Ghideon Posted March 21, 2019 Posted March 21, 2019 6 minutes ago, esposcar said: 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 repulsive force as its peer, 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 Ok! So my approach above fails to let you see where the idea is incompatible with Newton. Lets try another angle, two questions: 1: Can I assume you are convinced that the rig, as a whole, follows the Newton's laws of mechanics* ? 2: Do you have experience in applying the formula F=m*a (Force equals mass times acceleration)? *) Just need to make sure you are not trying to find evidence that Newton is wrong, I can't be of any help in that case.
esposcar Posted March 21, 2019 Author Posted March 21, 2019 (edited) 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?. 38 minutes ago, esposcar said: 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. Edited March 21, 2019 by esposcar
Ghideon Posted March 21, 2019 Posted March 21, 2019 (edited) 3 hours ago, esposcar said: 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?. Ok. Let’s combine some basic math and empirical setups to check the idea: We assume the propulsion device works as described and it’s internal parts are configured to make the device, mounted on a rig, accelerate to the right. Note: we do not have to agree yet about how the device is supposed to work internally or how to analyse it, we only assume that it is capable of accelerating to the right for some time*. Since it is a reactionless device the rig, as a complete system, is not applying any force to anything outside of the rig. Internally there are many forces such as fluid pressure and magnetic forces, some of which we may not agree upon yet. But again, from the outside, the propulsion device accelerates using zero force and it is applying zero force to its surroundings during acceleration. This is a central part. The propulsion device is mounted solid inside a cart. The total mass is m. Wheels are ideal; there is no friction. When the propulsion device is activated the cart will move to the right without anything pushing the cart. Also, the cart isn't pushing against anything. 1, First experiment: The device is activated and the rig behaves as expected and moves right. Newton’s 2nd law states that the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. In this case the sum of the forces F is zero so the acceleration according to Newton is zero. Remember, we look at the complete moving rig from the outside where no forces are acting**. Experiment says: F=0, m>0, a>0 which is a contradiction since Newton says F=0 means a=0. We have two options; the device idea is invalid or Newtons laws are wrong. 2, second experiment. Attach the cart to a Dynamometer (Spring Balance). The propulsion device is activated and the cart tries to accelerate to the right. Since the cart is held back by the spring there should be a reading on the scale. But then the only force acting on the rig is the dynamometer pulling to the left, there is no force pointing to the right. But if the dynamometer is reading >0 then according to Newton and F equals m times a the rig must accelerate to the left. It cannot be stationary while affected by only one force. Remember, the propulsion device should accelerate the the rig without any force. Again we have a contradiction; math shows that the idea is not compatible with Newton, or Newtons laws are wrong. I am pretty convinced that Newton is correct***. If you still are unable to spot any issues with the propulsion device, please submit the calculations as seen from the outside when the device is operating. (Or please supply proof that newton is wrong****). *) as we agreed when discussing the four scenarios earlier **) except for the force on the wheels / ground bot that force is not affecting the experiment. But if normal force should be an issue then we can make the experiment in space instead. ***) There's of course also the possibility that my analysis is incorrect; It's tricky to try to use descriptions from the (invalid) device idea together with mainstream concepts and still produce a logic outcome. ****) I assume Newton to be a good enough approximation at this point. At this low velocity (v=0) we do not have to use relativistic corrections. Edited March 21, 2019 by Ghideon minor grammar issue
esposcar Posted March 22, 2019 Author Posted March 22, 2019 (edited) 16 hours ago, Ghideon said: Ok. Let’s combine some basic math and empirical setups to check the idea: We assume the propulsion device works as described and it’s internal parts are configured to make the device, mounted on a rig, accelerate to the right. Note: we do not have to agree yet about how the device is supposed to work internally or how to analyse it, we only assume that it is capable of accelerating to the right for some time*. Since it is a reactionless device the rig, as a complete system, is not applying any force to anything outside of the rig. Internally there are many forces such as fluid pressure and magnetic forces, some of which we may not agree upon yet. But again, from the outside, the propulsion device accelerates using zero force and it is applying zero force to its surroundings during acceleration. This is a central part. The propulsion device is mounted solid inside a cart. The total mass is m. Wheels are ideal; there is no friction. When the propulsion device is activated the cart will move to the right without anything pushing the cart. Also, the cart isn't pushing against anything. 1, First experiment: The device is activated and the rig behaves as expected and moves right. Newton’s 2nd law states that the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. In this case the sum of the forces F is zero so the acceleration according to Newton is zero. Remember, we look at the complete moving rig from the outside where no forces are acting**. Experiment says: F=0, m>0, a>0 which is a contradiction since Newton says F=0 means a=0. We have two options; the device idea is invalid or Newtons laws are wrong. 2, second experiment. Attach the cart to a Dynamometer (Spring Balance). The propulsion device is activated and the cart tries to accelerate to the right. Since the cart is held back by the spring there should be a reading on the scale. But then the only force acting on the rig is the dynamometer pulling to the left, there is no force pointing to the right. But if the dynamometer is reading >0 then according to Newton and F equals m times a the rig must accelerate to the left. It cannot be stationary while affected by only one force. Remember, the propulsion device should accelerate the the rig without any force. Again we have a contradiction; math shows that the idea is not compatible with Newton, or Newtons laws are wrong. I am pretty convinced that Newton is correct***. If you still are unable to spot any issues with the propulsion device, please submit the calculations as seen from the outside when the device is operating. (Or please supply proof that newton is wrong****). *) as we agreed when discussing the four scenarios earlier **) except for the force on the wheels / ground bot that force is not affecting the experiment. But if normal force should be an issue then we can make the experiment in space instead. ***) There's of course also the possibility that my analysis is incorrect; It's tricky to try to use descriptions from the (invalid) device idea together with mainstream concepts and still produce a logic outcome. ****) I assume Newton to be a good enough approximation at this point. At this low velocity (v=0) we do not have to use relativistic corrections. 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. Edited March 22, 2019 by esposcar
swansont Posted March 22, 2019 Posted March 22, 2019 59 minutes ago, esposcar said: since it begins the propulsion up to it hit something. If it stops when it hits the wall (and this means what I think it means), that's not propulsion. That's rearranging internal mass. The CoM will not have moved.
