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Abeling Gravity Motor

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Thu Apr 09, 2009 3:41 pm PostPost subject: Abeling Gravity Motor
overconfident
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Any thoughts on the Abeling device?
http://www.overunity.com/index.php?topic=7150.0

Here's some of mine:

Take a wheel with balls at each segment (or dumbbells, or whatever) and spin it fast enough that the balls are held in place by centrifugal force. Let's say there is a trapdoor beneath each ball, so that when the trapdoor is opened, the ball can escape from the wheel and go wherever it's accumulated momentum sends it.

Now, beneath the wheel at the bottom is a fixed U-shaped ramp or tube that can catch a ball that flies downwards through the trapdoor and redirect it back towards the top of the wheel. At a certain wheel RPM, the ball will exit through the trapdoor at the bottom with enough momentum for it to reach the top again.

If there is a capture mechanism on the wheel, the ball can be reinserted at the top, past top dead center. Gravity takes over at this point and the side with all the balls will fall, keeping the wheel spinning.

What we have is an overbalanced wheel that only has balls on one side. Each ball is spit out at the bottom and recaptured at the top.

If properly designed, the ramp and recapture mechanism might also be able to use some of the ball's ballistic energy, for higher speed operation.

Issues:
1) The wheel would need to be started using an external source of energy to get it up to speed where the balls have enough momentum to do their own thing.

2) Once critical RPM is attained and balls start flying out the bottom, the wheel will be drastically overbalanced and may require some significant design considerations to allow for that.

3) Trapdoor mechanism should be designed so it does not begin actuation until a certain RPM is achieved.

4) Ramp should be designed to minimize travel and ballistic energy losses.

5) Recapture mechanism needs to be designed.

If I can find the time, I'll try to make a crude drawing.

Any thoughts?
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Thu Apr 09, 2009 5:49 pm PostPost subject:
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Upon further reflection, the wheel will probably need to have at least one missing ball, an empty slot to receive the ballistic ball ... unless the wheel can be spunn fast enough that the ball can return to the same slot it came from ... but if this is the case, we would have several balls in transit at any given moment.

Just some more food for thought.
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Thu Apr 09, 2009 6:03 pm PostPost subject:
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http://www.youtube.com/watch?v=cetnLKDqQJ4&feature=related
is some similar may be more simple may be more bad who known?
In @OC link i saw which the lines (ball's contures /rails are different Exzample first is different from second third is as first forth is as rayl number two etc etc.
This man is wery good profesionalist
What is the principle I not good was undestand but a wheel is rotating with more big speed in time befor a journalist Real (I think) the rotation must be a lot of slow.
Old man's toys as us:)))
Not not:))) OC MPMM is fammous devise this is not toy it is a new and very hard invention Big invention
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Thu Apr 09, 2009 8:26 pm PostPost subject:
Harvey
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@OC,

Regardless of how it is done, the leverage will always equal unless you can nullify the force by some means. The translator used the letters PM in the video relating to his research. It would seem that he has coupled magnetism and gravity to produce the effect. Using magnetism on one side to nullify the gravity and then again on the other side to add to it.

"Never say never" Wink

His D shaped approach is interesting, but if you plot the leverages what do you get? I don't have time today, perhaps tommorow.

Cheers,

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Thu Apr 09, 2009 9:02 pm PostPost subject:
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Harvey wrote:
@OC,

Regardless of how it is done, the leverage will always equal unless you can nullify the force by some means.

...

His D shaped approach is interesting, but if you plot the leverages what do you get? I don't have time today, perhaps tommorow.


I'm actually extending the "D" concept a bit. I'm talking about the weights (balls) actually departing from the wheel at the bottom (through a trapdoor, once it is up to speed) and rejoining it at the top (with a capture mechanism). This way there are no weights on one side of the wheel and numerous weights on the other side. No way that will equal out and "nullify the force". The wheel will be grossly unbalanced because all the weight is always on one side only.

(Where's that Alsetalokin guy when I need him. Crying or Very sad )
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Thu Apr 09, 2009 9:38 pm PostPost subject:
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Historically this approach always result in an energy conversion. You must use the energy produced (Kinetic) convert it (torque) and apply it to a lift mechanism (undetermined). The conversions to date always result in a loss.

