At my website http://bsandler.com there is a description with drawings of a machine designed to convert gravity to mechanical energy and heat.
The machine is patent pending.
That sure is a lot of work, in your designs and drawings. I commend your efforts, and your creativity.
Unfortunately you have failed to see a fundamental flaw. It is the same flaw that all other gravity wheel designs have.
You see, gravity is a conservative field. Your design depends upon the conversion of Potential Energy to Kinetic Energy and back again. But the potential energy of a mass depends only on its height in the gravitational field, not on how it got there. No matter how complex the geometry or how clever the mechanism, the potential energy at the top of the travel of the mass is the same. So your design depends on equal weights lifting, or trying to lift, equal weights through a different path than they descend, but to the same height. Thus, you only get out what you put in--that is, the initial energy to create the very first imbalance, is all you are going to get, and that will quickly be dissipated as heat, and the device will stop.
Your design is essentially the same as the "sponge wheel", even though you have a more complex mechanism for moving the water around.
As I said, I commend your efforts and creativity. But it's sort of like going to the drag races, isn't it? I mean, there's a terrible lot of effort just to get to the end of that dead-end quarter-mile. But that's as far as you can go.
In detail what phase of my machine won't work?
For a detailed answer, you could please read http://www.lhup.edu/~dsimanek/museum/themes/bellows.htm (http://www.lhup.edu/~dsimanek/museum/themes/bellows.htm)
Your machine is a more complex version of these:
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%7Edsimanek%2Fmuseum%2Fthemes%2Fchain4.gif&hash=5f17743302b19fa4c1186b00d1df5ed7147f974c)
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%7Edsimanek%2Fmuseum%2Fthemes%2Fstart1.gif&hash=c70da2658bd90c48803f3c9fd2d7f6177e8c38d0)
brian334
I commend you on your diagrams and such. But as TinselKoala has shown that it is not new. Out of all the fluid transfer designs I still think my second wheel design which is a fluid design is the best. And I am not in any hurry to build it either due to the natural problems of this approach.
Here is mine from Aug. 2007
I will ask the question again. What phase of MY invention won’t work?
I didn't went over your full design. But I commented on something similar. The part that will not work is the transfer part. At the bottom for instance the extended weight will try rock the whole thing back. Since it's too far out. Same goes for the top, but the reason is because it's too far in. So the displaced fluid doesn't affect it much.
Quote from: brian334 on October 05, 2008, 12:59:21 PM
I will ask the question again. What phase of MY invention won’t work?
Build it and find out. For it may haunt you until you do build it. It seem like you want to be proven beyond a shadow of doubt so you may not have to build it to finalize your thoughts on the idea. I build my wheels for the very same reason. To finalize if it will work or not. So this should tell you I am talking from experience. Heck! one of my wheels spinning at about 60 RPM nearly took off my thumb ( I was trying to figure out a shifting problem and ZAP!!!) I laughed about it for it was still another step in the correct direction.
All and all the fluid movement speed is sadly a big problem for it has to move up hill.
Broli, if you are talking about gravity machine # 1. The momentum of the extended weight will carry it past the low spot.
brian334
I went back and looked a little closer at you diagrams. (forgive my earlier approach) I tend to put fluid movements and buoyancy types in the same category (my point of view). Fluid movement = resistance due to the speed of the movement. Have you ever paddled a boat? Note how much force it takes to go through the fluid. Every movement encounters friction and in fluid there is allot of friction. Either it be fluid or air transfer it will still take time to do. Again you have a problem so, even if you can get it to work, it will have very little use than just a novelty.
When people are trying to get perpetual motion. They all seem to tend to try for the barely making it types. Re-do your math and try for the overwhelming types. Then if successful you will have something.
Dear Mr. Hammer
When the tanks fall in a continuous column what will the drag be? What gets moved.
Quote from: brian334 on October 05, 2008, 02:43:39 PM
Dear Mr. Hammer
When the tanks fall in a continuous column what will the drag be? What gets moved.
LOL I think you have a good head on your shoulders but lack in experience of reactions. So you reduce the paddle resistance by a straight column. This means you have a wider paddle at the shift, then it goes back to column. Don't you think you are loosing something there as well? Like buoyancy? Displacement? Have you ever hung a tow off the rack and 1/3 of it held the whole tow?
You will just have to build it as see for yourself. I have to get back to work. But I will tell you there are better ways of buoyancy devices and fluid devices to be discovered. ;D Study the reactions.
Quote from: brian334 on October 05, 2008, 12:59:21 PM
I will ask the question again. What phase of MY invention won’t work?
Virtually all of it. I suggest you buy a really good book on hydraulics. It appears to me from your design that you don't understand the fundamental principles of hydraulics and yet you are attempting to design a hydraulic system.
In order for your tank to expand the weight inside must be greater than the weight of the water it displaces. That means that the tank sinks, not floats. But that is only one fatal flaw, there are more.
Hans von Lieven
The falling 64 lb weight has enough force to overcome the external water pressure.
Quote from: brian334 on October 05, 2008, 05:06:15 PM
The falling 64 lb weight has enough force to overcome the external water pressure.
I don't see what you're trying to prove here. If you think these people are wrong then prove them wrong by at least an experiment or something. We are all after the same goal but these pointless discussion will lead to nowhere. Oh and the same goes for patents ;D.
Quote from: brian334 on October 05, 2008, 05:06:15 PM
The falling 64 lb weight has enough force to overcome the external water pressure.
a 64 lb weight cannot displace more than 64 lb of water no matter what the arrangement.
Hans von Lieven
A moving 64 lb weight has a lot of force.
The falling tank only weights 3 lb more than the liquid it displaces.
Quote from: brian334 on October 05, 2008, 05:41:50 PM
The falling tank only weights 3 lb more than the liquid it displaces.
So does the rising tank
Hans von Lieven
The rising tank does not weight 3 lbs more than the liquid it displaces.
what do you think makes it lighter? The maximum that it is capable of displacing in addition is 3 lbs, no more. Which still keeps it at the bottom.
Hans von Lieven
Dear Hans,
I don't think you understand the machine.
No my boy,
You don't understand hydraulics. There is nothing in your machine that an engineer cannot understand. Your underlying physics is wrong. The machine cannot behave like you say in this environment, howevermuch you wish.
This is a clone of a hundred similar designs I have seen, some of them patented, and none of them has ever worked. Do some simple experiments in a fish tank or bathtub and you will see where your errors are.
Hydraulics is a very well understood subject in physics. You really should study it.
Hans von Lieven
Dear Hans,
In detail please explain what phase of my machine won't work?
Greetings Hans
Thanks for stepping in. I don't think brian334 understood my crude blacksmith way of speech, and with your engineering of many years, might get him to understand that he needs to do some test and studies, if he wants to figure this out.
@brian334
Hans is someone who you should never take lightly of what he says.
Dear Hans,
Do you think phase # 1 of my machine will work? If not please explain in detail why not.
http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
Here is a page that deals with your design. It starts with buoyancy and continues to your design halfway down the page under the heading "modifications".
There is no comparison between the machine you cite and my machine.
The machine you cite has to lift the weight twice in each cycle. At the bottom the weight goes up than it falls down pushing the water up. The weight than has to be lifted again.
I am still waiting for someone to explain what phase of my machine won't work?
All I get are reasons other machines won't work.
Dear Mr. Rat2,
Even a fifth grader can understand why the machine you cite won’t work. In each cycle the machine has to lift the water from the bottom to the top. You gain nothing.
Can you explain why MY machine won’t work? What phase won’t work?
It should be easy to explain why MY machine won’t work.
I already told you what is wrong in your thinking. A weight cannot expand a cylinder sufficiently to cause buoyancy. Simple as that.
In order to expand the cylinder the weight has to be at least as heavy as the water it seeks to displace. You cannot displace 10 kg of water with a 5kg weight.
You probably don't know this, 1 liter of water weighs 1 kg at 4 degrees centigrade, a little less at other temperatures, lower or higher.
Hans von Lieven
Quote from: brian334 on October 06, 2008, 03:58:19 PM
Dear Mr. Rat2,
Even a fifth grader can understand why the machine you cite won’t work. In each cycle the machine has to lift the water from the bottom to the top. You gain nothing.
Can you explain why MY machine won’t work? What phase won’t work?
It should be easy to explain why MY machine won’t work.
Sorry Brian, my link was in response to TinselKoala's picture.
Dear Hans,
Not true - you can displace 10 kgs of weight with a 5 kg weight moving at a high rate of speed.
Like I say Hans you don’t understand my machine. That is a true statement.
@brian334,
Here is what I can't wrap my head around. I understand that you expect the cylinders on the weighted (dropping) side to accelerate and increase in kinetic energy due to gravity. The cylinders on the lightened (rising) side must also accelerate and increase in kinetic energy due to gravity.
So each cylinder is moving at a different speed. How could they all stay together as in the diagrams?
And at the bottom each cylinder must stop. Because the kinetic energy in the weight is what makes the cylinder expand, right? So it must begin to accelerate from a velocity of zero again.
I can't see this working as in the diagrams. Maybe a sim or animation would help?
M.
Ok Brian, your machine will work. ;D
But then... no-one will build it because no-one thinks it will work??? So unless you build it yourself, it definitely won't work.
Mondrasek,
At the bottom of the web page there is a diagram with two return lines.
For each set of falling tanks, there could be two or three sets of rising tanks.
It depends an how fast the expanded tanks are allowed to float up.
Mondrasek,
Unrestricted the expanded tanks will float up faster then they unexpanded tanks fall.
Unrestricted the expanded tanks will float up faster then the* unexpanded tanks fall.
Quote from: brian334 on October 06, 2008, 04:34:28 PM
Dear Hans,
Not true - you can displace 10 kgs of weight with a 5 kg weight moving at a high rate of speed.
Like I say Hans you don’t understand my machine. That is a true statement.
If the kinetic energy is high enough, yes, you can, for a moment. As soon as the kinetic energy is exhausted it will bounce back, like a spring. No gain there.
Hans von Lieven
Dear Hans,
The piston has a lock.
Hans, at my website bsandler.com the is a description of a machine.
Maybe you should go read it.
Brian,
I'm trying to follow the progress of one cylinder all the way around the "loop". I'll start at the bottom, where the cylinder has no velocity, but is now boyant since it has expanded and is filled with air. I'll call this cylinder #0. So it starts at zero velocity and begins to accelerate upwards. But the cylinder above it (let's call it #1) has already accelerated to a velocity greater than zero and is continuing to accelerate. So cylinder #0 will never catch cylinder #1. In fact, the gap between them will increase on the way up.
Likewise, the gap between cylinders will increase on the way down when they are heavier than the water they displace, right?
Maybe it is the diagrams with all the cylinders in a row, end to end, that is confusing me.
Shouldn't this work with just one cylinder (without the added efficiency of the reduced drag, due to "drafting" as it were, with the continuous loop of cylinders)?
M.
Brian,
I'm also wondering about how the air will flow down into the expanding cylinders at the bottom. The water pressure at the bottom must be many times greater than the pressure at the top of the loop or top of the water the entire system is in. So the air must have to compress as it goes down the tube and into the expanding cylinder, right?
M.
Mondrasek,
The acceleration of cylinder # 1 is restricted by the load. The acceleration of cylinder # 0 is unrestricted. Cylinder # 0 would quickly catch up to cylinder # 1.
On the way down the cylinders could be hooked together.
The answer to the last part of your question is yes, but it would work better when the tanks fall in a continuous column.
Brian,
What is the "load" on cylinder #1 that is restricting it and not cylinder #0?
If the cylinders on the way down are all hooked together how will they accelerate? They must move at a contstant velocity then?
M.
Mondrase,
Its all about getting the tanks to fall fast. The faster the tanks fall the better this thing will work.
Air does not cause the tank to expand. The air pressure inside the tank would be about 1 atmosphere.
I believe the "atmospheric" pressure in water increases as you go deeper. Every 33 feet (34 feet in freshwater) you go, the atmospheric pressure increases by one times. So at 33 feet, 2 atmospheres. At 66 feet, 3 atmospheres. If you do not increase the air pressure in the cylinders as they go deeper, they could implode. And at the bottom you will have to supply air at pressure equal or greater to the pressure of the surrounding water if it is to enter the cylinders as they expand.
Right now when your cylinders reach the bottom they will have a hefty internal vacuum with regards to the surrounding external water pressure. That will need to be equalized with equally high pressure air before taking on the additional high pressure air to allow for expansion.
M.
The tank needs to be strong enough to withstand the external pressure.
The breather tube to the top supplies air to the expanding tank. The pressure inside the tank is not the same as the pressure outside the tank.
And at the bottom you will have to supply air at pressure equal or greater to the pressure of the surrounding water if it is to enter the cylinders as they expand
NOT TRUE
Right now when your cylinders reach the bottom they will have a hefty internal vacuum with regards to the surrounding external water pressure. That will need to be equalized with equally high pressure air before taking on the additional high pressure air to allow for expansion.
NOT TRUE
Does anyone have anything nice to say about my machine?
Nice drawings and well displayed.
I think I will have draw up a buoyancy device or wheel. Just to see how close I may be able to get, knowing what I know about effects.
Quote from: brian334 on October 06, 2008, 08:09:19 PM
And at the bottom you will have to supply air at pressure equal or greater to the pressure of the surrounding water if it is to enter the cylinders as they expand
NOT TRUE
Right now when your cylinders reach the bottom they will have a hefty internal vacuum with regards to the surrounding external water pressure. That will need to be equalized with equally high pressure air before taking on the additional high pressure air to allow for expansion.
NOT TRUE
Brian, I think your cylinder mechanicals are very ingenious!
