Thanks! So now we have a complete but simpler picture describing the suggested propulsion mechanism, OK?

Some more questions:
-According to your idea; in which direction is the device supposed to move in the latest picture? 
-When will the device accelerate? The amount of acceleration and is not important yet, I just want to know at what events it accelerates. For instance; Does the device accelerate as soon as the pistons initially affect the fluids? Or does the device start to move when the rods finally hit the walls?  

I 've a few ideas about an (even more) simplified view of the setup, hopefully useful to highlight some issues. But I need the direction of expected movement to draw it correctly.

 

- Obviously on the direction that the input piston have more speed, no other way, so in this case to the right.

- It accelerate at any moment, it can accelerate for example as you see in the pic, in that moment, that the distance is the double before the rod hits in the smaller area input piston and the rod that belongs to the input piston with more area the rod is settled at half distance for the friction compensation, so meanwhile exist that distance difference, it will work at any point you want to produce an acceleration on both systems using the same force at both sides.

- About accelerating meanwhile the rods are contact with the wall, I dont think it would work if they need to have some distance to accelerate before the rod hitting.

- The direction will go in the direction of the smaller piston area, that means to the right.

-Piston is in constant contact with the fluid and have a pression on it

Edited by esposcar
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1 hour ago, esposcar said:

- Obviously on the direction that the input piston have more speed, no other way, so in this case to the right.

- It accelerate at any moment, it can accelerate for example as you see in the pic, in that moment, that the distance is the double before the rod hits in the smaller area input piston and the rod that belongs to the input piston with more area the rod is settled at half distance for the friction compensation, so meanwhile exist that distance difference, it will work at any point you want to produce an acceleration on both systems using the same force at both sides.

- About accelerating meanwhile the rods are contact with the wall, I dont think it would work if they need to have some distance to accelerate before the rod hitting.

- The direction will go in the direction of the smaller piston area, that means to the right.

-Piston is in constant contact with the fluid and have a pression on it

Ok! Do you then agree on the following description* of the propulsion?

  1. Starting position is the setup in the picture below.
  2. Push both the pistons with the same force F in opposite directions (Black arrows in the picture).
  3. Both the pistons will move, pushing liquid, but at different speeds: Sl and Sr (Speed Left and Speed Right, Direction shown by the same black arrows in the picture).
  4. The complete rig will accelerate to the right.
  5. Stop pushing the pistons (Force F=0) before rods hit the walls**. 
  6. The rig will not accelerate anymore but continue to move with constant speed to the right.

So:
-The propulsion of the rig depends only on the speed difference between Sl and Sr, as long as there is a speed difference while force F is equal in each direction? 
-As long as the piston areas are different on the left and right side there will be propulsion when identical force is applied, Ok? 

17 hours ago, esposcar said:

Paso1.png.ea8cf6d22be44da5c3fc286063849159.png

 

 

*) Note; this is not a description of what I think will happen. It is an attempt at simplifying the analysis.

**) there are other options but I chose this case for simplicity.

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22 hours ago, Ghideon said:

Ok! Do you then agree on the following description* of the propulsion?

  1. Starting position is the setup in the picture below.
  2. Push both the pistons with the same force F in opposite directions (Black arrows in the picture).
  3. Both the pistons will move, pushing liquid, but at different speeds: Sl and Sr (Speed Left and Speed Right, Direction shown by the same black arrows in the picture).
  4. The complete rig will accelerate to the right.
  5. Stop pushing the pistons (Force F=0) before rods hit the walls**. 
  6. The rig will not accelerate anymore but continue to move with constant speed to the right.

So:
-The propulsion of the rig depends only on the speed difference between Sl and Sr, as long as there is a speed difference while force F is equal in each direction? 
-As long as the piston areas are different on the left and right side there will be propulsion when identical force is applied, Ok? 

 

*) Note; this is not a description of what I think will happen. It is an attempt at simplifying the analysis.