How will you raise the balls back to the top without nullifying the gravitational force? Catch 22

Confused
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Thu Apr 09, 2009 9:50 pm PostPost subject:
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Harvey wrote:
Historically this approach always result in an energy conversion. You must use the energy produced (Kinetic) convert it (torque) and apply it to a lift mechanism (undetermined). The conversions to date always result in a loss.

How will you raise the balls back to the top without nullifying the gravitational force? Catch 22

:?


As I said above:

"Now, beneath the wheel at the bottom is a fixed U-shaped ramp or tube that can catch a ball that flies downwards through the trapdoor and redirect it back towards the top of the wheel. At a certain wheel RPM, the ball will exit through the trapdoor at the bottom with enough momentum for it to reach the top again."

There is no lift mechanism. The ball's own acceleration due to centrifugal force is what provides the lift, after it has accelerated through the trapdoor. It's like having a weight on a string and letting go. If the weight is spinning fast enough, it will have enough momentum to do a certain amount of work. The faster it is spinning, the more energy it contains. If you need more energy, spin it faster.
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Fri Apr 10, 2009 12:34 am PostPost subject:
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And therein lies the rub. the wheel cannot travel any faster than 9.8m/s for any radius. The larger the radius, the slower the RPM. Any energy imparted to the wheel by the falling ball will be removed from the ball when it leaves the wheel.

When we add the associated losses, the ball will never reach the same height that it was at when it started. Unless... (there is always a however eh?)

The gravitational force is nullified by some means on the trip back up.

Now, keep the 'trapdoor' on the moving wheel, open the door when you want the ball to be released. The ball spirals out away from the wheel into a tube designed to match that curve. Line the top of the tube with magnetic attractors of a force = 4g, to offset the gravitational force. The ball accelertates upward at 4g-1g = 3g. When the ball strikes the top plate at 3g force and is now 90 to the attractors, the impact moves the plates out of the way and latches them. The ball now rolls down to the wheel and unlatches the attractors.

That would be a nonconservative system and the energy source would have to be identified, but it would probably be found to reduce the earths gravitational field by that amount - IOW it is a ripple along the gravitational density for that distance from the center of the earth. It's ok to do that...tides do it twice a day.

Cool
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Fri Apr 10, 2009 1:13 am PostPost subject:
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@Harvey,

Take one of your WhipMags, stand it up vertically, magnetically attach spare rotor magnets to the outside plastic circumference of the rotor so they are only held in place by attraction to the rotor magnets. Now spin the rotor up with a motor or dremel until the magnets start flying off. Does the wheel slow down when magnets fly off? How much momentum do the flying magnets have when they depart?
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Fri Apr 10, 2009 1:41 am PostPost subject:
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Can anyone tell me how to do "chutes and ladders" or plumbing with Phun? I think I can figure out the wheel and balls OK.
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Fri Apr 10, 2009 8:36 am PostPost subject:
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OC, yes, when the magnet leaves the wheel the MOI of the wheel will change. The energy of the magnet however is less because it has transferred some of it's energy to the wheel. The reason for this is because during the spin up, the magnets added to the MOI of the whole system, so together they determine the RPM, but when the magnet separates and leaves contact (which gets a bit involved because the field isn't really shearing at this point like it would with a non magnetic connection), the RPM of the wheel must either give up energy to the magnet and reduce, or the magnet must give up energy so the RPM will stay the same. If the mass of both were equal, then the energy would split. But in this case the energy is transferred to the larger object, think of it as a reverse collision; instead of impacting they are separating and momentum is the big winner in both cases. Another example of that is rocket separation, but that actually involves additional energy that is transferrred totally to the craft taking everything from the booster and stopping its momentum. Also, keep in mind in your example that you are adding energy from the tool rather than gravity.

In Phun (I don't have the latest version yet) most of the vids I have seen, simply use blocks and freehand surfaces to make channels or flumes. One trick I learned, is to zoom way in for the freehand stuff and then zoom out to place it - it improves the resolution. I wish it had modeling functions like union, fragment and subtract - 'but what do we expect for free - a rubber bisquit?' Mr. Green

Cheers,

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Fri Apr 10, 2009 10:08 am PostPost subject:
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Just posted the following on the Overunity.com site.

If I'm right (and I'm sure I am Wink ) we won't be needing the Mylow Motor after all.


==============================================================
Well, I've worked out the power cycle for the Sjack Abeling Gravity Wheel - and here it is.



So it looks as though his gravity wheel really will work, though as he says, why someone didn't find this out a long time ago just by chance is a bit of a mystery.

