Why the potential energy is unlimited?
The definition of potential energy is not correct and the conservation law of mechanical energy is faultiness for the following reasons.
Potential energy can be converted into kinetic energy, the resistance removed. Otherwise, the kinetic energy can't be changed into potential energy at all.
Mechanical energy is just produced by the movement of objects according to the definition of it.
Force is the cause to change the movement of the objects by the first law of Newton. Therefore force is the real source of mechanical energy.
The force (outside force) is limited. However, gravity is a special force which is unlimited.
As a result, mechanical energy can be divided into two types; one is called kinetic energy (limited) the other potential energy (unlimited).
Potential energy of objects is unlimited because it got from gravity, and it is general knowledge that gravity is eternal.
Gravitational potential energy of objects is created by gravity. If there is no gravity, the potential energy will disappear.
At present, the potential energy is generally agreed that it comes from the movement of objects being taken from low to high.
But people have ignored that the kinetic energy has already been consumed during the period of movement.
The movement of objects takes place depending on the force acted on. Without force acting on the objects maintain static. Even with equal forces from opposite directions act on each other, they can't move as well. Gravity is just in this condition.
The potential energy is always equal to the weight of objects no matter the objects is low or high. Potential energy will be zero only when the gravity disappears.
In the absence of resistance, the potential energy will be released to do work.
The formula of potential energy: U = mgh, U represents work more suitable than energy.
In a word, force is energy itself or a form of expression of energy just like the electric energy, the heat energy and so on.
According to the following formulas: W=F•S, F=ma=1/S(E2-E1), Work or energy is made up of force while the quantity of force is measured by work or energy. The force consumed has turned into work; unconsumed, energy.
We can get potential energy by producing a vacuum without doing work.
Referring to the Torricelli experiment, We put a cylinder with water into a water basin ( set a splints in the upper part of the cylinder, switched by lines), get a inflated balloon with a hollow tail from the bottom of the cylinder, when the balloon floated up to top, release the air by pulling the switch wire of the splints, then pulling the deflated balloon out. In the course of the experiment, W1=G(water)h, W2=G(air)h, W3=G(water)h=P(atmospheric pressure)V . We will get W1 from W2, and W3 from the atmospheric pressure will compensate for W1.
Quote from: Zhang Yalin on November 10, 2008, 01:16:43 AM
Why the potential energy is unlimited?
The definition of potential energy is not correct and the conservation law of mechanical energy is faultiness for the following reasons.
Potential energy can be converted into kinetic energy, the resistance removed. Otherwise, the kinetic energy can't be changed into potential energy at all.
Mechanical energy is just produced by the movement of objects according to the definition of it.
Force is the cause to change the movement of the objects by the first law of Newton. Therefore force is the real source of mechanical energy.
The force (outside force) is limited. However, gravity is a special force which is unlimited.
As a result, mechanical energy can be divided into two types; one is called kinetic energy (limited) the other potential energy (unlimited).
Potential energy of objects is unlimited because it got from gravity, and it is general knowledge that gravity is eternal.
Gravitational potential energy of objects is created by gravity. If there is no gravity, the potential energy will disappear.
At present, the potential energy is generally agreed that it comes from the movement of objects being taken from low to high.
But people have ignored that the kinetic energy has already been consumed during the period of movement.
The movement of objects takes place depending on the force acted on. Without force acting on the objects maintain static. Even with equal forces from opposite directions act on each other, they can't move as well. Gravity is just in this condition.
The potential energy is always equal to the weight of objects no matter the objects is low or high. Potential energy will be zero only when the gravity disappears.
In the absence of resistance, the potential energy will be released to do work.
The formula of potential energy: U = mgh, U represents work more suitable than energy.
In a word, force is energy itself or a form of expression of energy just like the electric energy, the heat energy and so on.
According to the following formulas: W=F•S, F=ma=1/S(E2-E1), Work or energy is made up of force while the quantity of force is measured by work or energy. The force consumed has turned into work; unconsumed, energy.
We can get potential energy by producing a vacuum without doing work.
Referring to the Torricelli experiment, We put a cylinder with water into a water basin ( set a splints in the upper part of the cylinder, switched by lines), get a inflated balloon with a hollow tail from the bottom of the cylinder, when the balloon floated up to top, release the air by pulling the switch wire of the splints, then pulling the deflated balloon out. In the course of the experiment, W1=G(water)h, W2=G(air)h, W3=G(water)h=P(atmospheric pressure)V . We will get W1 from W2, and W3 from the atmospheric pressure will compensate for W1.
...so how can we acquire free electrical energy to run an electric vehicle for an example?
@ Zhang Yalin
can you make a sketch. I don't know exactly what you mean (ballon, cylinder, ...)