**) there are other options but I chose this case for simplicity.

Good morning. Well it is getting interesant, because me myself cannot believe that reactionless devices are possible, but I am finding in this devices, something that makes of this case special, is that in this last device, there are maths that support it and works excellent with the laws of energy conservation for example, because if both input pistons, would run at the same speed with the same force, then one of them (smaller area, input piston) would not respect the conservation of energy, because both of the have the same size of output piston, so in order that both output pistons comply with the conservation of energy, oone of the input pistons, if have half of the area of the other, will have to run the double to compensate its lack of area and push at the same rythim its output piston, so it will have to have double speed. You can also brake easly one of the input pistons, by setting a mass in the way of the output piston. There are plenty of options, and the most interesant issue here, is that all of them fits the maths unless I forgot something really important, and thats why I am here. But I liked your enthusiasm and must have a reason of beeing ;). Ok answering your questions:

1- Yes totally

2- Exactly, you dont need to apply more force in one than in another (Obviously it would not work), the conservation of energy of each system will compensate the output piston rate, with more speed fron the input piston with smaller area.

3- Yes exactly, thats the trick of the device and from where obtains the thrust. You have two input pistons with different areas and carrying different momentum, so that makes an unbalance of forces.

4- Exactly

5- Yes there are plenty of options. Or you can keep the force up to the end. If the device works, that wont matter. The important is that the different rodes strike with different speeds (more momentum more speed) the wall, and that proportionally one of them have to do the double of distance in the same time to compensate the more friction of the bigger area input piston.

6- Exactly, we have to make some concessions to Newton Lol. In an enviroment of space it will reload the system and repeat all again, so that would do it accelerate all the times you want. If the system takes 2 minutes to reaload and generate a constant speed in each load, then..... I know the consecuences of what I am saying, but if works, it means this.

* Note that it can even work perfectly with 2 extendible hydraulic rodes that push the pushers, even in that case (if that would not work, the device neither) it will work marvelous Lol

 

 

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34 minutes ago, esposcar said:

Good morning. Well it is getting interesant, because me myself cannot believe that reactionless devices are possible, but I am finding in this devices, something that makes of this case special, is that in this last device, there are maths that support it  

Since we know at a very broad level that this cannot work, the only valid conclusion is that the maths don't actually support it, and that you have erred in the application of the physics.

It's very easy to make claims that are not supported by physics, as you have shown several times already in this thread (To your credit, you have been open to correcting your mistakes)

On 3/19/2019 at 6:39 PM, esposcar said:

Yes exactly. I had some problems with the render. I post one clear pic that show what are the intention of the device. There is a new component. I have attached bars perpendicular to the piston-pushers, and as the pushers advance, so the bar will, and they will hit their "wall". But one bar will arrive with more speed than its counterpart and because of that will carry more momentum.

Paso1.png.ea8cf6d22be44da5c3fc286063849159.png

Put a heavy mass over the output piston of the left, and you will have even a bigger difference of speed ;)

 For this to work for propulsion, you need to also analyze what happens when the system reverses itself. You can make a box move by rearranging the contents, because the center of mass is what can't move without an external force. But that defines the limit of its motion. You can't end up with continual propulsion from that. All you get is the box rocking back and forth.

What part of your system isn't reversible, regarding momentum and mass location?

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18 minutes ago, swansont said:

Since we know at a very broad level that this cannot work, the only valid conclusion is that the maths don't actually support it, and that you have erred in the application of the physics.

It's very easy to make claims that are not supported by physics, as you have shown several times already in this thread (To your credit, you have been open to correcting your mistakes)

Please (no sarcasm on it) show me what rules have not been repected, thats what I want to know. Conservation of energy is respected in each device separatdley, one piston with half area will have to run the double than its peer in order to have the same filling rate of both output pistons (same area both). So obectivly and tell me why my maths are not right in this case for example? Considering that one of the input pistons will do the double of work than its peer (this is a fact), or they do the same work? Because this is the soul of the device, if both input pistons do the same work, it wont work, simple as this.

Edited by esposcar
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I am really expecting what you come up with. To be honest I am sure something must be missing, but I want an objective reason. I looking forward for it.

52 minutes ago, swansont said:

Since we know at a very broad level that this cannot work, the only valid conclusion is that the maths don't actually support it, and that you have erred in the application of the physics.

