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Hi,

I have made a drawing here. It contains four containers which has an altering volume of vacuum. To decrease or increase a volume of vacuum takes the equal amount of energy regardless of the initial volume.
If the containers was filled with air, it would take much more energy to decrease volume than increasing the volume.

The altering volume in the containers are controlled by spokes attached to a rod or axis outside the axis of the main wheel where the containers are attached. The pull force on each spoke are equal regardless of container position - this due to the vacuum inside the containers. So there is in sum, no forces at all required to alter the volume inside them. Vacuum is vacuum, hence force equals counterforce.

That said, what will happen if we put this wheel under water?

Will the container with most volume rise, the one with no volume go down?

Will the wheel find its rest at the very position in the drawing, or will it continue to rotate (counter clockwise in this case)?

Please discuss :)

Br.

Vidar

Vidar,

Looks interesting.  Long ago a similar idea was discussed inside a thread here: http://www.overunity.com/index.php?topic=388.0 (http://www.overunity.com/index.php?topic=388.0). Post #23

Instead of using a cam or offeset hub to displace the water it's proposed to use weights as close to the axis as possible.  I think these ideas are brothers.

I have been slowly working on a build of this.  It's very difficult to prototype.  Everything has to be sealed water tight and a suitable sized water container has to be found for testing.  This makes the scale I can build and test inside the house fairly small.

Your idea interests me.  Any next steps to a build in mind?

Keep us posted.

Thanks,

Hankinator

This is just an idea that crossed my mind in a hurry. I have to calculate any time- and distance issues in one round of the wheel to determind if there is a good idea building one. Time of travel, plus the Net density of left versus right side of the wheel, plus the amount of time during one round the density differences are present, must be calculated. One direct thought of an obstacle is the effect of the spokes and how they interfere with the energy required to pull out the membrane on different places in the wheels position. I will however assume that the energy spent to decrease and increase the volume in the boxes during one revolution are cancelled out, but some friction must be taken into account. If the thing works, it will be a matter of engineering how to reduce friction to a minimum, so the total torque is sufficient to make it work even with small friction.

The buoyancy of the highest and lowest box should be equal, but the force to achieve the same volume will be less at the highest box than the lowest one. This might be the "key" to a non working device. However, the spokes are fixed, and the question is if this wheel will just cog a bit, but if the sum of forces is zero during one round (As it in, my mind, should be), cogging should not be an issue to stop the device from working.

I think this is an interesting approach to a gravity wheel.

As a start one can try to use medical injectors and lubricate them with silicon oil to reduce friction. They are capable of sustaining total vacuum for a long time - just to test out the idea.

br.

Vidar

I have done some calculations if I use medical injectors. I have a pair of those, and the friction in them are quite low if I lubricate them with silicone oil. If I manage to make a system that makes at max 10ml of emty volume, and a diameter of approx 50cm, I will have an average torque of 0,025Nm with two small 10ml medical injectors. It takes at about 50g to move the piston in the medical injector due to friction. Devided by the relationship between length of stroke and the total radius, the energy required to move the pistons should be less than the energy coming out of the torque in the wheel. This experiment should work - if I haven't missed something.

Br.

Vidar

Unfortunately it will take slightly more force to pull the 10 cc of vacuum at the bottom of the wheel than the vacuum will return at the top. This is because of the additional pressure of the water column. So the system won't move on its own.

Besides, buoyancy is just gravity under water. Consider the case of a big globule of water floating in zero-g. Now introduce a bubble into the center (with your medical syringe?)  Will the bubble "float" to the surface by buoyancy? Of course not. (It will eventually reach the surface, but not by buoyancy.) Buoyancy is an effect of gravity, nothing more or less. So, if we can agree that gravity wheels can't work, because gravity is a conservative field, by extension, buoyancy wheels can't work either, no matter how clever the mechanism.
In fact, the drag of pushing stuff through the water is very difficult to overcome in a system that could only produce forces on the order of a tenth of a quarter-Newton--even if it could produce that torque.

