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General Discussion => New Physics for Space Technology => Topic started by: esposcar on 03/11/2019 01:40 am
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I have a pending patent of the invention of a reactionless device version that uses the principle of Pascal, to move moving mass to stationary mass. As you can view in this device example, the 3 pics that I uploaded, you can view in 3 steps, how does this device works. I would like to know, if it violates any Newtons or Pascal law, and if not, would it be considered a working reactionless device?
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I would like to know, if it violates any Newtons or Pascal law, and if not, would it be considered a working reactionless device?
The second half of your question can be reworded and has the exact same meaning as the following:
"if this device obeys the law of equal and opposite reactions, then is it a device that violates the law of equal and opposite reactions?"
Hopefully that rewording makes the answer clear, and independent of any details of your device.
Too many new posters who present ideas like yours react poorly to being told "no" for me to want to spend much time on a device specific answer, at least before having some evidence that you are interested in listening. It would do you more good to work it out yourself. (Hint: at some point to get from A to B, the fluid must have momentum that is at least partially pointed to the right.)
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Another way of looking at it might be the conservation of the center of mass of the system - before and after the net momentum returns to zero - after the shifting of the fluid.
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The Law of Conservation of Momentum predicts that for any system not subjected to external forces the momentum of the system will remain constant, which means the center of mass will move with constant velocity. This applies for all systems with classical internal forces, including magnetic fields, electric fields, chemical reactions, and so on. More formally, this is true for any internal forces that cancel in accordance with Newton's Third Law.[11]
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(https://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Orbit3.gif/180px-Orbit3.gif)
emphasis on new big mass and little mass.
Edit: in your case linear motion not circular motion.
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Another way of looking at it might be the conservation of the center of mass of the system - before and after the net momentum returns to zero - after the shifting of the fluid.
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The Law of Conservation of Momentum predicts that for any system not subjected to external forces the momentum of the system will remain constant, which means the center of mass will move with constant velocity. This applies for all systems with classical internal forces, including magnetic fields, electric fields, chemical reactions, and so on. More formally, this is true for any internal forces that cancel in accordance with Newton's Third Law.[11]
...
(https://upload.wikimedia.org/wikipedia/commons/thumb/5/59/Orbit3.gif/180px-Orbit3.gif)
emphasis on new big mass and little mass.
The device is linear and if the shifting of the fluid happens in an acceleration, like a rotational device, it would not work, because the lose of fluid would increase the acceleration. Just responding to the moon and the earth pic Lol.
In my next post I have done two animations (excuse me for the amateur flow of the animation) to show how it would work in my humble opinion, and I will give details to show, that the only forces present inside a confined and isolated fluid, the only force that exist is pression. About momentum, if you think about it, its the velocity of an object what gives sense to the momentum, not the mass or the object itself, so during the tranfer, velocity wont be transmited, and only mass will be transmited, because pression avoid the enter of more forces inside the fluid (exception of gravity), I mean that you can do a force to push a piston, but the piston will transmit this force as pression. Pression in a fluid is distributed every where, its vector force, goes everywhere, and that is the secret. Mainly, momentum will affect the pipe structure itself, the two containers structure (Walls that confine the fluid, pistons etc), but not the fluid itself. The pipe will be always full of fluid and no flow will happen, it will just translate the fluid mass, because fluids behave different than solids. Anyway I will add some examples to show it also. In 2 hours I will expose it. Thank you for the critics.
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PBS Spacetime did an episode about a week ago on perpetual motion machines. This proposal seems to fit into this category.
https://www.youtube.com/watch?v=rckrnYw5sOA
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PBS Spacetime did an episode about a week ago on perpetual motion machines. This proposal seems to fit into this category.
https://www.youtube.com/watch?v=rckrnYw5sOA
Can you tell me in which second appears something similar to this device? Which category? moving fluids responds like solids, that means obeys Newton laws, but I am talking about moving confined and totally pressed fluid with income and output pistons, that means using the principle of Pascal, and I did not see nothing similar. Just moving fluids in a rotational way which is not the way this device behaves if you take a better look to the description in the three pics. It has absolutly nothing to do with this. But I will settle in the next post some examples, more clear, because I have the sensation, that I have not explained good the device, despite of the visual. Its three pics, that show the beginning of the function of the device, the middle function and the end. And if you see the working way, its very simple so it dont let doubts.
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The device is linear and if the shifting of the fluid happens in an acceleration, like a rotational device, it would not work, because the lose of fluid would increase the acceleration. Just responding to the moon and the earth pic Lol.
In my next post I have done two animations (excuse me for the amateur flow of the animation) to show how it would work in my humble opinion, and I will give details to show, that the only forces present inside a confined and isolated fluid, the only force that exist is pression. About momentum, if you think about it, its the velocity of an object what gives sense to the momentum, not the mass or the object itself, so during the tranfer, velocity wont be transmited, and only mass will be transmited, because pression avoid the enter of more forces inside the fluid (exception of gravity), I mean that you can do a force to push a piston, but the piston will transmit this force as pression. Pression in a fluid is distributed every where, its vector force, goes everywhere, and that is the secret. Mainly, momentum will affect the pipe structure itself, the two containers structure (Walls that confine the fluid, pistons etc), but not the fluid itself. The pipe will be always full of fluid and no flow will happen, it will just translate the fluid mass, because fluids behave different than solids. Anyway I will add some examples to show it also. In 2 hours I will expose it. Thank you for the critics.
The device being linear is nothing to do with it.
Newton's law is universal and applies to anything with mass including fluids. Pascal's only applies to a non-moving fluid. (Note the phrase "at rest" in the definition on Wikipedia (https://en.m.wikipedia.org/wiki/Pascal's_law)) Fluid flowing through a bent pipe has a force applied to it by the pipe to cause the flow to change direction. You can easily feel this force if you are spraying water from a garden hose. The way you describe pipes is as if they are magical teleportation devices, where the fluid is not actually in motion, but that is simply wrong.
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The device is linear and if the shifting of the fluid happens in an acceleration, like a rotational device, it would not work, because the lose of fluid would increase the acceleration. Just responding to the moon and the earth pic Lol.
In my next post I have done two animations (excuse me for the amateur flow of the animation) to show how it would work in my humble opinion, and I will give details to show, that the only forces present inside a confined and isolated fluid, the only force that exist is pression. About momentum, if you think about it, its the velocity of an object what gives sense to the momentum, not the mass or the object itself, so during the tranfer, velocity wont be transmited, and only mass will be transmited, because pression avoid the enter of more forces inside the fluid (exception of gravity), I mean that you can do a force to push a piston, but the piston will transmit this force as pression. Pression in a fluid is distributed every where, its vector force, goes everywhere, and that is the secret. Mainly, momentum will affect the pipe structure itself, the two containers structure (Walls that confine the fluid, pistons etc), but not the fluid itself. The pipe will be always full of fluid and no flow will happen, it will just translate the fluid mass, because fluids behave different than solids. Anyway I will add some examples to show it also. In 2 hours I will expose it. Thank you for the critics.
The device being linear is nothing to do with it.
Newton's law is universal and applies to anything with mass including fluids. Pascal's only applies to a non-moving fluid. (Note the phrase "at rest" in the definition on Wikipedia (https://en.m.wikipedia.org/wiki/Pascal's_law)) Fluid flowing through a bent pipe has a force applied to it by the pipe to cause the flow to change direction. You can easily feel this force if you are spraying water from a garden hose. The way you describe pipes is as if they are magical teleportation devices, where the fluid is not actually in motion, but that is simply wrong.
Soory to contradict you but that is not true. A fluid engineer can tell you that, the fluid is transmited as a block. At a constant velocity there is no force applied to the moving container. And in the transfer during a constant velocity, there is NO FLOW inside the fluid, 0. A force by the way that is applied by the piston perpendicular to the motion of the container by the way. And about the garden pipe, of course you notice Newton force, because the fluid is not confined, and it behaves like Newton says. But please let me finish the animation and settle some examples, and you can answer.
You can move in a platform at constant velocity in the same way as if you are in a stationary platform. So the moving of the fluid will be done in that mode, not in acceleration. Your example is not good about the garden, because le fluid is not confined, there are no pistons that retain the water, and the water will spread with Newton laws, as a missile throws it mass, but in this case, the fluid dont go out anywhere. If you put a moving piston and another piston in the opposite side of a pipe full of fluid, and you push one of the pistons, you wont notice anything if you put your hand in the middle of the pipe. No flow
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Here is a link that answers how the pipe would behave during the transfer of fluids.
https://www.explainthatstuff.com/hydraulics.html
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Soory to contradict you but that is not true. A fluid engineer can tell you that, the fluid is transmited as a block.
You aren't contradicting me, you are contradicting every scientist and engineer on this planet. Any fluid engineer, or even a good plumber could tell you that you are wrong.
At a constant velocity there is no force applied to the moving container. And in the transfer during a constant velocity, there is NO FLOW inside the fluid, 0.
based on some Fench words you have slipped in, I am guessing that your native language is French, so run the next paragraph though a translator if you need to:
By definition, a fluid moving through a pipe from one container to another is flowing. There is nothing "constant velocity" about the fluid moving through the pipe, because velocity includes direction, and there are bends in the pipe which change the direction of the flow.
And about the garden pipe, of course you notice Newton force, because the fluid is not confined, and it behaves like Newton says. But please let me finish the animation and settle some examples, and you can answer.
Newton's laws are universal, that means that whether the fluid is confined or not is irrelevant to them. Pascal's law does not cover the case when fluid is moving from one container to another, it only covers the static case, when the fluid is moving from one container to another through a pipe, there are additional forces due to the motion of the fluid.