It's very easy to make claims that are not supported by physics, as you have shown several times already in this thread (To your credit, you have been open to correcting your mistakes)

 For this to work for propulsion, you need to also analyze what happens when the system reverses itself. You can make a box move by rearranging the contents, because the center of mass is what can't move without an external force. But that defines the limit of its motion. You can't end up with continual propulsion from that. All you get is the box rocking back and forth.

What part of your system isn't reversible, regarding momentum and mass location?

I have red your comments below just now, sorry. Ok you are describing a total Newton system, and for that, if a device operates on those rules it will happen what you say. But the original of this proposition is that if we think in Newton laws, it takes the input pistons as for example repel function, and the reality is that the magnet etc that do the force over the input piston, it will encounter more force that repels him than in the other side. Think that one magnet, will find more force against him than the other, and that combined makes a descompensation of forces. One magnet will have more repulsion against him, the one that moves the bigger area piston and the other will have less repel reaction, because does less resistances, so one magnet will push more the other. Try to imagine what I am saying applied to the device, and it have a lot of sense. A reactionless device that is supposed to work just with Newton rules, it will never work, asi most of the reactionless devices that have been invented, with the exception maybe of the EM Drive, but this proposition bring extra force to Newton laws departing for another law. And Pascal law we can see it everywhere. From the excavator working next to your neighbourhoood, to all the factories machines in the world.

Sorry I did not understood your last question, could you clarify it?

If you think about it, the machines that operates with pascal laws, they use less force to generate more force. Imagine that Pascal principle was not invented, and somebody brings the idea and says I have invented a machine that multiply forces. Everybody would come with, thats not possible, it does not respect Newtons law...

Edited by esposcar
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1 hour ago, esposcar said:

Please (no sarcasm on it) show me what rules have not been repected, thats what I want to know. Conservation of energy is respected in each device separatdley, one piston with half area will have to run the double than its peer in order to have the same filling rate of both output pistons (same area both). So obectivly and tell me why my maths are not right in this case for example? Considering that one of the input pistons will do the double of work than its peer (this is a fact), or they do the same work? Because this is the soul of the device, if both input pistons do the same work, it wont work, simple as this.

Conservation of momentum. Energy is not the only property that must be conserved.

F = dp/dt

There is no net external force on the box, therefore, its momentum can't change.

If you come to the conclusion that the momentum is changing, then you have misapplied physics. The only question is where the error is.

48 minutes ago, esposcar said:

 A reactionless device that is supposed to work just with Newton rules, it will never work, asi most of the reactionless devices that have been invented, with the exception maybe of the EM Drive, but this proposition bring extra force to Newton laws departing for another law. And Pascal law we can see it everywhere. From the excavator working next to your neighbourhoood, to all the factories machines in the world.

Pascal's law does not violate Newtonian physics.

Quote

Sorry I did not understood your last question, could you clarify it?

What happens when you run your system in reverse? i.e. when you return to the original state?

 

Quote

If you think about it, the machines that operates with pascal laws, they use less force to generate more force. Imagine that Pascal principle was not invented, and somebody brings the idea and says I have invented a machine that multiply forces. Everybody would come with, thats not possible, it does not respect Newtons law...

No. Newton's laws tell you how motion and forces are related. There is no conservation of forces. 

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49 minutes ago, esposcar said:

If you think about it, the machines that operates with pascal laws, they use less force to generate more force. Imagine that Pascal principle was not invented, and somebody brings the idea and says I have invented a machine that multiply forces. Everybody would come with, thats not possible, it does not respect Newtons law...

That is not accurate.  Using hydraulic pistons to increase the force is completely in line with Newtonian physics.  The work done by the 2 pistons will be equal.

F1D1 =  F2D2

The smaller piston will have a lower force but has a longer distance traveled that the larger piston.

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44 minutes ago, swansont said:

Conservation of momentum. Energy is not the only property that must be conserved.

F = dp/dt

There is no net external force on the box, therefore, its momentum can't change.

If you come to the conclusion that the momentum is changing, then you have misapplied physics. The only question is where the error is.

