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Author Topic: Under water gravity wheel  (Read 15140 times)

Low-Q

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Under water gravity wheel
« on: October 15, 2011, 11:37:04 PM »
Hi,

I've made a drawing of a gravity wheel idea. I like the idea, but I cannot imagine 100% how it should work, or not work.

There is two weights, one on each see saw. These are pivoted with the green dots.
In that position there is an expanded air volume at the left side, and no air volume on the right side. The configuration of the weights and the pivots makes sure that it happens. With this configuration there will be more buoyancy on the left side than the right side.

The arrangement can be copied in numbers in order to even possible uneven run.

What do you think?

Vidar

Low-Q

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Re: Under water gravity wheel
« Reply #1 on: October 16, 2011, 12:10:49 AM »
Here is an example with more of those arrangements.

CuriousChris

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Re: Under water gravity wheel
« Reply #2 on: October 16, 2011, 12:25:40 AM »
It is an intriguing concept well worth a deeper look. My initial objection was it needs to get over the 12 oclock position to continue, which I doubted it would, but maybe it will.

some points.
the water pressure is higher at the bottom. so pushing the air down will take more effort than appears in the diagram.
at the very apex (12 oclock) there is no energy available to push the air down.
at 12:01 the energy available is very small, will it be enough?

Redraw your diagram at 12 oclock and see if it still looks feasible.

onthecuttingedge2005

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Re: Under water gravity wheel
« Reply #3 on: October 16, 2011, 12:24:41 PM »

Low-Q

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Re: Under water gravity wheel
« Reply #4 on: October 16, 2011, 09:15:17 PM »
It is an intriguing concept well worth a deeper look. My initial objection was it needs to get over the 12 oclock position to continue, which I doubted it would, but maybe it will.

some points.
the water pressure is higher at the bottom. so pushing the air down will take more effort than appears in the diagram.
at the very apex (12 oclock) there is no energy available to push the air down.
at 12:01 the energy available is very small, will it be enough?

Redraw your diagram at 12 oclock and see if it still looks feasible.
At 12 o'clock the bottom cylinder will be "inflated" and the top will not. The main concept is to have the majority of air volume on the left side. If it is, there will be unbalance in the buoyancy effect which will force the wheel to rotate. However, the weights have to change positions all the time while following the rotation, so they could counter balance the wheel. I have started to make some simulations in Phun. I have not yet succeed to make an air-tight cylinder. I will figure something out.

Vidar

Low-Q

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Re: Under water gravity wheel
« Reply #5 on: October 16, 2011, 10:34:47 PM »
I have uploaded a video on youtube from Phun. The video is for visualizion and concept only.

There is somewhat cylinders in the videos which is increasing and decreasing volume due to weights.

http://www.youtube.com/watch?v=OLk8yQ8NO3k

Vidar

CuriousChris

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Re: Under water gravity wheel
« Reply #6 on: October 17, 2011, 12:56:05 AM »
At 12 o'clock the bottom cylinder will be "inflated" and the top will not. The main concept is to have the majority of air volume on the left side. If it is, there will be unbalance in the buoyancy effect which will force the wheel to rotate. However, the weights have to change positions all the time while following the rotation, so they could counter balance the wheel. I have started to make some simulations in Phun. I have not yet succeed to make an air-tight cylinder. I will figure something out.

Vidar

Actually at 12 o'clock the top will be inflated and the bottom will be deflated, otherwise why did the weight rise to the surface? It must be inflated. The video is misleading in that it shows the air exchange starting to take place at about 11 o'clock. but there is no actual reason for that exchange to take place.  In reality, at 12:01 the top will start to deflate forcing air through the connecting tube to the bottom. by 3:00 o'clock the process is complete. and the newly inflated reservoir is now dragging the weight to the top.

This is where the problem occurs, the weight lags the air reservoir by a few degrees. so when the air reservoir is at its highest position the weight is behind it and therefore cannot apply any pressure to the air reservoir to force the air into the lower reservoir.

There is a second problem, not quite so obvious. at the 12 o'clock position when drawn through the centre of gravity. (approximately through the centre of the top reservoir and through the central fulcrum) the top weight is behind the top reservoir by a few degrees as previously stated, but the bottom weight is almost at exactly bottom dead centre. therefore it provides no torque, it has already reached its lowest energy state. Therefore the top weight will actually cause the mechanism to rotate counter clockwise slightly .

The following may help or hinder, if it confuses ignore it, but it helped me a lot.
+++++++++
To help you visualise the problem draw a V. The top of the left arm of the V is the weight, the top right is the 'counter arm', and the space between is the air reservoir. The point of the v is the same, but compressed together as there is no air. Draw a line intersecting the point of the V and midway between the arms. This is the centre line and is pointing towards 12 o'clock when the system is at its highest point. An upside down capital A may be even better.

Now rotate the mechanism a couple of degrees anticlockwise, but leave the centre line vertical, that is the true centre line. it doesn't quite go through the middle of the bottom weight nor the centre of the top reservoir.
+++++++++

Thank you for the video, it helped a lot to isolate the issues I suspected would occur.

 CC

Low-Q

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Re: Under water gravity wheel
« Reply #7 on: October 17, 2011, 07:22:42 AM »
Actually at 12 o'clock the top will be inflated and the bottom will be deflated, otherwise why did the weight rise to the surface? It must be inflated. The video is misleading in that it shows the air exchange starting to take place at about 11 o'clock. but there is no actual reason for that exchange to take place.  In reality, at 12:01 the top will start to deflate forcing air through the connecting tube to the bottom. by 3:00 o'clock the process is complete. and the newly inflated reservoir is now dragging the weight to the top.

