Jump to content

Reactionless device using the principle of Pascal for fluids


esposcar

Recommended Posts

6 minutes ago, Ghideon said:

(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. 

 

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.

Edited by esposcar
Link to comment
Share on other sites

1 minute ago, esposcar said:

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.

Show the math of this specific part without any unnecessary things. Do not use numbers. Use letters to show the formulas for the two sides. Then we will check what happens when we remove the "loosing" side of the equation. 

Link to comment
Share on other sites

32 minutes ago, Ghideon said:

Show the math of this specific part without any unnecessary things. Do not use numbers. Use letters to show the formulas for the two sides. Then we will check what happens when we remove the "loosing" side of the equation. 

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.

 

15 minutes ago, esposcar said:

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:

    P1 = P2

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.

 

 

Edited by esposcar
Link to comment
Share on other sites

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. fotopresion.jpg.6217f961b7c3536ac3e936e0581e85a5.jpg

 

fotopresion2.jpg

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.

Edited by esposcar
Link to comment
Share on other sites

8 hours ago, esposcar said:

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.

The above is a key issue when analyzing this setup. The complete system is supposed* to accelerate because work done by one part is less than work done by the other part. Acceleration to the right requires: 
[math]W_{right}>0, W_{left}>0[/math]
[math]W_{right} - W_{left} > 0[/math]
Then by reducing the negative impact of the left cylinder so that 
[math]W_{left}=0[/math]
we have
[math]W_{right} - 0 > 0[/math]
[math]W_{left}=0[/math] could be achieved by [math]D=0[/math] so that [math]W = F * D = F * 0 = 0[/math]
Which implies that the system should work using one cylinder if it works with two. Working with the above might help to highlight issues with the basic assumptions. Note the earlier issue where a one cylinder setup was rejected (quoted below for convenience).

On 3/25/2019 at 10:02 PM, Ghideon said:

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.

CgcZuizC_Q6JaXhZMjrZKjhfS73RdxEwnGA2v4aiW42eGbOdUw4IO-OJzFOIE-WBJHBHQ2up3j1dXznSuMagQccIBHuEmury-eOzF4BLzluqUt-DRP0Cz1aBy5lYtZG3g7yA509g

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, according to interpretations of the descriptions, not according to Newton.

Link to comment
Share on other sites

On 3/26/2019 at 8:31 PM, esposcar said:

Let me know any mistake or what other equations should apply. Even which force is missing.

One* issue is missing the mass on top of each cylinder when calculating acceleration on each side. Those masses are accelerated upwards then pistons are moving. 

On 3/26/2019 at 8:31 PM, esposcar said:

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     

I repeat: remove the numbers and use symbols. 

 

*) There are  many more issues ...

Link to comment
Share on other sites

On 3/26/2019 at 8:10 PM, esposcar said:

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. 

Without the math, that’s incorrect.You make a quantitative prediction. It must be supported by analysis 

On 3/26/2019 at 8:10 PM, esposcar said:

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.

You can ignore the momentum if it’s small, which is unlikely to be true for the system you described, where the speeds you used were not small.

On 3/26/2019 at 5:13 PM, esposcar said:

What is speculation?

Your claim that this is a reactionless drive

 

On 3/26/2019 at 6:01 PM, esposcar said:

Which equation describes that flow?

Bernoulli and continuity, for starters

On 3/26/2019 at 6:01 PM, esposcar said:

because its an important force then and if even have momentum, why there is no equation in hydrostatic that describe it?

Static implies no flow. 

On 3/26/2019 at 6:01 PM, esposcar said:

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. .

There is no path to use Newtonian physics to get an answer that disagrees with Newtonian physics, unless the maths are wrong.

On 3/26/2019 at 6:16 PM, 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.

It violates conservation of momentum.

 

Also, and I will point this out yet again, it isn’t propulsion unless this works through a complete cycle, and I haven’t seen an analysis of that. What happens when the pistons go in the opposite direction?

Merely rearranging the mass is not propulsion. A person walking from one side of your device to the other will cause the box to move, but the motion ends when they reach the wall. When they walk back, the system returns to the original configuration.

 

Link to comment
Share on other sites

On 3/27/2019 at 12:10 AM, esposcar said:

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. fotopresion.jpg.6217f961b7c3536ac3e936e0581e85a5.jpg

 

I can't imagine a more fundamental mistake than this picture.

 

The picture on the right does not describe a force applied to any fluid.

 

It amounts to the simplest schoolboy misunderstanding of forces and Newton's Laws.

All the arrows except perhaps one are the wrong way round.

 

Many pages back I asked for a proper free body diagram of the fluid.