If we assume a cylinder that, when collapsed, has equal diameter and length and one square foot volume, then the diameter (and length) would be ~1.0839 feet. The surface area of one of the circular ends is then ~.9226 square feet or 132.86 square inches. If the bottom of the system loop where the cylinder is stopped to expand is at 33 feet deep, then the outside pressure on the tank is two atmospheres, or 2 x 14.7psi = 29.4psi. Distributed across the circular lower surface of the cylinder that is 29.4psi x 132.86 in = 3906 lbs of force pushing up on it from the outside. If it came down with only one atmosphere of pressure inside there would be 14.7psi x 132.86 = 1953 lbs of force pushing down on the same surface from the inside. That's a difference of 3906 - 1953 = 1953 lbs of force pushing up on the bottom surface! Or you can consider it as 1953 lbs of force pulling up from the inside due to the apparent vacuum inside relative to the outside pressure.
Compressed air must be supplied to divers underwater for this reason. Take the simple "Diving Bell" http://en.wikipedia.org/wiki/Diving_bell for example.
If you do not equalize the apparent vacuum created inside as the cylinders fall, and then introduce compressed air (equalized with the outside pressure at that depth) to fill the volume as the cylinder expands, you must be overcoming the 1953 lbs of force resisting the expansion as well as overcoming the drag of drawing in even more lower pressure air. I have trouble seeing how this can happen without your cylinders having reached speeds much greater than I can imagine due to gravity while sinking in water before being suddenly stopped.
Thanks,
M.
Mondrasek,
The distance the tanks fall is about 10 ft.
The diameter of the piston is adjustable , it can be made smaller.
Remember the falling tank only weights 3 lbs more than the liquid it displaces. The tank does not half to expand very much to be lighter than the liquid it displaces.
Mondrasek,
A second breather tube could be added.
@ everybody here
what do you think about this. Looks very promising to me
source: http://gravitator.org/
Hello
First time posting here for it is one heck of a hobby. I like the way brian334 stack but I am not sure how much friction will be caused by the division on the ascending side. As for the you posted AlanA I don't see a chance for the water pressure will hold it out and not shift correctly and balance out, but a nice try.
@AlanA: That's not Brians design. That design is in it's own thread started by Gravitator.
http://www.overunity.com/index.php?topic=5494.msg128709#msg128709
But then you already know that as you posted in that thread. ???
The weight at the bottom needs to be lifted twice. As the weight on top pivots forward it is not turning the wheel.. This machine clearly will not work.
Quote from: brian334 on October 07, 2008, 10:59:13 AM
Mondrasek,
The distance the tanks fall is about 10 ft.
The diameter of the piston is adjustable , it can be made smaller.
Remember the falling tank only weights 3 lbs more than the liquid it displaces. The tank does not half to expand very much to be lighter than the liquid it displaces.
With a 10 ft. difference in height the surrounding pressure would be 4.454psi greater at the bottom of the loop than at the top. So even with only a one inch diameter cylinder, with circular surface area of .7854 square inches, you would have 4.454 x .7854 = 3.5 lbs of force pushing against the cylinder expanding. If two inches in diameter the force against expanding is 14 lbs. At three inches in diameter it is 31.5 lbs pushing against the expansion. What diameter cylinder do you envision being the minimum size to make your devices and still enclose the working mechanisms and weight?
It does not matter how many breather tubes you have. The air pressure in the tubes is only at one atmosphere and will never equalize the pressure surrounding your cylinder at any depth. It will only equalize the pressure if the cylinder is also out of the water (at one atmosphere) or if the air has been likewise pressurized to be equal to the pressure at the depth of the cylinder it is supplying.
Mondrasek,
For simplicity reasons I left out some of the details of the machine. There would be a closed valve at the end of the combination stop and breather tube. When the falling tank crashes into the stop the valve would be forced open allowing air from the surface into the expanding tank.
The air in the tank and breather tube are never exposed to the external water pressure.
As to the size of the piston, I don't know how big the piston would be.
Hopefully big enought to displace more than 3 lbs. of water.
Mondrasek,
Hypothetically speaking, what if the falling tank only weighted 1 gram more than the liquid it displaces. The tank would fall very slow. Forget the momentum of the 64 lb weight.
When the tank got to the bottom the 64 lb weight would still be at the top of the tank.
As the 64 lb. weight falls from the top of the tank to the bottom how much work could it do. Enough to increase the displacement of the tank by more than 1 gram?
Brian,
On a cylinder 4.2733 inches in diameter and at a10 ft depth, the force due to the surrounding water on opposing the expansion of the cylinder on one circular end is 64lbs. Connecting the air in the cylinder that is at one atmosphere to the surface air also at one atmosphere does not change that one bit. The pressure of the water at 10 feet deep is 4.454 psi greater than the air in the cylinder and surface, always. Only a supply of compressed air to the cylinder will equalize and/or eliminate this opposing force.
So to have a cylinder that will expand under just a 64 lb weight as in your design at a depth of 10 feet it would need to have a diameter less than 4.2733 inches. BTW, at 4.2733 inches in diameter, a lead weight of 64 lbs would be 10.87 inches long. So the length of the cylinder must be longer than that at least.
No cylinder of a larger diameter could expand at 10 ft in depth. The deeper you go, the smaller that cylinder must become for the expansion to be possible.
One other thing to consider: For a cylinder to have a volume of 1 cubic foot and a diameter of 4.2733 inches it would need to be 10.02 ft long. I guess it would be at least 10.87 inches longer than that if it had the lead weight inside. Of course all the wall thicknesses, cables, pulleys, and other mechanisms are ignored for this hypothetical case.
M.
Mondrasek,
You lost me on that last post, I don't get any of it.
Don’t forget the falling 64 lb. Weight has momentum.
Mondrasek,
The momentum of the falling 64 lb weight needs to be considered in any estimation of how much the tank will expand.
The momentum of the falling 64 lb. weight is the invention.
Brian,
Lets take a tube of any diameter that is longer than ten feet. Cap one end. Now start pushing the capped end straight down into a body of water. It will resist this and push back up, right? The further down you push, the more force it takes to push it down. How much force? That is a function of the depth of the tube in the water and the diameter of the tube (and density of the water, ie. fresh or ocean).
The force that is pushing up on the submerged capped end is the surface area of the circular capped end times the pressure at that depth. The pressure of the water at 10 feet deep is 4.454 psi greater than one atmosphere. That means if the surface area of our tube end is exactly 1 square inch, it would be pushing our tube up with exactly 4.454 lbs of force. If the tube was bigger in diameter and the end cap was exactly 2 square inches the force pushing up would be exactly 2 times that or 2 x 4.454 = 8.908 lbs of force. So how many square inches of cap will push up with exactly 64 pounds of force? 64 / 4.454 = 14.3691 square inches. And the size tube with a 14.3691 inch cross section is expressed by pi * r2 = 14.3691. Solving for the radius r = 2.13865. Two time the radius is the diameter = 4.2773.
So if you take a capped tube of diameter 4.2773 inches and make it weigh exactly 64 lbs it will sink upright in water of it's own weight to exactly 10 ft. To make it go deeper you will need to add more weight or make it smaller in diameter.
Note in this example that the tube is open to the atmosphere at all times. And with 10 feet of tube in the water it is not even displacing one cubic foot. It needs to be 10.02 feet to displace 1 cubic foot.
M.
PS. Had to try and post a few times because of web errors and noticed your new posts. You said to ignore momentum for the hypothetical case.
Mondrasek,
Buoyancy is based on displacement not on depth. The amount of buoyancy a object has does not have anything to do with water pressure.
The general idea of the invention is that the momentum of the falling 64 lb. Weight would be enough to overcome the external water pressure.
Remember the weight falls about 11 inches inside the tank, also the moveable pulley gives the falling 64 lb. Weight a 2 to 1 mechanical advantage.
Notice in the drawing the weight moves about twice as far as the piston.
Quote from: brian334 on October 07, 2008, 03:38:11 PM
Mondrasek,
Buoyancy is based on displacement not on depth. The amount of buoyancy a object has does not have anything to do with water pressure.
Not True
Buoyancy
From Wikipedia, the free encyclopedia
In physics, buoyancy (BrE IPA: /ˈbÉ"ɪənsi/) is the upward force on an object produced by the surrounding liquid or gas in which it is fully or partially immersed, due to the pressure difference of the fluid between the top and bottom of the object.
Brian, the difference in water pressure at the top and bottom of your cylinders is also equal to the weight of the water they displace.
The buoyancy of a object is the same at a depth of 1 ft. or 100ft.
Or to put it your way, the pressure difference between the top of a submersed object and the bottom is the same at 1 ft or 100 ft.
Correct! The *difference* in the pressure on top and bottom is the same (bouyancy), but both pressures increase as you go deeper, right? So the force on the bottom side increases as you go deeper, right? And you must overcome that force to expand your cylinder, right?
I calculated the cylinder diameter that would have exactly 64 lbs of force pushing against it on the bottom at 10 feet deep. That diameter was 4.454 inches. I also calculated the length of the cylinder of that diameter that would give the 1 cubic ft of volume you wanted to displace. That length ended up being 10.02 ft. But that was rounding error. It would be exactly 10 ft, exactly the depth you prescribed, exactly equal to the weight in the column of water displaced above the cylinder bottom.
If you take a cylinder that displaces 64 lbs of water and sink it with a weight that weighs 64 lbs it will only sink until the top it is at the exact surface of the water. No matter what diameter and length combination this will be the case. And at the bottom, the force there holding the cylinder up will also be exactly 64 lbs.
You want to make the cylinder weigh 65 lbs with an internal weight that is 64 lbs. that displaces 64 lbs. So it will sink. But as soon as the top of the cylinder is below the surface of the water the force on the bottom opposing expansion will be greater than 64 lbs. The further down you go the greater the force opposing the possible expansion.
M.
Hi Brian,
Obviously you believe your invention will work. Who cares what anybody on this forum believes anyway? The only way to really know if it will work is to build it and test it out.
I wish you the best of luck, and I hope it works out for you.
M.
We are not on the same page.
Mr. Rat2,
I believe the invention might work, I do not know if it will work.
Its all on paper.
I posted the invention here to find out what other people think of it.
Brian,
Well I tried to explain as clearly as I could since you had asked repeatedly and politely how your machine could not work. I think my last post was the clearest, but I know from experience that it can take different explanations or views to make things click for everyone. Maybe they will sink in over time or with a second or third reading. If not, well, I tried.
I really do think it is quite ingenious!
FYI, I'm on the road tomorrow and will not be able to type at you unless later in the day. Be glad to have a go on Thursday if you are so inclined.
Later,
M.
Mondrasek,
I understand that as depth increases so does water pressure, the goal of this invention is to overcome the external water pressure and expand the tank.
The momentum of the falling 64 lb. Is used to overcome the external water pressure and expand the tank.
@ Brian,
The momentum of the 64 lb weight is is insufficient to expand the tank. Besides, the momentum is exhausted very quickly yet the outside pressure is still there. As well, in order to prevent water to leak into the tank you must have a very tight seal, that means an enormous amount of friction you have to overcome as well as the outside pressure. No internal mechanism can do this without input of extra energy.
Hans von Lieven
Hans,
How much momentum does the falling 64 lb weight have?
Is there a way to store the momentum of the falling 64 lb weight?
Brian,
How big is your tank, I mean what is the size of your bottom plate?
Hans von Lieven
Hans,
As the tank with the 64 lb weight free-falls the 64 lb weight builds momentum.
At the bottom the tank is abruptly stopped, the 64 lb weight keeps going pulling the cable that goes thru the moveable pulley that moves the piston out of the tank. After the piston is moved out of the tank by the momentum of the falling 64 lb weight it is LOCKED in position by a PISTON LOCK. The tank is than lighter than the liquid it displaces and will float-up doing work.
The size of the tank is posted at my website bsandler.com
At a depth of 10 m the outside water pressure (in all directions) is 1 kg per square cm. That means a bottom plate of 4 inches by 4 inches needs an internal force of 100 kg just to compensate for the pressure. In order to expand the tank in a reasonable time you need about twice that.
So you are looking at an internal force of 200 kg to expand a 4" by 4" tank. There is just no way something like this can float, momentum or not.
To expand a tank of 12" by 12" at that depth you need about 1800 kg of internal force, 1.8 metric tons !
Hans von Lieven
Hans,
The tank falls 10 FEET
LOL at y
It does not matter Brian, the ratios are the same. !0 feet is a little less than one third of 10 meters so you are still looking at about 500 kg, that is half a metric ton.
Hans von Lieven
Hans,
u r embarrassing yourself.
Brian,
As my examples have tried to show: Take a cylinder of any diameter and length combination that has an internal volume of 1 cubic foot (displaces 64 lbs of water). Volume of a cylinder Vc=L x pi x r^2 where L is the length and r is the radius (half the diameter) of the cylinder.
Once you have selected a diameter or length and calculated the other so that you have a cylinder with a displacement volume of one cubic foot, consider it placed in water so that the top surface of the cylinder is at exactly the top surface of the water. If you then calculate the force of the water pressure on the bottom surface of the cylinder, you will find that it is also exactly 64 lbs. Since your weight inside is also 64 lbs, it will not expand.
To gain momentum as you wish you must let the cylinder fall deeper. But the deeper you go, the higher the force on the bottom surface becomes. It starts at 64 lbs and increases very fast. This has also been calculated for you several times.
There is absolutely no way you will gain momentum that will overcome the force of the water that is opposing the cylinder expansion.
It is a wonderful idea and looks great on paper to anyone who does not completely understand that gravity is a conservative field (I have been guilty of that myself quite recently). It was so well thought out that it took me awhile to actually find the physical flaw, even though I knew it could never work from the start.
Please consider all of the posts explaining why this will not work again with an open mind. They are all correct in there own way.