I am always ready to be proven wrong...so when you build it, keep me posted on the progress.

Thanks for the reply @TinselKoala. You are right about the greater pull force at the bottom compared to the one on the top.

Say we have two cylinders placed right across eachother. Each cylinder displace 5cm³ each when the alignment is vertical. Say there is 10cm between the surfaces of the diaphragms, and the diaphragm area is 1cm².

Each diaphragm in air will provide 1 kilo pull, but there is two opposite forces, so the net pull force is zero.

In water, there will be approx 1 % more pull on the bottom diaphragm than the top diaphragm. The difference in pull force in grams is 10 grams.

Say the max displacement of water is 10cm³ (5 + 5 vertical, or 0 + 10 horizontal). this is a weight of 10 grams too. Just enough to make the wheel stay still in any position.

What if it's possible to keep the design except increasing total water displacement? What will then happen?

If the water displacement is greater than the 10g of pull force difference, shouldn't then the wheel start running?

So the next question is: How to design such wheel? Is it possible to increase water displacement without increasing the distance between each diaphragm, and witout increasing diaphragm area? Which means, increasing the diaphragm stroke only.

Br.

Vidar

OK,

Lots of replies from myself here, but I would again like to challenge the reply from @TinselKoala with a drawing.

This drawing shows two identical cylinders, in the same depth in a water tank. It also shows two ways to construct the piston that keeps the vacuum inside. The difference here is in what hight in the water the main surface area of the piston is present. They are 10cm apart in hight.
Say that both pistons (ALL the blue parts) have the same weight.

The question:
We know that in these two examples, the pull force is vertical. We also know that the pressure in water is greater at greater depths. So, taken into account that both cylinders of vacuum is at the same depth, what piston have the greatest pull force - A or B?

Br.

Vidar

It seems to me that the "pull" force will be greater on the B side.
In response to your question, let me ask you one: which one of your cylinders will experience the greatest relative change in its buoyancy?
All the blue parts may have the same weight (a neat trick, but OK), but the water they displace doesn't.

Here is the idea, that came out of the previous picture.
The idea is to have a linear travel of all pistons, and where there is no atmospheric pressure or water pressure that acts on the pistons differently depending of depth in the water.
All cylinders are fixed to another wheel which is not showing in the drawing.

Now @TinselKoala, what do you think?


(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Flyd-interior.no%2Fdiv%2FVacuum-motor.jpg&hash=76844542d3a6b0ece5b496ca5baed3fbda783925)
Br.

Vidar

I think it's getting pretty complicated.
;)
You make nice drawings, though. I wonder if it could be animated?
Do you know about Phun?
http://www.phunland.com/wiki/Home (http://www.phunland.com/wiki/Home)
You might be able to get this working in Phun, if you play around a little.

I still don't think it will work in real life, though, because I don't think the overbalanced (or overbuoyant) condition will continue all the way around the wheel. There will be a point (several actually) where all forces  balance and the wheel will stop, because inertia won't be able to carry it past that point.

I see that CLaNZeR is building a similar idea on another thread. If anybody can get one working, it will be he.

(EDIT to add there isn't going to be much power available from buoyancy changes, as you no doubt have already calculated. All that complex mechanism, and the drag of pulling stuff through the water, subtracts a lot of power from the system, and the bigger (more buoyant) it gets, the more drag there is...)

Say you werre to create hydrogen at the very bottom can using spark gap inside trapped water , would it change the pressure inside the can ?

Since you change the pressure it would try to go up to release it , once at the top it then release and loose its hydrogen where you can capture it to recycle it . Would these action yield enough energy to rotate the waterwheel ?

Well, if you have enough energy to make a spark why not just use that energy to turn the device? Or better yet, just use the spark energy for something else, and forget the buoyancy device...

No, that can't be what you meant.

