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Reactionless device using the principle of Pascal for fluids


esposcar

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16 hours ago, esposcar said:

I am preparing another example about how to use pascal laws to create throttle with mathematical proves and has nothing to do with this example that I posted.

Ok! Since the new example is very different from the one I commented on earlier I'll focus on the new one. 

 

6 hours ago, esposcar said:

I think I dont miss anything

When I look at the math it seems that for instance momentum (and/or kinetic energy) of the balls and other relevant parts is missing. How does conservation of momentum work while the setup generates thrust?

Quote

carry on 20 N of force to transmit to whiteball 1

What does the above mean? If the momentum of the black ball is transferred to the white ball the black ball will not push the piston any more? 

swansont's earlier comment still applies:

On 3/15/2019 at 12:13 PM, swansont said:

the situation you describe is not static.

 

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7 hours ago, esposcar said:

This is the first model in history of a workable reactionless device without any doubt. Tell me what you miss here. Thing to consider. The black balls are inelastic, so they will get attached to their correspondant piston-pushers after the hit. The white balls are elastic, and will get the force and the momentum fully, as when a ball hits another in billiards. And finally the white balls will hit a type of hydraulic stopper to stop the balls soft and absorb all the momentum. The balls are floating because there is no gravity, the enviroment is space. The hole system will have an extra force coming from no Newton reaction of 20 Newtons to the right. Of course it can generate much more force, just make one of the input pistons much more smaller than its pair and propulse the black balls much more fast to have a higher force of pushing. 

You now have a thread with two pages of nonsense and wasted effort.

If you put one tenth the effort of making these diagrams into learning some basic mechanics you would be much further forwards.

Properly learning this is urgent for anyone who offers the statement

"Blackball 1 already have a force of 60N before striking the piston pusher."

 

Do you know what force is and would you be willing to learn if someone told you?

 

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When you say one piston will do more work than another, that's wrong. The different areas means the force is different, but it also means that the distance you need to push is different as well. Energy is conserved, so the work done by each piston is equal if the input is equal.

It looks like your conclusion is based on a violation of conservation of energy.

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3 hours ago, swansont said:

When you say one piston will do more work than another, that's wrong. The different areas means the force is different, but it also means that the distance you need to push is different as well. Energy is conserved, so the work done by each piston is equal if the input is equal.

It looks like your conclusion is based on a violation of conservation of energy.

 

https://physics.stackexchange.com/questions/439325/why-pascals-law-is-true-and-what-is-the-mechanism-for-force-amplification-at-mo

I found this link usefull because talks about Conservation of Energy. It is not violated, but the smaller piston compared with its pair the piston with bigger area, to elevate their corresponding output pistons to the same level,  one input piston will have to be displaced more than the other using the same force. Each input piston to keep the conservation of energy will have to keep its corresponding output system on the same level so one piston will have more acceleration than the other. This is an objective fact and no conservation of energy is violated. And one piston of Area = 2 will do 3 metres to apply the conservation of energy for its corresponding output piston and the other will have Area= 4 and do 1.5 metres to elevate the same output piston area as its pair. So one piston moves the double than the other with the same force, because one have the double area than the other. This is a fact. So if we view it like reality, the work done by one input piston in perpective from the other input piston, will be more. But for their own hydraulic system no law is violated. Remember, we have two hydraulic systems in the platform, not one, and each system have its own conservation of energy, but we profit of their input pistons, that will move at different accelerations to cover different distances in the same time. So the white balls are just in the way of the input displacements, and are placed at the same distance from their corresponding pistons, so when both pistons moves and hit the white balls at the same time, one will carry more momentum than the other. P=mv because the piston structure will have a mass and it will carry a momentum. When each piston will hit its balls one will have more speed than the other. This is a fact, the rest you can imagine it.