And how can I deny such statement. Of course the laws have to be respected, its exactly why I want to detect. In this case applied to a physical problem(think about it of a physics teacher exam for the studients. What can merge from so many harvard brilliant brains to find the error? that would be great to see. But there is one thing clear, and thats why I red the forces and variables that the members here are describing, but if you read the literature, it applies to One enclosed system working entirely on Newtons laws. But what makes in my humble opinion this device interesant and I want to put a big enphasis on this statement.

This device in general its an enclosed system itself, but their components responds to actually, one general system that operates from Newton laws and 2 separate enclosed systems that responds to Pascal laws. You got a total of 3 systems in one device system. And the only forces of Newton laws that will operate through the input pistons, is the friction, thats why I put a lot of enphasis on that. The opossing resistance that each magnet will face, will be one more resistant than the other and more repulsion against, but the resistance is not provided by Newton laws, its provided by Pression (acts up as down, as right and everywhere with the same force) that is provided by Pascal principle. To be clear, literally a big part of the momentum pushing the pistons will literally get diluted in the pression, the rest goes to Newtons world, the friction force.

32 minutes ago, Bufofrog said:

That is not accurate.  Using hydraulic pistons to increase the force is completely in line with Newtonian physics.  The work done by the 2 pistons will be equal.

F1D1 =  F2D2

The smaller piston will have a lower force but has a longer distance traveled that the larger piston.

Exactly respect to their own systems, because the outer input piston of the other system, will do more or less work than its peer. I compare 2 different systems and a Input piston to the other. This I said it before. Again, question. Will one of the input pistons compared to its peer, have more speed to compensate its lack of area and comply with its own conservation of energy system?

- It was an example. I have the pics if you want to find the error in the device.

44 minutes ago, swansont said:

What happens when you run your system in reverse? i.e. when you return to the original state?

 

No. Newton's laws tell you how motion and forces are related. There is no conservation of forces. 

Ahhh ok, yes, well it will come back following Newton laws. The pistons get in their original positions, you can even pull the input pistons to their original positions, and you begin again the system. Another very nice option that I just thought about, is, that you return back the pistons pushing the output pistons down.

We are talking about resistance provided by pression, and acts evertywhere the same, so I would say, that the device operates as a  quasi Newton-Pascal system, to be more accurate.

Edited by esposcar
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Since we have agreed on the consequences of the idea so far, I’ll simplify the setup some more. The initial facts still apply:
The setup contains two parts, the left and the right cylinders.
Both cylinders are mounted solid on a common rig so the cylinders cannot move independently.
If the propulsion works* the rig will move as a complete system. Here are four scenarios: 

ifjxtRlnheATf-4679BvnuhjMQCzNYLtIsjIh4K8Lh3AVAeu3UiZM33wJKe-ZJ79bJmsMnP_7jEo66-RUrHzR1cC5fWAZee0k-gzAMOwxL5AXkp_854d6V7-DL7ALsN02I0WuNM1

Scenario I is the situation we agreed upon earlier. The rig will move to the right. Exact velocity is not important, only that it is greater than zero.

Scenario IV*
The left part of the rig is replaced by a copy of the right part. In this case the velocity is zero since any propulsion generated by the devices cancels out.

Scenario II: In this case the cylinder to the right is replaced with a larger one than the cylinder to the left. In this case the velocity is to the left. Exact velocity is not important, only that it is greater than zero.

Scenario III
The right cylinder is identical to the on used initially in Scenario I but the left one is replaced with an even smaller one. The result is that the rig will move to the left again. just as in scenario II.

Note that it is not necessary to know how the propulsion works at this time, we are just assuming it works as described.
Now we can see the left and the right parts of the rig both can cause movement. The rig moves in the direction of the “winning” cylinder. The “loosing” cylinder has a negative impact.  Therefore we can simplify it further. If the rigs in the scenarios above works then the rig will move even if one of the cylinders is removed from the rig. Only one cylinder is required, not two.

CgcZuizC_Q6JaXhZMjrZKjhfS73RdxEwnGA2v4aiW42eGbOdUw4IO-OJzFOIE-WBJHBHQ2up3j1dXznSuMagQccIBHuEmury-eOzF4BLzluqUt-DRP0Cz1aBy5lYtZG3g7yA509g

The magnet is pushing against the vertical bar it is mounted on, I choose to show that as force F pointing to the left.