Well people already create hydrogen , when it creates it the bubble need to "fight" trough pressurized water , the buyoancy system is a system that use the difference in pressure , so if you were to add a small buyoancy system to an already working hydrogen  system it's energy that is freely tapped ( the action of the hydrogen fighting trough water atom to reach the top of the can ). Yea i don't know if it would be worth it ...

Quote from: TinselKoala on November 11, 2008, 05:40:23 PM
I think it's getting pretty complicated.
;)
You make nice drawings, though. I wonder if it could be animated?
Do you know about Phun?
http://www.phunland.com/wiki/Home (http://www.phunland.com/wiki/Home)
You might be able to get this working in Phun, if you play around a little.

I still don't think it will work in real life, though, because I don't think the overbalanced (or overbuoyant) condition will continue all the way around the wheel. There will be a point (several actually) where all forces  balance and the wheel will stop, because inertia won't be able to carry it past that point.

I see that CLaNZeR is building a similar idea on another thread. If anybody can get one working, it will be he.

(EDIT to add there isn't going to be much power available from buoyancy changes, as you no doubt have already calculated. All that complex mechanism, and the drag of pulling stuff through the water, subtracts a lot of power from the system, and the bigger (more buoyant) it gets, the more drag there is...)

Hi,

I some times over do it with my drawings ;D. The idea is to have membranes/pistons that doesn't get influenced by waterdepth, as you pointed out in my first drawing. How the practical solution is, can ofcourse be different. Using large gears and tooth rods instead. Less loss.
Drag in water is mentioned. This drag is relativly low loss, even if it is drag. Drag isn't pure friction, and does not create loss as heat very much. It is just a little time consuming.

Anyway, I have tried to understand how this machine does NOT work. If it worked, it would be to good to be true - honestly. So I have twisted my mind to find the flaw in the design.

For those who doesn't understand the principle, try to look at the two pistons in the four vacuum chambres, both vertical one and horizontal one. As the red vacuum chambres rotat around its axis, another axis to the left of the vacuum chambre axis is controlling the movement of the pistons by the brown mechanism. Both vacuum chambres axis and piston control axis is linked so they both run in the same speed in the same directions.

Br.

Vidar

Now I have made a simple animation.

here is the construction:

The red tube contains only vacuum.
The black "thing" is a piston which is controlled by a mechanism that moves it inside the vacuum tube. That mechanism should be a magnetic hinge or lock as the vacuum tube is sealed.
It is a vacuum tube so the piston inside can move more freely. With air, it would cost a lot of friction due to unlinearity in the compression and decompression process.

Be focused:
Look close to the black piston. Its centre and axis is aways following the black circle.
Look close to the red vacuum tube. Its centre and axis is fixed at the rightmost side of the circle.
Look close to the sum of all movements, and how the free space inside the tube is distributed at any time.
The black circle is just the path of the magnetic positioning lock of the piston, where the piston centre is following this path, and are not fixed to the main axis where the vacuum tube are fixed.
The piston will therefor slide from one side to the other if the vacuum tube as it turns around.
When the vacuum tube is in vertical position there is the exact same volume of emty space at the top and the bottom. When it is horizontal, there is no emty space at the left side, and only emty space at the right side.

The idea of how it works:
The wheel is not working by the weight of the black piston alone - at least I assume it doesn't. It requires zero energy to complete one round, but it also don't give any energy.
The clue is now how this machine keeps the buoyancy on the rightmost side of the red vacuum tube, which means this machine must run in something thicker than air - water or other liquids.
As assumed, the piston does not provide any work at all. But try to see how a tube with lots of empty space on the right side will act in water - it will be forced to flip over counter clockwise due to buoyancy at the right side.
There is allways more empty space at the right side of the wheel, and it will therefor allways try to stabilize itself by flipping over to a vertical position.
The idea is then how to move mass to one side to another without using energy, so the final energy consumtion is therefor negative - which in turn means more energy to take out than what it consumes -> OU.

OK! This is not a claim, but If you can't find a reason why this thing shouldnt work, you can try building one and see what's happening.