Thanks
Alana
Zhang said:
"We put a cylinder with water into a water basin ( set a splints in the upper part of the cylinder, switched by lines), get a inflated balloon with a hollow tail from the bottom of the cylinder, when the balloon floated up to top, release the air by pulling the switch wire of the splints, then pulling the deflated balloon out."
I say:
How does the inflated balloon get to the bottom of the cylinder of water? Is it inflated first, then pulled down there, or perhaps it is inflated while already at the bottom of the cylinder of water?
Because this system, too, is conservative. Buoyancy is just gravity, hiding underwater!! You will get out only what you put in, less losses. The work you put in, inflating the balloon, or pulling the inflated balloon to the bottom of the cylinder, is all the work you will be able to get back out.
Gravity isn't a force, any more than a bank is money. You can do work against gravity, by raising up a weight, or by sinking a buoyant object. Gravity will then happily return this work, minus losses due to friction, etc.
Please copy the address below to see the picture.
http://picasaweb.google.com/lh/photo/oZYYjnAFMHDfVXtAeYfwGw
When your balloon goes up, an equal volume of water must go down.
And with it, goes your "extra" work.
Very nice forest road, and I also like the giant bamboo very much.
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So tell us again, please: Just where is energy supposed to come out of this system?
The energy used to put the balloon into the water is: W0=G(water)h(1m)
The buoyancy of the balloon will do work: W1=G(water)H(2-10m).
Releasing the air out of the balloon only costs us W2=G(air)H.
The vacuum produced by the balloon can absorb the water up, W3=W1=G(water)H,
The energy we get should be: E=W1-W2-W(W0 and other waste).
W2=775W1; W0=W1/10;
W2=P(atmospheric pressure)V(water or air)=100,000(J). η＞80ï¼.....
hmm i think i get it correct me if i am wrong.
there is little pressure at bottom of the basin so because of this it is a lot easier to pump air inside the balloon with no water leaks. you just have to make the cylinder high enough to harness the kinetic energy of rising/falling of balloon to overcome the energy to pump air into the balloon.
Pump air into the balloon after pulling it out of the water.
It’s easy to do this just like opening an umbrella.
if you're going to have the trouble of putting the balloon back inside the water why not have it already in the water?
Why do you need a balloon, splints, and wire at all? Why not just introduce some air into the bottom of the tube, let the bubble rise to the top, and then let the air out with a small valve at the top of the tube.
Isn't this completely dynamically equivalent, and with less losses, than using the balloon to contain the "bubble"?
Now, without repeating those calculations, please tell me how to get energy out of this system. Do you run a turbine on the air ejected from the balloon (or valve), or attach a string to the balloon and let it pull something up as it rises, or what?
(BTW, Yalin, there's something wrong with your calculation...as I tried to tell you before, for the balloon to go up, water must come down---the balloon does work all right: it moves the water...)
Is there anyone understood my meaning? Please try to do my experiment than ask me constantly. My English is rather poor, Who will translate my paper into pure English? I would much obliged to him.
Quote from: Zhang Yalin on November 14, 2008, 02:19:45 AM
Is there anyone understood my meaning? Please try to do my experiment than ask me constantly. My English is rather poor, Who will translate my paper into pure English? I would much obliged to him.
do you have more picture examples? please post them here.
I have no more pictures. If you have any trouble in doing my experiment, I would like to solve it.
solve it :)
Yalin, your "experiment" is incomplete.
Unless you specify how to get work out of the system, or how to measure any suspected "energy gain", we will just have to rely on ALL THE EXPERIMENTS ALREADY DONE that show your operating principle to be incorrect.
What's the matter--why don't you do your experiment yourself and report the results. Surely ballons and buckets of water can't be that hard for you to get.
And what about the idea of not using the balloon at all and putting a valve in the top of the cylinder--you haven't said whether you realize that is equivalent to the balloon situation--but of course without the complexities that lead one to make incorrect kinematic analyses.
If in the experiment a ball is used instead of a balloon,
The energy got by the buoyancy of the ball will be 9 times than that we paid.
but the ball can't be taken out because the air in the ball can't be released in no way.
So the experiment can't continue.
The point is how to take out the ball of the device without doing any work.
Therefor, a special balloon must be used but not a ball.
You cannot deflate the balloon without doing work. The water in the column tries to reach its own level with the water in the pan. It can't do so because the system is airtight. When you introduce the balloon into the column from the bottom, not only are you lifting up an equal volume of water to the top of the pan, thus doing work against gravity, but when the balloon reaches the top of the column the weight of the water in the column will create a "suction" that will resist the attempt to deflate the balloon with "splints" or otherwise. To deflate the balloon you will have to, once again, lift up an equal volume of water to the top of the column to reset the system to its initial state. It doesn't matter how much work you are getting out of the balloon, because you are forgetting about the work it takes to lift all that water, twice, per each cycle.
It’s easy to let the air out because the atmospheric pressure inside and outside of the balloon is the same.
