esposcar

You can also try it by yourself and show me wrong!!!!

I will send you the link of the video, when I will do the test, and then lets see who knocks the wall

It can be tested at earth, to see how works and I will have a direct prove. If works as you say, then excuses but You are giving opinions which is a way of calling it analyze in a subjective way. But in this case, we are just talking without any empirical prove from my side and from your side. But in the second device maths where done, and the way of proving me wrong is to turn out my results, or just indicate me what I have done wrong in the equations.Obviously if there is a stronger force that pulls more from one side than the other, well we can call it propulsion

What moves the fluid is not flow, its pression. If you have any doubt, take a look to the equations. In a moving fluid, there is pression of course but its a dynamic pression (for your own knowledge, Bernoulli applies to this kind of pression, not static pression) and in this case its an static pression. If flow was the responsable of moving the fluid, from static (piston inactive) to dynamic (piston active), then it would be not instantaneous, the flow would arrive some time later. Very strange to move a fluid like that. But its pression what moves the pistons not flow. Inside a pascal system, its an isolated system itself, and pression will take care that any force that touches the fluid, the force effect will be expanded to everywhere. So as you could see, you cannot even fit the variable of pression properly from a Newton equations, you have to know that variable through another equation of another system. So I insist, meanwhile there is pression in the fluid it will obey Pascal laws, outside, the hose etc, ok it will be moving but the fluid itself, will obey Pascal laws, expanding in every direction the force that receive through the hoses or the moving container, but inside the fluid, no flow, because I insist its static. An excavator runs a lot and still can move its devices perfectly, there is no strange flows inside the static fluids. STATIC MEANS THAT THE FLUID DONT ESCAPE, IF THERE ARE 2 PISTONS THE FLUID INBETWEEN IS STATIC. AGAIN YOU MOVE THE FLUID BY PRESSION NOT BY FLOW. Another important point to show you that its an islolated system. Pressure as force have vector direction everywhere, in Newton, that type of force does not exist, all the forces in Newton world have a direction. Well, let me explain you a little. All this friendly and furious debate have begun with a first device that consists basically (there are some pics) in fluid inside a moving container that has a hose conected to another stationary container, and would transfer its fluid mass at constant velocity. So the debate was, that I said that the fluid as it arrives to the stationary container it would arrive calm and would not transport its momentum when a part of the fluid was at constant velocity. And the people here says that once the transfer of fluid begins, the stationary container would begin to move if was having wheels. And I say that it can be proven easly, like having a water bumper placed in a pick up for example and you hold a long hose, so the pick up begins to move, and the water as it go out, will have the same rate it wont change, the water as it exits wont go 40 m/s ahead or back if the truck moves at 40 m/s, and people here says that the water will jump back or forward but with no proves at all. So I could not prove it mathematically but neither anybody could here. So I jumped to a second device, that is the last pic that I posted, and I showed it mathematically, and now we have this debate about what static and dynamic fluid means or that flow moves the piston or the pistons create flow. You talked about conservation of energy, of course it has to fit. In a pascal hydraulic system, the input piston will move relative to the output piston, this is what conservation of energy applies in a Pascal system, but my point is to profit about how a force that dont exist in Newtons world, and can multiply the force applied to it in a way that Newton dont contemplates it in his equations. So if you have two input pistons, one with the double of area, this one will move half of the distance than the thinner one, with the same force applied to both. So if one input piston do more work relative to its peer, you can profit of it, to create an unbalance of forces. Of course it moves, totally logic and even might be have some friction, but understand this, it dont moves as a river it moves as a block, as a solid and adapts its shape as a fluid to all the sizes, but moves as a solid.Pressure is everywhere and moves ther fluid like that. Thats the reason why the output piston feels inmediatley the pression if you move the input system, that can be 10 km away, and with flow, its not like that. This is a very very basic concept, its all around the net, I dont need to show you the links. Just surf, and if you find something that contradictes what I say show it, but please have clear how static fluids work.

