Bhaskara's Wheel

[vidbid]

Bhaskara's Wheel aka Overbalanced Wheel



Bhaskara

Bhaskara's wheel was invented in 1150 by Bhaskara, an Indian mathematician, in an attempt to create a perpetual motion machine. The Wheel consisted of curved or tilted test-tubes partially filled with mercury. Once in motion, the mercury would flow from one side of the tubes to another, thus forcing the wheel to continue motion.

Perpetual Motion - Free Energy - Home Made - YouTube

Perpetual Motion Machine - How it was made - Free energy - YouTube

Perpetual Motion Machine Update - Free Energy - Home Made - Longer Run Time - YouTube

You could call this device a Differential Density Drive.

[Peter Lindemann]

Center of Gravity Analysis

Hey vidbid et al,

Yes, the YouTube videos look good, but a simple "center of gravity" analysis of the device shows that the center of gravity moves in a tight, uniform oval just below the axle. It shows that the weight of the moving fluids are distributed evenly on either side of the axle, strongly suggesting that the wheel is never heavier on one side or the other. Here's an image of the simple analysis:



So, the wheel is most likely not "driven" by gravity. That leaves the possibility of the wheel being "driven" by the differential of the liquid "sloshing" to the outside on the left side of the wheel producing more force than the liquid "sloshing" backwards to the inside on the right side of the wheel. I consider this "unlikely" as well.

Considering these two realities, it strongly suggests that it "shouldn't work" because we cannot identify a non-uniform force acting more on one side of the wheel than on the other. In the absence of this force, I believe that it probably doesn't work.

If it doesn't really work, then the films are cleverly faked.

Peter

[citfta]

Actually it is pretty easy to fake. All it takes is an air hose and some simple editing of the sound track to edit out the sound of the blowing air. Simply erase the original sound track and insert a talk-over. In one of the videos showing this device you can even see the air blowing the shirt of the presenter.

[vidbid]

A Not-so-new Kid on the Block

Okay, PBR Streetgang, point taken, but here's a not-so-new video for you, showing another liquid overbalanced wheel built by a kid.

[VIDEO]https://www.youtube.com/watch?v=ePH6vkqAeB4[/VIDEO]

Perpetual Motion - Free Energy - YouTube

https://www.youtube.com/watch?v=ePH6vkqAeB4

Enjoy.

[RAMSET]

That is the one where the Boy has his shirt Blowing at 56-58

Chet K

[vidbid]

That is the one where the Boy has his shirt Blowing at 56-58

Chet K

VERY GOOD OBSERVATION!

The shirt is fluttering at that time index.

[BroMikey]

Mojo risin

I think in some cases light sensitive fluids can play a roll in these
perpetual motion tricks.

Here is a small collection of working units.

https://www.youtube.com/watch?v=zTIatEe86R8

[VIDEO]https://www.youtube.com/watch?v=zTIatEe86R8[/VIDEO]

[serendipitor]

There is one problem with all the over-balanced wheels of this type. They all try to take weight from the ascending side and throw it outwards on the descending side, using centrifugal force.

Think of figure skaters. When they throw their weight outwards, they *slow down*. When they pull inwards they speed up. In such a gravity wheel, the very mechanism intended to keep the wheel moving causes it to also slow down. Overall, the momentum exchange is a wash, and energy neutral at best.

[rickoff]

Hi folks,

Just though I'd revive this thread for the simple reason that the notion of an overbalanced wheel interests me. I tend to agree with Peter's analysis, except that if the wheel is equally balanced at the left and right sides then the lines drawn should converge at the axle, yet most of the convergence points are to the right of center. That would seem to indicate that the wheel is too out-of-balance to spin for very long before coming to a stop.

There does appear to be an overbalance at the left side if one refers to either of the below images, which I have labeled and explained, though this may not be entirely accurate either. I do think it is worth looking at, though, as a point for further discussion.


In the above photo you can see that the fluid in bottle D is poised to rush outward, while the fluid in bottle I will be rushing inward. In other words, this is where the "sloshing" effect is initiated, and the actual effect of that sloshing is seen below.

Of course the effect which bottle A has on rotation is very small at this point, when compared to bottles B or C, so shouldn't be thought of as a rotational asset, however neither is it a negative force to rotation. With A-F and D-I pairs pretty much canceling their effects, you still have two overbalanced pairs (B-G and C-H) and only one underbalanced pair (E-J) at the left side.

