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I agree 100% about the motion direction and the logic point of example 4 represent exactly what I mean. Its that exactly. If both input pistons have the same area, Newton wins, there is nothing special. It would be a Pascal system in potence but never in act Lol. Totally, its really as it is supposed to work, forces of repulsion, but please consider that the magnet that encounters more repel oposition, will be the larget input piston, so it would be what Newton requires for propulsion. You have to propulse from something and the input pistons are supported by pression and that only belongs to Pascal principle. Newton dont apply in the support, if not, it would not work.

And how can I deny such statement. Of course the laws have to be respected, its exactly why I want to detect. In this case applied to a physical problem(think about it of a physics teacher exam for the studients. What can merge from so many harvard brilliant brains to find the error? that would be great to see. But there is one thing clear, and thats why I red the forces and variables that the members here are describing, but if you read the literature, it applies to One enclosed system working entirely on Newtons laws. But what makes in my humble opinion this device interesant and I want to put a big enphasis on this statement. This device in general its an enclosed system itself, but their components responds to actually, one general system that operates from Newton laws and 2 separate enclosed systems that responds to Pascal laws. You got a total of 3 systems in one device system. And the only forces of Newton laws that will operate through the input pistons, is the friction, thats why I put a lot of enphasis on that. The opossing resistance that each magnet will face, will be one more resistant than the other and more repulsion against, but the resistance is not provided by Newton laws, its provided by Pression (acts up as down, as right and everywhere with the same force) that is provided by Pascal principle. To be clear, literally a big part of the momentum pushing the pistons will literally get diluted in the pression, the rest goes to Newtons world, the friction force. Exactly respect to their own systems, because the outer input piston of the other system, will do more or less work than its peer. I compare 2 different systems and a Input piston to the other. This I said it before. Again, question. Will one of the input pistons compared to its peer, have more speed to compensate its lack of area and comply with its own conservation of energy system? - It was an example. I have the pics if you want to find the error in the device. Ahhh ok, yes, well it will come back following Newton laws. The pistons get in their original positions, you can even pull the input pistons to their original positions, and you begin again the system. Another very nice option that I just thought about, is, that you return back the pistons pushing the output pistons down. We are talking about resistance provided by pression, and acts evertywhere the same, so I would say, that the device operates as a quasi Newton-Pascal system, to be more accurate.

I am really expecting what you come up with. To be honest I am sure something must be missing, but I want an objective reason. I looking forward for it. I have red your comments below just now, sorry. Ok you are describing a total Newton system, and for that, if a device operates on those rules it will happen what you say. But the original of this proposition is that if we think in Newton laws, it takes the input pistons as for example repel function, and the reality is that the magnet etc that do the force over the input piston, it will encounter more force that repels him than in the other side. Think that one magnet, will find more force against him than the other, and that combined makes a descompensation of forces. One magnet will have more repulsion against him, the one that moves the bigger area piston and the other will have less repel reaction, because does less resistances, so one magnet will push more the other. Try to imagine what I am saying applied to the device, and it have a lot of sense. A reactionless device that is supposed to work just with Newton rules, it will never work, asi most of the reactionless devices that have been invented, with the exception maybe of the EM Drive, but this proposition bring extra force to Newton laws departing for another law. And Pascal law we can see it everywhere. From the excavator working next to your neighbourhoood, to all the factories machines in the world. Sorry I did not understood your last question, could you clarify it? If you think about it, the machines that operates with pascal laws, they use less force to generate more force. Imagine that Pascal principle was not invented, and somebody brings the idea and says I have invented a machine that multiply forces. Everybody would come with, thats not possible, it does not respect Newtons law...