I will tell you what condition does not apply to Bernoulli equations. You said that if you push the piston there is flow. You know that Bernoulli belongs to the field of Hydrodynamic right? Let me let you clear for which conditions the formulas of Bernoulli are applied: FLOW ALONG A STREAMLINE (Maybe we should search a new definition for streamline?) - And the statement its not me who say it. Its just valid for Flow along a streamline and there is 0 streamline in a pascal system unless you want to reinvent static and dynamic in fluid. It dont work and its not me who say it. http://www.brown.edu/Departments/Engineering/Courses/en3-Advanced/Hydrostatics and Bernoulli Principle Slide Notes.pdf Note that you wanted to create flow in an hydrostatic system, by saying that pushing a piston creates flow because it creates work. Sorry but you are terribly in a mistake. Have you seen the tubes and hoses that surround an excavator. Have you seen the hoses move when the machine is working? where is your flow, because the pistons if they are pushing the flow must move the hose. As you have said with other words.

hahahahahahahaha you think I drawed this picture? This is the link from where the pictures come from https://www.hwhcorp.com/ml57000-012-ch1.html Its the link that Ghideon gave me. So this represents that you are not able to give any opinion in this matter, with all my respects. Why you dont try to expose your theory about pressure in a forum of Hydraulic topics, and see what you will be answered Lol

"Without the math, that’s incorrect.You make a quantitative prediction. It must be supported by analysis." I am ok with it, how can I not be ok with. But for example in the first device of the transfer, nobody could answer yes or no, to the question if the stationary container would move or not, once he receives fluid from a moving container. You can insist if the fluid moves or not, but you can shake a pascal system and this is one, it have a piston in one side and another in the other, and there will be no flow. Pressure is defined as force per unit area. Can pressure be increased in a fluid by pushing directly on the fluid? Yes, but it is much easier if the fluid is enclosed. The heart, for example, increases blood pressure by pushing directly on the blood in an enclosed system (valves closed in a chamber). If you try to push on a fluid in an open system, such as a river, the fluid flows away. An enclosed fluid cannot flow away, and so pressure is more easily increased by an applied force. What happens to a pressure in an enclosed fluid? Since atoms in a fluid are free to move about, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted undiminished. A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. Bernoully applies for not closed systems. If you dont agree show me a link about how Bernoully works in a Pascal system. You said that at low speed could have more chance to work? or I have missunderstood you? in your second reply. I will place you an interesting point of view that begins to have more sense. If you think about it, in Newtons world all the forces have a vector and all the quations are focused in those points, but pression, is also considered a force, but cannot fit in Newtons equations, because it has the vectors everywhere, with no concrete direction. And if you think about what defines an isolated system, since when you have to use equations of Pascal to have forces that belongs to Newton? And why Newton did not have never in consideration pression. Something normal unless you want to build a reactionless device Lol. I am not an engineer, but those are just technical matters, nothing that cannot be arranged with the best configuration. I just wanted to let clear the concept. Well to get the force of the pression, I have done it with an equation that is not Newton mechanics, and I insist that this makes the difference or that is what maths say, if I am wrong, then please highlight the error. Bernoully dont applies to any Pascal System, and we can insist if its hydrostatic or hydrodinamic, but its very clear that its said that Bernoully is not in this system. We can even apply the equation of Bernoully if you want, but its action-reaction, by itself it dont transmits any momentum and in opposite direction of the momentum fluid itself. You cannot know it. By the way, a moving pascal hydraulic system, will stay the same stationary or in movement. Applies just gravity force, no other Newtons force. It dont have vector the pression, its the secret of all. "There is no path to use Newtonian physics to get an answer that disagrees with Newtonian physics, unless the maths are wrong. " You can use Newton physics and even add a force that dont belong to Newtons variables. But the equation from where comes this force is calculated with Pascal equations, so its bizarre, to insist in a sole isolated system, and you have to use an equation that dont belongs to that system to have a complete information of the system itself. This is not a contradiction itself? Maybe we should begin by defining what isolated system means.

