Zeroing in on Bessler's wheel

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PeterAX
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re: Zeroing in on Bessler's wheel

Post by PeterAX »

Hi Sam,
You have a new idea? Congratulations! You are really a generator of new and non-standard original ideas! Some more details about your new concept?
Regards,
Peter
Sam Peppiatt
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re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Calloway,

It's very close to what you have described. Only instead of a wire I use a long spring / bungee cord that connects the two pend. together. With both pend. vertical, one at 12:00 and the other at 6:00 I have a pin 2" inches up on the one at 12:00 and 2" down on the one at 6:00, (which I call offset), for a difference of 4 inches, which will stretch the spring eight inches! The cord connects between these two pins.

The dam able thing is; it will rotate the wheel CW for 180 degrees, then CCW for the next 180 degrees. There has to be a better way to stretch out the spring. maybe you can think of a way-----------------
Thanks for the encouragement!! So much approved over ridicule-------------------------Sam
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re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Hi peter!!

The new idea was a bust! It simply deferred the forces from one to the other. What can I say, it was a step backwards, another part of the zigZag---------------------Sam
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re: Zeroing in on Bessler's wheel

Post by Georg Künstler »

Hi Sam,
correct. I have something that will hit the rim of the wheel and will produce in addition the upswing. But the hit is comming from inside of the wheel.

So the energy of the hit is not wasted, it is transformed in a downwards torque of the wheel. while the pendulums swing up.

The function is compareable like you have a compressed spring at the 3:00 o'clock position on the rim of the wheel.
When the spring will be release, it will shot up a weight on one side of the wheel and turn the wheel downwards.
Best regards

Georg
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re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Georg,

I'm catching on. Do you use the one way bearings to hold the pend. up?
Maybe you are on the right track.

I can't seam to resolve it--------------------------Sam
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re: Zeroing in on Bessler's wheel

Post by Calloway »

Hey Sam, I really think you may be on to something. I can set my wheel up with 2 pendulums pretty quick as I still have the one way bearings mounted on the pendulum bars. Maybe I can come up with a way to kick those pendulum pairs. Been thinking about this for a while now. You keep working on it also. A bunch of heads are better than one! That's what this group is about or should be!

Cheers
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agor95
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re: Zeroing in on Bessler's wheel

Post by agor95 »

Hello Calloway
You keep working on it also. A bunch of heads are better than one! That's what this group is about ...
It is :-)
[MP] Mobiles that perpetuate - external energy allowed
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re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Calloway,

I just thought of some thing. I used to work on large instrumentation tape recorders. They had a brake, I think it was on the take up reel, a band brake that was spring loaded. It would grab one way, but release the other.

Maybe that would keep the pendulums from swing back wards / out, on the up side. Who knows----------------------

Like you told me a long time ago, if I lift the pends. up, the wheel will run all day---------------------------------------Sam
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re: Zeroing in on Bessler's wheel

Post by Georg Künstler »

Hi Sam,
I have also tried to control the backswing of a pendulum.
I can control the backswing with a one way bearing but that will not lead to an OU effect.
It still will act as an normal pendulum.
The deflected pendulum leads to a single partly rotation of the wheel until the lowest point of the mass is reached.
There is no restore of the lost energy.
So in the normal construction you will always loose the potential energy in the form of m*g*h, in effet you loose (h)igh of the mass.
The wheel will stop.

To gain energy from gravity we have to use the impact for our advantage.
An impact/hit with an different acceleration than g.
I will transform the general formular E=1/2*m*v*v
so that you can understand what I am talking about.

When you drop a mass from any high m*g*h = 1/2*m*v*v.
For v we have the formular v=a*t (acceleration multiplied with the time)
In our formular m*g*h we have g which is a constant of 9,81...
An impact/hit will result in an high acceleration which is beyond g.

A pendulum which you move horizontal with the speed v will have the energy E=1/2*m*v*v
The pendulum has a suspension point. So the suspension point is first also travelling with the speed v.

When now the suspension point is stopped you get an upswing of the pendulum. The sidewards movement is transfered back to m*g*h.

So we must do a combination of the move.
We allow a sidewards move, an impact/hit and a upswing of the mass.
Best regards

Georg
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re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Hi Georg,

Fascinating!! I think I get the sense of it. You are using the RKE of the pendulum more than it's gravity component, if that's the right word. It's kind of like a reactionless drive, only in a circle. Seams like great speeds could be reached with almost unlimited power---------------------------Sam
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re: Zeroing in on Bessler's wheel

Post by Georg Künstler »

Hi Sam,
that is right.
Masses are always accelerated from gravity.
The acceleration multiplied by the time will give us the speed.

So the energy output which we will get is a function of the mass, the acceleration and the time.

see as a example
https://www.youtube.com/watch?v=0_Cgxy7N-V0&t=285s

Only holding the mass of the airplane above the ground will consume a lot of energy, the tank of the airplane will be emptied soon over the time.
How much gallons a second ?
The energy consumed is exactly the energy we can get.
It is a function of mass, acceleration and time.

As heavier the airplane will be, as more energy is used up per second to keep it away from the ground.
Best regards

Georg
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Re: Zeroing in on Bessler's wheel

Post by Calloway »

Sam, What about the drawing I show below? If it has a chance, I think a minimum of 4 pendulums would be required. MT13 with a twist.

Cheers
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MT13 Twist
MT13 Twist
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Hi Calloway!

I don't think it would work, because the stator is stationary. Any kind of stationary stop or guide, just kills it------------------------------Sam

PS However, It's the right idea. If the pendulums swing in and or out the wheel will turn.
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Calloway, (or any one),

If you can find a way to control the pendulums the wheel will turn. I beginning to realize that one spring won't work. Each pend. has to have two springs. One spring to flip then in on the up side and a 2nd spring to flip them out on the down side. Please let me know if you have any questions----------------------------------------Sam
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Re: Zeroing in on Bessler's wheel

Post by silent »

Now you're thinking! The key to the Bessler wheel is a harmonic oscillator in a rotating environment! The energy you lose in a spring, comes back 100% when you suck a little centrifugal force for the reset. Then build several of the mechanisms and stack them.

silent
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