M.
no Brian, you are.
You say: Water weights about 62 lbs/cu. ft.
OK. That means you have a pressure at the bottom of 62 lbs per square foot.
At a depth of two feet you have twice that because you need to take into consideration the foot of water above it. At a depth of three feet you have three times that and so forth. You forget in your calculations the weight of the water column above the tank.
As I said before you need to study hydraulics since you obviously don't understand it. My calculations are correct.
Hans von Lieven
@ mondrasek,
You are also forgetting the mass above the tank which exerts a pressure which has to be overcome.
Hans von Lieven
EDIT: the pressure at the bottom of a 10" tank is about 620 lbs per square foot this being the weight of a 10' column of water with a footprint of 1 square foot
Hans,
Water in water does not weight anything.
Mondrasek,
It is not the cylinder that builds momemtum, it is the internal weight that builds momentum.
I don’t think either one of you understands the invention.
I am not taking about density, though there is some increased density in water at extreme depth, I am talking about the weight of the water column. There is a difference.
Water density increases very little with depth since water is almost incompressible. It does increase in PRESSURE with depth, and THAT is what you have to overcome
Hans von Lieven
Brian,
You plan to build momentum in the weight by having the weight *and* cylinder fall together. When the cylinder is abruptly stopped, the weight should continue to move downward, extending the bottom telescoping cylinder through your clever cable and pulley system. The extended cylinder will then lock in place via a latch. The entire internals of the cylinder will be open to the surface atmosphere during the expansion process through a vent tube that is also coupled at the time the cylinder is stopped.
About right?
The internal pressure of your cylinder will never exceed that of the surface atmosphere on the way down and during the expansion process. That pressure is 14.7psi. However, at evey depth in water the pressure is greater than 14.7 psi, and doubles every ~33 feet.
As I have tried to get you to understand, the bottom surface of any, any, any, cylinder that displaces 64 lbs of water will have an equal and identical 64 lb force pushing up on it, aginst any expansion, just as the top surface of the cylinder is at the surface of the water. It will increase to very significant forces very quickly as it decends. You'll remember Hans mentioning values in tons?
You have only 64 lbs of force, plus any momentum you gain while dropping due to what? A few lbs of bouyancy difference? Have you ever watched how slowly things fall in water?
Some day, if you swim, try taking a small air inflated ball down under water. Watch how quickly it collapses under the pressure. This is the same reason your ears pop.
M.
@ Mondrasek
You said: The internal pressure of your cylinder will never exceed that of the surface atmosphere on the way down and during the expansion process. That pressure is 14.7psi. However, at every depth in water the pressure is greater than 14.7 psi, and doubles every ~33 feet.
This is not correct.
Pressure doubles with doubled distance.
At 10 cm depth the pressure is 10 grammes per centimeter squared, at 20 cm depth it is twice that and so forth.
The pressure is equal to the weight of the water column above. As the weight doubles, so does the pressure.
Hans von Lieven
So is it your position that a falling 64 lb weight can not expand a tank that is submerse 10 ft.? Not even a little bit?
Not even 1 cb. in.
This is getting silly.
Yes it is.
Study hydraulics and then you might understand what you are trying to do. Any book on elementary physics will do.
You could always try the library if you don't want to spend the money.
Hans von Lieven
Quote from: brian334 on October 09, 2008, 07:10:35 PM
This is getting silly.
People do laugh at what they don't understand. I think since you won't take Hans words of experience, and you seem to not understand hydraulics. You need to study the friction that submarines have to deal with to go through water, and then you might understand what Hans is saying.
Hans
I will be building my buoincy device sometime in the future. The designs are done.
Quote from: hansvonlieven on October 09, 2008, 06:48:22 PM
@ Mondrasek
You said: The internal pressure of your cylinder will never exceed that of the surface atmosphere on the way down and during the expansion process. That pressure is 14.7psi. However, at every depth in water the pressure is greater than 14.7 psi, and doubles every ~33 feet.
This is not correct.
Pressure doubles with doubled distance.
At 10 cm depth the pressure is 10 grammes per centimeter squared, at 20 cm depth it is twice that and so forth.
The pressure is equal to the weight of the water column above. As the weight doubles, so does the pressure.
Hans von Lieven
Hans, you are quite right to correct me there. I did not type what I meant to say well at all.
At 33 ft deep the pressure is 2 atmospheres. At 2 x 33 = 66 ft it is 3 atmospheres (not doubled to 4 as I misstated).
M.
Imagine a tank that displaces about 1 cb ft of water submersed to a depth of about 10 ft.
The dimensions of the tank are about 12 in x12in x 12 in.
At the top of the submersed tank is a 62 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 62 lb lead weight can fall 11 inches before it hits the bottom of the tank.
Now imagine a piston with a surface area of 1 square inch is attach to the bottom of the lead weight.
The external water pressure at a depth of 10 ft. is about 4.6 lb/ sq in.
Could 62 lb weight overcome the 4.6 lb/sq. in. external water pressure and push the piston with the 1 sq. in. surface area out of the tank? Yes it could. How many pistons with a surface area of 1 sq. in. could 62 lb. Push out of the tank?
62 lbs. Divide by 4.6 lb/sq in = 13.4
So, as the 62 lb weight falls a distance of 11 ins. It will push 13.4 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 ins.
What is the volume of the 13.4 pistons? 1 sq in x 11 ins x 13.4 pistons = 147 cubic inches
So, the displacement of the tank would increase by 147 cubic inches.
Water weights .036 lbs/cu. In.
So how much less would the submersed tank weight after the 13.4 pistons were pushed out of it when the 62 lb weight in the tank falls 11 in ?
147 cu. In. x .036 lbs/cu in = 5.3 lbs.
Now add to that the momentum of the 64 lb weight falling 10 ft.
brian334
It is time for you to build the design. It is the only way you will understand. If you are trying to interest other people to build it for you, good luck. I don't want to build it, for I see to many problems and don't want to work on something unless it passes a minimum of good possibilities.
@AB Hammer
QuotePeople do laugh at what they don't understand. I think since you won't take Hans words of experience, and you seem to not understand hydraulics. You need to study the friction that submarines have to deal with to go through water, and then you might understand what Hans is saying.
I don't think so, you have assumed the water in the larger tank is immobile like in your submarine analogy when it cannot be, the smaller tanks are moving thus the water is circulating at the same velocity with the tanks once in motion. You know---things in motion tend to stay in motion---momentum, there is not nearly as much friction as you have assumed.
@Brian334
This system is very interesting and as you say rather funny---nobody gets this in any way,shape or form. ;D I have a little saying---all forces being equal inertia can sway the balance, I wonder if this applies here.LOL
@ allcanadian,
you are correct by stating that the water in the tank will rotate with the mechanism. This does not free you of friction though as the water is forced to rotate against the outer tank's walls. Whichever way you turn it you are still losing energy because of it.
Hans von Lieven
Quote from: allcanadian on October 10, 2008, 03:47:18 PM
@AB HammerI don't think so, you have assumed the water in the larger tank is immobile like in your submarine analogy when it cannot be, the smaller tanks are moving thus the water is circulating at the same velocity with the tanks once in motion. You know---things in motion tend to stay in motion---momentum, there is not nearly as much friction as you have assumed.
Have you ever tested a boat motor in a 55gal. drum of water?
Now think vibration into the calculations. It can get worse. LOL
Mr. Hammer,
That is da-um.
Quote from: brian334 on October 10, 2008, 09:43:51 PM
Mr. Hammer,
That is da-um.
Brian
You haven't ever worked on a boat motor have you? Most boat motors are water cooled so you have to test it in water. Most people don't have there shop near water so you have to improvise.
This is what I always will laugh about. The lack of real world experience. The hands on versus the arm chair philosophers. All the book learning in the world, will not give you the whole story. So it is time to get your hands into it and learn.
And so you will under stand what I am saying, in other terms. A closed system of fluid with movement will cause vibration, and vibration will work against you as well. I can tell you how to correct it, but I am not going to, until you have tried to build it. It is now time to get to work. Good Luck
PS I use to be in the US Coast Guard (river navigation) and one of my best friends was an engineer with the US Corp of Engineers. And my fondest memory is when I boarded the Calypso, Capt Jacques Cousteau (He is sorely missed) original ship and shook his hand and talked to the crew as well. ( I had a chance to take a year a tour on it after the Coast Guard but I had to learn french) So you might say I know a little about water movement by real world experience. And some things I have seen on the Mississippi river would scare the hell out of you. For instance a harbor tug sucked down into the river and the tug had no leaks or anything wrong with it. Just pure water movement.
Nice one Alan.
D
Quote from: AB Hammer on October 10, 2008, 10:23:42 PM
Brian
And some things I have seen on the Mississippi river would scare the hell out of you. For instance a harbor tug sucked down into the river and the tug had no leaks or anything wrong with it. Just pure water movement.
People in Sydney are forbidden from swimming too close to the rear of ferries in case they drowned due to the air in the water in the form of bubbles. Plus, I remember on TV they tried to explain the sinking ships in the Bermuba Triangle due to the same effect? (bubbles/gasses) from the deep?
I think I remember watching it also on the "MythBusters" a while ago ???
good luck!
I posted this machine 5 days ago and I am still waiting for a rational explanation why the machine won’t work.
Greetings sevich
Not to many people pick up on my hints so fast. Very good
I was hoping to get Brian working before I told him this. Bubbles are the secret for a buoyancy device. But it is still tricky to make it work on its own. That is what my design is and it has the extra to do it. It is not expected to be as good as some of my other designs but it should still work. Brian's machine is over complicated.
I will post my device later on today, I have to get to work.
Brian,
Sorry for the delay. My wife and daughter both have a fever and I'm not able to get to the internet as often as before.
I like your response to the explanation that your oringinal design would not be able to extend the bottom of the cyclinder due to the pressure of the water. You reduced the surface area of the cylinder you wish to extend so that it did not exceed the force opposing it, right? I did not see anything wrong with that! Good job. Keep in mind that you must be coupled to the vent tube during the expansion process so you do not increase the opposing force by decreasing the pressure inside your falling object.
But your original design required both sides of your cylinder (or cube in your new hypothetic re-design) to be able to expand. So you must have expandable structures on both side to be able to push out, right?
With the weight in the middle of the two expandable structures you will not have 11 inches of stroke for the weight to fall through I think.
At the end of the fall you will need to abruptly stop the falling cylinder/cube/whatever, and then shift it to the side. What mechanism is doing that? Where does the energy for this/those actuators come from?
How are you coupling to a vent tube (internally pressurized below the surrounding water pressure)? How much force with that take? Where is that energy coming from?
M.
The tank flips at the top, putting the weight back on top.
Quote from: brian334 on October 11, 2008, 02:04:30 PM
The tank flips at the top, putting the weight back on top.
I am aware of that. But it addresses none of the questions in my last post.
#1:
If you need geometric shapes to extend on both ends of your cylinder/cube/whatever device, those geometric shapes (cylinders, cube,whatever) must also be able to colapse into the body of the device, right? That is the premise of the device displacing less volume on the way down, right? Where do these extensions and the weight all fit together when everything in not extended?
M.
PS. The lead weight need not be more than .76 inches thick. At 1 inch thick it weighs 82 lbs. So you have an extra .24 inches to work with inside your device.
The pistons extend out of the tank.
The impact between the tank and the breather tube will open the valve, a return spring will close the valve.
The transfer solenoid is powered by the machines generator, or a mechanical transfer device.
Quote from: brian334 on October 11, 2008, 02:50:44 PM
The pistons extend out of the tank.
The impact between the tank and the breather tube will open the valve, a return spring will close the valve.
The transfer solenoid is powered by the machines generator, or a mechanical transfer device.
When the pistons on both ends are not extended, they are inside the "tank", right? Please try and draw this tank device with both sets of pistons and the weight inside.
M.
M,
There is a drawing of the tank posted at my website.bsandler.com (http://bsandler.com)
Quote from: brian334 on October 11, 2008, 03:16:16 PM
M,
There is a drawing of the tank posted at my website.bsandler.com (http://bsandler.com)
Not True.
That picture shows a 'tank" that will not work. Please show us the re-designed tank.
M.
Quote from: brian334 on October 11, 2008, 09:09:33 AM
I posted this machine 5 days ago and I am still waiting for a rational explanation why the machine won’t work.
Rational explanations have been given since day one.
The first reply was the one you may hopefully, someday, understand as truth. TinselKoala was, and is, 100% correct it what he wrote there.
M.
I am new here, and I have rarely seen as hard of a hard head as brian334
Brian you have been given answers over and over and you seem to ignore them all ::). AB Hammer has given up one of his secrets to make buoyancy devices work, and you ignore that as well :o I am looking forward to his posting of the design. To see what it is, and how it works. I prefer open minded people to talk to.
M.
What re-designed tank? I did not change the tank.
Grayone,
I don’t ignore the answers, the problem is I don’t understand them.
Quote from: brian334 on October 11, 2008, 04:10:11 PM
Grayone,
I don’t ignore the answers, the problem is I don’t understand them.
Brian
Most of us who build these devises have thought we had it. And allot of us get the idea in our heads and think this is it. It is harder to admit a mistake, so we tend to try to over complicate what is being said and it comes out in a form of confusion. When we build these devices we see first hand the effects of the whys and why nots, which will allow us to do better designs. Real world experience. The biggest lesson is use the KISS method for every step we take in this game is extra friction.
I will ask the question agian, what phase of my machine won't work?
Brian,
The re-designed tank is the one where you are not expanding the entire bottom. The one from the only example you have shown calculations for here. The one where you show you can expand only 13.4 square inches. If you try and draw that tank you will see there is no way to fit those expandable members in both ends as well as your weight inside your 12 x 12 x 12 inch cube and have the 11 inches of stroke for expanding that you used in your calculation.