These pics are the best example to show the difference between forces applied to fluids and solids. The pics talks by themselves and describes my device perfectly. This last pic makes me think that the first device that opened this topic would work also. No momentum transmitted but no way I can show it mathematically.

For obtaining the force of pression that the input piston will face I use Pascal principle, which is simply a rule of three or a rule of proportions itself. Where we understand that the same pression is in the input piston than its output piston: P = P And pression is: P=F/A So we combine the area and the forces to know the magnitudes in this case of the force of pression that each input piston will face in its displacement. F1 will represent the input piston. F1/Area1 = F2/Area2 So as I knew that a mass of 200 Kg was pushing down the output piston and I knew both pistons area, I could calculate the force facing the input piston. And I used the equation 2 times for each hydraulic system. So because of the input piston area diference, there is going to be a force difference between the 2 systems. Then by using this equation I knew how many displacement each piston was going to do despite they are pushed by the same magnitude of force. Input piston Area * (Input piston displacement) = Output piston Area * (Output piston displacement) And this marvelous equation tells you that if the input piston have more Area it will travel slower than if you use the equation with another system with a smaller input piston Area. So again I used it 2 times. And with the distance traveled you know the work that each piston will do in its system. W = F * D And you obtain 2 different work results for the input piston comparing them together, and the device profits from that difference. The device dont violate any law, any at all.

Each side will do with 20.000 of force some work, but each input will respond to its own output piston about conservation of energy, so that is solved. But its the work diference of each device that will generate more speed in an input piston than in the other. Because the smaller piston have to compensate his output piston fluid filling. A smaller piston will have to move the double of speed to push the same area output than the other one, if not conservation of energy would be violated. I profit from this difference. Its just this what I pretend and its very logic, its so simple that maybe it was to simple to think about. But the equations fit marvelously and are applied good. If I miss a vital equation or a force that will arrage the pistons work, then thats what I search, something logic and empirical. Its very difficult to go against my argument, its very simple.

That you dont know it if works or not, that must give me the answer an Hydraulic engineer. Its just new, nobody new about this before. We are talking about an invention that dont contradicts any law. And needs two equations for the 2 hydraulic system not 1. Read again the maths. Where do you think that the throttle will come from? from the difference between the 2 pascal equations of the 2 systems that will give you different pressure numbers. You dont even know if its compatible or imcompatible if you dont have even the concept clear or how some laws of hydristatic applies. You cannot know it and I am realizing it now

But you know what equations are not good applied or yes? Which equation describes that flow? because its an important force then and if even have momentum, why there is no equation in hydrostatic that describe it? The difference is to small, and that is not important. There are 2 hydraulic devices, that for the pression dont have any any importance. Its even a tube extended horizontally, does not have any importance. The maths are right, I tell you. You still dont get it. That wont influence in the acceleration of the propellers, they will still face pressure ressistance. One more than the other, thats it, and no flow with momentum will travel along the fluid, if not show me the equation that describe that please. And I repeat the question: if one input piston is 6 kilometers away from his output piston, when you press the input piston the output piston feels the pressure at nearly the speed of the light, since when flow travels so fast? Hint: the gravity might create more pressure down, but its perpendicular to the motion of the device.

That is not flow. You have to get a little of base in this matter because you dont have it. I repeat if you put whereever you put holes in the system, the water will go out at the same intensity as the piston avances, no matter where. Speculations that I am beginning to support with maths as the speculation topic rule need, but if its without maths the arguments against, just missunderstanding how certain physics apply, then well, I cannot see a technical or scientific answer to my questions. If you cannot show that I am wrong in my maths, then sorry but, what you trying to defend or critic is not science, not even speculations, its talking just to talk. Just a small remark, if one input piston is 6 kilometers away from his output piston, when you press the input piston the output piston feels the pressure at nearly the speed of the light, since when flow travels so fast? Which equation describes that flow? because its an important force then and if even have momentum, why there is no equation in hydrostatic that describe it?