All this being said, it appears to me that the gaps between each bottle cap and the adjacent bottle's side are not equal, which could in fact help explain why the intersecting lines in Peter's analysis were not quite centered on the axle. This situation would set up an imbalance whether the unit is at rest or in motion, so doesn't seem helpful, and in fact makes me wonder how this could possibly run continuously. Before the unit is tested, it should be well balanced, and the only way to determine that would be to spin it and see where it comes to rest. If it always stops at the same location then the heavy point is at 6 o'clock and one would have to add a counterweight at 12 o'clock to balance the wheel. The counterweight could be a weight added to the wheel, such as a small magnet, or one could add more water to one or two bottles, but of course this second option would be more difficult in this build since the cap is pressed tightly against the perimeter of the wheel.

I noticed that none of the video links shown in the #1 post on this thread are working, so I captured the above still photos from video found at this link. The wheel does keep turning for 5 minutes, just as is claimed, however I noticed that there were two recurring chirping sounds (a higher pitched one followed by a lower pitched one 2.5 seconds later. This may have been a bird outside the garage, and it is not all that unusual for a bird to make two differing pitched chirps close together. What I did find unusual, though, was the fact that when I timed the interval starting immediately upon hearing the high pitched chirp, and ending immediately after each successive high pitched chirp, it always timed out at 19.6 seconds during this 5 minute run. This, of course, is pretty much a dead giveaway that the video was edited to loop within that 19.6 second time frame, which could have started at the chirp or anywhere in between chirps. It's funny that the editor didn't notice the chirps, though perhaps he did and simply thought no one else would notice. If balanced fairly well, it wouldn't be difficult to spin the wheel and have it rotate at a seemingly consistent rpm for just under 20 seconds. It's really too bad that some people will fake their results rather than just showing what actually happens and looking for ways to improve on those results if possible.

While this build is not a continuous runner, it does seem that the sloshing hammer effect, as shown above, does perhaps have potential, especially if a longer tapered bottle neck is used. The usefulness of such a device, if it could be made to continuously run, would probably be almost nil unless it could be constructed on a much grander scale, but one wouldn't even think of doing that unless a relatively small scale model such as this one could be made to function as desired. Certainly there are more promising projects to pursue, so there probably won't be much interest in starting with this build and somehow making it work, but then again there aren't that many projects one can work on that would cost so little.

Best regards to all,

Rick

[fan1701]

I believe that custom containers specially shaped could enhance this effect greatly.

[rickoff]

I believe that custom containers specially shaped could enhance this effect greatly.

Yes, I'm sure that the sloshing effect could be somewhat enhanced with an improved container, but how about redesigning the perimeter of the wheel itself to incorporate a tubular waterway? Containers moving counterclockwise from the bottom would be dumping their contents into the perimeter tube at position of bottle I in the below photo, thus going empty, while containers moving counterclockwise from the top position would be quickly filled by the water in the tube when at position of bottle D.


Of course, in such an arrangement, you would not want the water to freely circulate in the tube - the tube would be divided into sections with barrier walls, with enough space between barriers to hold the entire contents drained from each container at the right side of the wheel. That drained water, held at the perimeter between two barriers, would then travel up and over the top of the wheel and dump into each container vessel when the vessel reaches the D position. The design of the perimeter tube would need to be evaluated based on being either wide enough, or thick enough, to hold the full contents of a container vessel (or the desired amount for best performance) between barrier baffles. It probably wouldn't be necessary to have as many container vessels as are shown in the above photo. Perhaps six would be ideal, spaced at 60 degree intervals, and you wouldn't want to use anything other than consistent diameter container vessels (such as PVC pipe, for example) as this would allow quick filling as well as quick draining with no restriction in either direction.

Just picture it while looking at the above photo. With the contents of container vessels D, C, and B providing the counterclockwise rotational force, and E,F,G, H, and I emptied to the perimeter tube between their respective barrier baffles, only containers A and J would be resisting rotation as they move towards the container I position, where each is quickly drained to the perimeter tube. Thus, it seems that you would have a relatively strong overbalance condition at the left half of the wheel which would ensure continuous rotation. It's past my usual bedtime, and perhaps I'm too tired to be thinking straight, but at this time it all seems logical.