This is where the problem occurs, the weight lags the air reservoir by a few degrees. so when the air reservoir is at its highest position the weight is behind it and therefore cannot apply any pressure to the air reservoir to force the air into the lower reservoir.

There is a second problem, not quite so obvious. at the 12 o'clock position when drawn through the centre of gravity. (approximately through the centre of the top reservoir and through the central fulcrum) the top weight is behind the top reservoir by a few degrees as previously stated, but the bottom weight is almost at exactly bottom dead centre. therefore it provides no torque, it has already reached its lowest energy state. Therefore the top weight will actually cause the mechanism to rotate counter clockwise slightly .

The following may help or hinder, if it confuses ignore it, but it helped me a lot.
+++++++++
To help you visualise the problem draw a V. The top of the left arm of the V is the weight, the top right is the 'counter arm', and the space between is the air reservoir. The point of the v is the same, but compressed together as there is no air. Draw a line intersecting the point of the V and midway between the arms. This is the centre line and is pointing towards 12 o'clock when the system is at its highest point. An upside down capital A may be even better.

Now rotate the mechanism a couple of degrees anticlockwise, but leave the centre line vertical, that is the true centre line. it doesn't quite go through the middle of the bottom weight nor the centre of the top reservoir.
+++++++++

Thank you for the video, it helped a lot to isolate the issues I suspected would occur.

 CC
Thanks for correcting me. I meant the bottom is deflated and the top is inflated. Sorry my mistake. In the video there is no water, no pressure which deflate the bottom volume, no air drag, so it is an ideal situation to see the stability.

Vidar

onthecuttingedge2005

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Re: Under water gravity wheel
« Reply #8 on: October 17, 2011, 07:57:00 AM »
sorry but the dam Hydroelectric plant is the most efficient gravity fed power plant known to man. nobody has matched it so far.

CuriousChris

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Re: Under water gravity wheel
« Reply #9 on: October 17, 2011, 08:00:42 AM »
As I see it a pair of weights and air reservoirs wont work. but what about two pairs?

Normally I tend to believe if you cant get a gravity wheel to work with one pair of weights adding more does not help, Usually the inventor gets excited (often me) and thinks Eureka that's it. but usually its fools gold.

This case has me intrigued though, will two sets work? I am trying to visualise it.

The part I think that will cause the problem is the displacement. to pump the air from the upper chamber to the lower chamber you must displace the equivalent volume of water. I think with careful selection of volumes and weights the displacement problem may be addressed. trial and error is probably the best way to go.

Start small. To create a water tight container you could use two balloons connected by a straw (4mm garden riser would be perfect) a small can could contain the balloons, the balloons act as a membrane.

I'll be fascinated to see how you get along, I know at some point equilibrium must be reached. I just can't work out where it would be in my head.

With one set equilibrium is at 12 and 6, but with 2 sets when one set is at equilibrium, the other set is at 3 and 9 and going through maximum transfer. which should push the other pair over equilibrium. Is this Eureka?

CC




CuriousChris

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Re: Under water gravity wheel
« Reply #10 on: October 17, 2011, 08:08:15 AM »
sorry but the dam Hydroelectric plant is the most efficient gravity fed power plant known to man. nobody has matched it so far.

its also the most efficient solar powered power plant known :)

I might start another thread on this a little later I have been musing about using water to store energy in an unconventional way.

CC

Low-Q

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Re: Under water gravity wheel
« Reply #11 on: October 17, 2011, 09:38:05 AM »
sorry but the dam Hydroelectric plant is the most efficient gravity fed power plant known to man. nobody has matched it so far.
Agreed. It is also the most expensive power we have in Norway at the moment - even if all our hydro power plants is "green", litterally without any toxic waste. We sell the cheap excess power to other countries at summer time when the water reservoir is full, and buy expensive power from other countries in the winter when the water reservoir is almost emty - just to keep the cost within the borders high all year around. It does not make sense, but business is business.

Vidar

Low-Q

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Re: Under water gravity wheel
« Reply #12 on: October 17, 2011, 09:48:15 AM »
As I see it a pair of weights and air reservoirs wont work. but what about two pairs?

Normally I tend to believe if you cant get a gravity wheel to work with one pair of weights adding more does not help, Usually the inventor gets excited (often me) and thinks Eureka that's it. but usually its fools gold.

This case has me intrigued though, will two sets work? I am trying to visualise it.

The part I think that will cause the problem is the displacement. to pump the air from the upper chamber to the lower chamber you must displace the equivalent volume of water. I think with careful selection of volumes and weights the displacement problem may be addressed. trial and error is probably the best way to go.

Start small. To create a water tight container you could use two balloons connected by a straw (4mm garden riser would be perfect) a small can could contain the balloons, the balloons act as a membrane.

I'll be fascinated to see how you get along, I know at some point equilibrium must be reached. I just can't work out where it would be in my head.

With one set equilibrium is at 12 and 6, but with 2 sets when one set is at equilibrium, the other set is at 3 and 9 and going through maximum transfer. which should push the other pair over equilibrium. Is this Eureka?

CC
You're right. If one doesn't work, adding more non-workable ones doesn't help. If the consept doesn't work, it will never work.

A small model will help to understand how these mechanisms work. I have no place to make one, so Phun is the closest I can get...

Vidar

CuriousChris

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Re: Under water gravity wheel
« Reply #13 on: November 17, 2011, 09:40:17 AM »
Hi Low-Q

Thought you may be interested in this, I found it the other day

http://www.lhup.edu/~dsimanek/museum/patents.htm

search for patent 1330, about halfway down the page :)

CC