Here is a link that answers how the pipe would behave during the transfer of fluids.
https://www.explainthatstuff.com/hydraulics.html
No, that link simply does not discuss the the forces on the pipe through the bends. It assumes the pipe is held by something very heavy (you in the case of the water pistol, the main body of construction equipment in the other case.) Actually, it contradicts you by showing that there is nothing wrong with using something open on one end like a water pistol or garden hose for explaining these things.
Your animation will not help anything, it is already clear that your mistake is that you want to assume that pipes just teleport liquid from one place to another. No animation will change the fact that in real life, fluids are moving through pipes, and that those pipes have additional forces on them (and by Newton's laws, equal and opposite forces on the fluid) than solely what is described by Pascal's law.
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If, as you state, you have a patent pending, be prepared to build and submit a "working model". https://en.wikipedia.org/wiki/Perpetual_motion#Patents
. If you manage to make a working model, you've proven your invention and theory. If not, you've proven hundreds of years of physics correct.
Good luck.
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If, as you state, you have a patent pending, be prepared to build and submit a "working model". https://en.wikipedia.org/wiki/Perpetual_motion#Patents
. If you manage to make a working model, you've proven your invention and theory. If not, you've proven hundreds of years of physics correct.
Good luck.
I agree, btw I am working in two new concepts that came from this idea. The only problem with this version that I showed, is that with gravity and friction, it wont work good or nothing. Consider that if you take mass from moving, it will weight less, and friction will be less, also there is no constant velocity, because you need always acelleration in earth, and that makes this difficult to execute in earth sadly. But I am with another 2 version as I said, and I think I can have a working model.
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If, as you state, you have a patent pending, be prepared to build and submit a "working model". https://en.wikipedia.org/wiki/Perpetual_motion#Patents
. If you manage to make a working model, you've proven your invention and theory. If not, you've proven hundreds of years of physics correct.
Good luck.
Tell me something, if in my next post, i show that momentum is not transfered in anyway from the moving container to the stationary container through the pipe, it could begin to be considered, this deviced more interesant ;)?
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Tell me something, if in my next post, i show that momentum is not transfered in anyway from the moving container to the stationary container through the pipe, it could begin to be considered, this deviced more interesant ;)?
It is trivial to make an animation that ignores Newton's laws and conservation of momentum.
You have already stated that you built a version of this and it didn't work, for which you blamed "gravity and friction" with no detailed explanation. If you try enough times, it seems likely you could tune it to take advantage of the difference between static and dynamic friction, and get it to appear to move, but really it would just be pushing off of the surface it is sitting on. Or if you do get friction low enough it will move slightly to the side to keep its center of mass in place, and then move back as it resets to its original configuration, assuming it has a built in reset mechanism.
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Tell me something, if in my next post, i show that momentum is not transfered in anyway from the moving container to the stationary container through the pipe, it could begin to be considered, this deviced more interesant ;)?
It is trivial to make an animation that ignores Newton's laws and conservation of momentum.
You have already stated that you built a version of this and it didn't work, for which you blamed "gravity and friction" with no detailed explanation. If you try enough times, it seems likely you could tune it to take advantage of the difference between static and dynamic friction, and get it to appear to move, but really it would just be pushing off of the surface it is sitting on. Or if you do get friction low enough it will move slightly to the side to keep its center of mass in place, and then move back as it resets to its original configuration, assuming it has a built in reset mechanism.
Read again my message because you have not understood anything. This is from my last post...."The only problem with this version that I showed, is that with gravity and friction, it wont work good or nothing." Where did I said that I already tested a model and did not work? I never done a physical model.
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Read again my message because you have not understood anything. This is from my last post...."The only problem with this version that I showed, is that with gravity and friction, it wont work good or nothing." Where did I said that I already tested a model and did not work? I never done a physical model.
You said that you were working on a version 2, in response to a question about building a working model. That implies that there was a version 1. "But I am with another 2 version as I said, and I think I can have a working model." There is a bit of a grammar problem with that sentence, and since I assume there is a language barrier here, I can see where the miscommunication may have come from, but your previous post sounds like you built a model and it didn't work.
My point about not explaining why gravity and friction are a problem stands. Actually, all of my points stand, since you have simply ignored them and repeated assertions that pipes are magic devices that teleport liquids and violate conservation of momentum. You are the one who has some serious misunderstandings, and you should review my posts again, ask me if there are any parts that aren't clear enough and I can try to reword them.
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Read again my message because you have not understood anything. This is from my last post...."The only problem with this version that I showed, is that with gravity and friction, it wont work good or nothing." Where did I said that I already tested a model and did not work? I never done a physical model.
You said that you were working on a version 2, in response to a question about building a working model. That implies that there was a version 1. "But I am with another 2 version as I said, and I think I can have a working model." There is a bit of a grammar problem with that sentence, and since I assume there is a language barrier here, I can see where the miscommunication may have come from, but your previous post sounds like you built a model and it didn't work.
Well, possibly a bad communication. But I am from Spain, so its not the same. This 3 pics in the entry of the topic, is my version 1, but I have 2 other versions of the same invention.
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The device is linear and if the shifting of the fluid happens in an acceleration, like a rotational device, it would not work, because the lose of fluid would increase the acceleration. Just responding to the moon and the earth pic Lol.
In my next post I have done two animations (excuse me for the amateur flow of the animation) to show how it would work in my humble opinion, and I will give details to show, that the only forces present inside a confined and isolated fluid, the only force that exist is pression. About momentum, if you think about it, its the velocity of an object what gives sense to the momentum, not the mass or the object itself, so during the tranfer, velocity wont be transmited, and only mass will be transmited, because pression avoid the enter of more forces inside the fluid (exception of gravity), I mean that you can do a force to push a piston, but the piston will transmit this force as pression. Pression in a fluid is distributed every where, its vector force, goes everywhere, and that is the secret. Mainly, momentum will affect the pipe structure itself, the two containers structure (Walls that confine the fluid, pistons etc), but not the fluid itself. The pipe will be always full of fluid and no flow will happen, it will just translate the fluid mass, because fluids behave different than solids. Anyway I will add some examples to show it also. In 2 hours I will expose it. Thank you for the critics.
The device being linear is nothing to do with it.
Newton's law is universal and applies to anything with mass including fluids. Pascal's only applies to a non-moving fluid. (Note the phrase "at rest" in the definition on Wikipedia (https://en.m.wikipedia.org/wiki/Pascal's_law)) Fluid flowing through a bent pipe has a force applied to it by the pipe to cause the flow to change direction. You can easily feel this force if you are spraying water from a garden hose. The way you describe pipes is as if they are magical teleportation devices, where the fluid is not actually in motion, but that is simply wrong.
1 Soory to contradict you but that is not true. A fluid engineer can tell you that, the fluid is transmited as a block. At a constant velocity there is no force applied to the moving container. And in the transfer during a constant velocity, there is NO FLOW inside the fluid, 0.
*snip*
2 If you put a moving piston and another piston in the opposite side of a pipe full of fluid, and you push one of the pistons, you wont notice anything if you put your hand in the middle of the pipe. No flow
Fluid in, and moving in, a pipe always exerts pressure on the pipe. The only time there is zero pressure in a pipe is if the pipe is empty.
A pipe with two pistons DOES move the hydraulic fluid, yes, there is flow - so the second part is also wrong.
Are you familiar with Bernoulli's principle and the related equations?
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The device is linear and if the shifting of the fluid happens in an acceleration, like a rotational device, it would not work, because the lose of fluid would increase the acceleration. Just responding to the moon and the earth pic Lol.
In my next post I have done two animations (excuse me for the amateur flow of the animation) to show how it would work in my humble opinion, and I will give details to show, that the only forces present inside a confined and isolated fluid, the only force that exist is pression. About momentum, if you think about it, its the velocity of an object what gives sense to the momentum, not the mass or the object itself, so during the tranfer, velocity wont be transmited, and only mass will be transmited, because pression avoid the enter of more forces inside the fluid (exception of gravity), I mean that you can do a force to push a piston, but the piston will transmit this force as pression. Pression in a fluid is distributed every where, its vector force, goes everywhere, and that is the secret. Mainly, momentum will affect the pipe structure itself, the two containers structure (Walls that confine the fluid, pistons etc), but not the fluid itself. The pipe will be always full of fluid and no flow will happen, it will just translate the fluid mass, because fluids behave different than solids. Anyway I will add some examples to show it also. In 2 hours I will expose it. Thank you for the critics.
The device being linear is nothing to do with it.
Newton's law is universal and applies to anything with mass including fluids. Pascal's only applies to a non-moving fluid. (Note the phrase "at rest" in the definition on Wikipedia (https://en.m.wikipedia.org/wiki/Pascal's_law)) Fluid flowing through a bent pipe has a force applied to it by the pipe to cause the flow to change direction. You can easily feel this force if you are spraying water from a garden hose. The way you describe pipes is as if they are magical teleportation devices, where the fluid is not actually in motion, but that is simply wrong.
1 Soory to contradict you but that is not true. A fluid engineer can tell you that, the fluid is transmited as a block. At a constant velocity there is no force applied to the moving container. And in the transfer during a constant velocity, there is NO FLOW inside the fluid, 0.
*snip*
2 If you put a moving piston and another piston in the opposite side of a pipe full of fluid, and you push one of the pistons, you wont notice anything if you put your hand in the middle of the pipe. No flow
Fluid in, and moving in, a pipe always exerts pressure on the pipe. The only time there is zero pressure in a pipe is if the pipe is empty.
A pipe with two pistons DOES move the hydraulic fluid, yes, there is flow - so the second part is also wrong.
Are you familiar with Bernoulli's principle and the related equations?