I have only been showing the flaws in phase one! Those flaws clearly show that your original tank design would not work at all. It would never expand as you say it will.
If you are banking on momentum please consider what the terminal velocity in water will be for a 12 x 12 x 12 inch cube that only weighs one pound more than the water it is displacing. It will drift down ever so slowly due to the drag of the water on all sides. THAT is an experiment you can try without spending too much time or money! Don't even worry about getting the dimensions exact. Take a gallon milk jug, fill it with water and one extra pound of any weight you can find. Rocks or sand will do fine. Then drop it in water somewhere. Tie a string to it so you can see what speed it reaches while falling. If you can attach a scale to the string and make the string 10 feet long you can see what force you get from the momentum.
M.
M.
My advice is for you to go back to my website http://bsandler.com and re-read the entire post start to finish.
The points you make are addressed at my website.
Quote from: brian334 on October 12, 2008, 08:11:29 AM
I will ask the question agian, what phase of my machine won't work?
You will never know until you or someone else builds it. So why do you keep asking the same question?
The word of man can not bring forth fruit, but the work from his hands.
Quote from: brian334 on October 11, 2008, 04:10:11 PM
I don’t ignore the answers, the problem is I don’t understand them.
That is indeed the problem. Your website shows a device that clearly will not work as explained. Our attempts to answer your questions about why it will not work are not understood by you. Asking me to reread your incorrect assumptions will not make them become correct. Only if and when you understand any of the posts that explain the flaws in your concept will you see that it will not work as you envision. If you cannot understand the explanations, you will have to resort to experiments. I recommend simple and inexpensive ones. I'd hate to see you waste any money you could use for other projects or necessities. Experiments are a great way to learn and to better understand even concepts that you already grasp, so I highly recommend them if and when you can.
If you truly want to understand what phase will not work, you might want to reread any of the posts in this thread that you dismissed. Each one gives at least one fact that you are not accepting. Maybe in review you will be able to understand at least one that will begin to show you how this will not work.
Please remember that the majority of the people on this forum are also hoping or searching for a way to find cleaner or free energy. We are not attacking you or your device maliciously. We are only trying to help you understand the flaws as you have asked. A lot of people have spent a lot of time and energy trying to help you understand. If your idea was workable you would find much positive advice, offers to help, and even computer and physical sims and models, if not full size builds and replications.
M.
Quote from: brian334 on October 12, 2008, 08:11:29 AM
I will ask the question again, what phase of my machine won't work?
The part that makes excess energy. the energy to expand against water pressure is equal to the energy gained as the float rises.. do the numbers its obvious.
to expand a one square foot area to one cubic foot at a depth of ten feet, at 5 psi (approx.) takes
5pounds X 144 X 1foot= 720 foot pounds
the pressure upwards is the difference between the top and bottom of the float .5 psi.
the energy is.
.5pounds X 144 X 10feet = 720 foot pounds
this is what you have to get around.
Fritz,
The part that makes the excess energy is the momentum of the falling 65 lb. weight.
The position and the momentum of the 65lb weight is the energy adding to the system.
Quote from: brian334 on October 12, 2008, 12:46:41 PM
Fritz,
The part that makes the excess energy is the momentum of the falling 65 lb. weight.
The position and the momentum of the 65lb weight is the energy adding to the system.
and just where dose this energy come from? the only energy i see is that of the float rising and that is all accounted for. if the falling mass generates excess then get rid of the floats and concentrate on that.
more important if you are so right why have you not built a self runner and powered your house?
fritz,
Its important to remember that because of the moveable pulley the falling 65 lb weight has a 2 to 1 mechanical advantage over the piston.
The weight falls about 11 in. the piston moves about 5 in.
Fritz,
The energy comes from gravity.
Fritz,
The surface area of the piston is not written in stone.
The size of the piston should match the falling 65 lb weights ability to push it out.
Quote from: brian334 on October 12, 2008, 02:22:05 PM
fritz,
Its important to remember that because of the moveable pulley the falling 65 lb weight has a 2 to 1 mechanical advantage over the piston.
The weight falls about 11 in. the piston moves about 5 in.
i never gave a thought to the details of the mechanism, just how much energy to make the expansion
for so much as the float goes up. a 65 pound weight falls 11 inches=69.6 foot pounds , thats enough to move a piston 13
square inches 11 inches against 5 psi water. 143 cubic inches of exspansion. call it a column 12 inches square by 12 high. .5pounds X 12 X 10 = 60 foot pounds. which is what you put in with the weight.
how are you going to reset the weight without using all you gained from the float?
Fritz,
The weight in the tank floats to the top. At the top the tank flips putting the weight back on top ready for the next cycle.
Any calculation of how much the tank will expand needs to include the momentum of the falling 64lb weight.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water. At the top and bottom the tanks are pulled along by the draft of the preceding tanks.
Quote from: brian334 on October 13, 2008, 08:58:51 AM
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water. At the top and bottom the tanks are pulled along by the draft of the preceding tanks.
Which, sadly, once again, illustrates that you have very little experience or knowledge about how things really behave. Your first sentence assumes free fall of an unconnected weight. The rest of the statement refers to your device, where the falling tanks are also lifting the rising tanks, hence, very much slower than free falling, hence your momentum calculations are in error.
In addition, even though your tanks are falling in a column, there will be a lot of resistance from the water.
And in another addition, your pulley mechanism, like all pulley mechanisms, trades off force for distance. The quantity you are dealing with is foot-pounds, as several posters have pointed out. Your "mechanical advantage" comes with a price.
If anyone wants some casual amusement, you can compare Mondrasek's and my early dialog from a few months ago, concerning his magnet-assisted gravity wheel, with the current dialog between Mondrasek and brian334.
Quote from: mondrasek on October 06, 2008, 06:24:16 PM
Brian,
I'm trying to follow the progress of one cylinder all the way around the "loop". I'll start at the bottom, where the cylinder has no velocity, but is now boyant since it has expanded and is filled with air. I'll call this cylinder #0. So it starts at zero velocity and begins to accelerate upwards. But the cylinder above it (let's call it #1) has already accelerated to a velocity greater than zero and is continuing to accelerate. So cylinder #0 will never catch cylinder #1. In fact, the gap between them will increase on the way up.
Likewise, the gap between cylinders will increase on the way down when they are heavier than the water they displace, right?
Brian,
This is from my first reply to you. It explains how your cylinders will *not* be in a continuous column, end to end, either while falling or rising. There will always be a gap between them that will actually increase the further they drop or rise. There will be water on all sides creating drag. You will never come close to the 17mi/hr speed you are assuming.
The drag due to water only on the sides (if you could create a continuous column of tanks) is also quite real and large. That is why boats, that are neither sinking or rising, require so much energy to be pushed through the water. It is due to the drag of the water on the hull, very much like the drag your tanks will have on their side surfaces.
Consider a topedo or submarine, both shapes that are designed to be as fast as possible in water. How fast do they go, and how much thrust does it take to achieve those speeds? Your tanks are not similarly shaped to minimize the drag, any you only have one pound of "thrust". Your terminal velocity will be much, much lower than 17 mi/hr.
Have you ever considered how quickly a bullet slows when fired into water? It is again a very good shape to go quickly through a fluid. But the drag due to the water will slow it quite quickly to it's terminal velocity, in only a few feet! This is because the force due to drag is also proportional to the speed at which you are going through the fluid.
M.
Quote from: TinselKoala on October 13, 2008, 01:19:26 PM
If anyone wants some casual amusement, you can compare Mondrasek's and my early dialog from a few months ago, concerning his magnet-assisted gravity wheel, with the current dialog between Mondrasek and brian334.
TK, I totaly understand the frustration I brought to you back then! And I feel I owe you and this community for the help and knowledge you and others gave during that exchange as well as since while trying to understand other neat ideas posted here. Guess that is why I am trying to help Brian out similarly. Thanks again for sticking it out with me back then.
M.
TK
The falling tanks do not lift the rising tanks.
HOW MUCH WOULD THE TANK EXPAND?
Imagine a tank that displaces about 1 cb ft of water submersed to a depth of about 10 ft.
The dimensions of the tank are about 12 in x12in x 12 in.
At the top of the submersed tank is a 64 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 64 lb lead weight can fall 11 inches before it hits the bottom of the tank.
Now imagine a piston with a surface area of 1 square inch is attach to the bottom of the lead weight.
The external water pressure at a depth of 10 ft. is about 4.3 lb/ sq in.
Could 64 lb weight overcome the 4.3 lb/sq. in. external water pressure and push the piston with the 1 sq. in. surface area out of the tank? Yes it could. How many pistons with a surface area of 1 sq. in. could 64 lb. Push out of the tank?
64 lb. divide by 4.3 lb/sq in = 14.9
So, as the 64 lb weight falls a distance of 11 in it will push 14.9 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 in.
What is the volume of the 14.9 pistons? 1 sq in x 11 in x 14.9 pistons = 163 cubic inches
So, the displacement of the tank would increase by 163 cubic inches.
Water weights .036 lb/cu. in.
So how much less would the submersed tank weight after the 14.9 pistons were pushed out of it when the 64 lb weight in the tank falls 11 in ?
163 cu. in. x .036 lbs/cu in = 5.8 lb.
So what we can conclude is that a 64 lb weight falling a distance of 11 in can increase the displacement of a tank submersed in 10 ft of water by 5.8 lb.
The next question is what if the 64 lb weight falls 10 ft instead of 11 in ?
Would the falling 64 lb weight expand the tank by 10 ft x 5.8 lb/ ft = 58 lb. No it would not. The tank expanding 58 lb would be the maximum possible.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
So how fast would the 64 lb fall? For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water. Drag is based on the amount of liquid moved. When the tanks fall as a continuous column there is almost no liquid being moved. At the top and bottom the tanks are pulled along by the draft of the preceding tanks.
The maximum increase in tank displacement caused by the 64 lb weight falling 10 ft is
58 lb. What if we only got measly 50 %? 29 lb - 3 lb wt of the tank = a increase in the displacement of the tank 26 lb/cycle. We could probably do better than 50 % of the maximum.
Quote from: brian334 on October 13, 2008, 02:13:13 PM
TK
The falling tanks do not lift the rising tanks.
I stand corrected, sorry. Apparently you expect the rising tanks to be lifted by buoyancy.
But that's a minor point, really, compared to the others.
I see how you could believe this device could work, though. As I sit here thinking about it, I almost can convince myself that it would. I see that the tank's buoyancy might be varied by the method you describe. In the compression and expansion stages there will be frictional losses, and these losses will increase with speed and size and number of active elements. But the main problem (other than the various sources of drag and friction) is that you just won't be able to achieve or maintain the overbalanced condition without external energy input, even through your clever buoyancy changing system. Why? Because you are dealing with conservative fields. Just as the potential energy of a mass depends only on its height, and not how it got there, so too, in a "buoyancy field" encountered by displacing water, the energy (work) available from floating a mass is never greater than that required to sink it in the first place.
Or vice versa.
Or something.
"Drag is based on the amount of liquid moved."
Not quite. Drag is proportional to the square of the velocity, and is proportional to the wetted area in contact with the water (hence your simplification of "amount of liquid moved.") While you do get some advantage from moving a long narrow shape thru the water (think racing shells) it isn't as great as you seem to think, and your tanks and the suspension arrangement aren't nearly as streamlined as they need to be to get any significant drag reduction from the configuration during descent. And don't the tanks make drag all the way around, not just on descent?
You might be able to simulate the behaviour of your device in the "phun" simulation, if you play around enough. I know of at least one gravity wheel that does go continuously in "phun".
@Mond: Thanks, and hang in there!
@TinselKoala
QuoteWhy? Because you are dealing with conservative fields. Just as the potential energy of a mass depends only on its height, and not how it got there, so too, in a "buoyancy field" encountered by displacing water, the energy (work) available from floating a mass is never greater than that required to sink it in the first place.
I don't think you understand the principals involved, Let's do a simple thought experiment, I have a hollow tube full of air with a valve like the aperature of a camera on the bottom, this tube extends from just above the surface of the water to 30 feet under water. I take a 20lb torpedo shaped weight that will exert a boyant force of ten pounds under water---the weight is hollow and has air in the middle. I drop the weight in the tube and it freefalls in the tube full of air and accelerates downward, the tube wall is wide enough that air can move around the torpedo weight. It is easy to calculate the speed of the torpedo just before the aperature valve and a sensor detects the torpedo just before it reaches the valve. Next the valve opens presicely as the torpedo passes throught it with less than 0.01mm clearance between the torpedo and the aperature, the water below has mass thus momentum and cannot move instantly and the valve opens and closes in perfect harmony with the torpedo passing through so no water enters the tube. The impact of the torpedo against the water will also push some water downward and the torpedo floats to the surface performing work in some mechanism. In this case the velocity of the torpedo has displaced the water below--they are equal forces, but the boyant force which lifts the torpedo has no relation to the falling velocity, the inertia on impact nor the displacement of water. This is because the torpedo has moved through fluids of differing densities, water and air. The only reason this can work is because of the control function, that is intelligent control of the process through a boundary condition--water and air-----the valve makes it work ;D. LOL, it's funny I could sit here all day and just make up new devices that violate your cherished "laws" and it is actually quite easy once you understand the mechanism as to how the laws are violated. However do not misunderstand what I am saying here, the conservation of energy applies in every case. But in this case the weight falling through air versus water has caused it to gain energy in the form of velocity, If it cannot "gain" energy then this device cannot work. As well I am not saying anyone should build this nor would it be economical to do so, I am saying it does what you say cannot be done. ------- and it took me all of 30 minutes to invent it ;D
Regards
AC
Quote from: allcanadian on October 16, 2008, 03:24:12 AM
@TinselKoalaI don't think you understand the principals involved...