So if work=force * distance and we have two input pistons of 1/2 area and another with 1 Area with a force applied of 60 Newtons by the two black balls to their corresponding piston pushers that push at the same time the pistons. And we see that to one have done with 60N of force 3 metres and the other 1.5 metres, I understand that if you compare the work of both of the, one have done more work than the other, right? this is 1 + 1=2, here there is no argument, its a fact.

"When you say one piston will do more work than another, that's wrong. The different areas means the force is different, but it also means that the distance you need to push is different as well. Energy is conserved, so the work done by each piston is equal if the input is equal." I totally agree with you, the work of the smaller input is the same as the work done by its output input, its conservation of energy for a closed system, but the work I compare it for its other pair, the other input piston of the other system. My device is a platform and 2 systems are placed.

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24 minutes ago, esposcar said:

And we see that to one have done with 60N of force 3 metres and the other 1.5 metres

How is the force applied by the black ball constant=60 N? Once the white ball is affected by the movement, what happens? 

The picture uses "force" in a way that makes it hard to discuss the proposed setup; the following seems incorrect:

image.png.d6fb64fd17f90ebf92eef31d6c8904f5.png

There seems to be confusion about force, momentum and kinetic energy in the picture? How about the momentum of the moving parts?

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System 1 (right) composed by an input piston of 2 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*3 = 180          Output piston W=F*D = 600*0.3 = 180 

 

System 2 (left) composed by an input piston of 4 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 1.5 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*1.5 = 90       Output piston W=F*D = 300*0.30 = 90

 

Input piston work of system 1          COMPARED TO        Input piston work of system 2

    W= 180                                                           W= 90

I compare input with input, not input with output. There is no violation of conservation of energy, because I compare the input pistons work with each other and both belong to another system, placed in the same platform. Glued however you want to imagine it. I just can show you maths.

 

There is no violation of energy in this system, But I compare the work done by one input system with the other.

46 minutes ago, Ghideon said:

How is the force applied by the black ball constant=60 N? Once the white ball is affected by the movement, what happens? 

The picture uses "force" in a way that makes it hard to discuss the proposed setup; the following seems incorrect:

image.png.d6fb64fd17f90ebf92eef31d6c8904f5.png

There seems to be confusion about force, momentum and kinetic energy in the picture? How about the momentum of the moving parts?

Well you are right, we will consider that the black ball after the acceleration will have a constant speed, and will arrive with P=V*M, its not a potential force once hits something, in the motion it has no force applied but carries momentum, with a vector, so that vector, is a type of force because will push the input cylinder. Technically maybe was not right but the context I expect is more understandable.

35 minutes ago, esposcar said:

System 1 (right) composed by an input piston of 2 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*3 = 180          Output piston W=F*D = 600*0.3 = 180 

 

System 2 (left) composed by an input piston of 4 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 1.5 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*1.5 = 90       Output piston W=F*D = 300*0.30 = 90

 

Input piston work of system 1          COMPARED TO        Input piston work of system 2

    W= 180                                                           W= 90

I compare input with input, not input with output. There is no violation of conservation of energy, because I compare the input pistons work with each other and both belong to another system, placed in the same platform. Glued however you want to imagine it. I just can show you maths.

 

There is no violation of energy in this system, But I compare the work done by one input system with the other.

Well you are right, we will consider that the black ball after the acceleration will have a constant speed, and will arrive with P=V*M, its not a potential force once hits something, in the motion it has no force applied but carries momentum, with a vector, so that vector, is a type of force because will push the input cylinder. Technically maybe was not right but the context I expect is more understandable.

Answering to your questions. I will answer your other posts, but just dont have the time now. But to your last post ok. I did not do the maths up to how arrives the black ball to its pusher with 60 N, but it can arrive with whatever the magnitude you want, even it can strike the pusher applying a force of 10.000 N, the fact is for the sake of the example a symbolic force when the black ball push the piston pusher. Lets say 60 Newtons before the displacement begins, and lets say that advances 3 metres, but in the advance we settle a whiteball just 1 metres further from the piston pusher that will advance 3 metres also, so if we divide the force by the distance.