1. Pascals Law
This states that the fluid transmits equal pressure in all directions. This law is also responsible for the derivation of the hydrostatic law
Dell P = rho*g*h.
2. Bernoulli's principle
This is an application of conservation of energy. This is normally used to find conditions at a particular point when we know the conditions at some other point. But be careful. This is applicable only for non-viscous, incompressible and steady flows. Put mercury in the as the fluid, if you want to be sure that dont applies Bernoulli principle in this case.
Bernoulli's equation is about energy conservation within a fluid in motion and is not applicable in hydrostatic situations where pressure differences dominate. Lets begin to give proper arguments.
By the way, pressure is in all the fluid equal. All the infinite directions, its not a vector force. Very important point.
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Well, possibly a bad communication. But I am from Spain, so its not the same. This 3 pics in the entry of the topic, is my version 1, but I have 2 other versions of the same invention.
Your profile says Canada, and you mixed in a couple French words, so French was my assumption. While your English seems better than my out of practice Spanish, if anything I said is particularly confusing to you, I can attempt a Spanish translation to clarify.
1. Pascals Law
This states that the fluid transmits equal pressure in all directions. This law is also responsible for the derivation of the hydrostatic law
Dell P = rho*g*h.
You left off the fact that this is only true when the fluid is stationary, not while it is transferring from one container to another. When flowing through a bent pipe, there are additional forces present beyond what is stated in that equation.
2. Bernoulli's principle
This is an application of conservation of energy. This is normally used to find conditions at a particular point when we know the conditions at some other point. But be careful. This is applicable only for non-viscous, incompressible and steady flows. Put mercury in the as the fluid, if you want to be sure that dont applies Bernoulli principle in this case.
Mercury is a strawman and irrelevant. The issue in that case is friction with the side of the pipes, which is again an additional force, it doesn't make forces due to other things go away, and to some extent is present with any fluid.
Bernoulli's equation is about energy conservation within a fluid in motion and is not applicable in hydrostatic situations where pressure differences dominate. Lets begin to give proper arguments.
Yes, it would be nice if you could begin to give proper arguments. You have not described a hydrostatic situation, you have described a situation where fluid is moving from one container to another, and there are forces involved due to the fluid motion that you are ignoring.
By the way, pressure is in all the fluid equal. All the infinite directions, its not a vector force. Very important point.
Pressure is not equal at all points in a moving fluid (Also, gravity, but it is easy to make that irrelevant by placing your device on its side). Also, while pressure is not a vector quantity, when it acts on a surface, it creates a force in a direction normal (perpendicular) to that surface, which does have direction.
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Well, possibly a bad communication. But I am from Spain, so its not the same. This 3 pics in the entry of the topic, is my version 1, but I have 2 other versions of the same invention.
Your profile says Canada, and you mixed in a couple French words, so French was my assumption. While your English seems better than my out of practice Spanish, if anything I said is particularly confusing to you, I can attempt a Spanish translation to clarify.
1. Pascals Law
This states that the fluid transmits equal pressure in all directions. This law is also responsible for the derivation of the hydrostatic law
Dell P = rho*g*h.
You left off the fact that this is only true when the fluid is stationary, not while it is transferring from one container to another. When flowing through a bent pipe, there are additional forces present beyond what is stated in that equation.
2. Bernoulli's principle
This is an application of conservation of energy. This is normally used to find conditions at a particular point when we know the conditions at some other point. But be careful. This is applicable only for non-viscous, incompressible and steady flows. Put mercury in the as the fluid, if you want to be sure that dont applies Bernoulli principle in this case.
Mercury is a strawman and irrelevant. The issue in that case is friction with the side of the pipes, which is again an additional force, it doesn't make forces due to other things go away, and to some extent is present with any fluid.
Bernoulli's equation is about energy conservation within a fluid in motion and is not applicable in hydrostatic situations where pressure differences dominate. Lets begin to give proper arguments.
Yes, it would be nice if you could begin to give proper arguments. You have not described a hydrostatic situation, you have described a situation where fluid is moving from one container to another, and there are forces involved due to the fluid motion that you are ignoring.
I live in Canada, but mixing french and english, make it not easy when comes to describe complicate arguments. I will show you in short
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I live in Canada, but mixing french and english, make it not easy when comes to describe complicate arguments. I will show you in short
Luckily there is no complicated argument needed:
-Pressure being equal at all points in a fluid only applies if the fluid is not moving.
-You are describing fluid transferring from one container to another.
-To transfer from one container to another the fluid moves through a pipe.
These are simple facts. Please acknowledge them.
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I live in Canada, but mixing french and english, make it not easy when comes to describe complicate arguments. I will show you in short
Luckily there is no complicated argument needed:
-Pressure being equal at all points in a fluid only applies if the fluid is not moving.
-You are describing fluid transferring from one container to another.
-To transfer from one container to another the fluid moves through a pipe.
These are simple facts. Please acknowledge them.
And an hydraulic system that operates with Pascal law, how does it operate? Pascal principle talks about moving a fluid from one side to the other. How do you think all pascal inventions, like compressors etc work. They move the fluid, but as a block. And if you insist about the friction inside the pipe, show me a link that talks about Pascal principle and which forces acts inside the fluid. Find me a link that mentions friction. The fluid is not moving, its like you inside a bus at constant speed, you can walk inside the bus as if you are at earth. And the pipe will have in the exterior of the pipe the same Newton forces as the pipe in the other side. If they weight with the fluid each pipe alone, 4 kilos each, they will keep the same weight no matter what
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I live in Canada, but mixing french and english, make it not easy when comes to describe complicate arguments. I will show you in short
Luckily there is no complicated argument needed:
-Pressure being equal at all points in a fluid only applies if the fluid is not moving.
-You are describing fluid transferring from one container to another.
-To transfer from one container to another the fluid moves through a pipe.
These are simple facts. Please acknowledge them.
And an hydraulic system that operates with Pascal law, how does it operate? Pascal principle talks about moving a fluid from one side to the other. How do you think all pascal inventions, like compressors etc work. They move the fluid, but as a block. And if you insist about the friction inside the pipe, show me a link that talks about Pascal principle and which forces acts inside the fluid. Find me a link that mentions friction. The fluid is not moving, its like you inside a bus at constant speed, you can walk inside the bus as if you are at earth. And the pipe will have in the exterior of the pipe the same Newton forces as the pipe in the other side. If they weight with the fluid each pipe alone, 4 kilos each, they will keep the same weight no matter what
https://www.quora.com/Is-Pascals-law-valid-under-a-dynamic-condition this link of quora talks about static water, and explains exactly what I mean. CONSTANT VELOCITY TO WORK.
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And an hydraulic system that operates with Pascal law, how does it operate? Pascal principle talks about moving a fluid from one side to the other.
No, Pascal's law applies to the static case where the fluid is not moving. I gave you a link many posts ago to the definition on wikipedia that clearly says "at rest."
How do you think all pascal inventions, like compressors etc work.
They work by fluid dynamics. As you have been told repeatedly in this thread, there are forces in a moving fluid beyond those described by Pascal's law. The forces described by Pascal's law don't just suddenly disappear, but other forces are added on top, so the statement that "pressure is the same at all points" is no longer true.
They move the fluid, but as a block.
different parts of the fluid are moving in different directions, please stop repeating this nonsensical statement.
And if you insist about the friction inside the pipe, show me a link that talks about Pascal principle and which forces acts inside the fluid. Find me a link that mentions friction.
The forces I am talking about are not friction. They are pressure gradients in the fluid due to the motion of the fluid, present even if there is no friction at all.
You apparently still haven't read the definition of Pascal's law that I previously linked you to, but here are the first 2 links from a google search of "pressure in fluid moving through a bent pipe."
https://www.tutorialspoint.com/fluid_mechanics/force_exerted_by_a_flowing_fluid_on_a_pipe_bend.asp
https://physics.stackexchange.com/questions/130691/pressure-at-a-bend-in-a-pipe
The fluid is not moving, its like you inside a bus at constant speed, you can walk inside the bus as if you are at earth.
The fluid is moving, otherwise it can't get from one container to the other. You can only walk on a bus by exerting a force on the floor of the bus. Either the force the bus is exerting on the ground also changes to match, or the outside of the bus changes velocity (by a small amount since the bus is very heavy.)
And the pipe will have in the exterior of the pipe the same Newton forces as the pipe in the other side. If they weight with the fluid each pipe alone, 4 kilos each, they will keep the same weight no matter what
The mass of the fluid doesn't change but its velocity does. The forces would not be equal, because the fluid has changed speed. At the first bend, forces need to change the velocity from the speed of box A pointed to the left, to some velocity pointed to the right to match the fluid flow rate. Since box B is stationary in the lab frame, the fluid only needs to slow to a stop not switch directions, so less force is exerted there. This is all nicely balanced and obeys Newton's laws. (I am ignoring the vertical components of velocity which are unimportant for this discussion, though necessary when considering conservation of energy.)
Of course to actually calculate these forces, first you need to acknowledge their existence.
https://www.quora.com/Is-Pascals-law-valid-under-a-dynamic-condition this link of quora talks about static water, and explains exactly what I mean. CONSTANT VELOCITY TO WORK.
No, it says the exact opposite of that. It uses the phrases "at rest" and "not moving." You have described a situation where one container is emptying of liquid, and another is filling. That is by definition a dynamic situation. It cannot be considered static. More so, the water is moving at different velocities in different locations, therefore, no matter what frame you pick, you will always see water moving.
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And an hydraulic system that operates with Pascal law, how does it operate? Pascal principle talks about moving a fluid from one side to the other.
No, Pascal's law applies to the static case where the fluid is not moving. I gave you a link many posts ago to the definition on wikipedia that clearly says "at rest."
How do you think all pascal inventions, like compressors etc work.