Still, I suppose before knowledge of the Carnot cycle and the Leibniz calculus it would have been a bit of a mental stretch.

When I first recognised that the acceleration leg (d2x/dt2)and the rate of change of acceleration leg (the adiabatic leg - d3x/dt3) gave the potential for a power cycle I couldn't see where the other return legs were going to come from. But then I realised that cos everything is turning round the returning legs are the same as the outgoing legs. Tricky, eh!

But what on earth has this got to do with a Carnot cycle you may ask?

Well, think of the balls being thrown around as monster molecules and the structure enclosing them as the cylinder. The equivalent of the thermal potential change is the gravitational potential change and the adiabatic leg (the balls rising up) is where the motion energy is exchanged for gravitational energy.

If you don't understand what I'm on about (or even what I'm smoking Wink ) I can't say I blame you.
And if you want to ask questions I'll do my best to answer them.

Oh, and one more think to note. The containing structure can be made as stiff as needs be so no significant power is lost to the structure.
==============================================================

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Upon the place beneath: it is twice blest;
It blesseth him that gives and him that takes:
'Tis mightiest in the mightiest: it becomes
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His sceptre shows the force of temporal power,
The attribute to awe and majesty,
Wherein doth sit the dread and fear of kings;
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Fri Apr 10, 2009 10:28 am PostPost subject:
Frank
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Of course, just between us on Fizzx, I'm not really that sure I'm right. Rolling Eyes
I'm relying on the fact that Sjack seems pretty confident and it's difficult to see how one could be mistaken about something as simple as his device. Either it goes round delivering power or it don't. If it don't then it's some unbelievably stupid scheme for defrauding investors.

But it has been confirmed that Henkel are indeed involved so I reckon it's simply been missed. After all, I have personal experience of something that has been missed for the best part of a century, viz. the equations of state for water vapour.

Why on earth has no one but me made any attempt to interpret the significance of those laws?

I believe it's because they create too much cognitive dissonance and people are incapable of interpreting them without doing serious damage to the existing canon.

In particular, the notion that zero degrees Kelvin is absolutely absolute and not relatively absolute, i.e. relative to one class of motion.
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Fri Apr 10, 2009 2:44 pm PostPost subject:
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Harvey wrote:
And therein lies the rub. the wheel cannot travel any faster than 9.8m/s for any radius. The larger the radius, the slower the RPM. Any energy imparted to the wheel by the falling ball will be removed from the ball when it leaves the wheel.

When we add the associated losses, the ball will never reach the same height that it was at when it started. Unless... (there is always a however eh?)

The gravitational force is nullified by some means on the trip back up.

Now, keep the 'trapdoor' on the moving wheel, open the door when you want the ball to be released. The ball spirals out away from the wheel into a tube designed to match that curve. Line the top of the tube with magnetic attractors of a force = 4g, to offset the gravitational force. The ball accelerates upward at 4g-1g = 3g. When the ball strikes the top plate at 3g force and is now 90 to the attractors, the impact moves the plates out of the way and latches them. The ball now rolls down to the wheel and unlatches the attractors.

That would be a non conservative system and the energy source would have to be identified, but it would probably be found to reduce the earths gravitational field by that amount - IOW it is a ripple along the gravitational density for that distance from the center of the earth. It's OK to do that...tides do it twice a day.

Cool


I like that last bit. One can think of it as a mini tide machine. This gives people a mental crutch to hang on to. After all everyone has been to those swimming poles with artificial waves so the idea of a tide machine will not be entirely foreign to them, eh.

And I'm delighted to see that "(there is always a however eh?)". A refreshing open minded attitude. How different from the Steorn Forum peanut gallery who instantly dismiss anything resembling a perpetual motion machine as a scam or delusion.


"There is a tide in the affairs of men which taken at its flood leads on to fortune".

-- William Shakespeare
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Fri Apr 10, 2009 7:57 pm PostPost subject:
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@All

Evening chaps.

In conclusion then, are we saying that the Abeling GM is a rotating momentum mass gravity driven/magnetically assisted device?
IOW gravity driven on the extended mass rotation and magnetically lifted on the assending rotation.

If this general explanation is accurate then blow me down with a large feather-
it's the dead copy of my 2007 gravity wheel idea i penned and drew in some detail.