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

Good morning. Well it is getting interesant, because me myself cannot believe that reactionless devices are possible, but I am finding in this devices, something that makes of this case special, is that in this last device, there are maths that support it and works excellent with the laws of energy conservation for example, because if both input pistons, would run at the same speed with the same force, then one of them (smaller area, input piston) would not respect the conservation of energy, because both of the have the same size of output piston, so in order that both output pistons comply with the conservation of energy, oone of the input pistons, if have half of the area of the other, will have to run the double to compensate its lack of area and push at the same rythim its output piston, so it will have to have double speed. You can also brake easly one of the input pistons, by setting a mass in the way of the output piston. There are plenty of options, and the most interesant issue here, is that all of them fits the maths unless I forgot something really important, and thats why I am here. But I liked your enthusiasm and must have a reason of beeing ;). Ok answering your questions: 1- Yes totally 2- Exactly, you dont need to apply more force in one than in another (Obviously it would not work), the conservation of energy of each system will compensate the output piston rate, with more speed fron the input piston with smaller area. 3- Yes exactly, thats the trick of the device and from where obtains the thrust. You have two input pistons with different areas and carrying different momentum, so that makes an unbalance of forces. 4- Exactly 5- Yes there are plenty of options. Or you can keep the force up to the end. If the device works, that wont matter. The important is that the different rodes strike with different speeds (more momentum more speed) the wall, and that proportionally one of them have to do the double of distance in the same time to compensate the more friction of the bigger area input piston. 6- Exactly, we have to make some concessions to Newton Lol. In an enviroment of space it will reload the system and repeat all again, so that would do it accelerate all the times you want. If the system takes 2 minutes to reaload and generate a constant speed in each load, then..... I know the consecuences of what I am saying, but if works, it means this. * Note that it can even work perfectly with 2 extendible hydraulic rodes that push the pushers, even in that case (if that would not work, the device neither) it will work marvelous Lol

- Obviously on the direction that the input piston have more speed, no other way, so in this case to the right. - It accelerate at any moment, it can accelerate for example as you see in the pic, in that moment, that the distance is the double before the rod hits in the smaller area input piston and the rod that belongs to the input piston with more area the rod is settled at half distance for the friction compensation, so meanwhile exist that distance difference, it will work at any point you want to produce an acceleration on both systems using the same force at both sides. - About accelerating meanwhile the rods are contact with the wall, I dont think it would work if they need to have some distance to accelerate before the rod hitting. - The direction will go in the direction of the smaller piston area, that means to the right. -Piston is in constant contact with the fluid and have a pression on it

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

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

You are absolutly right, but it would require the input pistons to have a different dispacement length to compensate the friction and yes, could work also, of course. But I thought that maybe it would be more descriptive. I will post three frame pic without ball and without maths. Maths are already done, and we will see which piston pusher arrives with more speed to its corresponding top, that would do the picture more clear for analisis.

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

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

Not at all, I have placed the magnets, so we forget about any mechanical propulsion. In my older version, the force would not be constant or accelerating and it would have serious doubts to work in my opinion due to friction and area of friction. So I posted a much more clean example, and showing mathematically the basic working of the device. Anyway the first device it still has not been proven right or wrong Lol. But this version yes I could show it mathematically and that gives me some hopes!!!! Ohhh sorry, I have realized now what you said. Yes it applies, the most important is that one pusher with the same foce pushing it, have more speed than the other due to the input piston area differences. Sorry again for the lapsus Lol

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

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

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

Ok I have taked your hint and I have done it more simple and clear. The propulsion system, this time I make it more simple with a constant force produced by an electric magnet of Neodymium to create a repulsion of 60 N against its piston pusher, to push with a constant velocity the input piston. The black balls are eliminated and a lot of useless text. To everybody, just to say that conservation of energy or mechanical enery etc etc dont apply to this system, because its 2 enclosed systems settled and placed in the same common platform. I think this solves propulsion problems and it creates a constant force. It can work also with extendable hydraulical bars to push the piston pushers, but I thought about this idea, because its more easy to analyze for mistakes or missing forces. *IMPORTANT Whiteball 2 will be placed the double of distance from Piston-pusher 2 than Whiteball 1 due to the double size of friction area that Input piston 1 will have compared to input piston 2

