Twist to Classic Wheel

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beapilot1
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Twist to Classic Wheel

Post by beapilot1 »

Please see the images.

This one is the classic wheel gravity lifting ramp:
https://imgur.com/a/hvN7U


This one corrects the problem by utilizing magnets to pull instead of the ramp pushing:
https://imgur.com/a/3PCtA
(in the image, I meant gravitational radial pull)

There are various ways to make it spin automatically and correctly, but the basic principle is to show that it looks possible. If magnets of various strengths and distances, in such an arrangement, then the steel will always be in equilibrium during lift in the radial direction, not effecting the wheel properties. A slight touch to the steel will cause the steel to disconnect from the wall. For stake of clarity of reset, simply at horizontal the steel can be placed to the left with minimal effort. Then again, there are different designs of which this wouldn't be needed.
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jonnynet
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Post by jonnynet »

I think I got it. So the steel ball can't push the wheel back when on the ramp because magnets pulling it upwards? If this is true, does the steel ball has still a mass for the wheel? If magnets are all around then the ball also has no mass for leverage? If magnets are on the left side only, ball could have leverage on right side but can't overcome the end of the magnetical track? These are the problems I see with this concept.
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re: Twist to Classic Wheel

Post by beapilot1 »

@ jonnynet

The images I showed are fundamentals. This allows you to apply the ideal situation to the PMM world- friction-less, mass-less wheel, and ignore the shape. All of those situations will come into consideration with design. That is why it is strictly stated as "steel." The steel are the weights of the wheel.

The magnets are already defined as such an arrangement, that wherever the steel is touching the black wall, it is in equilibrium.

This design is to allow the steel to be in equilibrium via radial direction of the wheel. In other words, to face the center axis of the wheel along the rods. If forces are exerted anywhere else, the wheel will be effected. The radial forces do not effect the wheel.

Please look at the images shown from the links that are large, clear, and provide instructions of what is going on. The image upload on here is bad quality due to compression requirements from this website.
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Mikhail
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re: Twist to Classic Wheel

Post by Mikhail »

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re: Twist to Classic Wheel

Post by beapilot1 »

@ Mikhail

Thank you for the video.

The video concept don't work because the magnetic pull is pulling via radial direction only, of which the radial forces do not effect the wheel. The concept I show, you have another weight on the other end further away from the center, causing an off balance. The key point though, is that the steel must be in equilibrium when it is being pulled by the magnet, only in radial direction. If any other force is being exerted by the magnet, it will effect the wheel.
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re: Twist to Classic Wheel

Post by beapilot1 »

The strength the magnet is needed is based on the following equation

magnet strength at degree of radial = abs(mg*sin(deg))

mg = steel
sin(deg) = the radial direction forces (actually absorbed by the wheel when stationary and are typically not considered when calculating the torque of a PMM wheel)

To see this, have a pendulum. Let it rest. It rest at 270* (6 o'clock). This is a abs(mg*sin(270)) force counter forced by the wheel which simply it hangs there. When you give it a little swing, it rocks a little, because there is not much of a drop, as most of the force is being absorbed by the radial direction. Now, if you go completely horizontal, then you have 100% of the force acting on the wheel and no force acting in the horizontal direction.

I hope that clears up some thought.
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re: Twist to Classic Wheel

Post by rlortie »

beapilot1,

I presume you are using an engineers method of initiating degrees of a circle. That is with nine o'clock being 0-360*, three o'clock being 180* and six being 270*

Some here may find this confusing as most associate 180* as six o'clock.

Ralph
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re: Twist to Classic Wheel

Post by beapilot1 »

@ rlortie

I am using the actual polar coordinate system.

I tried making it clear by using a clock, of which 9 o'clock is actually 180* and 3 o'clock being 0*.

I'd rather stick with engineering polar coordinates because I rather be right than trying to suit a cultural coordinate system that was developed on here.

Thank you though for letting me know.
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re: Twist to Classic Wheel

Post by beapilot1 »

Anyone disagree?
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re: Twist to Classic Wheel

Post by beapilot1 »

There are hundreds of people reading this thread. Very few actually say something.

