Zeroing in on Bessler's wheel

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johannesbender
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Re: Zeroing in on Bessler's wheel

Post by johannesbender »

Sam Peppiatt wrote: Tue Dec 05, 2023 12:43 am Tarsier,
I was referring to shifting weights in & out on a radius. But no matter, either way it doesn't work. however , if a scientist says it's because of minor losses due to friction; I say he doesn't have a clue----------------------Sam
If a weight is fixed to a wheel and you let it go it will get back up to about eleven and no more. That's the zero or less sum game, right? And, I agree it lacks the energy to get all the way back to 12. However, if I try to shift a weight out on the down side and back in on the up side, I'm lucky to get back to 8.
Imagine a pendulum bob on a string , it falls from 12 and reaches about 11 , this is already a singular demonstration of losses of which one example is the bob colliding with air molecules and the larger the surface of the bob the more air molecules collide with it (air drag / resistance ) because air has mass (think of a tub filled with balls as the air molecules and the pendulum bob swings through it) , a kite being pushed up with the wind or a windmill turning is already an example of the air's ability as a force and mass.

Imagine your car stranded by the side of the road which is a flat horizontal surface, you have to push it out of the road and the push across the distance it has to move takes work , the work is the force you have to apply times the total distance you have to apply that force across ,to do that work you have to transfer energy , if you have no energy you cant push it (imagine running a mile and trying to then push a vehicle) but if you have energy you can push it .

The car has mass and the larger the mass the larger the inertia , inertia is just how much something resists change , so the more mass or inertia an object has the harder it is to get it to change direction or start moving .

Imagine your pendulum's ability to reaching 11 o' clock , as a car being able to travel 11 miles/km on its fuel , but by taking extra trips to other places it burns off fuel , which means less fuel is less distance it can travel in the end it only reaches 8 miles/km.

With this in mind as an example of the pendulum having energy losses , if the pendulum goes out on the downside and in on the upside , these in and outwards motion costs energy usage (like you needing energy to push the heavy car or the fuel being burned by extra trips) , and the only energy you have available is that from the falling weight , when that weight is being forcefully moved it costs energy which would be subtracted from the total energy it had (when you pushed the car you burned energy or when the car travels extra trips it burns fuel) , before it would reach 11 o clock and now it would reach for example lets say about 8 o clock as your given example , if you run a mile and then push a vehicle you cant push very far because you don't have enough energy but if you did not run a mile and then push the vehicle you can push it further , or if the car has not burned off fuel from extra trips it could reach 11 miles/km instead of 8 miles/km.

There are also many ways that energy is lost in a mechanical system apart from air resistance and friction too , all the losses and expenditure of energy adds up .
Last edited by johannesbender on Tue Dec 05, 2023 8:31 am, edited 1 time in total.
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Sam Peppiatt
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

jb,
That's the point I was trying to make-----------------Sam
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Re: Zeroing in on Bessler's wheel

Post by andyblues »

jb hi your end point that all the transference of forces and frictions all add up is so on the point ,i have come to the under standing now that there has to be a spring that is compressed and releases putting the wheel back in to balance ,which in its self is drive to get these wheels to work, as i see it nothing being moved and allowing to naturally reset works so banging on a spring a bow has to be the answer and yes in on the lift and out on the fall together looks like the only place to really get the best torque but if you don't reset after 45 degrees it aint going nowhere the reset can go from 45 to 90 and i think the momentum from the reaction could take it on to there but really it looks like a keel again un less the weights them selves actually carried energy to help this 45 degree conflict, it certainly could do when moving fairly fast, but you got to get it there first and that looks tricky but i do agree that radial exchange on both sides above the axis looks like the best place .

The thing that always throws me all over the place when thinking about where to move weight is between the zenith and the right angle i think the answer may lay in both, but it looks like they can both be as rewarding and unrewarding which is why i think a spring is the leaver, the system has to reset with force with in 90 degrees as my little brain sees it today ,thank god there is tomorrow Andy
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Re: Zeroing in on Bessler's wheel

Post by johannesbender »

Imo , ultimately it all boils down to reset , you can go sideways in and out or vertical in and out movement , you can pendulum or spin or collide or bounce ... I think whatever the method used by the design , if you don't manage to accomplish the reset its just not going to cycle , of course I am excluding the notion of whether it would have the ability to keep cycling or not .