Now that's a fascinating configuration.
I disagree, though, that it is still a buoyancy device, because the total buoyancy is not changing, it's just moving from side to side. The total volume displaced by the red tube is constant, as is its total mass. It is only the mass (or buoyancy) that is shifting position.
So, if it would run at all, it should run in air, or vacuum, because it is really a gravity wheel "powered" by the weight shift.
Your animation makes that very clear. It's in effect a single tube with a sliding weight in it, and instead of sliding the weight magnetically or with air pressure or some other way, you are sliding it mechanically with the cycloidal linkage.
So this one won't work for the same reasons other simple gravity wheels won't work: friction in the mechanism, combined with inability to maintain imbalance for more than a single cycle.

Gravity is conservative; many of the gravity motor designs are so efficient that, if only for that tiny bit of friction, they might go on running for a while under no-load conditions. But gravity only gives back what you put in, so as soon as there's the tiniest drag, or power-take-off, the system rapidly grinds to a halt--because it only possesses the energy you put into it to start it in the first place.

Quote from: TinselKoala on November 16, 2008, 05:54:14 PM
Now that's a fascinating configuration.
I disagree, though, that it is still a buoyancy device, because the total buoyancy is not changing, it's just moving from side to side. The total volume displaced by the red tube is constant, as is its total mass. It is only the mass (or buoyancy) that is shifting position.
So, if it would run at all, it should run in air, or vacuum, because it is really a gravity wheel "powered" by the weight shift.
Your animation makes that very clear. It's in effect a single tube with a sliding weight in it, and instead of sliding the weight magnetically or with air pressure or some other way, you are sliding it mechanically with the cycloidal linkage.
So this one won't work for the same reasons other simple gravity wheels won't work: friction in the mechanism, combined with inability to maintain imbalance for more than a single cycle.

Gravity is conservative; many of the gravity motor designs are so efficient that, if only for that tiny bit of friction, they might go on running for a while under no-load conditions. But gravity only gives back what you put in, so as soon as there's the tiniest drag, or power-take-off, the system rapidly grinds to a halt--because it only possesses the energy you put into it to start it in the first place.

Thanks for the input.

I will calculate the torque provided by the weight of the pistons and buoyancy by using vectors. I havent done that this far. I guess the answer is zero, but hope for more. To explain it doesn't work because it can't, isn't an answer I am looking for. So please provide calculations if you, or anyone for that matter, are willing to spend more time on this :)

If it can accelerate with no load, I have found what I'm searching for.

br.

Vidar

So, if you still think buoyancy is important to the most recent design, answer me this: when the empty part of the tube is at the top of its travel, it has displaced an equal volume of water downward. This is where the buoyancy comes from! So when the tube rotates around the cycle, at some point it must raise this same amount of water back up, in order for the cycle to repeat.
Where does the work come from to raise this water up?

(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%257Edsimanek%2Fmuseum%2Ffloat-pm.gif&hash=6c5180edcf9e609a5ea7de951bfc1edd3e046ad9)

I believe your current design is almost the same in principle as the above, "Buoyancy Motor #3" at
http://www.lhup.edu/~dsimanek/museum/unwork.htm#wheels (http://www.lhup.edu/~dsimanek/museum/unwork.htm#wheels)
And here's a mathematical analysis of why it can't work, which I believe applies exactly to your design as well.
http://www.lhup.edu/~dsimanek/museum/bmitch.htm (http://www.lhup.edu/~dsimanek/museum/bmitch.htm)

But I will be interested to see your calculations, as well, of course.