The same work is done by the input piston as the output piston. What are you talking about? Do you know conservation of energy or you just post without knowing what we are talking about???? What question have you formulated? Ahhh ok yes, I viewed. I will answer you later

System 1 (right) composed by an input piston of 2 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 3 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is: Input piston W=F*D = 60*3 = 180 Output piston W=F*D = 600*0.3 = 180 System 2 (left) composed by an input piston of 4 square metres of area and an output piston of 20 square metres area. The input piston have advanced with 60N of force applied 1.5 metres up it will stop, because friction and pression have stop it and it has pushed 0.30 metres the output system. So the work done by each piston is: Input piston W=F*D = 60*1.5 = 90 Output piston W=F*D = 300*0.30 = 90 Input piston work of system 1 COMPARED TO Input piston work of system 2 W= 180 W= 90 I compare input with input, not input with output. There is no violation of conservation of energy, because I compare the input pistons work with each other and both belong to another system, placed in the same platform. Glued however you want to imagine it. I just can show you maths. There is no violation of energy in this system, But I compare the work done by one input system with the other. Well you are right, we will consider that the black ball after the acceleration will have a constant speed, and will arrive with P=V*M, its not a potential force once hits something, in the motion it has no force applied but carries momentum, with a vector, so that vector, is a type of force because will push the input cylinder. Technically maybe was not right but the context I expect is more understandable. Answering to your questions. I will answer your other posts, but just dont have the time now. But to your last post ok. I did not do the maths up to how arrives the black ball to its pusher with 60 N, but it can arrive with whatever the magnitude you want, even it can strike the pusher applying a force of 10.000 N, the fact is for the sake of the example a symbolic force when the black ball push the piston pusher. Lets say 60 Newtons before the displacement begins, and lets say that advances 3 metres, but in the advance we settle a whiteball just 1 metres further from the piston pusher that will advance 3 metres also, so if we divide the force by the distance. - At 0 metres it has 60N of force pushing, at 1 metres it will rest 40N (it has gone in friction and pushing against pression) so like a billiard ball, once the pusher encounters the white ball, it will strike it as a billiard ball and stop advancing and the white ball have been propulsed by a force of 40 Newtons and on the other side, due to the piston area size, the piston pusher it will arrive to its corresponding white ball placed one metre from it with a force of 20 N.

https://physics.stackexchange.com/questions/439325/why-pascals-law-is-true-and-what-is-the-mechanism-for-force-amplification-at-mo I found this link usefull because talks about Conservation of Energy. It is not violated, but the smaller piston compared with its pair the piston with bigger area, to elevate their corresponding output pistons to the same level, one input piston will have to be displaced more than the other using the same force. Each input piston to keep the conservation of energy will have to keep its corresponding output system on the same level so one piston will have more acceleration than the other. This is an objective fact and no conservation of energy is violated. And one piston of Area = 2 will do 3 metres to apply the conservation of energy for its corresponding output piston and the other will have Area= 4 and do 1.5 metres to elevate the same output piston area as its pair. So one piston moves the double than the other with the same force, because one have the double area than the other. This is a fact. So if we view it like reality, the work done by one input piston in perpective from the other input piston, will be more. But for their own hydraulic system no law is violated. Remember, we have two hydraulic systems in the platform, not one, and each system have its own conservation of energy, but we profit of their input pistons, that will move at different accelerations to cover different distances in the same time. So the white balls are just in the way of the input displacements, and are placed at the same distance from their corresponding pistons, so when both pistons moves and hit the white balls at the same time, one will carry more momentum than the other. P=mv because the piston structure will have a mass and it will carry a momentum. When each piston will hit its balls one will have more speed than the other. This is a fact, the rest you can imagine it. So if work=force * distance and we have two input pistons of 1/2 area and another with 1 Area with a force applied of 60 Newtons by the two black balls to their corresponding piston pushers that push at the same time the pistons. And we see that to one have done with 60N of force 3 metres and the other 1.5 metres, I understand that if you compare the work of both of the, one have done more work than the other, right? this is 1 + 1=2, here there is no argument, its a fact. "When you say one piston will do more work than another, that's wrong. The different areas means the force is different, but it also means that the distance you need to push is different as well. Energy is conserved, so the work done by each piston is equal if the input is equal." I totally agree with you, the work of the smaller input is the same as the work done by its output input, its conservation of energy for a closed system, but the work I compare it for its other pair, the other input piston of the other system. My device is a platform and 2 systems are placed.