Maybe it is too complex to you? Or it looks too simple to work?
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Post by eccentrically1 »

I think it's not getting any replies because it's been tried. Like this:
If we can't block the force of gravity on one side of the wheel, then surely if we can block the force of magnetism on one side of the wheel it will work.
But that's not how it works (or doesn't) :)
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re: Twist to Classic Wheel

Post by beapilot1 »

I think it requires more experienced PMM enthusiasts to gauge the concept. If you have a physics background, it will be easier to grasp.

Gravity is not blocked in this design. Gravity is needed in this design to work.

Because the forces are pushing toward the axis, it is unlike the video posted a few posts before this because that video is has a pull. Think of it like a string. You pull a bike tire with a string outward and try to spin the wheel. The wheel is stuck, because of the external forces of the magnet acting on the wheel.

Now, have the string pull through the axis instead, the wheel is not effected. Therefore, replace the string with a magnetic force. I hope this helps.
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Post by ME »

Unfortunately:
1: The negative - It can't work;
2: The positive - Try it out.

Option 1:
Work: The result of a force on a point that moves through a displacement, δW =F⋅s = F⋅ (v dt).

The energy gained by moving in the direction of a force (eg. a drop in a gravitational field) equals the hindrance to go against this force (eg. a lift in a gravitational field)... The same issues for the magnetic field.
And it works in relation to a single point (gain=-hindrance). It's cumulative, so moving through a changing force field looks like it's different at different positions but it's still an accumulation of many single points. No matter where you put your magnet (F⋅s) or its relative motion (v dt)
Yes, at some point gravitational force is counteracted by magnetic force...

To pull a weight in when at the center of a carousel is the same as working against centrifugal (or with centripetal)... release that weight (as like dropping it in gravity, but less uniform) and it goes back to the rim. Here too: gain+hindrance=0

And then there's option 2.
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-- May the force lift you up. In case it doesn't, try something else.---
beapilot1
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re: Twist to Classic Wheel

Post by beapilot1 »

@ ME

Thanks for trying to solve it. It seems like you tried to consider the wheel to move at a fast speed because you considered the centrifugal force. The reality is, in order to maintain equilibrium (stick to the wall such that such little touch will cause the steel to disconnect), the wheel must spin slowly.

The math you used don't add up correctly. Magnets are not simply formulated by force times distance. That is simply the definition of work. In this case, the magnet is not moving and neither is the steel as it crosses the single point.

What people don't realize is:
Magnets are at a potential when the magnetic field is not encountering any material. As the material approaches the magnet, kinetic energy appears. It is the opposite of a spring. When you pull back the material from the magnet, you are adding energy to the magnet, therefore having potential energy again.

Maybe I am not understanding your explanation. What I can say is that the reason why many of the ramp lift wheels don't work is due to the fact of the forces caused by external normal forces such as the ramp itself. The "ramp" in this case is the wall, of which is the opposite, allowing no external normal forces to counter the wheel.

The magnet must be in equilibrium or it will cause gate issues when escaping and it will cause a run to the strongest magnet situation.

When you said it moves through a changing field, when you move up, or when you move down the wall, all points are the same = equilibrium in the radial direction. The wall line is the equilibrium line of the magnetic strength. Just remember that the radial pull decreases as it nears the horizontal plane. The wheel is never effected because no external forces are acting on the wheel other than the axial normal force (magnetic force pulling the steel toward the axis which generates the normal force due to the axial of which cancels out the forces).

The reason why I put this on here is so that someone will be interested enough to try to build it.

Edit: if the wall was a perfect circle like the wheel and the magnetic strength vary, the steel never loses its position in relation to the radius. The entire wheel will be in equilibrium as if the magnets are not there. Now we have a wrap wall that closes in towards the center, there is a normal force acting on the wheel caused by the wall in the direction of where the wheel spins (clockwise) of which supports the spin of the wheel. Because the pull is designed to make the steel in equilibrium, the steel can move along the wall in both directions. Increasing magnetic force all together will force the ball up the wheel with support of the counter weight but the catch is the gate and the inability to pull apart to reset. That is the reason for the magnets to be in equilibrium.
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Post by eccentrically1 »

The gain/hindrance = 0 part of ME's reply is the relevant bit. Magnetism is a conservative force, just like gravity. It can't be exploited for an energy gain. The magnet is actually going to stop this design before it would even make one rotation. Maybe someone will build it here, but I'm pretty sure it's been tried.
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