;D
Quote
Let's do a simple thought experiment, I have a hollow tube full of air with a valve like the aperature of a camera on the bottom, this tube extends from just above the surface of the water to 30 feet under water. I take a 20lb torpedo shaped weight that will exert a boyant force of ten pounds under water---the weight is hollow and has air in the middle. I drop the weight in the tube and it freefalls in the tube full of air and accelerates downward, the tube wall is wide enough that air can move around the torpedo weight. It is easy to calculate the speed of the torpedo just before the aperature valve and a sensor detects the torpedo just before it reaches the valve. Next the valve opens presicely as the torpedo passes throught it with less than 0.01mm clearance between the torpedo and the aperature, the water below has mass thus momentum and cannot move instantly and the valve opens and closes in perfect harmony with the torpedo passing through so no water enters the tube. The impact of the torpedo against the water will also push some water downward and the torpedo floats to the surface performing work in some mechanism. In this case the velocity of the torpedo has displaced the water below--they are equal forces, but the boyant force which lifts the torpedo has no relation to the falling velocity, the inertia on impact nor the displacement of water. This is because the torpedo has moved through fluids of differing densities, water and air. The only reason this can work is because of the control function, that is intelligent control of the process through a boundary condition--water and air-----the valve makes it work ;D.
;D ;D
Quote
LOL, it's funny I could sit here all day and just make up new devices that violate your cherished "laws" and it is actually quite easy once you understand the mechanism as to how the laws are violated. However do not misunderstand what I am saying here, the conservation of energy applies in every case. But in this case the weight falling through air versus water has caused it to gain energy in the form of velocity, If it cannot "gain" energy then this device cannot work. As well I am not saying anyone should build this nor would it be economical to do so, I am saying it does what you say cannot be done. ------- and it took me all of 30 minutes to invent it ;D
Regards
AC
Great entertainment, thanks.
Cheers!
Quote from: allcanadian on October 16, 2008, 03:24:12 AM
@TinselKoalaI don't think you understand the principals involved, Let's do a simple thought experiment, I have a hollow tube full of air with a valve like the aperature of a camera on the bottom, this tube extends from just above the surface of the water to 30 feet under water. I take a 20lb torpedo shaped weight that will exert a boyant force of ten pounds under water---the weight is hollow and has air in the middle. I drop the weight in the tube and it freefalls in the tube full of air and accelerates downward, the tube wall is wide enough that air can move around the torpedo weight. It is easy to calculate the speed of the torpedo just before the aperature valve and a sensor detects the torpedo just before it reaches the valve. Next the valve opens presicely as the torpedo passes throught it with less than 0.01mm clearance between the torpedo and the aperature, the water below has mass thus momentum and cannot move instantly and the valve opens and closes in perfect harmony with the torpedo passing through so no water enters the tube. The impact of the torpedo against the water will also push some water downward and the torpedo floats to the surface performing work in some mechanism. In this case the velocity of the torpedo has displaced the water below--they are equal forces, but the boyant force which lifts the torpedo has no relation to the falling velocity, the inertia on impact nor the displacement of water. This is because the torpedo has moved through fluids of differing densities, water and air. The only reason this can work is because of the control function, that is intelligent control of the process through a boundary condition--water and air-----the valve makes it work ;D. LOL, it's funny I could sit here all day and just make up new devices that violate your cherished "laws" and it is actually quite easy once you understand the mechanism as to how the laws are violated. However do not misunderstand what I am saying here, the conservation of energy applies in every case. But in this case the weight falling through air versus water has caused it to gain energy in the form of velocity, If it cannot "gain" energy then this device cannot work. As well I am not saying anyone should build this nor would it be economical to do so, I am saying it does what you say cannot be done. ------- and it took me all of 30 minutes to invent it ;D
Regards
AC
I think if you add a valve at the top of the tube which seals airtight after the hollow weight has entered, then you could count on air pressure to prevent water from entering at the bottom, thus eliminating the need for such a fancy valve and tight tolerances at the bottom. This might make a functional experiment more obtainable. Although I'm still not sure as to the best configuration for the exit valve. I tend to believe using multiple forces is a key to producing free energy, in this case gravity and buoyancy.
Regards,
jeffc
You are wrong in your conjecture that I don't understand the principles involved. Nay, I submit that I understand them better than you do (and I can spell them, as well.)
I have no doubt that the device you describe would float--once. That is, given the functioning of your miraculous intelligent leakproof and instantly-acting valve.
But what, pray tell, resets the weight at the top of the tank, so the cycle can repeat, and a chain of these devices continue forever?
And, now that we've asked that fundamental question having to do with those pesky conservative fields and forces, what about that valve? You design it, I'll build it, and together we'll get richer than Croesus.
Can we stick to the original topic?
@TinselKoala
I remember when I thought I understood things pretty well, but learning is a never ending process, it only stops if you decide to stop learning by closing your mind to new thoughts. I will put this in terms you may understand better----no valves. In one of Nicola Teslas interviews he proposed a simple thought experiment, put a pressure tank at the bottom of a lake and this tank will have a hole in it. The water under pressure surrounding the tank could rush in through the hole and perform work in the process. Tesla proposed that the same amount of work would be required to remove the water in the tank as the work gained when the water rushed in ------however this equality of work only applies if the water remains in its original state. Tesla stated that if the water could be transformed into something else then the the work in versus work out would not be equal. I understood that Teslas thought experiment should be taken literally and have proven by calculation many years ago that he is correct.
Below is a tank submerged under water, the pressurized water operates a turbine or rotary motor which drives a generator. The generator powers an electrolizer which splits water into Hydrogen/oxygen gas, this gas is lighter than air thus it easily travels up the hollow tube to be utilized in any number of combustion devices. It should be understood that the deeper the tank is in the water the greater the pressure difference between the conditions in the tank and external to it---across the turbine, but this pressure difference does not apply to the the hydroxy gas open to atmospheric pressure which is in fact boyant to some extent which I did not use in my initial calculations. It should be obvious that the pressure difference is the energy gain and this gain grows with depth or pressure provided by gravity--it is powered by gravity and the hydroxy gas is free. I will state again that this does not violate the conservation of energy but it should be understood that if a media is transformed many rules we take for granted no longer apply. I have proven Teslas thought experiment by calculation for myself and have no need to debate this issue, if you need proof you can easily do the calculations and prove it for yourself, I used an average 60% efficiency for all mechanical components which is reasonable I think. It should also be understood that if this device is "possible" many other variations of this principle are possible. The mechanical aspects of this process can be applied in electrical terms, that is pressure to electrical potential, transformation of electrical current into an inductive discharge current utilizing an inductor. When you understand this you should realize that all of the devices depicted in this thread are to some extent pointless as a small circuit utilizing the proper components utilized in the correct manner can accomplish the same thing--- an energy gain.
@brian334
Sorry, this is my last off topic post
Regards
AC
Can we stick to the original topic?
Brian,
Your entire premise for successful operation relies on being able to achieve great forces from momentum due to the falling weight accelerating to a high velocity. You are using equations for a weight dropping in a vacuum to show what velocities you believe can be attained. Several facts have been discussed that show this is simply not the case. If you do not believe these facts, you should show some evidence/experiment that proves them wrong. If not, you need to show how your system could work without the extra force due to momentum since there have also been many reasons already given why it cannot work if momentum is not considered.
I submit that the problem with your idea lies in your *assumption* that the falling tanks can accelerate to a velocity large enough to have enough momentum for the internal weight to overcome the external water pressure opposing expansion of your tank. What leads you to believe it can?
M.
Hypothetically speaking, what if the falling tank only weighted 1 gram more than the liquid it displaces. The tank would fall very slow. Forget the momentum of the 64 lb weight.
When the tank got to the bottom the 64 lb weight would still be at the top of the tank.
As the 64 lb. Weight falls from the top of the tank to the bottom how much work could it do. Enough to increase the displacement of the tank by more than 1 gram?
HOW MUCH WOULD THE TANK EXPAND?
Imagine a tank that displaces about 1 cb ft of water submersed to a depth of about 10 ft.
The dimensions of the tank are about 12 in x12in x 12 in.
At the top of the submersed tank is a 64 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 64 lb lead weight can fall 11 inches before it hits the bottom of the tank.
Now imagine a piston with a surface area of 1 square inch is attach to the bottom of the lead weight.
The external water pressure at a depth of 10 ft. is about 4.3 lb/ sq in.
Could 64 lb weight overcome the 4.3 lb/sq. in. external water pressure and push the piston with the 1 sq. in. surface area out of the tank? Yes it could. How many pistons with a surface area of 1 sq. in. could 64 lb. Push out of the tank?
64 lb. divide by 4.3 lb/sq in = 14.9
So, as the 64 lb weight falls a distance of 11 in it will push 14.9 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 in.
What is the volume of the 14.9 pistons? 1 sq in x 11 in x 14.9 pistons = 163 cubic inches
So, the displacement of the tank would increase by 163 cubic inches.
Water weights .036 lb/cu. in.
So how much less would the submersed tank weight after the 14.9 pistons were pushed out of it when the 64 lb weight in the tank falls 11 in ?
163 cu. in. x .036 lbs/cu in = 5.8 lb.
So what we can conclude is that a 64 lb weight falling a distance of 11 in can increase the displacement of a tank submersed in 10 ft of water by 5.8 lb.
The next question is what if the 64 lb weight falls 10 ft instead of 11 in ?
Would the falling 64 lb weight expand the tank by 10 ft x 5.8 lb/ ft = 58 lb. No it would not. The tank expanding 58 lb would be the maximum possible.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
So how fast would the 64 lb fall? For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water. Drag is based on the amount of liquid moved. When the tanks fall as a continuous column there is almost no liquid being moved. At the top and bottom the tanks are pulled along by the draft of the preceding tanks.
The maximum increase in tank displacement caused by the 64 lb weight falling 10 ft is
58 lb. What if we only got a measly 50 %? 58 lb divided by 2 = 29lb Than 29 lb - 3 lb wt of the tank = a increase in the displacement of the tank 26 lb/cycle. We could probably do better than 50 % of the maximum.
When your freefalling mass strikes the bottom of the tube it will share its momentum with a certain mass of water. The water moved will probably have a greater mass than the falling object; this mass ration would depend upon the experimental design.
Lets say your falling mass is one kilogram moving 10 m/sec, that would be 50 joules of energy. As it strikes; the one kilogram shares its momentum with (lets say) 9 kilograms of water. Now we have 10 kilograms moving 1 m/sec, this is from the Law of Conservation of Momentum which has never been violated. 10 kilograms moving 1 m/sec is only 5 joules of energy. (1/2mv²) You can not expect 5 joules of energy to do more work than 50 joules.
You are doing the problem backwards. You should start with 10 kilogram moving 1 m/sec and give all the motion to one of those ten kilograms. Then you would have a ten fold energy increase.
Quote from: brian334 on October 17, 2008, 07:23:42 PM
HOW MUCH WOULD THE TANK EXPAND?
Imagine a tank that displaces about 1 cb ft of water submersed to a depth of about 10 ft.
The dimensions of the tank are about 12 in x12in x 12 in.
At the top of the submersed tank is a 64 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 64 lb lead weight can fall 11 inches before it hits the bottom of the tank.
Now imagine a piston with a surface area of 1 square inch is attach to the bottom of the lead weight.
The external water pressure at a depth of 10 ft. is about 4.3 lb/ sq in.
Could 64 lb weight overcome the 4.3 lb/sq. in. external water pressure and push the piston with the 1 sq. in. surface area out of the tank? Yes it could. How many pistons with a surface area of 1 sq. in. could 64 lb. Push out of the tank?
64 lb. divide by 4.3 lb/sq in = 14.9
So, as the 64 lb weight falls a distance of 11 in it will push 14.9 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 in.
What is the volume of the 14.9 pistons? 1 sq in x 11 in x 14.9 pistons = 163 cubic inches
So, the displacement of the tank would increase by 163 cubic inches.
Water weights .036 lb/cu. in.
So how much less would the submersed tank weight after the 14.9 pistons were pushed out of it when the 64 lb weight in the tank falls 11 in ?
163 cu. in. x .036 lbs/cu in = 5.8 lb.
So what we can conclude is that a 64 lb weight falling a distance of 11 in can increase the displacement of a tank submersed in 10 ft of water by 5.8 lb.
However, your tank design requires that the expandable piston(s) be on both ends. In your drawings on your web page you show the weight nested inside these hollow expandable pistons of ~144 square inches surface area. In this theoretical model you show that you are only able to expand 14.9 square inches. I have asked you to show a diagram of a design of a tank where you have the two sets of 14.9 square inch pistons on each end and the weight inside. If you try you should see that the weight cannot be at one end so as to be able to fall 11 inches. There are pistons on the top side of your tank that are in the way of your weight being there. Another consideration is that you have your pulley system that requires your weight to move at a 2:1 ratio to the piston(s).
This theoretical model you keep repeating does not represent a model of the tank you need to run your entire cycle. You are only analysing a model of a tank that can sink and then rise again. Just like AllCanadian's falling torpedo, it could go down one time and then rise again. Nothing magical there. But your model eliminates the features you require for the system to cycle continuously.
Quote from: brian334 on October 17, 2008, 07:23:42 PM
The next question is what if the 64 lb weight falls 10 ft instead of 11 in ?
Would the falling 64 lb weight expand the tank by 10 ft x 5.8 lb/ ft = 58 lb. No it would not. The tank expanding 58 lb would be the maximum possible.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
The maximum impact velocity *for an unrestricted object accelerating due to gravity near the surface of the Earth* is about 17 mi/hr. Your tanks are restricted by the drag of water. Your statement about 17 mi/hr impact velocity is not relavant.