They work by fluid dynamics. As you have been told repeatedly in this thread, there are forces in a moving fluid beyond those described by Pascal's law. The forces described by Pascal's law don't just suddenly disappear, but other forces are added on top, so the statement that "pressure is the same at all points" is no longer true.
They move the fluid, but as a block.
different parts of the fluid are moving in different directions, please stop repeating this nonsensical statement.
And if you insist about the friction inside the pipe, show me a link that talks about Pascal principle and which forces acts inside the fluid. Find me a link that mentions friction.
The forces I am talking about are not friction. They are pressure gradients in the fluid due to the motion of the fluid, present even if there is no friction at all.
You apparently still haven't read the definition of Pascal's law that I previously linked you to, but here are the first 2 links from a google search of "pressure in fluid moving through a bent pipe."
https://www.tutorialspoint.com/fluid_mechanics/force_exerted_by_a_flowing_fluid_on_a_pipe_bend.asp
https://physics.stackexchange.com/questions/130691/pressure-at-a-bend-in-a-pipe
The fluid is not moving, its like you inside a bus at constant speed, you can walk inside the bus as if you are at earth.
The fluid is moving, otherwise it can't get from one container to the other. You can only walk on a bus by exerting a force on the floor of the bus. Either the force the bus is exerting on the ground also changes to match, or the outside of the bus changes velocity (by a small amount since the bus is very heavy.)
And the pipe will have in the exterior of the pipe the same Newton forces as the pipe in the other side. If they weight with the fluid each pipe alone, 4 kilos each, they will keep the same weight no matter what
The mass of the fluid doesn't change but its velocity does. The forces would not be equal, because the fluid has changed speed. At the first bend, forces need to change the velocity from the speed of box A pointed to the left, to some velocity pointed to the right to match the fluid flow rate. Since box B is stationary in the lab frame, the fluid only needs to slow to a stop not switch directions, so less force is exerted there. This is all nicely balanced and obeys Newton's laws. (I am ignoring the vertical components of velocity which are unimportant for this discussion, though necessary when considering conservation of energy.)
Of course to actually calculate these forces, first you need to acknowledge their existence.
https://www.quora.com/Is-Pascals-law-valid-under-a-dynamic-condition this link of quora talks about static water, and explains exactly what I mean. CONSTANT VELOCITY TO WORK.
No, it says the exact opposite of that. It uses the phrases "at rest" and "not moving." You have described a situation where one container is emptying of liquid, and another is filling. That is by definition a dynamic situation. It cannot be considered static. More so, the water is moving at different velocities in different locations, therefore, no matter what frame you pick, you will always see water moving.
If you are at a train at constant velocity what diference you see, when its accelerating? Your glass of water in the train is moving out of the glass, when you are at constant speed or does that when accelerating. About the pipe how it will work, if you got a container of 10.000 kilos and a pipe of 4 kilos, for example to bring it to an extream, the pipe would be something with no sense. What is important, is that the container that will transfer the fluid is in a constant velocity and the container that receive it, is stationary. If you still say that at constant velocity, the water is jumping out of your glass, then sorry, but I cannot believe it.
And this works in space, so at a constant speed, sorry, but if there is water over it, it will be the most quiet water, that you will ever see. The problem is that mathematicaly, this two laws have never been together.
About the link, in any moment talks about a closed statical fluid system, as Pascal law. It dont applies. Not valid. In any moment talks about Pascal law sorry. Not even once, they talk about a piston for example. Read more about Pascal, because you seem lost. We are talking about fluids at rest, thats why I mention always constant speed. CONSTANT SPEED. If you travel in a spaceship at 1000 km second, you wont notice the diference even if you are at stop. You will move the same, and the fluid will be quiet quiet. Newton laws will work, but not during the tranfer at constant velocity
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If you are at a train at constant velocity what diference you see, when its accelerating? Your glass of water in the train is moving out of the glass, when you are at constant speed or does that when accelerating. About the pipe how it will work, if you got a container of 10.000 kilos and a pipe of 4 kilos, for example to bring it to an extream, the pipe would be something with no sense. What is important, is that the container that will transfer the fluid is in a constant velocity and the container that receive it, is stationary. If you still say that at constant velocity, the water is jumping out of your glass, then sorry, but I cannot believe it.
And this works in space, so at a constant speed, sorry, but if there is water over it, it will be the most quiet water, that you will ever see. The problem is that mathematicaly, this two laws have never been together.
Language barrier or not, this is inane. You have not shown anything mathematically or logically about what you are suggesting. Meberbs has done all the work here. The system you show in your diagram will not behave as you say for the reasons you state. ( 100 m/s constant velocity!? no accelerating?) Nor will it behave differently for the reasons, as given by Meberbs & others, that you are ignoring. In sum, it is an uninteresting system with inconsequential implications. Fluid flows under a pressure gradient, electrons flow within an electric field gradient, apples fall from tree's in a gravitational field, so what? You are not violating Pascal's law because it does not apply to your system. You are not violating conservation of momentum either, & I haven't seen any math showing how you think the momentum & energy put into the system are accounted for.
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If you are at a train at constant velocity what diference you see, when its accelerating? Your glass of water in the train is moving out of the glass, when you are at constant speed or does that when accelerating. About the pipe how it will work, if you got a container of 10.000 kilos and a pipe of 4 kilos, for example to bring it to an extream, the pipe would be something with no sense. What is important, is that the container that will transfer the fluid is in a constant velocity and the container that receive it, is stationary. If you still say that at constant velocity, the water is jumping out of your glass, then sorry, but I cannot believe it.
And this works in space, so at a constant speed, sorry, but if there is water over it, it will be the most quiet water, that you will ever see. The problem is that mathematicaly, this two laws have never been together.
Language barrier or not, this is inane. You have not shown anything mathematically or logically about what you are suggesting. Meberbs has done all the work here. The system you show in your diagram will not behave as you say for the reasons you state. ( 100 m/s constant velocity!? no accelerating?) Nor will it behave differently for the reasons, as given by Meberbs & others, that you are ignoring. In sum, it is an uninteresting system with inconsequential implications. Fluid flows under a pressure gradient, electrons flow within an electric field gradient, apples fall from tree's in a gravitational field, so what? You are not violating Pascal's law because it does not apply to your system. You are not violating conservation of momentum either, & I haven't seen any math showing how you think the momentum & energy put into the system are accounted for.
You have not understood anything. Its acellerated and then what happens after acceleration, what comes????? constant speed. I find here a big lack of knowledges about basic things. Understand the device and the make the critic, I would suggest you. The clue of the system is the transfer. And maths are on the pics, so you did not red anything.
By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
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*snip* We are talking about fluids at rest, thats why I mention always constant speed. CONSTANT SPEED. If you travel in a spaceship at 1000 km second, you wont notice the diference even if you are at stop. You will move the same, and the fluid will be quiet quiet. Newton laws will work, but not during the tranfer at constant velocity
If you are transferring fluid - even at a constant velocity - then it is not at rest. It is moving.
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By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
I'm not a scientist so I'm not passing judgement on what you are presenting, but I just wanted to point out that YOU came here with an unsolicited proposal, and now you are not happy because no one is agreeing with you.
It's not our job to agree with you, and you have received a LOT of free advice.
So if your goal in coming to NASASpaceFlight.com was not to learn from its members, why are you here?
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You have not understood anything. Its acellerated and then what happens after acceleration, what comes????? constant speed. I find here a big lack of knowledges about basic things. Understand the device and the make the critic, I would suggest you. The clue of the system is the transfer. And maths are on the pics, so you did not red anything.
By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
You are the one who has been repeatedly claiming that fluid somehow goes from a moving box to a stationary box (or a stationary box to a moving box if you pick a different frame) without moving, accelerating, or feeling any net force. In no case does this make the slightest bit of sense. Your claims have been self-contradictory from the very first post, as I pointed out in my first post.
You have not listened to a single thing for this entire thread (my first post in this thread stated a fear that you would react that way to hearing "no"), ignoring the definitions of terms that have been given to you. It seems to me that your questions in your first post were insincere. You never acknowledged the inherent contradiction I pointed out in my first post. You have refused to accept basic facts about fluid flow, despite one of your slides asking if there are any mistakes in your analysis.
...The problem is that mathematicaly, this two laws have never been together.
About the link, in any moment talks about a closed statical fluid system, as Pascal law. It dont applies. Not valid. In any moment talks about Pascal law sorry. Not even once, they talk about a piston for example. Read more about Pascal, because you seem lost. We are talking about fluids at rest, thats why I mention always constant speed. CONSTANT SPEED. If you travel in a spaceship at 1000 km second, you wont notice the diference even if you are at stop. You will move the same, and the fluid will be quiet quiet. Newton laws will work, but not during the tranfer at constant velocity
Newton's laws work in all cases without caveat (even in relativity if you define force and momentum correctly.) Pascal's law is a special case (actually derivable from Newton) That only works in a very specific case.
You are the one repeating a mantra here of "constant velocity" when by the definition of your setup, the fluid has different velocities at different locations. For a fluid to be not moving, one basic condition is that the shape of the volume occupied by the fluid can't be changing. That is simply untrue in your system. (Of course even if the shape of the volume was constant, the fluid still could be moving within that volume.) As long as you continue to deny basic properties of your system as described by you, productive conversation can't happen.
Your complaints about the links I provided don't even make sense. The links describe basic facts about fluid flow that are true regardless of where the fluid is coming from or where it is going. Pascal's law is not discussed, because as you have been told repeatedly, it does not apply, a more general approach needs to be used.
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You have not understood anything. Its acellerated and then what happens after acceleration, what comes????? constant speed. I find here a big lack of knowledges about basic things. Understand the device and the make the critic, I would suggest you. The clue of the system is the transfer. And maths are on the pics, so you did not red anything.