I cannot see how you can perform any Mathematics or apply any laws successfully without one.

Link to comment
Share on other sites

"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.

Edited by esposcar
Link to comment
Share on other sites

1 hour ago, esposcar said:

"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. It just dont apply. Dont get it wrong, no matter how the system will move or stop, it will do it outside the fluid, but the pression is the same in all the system, and you wont it or not the fluid itself is transmitted nearly as a block, because the pression is transmitted nearly at the speed of light, dont get wrong flow with pression. 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.

 

12 hours ago, studiot said:

 

I can't imagine a more fundamental mistake than this picture.

 

The picture on the right does not describe a force applied to any fluid.

 

It amounts to the simplest schoolboy misunderstanding of forces and Newton's Laws.

All the arrows except perhaps one are the wrong way round.

 

Many pages back I asked for a proper free body diagram of the fluid.

I cannot see how you can perform any Mathematics or apply any laws successfully without one.

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

Edited by esposcar
Link to comment
Share on other sites

6 hours ago, esposcar said:

if I am wrong, then please highlight the error.

By now several different*  issues highlighted, I have tried to use both logic and some Math. May i suggest that you address those issues before asking for more issues to be pointed out? Note my earlier comments, the device can not work as described and there are issues independent of how the cylinders works internally. 

 

18 hours ago, studiot said:

fundamental mistake

Sorry, my bad, I used a small quote from the page and included a link as reference. My intention was not to suggest that all content or pictures on the page was of good quality. I should probably have used another source in this case or checked the whole page for quality issues before referencing it, I did not intend that other parts of the page should be used. Sorry @studiot for confusing the discussion.

 

Including the obvious one that reactionless device is not possible, regardless of how it is constructed.

Edited by Ghideon
Spelling
Link to comment
Share on other sites

8 hours ago, esposcar said:

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

Reported.

Link to comment
Share on other sites

10 hours ago, esposcar said:

"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.

I did. I said if you change the mass distribution, the container would move, but the CoM would remain stationary. This doesn’t seem to register with you.

10 hours ago, esposcar said:

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.

If a piston doesn’t move, it can’t do work. If it does, fluid must flow.

10 hours ago, esposcar said:

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.

If you understood the physics, you’d know that Bernoulli would say the pressure is equal everywhere in a static system, with no height differences.

 

10 hours ago, esposcar said:

You said that at low speed could have more chance to work? or I have missunderstood you?

You misunderstood 

10 hours ago, esposcar said:

 I am not an engineer,

That seems obvious. You post like you know what you’re talking about, and you don’t.

 

Link to comment
Share on other sites

4 hours ago, swansont said:

I did. I said if you change the mass distribution, the container would move, but the CoM would remain stationary. This doesn’t seem to register with you.

If a piston doesn’t move, it can’t do work. If it does, fluid must flow.

If you understood the physics, you’d know that Bernoulli would say the pressure is equal everywhere in a static system, with no height differences.

 

You misunderstood 

That seems obvious. You post like you know what you’re talking about, and you don’t.

 

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.

Edited by esposcar
Link to comment
Share on other sites

1 hour ago, esposcar said:

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.

When the excavator piston is extended, how do you think the hydraulic oils is moved inside the system? Teleportation? 

How about addressing the issues I've already told you about? 

Edited by Ghideon
grammar
Link to comment
Share on other sites

2 hours ago, esposcar said:

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

“fluids at rest  (hydrostatics) or in motion (fluid dynamics)”

Tell me, are any fluids moving in your system? If not, how do you transfer fluids from one container to another?

2 hours ago, esposcar said:

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.

So the piston moves but the fluid doesn’t? How does that happen? Is the piston porous? 

2 hours ago, esposcar said:

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.

Moving fluid doesn’t require that hoses move. 

Link to comment
Share on other sites

2 hours ago, dimreepr said:

Sometimes pandering creates a monster...

does a reactionless device deserve five pages?

 

Sometimes pandering creates a monster...

does ^ (not) ^ a reactionless device deserve five pages?

 

+1

Link to comment
Share on other sites

On 3/6/2019 at 5:14 AM, esposcar said:

Perfectly clear. But with all respects, if you have not even taken a second about this, how can you quote already what I have done? You just view in the video how it works and detect that if there is a second that the device violates Newton laws or Pascal laws, actually you just need a diagram of forces, not any exhotic quantum equation, I will excuse myself for making you lose your time, despite this time is work ;)

But anyway I am not here to discuss, I just wanted an answer, and the animation is very clear, even a child can view it.

it is an animation, not a model and to "prove" something without making it you either have to have the equations to back it up and let us know how it lines up with popular views on physics/proven laws of physics. I am interested but I would like to see something that really proves it.