- At 0 metres it has 60N of force pushing, at 1 metres it will rest 40N (it has gone in friction and pushing against pression) so like a billiard ball, once the pusher encounters the white ball, it will strike it as a billiard ball and stop advancing and the white ball have been propulsed by a force of 40 Newtons and on the other side, due to the piston area size, the piston pusher it will arrive to its corresponding white ball placed one metre from it with a force of 20 N.

Edited by esposcar
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46 minutes ago, esposcar said:

System 1 (right) composed by an input piston of 2 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*3 = 180          Output piston W=F*D = 600*0.3 = 180 

This is complete and utter nonsense in that it utterly ignores the laws of Mechanics.

I asked a simple question (Hint the sentence ending in a question mark).

The Laws of ScienceForums expect you to answer.

 

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The same work is done by the input piston as the output piston. What are you talking about? Do you know conservation of energy or you just post without knowing what we are talking about???? What question have you formulated?

Ahhh ok yes, I viewed. I will answer you later

Edited by esposcar
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1 hour ago, esposcar said:

The same work is done by the input piston as the output piston. What are you talking about? Do you know conservation of energy or you just post without knowing what we are talking about????

 

I know you have ignored the law of conservation of mechanical energy by introducing friction and some other magic force called 'pression' and then failed to account for their energies in your calculations.

The hydraulic systems you have drawn cannot act in the way you describe.

You need to know enough Mechanics to understand what they will actually do before putting numbers to things.

Edited by studiot
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See the following (Bold added by me) 

3 hours ago, esposcar said:

The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system.

and

3 hours ago, esposcar said:

- At 0 metres it has 60N of force pushing, at 1 metres it will rest 40N (it has gone in friction and pushing against pression) so like a billiard ball, once the pusher encounters the white ball, it will strike it as a billiard ball and stop advancing and the white ball have been propulsed by a force of 40 Newtons and on the other side, due to the piston area size, the piston pusher it will arrive to its corresponding white ball placed one metre from it with a force of 20 N.

Doesn't those contradict each other? 

3 hours ago, esposcar said:

Well you are right, we will consider that the black ball after the acceleration will have a constant speed, and will arrive with P=V*M, its not a potential force once hits something, in the motion it has no force applied but carries momentum, with a vector, so that vector, is a type of force because will push the input cylinder. Technically maybe was not right but the context I expect is more understandable.

Ok, but unfortunately that makes the calculations wrong. And how should I now read & interpret your image and notes? Should all occurrences of "force" be replaced with "momentum"?
 

On 3/6/2019 at 12:12 AM, esposcar said:

I have a good knowledge of Newton Mechanics ;)

Ok, maybe you can apply some of that knowledge and provide calculations compatible with Newton?
Hint: it seems easy to simplify the setup to allow for some easier analysis. But since this is speculations, that's your task :-)

 

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22 hours ago, Ghideon said:

Ok, maybe you can apply some of that knowledge and provide calculations compatible with Newton?
Hint: it seems easy to simplify the setup to allow for some easier analysis. But since this is speculations, that's your task :-)

 

Ok I have taked your hint and I have done it more simple and clear. The propulsion system, this time I make it more simple with a constant force produced by an electric magnet of Neodymium to create a repulsion of 60 N against its piston pusher, to push with a constant velocity the input piston. The black balls are eliminated and a lot of useless text. To everybody, just to say that conservation of energy or mechanical enery etc etc dont apply to this system, because its 2 enclosed systems settled and placed in the same common platform.

Modelo2propforos.png.9e8018bb256ca19f142ca9ab50881a43.png

I think this solves propulsion problems and it creates a constant force. It can work also with extendable hydraulical bars to push the piston pushers, but I thought about this idea, because its more easy to analyze for mistakes or missing forces.