Does this simplification of the setup show the issues with this specific propulsion idea? If not, I’ll post some more details in a followup.

 

*) Again, this is not about what I believe or how Newton mechanics works. It is an attempt to use the consequences of the setup in an analysis.

**) I screwed up the numbering

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2 minutes ago, Ghideon said:

 

 

Since we have agreed on the consequences of the idea so far, I’ll simplify the setup some more. The initial facts still apply:
The setup contains two parts, the left and the right cylinders.
Both cylinders are mounted solid on a common rig so the cylinders cannot move independently.
If the propulsion works* the rig will move as a complete system. Here are four scenarios: 

ifjxtRlnheATf-4679BvnuhjMQCzNYLtIsjIh4K8Lh3AVAeu3UiZM33wJKe-ZJ79bJmsMnP_7jEo66-RUrHzR1cC5fWAZee0k-gzAMOwxL5AXkp_854d6V7-DL7ALsN02I0WuNM1

Scenario I is the situation we agreed upon earlier. The rig will move to the right. Exact velocity is not important, only that it is greater than zero.

Scenario IV*
The left part of the rig is replaced by a copy of the right part. In this case the velocity is zero since any propulsion generated by the devices cancels out.

Scenario II: In this case the cylinder to the right is replaced with a larger one than the cylinder to the left. In this case the velocity is to the left. Exact velocity is not important, only that it is greater than zero.

Scenario III
The right cylinder is identical to the on used initially in Scenario I but the left one is replaced with an even smaller one. The result is that the rig will move to the left again. just as in scenario II.

Note that it is not necessary to know how the propulsion works at this time, we are just assuming it works as described.
Now we can see the left and the right parts of the rig both can cause movement. The rig moves in the direction of the “winning” cylinder. The “loosing” cylinder has a negative impact.  Therefore we can simplify it further. If the rigs in the scenarios above works then the rig will move even if one of the cylinders is removed from the rig. Only one cylinder is required, not two.

CgcZuizC_Q6JaXhZMjrZKjhfS73RdxEwnGA2v4aiW42eGbOdUw4IO-OJzFOIE-WBJHBHQ2up3j1dXznSuMagQccIBHuEmury-eOzF4BLzluqUt-DRP0Cz1aBy5lYtZG3g7yA509g

The magnet is pushing against the vertical bar it is mounted on, I choose to show that as force F pointing to the left.

Does this simplification of the setup show the issues with this specific propulsion idea? If not, I’ll post some more details in a followup.

 

*) Again, this is not about what I believe or how Newton mechanics works. It is an attempt to use the consequences of the setup in an analysis.

**) I screwed up the numbering

I agree 100% about the motion direction and the logic point of example 4 represent exactly what I mean. Its that exactly. If both input pistons have the same area, Newton wins, there is nothing special. It would be a Pascal system in potence but never in act Lol.

Totally, its really as it is supposed to work, forces of repulsion, but please consider that the magnet that encounters more repel oposition, will be the larget input piston, so it would be what Newton requires for propulsion. You have to propulse from something and the input pistons are supported by pression and that only belongs to Pascal principle. Newton dont apply in the support, if not, it would not work.

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18 minutes ago, Ghideon said:

Ok!
So, just to be 100% clear: According to your idea, the one-cylinder-rig in my last picture will accelerate to the right?

In theory yes but to the left, because the force that the piston will face it will be supported in a percentage by the pression, so the magnet would have 100% Newton laws and the piston a mix of friction and pression, so it would transmitted to the hole fluid in the same area with no vector contained on it., but I would set up symetry to be sure. I cannot show it mathematically your last pic. I know that in theory would work, but cannot show it empirically, only if there are two opposites.

The 2 input and output pistons have the same area?

No matter what, it would work as I explained above. And always in the direction opposite to the Pascal system.

Edited by esposcar
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(bold by me)

4 minutes ago, esposcar said:

In theory yes, but I would set up symetry to be sure. I cannot show it mathematically your last pic. I know that in theory would work, but cannot show it empirically, only if there are two opposites.

Maybe that indicates that the idea is incompatible with newtons laws? Are you at this point able to see the issues or are some more details required? There is still room for more simplifications.