By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
You keep using the word "math", & don't think we share a common definition. Your step 1 has a semblance of a free body diagram with static forces displayed at time T=+0, but there is a huge gap to step 2 & 3. Show me your math of how you get to 100 m/s instantaneously? Tell me mathematically what is happening in that bendy pipe of your diagram that allows it move with the system to get to 100m/s. Show me your math for the work done by the piston imparting momentum into the fluid flow out of the cylinder and into the bendy pipe. Please do so in vector form for x/y/z components of the fluid flow. Show me that & I'll retract my math criticisms.
I may have slept through my college history class when they taught me about the Germans bombing Pearl Harbor, but I was pretty awake through my engineering calculus/differential equations, statics, dynamics, fluid dynamics, materials, & thermodynamics that I think I still recognize "math" when I see it. Your steps 1-3 do not contain consequential math.
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I think you may have invented a double barreled stomp rocket, with one barrel rigged with hold down clamps (your cylinder C)
https://www.youtube.com/watch?v=f2ao_EUhw88
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You have not understood anything. Its acellerated and then what happens after acceleration, what comes????? constant speed. I find here a big lack of knowledges about basic things. Understand the device and the make the critic, I would suggest you. The clue of the system is the transfer. And maths are on the pics, so you did not red anything.
By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
You keep using the word "math", & don't think we share a common definition. Your step 1 has a semblance of a free body diagram with static forces displayed at time T=+0, but there is a huge gap to step 2 & 3. Show me your math of how you get to 100 m/s instantaneously? Tell me mathematically what is happening in that bendy pipe of your diagram that allows it move with the system to get to 100m/s. Show me your math for the work done by the piston imparting momentum into the fluid flow out of the cylinder and into the bendy pipe. Please do so in vector form for x/y/z components of the fluid flow. Show me that & I'll retract my math criticisms.
I may have slept through my college history class when they taught me about the Germans bombing Pearl Harbor, but I was pretty awake through my engineering calculus/differential equations, statics, dynamics, fluid dynamics, materials, & thermodynamics that I think I still recognize "math" when I see it. Your steps 1-3 do not contain consequential math.
Acelleration from a type of spring, you can have the acceleration you want. I am not here to discuss about engineer problems, thast not my issue. I just give a concept of a device. I find difficult to find equations that describes how the mass is transfered. Do you have any clue?. Step one goes in Newton, step two also, step three of course, the only thing that in my idea makes it reactionless is the transfer of fluid, once the container have a final velocity, say 5 metres per second or whatever, the fluid aboard is at rest. NEVER IN ACCELERATION the transfer of fluid must happen. Now I post you two systems of the same version so you can view it, and tell me if you see at first sight something that can work in reality, and I will post you a pic, with a problem. Find me the equations, because it does not exist, to explain how is the fluid transfer done.
I was meaning the device knowledge, not your personal knowledges, sorry.
The links are:
https://youtu.be/Vz8R1et6BRA This one is like the pics
https://youtu.be/aIZN5zDDXEo This one begins the animation at second 37.
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By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
I'm not a scientist so I'm not passing judgement on what you are presenting, but I just wanted to point out that YOU came here with an unsolicited proposal, and now you are not happy because no one is agreeing with you.
It's not our job to agree with you, and you have received a LOT of free advice.
So if your goal in coming to NASASpaceFlight.com was not to learn from its members, why are you here?
Sorry but I dont have the same opinion and the forum is to exchange opinions, right? I dont blame anybody, but no god is able to tell me why it would not work EMPIRICALLY, and I say why.
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If you travel in a spaceship at 1000 km second, you wont notice the diference even if you are at stop. You will move the same, and the fluid will be quiet quiet.
This is the source of your misunderstanding. You're attempting to treat a system of moving parts as if it were a single monolithic entity.
The definition in wikipedia states:
A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid.
Bold mine. Here's the issue. When you read this, you're equating at rest with "constant velocity." This is incorrect. What you should be doing, is asking yourself "at rest relative to what?". The answer is given earlier in the definition; enclosed. This means that the fluid must be at rest relative to whatever is enclosing it. In this case, the piping in your system. So:
Piping moving at 0m/s, fluid moving at 0m/s: Pascal's law works.
Piping moving at 100m/s, fluid moving at 100m/s: Pascal's law works.
Piping moving at 0m/s, fluid moving at 100m/s: Pascal's law breaks down; and a more general approach must be used.
Piping moving at 100m/s, fluid moving at 0m/s: Pascal's law breaks down; and a more general approach must be used.
Any mismatch between piping velocity and fluid velocity invalidate the approximations made in Pascal's law. Here on Earth, the reactive motion of the piping (and whatever it's attached to!) is usually too small for us to care about; but it will definitely be noticed by a system floating up in space. You can move your blocks and fluids with whatever acceleration profile you want. Your center of mass will not change, and the drive will go nowhere.
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If you travel in a spaceship at 1000 km second, you wont notice the diference even if you are at stop. You will move the same, and the fluid will be quiet quiet.
This is the source of your misunderstanding. You're attempting to treat a system of moving parts as if it were a single monolithic entity.
The definition in wikipedia states:
A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid.
Bold mine. Here's the issue. When you read this, you're equating at rest with "constant velocity." This is incorrect. What you should be doing, is asking yourself "at rest relative to what?". The answer is given earlier in the definition; enclosed. This means that the fluid must be at rest relative to whatever is enclosing it. In this case, the piping in your system. So:
Piping moving at 0m/s, fluid moving at 0m/s: Pascal's law works.
Piping moving at 100m/s, fluid moving at 100m/s: Pascal's law works.
Piping moving at 0m/s, fluid moving at 100m/s: Pascal's law breaks down; and a more general approach must be used.
Piping moving at 100m/s, fluid moving at 0m/s: Pascal's law breaks down; and a more general approach must be used.
Any mismatch between piping velocity and fluid velocity invalidate the approximations made in Pascal's law. Here on Earth, the reactive motion of the piping (and whatever it's attached to!) is usually too small for us to care about; but it will definitely be noticed by a system floating up in space. You can move your blocks and fluids with whatever acceleration profile you want. Your center of mass will not change, and the drive will go nowhere.
So if you have viewed my last pic, fluid will move container B as enters, right?
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So if you have viewed my last pic, fluid will move container B as enters, right?
If your piping is rigid, the the transfer of fluid from container A to B essentially moves fluid mass to the right. This is countered by motion of both containers to the left. The center of mass of the entire system, however, remains unchanged. In moving the fluid from container B back to container A, both motions are reversed, and the entire container/fluid/pipe system returns to its original location. It does not matter how much time either transfer takes to occur.
If your piping is flexible, the container B could move left, could move right, or could even stand still depending on what quantity of fluid+piping moves beyond the right edge of the container. Difficult to tell what would happen from your drawing, but in all cases, center of mass would remain unchanged.
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So if you have viewed my last pic, fluid will move container B as enters, right?
If your piping is rigid, the the transfer of fluid from container A to B essentially moves fluid mass to the right. This is countered by motion of both containers to the left. The center of mass of the entire system, however, remains unchanged. In moving the fluid from container B back to container A, both motions are reversed, and the entire container/fluid/pipe system returns to its original location. It does not matter how much time either transfer takes to occur.
If your piping is flexible, the container B could move left, could move right, or could even stand still depending on what quantity of fluid+piping moves beyond the right edge of the container. Difficult to tell what would happen from your drawing, but in all cases, center of mass would remain unchanged.
Then what happens if one of the container pulls the system with more force than the other after the transfer, because have more fluid in it? the center of mass, wont move? I really see it difficult that stays the same. Then Newton third law would not work. Since when two opposite mass forces, both tied up, the opposite object with more weight, will not able to pull the other object and the system to its side, if both have the same speed? its not logic. Explain me visually how it would be arranged, that no force win. What would be the momentum of one deposit full of fluid respect to the other deposit going light?
Can you show what you say with an equation? So it will move container B to the same direction as container A has. Sorry but you cannot show that.
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So if you have viewed my last pic, fluid will move container B as enters, right?
If your piping is rigid, the the transfer of fluid from container A to B essentially moves fluid mass to the right. This is countered by motion of both containers to the left. The center of mass of the entire system, however, remains unchanged. In moving the fluid from container B back to container A, both motions are reversed, and the entire container/fluid/pipe system returns to its original location. It does not matter how much time either transfer takes to occur.
If your piping is flexible, the container B could move left, could move right, or could even stand still depending on what quantity of fluid+piping moves beyond the right edge of the container. Difficult to tell what would happen from your drawing, but in all cases, center of mass would remain unchanged.
It can be like a garden pipe, and very long, the two deposits will hit their respective wall, one with more fluid and the other with less fluid, at the same speed, because the transfer wont affect the their final speed, so you can see it as you want, but there will be an unbalance of momentum, and with that and enough force, it can move the hole center of mass.
By the way, the piping or better the flexible garden pipe, will be affected by the same forces as in the other pipe, going opposite direction, Newton will be there, but that dont affect the moving containers. You dont see it ehhh?
BY THE WAY THE THIRD LAW OF NEWTON THAT RELATES TO THE CENTER OF MASS, EXPLAIN THAT NO INTERNAL FORCE CAN MOVE THE CENTER OF ITS OWN MASS BECAUSE OF ACTION-REACTION, BUT IN THIS DEVICE HOW IT LOSES MASS THE CONTAINER B, IT HAS NOTHING TO DO WITH NEWTON.
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*snip* there will be an unbalance of momentum, and with that and enough force, it can move the hole center of mass.
Where is the unbalanced momentum?