esposcar Posted March 22, 2019 Author Posted March 22, 2019 (edited) 42 minutes ago, swansont said: If it stops when it hits the wall (and this means what I think it means), that's not propulsion. That's rearranging internal mass. The CoM will not have moved. 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? Edited March 22, 2019 by esposcar
Ghideon Posted March 22, 2019 Posted March 22, 2019 8 hours ago, esposcar said: 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. Please do not use numbers, use symbols such as forces F1 and F2 and the math required to describe relations, for instance F1=F2, F1=2 * F2 or similar. When describing where and when the formulas apply, please get the details right: (bold by me) 8 hours ago, esposcar said: 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. That contradicts your earlier explanations of the setup. I specifically asked about the timing of the acceleration: On 3/20/2019 at 7:47 AM, Ghideon said: -When will the device accelerate? The amount of acceleration and is not important yet, I just want to know at what events it accelerates. For instance; Does the device accelerate as soon as the pistons initially affect the fluids? Or does the device start to move when the rods finally hit the walls? The answer: On 3/20/2019 at 3:37 PM, esposcar said: - 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. I followed up to be sure before attempting further analysis: On 3/20/2019 at 5:26 PM, Ghideon said: Ok! Do you then agree on the following description* of the propulsion? Starting position is the setup in the picture below. Push both the pistons with the same force F in opposite directions (Black arrows in the picture). Both the pistons will move, pushing liquid, but at different speeds: Sl and Sr (Speed Left and Speed Right, Direction shown by the same black arrows in the picture). The complete rig will accelerate to the right. Stop pushing the pistons (Force F=0) before rods hit the walls**. The rig will not accelerate anymore but continue to move with constant speed to the right. Your answer: On 3/21/2019 at 3:26 PM, esposcar said: 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 Please supply a consistent description of how you expect the device to work.
Ghideon Posted March 25, 2019 Posted March 25, 2019 (edited) While waiting for an updated model with math to investigate I reread some posts. Here is another source of confusion that needs to be addressed. The one-cylinder analysis I used to highlight issues: (bold by me in the quotes) On 3/21/2019 at 5:44 PM, Ghideon said: Note that it is not necessary to know how the propulsion works at this time, we are just assuming it works as described. Now we can see the left and the right parts of the rig both can cause movement. The rig moves in the direction of the “winning” cylinder. The “loosing” cylinder has a negative impact. Therefore we can simplify it further. If the rigs in the scenarios above works then the rig will move even if one of the cylinders is removed from the rig. Only one cylinder is required, not two. The magnet is pushing against the vertical bar it is mounted on, I choose to show that as force F pointing to the left. Does this simplification of the setup show the issues with this specific propulsion idea? If not, I’ll post some more details in a followup. First an agreement: On 3/21/2019 at 5:51 PM, esposcar said: Totally, its really as it is supposed to work confirmation requested On 3/21/2019 at 5:58 PM, Ghideon said: Ok! So, just to be 100% clear: According to your idea, the one-cylinder-rig in my last picture will accelerate to the right? response: On 3/21/2019 at 6:03 PM, esposcar said: In theory yes but to the left, How can the standalone cylinder move in the opposite direction? Then some hesitation: On 3/21/2019 at 6:26 PM, esposcar said: Maybe it wont work individually, And then rejection of the one-cylinder case: On 3/21/2019 at 6:26 PM, esposcar said: there are a reason for the symetry, its OBLIGATORY and On 3/21/2019 at 6:26 PM, esposcar said: Its 2 hydraulic separated system. 2 not 1. Again; please supply a consistent description of how the device is designed to work. Of course devices like this cannot work. To be able to answer what is wrong with this specific idea one has to have a consistent description of the idea. Edited March 25, 2019 by Ghideon grammar
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 8 hours ago, Ghideon said: While waiting for an updated model with math to investigate I reread some posts. Here is another source of confusion that needs to be addressed. The one-cylinder analysis I used to highlight issues: (bold by me in the quotes) First an agreement: confirmation requested response: How can the standalone cylinder move in the opposite direction? Then some hesitation: And then rejection of the one-cylinder case: and Again; please supply a consistent description of how the device is designed to work. Of course devices like this cannot work. To be able to answer what is wrong with this specific idea one has to have a consistent description of the idea. 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. Edited March 26, 2019 by esposcar
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 15 hours ago, esposcar said: 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. 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. Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 48 minutes ago, esposcar said: I forgot to draw it, but the hole device is enclosed, so the propellers will use the air inside the device only. Why use propellers? How will the analysis of the device differ from when magnets were used?
esposcar Posted March 26, 2019 Author Posted March 26, 2019 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.