Here's a suggestion as to how the container vessels could be made from PVC pipe and attached to either or both sides of a rectangular cross section perimeter tube. This is of course a very crude and simplified drawing, and is only offered to illustrate the basic idea, without anything being drawn to an exacting scale. I don't know if I can find the time to make a better drawing, so if anyone would like to give that a shot then feel welcome to do so.

[rickoff]

A basic understanding of my water wheel concept

I finally found enough time to make some improved drawings to further explain my concept for an improved overbalanced water wheel. From a visual aspect, I believe this is workable, and that my design concept should solve all the problems which made the soda bottle water wheel incapable of continuous rotation. Actually, the only two factors that had any positive effect in the soda bottle water wheel were as follows:
1. The angling of the soda bottles allowed for outflow of the liquid to proceed soon after each bottle passed over the top, and at a position between bottles E and D in the top photo of my previous post. As rotation continued past the bottom of the wheel, this same angling allowed the liquid to flow inward towards the wheel's rim much sooner than if the bottle had been oriented 90 degrees to the wheel rim. Thus, we want to make use of similar but improved angling.
2. The sloshing effect, as shown in bottle D of that same photo, was stronger in bottle D with the liquid moving outward than in bottle I (D's counterpart) and shows the potential for positive use of this sloshing hammer effect. Thus an improved design should make even better use of this effect by maintaining a strong hammer effect at one side of the wheel while eradicating (or at least minimizing) the hammer effect at the other side.

In looking at the bottom diagram of my previous post, you see a very basic drawing of my idea for incorporating a circular waterway at the perimeter of the wheel which is divided, by barriers, into as many sections as the number of tubes that would be incorporated. One tube is shown at the left side of that drawing, for simplicity's sake, as I didn't have enough time to show a fully decked out wheel. I didn't give the waterway enough height in that drawing, and didn't have enough time to alter it, but the figure at the right side of that illustration shows the rectangular shaped waterway at a more realistic height. This waterway section, or containment/feeder vessel, with a length of tubing (or pipe) attached to the side of the vessel by means of a 90 degree elbow is very important for taking full advantage of the sloshing hammer effect during feeder vessel outflow to the tube, as well as avoiding that effect during flow back into the vessel. To better understand how this will work quite well in my concept, I'll show the feeder vessel feeding the tube on the left side of the below drawing, and then show, at the right side, what happens when the vessel and tube are inverted. What you will readily notice is that the liquid from the feeder vessel is able to fall rapidly to the bottom of the tube, creating a nice hammer effect which aids rotation, while there is practically no counter-productive hammer effect created by the water which is being sent back to the vessel from the inverted tube. That's because the 90 degree elbow, which is of course curved on the inside, smoothly redirects the flow of incoming water against the side of the vessel, rather than it's bottom (which is actually the inverted vessel's top).



Now of course the above figures are drawn as simply as possible for the purposes of illustrating the hammer and non-hammer effects of the design, and this is why I have shown these effects in a straight-on view. In reality, the hammer effect shown above occurs at the angular position occupied by soda bottle D in the photo shown at top of my previous post (if rotation is counter-clockwise), while the non-hammer effect shown above can be related to the position of bottle I. If rotation is clockwise then of course the opposite would be true.

I believe that this validates the usefulness of these design features, and tomorrow I will offer an illustration of the concept wheel showing all the containment/feeder vessels and their related tube mount borings.

Best to all,

Rick

[rickoff]

The wheel with perimeter vessels shown.

As promised, the below drawing further illustrates my overbalanced water wheel concept, showing the wheel and all feeder/receiver vessels in place at the perimeter. Each vessel is separated from the next one by a dotted line which represents a barrier, and thus each vessel is fully contained between two barriers, as well as by its top and bottom surfaces. The only opening to each vessel is a single hole bored in the side of the vessel, as close as possible to the vessel's bottom and left barrier surfaces. These holes represent the best possible locations for side-mounting the tubes or pipes which will interact with the feeder/receiver vessels. As you can see, I chose to divide the perimeter of the wheel into twelve equal sections in order to maximize the potential of the overbalancing effects.