I'll post my sketch here (or somewhere near) so everyone can see
if its matches up with everyones expectation of the Abeling.??!!
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Fri Apr 10, 2009 9:38 pm PostPost subject:
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@Frank,
If gravity pulls on the 'Wheel Side' of the 'D' then its acceleration is not linear from top to bottom, you have to integrate the curve from 0 (top perpendicular force, toward axel) to 90 (tangent force) and back to 0 (bottom perpendicular force, away from axel). Because gravity is conservative, the straight path down the vertical drop from top to bottom will have the same overall force as the integrated curved path independant of time. So, if the force is identical for each path taken, how does a differential in time effect the system? Velocity. For the vertical drop, the velocity increases throughout the path until it reaches the bottom. For the curved path the velocity is split between the two axis (x & y, with y being vertical). It takes time to move in the x direction, and time to move back, but the velocity for x will be negated, or net to zero. Therefore we have an extension of time, and we have a vertical velocity which ends at zero. The force is the same and therefore the energy is the same, but the action and timing are significantly different. In one case the 'ball' (integrating OC's concept here) has all of the energy stored kinetically. In the other case that energy is transferred to the wheel.
So I think a comparison chart between the two paths for dv/dt would be good to see. And then, if we compare each of those to the energy curve and the energy curves to each other, then we will have comprehensive way of looking at this. I would set the start criteria as "motionless" with potential energy at the top.

@chrisbis,
The magnetic assistance is a huge guess on my part, or as they say 'WAG'. My only bases for that is a reference by a translator that typed some text in a video that was "PM", which I took for permanent magnet. It may have nothing to do with magnets; that blurb related to the research that Abeling had done prior to creating his device. But from what we know of physics, gravity alone will not power the device. Some time back I developed some dialog here regarding gravity and escape velocity and such. It was determined mathematically that a falling object coming at the earth with no friction (wind resistance) and total elasticity (or direction reversal) would have to start at a distance near the moon in order to reach escape velocity on the rebound. IOW, it becomes impractical to think we can use just the earths gravity alone as a force. There must be another force at play in conjuction with it if we truly want to utilize it to produce energy. The moon's gravitational force is such a force and we have extracted energy from these to fields interacting.

Yes, magnets make a good form of counterforce to offset gravity. But to be effective, there must be a means to remove them from the equation (turn them off) when you reach the top, otherwise we have only traded one trap for another. I have considered the variance of inverse cube verses inverse square and wondered if this force / distance asymmetry could be exploited for energy extraction. Perhaps there is a way to resonate between the two forces where energy extraction keeps the system at an imbalance. Just one of the multihundred concepts bouncing around in this cranium Confused

Cheers,

Cool
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Fri Apr 10, 2009 10:37 pm PostPost subject:
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Harvey,

I struggle sometimes with the theoretical side of things compared to u scientists.

So, forgive me for beinng exceedingly thick here, but wont the rotation speed be an almost constant on a device thats has mass weights on the periphery on any mounting wheel/disk? Rather like a ferris wheel?
As each mass comes into play, all the others masses will have done, or part done
their vector momentum thus putting the wheel into almost equilibrium motion.
If u where taking the action of one mass on one arm say at one starting position on the disk, then yes i agree it would be a more complex model to anaylise thro a inclination of angles/indices, as it rotates.

What really worries me is this notion of a trap-door item to release the rollers/balls from their current position to a lesser influencing position within the device.


BTW, i took PM to nearly always mean Perpetual Motion, and any mention
of mags would have been PMM as in Permanent Magnetic Motor.
Perhaps we should start a glossary of absolute terms for everyone to use when mentioning anything along these lines.
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Sat Apr 11, 2009 12:44 am PostPost subject:
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chrisbis wrote:
What really worries me is this notion of a trap-door item to release the rollers/balls from their current position to a lesser influencing position within the device.


In the idea I tried to describe above, the trapdoor does not "release the rollers/balls from their current position to a lesser influencing position within the device". I intend for the rollers/balls to actually exit the device entirely, to be recaptured by the wheel when their presence will benefit rotation instead of retard it.

If we have several cups around the perimeter of the wheel, like a water wheel that can each hold a ball. The cups are positionsed so they are upright on the right side of the wheel and inverted on the left side. So when the cups are filled on the right side, they will empty at the bottom, and the wheel will spin clockwise.

Now take a vertical tube of balls and a feed mechanism that fills each cup as it passes, as soon as the cup is in a position it can hold a ball (just past top dead center). The balls are fed into each cup as it approaches, and the balls roll out of the cups just past bottom dead center.