*IMPORTANT Whiteball 2 will be placed the double of distance from Piston-pusher 2 than Whiteball 1 due to the double size of friction area that Input piston 1 will have compared to input piston 2

Edited by esposcar
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11 minutes ago, esposcar said:

Ok I have taked your hint and I have done it more simple and clear. The propulsion system, this time I make it more simple with a constant force produced by an electric magnet of Neodymium to create a repulsion of 60 N against its piston pusher, to push with a constant velocity the input piston. The black balls are eliminated and a lot of useless text. To everybody, just to say that conservation of energy or mechanical enery etc etc dont apply to this system, because its 2 enclosed systems settled and placed in the same common platform.

Modelo2propforos.png.9e8018bb256ca19f142ca9ab50881a43.png

I think this solves propulsion problems and it creates a constant force. It can work also with extendable hydraulical bars to push the piston pushers, but I thought about this idea, because its more easy to analyze for mistakes or missing forces.

9

sometimes, being wrong is the answer.

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3 minutes ago, dimreepr said:

sometimes, being wrong is the answer.

Show me wrong mathematically and not in words. As the boss said, this is a science forum. I am giving now empirical arguments, show me wrong in the same way.

Edited by esposcar
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6 minutes ago, dimreepr said:

do I have too... :rolleyes:

With bla bla bla you wont do it ;) Show me equations that fit in this excercise and explain me if one of the white balls will receive more force than the other. Or what will do the system mathematically to avoid the maths I showed if its not an isolated system.

Edited by esposcar
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50 minutes ago, esposcar said:

Ok I have taked your hint and I have done it more simple and clear. The propulsion system, this time I make it more simple with a constant force produced by an electric magnet of Neodymium to create a repulsion of 60 N against its piston pusher, to push with a constant velocity the input piston.

Why is the velocity constant when the force is constant?

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29 minutes ago, swansont said:

Why is the velocity constant when the force is constant?

 

29 minutes ago, swansont said:

Why is the velocity constant when the force is constant?

Ahhh you are right, my fault, with constant acceleration the force is constant. Anyway even on acceleration it will still apply more work in one input piston than the other. But to keep a constant velocity, it is also possible, if the electrical magnet, keep and takes more or less repulsion in order to keep the input piston at a constant speed. If both piston-pushers receive the same amount of repulsion at the same rate of changes in the repulsion produced by the magnets to keep a constant velocity of the pistons. Then also it could work. The most important is to keep the same force applied to the pistons in both sides.

Edited by esposcar
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26 minutes ago, esposcar said:

Ahhh you are right, my fault, with constant acceleration the force is constant. Anyway even on acceleration it will still apply more work in one input piston than the other. But to keep a constant velocity, it is also possible, if the electrical magnet, keep and takes more or less repulsion in order to keep the input piston at a constant speed. If both piston-pushers receive the same amount of repulsion at the same rate of changes in the repulsion produced by the magnets to keep a constant velocity of the pistons. Then also it could work. The most important is to keep the same force applied to the pistons in both sides.

Ok! 
I have a few follow-up questions, but can you first update the calculations to show the corrections mentioned above?

It's a little tricky to follow the descriptions. For example, does the following statement still apply?

On 3/18/2019 at 3:54 PM, esposcar said:

once the pusher encounters the white ball, it will strike it as a billiard ball and stop advancing

 

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16 minutes ago, Ghideon said:

Ok! 
I have a few follow-up questions, but can you first update the calculations to show the corrections mentioned above?

It's a little tricky to follow the descriptions. For example, does the following statement still apply?

 

Not at all, I have placed the magnets, so we forget about any mechanical propulsion. In my older version, the force would not be constant or accelerating and it would have serious doubts to work in my opinion due to friction and area of friction. So I posted a much more clean example, and showing mathematically the basic working of the device. Anyway the first device it still has not been proven right or wrong Lol. But this version yes I could show it mathematically and that gives me some hopes!!!!

23 minutes ago, Ghideon said:

Ok! 
I have a few follow-up questions, but can you first update the calculations to show the corrections mentioned above?

It's a little tricky to follow the descriptions. For example, does the following statement still apply?