Instead I will post the other version of my device showed mathematically. This time there is a scientific rigor. Equations etc. I think I dont miss anything. Friction was not introduced in the forces, because as you can see it dont need it. 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.

Well, it partially works with Pascal system, you are right but in our case it might (Bulldozers etc etc the prove you have it in how those machines hold up and down weights moving on acceleration, dessacelerations and all kind of movements) rest efficiency but it would still work. And well yes, I would love to find the right equations that describes my draw, but despite this might sound very very idiot, but how about, if an equation that fits this purpose has not been invented or simply does not exist because it dont have sense to solve a different problem that was never placed? I can post a youtube video with the hose and the truck showing how the water will go out, its my only prove or I can ask the fireman of my place Lol. But if works, my idea could work. Anyway I am working in another example with mathematical proves as I will post and this example dont uses fluid transfer but still uses Pascal principle.

Good morning, let me first answer Mr. Swansont, in real life the machines that uses Pascal principles are in constant movement and with gravity applied to them, and they are still very efficient. Inside the containers and the hose or pipe that link them, will be an static fluid. Please note that pressure in an enclosed fluid can be considered uniform throughout a practical system. There may be small differences arising from head pressures at different heights, but these will generally be negligible compared with the system operating pressure. And we talk about a system in earth considering gravity. But we can go to practical examples, for example us, our circular system (blood), works as an enclosed Pascal system, and the prove is that we can be runing, moving and dancing, and if we take the pressure of one point of our body, it will represent the hole body pressure. We can turn around in a rotatory motion, to have the centrifugal force acting on our body like the NASA tests for astronauts, and still, taking the pressure of one point of the body like the nake, they will know the pressure in all the body, meanwhile the astronaut is having a constant acceleration. About the draw, it can actually be taken as a physical problem and even be tested at earth, as I posted before, and the transfer would still happen, and if not, my device would not work, so as a pragmatical physicial problem, we can give numbers to this question, but the heart of the question would be. If the container A with wheels, is going away at a constant velocity from the stationary container B and we do the transfer at a constant speed with no force pushing on it, when the fluid is arriving and filling the stationary container B, will it begin to move in direction to the container B and increase its acceleration as more fluid arrives on it? easy answer, yes or no. If container B does not move then the device works and if it moves gradually as the fluid arrive, then the device does not work at all. So show me mathematically giving numbers to the weights of the containers, and the total fluid and the fluid transfered, and find any equation that fits the purpose, and you will see that not equation can give you and answer unless its tested as I posted, with a pumper of water placed in a truck and on earth a container with wheels and lets see if it moves as the truck in movement makes the transfer bumping the water. You can hold the hose as the truck moves and you can see by yourself that the water will still go out normally as if it was static. I tried to find similar problems, but have not found any example to bring it to the table. Somebody have a solution to the problem? And answering to Mr. Ghideon to his first question, well you have to consider that on the right there is exactly the same thing, a pipe of exactly the same length and container C full of fluid, so the forces are the same, but it can be more efficient, if during the constant velocity the transfer is madebeacuse container A will lose mass and it will slow down more than its opposite container C, so Container C will be able to have much more force than container A before hitting its wall. But if that dont convince you, we can put extendable devices that hold some parts of the pipe and moves at the same speed as container A and B, so there is no drag at all, every thing is as Newton describes at the point. Consider that the transfer will occure when the pipe will be totally extended also, if that helps the bending problem that you have placed. And the bending will not affect the pressure, I mean if there is a lot of bending it will break up, before losing pressure. Its a liquid and there is no way to compress it in any mechanical way. But you can see pipes and hoses all around machines that operates with Pascal laws like bulldozers excavators etc etc, and the pipes or tubes are bended , so from an engineering point of view, this would not would be a problem at all. Even the hole process could be done in 3 seconds or less. Imagine how an Airbag works on that speed, so the fluid could be done in second, and the pipe with the fluid inside can represent also the 10% of the total weight of the moving container A, so that wont be also a big problem. But anyway, I am preparing another example about how to use pascal laws to create throttle with mathematical proves and has nothing to do with this example that I posted. Please not that I have edited my post, because I found mistakes. Sorry for the language problem.