The wheel continues to spin clockwise and continues to gain speed as long as there are balls that can be fed to the cups on the wheel until the wheel achieves a steady speed, a speed proportional to the gravitational acceleration of the falling balls. The wheel is obviously heavier on the right side than the left.

The difference in my idea above is that the balls experince two separate forces (or accelerations), a downward force due to gravity, and an outward (centrifugal) force from rotation.

If the wheel is spinning fast enough, the balls at the bottom will have a combined force vector from centrifugal force and gravity and will be ejected with more force than gravity alone would allow. This additional force can be used to return the ball to the top of the wheel. A spring or chute mechanism can be used to recover the energy and bounce or roll the ball as needed.

So now, instead of needing a constant supply of balls at the top, all we need to do is redirect the balls from the bottom back up to the top and continuously recycle them.

Also note that if the balls still have some kinetic energy remaining when they reach the top, they can be redirected again and can transfer some of that additional energy to the wheel, allowing the wheel to spin faster than gravity alone would allow.

This is not a self starting device. Using gravity alone, the balls will never depart from the wheel with enough energy to get back up to the top. The wheel needs to be spun up to a speed where it can cycle continuously.

(Hey, I'm just tossing out some wacko ideas here.)
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Sat Apr 11, 2009 8:33 am PostPost subject:
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I guess the real question you have raised here is:
How is energy distributed when two moving masses separate in a gravitational field?

Naturally the force of gravity here on the surface of the Earth has a primarily unidirectional vector - down.

So, when the joint that holds the two masses together is broken (aka trapdoor) then the applicable forces will act on the separating projectile (aka ball). As you have stated, there are two primary forces to consider. Gravity and Momentum. Now in the case of a wheel, the momentum is tied to the central axis of rotation and becomes a modified version of Newtons law wherein the momentum is now considered Angular Momentum. There is a third force, that which holds the projectile to the wheel and this is centripetal force. Centrifugal and Centripetal forces are inverse of each other, and often Centrifugal force is considered imaginary in the sense that it is a percieved force caused by the curvature created by centripetal force that prevents the momentum from following its preferred straight path. These two forces run perpendicular to the tangent of the wheel, while the momentum runs along the tangent line at all points in time. When the centripetal force is dropped to zero, the imaginary centrifugal force drops to zero with it, each moving in opposite directions and ending at the center of the projectile. At that zero moment, the momentum of the projectile itself carries it in a straight tangent line - Momentum (p) = Mass (m) x Velocity (v), and this is a vector so it has direction. The kinetic energy of the projectile is discovered as `E_k = 1/2mv^2` where m is Mass and v is Velocity. Again, this is a vector so it has direction. Pay attention to the square, because that means if it's velocity is doubled it's energy will be quadrupled.

So, let us imagine that we hold our projectile to the outside of the wheel with magnet. And let us suppose that as it reaches the bottom it passes through a stator field that causes it to no longer be held on, a neutralizing field of the same polarity that is holding it in place to the wheel. And let's suppose at that very place there is a track for the projectile to ride on which is very rigid and absorbs no energy, and it has a reasonable curve to it so that the projectile is changed from horizontal travel to vertical travel. Let us suppose that the projectile is round like a ball and it travels along the track. The curve in the track acts as an externally applied centripetal force, so the horizontal momentum becomes angular momentum for this 90 upturn at the end of which it will again be vector momentum. With careful planning, the release from the wheel and entrance to the upturn can be noncontact 'fly through the air' low friction and also, after leaving the upturn and flying straight up we can pull the track wide so that no contact occurs. So what do we have then as forces? Now the momentum carries it in a straight line and Gravity pulls it down. So we have to determine if friction is better than losing height. Perhaps a really slippery track will be advantageous over the horizontal drop through the air. The vertical climb out of the upturned track is pure gravity working against the remaining momentum (some is lost in the friction of the upturn). So we can figure minus 9.8 meters per second for each second it travels upwards. Notice it does not matter how much mass it has, the negative force is there for 1 gram or 1 kiloton the same. But lets suppose we have enough energy to make back up to the top. You can calculate what is needed from the equations above by choosing your own dimensions for wheel diameter and projectile mass. Once the diameter is known, you can work out the desired kinetic energy to get the projectile up to the top for a given mass. And once you know the desired energy, you can then determine the angular velocity of the wheel, and that is very important for the next step.