Quote from: brian334 on October 17, 2008, 07:23:42 PM
So how fast would the 64 lb fall? For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water.
This has been repeatedly explained as not true/possible.
Quote from: brian334 on October 17, 2008, 07:23:42 PM
Drag is based on the amount of liquid moved.
This has been explained as not true.
Quote from: brian334 on October 17, 2008, 07:23:42 PM
When the tanks fall as a continuous column there is almost no liquid being moved.
It has been repeatedly explained that your tanks cannot fall (or rise) as a continuous column if they are to accelerate.
M.
The initial velocity of the falling tank is not zero.
The initial velocity of the falling tank is not zero.
The tanks would be feed into the top of the machine at the same rate as they rise.
I posted this invention hear 60 days ago and nobody has explained why it will not work.
Because no one explained why it won’t work I will assume that it will work.
This machine is posted at my website http://bsandler.com if you would like to explain why this machine won’t work please visit my website, when you get there click on the tab at the top of the page marked GRAVITY MACHINE # 2
I look forward to your criticism
If you think it will work.. then build it! Please post a video of your results.
Jason
Quote from: brian334 on December 06, 2008, 03:22:37 PM
I posted this invention hear 60 days ago and nobody has explained why it will not work.
Because no one explained why it won’t work I will assume that it will work.
This machine is posted at my website http://bsandler.com if you would like to explain why this machine won’t work please visit my website, when you get there click on the tab at the top of the page marked GRAVITY MACHINE # 2
I look forward to your criticism
Actually several people have explained exactly why it won't work. You just don't believe them.
Why do you invite criticism if you just ignore it?
Mr. Koala,
The problem is you are not criticizing my invention. You post junk inventions and use them as a argument to prove my invention won’t work. Than you try to give me a lesson on buoyancy. Still not criticizing my invention. Than you try to give me a lesson on a closed gravity field, and still don’t criticize my invention.
Be bold criticize my invention, and spot posting stuff that does not have anything to do with my invention.
I never claimed this machine could create energy, this machine is designed to convert one form of energy into another.
Quote from: brian334 on December 07, 2008, 03:44:55 PM
Mr. Koala,
The problem is you are not criticizing my invention. You post junk inventions and use them as a argument to prove my invention won’t work. Than you try to give me a lesson on buoyancy. Still not criticizing my invention. Than you try to give me a lesson on a closed gravity field, and still don’t criticize my invention.
Be bold criticize my invention, and spot posting stuff that does not have anything to do with my invention.
Yes, I am and have been criticizing your invention. I have posted pictures and links of "junk" inventions to show you that your "invention" has been invented many times before, and didn't work then either. I have given you links to Simanek's website where exact mathematical analyses of buoyancy drives identical and similar to yours are given in detail.
I have tried to explain buoyancy to you, even to the point of making 2 illustrative videos. You refuse to understand the explanations or see how they apply to your device. But apparently everyone else does.
I have pointed out to you that buoyancy is just an effect of gravity, and since gravity is conservative, your device, which is a buoyancy drive and depends on "gravity" as its only source of energy, will not work.
How does all of this not have to do with your invention?
How is it not criticizing your invention?
Is English not your first language?
Not only that, but Mondrasek has been trying to explain another fundamental flaw in your reasoning, having to do with the acceleration. You also refuse to believe his analysis, which is so simple and obvious.
You also made an incredible logical error when you said that, since no one criticized your idea, it must be workable.
But you don't have to believe me. Or even Mondrasek.
Just go build your own model. It's pretty simple, right? You could probably get a high-school science student to build it for her science fair.
Build it.
PROVE ME WRONG.
Or, even, address the points made in Simanek's work on buoyancy, and tell us why he and his colleagues are wrong.
Quote from: brian334 on October 06, 2008, 03:33:48 PM
I am still waiting for someone to explain what phase of my machine won't work?
All I get are reasons other machines won't work.
It's the burden of the inventor to prove the invention works... I've read many of your posts and inventions, are they some kind of joke? The Hurricane destroyer his hilarious, do you actually know how large a hurricane is? And the gravity stuff, where did you go to school? Did you graduate?
Quote from: brian334 on October 11, 2008, 04:10:11 PM
Grayone,
I don’t ignore the answers, the problem is I don’t understand them.
Another good example of why school is important.
I just went to the Symantec website, those machines are junk. Even a fifth grader could explain why they won’t work. Can you explain why my machine won’t work? It should be easy for a intelligent man like your self.
That's Simanek, not Symantec.
Perhaps you could now take Simanek's analysis, point by point, and explain how your device is not similar to those shown and analyzed by him.
And I have already explained, several times and in several ways, why and how your machine cannot work.
If you do not understand or believe my explanations, that's fine. But it won't make them go away, as you seem to wish.
Now, go to your hobby shop, buy some parts, and PROVE ME WRONG.
It should be easy for an intelligent man like yourself.
Simanek machine won’t work because.
1. All the water would leak out the hole in the bottom.
2. Even if some magic valve could be invented that would allow the ball to go thru it without all the water leaking out, the ball would need to be pushed into the column with a force sufficient to lift the entire column of water. Nothing is gained.
O.K. now you explain why me machine won’t work.
The momentum of the falling 65 lb weight is the force used to expand my submersed tank.
Once again: When you expand your submerged tank, you are displacing water. This takes work. The deeper the column, the more work it takes. But the buoyant force on an expanded tank doesn't depend on depth, it's relatively constant. But the deeper you go the greater the water pressure you must overcome to expand your tanks. Your chain of tanks won't accelerate, for the reasons Mondrasek has tried to explain to you. Your idea about using the momentum of the falling tanks to expand the submerged tanks won't work, because there isn't any acceleration and the force (not momentum) needed to expand the tanks is too great. In addition, the cable and pulley system adds drag and doesn't give you the mechanical advantage you would need.
Your device fails at every stage.
But by all means, go and build it and see for yourself.
After all, it's simple enough.
(But I really don't see how your invention is in any significant way different from the below "junk" invention that is thoroughly analyzed on Simanek's site.)
The velocity of the tanks being feed into the top of the continuous column is the same as the velocity of tanks falling in the continuous column.
A tank is feed into the top of the continuous column, the tank is than attached to the continuous column. The continuous column is heavier than the liquid it displaces and will sink. As the continuous column sinks it will accelerate do to the force of gravity, how can it not accelerate? As the tanks in the continuous column accelerate the 65 lb weight in the tank builds momentum. How could the 65 lb in the tank not build momentum as it accelerates?
As to the loss of energy because of friction I don’t see it as a issue. If the tanks double in volume at the bottom relatively small loses from friction won’t matter.
You're making that same mistake about acceleration again.
The first sentence, about the velocity of the tanks being fed in is the same as the velocity of tanks falling--
is incompatible with the tanks accelerating on the way down. If they are attached to a belt and maintain the same distance apart, that is.
Oh, perhaps you mean that the whole belt accelerates, as in goes around faster and faster. Nope, not with the expansion and contraction mechanisms trying to operate, it won't.
Have you ever heard of "terminal velocity?" Even in air, things don't accelerate without limit. Your tanks will have a quite low terminal velocity in the water, and the whole assembly won't go faster than that. No, the tanks won't clear a path for one another, unless they are very close together. Then you have problems with flipping and articulation. They can't be that much heavier than the water anyway, if the expansion at the bottom is to make them float up--you don't have the luxury of excess mass, that's for sure. Anything that doesn't contribute to the volume change is a drag on the system.
When you somehow extract the momentum from the moving weight to expand the deeply submerged tank against the water pressure (about 15 psi per every 33 feet of depth), that momentum will be gone from the motion of the belt. Where do you think it goes, since momentum is conserved? It goes into displacing the water and moving it around. There is no mechanism for momentum gain in your system.
You seem to think that the tanks can be expanded instantaneously and at a single location. But even if the rest of your mechanism worked as you think, it still takes time to expand and compress the tanks. If the belt is moving along while this is happening, there goes your presumed overbalanced situation. Now you have no overbalance from the buoyancy, a slow starting belt motion, and massive losses from friction and operating the pulley system.
You are relying on water pressure to re-collapse the tank at the top of its travel. How do you think that can possibly happen, if the water pressure at the bottom of the travel is small enough for your 64-pound weight to push against it successfully? The pulley system doesn't give you the advantage you need, because compound pulleys trade off distance for force.
Your system simply has too many losses, and no way for momentum gain.
But you don't have to believe me. I encourage you to get up, get out of your armchair, or if you're in a wheelchair, send somebody else out, and build a model yourself. Get your hands dirty. Maybe the experience will teach you something.
PROVE ME WRONG.
Have you ever actually built anything? If you have built some models that work, or demonstrate a principle, let's see them. Please.
Three things I learned about posting here.
1. Don’t post whilst drunk.
2. It’s no fun to post hear unless you are drunk.
3. Either way it does not matter.
Kola,
Does 2 + 2 = 4 or,
Does 2 + 2 = 22
By your way of reckoning,
2 + 2 = 6
because you are trying to add one of your 2's twice.
2 + 2 = 4, the question is where do the 2’s come from? I say you can get them from gravity.
1. The free falling tanks in the continuous column are not attached to the track.
The free falling tanks in the continuous column are attached to each other.
2. The tanks are abruptly stopped at the bottom, when the tanks are abruptly stopped the 65 lb weight keeps going.
3. The external water pressure at the top would be sufficient to push the piston back into the tank.
Quote from: brian334 on December 10, 2008, 12:39:52 PM
1. The free falling tanks in the continuous column are not attached to the track.
The free falling tanks in the continuous column are attached to each other.
How can the falling tanks be attached to each other in a continuous column? That would mean all the tanks on the falling side are going the same speed, ie. NOT accelerating.
The "continuous column (that) are attached to each other" must be going the same speed as the tanks on the pulley at the top of the system. But those tanks are supposed to be collapsing. How do they do that? Do they a) collapse and leave a gap between each other the size of the collapsing piston, or b) move closer together as the piston collapses? If you answer "a", then they are no longer in a continuous column. If you answer "b", what moves them closer together?
Now the tanks on the upward rising column cannot be attached to each other, right? Because the bottom tank in that column is moving at a vertical velocity of zero, having been stopped for the expansion process to occur, right? So attaching anything to that tank would mean all the tanks stop, since it is effectively a zero velocity anchor.
M.
Mond, I don't think Brian is actually reading our posts.
Brian, if the free-falling tanks aren't attached to the track, how do you get any "gravity" power from them? How fast do you expect them to fall, since they are just barely heavier than the water they displace? When the tanks stop abruptly at the bottom, the whole column must stop too, since the tanks are attached to each other. There goes all your momentum. The next-to-the-bottom tank only can fall one tank-length from its stopped position (waiting for the tank ahead of it to be expanded) to the position where it in turn will be expanded. Where's the momentum you were going to use? It's not there.
Quote from: brian334 on December 10, 2008, 12:39:52 PM
2 + 2 = 4, the question is where do the 2’s come from? I say you can get them from gravity.
1. The free falling tanks in the continuous column are not attached to the track.
The free falling tanks in the continuous column are attached to each other.
2. The tanks are abruptly stopped at the bottom, when the tanks are abruptly stopped the 65 lb weight keeps going.
3. The external water pressure at the top would be sufficient to push the piston back into the tank.
You machine will work. Is that what you want to hear? OK, now you have heard it. Good luck with building it.
Brian, if the free-falling tanks aren't attached to the track, how do you get any "gravity" power from them?
The tanks expand at the bottom, after they expand they are lighter than the liquid they displace and float up doing work. Gravity forces the tanks up.
How fast do you expect them to fall, since they are just barely heavier than the water they displace?
I don’t know how fast the tanks will fall, when the tanks fall in a continuous column there is all most no drag with the water.
When the tanks stop abruptly at the bottom, the whole column must stop too, since the tanks are attached to each other.
At the bottom the tanks are quickly pushed over to the stop by the transfer solenoid.
The tanks need to be removed from the bottom as quickly as they fall.
I just wanted to thank you Brian,
I have read this entire thing, and the thing about the bubble tube, and your website.
I learned a few things.
1. Patents should not have dimensions on them.
2. Hurricanes are not something to play with.
3. The failure to listen to criticism that is designed to help you is embarrassing.
4. Defending a position after that position is proved to be false is embarrassing.
5. The only thing worse than being proved wrong is being proved wrong hundreds of years ago.
Why don't you just build the machine yourself?
Again, thanks for teaching me a few things.
Truth,
The truth is no one has explained why either the hurricane machine or GRAVITY MACHINE # 2 won’t work.
Every day my confidence in both machines increases.
Dear Brian,
How tall is a hurricane? How tall could a machine/ship be built? How many MILES across is the eye of a hurricane?
Did you see the Katrina photos of the OIL TANKER that was BLOWN up on a FREEWAY far from the beach?
I would title that contraption the "Hand of God" if I were you.
Now, more seriously, MANY others have explained the problems with your momentum theory much better than I ever could.
For the moment let us assume that part did work, and the machine turned and moved very fast to achieve momentum, and was strong enough to withstand the fatigue of that sudden stop constantly working on your frame, and you found a way to arrange and build all of those connecting tubes and locking devices. All of that is a stretch, but here we go. How often is this thing going to need repairs? My estimation is that the duty cycle would be very low due to mechanical failures requiring removal from the liquid to effect repairs. How often would parts need to be replaced due to wear and tear?
Now for the REAL test: Will your machine be more cost effective than hydrocarbons?
If ALL of these reasons don't help you understand the reasons that the device is unworkable I do have a suggestion for the next thing you should invent or improve. Please do work on those construction hats the men use for safety from falling objects. I am sure that you will do great because of your apparent personal expertise on the subject.