By the way, the piping or better the flexible garden pipe, will be affected by the same forces as in the other pipe, going opposite direction, Newton will be there, but that dont affect the moving containers. You dont see it ehhh?
BY THE WAY THE THIRD LAW OF NEWTON THAT RELATES TO THE CENTER OF MASS, EXPLAIN THAT NO INTERNAL FORCE CAN MOVE THE CENTER OF ITS OWN MASS BECAUSE OF ACTION-REACTION, BUT IN THIS DEVICE HOW IT LOSES MASS THE CONTAINER B, IT HAS NOTHING TO DO WITH NEWTON.
You are quite right, no internal force can move a center of its own mass. Your device is very much proving this. Newton is still there.
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This I found it in the net:
Why does a system need an external force to shift its center of mass?
It's a consequence of conservation of momentum, or alternatively, of Newton's third law. Any force that the boy exerts on the raft is countered by an equal but opposite force that the raft exerts on the boy. Those internal forces cancel, and therefore play zero role in the motion of the center of mass relative to an inertial frame.
Well, if one of the Containers lose mass in the transfer and no momentum will be transmitted to the stationary container, then container A will lose momentum because of mass. And the fluid will land as it departed, in total rest. And this have nothing to do with center of mass itself. If the force have not done reaction it works, and if it loses mass it loses momentum.
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*snip* there will be an unbalance of momentum, and with that and enough force, it can move the hole center of mass.
Where is the unbalanced momentum?
By the way, the piping or better the flexible garden pipe, will be affected by the same forces as in the other pipe, going opposite direction, Newton will be there, but that dont affect the moving containers. You dont see it ehhh?
BY THE WAY THE THIRD LAW OF NEWTON THAT RELATES TO THE CENTER OF MASS, EXPLAIN THAT NO INTERNAL FORCE CAN MOVE THE CENTER OF ITS OWN MASS BECAUSE OF ACTION-REACTION, BUT IN THIS DEVICE HOW IT LOSES MASS THE CONTAINER B, IT HAS NOTHING TO DO WITH NEWTON.
You are quite right, no internal force can move a center of its own mass. Your device is very much proving this. Newton is still there.
Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container. I post you again the pic, and you tell me what happens with the stationary container. First view how the device works and then you can critic me, without generalities, that you can not show in the device, with all my respects.
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I find difficult to find equations that describes how the mass is transfered. Do you have any clue?.
It seems that everyone in this thread except you has a clue on this. I posted a link to a video on how to calculate force through a bent pipe. Others pointed out Bernoulli's equation for fluid motion, which is also relevant. (See the other link I had provided that points out that fluid flow speeds would be different between the inside and outside curves of a curved pipe.)
Step one goes in Newton, step two also, step three of course, the only thing that in my idea makes it reactionless is the transfer of fluid, once the container have a final velocity, say 5 metres per second or whatever, the fluid aboard is at rest. NEVER IN ACCELERATION the transfer of fluid must happen.
It is reactionless when you transfer fluid, because you simply are choosing to ignore what everyone here is telling you: there would be a force present. You are saying that the fluid starts in container A which is moving away from container B. Somehow you claim that the fluid will end up in container B, and also be at rest. By definition the velocity of the fluid is different between the initial and final conditions. This cannot happen if there is no acceleration. The velocity of the fluid changed therefore it must have accelerated this is simply a definition.
Then what happens if one of the container pulls the system with more force than the other after the transfer, because have more fluid in it?
Your question is invalid. You are assuming the result you want (unbalanced forces). The if part of the question is phrased wrong. the question you should be asking is "what would happen?"
the center of mass, wont move? I really see it difficult that stays the same. Then Newton third law would not work.
According to Newton's laws, the center of mass of a system at rest will remain at rest unless an external force acts on the system. There are no external forces on the system. (If the center of mass is not at rest, which is the case for some of the diagrams you posted, then the center of mass instead moves with constant velocity, but you can simplify things by moving to the center of mass frame so my previous statement holds.)
Since when two opposite mass forces, both tied up, the opposite object with more weight, will not able to pull the other object and the system to its side, if both have the same speed? its not logic. Explain me visually how it would be arranged, that no force win. What would be the momentum of one deposit full of fluid respect to the other deposit going light?
No clue what any of what you are saying here is supposed to mean.
Can you show what you say with an equation? So it will move container B to the same direction as container A has. Sorry but you cannot show that.
Asserting that others can't show something is not going to help you.
Here is some math for you:
Mc = mass of empty cart (both carts are the same)
Mf = Mass of fluid that fits in 1 cart
Vi = initial velocity of cart A.
rho = fluid density
Total initial momentum of the system:
(assume the hose is made of some relatively super lightweight material, so we can ignore that part of it is moving, and part isn't)
Ptot = (Mc+Mf)*Vi
Let us assume that it is a long flexible tube as you suggest. For some length of hose, if the direction and velocity of fluid flow going in and out of the tubing is the same then, (ignoring fiction losses) there is no net momentum change in the fluid, so we can just ignore that section. To take advantage of this, assume a 90 degree bend fixed to the top of each cart and pointing to each other. Any net forces experienced by the piping will therefore be only from those bends. (You could drop this assumption and get the same results I am about to show, but then you would have to take into account the changing shape of the hose as the carts move.)
assume the fluid flow during the transfer happens at a constant rate dM/dt and the cross sectional area of all piping including the 90 degree bends is a constant area, A.
The cart will empty in time T = Mf / (dM/dt). The velocity of the fluid in the pipe is vt = dM/dt /(rho*A).
The fluid enters the pipe bend with upwards velocity, but leaves with velocity pointed to the right. Ignoring the upwards component of the force (which wouldn't end up relevant in this setup), then in the reference frame of cart A, the fluid after the bend must be moving to the right at vt, so it is carrying momentum to the right away at a rate of vt*dM/dt. By Newton's second law, that means it had a force being continuously exerted on it by the pipe bend of F1 = vt*dM/dt pointed to the right. by Newton's third law, it exerts a force on the pipe equal in magnitude to F1, but pointed to the left.
This is a force effectively applied to cart A through the pipe bend during the transfer. the final velocity of cart A is therefore:
vi + F1*T/Mc =vi + vt*dM/dt * Mf/ (dM/dt) /Mc = vi + vt*Mf/Mc
In other words, as the mass in the first cart decreases, the velocity of the cart actually increases.
More work can be done to show that a comparable force is applied to cart B, but making it move to the right so when you add everything together, the net momentum of the system is unchanged. Also, there are details about what happens when the cart is partially full, so to actually get everything correct, you would need to account for the variable effective mass of each cart with time (Go look up a derivation of the rocket equation for details). Plus, there are some details I glossed over about the hose. Since it is attached to 2 carts moving at different velocities, it will move itself, along with all of the fluid in it, and this will be balanced out when you consider that if the velocity of the fluid relative to the tubing is the same at either end, but the ends of the tubing are moving at different speeds, then the fluid must be accelerating (speeding up or slowing down) relative to the other frames, and therefore actually applying forces, as well the changing velocities of the carts dragging the hose with them.
Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container.
Too bad pipes don't operate by teleporting, but by having fluid flow through them and apply forces.
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this? And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in it ;)
By the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics. Maybe it is what happens when you arrive in new phisics topics Lol
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I find difficult to find equations that describes how the mass is transfered. Do you have any clue?.
It seems that everyone in this thread except you has a clue on this. I posted a link to a video on how to calculate force through a bent pipe. Others pointed out Bernoulli's equation for fluid motion, which is also relevant. (See the other link I had provided that points out that fluid flow speeds would be different between the inside and outside curves of a curved pipe.)
Step one goes in Newton, step two also, step three of course, the only thing that in my idea makes it reactionless is the transfer of fluid, once the container have a final velocity, say 5 metres per second or whatever, the fluid aboard is at rest. NEVER IN ACCELERATION the transfer of fluid must happen.
It is reactionless when you transfer fluid, because you simply are choosing to ignore what everyone here is telling you: there would be a force present. You are saying that the fluid starts in container A which is moving away from container B. Somehow you claim that the fluid will end up in container B, and also be at rest. By definition the velocity of the fluid is different between the initial and final conditions. This cannot happen if there is no acceleration. The velocity of the fluid changed therefore it must have accelerated this is simply a definition.
Then what happens if one of the container pulls the system with more force than the other after the transfer, because have more fluid in it?
Your question is invalid. You are assuming the result you want (unbalanced forces). The if part of the question is phrased wrong. the question you should be asking is "what would happen?"
the center of mass, wont move? I really see it difficult that stays the same. Then Newton third law would not work.
According to Newton's laws, the center of mass of a system at rest will remain at rest unless an external force acts on the system. There are no external forces on the system. (If the center of mass is not at rest, which is the case for some of the diagrams you posted, then the center of mass instead moves with constant velocity, but you can simplify things by moving to the center of mass frame so my previous statement holds.)
Since when two opposite mass forces, both tied up, the opposite object with more weight, will not able to pull the other object and the system to its side, if both have the same speed? its not logic. Explain me visually how it would be arranged, that no force win. What would be the momentum of one deposit full of fluid respect to the other deposit going light?
No clue what any of what you are saying here is supposed to mean.
Can you show what you say with an equation? So it will move container B to the same direction as container A has. Sorry but you cannot show that.
Asserting that others can't show something is not going to help you.
