Bessler's use of Gravity

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james kelly
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Bessler's use of Gravity

Post by james kelly »

I hope that I manage to start something here that will benifit all. Bessler made use of gravity. Now there is a discussion of gravity as a conservative force. Some have said that you cannot increase gravity.I ask a simple question of Ken about what happened to the law of acceleration in a free fall. whoooose ..... right over his head. When Bessler wanted to increase power, speed suffered. he simply made the same weight fall farther. I know that I am not wrong in this. I will put it up for discussion. jim kelly
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re: Bessler's use of Gravity

Post by ken_behrendt »

Jim Kelly...

I think I recall your mention of whatever happened to "free fall" a ways back. Perhaps I did not address it because I was distracted by other matters.

In my concept of how an overbalanced gravity wheel works, the eccentric location of the CG of the wheel's driving weights leads to a difference between the average vertical ascent and descent velocities of the weights during wheel rotation with the average vertical descent velocity being greater in magnitude. When one adds these two velocities in opposite directions together, one finds that all of the driving weights in the rotating wheel behave, on average, as though they were continuously descending! And, apparently, this situation then determines how the weights will interact with a planet's gravity field.

The result is that the rotating weights within the running overbalanced wheel will, even though they move through a closed path and rise and fall vertically through the same distance, continuously lose rest mass whose associated energy then is converted into the kinetic energy that the wheel outputs and which accelerates all rotating parts of the wheel. Some of this outputted kinetic energy can, as was demonstrated by Bessler, be used to perform useful work in the wheel's environment.

However, it is important to realize in this model that no free falling of the weights is involved. That is, although they are effectively continously falling as far as gravity is concerned, they do so with a constant velocity given by that equation that I show in the signature of my posts.

From Bessler's 4th Law of Motion, it is possible to derive an expression for the power output, P, of a running overbalanced gravity wheel and the expression for that is:

P = - 2 √2 π M g d ω cos φ

where π is the constant pi with value of about 3.1415927, M is the total mass of the wheel's driving weights, g is the acceleration due to the planet's gravity at the elevation of the wheel, d is the distance between the wheel's axle and the CG of the rotating weights, ω is the rotation rate of the wheel (i.e., number of rotations per unit of time), and φ is the angle that the CG makes with a horizontal line passing through the wheel's axle.

We see from this that, to increase the power output of a one-directional wheel, one must increase the total mass of the wheel's driving weights, locate the wheel in a stronger gravity field, increase the distance between the wheel's axle and the CG of its weights, or increase the rate of rotation of the wheel.

I suspect that the biggest obstacle that Bessler faced as he tried to increase the power output of this wheels was the CF that was created as the wheels began to accelerate. As a wheel accelerated, it's power output would have tended to increase, but the wheel's resulting increased rotation rate would also increase CF. As CF increased, it would have began to interfere with the automatic weight shifting action being performed by the wheel's internal mechanisms and this would, most likely, have caused the CG of the wheel's driving weights to drop down toward the location directly below the wheel's axle (the punctum quietus) which, as we see from the above equation for power, will cause cosφ to go to zero and, thus, also cause P to go to zero power output.

Thus, as a result of all of this, a running overbalanced gravity wheel will have its maximum power output upon startup and, thereafter, the power will decrease with increasing wheel rotation rate, ω, and increasing CF this causes.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Bessler's use of Gravity

Post by LustInBlack »

Ken your posts are soooooooooo long to read...

My eyes hurt . ..

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re: Bessler's use of Gravity

Post by james kelly »

KEN; Just pretend the the effort to raise the weights back to the top is zero, and you have Bessler's largest wheel. Calculate the falling weights effect. jim kelly This is not a trick
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re: Bessler's use of Gravity

Post by james kelly »

What has happened? Has everyone given up on the answer to this? It is certainly not hard to calculate. All you have to do is to figure what you Assume that you know about Bessler's largest wheel and what you assume the weights weighed. Zero up. HOW MUCH DOWN?
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re: Bessler's use of Gravity

Post by ken_behrendt »

James Kelly...

I'm not sure exactly what you mean by saying that no effort is required to raise the weights from their bottommost point within a rotating chronically overbalanced wheel to their topmost point.

However, Bessler's 4th Law of Motion does indicate that, as far as the energy released from the weights in such a wheel is concerned, we can, indeed, consider the weights to be effectively continously dropping and not rising at all!

