Pendulum raising com, any precedents ?

[daanopperman]

Nick,
I know you will not like what i am going to say but just think of what we both say.Put your water in a sphere and change the com by tilting the sphere, it is the container that is confusing you.Fill a straw fully with water and tilt it and you will see what i mean, the com stay put.

[nicbordeaux]

It's not a question of liking or not liking :)

here are my facts:

1) If a half-full oblong tank of water starts 15° left of horizontal and is tilted right 15° over the horizontal, the water will slide to the right side. The CoM is identical. Total travel 30° requiring force to initiate, no change of CoM. And this is your "objection".

2) If you actually take a tank as described, put it on a central pivot (a log might do even...), when it is set to the left position and the water is at quietus punctus or something clever, draw a line on the water level of the tank: "wedge" shape. Lift the tank and tilt it over. Disregard the force required for the moment, that is another area of mechanics. Wait for that quietus punctus stuff to happen. Now tilt the tank back towards the initial left position. You can not unless you are blind drunk fail to notice that a fair part of the water content will rise quite significantly above the line you drew. Then the waters will subside a fair amount. If you have your pinkie under the tank, you will notice that the force exerted on it dimishes considerably as the water rushes back up the tank with this "counter wave. It is only when it has fully subsided to keel end that the full weight will be continuously bought to bear on your finger.

As the water rushes back up the tank, the force required to initiate the return flip will be dimished.

As the water hits the first time your pinky will bear the full weight of the water. Once. As it rushes back to where it came from then back down, you will feel mild relief, then full pressure again.

If you have an outlet from your tank at the highest water level (surge level), water will spurt or pour or drip or trickle from a higher point than that from which it started. Your surge level is above the line you drew.

This is the sketch of the picture. If you disagree with this, then please tell me where I am wrong. And I am quite prepared to admit I am wrong.

Until such time as I reveal to an astounded world the great interacting pendular gizmo on parralelogram with pendulum drive feedback from the water, sumbmersible positive buoyancy slop height enhancers and other such trickery.

[jim_mich]

1) If a half-full oblong tank of water starts 15° left of horizontal and is tilted right 15° over the horizontal, the water will slide to the right side. The CoM is identical. Total travel 30° requiring force to initiate, no change of CoM. And this is your "objection".

Ah, but there is a change to the CoM!

Initially the CoM of the water is off to one side. Then as you tilt the tank you are lifting the water higher. Then the water runs down to the other side. The CoM is now at the other side of the pivot. Otherwise there would be no pressure on your pinky when the water is at that end.

The water height might be the same when tilted to the left as when titled to the right, but the CoM of the water has shifted from one side to the other side. And the CoM was lifted upward as it passed over the center. Otherwise there would be no pressure to move the water from side to side.

Or I could be wrong.

[nicbordeaux]

Many respectful thanks for your input Jim. 'Tis a rather complicated state of affairs, and describing it without knowledge of the proper terms (that's me) is rather bothersome. Not to mention that it's only through getting my socks wet that I can get a rough picture of what's actually happening.

So, agreed, the center of gravity has changed laterally. Whether this can be called the CoM I don't know. When I refer (probably mistakenly) to CoM I really mean the averaged out height of the mass.

In the description given just before your post, which was verified very early on by some holes drilled in the plastic tank above the "punctus quietus" line , a reasonable amount of the mass comprised of water surges through the holes. If captured at that high point (which means further lateral travel because of necessity of very shallow container to retain height gain), there is elevation of mass. Whether that raised height is sufficient pe to equate to and surpass force required to initiate the tip is another problem altogether. Dependant on two factors : how high you can get the water to "slop" or "slide" by means of baffles or styrofoam incline shapes glued to steel plates and held by mags located under tank; how energy cost effective a system you can build to initiate the tilt of tank. Definitely, tilting the tank from "keel left to keel right" by lifting it by direct non-leveraged force is a no-brainer, COE rules. Edited : leverage should have no incidence on CoE, but as the fluid dynamics are pretty unlike those of more stable forms of mass, leverage might buy you an advantage ?

As to the mass being raised as it flattens at horizontal in travel from one end to the other (it dosn't actually ever level to further complicate matters), that's another can of worms. Even if the water were to be level I'm not sure what the compared CoM of "tilted" as opposed to horizontal level would be.

To quote you "the CoM was lifted upward as it passed over the center. Otherwise there would be no pressure to move the water from side to side". I'll take your word for that. And will refrain from even considering mentionning ke of the moving water mass as a whole or seperated into "slices" pushing or braking each other, let alone ke of the quite substantial tank and tank supporting assembly, and mutual influence of this predicatable latter mass and the previously mentionned water.