Ghideon Posted March 26, 2019 Posted March 26, 2019 3 minutes ago, esposcar said: Its just for making it more easy. Isn't it more obscure? Because in a real experiment one would have to account for the turbulence in the system. That is probably not an easy task to calculate. Next: F2 = Output piston force = Mass*Gravity = 1.960 N Is the fluid in the cylinders massless?
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 3 minutes ago, Ghideon said: Isn't it more obscure? Because in a real experiment one would have to account for the turbulence in the system. That is probably not an easy task to calculate. Next: Is the fluid in the cylinders massless? 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. Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 7 minutes ago, esposcar said: No no, its all real, but you can put 100 litres of water and it will be the equivalent to 100 kilos. Then you will have to account for that: -The level of fluid will rise in the cylinders and since gravity is included this affects the force required. -When force is applied the fluid will have to be accelerated since it is moving through the cylinder. The fluid will gain momentum. Or, explain why these effects can be neglected.
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 33 minutes ago, esposcar said: 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. 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. Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 9 minutes ago, esposcar said: 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. No, this is speculation. You have to show how the system works. When the pistons reach the stoppers are the fluid in the same location as initially, or has the fluid moved? But Ok, here is one: Quote In hydraulic systems, energy in the form of pressurized liquid flow is controlled and transmitted through a piping system, to a hydraulic actuator from: https://www.hwhcorp.com/ml57000-012-ch1.htm
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 6 minutes ago, Ghideon said: No, this is speculation. You have to show how the system works. When the pistons reach the stoppers are the fluid in the same location as initially, or has the fluid moved? But Ok, here is one: from: https://www.hwhcorp.com/ml57000-012-ch1.htm 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 Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 Here is another description of the flow: http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html: Quote A hydraulic lift for automobiles is an example of a force multiplied by hydraulic press, based on Pascal's principle. The fluid in the small cylinder must be moved much further than the distance the car is lifted. For example, if the lift cylinder were 25 cm in diameter and the small cylinder were 1.25 cm in diameter, then the ratio of the areas is 400, so the hydraulic press arrangement gives a multiplication of 400 times the force. To lift a 6000 newton car, you would have to exert only 6000 N/400 = 15 N on the fluid in the small cylinder to lift the car. However, to lift the car 10 cm, you would have to move the oil 400 x 10cm = 40 meters. This is practical by pumping oil into this small cylinder with a small compressor. 6 minutes ago, esposcar said: What is speculation? You have posted in speculations section of this forum, where certain rules apply.
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 23 minutes ago, Ghideon said: Here is another description of the flow: http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html: You have posted in speculations section of this forum, where certain rules apply. 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? Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 (edited) 22 minutes ago, esposcar said: 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. If you drill a hole in the bottom the cylinder will be drained faster than if you drill for instance halfway up on the side. Gravity and density of the fluid have an impact Hint: https://en.wikipedia.org/wiki/Pascal's_law 22 minutes ago, esposcar said: 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. First one have to understand enough physics and what the math is supposed to describe. What if the mathematic formulas and calculations are correct but applied incorrectly? Edited March 26, 2019 by Ghideon Formula not displayed correctly
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 16 minutes ago, Ghideon said: If you drill a hole in the bottom the cylinder will be drained faster than if you drill for instance halfway up on the side. Gravity and density of the fluid have an impact Hint: https://en.wikipedia.org/wiki/Pascal's_law First one have to understand enough physics and what the math is supposed to describe. What if the mathematic formulas and calculations are correct but applied incorrectly? 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. Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 Please respond to the issue regarding the drilled hole; why does it not matter where the hole is drilled? 2 minutes ago, esposcar said: The maths are right, I tell you Then why is the result not compatible with mainstream physics? Please also review the issues I raised earlier. The rig should work with only one cylinder if your idea is working as you describe.
esposcar Posted March 26, 2019 Author Posted March 26, 2019 (edited) 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 Edited March 26, 2019 by esposcar
Ghideon Posted March 26, 2019 Posted March 26, 2019 7 minutes ago, esposcar said: 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. 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 (bold by me) This is a more interesting discussion, it can be analysed without going into details about the internals of the cylinders. If i get the math correctly there is a left and a right side and the amount of acceleration the system will generate is based on the difference of the "work" done on each side. So if one side does zero work then you have the maximum amount of acceleration. Or from the bold text above, of one pascal equation is zero.
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