Well, I have making some research about some equations, to try to explain how the transfer would not have momentum with it. And well, does not seem to be nothing about it. This might sound very stupid, but maybe there was never thought about a problem like that. About flow, it would be 0, there is no flow, inside a Pascal Hydraulic system, and that is also why it belongs to static fluids and no Dynamical fluids. And as it is stated for Pascal principle: In any fluid, the static pressure is exerted on the walls of the container and in the fluid. These forces act perpendicular to the walls of the container. When an external pressure is applied to the fluid, the pressure is distributed uniformly in all parts of the fluid and this is known as Pascal’s principle named after the Physicist Blaise Pascal.The Pascal’s principle refers to only the external pressure and within the fluid the pressure at the bottom is greater than the top. According to Pascal’s principle, the force per unit area describes an external pressure which is transmitted through fluid and the formula is written as, This is mainly how the transfer of the fluid will be done with the push of an input piston in a moving container to an stationary container that will have an output piston united by a very long hose or pipe. And the pression will be the same in all the interior of the system. If you think about it, if you have a bottle fulled up the top of water, and you hammer something, it will be as consistent as a real hammer, because the forces are transmitted to the interior of the fluid as pression. In the pic below I show an example. As it is viewed, the transfer of the fluid would work in the same way despite of the velocity diference between the containers. In this other pic, I show a type of rolling container, going at 100 m/s and transfers at it is shown, its fluid to an stationary container. My question would be, if the container A will transmit its momentum to the container B as the fluid arrives? So if its like that, the container B will move in the direction of container A, accelerating? It would be very bizarre. It would be as in a pick up, you place a water dumper, and its connected to a very long hose. The pick up goes away opposite to you and you hold the hose at its end. When the water would go out of the hose, it would just jump back in direction to the pick up, because the momentum was transmitted? I dont think so. What makes momentum is the velocity, without velocity momentum does not exist and the only thing that is stealed in the transfer from the equation is mass. Thats my personal opinion, but I cannot show it mathematically. But there is another version for this invention and I can show mathematically that works. It uses the principle of Pascal of course and profits from this principle in another way. But if there is some help, it would be appreciated as your opinion.

But just between you and me Lol, the concept at least looks interesant ;). I will do it, thank you.

Good morning!!! I have another more clear example of forces, about the fluid transfer, to settle it as a better example to understand the reactionless device example of above. I know it goes against any logic, but if there is no mistake, it works, if it obeys Newton laws and Pascal principle for fluids, then what are the consecuences in the device, as I show? In the equation of momentum, if you take away mass, it will decrease the momentum, as far as I am concerned. Do you have any empirical critic? please dont let me in the indiference Lol. The topic maybe was bad introduced. It does nothing to do with peryscopes and other inventions, that try to unbalance forces inside a closed system that works with Newtons laws, that is impossible, but with this device, I want to find the mistake. Some clues? or maybe I should show you a further example about how the transfer of fluids, wont affect constant speed devices (containers etc) were they are transfered from to stationary devices (containers etc).