Under water gravity wheel

[Low-Q]

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]

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]

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]

sorry but the dam Hydroelectric plant is the most efficient gravity fed power plant known to man. nobody has matched it so far.

[CuriousChris]

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