Aspirin. Or maybe another glass of wine.

[rlortie]

The water height might be the same when tilted to the left as when titled to the right, but the CoM of the water has shifted from one side to the other side. And the CoM was lifted upward as it passed over the center. Otherwise there would be no pressure to move the water from side to side.

To quote you "the CoM was lifted upward as it passed over the center. Otherwise there would be no pressure to move the water from side to side". I'll take your word for that. And will refrain from even considering mentionning ke of the moving water mass as a whole or seperated into "slices" pushing or braking each other, let alone ke of the quite substantial tank and tank supporting assembly, and mutual influence of this predicatable latter mass and the previously mentionned water.

Water seeks its on level via gravity. you are not lifting the water or the COM but rather shifting it. The only thing being lifted is the low end of the container bottom which if properly centered is balanced. As such it will will not take any effort to lift it in an empty state.

The force required is to change the COM position laterally creating a miniature tsunami.

Lets use a hydro electric dam as an example. We know that the static water creates X PSI on the face of said dam relative to depth. If we raise the bottom of the river the water is forced back upstream. We then lower the bottom and the water rushes back with acceleration hitting the dam face. The force gained via kinetic and inertial impact PSI is more than it was when static.

The surface level of the water never changes as you have not varied the displacement quantity in the container.

Now the question; is the impact energy created more than that which it took to raise the bottom of the river bed or Nick's container.

Ralph

[nicbordeaux]

"Now the question; is the impact energy created more than that which it took to raise the bottom of the river bed or Nick's container.

Ralph"

Well Ralph, you've been playing this OU or PM search game for much longer than most of us, and with real mechanics, not calulations or theories. So I guess you might have some ideas about this, or have tried it even.

There is more energy (impact ke or final weight height equating to greater pe) to be harvested in total upon tilting the tank if all the energy is collected. Devising a way of collecting all of it is not easy. And obviously you wouldn't be using a stick under tank as a pivot, you want the tank in a supended cradle type hanger. Travel limits, of course. The easiest tilt possible with the most possible amount of movement down, up, down etc the tank until it all subsides.

[Unbalanced]

I'm not certain what this is or how it works but I can't fault the guy's optimism and I think its somewhat relevant to this thread. The guy needs to have someone do his math me thinks.

http://www.youtube.com/watch?v=zd3B-EHfzLA

[nicbordeaux]

Had to watch it w/o sound because of sleeping family (they're not insane like me with the PM disease). Not sure how it works but certainly the camera shots made me seasick.

Still, think that everytime you pull the chain 7 liters of water and other stuff of (unless you are unlucky) higher desity stuff shoots down the sewage piping. Now 7+ kgs is a decent amount of mass to use to upset a seesaw device. If you have a bathtub half full of water and tilting as previously discussed, actuated or set in motion by some OB container or pendulum, you have pm if you have running water and beans. Produce some juice to offset the water bill.

[rlortie]

Nick,

Have not built a rocking tank but have researched the concept for others. Including but not limited to floating platforms with round weights osculating back and forth upon said platform. Problem with this is that the displacement never changes.

The idea of using the moving water itself for gained energy was not considered. I believe the idea worthy of farther research as you have some pluses in your favor.

Water has viscosity and the molecular properties to cling to itself. Dynamic inertia means that the water will continue to flow toward the low end of your tank. This was proven during the Japanese earthquake. The following tsunami did not create large waves here on the west coast. In fact it was the opposite, the water was sucked or forced out of the tidal areas and boat basins. Many boats and floating docks were left sitting in the mud.

I would think that after the moving water ceases from hitting the end of your tank there would be a considerable recoil as found in wave motion.

As for capturing any gain in force I would suggest bellows with flapper valves, not in the bottom of the tank (as per the rising river bed) but vertically on the ends of the tank. Use the inertia and KE rather than the compressive force of water density.

One might think that using a liquid with a higher viscosity would be to an advantage. Maybe it would and maybe not. It takes as much to get it moving as it does to stop it.

Ralph

[nicbordeaux]

Thx Ralph, I appreciate the input and encouragement.

[nicbordeaux]

As a start mech I might commend this rotary device (used in other builds...). I'd also suggest it as a reset mech, though the "engineering" is a mite complex and requires understanding the system. :D

The crude and not at all to scale sketch shows a tilted tank containing water as per previous spec. The problem is to initiate the tilt of the tank from left end keel to right end keel with as little possible effort.

