This is a build I did years ago. It is a variation of Mt 31. And in a way, it was a crude build. But then, I have been learning wood working.
The math is fairly simple. If a .5 Kg weight is 1 meter from center and it drops 15 cm's, it can lift another .5 Kg weight 15 cm's. And it is for this reason why perpetual motion is considered impossible. One weight can not lift another weight higher, it just won't happen.
With Mt 31 (in the manner of), if a .5 Kg weight drops 15 cm's, it will require
0.735499 n-m's of force to compensate for it. By using a 2nd lever, an extra
0.735499 n-m's of force is available. And if 1/2 of the 0.735499 is used to lift 2 weights equaling .5 Kg's, then about 1 n-m of work will have been performed.
This is because 2 - 0.5 Kg weights dropping 15 cm's each perform a total of
1.470998 n-m's of work will have happened. But since only one weight at the end of a lever is out of balance, then only it's imbalance will need to be compensated for.
edited to add; in reality, what I built was wrong. The levers do not have a fulcrum near the opposing weight. Because of this, a lot of force was lost. And if anyone is interested, the way I arranged the levers and chords, the weights actually lost force. Anyway, I know what I'm doing now and with any build, some trial and error might be necessary but if a person can't play around with a build, not sure but that should be half the fun. You know, doing.
Mt 31
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Dave Roberts
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- Location: Spencer, Indiana
re: Mt 31
I too built this one years ago with no success so I modified it by making it a solid wood disc and mounted short shafts at the pivot points. On the shafts, I put small sprockets on bearings with chains driving larger center mounted sprockets mounted on a clutch bearing driving a main shaft one direction, freewheeling in the opposite. This utilized both sides of the disc and I was very surprised how hard this wheel tried to continue rotating! It didn't, but caused me to devote considerable time trying to make it work. I finally went to other designs. Maybe I missed something and admittedly, it was very complicated.
Seems I commented on this before. It had a lot more potential than a lot of others I tried.
Seems I commented on this before. It had a lot more potential than a lot of others I tried.
Dave
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james.lindgard
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re: Mt 31
Dave,
This is something I think will work. Do you have any pictures of what you built ? Also, any math ? with this, to keep it simple, if all 4 weights have the same mass it simplifies things.
What this does is to give a frame of reference. If both levers go from one side of the wheel to the other, then they'll work better. And if a weight on one lever drops as an example 4 inches, then 2 weights can be lifted 3 inches.
The 6 inches of lift would be 75% of the work being done by 2 levers working together.
To get an understanding of if this will work, math will show if a net balance exists. The diagram shows it's basic algebra and trig. Basically is A+B < C+D. And the force is always calculated at 90° to the line A-C.
This is something I think will work. Do you have any pictures of what you built ? Also, any math ? with this, to keep it simple, if all 4 weights have the same mass it simplifies things.
What this does is to give a frame of reference. If both levers go from one side of the wheel to the other, then they'll work better. And if a weight on one lever drops as an example 4 inches, then 2 weights can be lifted 3 inches.
The 6 inches of lift would be 75% of the work being done by 2 levers working together.
To get an understanding of if this will work, math will show if a net balance exists. The diagram shows it's basic algebra and trig. Basically is A+B < C+D. And the force is always calculated at 90° to the line A-C.
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james.lindgard
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re: Mt 31
To give you a better idea, if the wheel rotates 90°, then the weighted lever at top is out of balance. If it is 15° out of balance, then it requires about 1/4 of it's mass to be over balance as realized by the A-C shift.
And if weight D dropped 4 inches, then weights A & C shift 3 inches. this means going from 16 away from center to 13. And the net force by such a shift is 3 in. lbs. And 4 oz's @ 16 inches (out of balance D weight) is 4 in. lbs.
This is almost funny. It seems to explain why using bellows will work. with bellows, the mass (weight) is shifted from one side of the wheel to the other.
One thing that might be worth considering is having the weights A & C shift after passing the A-C line. At a 45° angle they will require 30% less force to cause then to shift. Also, the use of a scissor will get extra distance in the shift. This is kind of why I think playing around with Bessler's work should not be frowned upon.
And if weight D dropped 4 inches, then weights A & C shift 3 inches. this means going from 16 away from center to 13. And the net force by such a shift is 3 in. lbs. And 4 oz's @ 16 inches (out of balance D weight) is 4 in. lbs.
This is almost funny. It seems to explain why using bellows will work. with bellows, the mass (weight) is shifted from one side of the wheel to the other.
One thing that might be worth considering is having the weights A & C shift after passing the A-C line. At a 45° angle they will require 30% less force to cause then to shift. Also, the use of a scissor will get extra distance in the shift. This is kind of why I think playing around with Bessler's work should not be frowned upon.
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Dave Roberts
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re: Mt 31
Sorry James. That was several years ago and I kept nothing. It was one design that showed a lot of promise but was very complicated. In no way would I discourage anyone from working on this approach. Bessler indicated he would sell one design but had others that would work. Maybe this was one?
Dave
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james.lindgard
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Dave,
I am starting to make some scissors to demonstrate a specific set of mechanics. With something like this, a 3 inch shift might become a 6 inch shift with no extra work being performed. The link is to a quick video I made. It shows a 1 lb. weight generating 3.79 lbs. of force. What this means is that if the 1 lb. weight drops 3 inches, it might be able to lift a different 1 lb. weight over 3 1/2 inches. If so, then a slight advantage could be gained using this design without scissors.
https://youtu.be/Lo6jysYLAoA
I am starting to make some scissors to demonstrate a specific set of mechanics. With something like this, a 3 inch shift might become a 6 inch shift with no extra work being performed. The link is to a quick video I made. It shows a 1 lb. weight generating 3.79 lbs. of force. What this means is that if the 1 lb. weight drops 3 inches, it might be able to lift a different 1 lb. weight over 3 1/2 inches. If so, then a slight advantage could be gained using this design without scissors.
https://youtu.be/Lo6jysYLAoA