What is speculation? Its called hydrostatic, and the name says it all. By the way this is a manual of hydraulics, but I know already about it. Show me in which chapter talks about flow or momentum? Have you seen the maths? If the piston moves then the fluid is moving, but not as a flow, as a unit. I think you dont have it clear. If the arrows that you see in the first pic of the manual, is flow, then I think you should take a course about hydrostatic. The arrows represent the force, but its transmitted instantanley, its not the flow like a river. You have missunderstood the pic

Gain momentum? It will work as any hydraulic system of Pascal in earth, in a 90% efficency. If not, you would not see how car garage elevators would work. When the pistons push there is no flow of any nature generated. Show me a link that says the opposite. And you are concentrating in things that are not the issue. You have to view what maths says. You can put holes in all the hydraulic system, and I assure you that as the piston pushes the fluid, the fluid will go out in all the holes with the same flow and speed no matter where you do the hole, in the ceiling or in the bottom, the pression is a force distributed everywhere inside the fluid.

Well, if you have in account that with the same force, the air will go out in the same amount and at the same speed in both directions, so all gets annulated. Action-reaction, this dont need any analysis, its like that. No no, its all real, but you can put 100 litres of water and it will be the equivalent to 100 kilos.

Well, just another idea, but the propellers will use the air inside the device and are more clear to visualize than with magnets. Its just for making it more easy.