That's simply wrong. The weight of the water in the column "pulls a vacuum" and the water wants to fall until the level inside the tube is the same as the level in the pan. The sealed tube prevents this. As soon as you put the balloon in there, it is the same as putting a bubble, or opening a valve at the top of the tube for a moment. The water level drops by the same volume as the air in the balloon. Now, to get that air out of the balloon, you have to RAISE THE WATER BACK UP. No matter how you do it.
Don't believe me? Get a water glass, a pan of water, and a piece of tubing. Set up the full glass column and pan with water as in your diagram. Now introduce one end of the tubing into the column, leaving the other end open to the air. What happens? How do you get the water back up into the glass?
Perhaps this little video will help with understanding:
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http://www.youtube.com/watch?v=rPEOPWG_gh8
No need to make a hole in the device to release the air.
OK, Yalin, it is now up to you.
Please explain how to release the air from the balloon without raising an equal volume of water back up into the column.
Or, if you simply withdraw the balloon while full, how do you do this without raising up that volume of water.
Or, if you have an opening into the balloon as shown in your first diagram, how do you keep the balloon from expanding from the suction, just as in my video above...and how could your "splints" possibly deflate the balloon without also pulling up an equal volume of water which weighs much more than your balloon...
Do you mean the following ideas:?
1. In the device
W=G(water)H(10m) ; W(negative)= 2G(water)h(1m)
2. The result is just the same by using a ball or a balloon.
W= G(water)H(10m) ; W(negative)= G(water)H(10m)
If so, the first one is correct, the second is wrong.
Perhaps we are encountering a language barrier.
I certainly do not understand how my questions are answered, or even addressed, by your cryptic post, Yalin.
Maybe we need an interpreter...
I ask you again, very plainly:
How do you intend to deflate or withdraw the balloon, which must inevitably raise up an equal volume of water?
How do you overcome the suction of the water column, that I have very clearly demonstrated for you in my video?
Yes, allowing the balloon to ascend in the column "releases" energy which can do work. But it takes exactly this much energy to reset the system to its initial state, and there isn't any left over. Remember, you have to raise that same volume of water back up, that the balloon displaced downward in its rise to the top.
No matter how you deflate or withdraw the balloon, this fact remains.
Maybe my device will not do, but the theory is correct. The way to use the potential energy without doing work will be found in the future.
what if you have the basin sealed as well? wouldn't that eliminate leaks?
No, Yalin, your theory is not correct, sorry. There will not be found any way to use potential energy without doing work, and potential energy isn't unlimited.
And, FreeEnergy, what does sealing the basin accomplish, except make it harder to introduce the balloon? Leaks don't keep this system from working. What keeps it from working is the mass of the water that must be raised back up to reset the system to its initial state (a requirement for any kind of cycle). Sealing the basin won't affect that.
Please copy the following address to see the picture.http://picasaweb.google.com/001zyl/UntitledAlbum
1. Draw water up as a doctor does medicine with a syringe by valve A, meantime valve B open, valve C closed.
2. Let the water in the cylinder out. (valve B closed, valve C open)
3. Press valve A to the cylinder bottom, close valve C, open valve B.
4. Recycle the operation continually.
The experiment suggests E1=W(up)=G(water)h(1m),
E2=W(down)=G(water)H(9m),
The energy we get should be: E=E2-E1-W(other waste).
E2=9E1; W(other waste)=E1/1000
Wrong.
You have to pump the water back up for the cycle to repeat.
No excess energy.
Please watch this video:
http://www.youtube.com/watch?v=EFO7OvcvUl0
Your machine is functionally identical to many other buoyancy drive ideas. You are lifting water with a plunger/pump, others are lifting it with a displacement volume in a cylinder or tank. Regardless, the work output of these systems comes from FALLING WATER which is the ultimate source of buoyancy, and the work required to lift the water back up to reset the system is ALWAYS EQUAL to the work output from the falling water. And there are always losses. So the systems cannot work, and for the exact same reason: They are GRAVITY ENGINES and gravity is conservative. You get out, what you put in, minus losses.
I don’t mean what you do.
In your opining: 1(gained)=1(paid)
In mine: 9(gained)=1(paid)
Unfortunately, in this case, I am right and you are wrong.
Your device is just a pump. If your device was overunity, all water pumps would be. But they aren't, are they.
If you don't believe me, then build your device and PROVE ME WRONG.
Or watch my videos on Buoyancy, over and over, until you get that "Eureka!" moment.
?
http://www.metacafe.com/watch/yt-gxnHJoyrQpM/energy_from_water_floatation/
That's pretty hilarious, not to mention annoying.
Read the comments below the video. Most of them are from at least a year ago.
Many of the commenters are saying the same things that I and others have said in this thread and other buoyancy drive threads here.
It won't work!! The reasons are clearly given.
If it did work, a year ago, don't you think we would all be using this system by now? Or is it another case of government suppression----yeah, sure. That must be it.