 

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Maybe it wont work individually, its one system, it would respect only action-reaction, its not a good example. Put it with two, because then you have 2 enclosed systems and like that works. After viewing it, presented like that it wont work

19 minutes ago, Ghideon said:

(bold by me)

Maybe that indicates that the idea is incompatible with newtons laws? Are you at this point able to see the issues or are some more details required? There is still room for more simplifications.

 

You wanted to simplify it to one enclosed system, and no, there are a reason for the symetry, its OBLIGATORY. Your example is presenting the car without wheels. Its 2 hydraulic separated system. 2 not 1. Lets begin from here. One do more work than the other and its shown mathematically, and thrust comes from that difference. You compare more work in a side than the other. As you present it, it breaks the symetry and there is no WORK difference. You can imagine the system as you as person duplicated. You are pushing in one side and on the other with the same force. In the larger input piston, you push and moves little but the counter reaction is your feets in the platform and on the other side you push a smaller input piston, so you can do the same force without making a big counterforce with your feets, so if you imagine all that, you will have to run pushing the smaller piston a lot with the same force. So if we put a fat man in front of you in both sides, you will hit the fat man in the side of the smaller piston with more strenght due to your speed than the other fat man placed in the bigger input piston. That is the idea and both of you have used the same action-reaction force to hit the fat man.

Edited by esposcar
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17 minutes ago, esposcar said:

You wanted to simplify it to one enclosed system, and no, there are a reason for the symetry, its OBLIGATORY. Your example is presenting the car without wheels. Its 2 hydraulic separated system. 2 not 1. Lets begin from here.

Ok! I'll try to find the correct level of details to allow you to spot the issues. We agree on the four scenarios I posted but not on the one-cylinder setup? Two questions:

1: Can we use a very large cylinder on the left side? I mean, a really large. The force is still "F" but the velocity of the rod doing the pushing will be close to zero. Will the rig move to the right?

2: Can we plug the top of the left cylinder so that the level of liquid cannot rise at all? Force is "F" on both left and right but the left cylinder is not moving. There will be pressure from the liquid inside the left container but no movement in the left cylinder. Will the rig move to the right?

In both cases above the right cylinder is identical to the scenarios I-IV where we agreed.

Edited by Ghideon
grammar
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25 minutes ago, Ghideon said:

Ok! I'll try to find the correct level of details to allow you to spot the issues. We agree on the four scenarios I posted but not on the one-cylinder setup? Two questions:

1: Can we use a very large cylinder on the left side? I mean, a really large. The force is still "F" but the velocity of the rod doing the pushing will be close to zero. Will the rig move to the right?

2: Can we plug the top of the left cylinder so that the level of liquid cannot rise at all? Force is "F" on both left and right but the left cylinder is not moving. There will be pressure from the liquid inside the left container but no movement in the left cylinder. Will the rig move to the right?

In both cases above the right cylinder is identical to the scenarios I-IV where we agreed.

1 - No doubt, it will always go in the direction of the system that do more work. You have to consider that the rod moving the larger piston area will have also a ver high opposed force as pushes the rod and on the other side, there will be not a lot of resistance by the rod of the smaller piston, so the rod will push the piston and will have more pushing force as its peer the other rod, and its peer will have the force of the other rod (bigger piston) pushing on its side and the rod will oposse less because of its piston smaller area and will push the piston further and with more momentum than its peer.

2 - That would be a problem, because it would move the hole fluid as a solid and would not work. It could be slowed down putting a mass over the output piston, but no need of it, to dessacelerate the input piston, but if the fluid dont move it will act like a Newton mass. Not a good idea.

Ideally is the area difference what will make the difference. With that I feel more comfortable, its not necesarry to go to other more complicated scenarios and try to find the error dialectically.

This is a very simple solution, show me mathematically and with equation that the final velocity of the rig will be 0 and I will be satisfied. I have showed why, now convince empirically. What would be the final velocity of the rig giving imaginary numbers to the variables to solve the final result. When someting scientific with proves more than bla bla bla, I will be convinced. You convince people in such forums with numers not words. Dont take it personal Lol, its just a general answer...

Edited by esposcar
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