The device is operating with gravity and on earth. I forgot to draw it, but the hole device is enclosed, so the propellers will use the air inside the device only. Description of the pic: The device have two sides (1 & 2). It will be composed each side by an static hidraulic system, with an output piston that will have a mass of 200 kg pushing down the output piston. And the input piston system will be composed by a propeller system that will propel back the air inside the device to propulse the input piston further in its displacement and finally the box that is attached to the rod of the input piston will hit the cylinder stopper. * Note that we use the same equation of Pascal to obtain different results. That by itself talks that conservation of energy is applied to each system.Another interesant fact, is that to know the pressure of each side (1,2), we will have to use an equation that is not contemplated in Newtons equations.That talks about the fact that we are talking about different systems that are in the same device. If it was the same enclosed system, we would not need to apply to different equations (Newton - Pascal) or obtain different with the same equation (Pascal equations). - First thing we will do is to find out the pressure forces that applies at each input piston of both sides (1,2) Details: SIDE 1 SIDE 2 F2 = Output piston force = Mass*Gravity = 1.960 N Output piston force = Mass*Gravity = 1.960 N A2 = Output piston Area = 20 m2 Output piston Area = 20 m2 A1 = Input piston Area = 0.2 m2 Input piston Area = 15 m2 P = P P = P F1/A1 = F2/A2 F1/A1 = F2/A2 F1/02 = 1.960/20 F1/0.2 = 1.960/20 F1= 1.960*0.2/20 F1 = 1.960*15/20 F1= 19,6 Newtons F1 = 1.470 Newtons Here begins the first unbalance of forces, because the input piston of Side 2 will have to face an opposition force in pression of 1.470 Newtons meanwhile its peer in side 1 will face a pression force of just 19,6 Newtons due to its smaller area. Lets discover the amount of stationary friction that each input piston system of side 1 & 2 will have and add the pression force, to know the exact minimum amount of force needed by the input pistons to begin to move. - The minimum force with which the Propeller input system will begin to move coincides exactly with the maximum static friction force, whose mathematical expression is: F=Fre(max)=μe⋅N In our case, as the input pistons are on a horizontal plane, and does not move vertically (a = 0): ΣF = m⋅a ⇒ N-P = m⋅0 ⇒ N = P ⇒ N = m⋅g So: F = μe⋅m⋅g Details: SIDE 1 SIDE 2 Propeller input piston system mass = 200 kg Propeller input piston system mass = 200 kg Static coefficient of friction = 0.6 Static coefficient of friction = 0.6 Kinematic coefficient of friction = 0.25 Kinematic coefficient of friction = 0.25 Substituting the values that we know, we obtain that the necessary force in each side (1,2) are exactly the same: F = 0.6*200 kg⋅9.8 m/s2 ⇒ F = 0.6*200 kg⋅9.8 m/s2 ⇒ F = 1.176 N F = 1.176 N But as Newton does not contemplate the pression force, we have to add it to the forces against the pistons, because the pressure vector is pointed perpendicular to the wall or the piston that the pression touches. Having this in consideration and the different pressures that exist between side 1 and 2, the final total forces that will oppose to the propellers input piston system are: F. Stationary = 1.176 N + F1 Pressure = 19,6 N F. Stationary = 1.176 N + F1 Pressure = 1.470 N Total Stationary forces = 1.195,6 N Total Stationary forces = 2.646 N ***Note that you can increase and multiply the pressure forces difference by increasing the input piston area more on side 2 and decrease more the input piston area in side 1. As the force applied by the propellers will be greater than their own static friction forces + pression, the input piston propeller system will be moved, and therefore the frictional force in this state is the kinetic friction force: Propeller Input piston system mass = 200 Kg Kinematic coefficient of friction = 0.25 Frc = μc⋅N ⇒ Frc = μc⋅N ⇒ Frc = μc⋅m⋅g ⇒ Frc = μc⋅m⋅g ⇒ Frc = 0.25⋅200 kg⋅9.8 m / s2 ⇒ Frc = 0.25⋅200 kg⋅9.8 m / s2 ⇒ Frc = 490 N Frc = 490 N So we added to the pression force for each side and we have a difference of: Frc + F1 Pression = 490 N + 19,6 N = 509,6 N Frc + F1 Pression = 490 N + 1.470 N = 1.960 N Once we know the friction & pression forces, we can determine what acceleration the propellers input piston system acquires at each side (1,2). The direction of friction & pression forces are always contrary to the propellers input piston system movement. Applying the fundamental principle or Newton's second law: Each Propeller input piston system will have a pushing force of 20.000 N so with that in mind we calculate the acceleration at each side (1,2). We add the pressure force to the total forces operating the system. Propeller propulsion force = 20.000 N SIDE 1 SIDE 2 ∑F = m⋅a ⇒ ∑F = m⋅a ⇒ F−Frc−F1 Pression = m⋅a ⇒ F−Frc−F1 Pression = m⋅a ⇒ 20.000 N−490 N−19,70 N = 200 Kg ⋅ a ⇒ 20.000 N−490 N−1.470 N = 200 Kg ⋅ a ⇒ a = 97.45 m/s2 a = 90.2 m/s2 So if we translate it to Newtonian forces each propeller input piston system would generate in one second of acceleration: F = M*A F = M*A F= 200*97.45 = 19.490 N F= 200*90.2 = 18.040 N Let me know any mistake or what other equations should apply. Even which force is missing.

I will have it late on tomorrow. Its quite complete mathematically for this case and the only thing it might look complicate is the pic. But the maths will be very clear. In any case I will post it and if you prefer I will answer you after you view it and have more questions.

Ok, I am working with the maths, I will have in account even small details like statical and cinetical friction, I will have in consideration all forces, so there is a more clear idea if it works or not. Then if there is a force I missed, this force should be more easy to visualize and maybe that force makes no rig movement, but up to date, there is no evidence. Lets have a more empirical prove and it will be more easy to make critic. It can perfectly have the same propulsion intensity to the end, but one factor is clear, and this is the point that poeple miss. You got a propulsor with a weight in each side, that can throw back fluid por example to propulse itself, but the rythim will be settled by the areas of the pistons. So I have clear that if weights each piston system 100 kilos and is at 10 metres per second in the displacement and the other due to its area goes at 5 ms of constant acceleration for example, one will hit with more momentum that the other. But this point everybody up to know have not been taken in consideration. In which point have been any change of center of masses. just when one piston moves slower than the other with the same force applied, actually it will hit the wall one side the double of force than the other. How does this rearrangment of mass happened and what the system will do to avoid this reality?