Here is some math for you:
Mc = mass of empty cart (both carts are the same)
Mf = Mass of fluid that fits in 1 cart
Vi = initial velocity of cart A.
rho = fluid density
Total initial momentum of the system:
(assume the hose is made of some relatively super lightweight material, so we can ignore that part of it is moving, and part isn't)
Ptot = (Mc+Mf)*Vi
Let us assume that it is a long flexible tube as you suggest. For some length of hose, if the direction and velocity of fluid flow going in and out of the tubing is the same then, (ignoring fiction losses) there is no net momentum change in the fluid, so we can just ignore that section. To take advantage of this, assume a 90 degree bend fixed to the top of each cart and pointing to each other. Any net forces experienced by the piping will therefore be only from those bends. (You could drop this assumption and get the same results I am about to show, but then you would have to take into account the changing shape of the hose as the carts move.)
assume the fluid flow during the transfer happens at a constant rate dM/dt and the cross sectional area of all piping including the 90 degree bends is a constant area, A.
The cart will empty in time T = Mf / (dM/dt). The velocity of the fluid in the pipe is vt = dM/dt /(rho*A).
The fluid enters the pipe bend with upwards velocity, but leaves with velocity pointed to the right. Ignoring the upwards component of the force (which wouldn't end up relevant in this setup), then in the reference frame of cart A, the fluid after the bend must be moving to the right at vt, so it is carrying momentum to the right away at a rate of vt*dM/dt. By Newton's second law, that means it had a force being continuously exerted on it by the pipe bend of F1 = vt*dM/dt pointed to the right. by Newton's third law, it exerts a force on the pipe equal in magnitude to F1, but pointed to the left.
This is a force effectively applied to cart A through the pipe bend during the transfer. the final velocity of cart A is therefore:
vi + F1*T/Mc =vi + vt*dM/dt * Mf/ (dM/dt) /Mc = vi + vt*Mf/Mc
In other words, as the mass in the first cart decreases, the velocity of the cart actually increases.
More work can be done to show that a comparable force is applied to cart B, but making it move to the right so when you add everything together, the net momentum of the system is unchanged. Also, there are details about what happens when the cart is partially full, so to actually get everything correct, you would need to account for the variable effective mass of each cart with time (Go look up a derivation of the rocket equation for details). Plus, there are some details I glossed over about the hose. Since it is attached to 2 carts moving at different velocities, it will move itself, along with all of the fluid in it, and this will be balanced out when you consider that if the velocity of the fluid relative to the tubing is the same at either end, but the ends of the tubing are moving at different speeds, then the fluid must be accelerating (speeding up or slowing down) relative to the other frames, and therefore actually applying forces, as well the changing velocities of the carts dragging the hose with them.
Have you viewed the pics? I think its clear, in the transfer of fluid from container A to B.
Teletransporting mass literally as I show, does not do any force but moving the piston but changes the momentum of the container.
Too bad pipes don't operate by teleporting, but by having fluid flow through them and apply forces.
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this?
The one thing that is clear in what you just wrote is that you are completely wrong and have no clue what you are talking about. Imagine if the hose wasn't there, just a 90 degree bend. Pushing the piston up would force water up through the bend and then out the back, accelerating the cart. Attaching a hose changes nothing about these forces. other forces may act on the hose itself, or whatever is on the other end, and dragging the hose increases the mass the cart has to deal with, but, the cart most certainly can accelerate like that.
And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in it ;)
Again, there are flows happening in the system you described, so you need hydrodynamics, not hydrostatics, and your blind application of Pascal is wrong.
By the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics.
You appear to be defining an expert as someone who agrees with you. You have been given multiple proofs, starting in my first post where I pointed out that your question was essentially contradictory.
I have lost track of how many ways people have tried to explain to you that the fluid must move to get from one box to the other. So many simple proofs of it:
-the boxes are moving at different velocities, so the fluid must accelerate to get from one to the other
-the fluid starts moving away from the second box, so to get to the second box, it must move towards it at some point.
-etc. (re-read this thread if you need to see some others)
If you can't even accept such simple proofs, then there is no point in discussing anything with you, and there is really no point in going through a detailed proof of basic physical facts such as the equivalence between Newton's laws and conservation of momentum.
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
In addition to what is stated in the answer to that question, the fluid can't be moving for Pascal's law to apply. Please stop ignoring this statement. (And the fluid is moving in your device, stop denying that.)
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Have it clear, in a closed hydraulic system there is no flow, NO FLOW, ask and hydraulic engineer or go in the net. I found a link just in a second. NO flow. You have got it wrong.
Second, the impulse of the container up to its final velocity will move the pipe structure, so if the container loses weight, the container will not accelerate by the pipe structure, because the momentum was donated by the container itself in the acceleration and final speed. So what can happen in any case it that will slow down.
the transfer ocurrs under pascal laws, and no flow happen. Show me a good link, no flow, o. If there is flow, it will rebound back and forward inbetween the pistons, so even in that fantasy case, it would be useless flow. No flow, closed system. If you are not able to understand this then I cannot do anything for you. You dont have any idea how pascal laws operates by insisting in flows.
And so about the pic, if you have a bumper machine, and you install it in a car and you live the end of the pipe in earth, a big long pipe of 400 hundred meters and you begin to bump the water from the track. If there is somebody in earth holding the pipe, it will jump the water, back at 100 km per hour as it goes out of the pipe? come on man Lol, it will go out as if the bumper was next to you hahaha
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Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
I'm not sure why I'm bothering to post this as you are clearly either not listening, ignoring or not understanding anything that is being posted. The information you have linked to states for Pascal's principle that:
"the pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid"
Your system is clearly not under equilibrium conditions as you required the fluid to move from one place to another. By definition that is not in equilibrium. What the statement above says is that if you apply pressure at one point and there is no movement of the fluid in the system then the pressure is the same at all points.
In other words Pascal's principle does not apply to your system. It does not apply when fluid is moving from one place to another. For there to be movement there has to be a pressure difference. In the case of a hydraulic jack Pascal principle only applies when there is no movement of the fluid.
It would seem that you are not clear about what you are talking about.
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Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
I'm not sure why I'm bothering to post this as you are clearly either not listening, ignoring or not understanding anything that is being posted. The information you have linked to states for Pascal's principle that:
"the pressure applied at one point in an enclosed fluid under equilibrium conditions is transmitted equally to all parts of the fluid"
Your system is clearly not under equilibrium conditions as you required the fluid to move from one place to another. By definition that is not in equilibrium. What the statement above says is that if you apply pressure at one point and there is no movement of the fluid in the system then the pressure is the same at all points.
In other words Pascal's principle does not apply to your system. It does not apply when fluid is moving from one place to another. For there to be movement there has to be a pressure difference. In the case of a hydraulic jack Pascal principle only applies when there is no movement of the fluid.
It would seem that you are not clear about what you are talking about.
In constant speed they are in equilibrium the liquids settled in the container in constant velocity. This I have repeated thousands of times, and you insist in no equilibrium, and I tell you why not, and you insist one and another, get it, I have explained it but no way. I repeat read about Pascal because you dont have it at all, in Pascal systems. The jack hydraulic operates in the same way, there is no flow inside, in a compressor system, there is a pipe that have liquid and move it, to make it work. Please have it clear.
http://aplusphysics.com/courses/honors/fluids/Pascal.html
dont mix up continuity of fluids with statical fluids. Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
http://aplusphysics.com/courses/honors/fluids/continuity.html
Hey you know the difference between constant velocity and acceleration? we can begin by there.
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https://www.myodesie.com/wiki/index/returnEntry/id/2986 here explains on the milimeter how it works. And to the next one that comes again with the equilibrium, just to tell him, that sitting in an airplane is nearly like sitting at home
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Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
You seem to be contradicting yourself. In order for mass to arrive in the containers, mass must flow through the pipe. See the animation on the "continuity" website.
Also, as mass flows through the pipe there is resistance. That means there is less force at the outflow than it took to push the mass into the pipe. A net loss in energy.
And again, you don't seem to want to gain knowledge from the NASASpaceFlight members, so I'm not sure what your goal is here. No one seems to be impressed by what you are proposing, especially since it seems like some sort of perpetual motion machine, and science has proven that those don't exist.
My $0.02
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Remeber the pipe is always full, no new flow arrives, just flow levels go up and down in the containers, view it like that
You seem to be contradicting yourself. In order for mass to arrive in the containers, mass must flow through the pipe. See the animation on the "continuity" website.
Also, as mass flows through the pipe there is resistance. That means there is less force at the outflow than it took to push the mass into the pipe. A net loss in energy.
And again, you don't seem to want to gain knowledge from the NASASpaceFlight members, so I'm not sure what your goal is here. No one seems to be impressed by what you are proposing, especially since it seems like some sort of perpetual motion machine, and science has proven that those don't exist.
My $0.02
The members of a forum, dont make it sound like god. No specialist, have been talking with me, because they acuse me of not having maths proves, and they insist on a thing that is not true. Since when they have becomed specialist of Pascal law? come on, lets be honest.
Between two pistons that there is only fluid, if you apply force to one piston it pushes the other piston and inbetween, force will be in all place the same. What kind of flow we are talking about? in a closed hydraulic system that works through Pascal laws, have the same charachteristics as the transfer of fluid in my device does. There is no lose, the force of the piston will be translated through all the fluid in the way of pression instantanly.
Science have not proven that it dont exists, again you mistake, because why that interest by the EM Drive and all that, for something that you say that is proved, as if you say is the law Lol. Be accurate and show me wrong empirically if not, well do another activity
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I will look with detail all what you wrote, bit please understand that the fluid inside the containers will be calm, the pipe will move also in a moment at constant velocity, and as the containers gets rid of fluid, ok, the pipe will not push further the empty container to compensate, becuase it travels already at 100 m/s and it cannot give more speed with the same speed of 100 m/s, because it ends are pulled also at 100 m/s and all the pipe structure in a moment will have 100 m/s, but the container will not accelerate, because the pipes cannot acellerate over 100 m/S. I dont know if I am clear at this?