The Catch 22:
We have supposed now, that the wheel is moving at a particular RPM needed to let a projectile leave it's periphery and reach the top of the wheel again. So far so good. We have done diligent work to ascertain a very carefully designed wheel to maximize the gravitational imbalance we produce for the mass and diameter chosen. And now we realize that the wheel is moving quite fast, but when the ball reaches the top it has lost all of it's energy and is all but stopped. We are now faced with the problem of trying to accelerate the ball to the wheels circumferential speed so it can join it at the top Sad

What do we do now?

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Sat Apr 11, 2009 8:50 am PostPost subject:
Harvey
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@Chrisbis,

Well done, Wink I think you are absolutely correct that the translator intended for us to read "Perpetual Motion" rather than 'Permanent Magnet'. Yep, makes perfect sense in that context Very Happy

So we are back to not knowing how he is doing what he claims to be doing, but Orffyreus may have perhaps known.

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Sat Apr 11, 2009 9:37 am PostPost subject:
chrisbis
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Harvey, overconfident,

Im imagining this simple (simplified) sketch just to get us all started on the same page.

IMME, (in my minds eye)
Take two disks- prob quite large in dia.
Mount them vertically, on the same shaft.
Seperate them with a spacer (say making the gap one tenth of the disk dia)
Also mounted, between each disk and this spacer, is a simple device like that that
changes the pitch on helicopter rotor blades, (poss two wedged shaped spacers)

(To be finished- need to see to family)
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Sat Apr 11, 2009 10:23 am PostPost subject:
Harvey
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Ok, I need sleep - BBL Wink
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Sat Apr 11, 2009 11:04 am PostPost subject:
Frank
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Harvey wrote:
@Frank,
If gravity pulls on the 'Wheel Side' of the 'D' then its acceleration is not linear from top to bottom, you have to integrate the curve from 0 (top perpendicular force, toward axle) to 90 (tangent force) and back to 0 (bottom perpendicular force, away from axle). Because gravity is conservative, the straight path down the vertical drop from top to bottom will have the same overall force as the integrated curved path independent of time. So, if the force is identical for each path taken, how does a differential in time effect the system? Velocity. For the vertical drop, the velocity increases throughout the path until it reaches the bottom. For the curved path the velocity is split between the two axis (x & y, with y being vertical). It takes time to move in the x direction, and time to move back, but the velocity for x will be negated, or net to zero. Therefore we have an extension of time, and we have a vertical velocity which ends at zero. The force is the same and therefore the energy is the same, but the action and timing are significantly different. In one case the 'ball' (integrating OC's concept here) has all of the energy stored kinetically. In the other case that energy is transferred to the wheel.
So I think a comparison chart between the two paths for dv/dt would be good to see. And then, if we compare each of those to the energy curve and the energy curves to each other, then we will have comprehensive way of looking at this. I would set the start criteria as "motionless" with potential energy at the top.


Thank you for your interesting treatise, Harvey. I'm afraid my forte is strategy, not tactics so I can't comment in any depth on the above (added: but it sounds good to me Smile ).

Having read the history of the Bessler wheel I am convinced that Johann had what he claimed.
Having seen the attempted reconstructions of the Sjack machine on the Overunity.com thread I can see that the source of power is the combination of an acceleration leg as the weights descend on the right hand perimeter of the wheel and the rate of change of acceleration leg as the weights rise up vertically past along the straight vertical of the D and past the left hand side of the axle. This gives me the gravitational equivalent of the Carnot cycle.

Frankly the small image of the machines action at the left hand top of Abeling's main page says it all. Exclamation



He shows the weights going round the right hand perimeter (d[sup]2[/sup]x/dt[sup]2[/sup]).

He shows the weights being driven toward the centre as they rise up the vertical straight of the D (d[sup]3[/sup]x/dt[sup]3[/sup]).

He shows that the straight vertical is to the left of the axle with all that implies about the unbalanced weight driving force.

As far as I'm concerned he had shown everything. He has shown me the Sjack cycle Idea and the energy area enclosed on the length entropy vs, time entropy graph.