Seriously, letting you continue to believe this will work would be the most unkind thing that could happen to you on this forum.
Follow the Nike slogan brian334...
Just Do It.
Why continue to question something if your sure it is correct?
All you are doing is wasting precious time which could be used engineering your device and discovering the truth are you not.
During the course of your posting and the hours invested in discussions you could have achieved a full product with the full answer staring you in the face.
Not inspired yet ?
Then you will never do so.
Just Do It.
-infringer-
truth said:
"5. The only thing worse than being proved wrong is being proved wrong hundreds of years ago."
And that's the Truth!!
In general the diameter of a hurricane eye is 10 to 15 miles, some are bigger and some are smaller. As far as the height of a hurricane it does not matter. The air diverted into the eye would almost completely stay at sea level.
As far as the tank being strong enough to withstand repeated sudden stops.
A few shock absorbers would solve the problem.
Is this cheaper than hydrocarbons?
The problem is the hydrocarbons will some day run out and when they do we will be screwed.
The reason I don’t build the machine and test it.
The machine is not complicated but its not something a amateur could build in there back yard.
Brian334,
So you consider yourself an in this classification...
Makes perfect since I appreciate your effort to bring us a solution as does everyone here in this thread I assume!
But in order to further prove this device it will have to be built by yourself.
Unfortunate well yes and no.
In order to be a back yard builder and no longer be an amateur you must do things such as build things and it appears this is the ordinary evolution of growth...
If you have ever seen some of the builds on here you would know that building can be done by anyone with nearly anything as long as you attempt... Excommon is a perfect example of the guy who can build something with what appears to be nothing!
Think of now the million ways you could build this device and what is humanly possible for you at the current time then achieve the build.
Some things to consider are resistance during the fall the shape of a bladder is most likely less wind resistant when empty which will provide just one of the many resistances to equate for .
I'm not saying your device will not work with 100% certainty but present it in a manner that is very mindful equating for every piece of the puzzle and maybe then you will find a builder of your device or possibly you will find that it is not worth building and in fact you may be incorrect.
It is obvious you are presenting an idea and expecting someone else to build and prove your idea as correct in order for this to happen you need substantial evidence to support your devices work.
Good Luck with your endeavor ...
-infringer-
Truth,
The drawings posted at my website are not blueprints, the only purpose of the drawings is to illustrate general ideas.
The tanks with the 63 lb weight fall in a continuous column.
The cintinuous column is always getting longer at one end and shorter at the other end.
When the tanks are added to the top of the column it gets longer, and when the tanks are removed from the bottom of the column it gets shorter.
The continuous column always stays the same size and weight and always accelerates down do to the force of gravity.
I stay the 63 lb weight will have enough momentum to at least double the displacement of the tank.
What do you stay, and please back it up with numbers.
@ Brian,
How about his number?
The total amount of people confident enough in your project to actually do any further work on it including yourself equals zero.
Sorry, but that is just the truth.
Lies,
That’s a silly post, I applied for a patent on this invention. I paid $515. To the United States Patent Office for the application fee.
If you think I am not serious about this invention you are kooksville.
brian
Quote from: brian334 on December 14, 2008, 08:08:54 PM
Lies,
That’s a silly post, I applied for a patent on this invention. I paid $515. To the United States Patent Office for the application fee.
If you think I am not serious about this invention you are kooksville.
brian
OK, so the number may need to be adjusted to 1. Let us know when you get it running.
You are really wasting your money. But it's your money, so OK.
But you really must deal with this: You think your descending column will be a continuous column. But you know that you have to abruptly stop a tank at the bottom, so the 64 pound weight can continue moving and expand the tank by its "momentum."
But since the tanks are in a continuous column, this means that ALL THE TANKS must also stop. So, once the tank at the bottom is expanded, the next tank can only fall ONE TANK LENGTH before it too must stop and (try) to be expanded. And so forth. You don't have any momentum, and your column cannot accelerate--because it has to stop each time a tank is expanded.
Not only that, even if there was enough momentum from the falling weight (which there cannot be, as the weight only falls one tank length, see above), that momentum goes into lifting the water--the same water that then "falls under" the rising buoyant tanks to give them their lift. Thus there is no energy gain, no momentum gain, possible in the system.
There is no need to put numbers to any of this. When you think about it clearly you can see that the mechanism won't do what you think it will, and even if it did, it still wouldn't work because you have omitted thinking about all the water that must be LIFTED when your tanks expand. One atmosphere of pressure every 33 feet of depth! That's a lot of water pushing against your tanks, keeping them from expanding. You think lifting a tank's volume of water, in water, is free because the water has the same buoyancy as the water surrounding it. But you are forgetting about the pressure of the water above. You must do work against this pressure. The same amount of work that you get back when the expanded tanks rise. Less the large losses in your mechanism.
@ Tinsel Koala
You are wasting your breath. I gave him the figures at 10 ft depth and advised him to study some hydraulics before embarking on designing a hydraulic system. His answer was "You are embarrassing yourself" when he saw my calculation. I challenge anyone to prove my figures wrong.
There simply is no substitute for an education. Postulating garbage just doesn't get you there.
Hans von Lieven
Brian, before you spend any more money on this machine you should take a good look at the maths and physics involved, which is all high school level stuff.
Work out the energy losses and gains for a single bucket at each stage of the loop. If you know how to use Excel you could make a model of it which will allow you to easily change the parameters of the machine to see the effects this would have.
This will save you alot of money and time in the future.
Jandell,
Thanks for the advice, I have limited math skills. I would appreciate your help.
Please explain to me mathematically why this machine won’t work.
Thanks,
Brian
brian334
Take all your thoughts on this design and reverse them. We who build become our worst critics for good reason so, that is what you need to learn and you will see for yourself. This will also tell you what you have to improve on. Don't get tunnel vision on what you are doing.
Quote from: brian334 on December 16, 2008, 03:45:35 PM
Jandell,
Thanks for the advice, I have limited math skills. I would appreciate your help.
Please explain to me mathematically why this machine won’t work.
Thanks,
Brian
You appear to have some kind of chip on your shoulder. A full mathematical analysis of your device is quite a bit of work, which no one likes to do for free, but it is obvious to most that at each and every stage, given frictional losses, your device would consume more energy than it produces.
Most likely no one will take the time and energy to prove that your machine cannot work, to your satisfaction. Why bother? As soon as you try to build your device, you will realize the obstacles involved in trying to extract energy from it.
Jandell,
Thanks for the advice, I have limited math skills. I would appreciate your help.
Please explain to me mathematically why this machine won’t work.
Thanks,
Brian
@Jandell: Then you can go beat your head against a brick wall. It will have just about the same effect.
Brian, you have yet to explain how, at the bottom of the continuous descending column of tanks, you can bring one tank to an abrupt halt, in order for its weight to keep moving, while not also bringing all the other tanks, which are following it without gaps, also to an abrupt halt. You have also not explained how, if the water pressure at, say, 30 feet is small enough to be able to be overcome by the momentum of the falling weight, yet the pressure at or near the surface is great enough to push the expander back in. There is no mechanical advantage from your pulley system. The chain of tanks must be moving as fast at the top as it is at the bottom, have you considered that? How do you deal with the momentum of your weight, at the TOP? It's going in the wrong direction.
You see, until you clear up your misconceptions about how things actually work, trying to make an exact mathematical analysis is like, well, beating your head against a brick wall.
But don't let me stop you, I'm the Evil Skeptic after all, and I get my jollies from seeing people waste their time. Just go build the thing. Oh, wait, you have no building skills or materials. Well, then use a spreadsheet to analyze the system properly. No, wait, you have limited math skills. Well, then, you could seek advice from those who have those skills--but clearly you don't have to take it.
Jandell,
Thanks for the advice, I have limited math skills. I would appreciate your help.
Please explain to me mathematically why this machine won’t work.
Thanks,
Brian
Brian,
I thought of a simple experiment that you may be able to try to show the effects of the drag of water on your system. It approximates your concept of the tanks falling in a continuous column. Find yourself a small wheel with a solid rim, ie. no spokes. You want the wheel and rim to be smooth so that any cross section taken through the center is equal to any other. These are common on children's push and ride toys as well as some strollers and wagons. When you spin this type of wheel it approximates your continuous column, since there are no gaps between sections in the wheel and the only place water resistance can exist is on the sides. So spin this wheel in air and note how much it runs. The only resistance is the friction from the bearings and air resistance. Then dunk the wheel underwater and spin it again. You will see that it does not spin much at all. This will show how much drag water has on an object moving through it, even if falling or rising in a continuous column. It may lead you to understand that your falling and rising tanks could only move very slowly, thus eliminating the momentum you envision being built up to do the work of expanding your tanks.
Also, a complete mathematical analysis of your system would be not only very laborious, but still not conclusive. With fluids you have to deal with turbulence, something very hard to model. This is one of the reasons that weather modeling is so difficult and imprecise. Back in the day the first super computers were used to solve the complex equations governing turbulence in a fluid and it's effects on drag. Now a modern computer could likely model simple systems in near real time, but only to a gross approximation. This is the reason that real wind tunnel testing exists: the math and complexity of trying to model every molecule of a fluid and all the forces governing it's interactions are beyond what we can do even with the fastest computers. But approximations do exist and lead those of us who have studied or worked in the field to know certain phenomenon will certainly exist, like drag. I personally cannot calculate exactly how much drag there will be, but can tell you with certainty that it will be significant. The above recommended experiment will show that.
Thanks,
M.
M.
If it took 2 weeks for the tanks to fall this machine would still work.
But the faster the tanks fall the better it will work.
Jandell made a statement that using high school math and science it would be easy to prove this machine won’t work, so what’s the proof ?
Quote from: brian334 on December 17, 2008, 10:30:59 AM
M.
If it took 2 weeks for the tanks to fall this machine would still work.
But the faster the tanks fall the better it will work.
Jandell made a statement that using high school math and science it would be easy to prove this machine won’t work, so what’s the proof ?
Figure it out for yourself. Your lack of understanding of basic physics and math does not have to be a permanent condition.
M.
When you calculate how fast the tanks will fall make sure you consider water temperature.
Hot water moves easier than cold.
Try this simple test.
Fill a bucket full of water, get a glass, put enough weight in the glass so that it almost sinks. Put the glass in the bucket. Spin the glass. Notice how long the glass spins.
Quote from: brian334 on December 17, 2008, 11:18:44 AM
M.
When you calculate how fast the tanks will fall make sure you consider water temperature.
Hot water moves easier than cold.
Try this simple test.
Fill a bucket full of water, get a glass, put enough weight in the glass so that it almost sinks. Put the glass in the bucket. Spin the glass. Notice how long the glass spins.
And now take the same glass (without water inside) and tie a thread around it so you can suspend it from a tall beam in air. This is an imperfect example because the thread will also resist twisting but that is insignificant. Spin your glass now. See how it does not appear to slow down for a much much much greater time?
Your glass in water experiment shows exactly what I was explaining. The glass refuses to spin FAST. It slows quite quickly. It spins SLOWLY for a long time, just as your tanks will only fall and rise slowly.
No speed equals no momentum. Please tell us how your device works without your assumed momentum? You have stated that it will work just fine. How? We can show you the math for the "no momentum" case. You need to understand that momentum will play an insignificant part in the expansion of the tanks for us to continue. Can you understand this yet?
M.
M.
I am not staying there is not any resistance from the water, assume the tanks only reach 50% of the maximum possible velocity.
When you spun the glass in the water did it wobble? Mine did.
M.
When the tank gets to the bottom the 63 lb lead weight is still at the top of the tank.
The weight will fall about 11 inches. When a 63 lb weight falls 11 inches it can do enough work to increase the displacement of the tank by about 5 lbs. At a depth of 10 ft.
I thought of another high tech test. No wobble in this one.
Fill a bucket full of water, the bucket has to be smooth inside. With a stick get the water spinning inside the bucket as fast as you can, than drop the stick into the bucket, watch the stick spin. Notice how long it spins. If you use hot water it will spin longer.
Hi Brian
I wouldnt actually mind doing the maths, its all stuff i know pretty well. I'll sketch out and explain the equations and diagrams on 2-3 A4 sheets of paper and email them to you tomorrow.
I also had an idea for a PM machine a few months ago and spent about $100 building it before i did the calcs and realized it wouldn't work. Like Tesla said, if you do the calcs first then youll save yourself alot of time and effort instead of just going ahead and trying to design your machines by experiment.
Brian,
Previously we have discussed that the 63lb weight cannot expand your tank if the expansion area is equal to the cross section of your tank (1' x 1' x 1'). It can only expand a piston with ~14 sq inches (not 144 sq inches). I have asked repeatedly for you to show a design of a tank with only a 14 sq. in. piston on each side and the weight in the middle. The purpose is to show you one cannot be made. You continue to avoid doing any of the work to teach yourself what you are missing.
I have to ask: Is your purpose here to try and win an argument/debate, or to learn the truth about your concept? If it is the former, I will gladly stop trying to help you learn and admit defeat for your arguments. You can then include me with the others that have "failed to prove" that your idea will not work. But if you wish to learn the truth then get off your ass and show the design!
You are avoiding doing any of the experiments/exercises we have suggested that should show you the flaws in your concept. Instead you have repeated statements time and again that have been shown to more knowledgeable persons to be false. You repeat these false statements as proof that your concept will work. Repeating a falsehood over and over does not make it true.
Show us the design of a tank that can expand only ~14 sq. in. pistons on each side. This exercise is designed to help YOU understand.
Also, my glass did not wobble noticeably (better glass or spin, who knows or cares?). It did however slow quite rapidly due to the known effect of the drag from the surrounding water. The water adhered to the sides of the glass in a relationship including the constant known as the Reynolds number for that surface. While I might have used a glass with a surface who's Reynolds number was more advantageous to your arguments, there is no known surface that will let it reach a terminal velocity approaching 50% of what it could reach in air. 5%, maybe, but that is only and educated guess.