Quote from: TinselKoala on November 16, 2008, 06:41:08 PM
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fwww.lhup.edu%2F%257Edsimanek%2Fmuseum%2Ffloat-pm.gif&hash=6c5180edcf9e609a5ea7de951bfc1edd3e046ad9)

I believe your current design is almost the same in principle as the above, "Buoyancy Motor #3" at
http://www.lhup.edu/~dsimanek/museum/unwork.htm#wheels (http://www.lhup.edu/~dsimanek/museum/unwork.htm#wheels)
And here's a mathematical analysis of why it can't work, which I believe applies exactly to your design as well.
http://www.lhup.edu/~dsimanek/museum/bmitch.htm (http://www.lhup.edu/~dsimanek/museum/bmitch.htm)

But I will be interested to see your calculations, as well, of course.

Hi

This inventon looks nice :) It seems to work at first, but doesnt after some calculations. I also believe they assumed the bubble wasn't filled with vacuum, in order to add the mass of air in the bubble. Would the facts change if there was vacuum in the bubble? After some thoughts, I don't think so. As the bubble is fixed on one or another side, the bubble will be forced upwards in any case.

In my design it might be the same deal, but the "bubble" in the vacuum tube is allways biggest at the right side - this is what made me blind about that design. And if the displacement of water is allways equal on both sides, as opposed to the invention above, the buoyancy on the right side should also be greatest. The flaw, as I see it, is how the wheel itself holds back the piston, so at the end the buoyancy on both sides are equal anyway. Case closed I guess...

Br.

Vidar

I have done some calculations.

Piston data:
length: 1m
weight: 1kilo

Vacuumtube data:
Length: 2m
Volume: 2 litres
Vacuum volume (Volume minus piston). 1 litre


Torque by the piston at:
0^o : -4,91 Nm
15^o : -4,74 Nm
30^o : -4,25 Nm
45^o : -3,47 Nm
60^o : -2,45 Nm
75^o : -1,27 Nm
90^o : 0,00 Nm
105^o : 1,27 Nm
120^o : 2,45 Nm
135^o : 3,47 Nm
150^o : 4,25 Nm
165^o : 4,74 Nm
180^o : 4,91 Nm
195^o : 4,74 Nm
210^o : 4,25 Nm
225^o : 3,47 Nm
240^o : 2,45 Nm
255^o : 1,27 Nm
270^o : 0,00 Nm
285^o : -1,27 Nm
300^o : -2,45 Nm
315^o : -3,47 Nm
330^o : -4,25 Nm
345^o : -4,74 Nm

Average piston torque: 0,00 Nm


Torque of the vacuum volume at:

0^o : 0 Nm
15^o : 0,52Nm
30^o : 1,54Nm
45^o : 2,56Nm
60^o : 3,59Nm
75^o : 4,55Nm
90^o : 4,91Nm
105^o : 4,55Nm
120^o : 3,59Nm
135^o : 2,56Nm
150^o : 1,54Nm
165^o : 0,52Nm
180^o : 0Nm

Average buoyancy torque of the vacuum volume in one revolution is 2,53Nm

Total average torque of the system is:
Vacuum Torque - Piston torque =
2,53Nm - 0 Nm = 2,53Nm

Not finished yet:
There is a link between the wheel and the mechanism that moves the piston. 270^o is the very bottom of the circular piston path, while 90^o is the very top ofthe circular piston path.
The pistons weight will force the wheel backwards from <270^o to >90^o. The pistons weight will also force the wheel forward from <90^o to >270^o with the same force.

Hmmmm ... my head hurts. Please help!!

I still, after explanations, have changed my mind, and cannot figure out how this wheel don't work.

I am sitting here with 2,53Nm left after the calculations - I know I shouldn't accodring to the laws of physics. I have concidered that the centre of each vacuum volume is closer to the axis than the ends of the tubes at any time, and that this centre is changing all the time as the wheel turns around. Just to be sure that the torque in Nm should be correct for each 15^o steps. The pistons torque is only affected by the angle, and where on the circular path the centre of the piston is present, in proportion to gravity direction (which is pretty much vertical).

If we use  several cylinders, say 12 of them, placed at every 30^o to reduce cogging. Say if we doubble the length of tube and piston, and increase the diameter to 50cm. The total torque will be 48 000Nm - What a dream :)

Br.

Vidar