As for the detailed implications, you are far better than me in working those out. I've done my bit. Cool

So good luck. I look forward to the result of your calculations. Very Happy

P.S. And by the way, gravity doesn't pull, it pushes. A correspondent of mine has build an experimental demonstration of this.

http://s136.photobucket.com/albums/q171/frank260332/?action=view&current=X1.flv

I'll try and find the details of the explanations behind it.
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Last edited by Frank on Sat Apr 11, 2009 12:53 pm; edited 1 time in total
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Sat Apr 11, 2009 12:38 pm PostPost subject:
Frank
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I'm reading up all I can about JB. The following site has a useful long paper on his life and work which I have downloaded as a .pdf.

http://www.scribd.com/doc/259189/Orffyreus-Wheel

I'm up to page 13 and it strikes me the the chief stumbling block in recognising the Bessler Wheel for what is, is people's inadequate view of the nature of gravity. If you recognise gravity as a powerful wind blowing steadily downwards then the Wheel is simply a gravitational windmill.

It's not getting energy from nowhere. It's getting energy from differential Delta-atmosphere pressure just at a Alpha-atmosphere windmill gets energy from differential air pressure. One is just tapping a source of energy on a continuous basis which up to now has only been tapped on a discontinuous basis. And it makes a damn site more sense than all that bollocks about gravity being some weird distortion of space. My goodness, there's soon going to be a lot of ground shifting among the academic community. Cracks will open and swallow some of them completely. Shocked
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Sat Apr 11, 2009 2:25 pm PostPost subject:
Frank
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I've come across quite a nice suggestion of how the Bessler wheel could work

http://www.orffyre.com/speculation.html

This represents the essence of the device I feel. If the attachment of the weights was more flexible and a barrier hangs down so they are drawn up vertically to get the full d[sup]3[/sup]/dt[sup]3[/sup] action, then we have the Abeling arrangement.

Two comments worthy of note:

His Highness, who possesses all the qualities that a great prince should have, has always had consideration for the inventor, and will not use the machine in any way for fear of the secret being discovered before the inventor had received a reward from foreigners. His Highness, who has a perfect understanding of mathematics, assured me that the machine is so simple that a carpenter's boy could understand and make it after having seen the inside of this wheel, and that he would not risk his name in giving these attestations, if he did not have knowledge of the machine...' - letter from Joseph Fischer to J.T. Desaguliers, 1721.

Orffyreus commented that when the secret is revealed, he is afraid that people will complain that the idea is so simple it is not worth the asking price.


This latter comment has been echoed by Sjack Abeling.
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Sat Apr 11, 2009 8:37 pm PostPost subject:
bano
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[quote="Frank"]
Harvey wrote:
@Frank,
If gravity pulls on the 'Wheel Sid
.........................................................................
demonstration of this.
http://s136.photobucket.com/albums/q171/frank260332/?action=view&current=X1.flv

I'll try and find the details of the explanations behind it.


By this moment I was think "I am a big deal"Smile))
Thanks Frank Nice pic's
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Sat Apr 11, 2009 9:07 pm PostPost subject:
bano
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[quote="bano"]
Frank wrote:
Harvey wrote:
@Frank,
If gravity pulls on the 'Wheel Sid
.........................................................................
demonstration of this.
http://s136.photobucket.com/albums/q171/frank260332/?action=view&current=X1.flv

I'll try and find the details of the explanations behind it.


By this moment I was think "I am a big deal"Smile))
Frank what mean a devise on video and what are these "Sean's bulbs"?
Thanks Frank Nice pic's

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Sat Apr 11, 2009 9:43 pm PostPost subject:
Frank
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[quote="bano"]
Frank wrote:
Harvey wrote:
@Frank,
If gravity pulls on the 'Wheel Sid
.........................................................................
demonstration of this.
http://s136.photobucket.com/albums/q171/frank260332/?action=view&current=X1.flv

I'll try and find the details of the explanations behind it.


By this moment I was think "I am a big deal"Smile))
Thanks Frank Nice pic's


I've found the letter.

==================================================
Hello Frank,

Hope you are fine, as you seem to be absent from the most interesting
places, including this group. I know you will love this one:

http://www.blazelabs.com/e-exp21.asp

remember to check out the movie on the same page.

We exposed some more lies in mainstream teaching, proved the existence
of the aether wind in terms of ultra-cosmic radiation, the pushing
force of gravity, and the possibility of making a flag which responds
to this wind. Next job is to work on the windmill.