M.
Here's my hypothesis:
Jandell will do the math on the spreadsheet and will demonstrate that Brian's machine cannot work.
Brian will then inform us that Jandell did not do the math model correctly, and will repeat his claim that nobody has yet shown how his device won't work. Or Brian will come up with some mechanical modification of the pulleys or transfer solenoid or other part of his mechanism, and claim that it will now work as designed, and the whole silly process will begin again.
PROVE ME WRONG!!
Please.
I will post Jendells numbers as soon as I get them.
M.
Go to my website and check out GRAVITY MACHINE # 1
It will help if you print out the drawings.
Quote from: brian334 on December 17, 2008, 01:49:14 PM
M.
Go to my website and check out GRAVITY MACHINE # 1
It will help if you print out the drawings.
Brian,
Sadly, you have once again sidestepped the issue. Again (fourth or fifth time now?), I ask you to draw the tank described. Directing me to a different concept does not fix the flaws in this one.
Also, I ask again: Do you only want to win the argument/debate, or do you truly want to learn why your concept will not work? So far you have shown little effort to learn. And when we abandon the discussion due to you repeating falsehoods over and over, you claim that means we have not disproved your concept and therefore it must work, ie. you win! What do you really want? The truth, or agreement with your misunderstanding of reality?
M.
M.
Gravity Machine # 1 is the best answer I can give you.
You asked for a different design, Gravity Machine # 1 is a different design.
I am not side stepping your question, I am answering your question.
At my website http://bsandler.com click on the tab at the top of the page marked Gravity Machine # 1
Quote from: brian334 on December 17, 2008, 02:50:13 PM
M.
Gravity Machine # 1 is the best answer I can give you.
You asked for a different design, Gravity Machine # 1 is a different design.
I am not side stepping your question, I am answering your question.
At my website http://bsandler.com click on the tab at the top of the page marked Gravity Machine # 1
Brian,
Seriously, please start with your tank design from Gravity Machine #2. We have done analysis already on that design assuming a cube shape with dimensions of 1' x 1' x1', so use those dimensions again. In that original design you proposed to expand each end by half the height, or six inches. The surface area of the expanded portion was 1' x 1' or 144 sq. inches. We have shown that this is impossible with only the force due to gravity of an internal weight of ~63lbs at a depth of 10 ft.. You countered that argument by showing calculations that the ~63lb weight could expand a piston of a mere ~14sq. in. and I agreed. I have repeatedly asked you to draw such a tank. Let me be clear: Please draw a tank that is 1' x 1' x 1' with expandable pistons on opposing ends that have a surface area of 14 sq. inches. Inside must be a ~63 lb weight that causes the expansion of the lower piston when the tank system is in the proper orientation.
No sidestepping.
Draw it.
Start with your drawings of the tank in Gravity Machine #2. Then reduce the size of the expanding piston to only have a ~14 sq. in. surface area. Fit the weight and the rest of your mechanisms inside.
Then (and only then) let's review your design. Okay?
M.
M.
Gravity Machine # 1 is the solution to your problem.
GM#1 does not use the momentum of the falling 63 lb weight to expand the tank.
It only uses the weight of the 63 lb lead weight to expand the tank.
And don't forget that a 14 square inch surface is going to experience, at ten feet depth, about 14x5 or 70 pounds of water pressure.
And also don't forget that your gravitymachine#1 is identical in function (or non-function, rather) to several of those also-patented "junk inventions" as you call them that are analyzed on Simanek's site.
Kolala.
Don’t forget the lead weight weights 63 lbs.
How in the world could 63 lbs of lead overcome 4 lbs/sq. in. of water pressure.
Its just not possible. Especially if the 63 lb lead weight is moving.
Quote from: brian334 on December 17, 2008, 03:39:36 PM
M.
Gravity Machine # 1 is the solution to your problem.
GM#1 does not use the momentum of the falling 63 lb weight to expand the tank.
It only uses the weight of the 63 lb lead weight to expand the tank.
Brian,
GM #1 is not the solution to anyone's problem. Please stop pointing to it and stay focused on the discussion of GM #2.
Many device concepts can appear to work when real dimensions and material properties are ignored. This is sometimes why some OU devices even work in simulations. For example, in a drawing or sim I can say a 1 cubic inch object weighs 2 tons. But in reality I do not have access to a material that is dense enough to construct such an object.
We have mathematically analyzed your tank design from GM #2 assuming the tank is a cube with dimensions of 1' x 1' x 1'. We have agreed that an internal 63 lb weight can extend a piston of surface area ~14 sq. in. (3.74 in. x 3.74 in.) at 10 ft. depth in water. Your system requires that there be two pistons on opposite sides of the tank such that the lower one will extend when the tank is stopped at the bottom of the 10 ft fall. I say you have never shown a device that fits those two pistons, the weight, and the pulley system inside such a tank. Please do so. The drawings of the tank on your web site do not meet those specific conditions.
I have attached a drawing of the required elements drawn to scale. I have given you three weight options and can draw more if you need. Each one would weigh 63 lbs if made of solid lead. You can use these elements to show the correct tank design. It does not need to be a perfect graphics representation. If it would save time you can simply print it, cut the pieces out, assemble, and draw the pulley system and other elements on them.
If you want to discuss GM #1, I would suggest you start an new thread. It would get difficult to follow discussions on these two different but similar concepts in a single thread.
Thanks,
M.
M.
Thanks for doing the work.
Substitute depleted uranium for the lead and cut some holes in it for the cables and pulleys.
Brian,
Would you like me to give you another drawing with the weights thinned by 40% to represend 63 lb weight of DU? Would you then be willing to show a design that fits all the elements inside the 1 sq ft. tank? If not, I will stop wasting my time with you. It is becoming clear to me that you don't really want to discuss your idea with the goal of learning anything. Not that there is anything really wrong with that, only it's just not the reason I am participating in this exchange.
M.
Hey
I did a few quick calcs but according to them it works, so obviously I made a mistake somewhere. I only considered four stages where energy could be lost or gained:
1) Energy is gained when the water pressure pushes the buckets into the air chamber
2) Energy is gained as the tanks fall in the air chamber
3) Energy is lost as the expanded tanks are pushed back into the water
4) Energy is gained as the expanded tanks float upwards
Think of a similar machine to Brians, but with cubes that dont change size. If this machine runs clockwise, and the cubes enter the air chamber at a low point, and exit at the top, say 2m higher, more work will be done by water pressure pushing the cubes in than pushing them out again.
If the cubes are 1m X 1m X 1m, and enter the air chamber at 10m depth, the work done pushing them in will be
pressure X area X distance = (1000 x 10 x 10) x (1 x 1) x (1) = 100 kJ.
The work done pushing them out at the top, at 8m depth, will be (1000 x 10 x 8 ) x (1 x 1) x (1) = 80 kJ.
From this it seems that work would be done by the machine.
Starting at the bottom the expanded tanks weight less than the liquid they displace and will float up. At the bottom the tanks have enough potential energy to lift the 63 lb weight all the way to the top. As the tanks rise they lose there potential energy and the 63 lb weight gains potential energy.
Because the tanks are accelerating as they rise the 63 lb weight is also gaining kinetic energy.
At the top the 63lb weight has both potential energy and because it is moving it also has kinetic energy.
Next at the top the piston lock is released and the external water pressure pushes the piston back inside the tank. Also at the top the tank flips putting the 63 lb weight back on top. The tank is now heavier than the liquid it displaces and will sink. As the tank sinks its potential energy turns into more kinetic energy for the 63 lb weight. When the tank gets to the bottom it is quickly pushed over by the transfer solenoid and abruptly stopped,
All the kinetic energy the 63 lb weight gained on the way up and down is converted to potential energy.
The 63 lb weight falls inside the tank pulling the cable moving the piston out of the tank. The tank is now lighter then the liquid it displaces and will float up doing work.
Also when the tank gets to the bottom the 63 lb weight still has about 11 inches of potential energy as it falls to the bottom of the tank.
The 63 lb weight falling 11 inches will increase the displacement of the tank by about 5 lbs.
At my website http://bsandler.com there is a complete description of this machine, click on the tab at the top of the page marked Gravity Machine # 2
Brian how much energy do you think it takes to flip one of your tanks end for end?
how dose that compare with the energy of the falling 63 pound weight?
:P
@ jandell254,
Sir, your calculations appear to be made based on some other machine that falls in an AIR CHAMBER.
Perhaps actually reading some of the problems that have been repeatedly explained would assist you in understanding the multiplicity of reasons this design has been proved not to work. We ALL are here to find a way to do the impossible, but miscalculations, deceit, and stubbornness are obstacles not solutions.
@ Brian,
Do you really think that elementary school yard name calling makes you right or even appear to be an adult?
@ everyone,
The math we all face is this equation:
Energy output equals Energy input minus friction loss.
In most cases (allowing maximum efficiency) that is 1 minus greater than 1 resulting in a net loss.
1 = 1 - (>0)
I apologize for restating the obvious!
"Also at the top the tank flips putting the 63 lb weight back on top."
From where comes the mysterious force that lifts the 63 pound weight, in the stopped tank, back up the 11 inches (or whatever) inside the tank?
All tanks must be stopped in order for the top and bottom tanks to flop over, expand or compress.
The rising tanks must be hooked to a belt, even if the descending ones aren't. Hence we are back to the problem that Mondrasek has been trying to get across. Either the tanks cannot maintain the same spacing (hence can't be attached to a belt), or the tanks cannot accelerate but must move at a constant speed.
So, it wouldn't work even if there was the momentum of the weight to use, and the momentum isn't there to begin with, because the whole assembly will have to move in little jerks, because it has to stop to allow tanks to flip/expand/contract.
Plus there must be an input of energy for each tank, at the top, to raise up the weight inside the tank.
The acceleration of the top tank is restricted by the generator allowing the tanks beneath it to catch up.
Brian, have you tried to get your mechanical arrangement to run in a simulator of some kind? I suggest you download and play with the Phun simulator for a while. It's easy to use and models gravity. And it's phun!
Since it is possible to get a Bessler-type gravity wheel to run in Phun, you might be able to get your system or a subset of it to work as well--if the mechanism is even possible, which I and others doubt strongly.
So, the point of the Phun exercise, for you, would be to see if your basic mechanical layout will even turn, start, stop, expand, accelerate, etc. like you think, regardless of whether it will extract free momentum from somewhere. You can power it with a Phun motor, for example, to test the mechanisms.
http://www.phunland.com/wiki/Home
and here is a file that is a working Bessler-style gravity wheel, thanks to our invisible rabbit friend Harvey.
http://www.mediafire.com/?kdnyloqmjhx
Have Phun!
TK,
Check out the Hinge at the center of that Bessler demo. It's a motor set at 15 RPM. Turn that off and the Bessler demo will not self run.
M.
Koala,
Thanks for the info, its looks like Phun could be great fun.
It will take me some time to learn how to use it.
Quote from: mondrasek on December 21, 2008, 11:33:25 AM
TK,
Check out the Hinge at the center of that Bessler demo. It's a motor set at 15 RPM. Turn that off and the Bessler demo will not self run.
M.
Darn that silly rabbit, anyway! I looked thru that file and didn't find the motor. Harvey is playing tricks again...
Oh, well, back to the drawing board...
Still, it should help Brian try out his mechanism.
EDIT
I just checked--that hinge motor is turned off, even though its speed is 15 rpm. You can set the speed to zero and the simulation still turns.
Koala.
When I looked at Phun I looked for a tool that would help with my invention. Phun did not help.
I am not saying its not a nice toy, but it is not any help to me.
Quote from: brian334 on December 31, 2008, 04:00:49 PM
Koala.
When I looked at Phun I looked for a tool that would help with my invention. Phun did not help.
I am not saying its not a nice toy, but it is not any help to me.
Is that because it confirmed you are wrong?
U
If you or anyone else can prove my invention will or will not work using Phun go do it.
Put-up or shut-up.
Than post the file name. I would be more than happy to look at it.
Quote from: brian334 on January 01, 2009, 01:11:14 PM
U
If you or anyone else can prove my invention will or will not work using Phun go do it.
Put-up or shut-up.
Than post the file name. I would be more than happy to look at it.
I don't care enough about your device to go through that effort, and I doubt anyone else does either.
U
You appear to care about it enough to criticize it.
Like I said, put-up or shut-up.
DOG GONE brian334
When are you going to build your device? I am almost finished with one of mine and If I can figure out a slight timing problem it will accelerate. Using my finger to add a little pressure for timing I had it running at approximately 200 rpm. Building is the only way. ;)
OBTUSE= Lacking acuity of sensibility or perception, having neither delicate feeling nor alert awareness: DENSE I cant believe this topic went on for 16 pages!!!!! ???
Quote from: Dr on June 14, 2010, 05:13:42 PM
OBTUSE= Lacking acuity of sensibility or perception, having neither delicate feeling nor alert awareness: DENSE I cant believe this topic went on for 16 pages!!!!! ???
Well Dr
We use to talk and discuss things. Not keep claiming that someone else is wanting what or is jellous of (IMO nothing) they have, no matter how many times it is said they don't want anything but the claimer keeps making stories and lies and never lets it alone. Now not allot of people here now. It is sad :'(
Alan: Yes the gravity wheel section has taken a serious black eye, I have a really cool Idea I wanted to discuss, but these forums seem to all be the same lots of personal attacks for no GOOD reason. Well I am still searching back threads to make sure my Idea has not been presented already. Nothing worse than plowing the seeds back up. :)
dr
Why did you waste your time reading 16 pages of gibberish?