The aim of this design concept, of course, is to provide a sufficient volume of water in each vessel to fill its attached tube to a specific desired level, by gravity flow, when the tube is lower than the vessel. I'm going to start out by assuming that we'd want to fill a tube no more than half full along its length, and that the highest water mark of a thusly filled tube should not be above the outermost perimeter of a vessel. This would ensure that all the weight of the water contained in a tube would be overbalancing the weight of water contained in a counterpart vessel at the opposite side of the wheel. The longer a tube is, the greater the sloshing water hammer effect will be, and the greater the leverage its water fill will be able to provide for turning the wheel, but of course an overly long tube would take longer to empty its contents back into the receiver vessel than a tube of shorter length, so that would be an important design consideration. Naturally, the inside diameter of a tube, if chosen as large as possible for the build, would hold more water weight when filled to its desired level, and also empty its contents faster than a smaller diameter tube. The appropriate length and inside diameter of each tube, and the length, height, and width of each feeder/receiver vessel will of course be dependent upon the diameter chosen for the wheel. Certainly a small wheel having a diameter of 12 inches could be constructed to test the concept, though of course a wheel capable of being harnessed to provide any truly useful amount of leverage and work would probably need to be at least 3 to 4 feet or more in diameter.

I hope this offers readers an easily understandable explanation of the basic design principles involved in my overbalanced water wheel concept. In my next post I will show the same wheel with the tubes added and oriented at the best possible angle.

Should any questions arise, please feel free to ask them of me and I'll do my best to offer a satisfactory answer.

Best regards,

Rick

[rickoff]

Addition of the tubes to the design concept

Below is my drawing of the overbalance water wheel as shown in my previous post, but with the tubes added. This should make it much easier for anyone to envision what would occur at any point in the rotation. In the drawing's current state, no water has been added. The idea here is to show the completed, or near complete, build overview.



With no water added, and if all the tubes are cut to the same length and fastened to the feeder/receiver vessels at precisely the same angle, the wheel should be in perfect balance. Of course finding the wheel to actually be in perfect balance after assembly would be quite rare. At this stage of the build we would want to test for such balance and make any adjustments necessary to achieve this. By spinning the wheel, allowing it to come to a stop on its own, and noting the position where it stops by applying a mark at the 6 o'clock position of the perimeter, we can then spin the wheel again to determine whether it stops at the same location. If so, this would tell us that this is the heaviest point of imbalance. To correct such an imbalance, we would need to add just enough weight at the 12 o'clock position of the wheel's perimeter to provide counterbalance to eliminate the out of balance effect. When correctly balanced, the wheel should rarely stop at the same place.

Once this balancing is properly accomplished, we would be ready to add the desired amount of water to each vessel so that, when its tube is in a vertical position at the left side of the wheel, the exact desired fill level within the tube will be achieved. Knowing what that level should be, when properly filled, is calculated by figuring the necessary volume, in cubic inches or cubic centimeters, according to the tube's inside diameter and height of the column of water. The formula for volume of a cylinder is Pi radius squared times height. For example, if a tube had a 1 inch inside diameter then its radius would be 1/2 inch, or 0.5 inch. The radius squared would be 0.25, so multiplied by Pi (roughly 3.143) would equal 0.786 when rounded up. So we take 0.786 and multiply that by the desired height of the water column. Let's for the moment assume that might be ten inches. The required volume would then be 10 times 0.786, which equals 7.86 cubic inches. What does 7.86 cubic inches of water weigh? The weight of water is 0.58 ounces per cubic inch, so 7.86 cubic inches would weigh 4.56 ounces.

In another example, suppose we used 2.5 inch schedule 40 PVC pipe for the tubes, which has an actual inside diameter of 2.469 inches. For the same 10 inch fill height, the volume would be 47.88 cubic inches and this amount of water would weigh 27.77 ounces, which is equivalent to about 1.74 pounds. So, increasing the diameter 2 and 1/2 times gives us 6.1 times the water weight.

Schedule 40 PVC pipe has a wall thickness of 0.203 inch, as it is meant to withstand a pressure of 300 psi (pounds per square inch), so of course this would be quite an overkill for what is needed in this build as pressure will be quite minimal even at the moment of the water hammer effect. Clear PVC pipe, which would be nice to use for the visual aspect of watching the water in motion, has the same dimensions as Schedule 40 PVC and is rated for half the pressure. The cost, though, is rather prohibitive, at around $16 to $22 per foot. I suppose one might consider using just one clear tube to study the visual aspect, as it would be the same in all tubes.