Thus, we have a closed tube containing water ballast (say 25% full) resting at horizontal and extending beyond left end. It is pivotting in a horizontal plane. It is also set quite high off the tank so as to avoid colliding with tank sides (live and learn).

Turning that ballast tube in a horizontal requires very little effort. It proceeds to rotate round to 180° or "keel right". In doing so, it initiates tilt of the main tank. The reason of course is shift of cg, and don't start hollering that the force is being applied to left end of tank so it won't work, that argument is only for people who never get to the build stage and know nada about mechanical systems in the realworld. :)

Obviously, as the tank starts to tilt after the water mass has upset the keel right stuff, the water in the rotary tube also sloshes at keel right, adding F.

If you wan't to know how to null out most of the disadvantage that arises with the lowering of the CoM of the ballast pipe (which is not horizontal to tank), you will need a brain (meaning Jim has not has you beheaded).

You can also have the tube pivot in a vertical plane much like a pendulum, but this is incredibly less effective.

Edit : to make for an easier test build were anybody to wish to build a test of the rotary weight shift thing, just use a bob on the end of a stick, and have the stick pivotted.

So, Ralph ?

[Gravitronic]

I think that the part about slop being chaos quite simply to be a part of a bigger formula we have yet to figure out. We just call it chaos because we don't understand it yet. As for being able to use the slop... it will just take some clever person to come across the right series of parts in the right configuration and voila!

Yes, Newton was certainly correct for the time when he proposed the 2nd law and for the most part humanity agrees. Frankly, I am a believer, all Newton's Laws are correct as stated and inviolate. Nevertheless, it is possible to achieve a higher CM in a system. What is sometimes forgotten, and we have probably all been guilty of that mental lapse on occasion, is that system is NOT a closed (isolated) system. The machine which produces the rise, was initially given the energy from outside the system (raising of a stone, raising of a pendulum, rotation from keel position of a wheel, etc.).

Example 1: Start with a balanced wheel, attach a pendulum pivoted near the wheel mid radius at the wheel 1:30 clock position (pendulum addition makes wheel out of balance). Raise the pendulum nearly vertical, ensure it is not completely vertical (12 clock) and release the system. Observe the pendulum as it completes a few multiple complete rotations, and depending on the configuration, the wheel, which is now out of balance, will also rotate temporarily. This is bidirectional momentum exchange between the wheel and pendulum and the system CM will at times be temporarily higher than the system CM position prior to system release.

Example 2: Bessler's bidirectional wheel was initially "given a gentle push" to start the continuous rotation. Well heck... that gentle push moved a weight within its mechanism into position to be acted upon by gravity. Gentle push to start = not a closed system. On his uni-directional wheels, a spring(s) was, gently, continuously, preloaded which explains the self start. But, the spring(s) had to have been preloaded by Bessler initially = not a closed system. Hmm... the law of Conservation of Energy = closed system.

All wheel designers should consider this in their designs and be aware that you can't have an operating wheel without an initial (small) force applied. Ok, I said it, that probably blows it for me. Someone will have an operating wheel patented within the month.

Mankind has written many laws. Societies change, our collective knowledge increases, and laws, valid at the time of their proposal have been changed throughout history. We who believe Bessler designed and built an operating wheel must also believe in 'shades of gray' in current physics beliefs. Some areas may soon change, and written material in the future may need to be more specific and definitive.

You might be interested to know, if you are working to design a self-running wheel, there is a clock ticking down on the time remaining for you to succeed. When the "Singularity" occurs in 2045, it will be accomplished. Yes, I will have to admit, I am a singularitarian.

Originally published in TIME magazine, here's a reprint:

http://wwsg.com/kurzweil-the-singularit ... s-immortal

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Nick,

The oblong washing up bowl, I like it, but don't put any fish in, when they jump they'll be changing the center of mass. Seriously, I'm really thankful you explained your idea further. A slow person like me can understand this. I'm thinking you may have something here. To elaborate on your half pipe (or gutter) concept, you probably can't prove this with any physics maths. Thankfully... I can't do them, those maths thingies. Well, actually I'm too lazy. Anyhew, physics defines slop, splash, etc. as chaotic and has no great laws or equations for those types of wet socks. However, you may have fun and be challenged by further experimentation to determine how much slop to a higher elevation you need. The key may be a calibrated gutter hole to release the necessary mass of water, at the correct time, back into the system. So, your three simple machines here, your lever (seesaw), your 1st inclined plane, and your 2nd inclined plane (bowl sidewall producing the lift) sounds like a system that should give you a reason to get your socks... ahh... wet for.

Good Luck !

Regards,
Chris