Whatever the method of choice ,(if its really true) Bessler had something that was not stopped by average losses and work output.
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

It's always been a two part problem. I finally figured out the best way to drive the wheel is, with a binary driver,(see top of page 102), because you never have to lift the weights / rollers. Since you don't lift them you avoid all the losses associated with that.
Of coarse some thing has to lift the weights that drive the binary driver. The best way, and probably the only way, is with a binary lifter. In my case it lifts 32 lbs. 3 inches every revolution, all the while retaining prefect balance. with little or no losses.
The next step is to hook the two together, with a toggle link. Before the rollers would just roll backwards. Hopefully the back stop clutch will prevent this. What it will do now, I do not know-----------------Sam
Last edited by Sam Peppiatt on Wed Dec 06, 2023 2:26 pm, edited 1 time in total.
andyblues
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Re: Zeroing in on Bessler's wheel

Post by andyblues »

Hi Sam i see your approach there, my issue is how do you create lift, some thing has to take the wheel out of balance and your approach seems to be based on rotation, as i see it, i have built inner rims and made them heavy as you are showing in your approach, but it has to back keel even if its light as you say over the whole rotation , so it does not do what besslers wheel did which was start with a heavy force and it does not reset with force at the right angle , what i will say i think you have there is a good store of inertial momentum not useful when static but well useful when moving, i found this approach very interesting and came as close to any thing i have ever built but its forces were to weak to really keep looking at it so they went in my very large pile of attempts ,good luck there if you feel i have missed the point please have another go at explaining nice work and good luck there Andy
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Hi andyblues!
You are the first person to show any interest in it, which I'm very thank full!! There are two sets of rollers connected together by a cross bar, a little over a foot apart. The vertical set,(see top of page 102), does the heavy lifting. Each roller weighs 16 lbs., together 32 lbs. That is to say, they lift and lower 32 pounds each revolution of the the wheel. They don't turn the wheel. As the bar moves up and down, with 32 pounds of force, it operates a toggle linkage.

This linkage is connected to the horizontal bar, that connects the horizontal set of rollers together. Pushing on them side ways makes the wheel turn. If pushed to the left the wheel will turn CCW, if to the right it will turn CW.

The giant problem I had; the first set of rollers would roll backwards instead of pushing the other set sideways. Hence, the need for the backstop clutch,(one-way bearing), It could be a runner, I just don't know.

The toggle link is required to change the vertical forces to horizontal and, to reverse forces every 1/2 turn of the wheel.

If you have any questions, any thing at all, please let me know-----------------------Sam
Last edited by Sam Peppiatt on Fri Dec 08, 2023 8:55 pm, edited 2 times in total.
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

andyblues,
You said some thing that took me a while to get through my thick scull. Some thing about the wheel won't start / won't turn. But it will turn. Please go to page 102, the top of page 102, and look at the drawing that mrry made. If both rollers are pushed or rolled, both at the same time, to the left, the wheel will turn CCW. If you push, again, both at the same time to the right, the wheel will turn CW. I think that's the big problem. No one on this forum can figure that out.

Not even Fletcher, he can't figure that out either--------------------------Sam
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

Correction,
I've been saying that the rollers lift 32 lbs., 3 inches every revolution of the wheel. It's actually 32 lbs., 3 inches every 1/2 turn of the wheel. A lot more than what I was thinking------------------Sam
Last edited by Sam Peppiatt on Sun Dec 10, 2023 7:39 pm, edited 1 time in total.
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

A mini update,
Yesterday completed the new backstop assys. What's interesting is the assy. or link, has translating motion. That is to say, it goes around and around with the wheel but, doesn't rotate.

This is important because, one-way clutches don't work if they rotate with the wheel------------------Sam

ETA
The link connects the small friction rollers, to the directional bearing mounted on the axle of the heavy disks.
Last edited by Sam Peppiatt on Fri Dec 15, 2023 9:06 am, edited 1 time in total.
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

spinner361, I have absolutely no knowledge of copy & paste, what it means or, how it might be preformed. I'm not a computer guy, I have no clue how you would do that-------------------------Sam Peppiatt
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Re: Zeroing in on Bessler's wheel

Post by spinner361 »

Ah. Understood.
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Fletcher
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Re: Zeroing in on Bessler's wheel

Post by Fletcher »

Sam Peppiatt
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Re: Zeroing in on Bessler's wheel

Post by Sam Peppiatt »

For what ever the reason, probably do to your superior intellect, you and others, are able to understand and follow the instructions. What ever that ability is, I don't have it; the instructions become meaningless to me. I don't knew what to do, or how to proceed. I'm freqing lost!

It must be nearly impossible, for any of you to understand that--------------------------Sam
Last edited by Sam Peppiatt on Sat Dec 16, 2023 1:55 pm, edited 1 time in total.
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Re: Zeroing in on Bessler's wheel

Post by Robinhood46 »

There are things that can cause copy and paste to not function. In this case you can do everything you are told/advised, perfectly well, and you will not be able to do it, or understand why you cannot do it. It is the sort of thing that can be very frustrating.
If there is someone you trust enough to let them access your computer from a distance, which means anyone you trust enough, anywhere in the world who has an internet connection, they can help you with these sort of problems. It must be someone you know you can trust because they have the possibility of doing whatever they wish on your computer.
My son frequently accesses my computer, to help me understand the stuff i can't get my head around, and he lives 150 miles away.
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