Best Regards
Xavier Borg
==================================================

Mmmm.... he's a bit late. The Bessler wheel responded to the gravitational wind four centuries ago Rolling Eyes
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Sat Apr 11, 2009 10:38 pm PostPost subject:
overconfident
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And itanimuLLi, over at ou.com, has come up with a mechanism to move the balls back to the top of the wheel. Just place a wheel in between the chute and the teeter totter so the balls roll down the chute into a cup on the wheel and then get forcefully ejected onto the teeter totter to lift the stack of balls another increment. The wheel will need to spin fast enough for the ball to transfer enough momentum.

http://www.overunity.com/index.php?topic=7150.msg170654#msg170654
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Sun Apr 12, 2009 6:49 am PostPost subject:
Harvey
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overconfident wrote:
And itanimuLLi, over at ou.com, has come up with a mechanism to move the balls back to the top of the wheel. Just place a wheel in between the chute and the teeter totter so the balls roll down the chute into a cup on the wheel and then get forcefully ejected onto the teeter totter to lift the stack of balls another increment. The wheel will need to spin fast enough for the ball to transfer enough momentum.

http://www.overunity.com/index.php?topic=7150.msg170654#msg170654


Well lets see what we have here:

Well first there is a glitch in the sequence where he gets a free ball on the top by some sort of reset in the vid, but thats just a detail.

The intention seems to be that a ball impacting a lever and moving it 1r (the radius of the ball) on the high side, will provide enough gain in the leverage such that the much smaller movement on the low side will have great force and move 9 balls greater than 1r where they will be latched. In addition to the 9x force needed, we also have an additional 1.?x force of a spring that must be compressed so that the mover can be properly positioned for a new strike.

Lets say the mass of the ball is m. Lets say the radius of the ball is r. Lets say the Lever is selected at 10:1 so a mass of m will lift 10m. And lets suppose the spring is exactly 1m after being loaded with the lever, so it will lift the ball to exactly the highest point, no further, no lesser.

So the lever then will move 1r on the long side and r/10 on the short side. This is important since the energy of r/10 distance must be enough to move 9m a distance of 1r so it can be latched. Therefore, we need to supply enough energy so that 9m will accelerate in a distance of r/10 and have sufficient momentum to travel the other 0.9r to be latched (or perhaps a bit more to clear the latch properly). So to determine this, what velocity do we need to move in order to offset the -9.8m/s negative acceleration (deceleration) gravity will impose during this interval? If r is in meters, then we can use (9.8 * 0.9r) /s, which means for each second we are raising the stack we must add an additional 8.82r meters per second velocity. Naturally your thinking now, how fast will we raise the stack? That depends on r. Because r sets the height, which sets how long the ball freefalls which determines how fast it will be moving the lever down etc. Whatever size is chosen, the fall distance will always be 8r because the lever is up by 1r and the stack is 9r at rest.

So we are building an equation. On the left we want to know how much energy it takes to move 9m balls a distance of 0.9r in t_Lever seconds against a force of g. On the right side of the equation we want to know how much energy a ball of 1m has after freefalling for a distance of 8r under the force of g, where g = 9.8meters/second. And we are hoping that `E_L< E_R`.

`E_L = (1/2)9m(g * t_[Lever])^2` and `E_R = (1/2)m(g * t_[fall])^2 - (1/2)m(g * t_[Lever])^2`

`g = 9.8[meters]/s^2`

`t_[Lever] = sqrt([0.9r]/(0.5g))`

`t_[fall] = sqrt([8r]/(0.5g))`

So lets plug in some values and see how things equate:

For round numbers, suppose we choose a value for r of one meter, so r = 1 and the mass is 1 kilogram:

`t_[Lever] = sqrt([0.9 * 1]/(0.5g)) = 0.429`

`E_L= (1/2)9 * 1(g * 0.429)^2 = 79.54` Joules

`t_[fall] = sqrt([8*1]/(0.5g)) = 1.278`

`E_R = (1/2) * 1(g*1.278)^2 - (1/2)* 1(g * 0.429)^2 = 69.59` Joules

So, it looks as though we will have a 9.95 Joule loss in using the spring to position the ball.

There is the math; you can try different mass and different radius values to see if there is a height and weight that produce different results.

Also, my choice for the lever was a mental optimum because I imagined the farther the ball can fall the more energy it will have, but you can try other values for that and change up the ratio. Perhaps you would prefer the short side to be 1r and the long side to be 10r and use a 100r stack. Play around with the numbers and see what you can discover.

Also, I make mistakes all the time, so triple check my work Wink

Cheers,

Cool
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