Link to comment
Share on other sites

3 hours ago, esposcar said:

Note that you wanted to create flow in an hydrostatic system, by saying  that pushing a piston creates flow because it creates work.

The only way the piston can move is if there is more fluid put into the cylinder.  Does that make sense to you?

Link to comment
Share on other sites

Here is an attempt at showing some basic flow of fluid in hydraulics. Two pistons with different area in a simplified setup, connected via a thinner pipe. Somewhere along the pipe there is a flow gauge. In the right piston there is a pressure gauge*.

IMG_9530.thumb.jpg.88e677a85ebdb74db40ae0c924f93cbe.jpg

There are three scenarios**. The first scenario is stationary. In the second scenario enough force is used so that the mass m is moving vertically at constant speed. The third scenario is again stationary. Any fluid is flowing from left to right cylinder is passing the flow gauge. In scenario 1 and 3 the registered flow will be zero, forces balances the mass m. The gauge will register a flow > 0 in scenario 2.

@esposcaraccording to you there is no flow of fluid in scenario 2, the gauge should register no fluid passing, how is that possible?  
 

 

*) Not yet used, added for later use depending on answers.

**) I try to not include unnecessary details at this point, such details include, but are not limited to: Between 1 and 2 (and also between 2 and 3) there's acceleration of the mass m. Forces F are not equal. Pressures P are not equal. Mass of fluid os not shown. 

 

Edited by Ghideon
misplaced text moved
Link to comment
Share on other sites

5 hours ago, swansont said:

“fluids at rest  (hydrostatics) or in motion (fluid dynamics)”

Tell me, are any fluids moving in your system? If not, how do you transfer fluids from one container to another?

So the piston moves but the fluid doesn’t? How does that happen? Is the piston porous? 

Moving fluid doesn’t require that hoses move. 

 

5 hours ago, swansont said:

“fluids at rest  (hydrostatics) or in motion (fluid dynamics)”

Tell me, are any fluids moving in your system? If not, how do you transfer fluids from one container to another?

So the piston moves but the fluid doesn’t? How does that happen? Is the piston porous? 

Moving fluid doesn’t require that hoses move. 

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.

4 hours ago, peterwlocke said:

it is an animation, not a model and to "prove" something without making it you either have to have the equations to back it up and let us know how it lines up with popular views on physics/proven laws of physics. I am interested but I would like to see something that really proves it.

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.

2 hours ago, Ghideon said:

Here is an attempt at showing some basic flow of fluid in hydraulics. Two pistons with different area in a simplified setup, connected via a thinner pipe. Somewhere along the pipe there is a flow gauge. In the right piston there is a pressure gauge*.

IMG_9530.thumb.jpg.88e677a85ebdb74db40ae0c924f93cbe.jpg

There are three scenarios**. The first scenario is stationary. In the second scenario enough force is used so that the mass m is moving vertically at constant speed. The third scenario is again stationary. Any fluid is flowing from left to right cylinder is passing the flow gauge. In scenario 1 and 3 the registered flow will be zero, forces balances the mass m. The gauge will register a flow > 0 in scenario 2.

@esposcaraccording to you there is no flow of fluid in scenario 2, the gauge should register no fluid passing, how is that possible?  
 

 

*) Not yet used, added for later use depending on answers.

**) I try to not include unnecessary details at this point, such details include, but are not limited to: Between 1 and 2 (and also between 2 and 3) there's acceleration of the mass m. Forces F are not equal. Pressures P are not equal. Mass of fluid os not shown. 

 

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.

Edited by esposcar
Link to comment
Share on other sites

41 minutes ago, esposcar said:

What moves the fluid is not flow, its pression.

Moving fluid is flow.

41 minutes ago, esposcar said:

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.

If you take Bernoulli’s equation and set v=0, you get Pascal’s law.

41 minutes ago, esposcar said:

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.

You insist it’s static and yet it’s not. I think this is part of the problem. You insist on things rather than analyze them.

41 minutes ago, esposcar said:

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.

You can have a dynamic system where fluid doesn’t escape. Your idea of static vs dynamic is just wrong.

You don’t have to worry about flow in what you are calling a “pascal system” because you really only care about the initial and final state, when it is indeed static. But you’re talking about a dynamic system when you claim propulsion.

41 minutes ago, esposcar said:

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.

Which is flow. It has energy, and needs to be accounted for. When fluids move, the pressure drops.

41 minutes ago, esposcar said:

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.

If the fluid is moving it has momentum, so this is nonsense. You are hand-waving instead of applying physics.

 

Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.