Good morning. 1- The first part when both fluid propulsors will do it the left, and lets say it have F=+X and the other because the force of the propulsor is pointed opposite will have F-X, so both of them added goes as a net force to 0. That mean at that point Newton works excellent, the rig at this point wont move any center of mass, but what makes the device special is after the propulsion, what it faces and which forces are in roll. So what I will do is show mathematically giving mass number to the objects, so I can have an outcome and people here, if they say I missed something, its more easy to show. 2- I think you missed the hole idea. Obviously its a force that will move the rig, and I thought I explained very good from were that extra force will merge. My device stays stationary up to the moment one of the piston components will hit something as the other, with the dfference that one will hit it with more force than the other (because one travels more fast than the other when they begin to push their input pistons. Because you dont seem to understand that one of the propulsors is attached to the piston and with the same force like its peer, one will travel through the piston displacement more quick than the other. That is shown mathematically and who dont want to see the evidence, could present an evidence against that evidence and dont get lost in generalities and literature. I will post later or tomorrow a total mathematical prove since it begins the propulsion up to it hit something. So with the evidence lets see who is able to find the error.

1-totally, if not it do not work. Totally, but have in mind how exist in the same system 3 different emclosed system. This is the clue. 2-Yes some Lol, if you take F by P and you take A by V you got a momentum equation. But if every body got the concepts so clear, why there is no empirical solution just bla bla?. I will post you a very very easy device. Lets say that we put two water propulsors in both sides with the same potence. Will as the piston displacement begins, the propulsor and the piston with smaller area, have more momentum if has more acceleration? And forget about action reaction. The two water propulsors throws the same amount of water at the same rate so no movement there. Think about the advance of the propulsors. I expect this is enough simple to explain you the target.

1 - No doubt, it will always go in the direction of the system that do more work. You have to consider that the rod moving the larger piston area will have also a ver high opposed force as pushes the rod and on the other side, there will be not a lot of resistance by the rod of the smaller piston, so the rod will push the piston and will have more pushing force as its peer the other rod, and its peer will have the force of the other rod (bigger piston) pushing on its side and the rod will oposse less because of its piston smaller area and will push the piston further and with more momentum than its peer. 2 - That would be a problem, because it would move the hole fluid as a solid and would not work. It could be slowed down putting a mass over the output piston, but no need of it, to dessacelerate the input piston, but if the fluid dont move it will act like a Newton mass. Not a good idea. Ideally is the area difference what will make the difference. With that I feel more comfortable, its not necesarry to go to other more complicated scenarios and try to find the error dialectically. This is a very simple solution, show me mathematically and with equation that the final velocity of the rig will be 0 and I will be satisfied. I have showed why, now convince empirically. What would be the final velocity of the rig giving imaginary numbers to the variables to solve the final result. When someting scientific with proves more than bla bla bla, I will be convinced. You convince people in such forums with numers not words. Dont take it personal Lol, its just a general answer...

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

In theory yes but to the left, because the force that the piston will face it will be supported in a percentage by the pression, so the magnet would have 100% Newton laws and the piston a mix of friction and pression, so it would transmitted to the hole fluid in the same area with no vector contained on it., but I would set up symetry to be sure. I cannot show it mathematically your last pic. I know that in theory would work, but cannot show it empirically, only if there are two opposites. The 2 input and output pistons have the same area? No matter what, it would work as I explained above. And always in the direction opposite to the Pascal system.