There are other materials that could be used for the tubes, of course. For example, Class 200 2.5 inch pipe has a wall thickness of 0.137 inch, an inside diameter of 2.6 inches, and has the same outside diameter as Schedule 40 pipe. Rated at 200 psi, this is still considerable overkill, though the cost is about 1/3 less than schedule 40 pipe. There is surely a material more suitable and less costly which could be sourced, and one such example would be 3 inch diameter PVC DWV (Drain-Waste-Vent) pipe. This comes in ten foot long pipes that sell for only $12, and can be found at any Lowes or Home Depot store. These are considerably less expensive than regular schedule 40 pipe of the same size because they are rated for low pressure applications, so would be ideal. A 3 inch 90 degree street elbow, which would also be ideal for the build, goes for about $4.40, and the pipe could be capped off with a removable PVC clean-out plug fitting that sells for $1.88, so is definitely worth considering. By the way, the 3 inch pipe filled to a 10 inch height would have a water volume of 70.686 cubic inches, which would weigh 41 ounces. That is 2.56 pounds. If 4 inch DWV pipe could be used, the same fill height of 10 inches would offer 4.56 pounds of water weight, and would cost little more than the 3 inch DWV pipe.

[rickoff]

Addition of water to the design concept

Okay, so let's see what happens when we add in the water. Remember that each tube and its companion feeder/receiver vessel is completely independent of all remaining pairs. In other words, the dotted lines which show the location of the barriers are simply dotted lines for illustration purposes. There are no perforations in the barriers that would allow any flow of water from one feeder/receiver vessel to an adjacent vessel.

I'm using a blue color, as I did in an earlier drawing, to show how each tube and vessel pair would be affected during rotation. While obviously not a perfect visualization of what would be occurring, I think anyone would agree that what is shown, as far as circulation of the water is concerned, does generally conform to what would be happening at this particular stage of rotation, and that the same thing would be happening when the wheel's rotation is advanced each successive 30 degrees. I say 30 degrees because of course there are 12 equal sectors, and a 360 degree circle divided by 12 equals 30 degrees. Thus, each barrier is 30 degrees apart, as are each of the tubes.


As can be seen, there are four water filled tubes (A, B, C, and D) which are providing overbalanced weight at the left side of the wheel, and the water contained in each of these tubes lies beyond the wheel's outer perimeter. Thus, each of these four tubes is providing leverage for rotation that would overcome the applied force of water enclosed within four or more of the six perimeter vessels on the right side. There are two water filled tubes (E and F) on the right side, however, which are acting in opposition to said leverage, thus negating the leverage of two tubes on the left side, and all perimeter vessels on the left side are empty while all at the right side are filled.

The question that arises at this point, of course, is whether or not there is enough overbalancing of water in the tubes at the left side of the wheel to overcome all applied force of water weight on the right side. As I stated, since there are two water filled tubes (E and F) on the right side, those tubes are counterbalancing the effects of two tubes at the left side (E is counterbalancing D, while F is most closely, though more than, counterbalancing A). Thus, if we take all four of these tubes out of the equation, that leaves just B and C to provide enough leverage to overcome all water weight enclosed in the perimeter vessels on the right side. Is that possible? Probably not in this drawing, but it could be, provided that the tubes are made long enough. A long lever provides far greater torque than a short one, for the same amount of applied force, which in this case is merely water weight, and making the tubes longer would give that advantage to the left side, while lending no counter-effect to the right side. In this case, the wheel would turn counter-clockwise. Conversely, though, if the tubes were made a little shorter than shown (which would make for an easier and less costly build) then they would offer less leverage than they do now. If in fact the applied force of water is actually greater at the right side, or could be made greater by shortening the length of the tubes, it would then appear that the right side would be in an overbalance condition in relation to the left side. If this currently is, or could be made the case, then of course the wheel would turn clockwise. If clockwise rotation is desired then adding a 45 or 90 degree elbow to the outermost end of each tube, with the elbow's outlet end facing downwards before being capped off, would be a good idea, as otherwise there would be a counter-rotational water hammer effect. Here's an example:



Notice that some amount of water is shown in tube L in the upper drawing. This wheel position would be the earliest stage of transition, or transfer of water, from a vessel to a tube if rotation is counter-clockwise, or it would be the final stage in transfer from a tube to a vessel if rotation is clockwise. In my next drawing I will show tubes L and F at the mid point of transition, when they are parallel to the horizontal plane.

Still no questions or comments, so I'll have to assume that all who are following along, watching and reading this, understand and agree with what I have presented so far.

Best to all,

Rick