 

Ohhh sorry, I have realized now what you said. Yes it applies, the most important is that one pusher with the same foce pushing it, have more speed than the other due to the input piston area differences. Sorry again for the lapsus Lol

Edited by esposcar
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51 minutes ago, esposcar said:

Not at all, I have placed the magnets, so we forget about any mechanical propulsion. In my older version, the force would not be constant or accelerating and it would have serious doubts to work in my opinion due to friction and area of friction. So I posted a much more clean example, and showing mathematically the basic working of the device. Anyway the first device it still has not been proven right or wrong Lol. But this version yes I could show it mathematically and that gives me some hopes!!!!

Ohhh sorry, I have realized now what you said. Yes it applies, the most important is that one pusher with the same foce pushing it, have more speed than the other due to the input piston area differences. Sorry again for the lapsus Lol

I disagree with you posted mathematics for the output pressures.

The full Pascal pressure can only be developed if you apply suitable opposing forces against the output piston.
Otherwise there is no reason for the stated increase in pressures.

On ‎3‎/‎18‎/‎2019 at 2:54 PM, esposcar said:

System 1 (right) composed by an input piston of 2 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is:

Input piston W=F*D = 60*3 = 180          Output piston W=F*D = 600*0.3 = 180 

And I am reporting your lack of reply as against the rules of this forum.

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17 minutes ago, studiot said:

I disagree with you posted mathematics for the output pressures.

The full Pascal pressure can only be developed if you apply suitable opposing forces against the output piston.
Otherwise there is no reason for the stated increase in pressures.

And I am reporting your lack of reply as against the rules of this forum.

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.

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1 hour ago, esposcar said:

Not at all, I have placed the magnets, so we forget about any mechanical propulsion. In my older version, the force would not be constant or accelerating and it would have serious doubts to work in my opinion due to friction and area of friction. So I posted a much more clean example, and showing mathematically the basic working of the device. Anyway the first device it still has not been proven right or wrong Lol. But this version yes I could show it mathematically and that gives me some hopes!!!!

So what happens to the forces and the velocities when the white balls are hit? Are the white balls massless? 

1 hour ago, esposcar said:

Ohhh sorry, I have realized now what you said. Yes it applies, the most important is that one pusher with the same foce pushing it, have more speed than the other due to the input piston area differences. Sorry again for the lapsus Lol

Ok! So the pusher stops when the white ball is hit, and therefore the pusher does not reach as far as the image says?  Can you update the calculations to and image show? 

 

(Note that I'm probably moving forward at a much slower pace than several other members; the issues raised by @studiot and others of course apply. If the thread remains open I'll address the pistons and fluids etc. But that requires a consistent description of the setup first.)

Edited by Ghideon
Grammar and clarification regarding stopping pusher
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7 minutes ago, Ghideon said:

So what happens to the forces and the velocities when the white balls are hit? Are the white balls massless? 

Ok! So the pusher stops when the white ball is hit, and therefore does not reach as far as the image says?  Can you update the calculations to and image show? 

 

(Note that I'm probably moving forward at a much slower pace than several other members; the issues raised by @studiot and others of course apply. If the thread remains open I'll address the pistons and fluids etc. But that requires a consistent description of the setup first.)

The balls are full of mass and elastic or inelastic, they can be even accelerating at the same time as the pusher, and suddenly the magnets stop working when the strike the white ball or stopping gradually as the white ball that begun the acceleration with, will continue with its momentum, all options are good. The picture is a reference, just view the context and the equations. Try if you have doubts about a component to fix it the most positive way, specially if it is things that technology can do much more efficiently than me with other ideas that I probably havent thought about, but the important is the interaction between Pascal principle and Newton laws.

Edited by esposcar
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12 minutes ago, esposcar said:

The balls are full of mass and elastic or inelastic, they can be even accelerating at the same time as the pusher, and suddenly the magnets stop working when the strike the white ball or stopping gradually as the white ball that begun the acceleration with, will continue with its momentum, all options are good.

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?

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