I think you are going to have to provide some more clarification on what you are suggesting.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Bessler's use of Gravity

Post by james kelly »

Ken; You stated a fact of besslers. In this wheel, no weight are raised. ALL are descending.......jim kelly
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re: Bessler's use of Gravity

Post by LustInBlack »

How could that be possible ..
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re: Bessler's use of Gravity

Post by bluesgtr44 »

Uhhh....a good start would be to stop pinning your weights to the wheel.
When Bessler wanted to increase power, speed suffered.
Not exactly true, Jim K...The second wheel, the Dratschwitz, had more power than the first one, but was almost the same speed. Both are noted to have exceeded 50 RPM's. The big difference was the diameter of the wheel itself....not the RPM's. The third wheel, Merseberg, was said to have rotated at over 40 RPM's and was able to lift almost twice as much as it's predecessor. Now, the Kassel wheel is the big difference as far as RPM's go...my take is that the speed did not have as much to do with the amount of work the wheel could perform as the diameter of the wheel did. And if this is so....then the increase in power would be associated with the amount of...maybe, leverage that can be applied to one side of the wheel.


Steve

P.S. I won't even go anywhere near the 4th law thing.
Finding the right solution...is usually a function of asking the right questions. -A. Einstein
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re: Bessler's use of Gravity

Post by james kelly »

Steve: You are on the RIGHT track! Keep thinking about what I am saying. jim kelly
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re: Bessler's use of Gravity

Post by ken_behrendt »

Steve wrote:
Uhhh....a good start would be to stop pinning your weights to the wheel.


This phrase is bandied about on this board on occasion, but, so far, to my recollection it is never explained in detail.

If one has weights driving a wheel, then, at some point, they are going to have to attached to either the wheel or its rotating axle. That is, there must be some physical contact between the driving weights and the drum that contains them or no torque will be applied to the wheel.

Rather than vague phrases, it would be nice if members could post a sketch now or then of what they mean by not "pinning" one's weights to the wheel.


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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re: Bessler's use of Gravity

Post by james kelly »

I think that the statements on here from others explain why that more progress is not made on ou. The method was spelled out by Bessler. The comments: Spell it out! That is impossible! Prevent them and others from accomplisment. I usually stay Pretty Quiet. I know by my own experience, the ideas projected, , will produce a superbly working unit. jim kelly
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re: Bessler's use of Gravity

Post by graham »

Quote from Steve :
then the increase in power would be associated with the amount of...maybe, leverage that can be applied to one side of the wheel.
Another factor would be the mass of the weights used ,ie heavier weights for more power.

I would also like a explanation of the phrase :
stop pinning your weights to the wheel
As it stands it makes no sense to me either.
Steve could you expand on this statement please.

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re: Bessler's use of Gravity

Post by bluesgtr44 »

I really thought that this had been covered...evidentally not.

This is not to pick on Ken...but, here is one his oldie but goodies. I have drawn the lines around the concentric paths of all of his points. No matter how you tweak the weights or the spring tensions...you have to tweak them all to the same. Basically, what have you changed? Nothing...maybe the keel point.

If this set up were to revolve, can you see where the forces would apply themselves to the concentric points of restraint...considering the reaction forces on the ascending side, how many revolutions does anyone think it would take before those reaction forces not only will cause the wheel to stop, but actually create enough counter force that the wheel stops dead and immediately starts spinning in the reverse direction. Those reaction forces, if left unchecked...will doom a design of this type.

AP pg. 295...J. Collins...
"Many would-be mobile makers think that if they can arrange for some of the weights to be a little more distant from the centre than the others, then the thing will surely revolve. A few years ago I learned all about this the hard wy. An then the truth of the old proverb came home to me that one has to learn through bitter experience...."
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No matter how you look at this, the forces will be applied at the fixed points of the design...and the points are all concentric.
No matter how you look at this, the forces will be applied at the fixed points of the design...and the points are all concentric.
Finding the right solution...is usually a function of asking the right questions. -A. Einstein
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re: Bessler's use of Gravity

Post by ken_behrendt »

Steve...

You still have not defined what you mean by "a good start would be to stop pinning your weights to the wheel". If the weights are to apply any torques to the wheel, then there must be some sort of a physical connection between the weights and the wheel.

The model wheel you attached is, indeed, one of my earlier designs using restoration springs. It was a kind of "pullover" design in which I hoped that as the wheel turned CW, the springs would be able to suddenly pull the weighted levers that they were attached to over toward the rim of the wheel at around the 1:00 position on the wheel. The idea was that this would cause a sudden shifting of the CG of all of the weights to the right of the wheel's axle which would then, of course, imbalance it and allow it to acclerate more than enough to bring the next weighted lever into position to repeat the process. Sadly, as can be seen from the attached graph, the weight's CG dropped below the wheel's axle, there was insufficient kinetic energy in the wheel to lift the CG high enough to create continuous motion.

My design did "pin" the weights to the wheel via the stops, springs, and lever pivots. What other alternatives does one have?


ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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