The one thing that is clear in what you just wrote is that you are completely wrong and have no clue what you are talking about. Imagine if the hose wasn't there, just a 90 degree bend. Pushing the piston up would force water up through the bend and then out the back, accelerating the cart. Attaching a hose changes nothing about these forces. other forces may act on the hose itself, or whatever is on the other end, and dragging the hose increases the mass the cart has to deal with, but, the cart most certainly can accelerate like that.
And about the water flow in a closed system ruled by Pascal, the forces are the same everywhere, no flow exist in Pascal equation. The links that you showed me before dont apply to Pascal principle, they talk about flow, but flow like river, I dont know what flow will do an input piston into the fluid.
The two links that you sent me, talks about hydrodynamics, like constant flow, and dont apply to Pascal laws, that define hydrostatic. An interesant name, that you want to put flow in it ;)
Again, there are flows happening in the system you described, so you need hydrodynamics, not hydrostatics, and your blind application of Pascal is wrong.
By the way, I need advices from experts, not people like me and honestly, very few I saw here. Experts come with proves to discredit somebody, but in this case it did not happen in any away, just moving on with topics.
You appear to be defining an expert as someone who agrees with you. You have been given multiple proofs, starting in my first post where I pointed out that your question was essentially contradictory.
I have lost track of how many ways people have tried to explain to you that the fluid must move to get from one box to the other. So many simple proofs of it:
-the boxes are moving at different velocities, so the fluid must accelerate to get from one to the other
-the fluid starts moving away from the second box, so to get to the second box, it must move towards it at some point.
-etc. (re-read this thread if you need to see some others)
If you can't even accept such simple proofs, then there is no point in discussing anything with you, and there is really no point in going through a detailed proof of basic physical facts such as the equivalence between Newton's laws and conservation of momentum.
Here I got from Physycs stack exchange. Please have it clear about what we are talking.
https://physics.stackexchange.com/questions/434907/why-does-pascals-principle-apply-to-a-hydraulic-jack-but-not-to-stream-lines-in
In addition to what is stated in the answer to that question, the fluid can't be moving for Pascal's law to apply. Please stop ignoring this statement. (And the fluid is moving in your device, stop denying that.)
There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
So you agree that bumping fluid from a car on 100 km/h, if I hold the pipe at earth, the water will go back at 100 km/h on the direction of the car, as the water drops go out of the pipe? if you tell me yes, then sorry, but you have discovered today Pascal laws
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For the transfer of fluids, you only need this equation to move the fluid.
P1=P2
F1/A1 = F2/A2
When you push the piston you Only need to know this. No flow in the equation to have in consideration, so no flow exists in this system, not at least continous flow. Get it clear everybody. And the fluid is stable at a constant speed, so stop arguing about something that is not true.
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A definition of what means water at rest in a confined area under Pascal law. This let clear that then Pascal law applies and a lot of people here will have to reconsider their positions.
https://en.wikiversity.org/wiki/Hydrostatics:_Fluids_at_Rest
A fluid in rest, is a fluid that is not moving, so this applies to my device. The fluid dont goes out anywhere, its contained, so its at rest, as what it means. Bring me if not a link that contradict what I said, and then reconsider your error. Who was right ;)
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There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
You claim in your very first post that fluid transfers from A to B. Yet you also claim "flow does not exist" between them, and in the original diagram, you claim the fluid is stationary.
In other words, you are claiming the volume of fluid inside A and B changes without fluid moving moving into or out of them. This is just utter nonsense. There's no need for further analysis or consideration, you can send your device directly to the museum (https://www.lockhaven.edu/~dsimanek/museum/unwork.htm).
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There is no flow, show me a real link that shows that, because again your two other links, in any video I saw a mention of Pascal by the way. Proves, I show you usefull links to confirm what I say, do the same. Flow does not exist
How about you actually read the links that have already been posted in this thread, here is one you posted:
https://www.myodesie.com/wiki/index/returnEntry/id/2986 here explains on the milimeter how it works. And to the next one that comes again with the equilibrium, just to tell him, that sitting in an airplane is nearly like sitting at home
It says:
In the operation of hydraulic systems, there must be a flow of fluid. The amount of flow will vary from system to system. To understand power hydraulic systems in action, it is necessary to become familiar with the elementary characteristics of fluids in motion.
Emphasis added.
So as everyone has been trying to tell you, you are wrong.
So you agree that bumping fluid from a car on 100 km/h, if I hold the pipe at earth, the water will go back at 100 km/h on the direction of the car, as the water drops go out of the pipe? if you tell me yes, then sorry, but you have discovered today Pascal laws
I honestly have no clue what this setup is supposed to look like. by "bumping" I suppose you mean pumping. I have no clue what "hold the pipe at earth" means, but the answer is that Pascal's law does not apply because you are pumping fluid, and the exit velocity depends on the pump.
For the transfer of fluids, you only need this equation to move the fluid.
P1=P2
F1/A1 = F2/A2
When you push the piston you Only need to know this. No flow in the equation to have in consideration, so no flow exists in this system, not at least continous flow. Get it clear everybody. And the fluid is stable at a constant speed, so stop arguing about something that is not true.
Completely gibberish. You have been given at least half a dozen definitions of Pascal's law. Every single one states that Pascal's law does not apply if there is any motion, flow, or transfer of fluid. It only applies to hydraulic systems that are stationary. You cannot conclude that there is no flow just because you are trying to apply Pascal'slaw in a situation where it does not apply. That amounts to simply assuming the thing you want to prove, which fails because your assumption is wrong.
A definition of what means water at rest in a confined area under Pascal law. This let clear that then Pascal law applies and a lot of people here will have to reconsider their positions.
https://en.wikiversity.org/wiki/Hydrostatics:_Fluids_at_Rest
A fluid in rest, is a fluid that is not moving, so this applies to my device. The fluid dont goes out anywhere, its contained, so its at rest, as what it means. Bring me if not a link that contradict what I said, and then reconsider your error. Who was right ;)
There is no definition of what "at rest" means on that page, it just again says that Pascal's law only applies when the fluid is at rest which it is not true in your device.
No specialist, have been talking with me, because they acuse me of not having maths proves, and they insist on a thing that is not true. Since when they have becomed specialist of Pascal law? come on, lets be honest.
You are new here, so here is a warning: insults are not an accepted form of communication on this site. You claim that no specialists have been communicating with you, despite one poster having listed a bunch of relevant engineering courses they have taken. (And I can assure you from past post history of many of the posters on this thread, quite a few of them are far more of an expert in this field than you are.) On the other hand, you have had to do research just to understand Newton's laws which are taught in any introductory physics class.
The only one insisting on things that are not true here is you. The fact that every single poster in this thread besides you is in agreement should be a clue for you. The real question is where did you decide that you know the what is "true" despite being contradicted repeatedly by the sources you yourself have provided.
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esposcar: I believe you are entirely wrong. I believe that virtually every word you have said is wrong. So does every single other poster who has responded here.
Your responses haven't convinced anyone, they just make us think you continue to be out-of-touch with reality. And nothing anyone else has said has convinced you of anything.
Is there any point in continuing? I don't see any point.
Lets just agree to disagree and all move on.
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The second half of your question can be reworded and has the exact same meaning as the following:
Too many new posters who present ideas like yours react poorly to being told "no" for me to want to spend much time on a device specific answer, at least before having some evidence that you are interested in listening. It would do you more good to work it out yourself. (Hint: at some point to get from A to B, the fluid must have momentum that is at least partially pointed to the right.)
You nailed it.
And just after that, you contradicted yourself, because you have posted a zillion times, wasting your precious time, with someone CLEARLY not interested in listening.
Can this thread please be closed before Meberbs loses more of his time (because he feels compelled to correct Espocar's nonsense) for no good reason at all?
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You nailed it.
And just after that, you contradicted yourself, because you have posted a zillion times, wasting your precious time, with someone CLEARLY not interested in listening.
Can this thread please be closed before Meberbs loses more of his time (because he feels compelled to correct Espocar's nonsense) for no good reason at all?
Good point, I saved time on my initial response, but failed to follow through when my prediction came true. No more posting in this thread for me, unless there is a radical change in Esposcar's behavior. (I reserve the right to make a short response like this though.)
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Sorry, esposcar, but your device and your description of the device does not make any sense. Build it, and you will find that it does not work as you describe.
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You have not understood anything. Its acellerated and then what happens after acceleration, what comes????? constant speed. I find here a big lack of knowledges about basic things. Understand the device and the make the critic, I would suggest you. The clue of the system is the transfer. And maths are on the pics, so you did not red anything.
By the way, what empirical or mathematical argument have your colegge given? no one, just insisting like a mantra one and another time, that it violates a device like that Newton law and is unable to show me why not
You keep using the word "math", & don't think we share a common definition. Your step 1 has a semblance of a free body diagram with static forces displayed at time T=+0, but there is a huge gap to step 2 & 3. Show me your math of how you get to 100 m/s instantaneously? Tell me mathematically what is happening in that bendy pipe of your diagram that allows it move with the system to get to 100m/s. Show me your math for the work done by the piston imparting momentum into the fluid flow out of the cylinder and into the bendy pipe. Please do so in vector form for x/y/z components of the fluid flow. Show me that & I'll retract my math criticisms.
I may have slept through my college history class when they taught me about the Germans bombing Pearl Harbor, but I was pretty awake through my engineering calculus/differential equations, statics, dynamics, fluid dynamics, materials, & thermodynamics that I think I still recognize "math" when I see it. Your steps 1-3 do not contain consequential math.
This is another version of the invention that has nothing to do with fluid transfer. 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.
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The black ball transfers momentum to the combination of (white ball + fluid), your calculations don't seem to be including the resulting momentum in the fluid.