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Thread: Bessler's Wheel

  1. #61
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    Smile

    Hopefully things will go a little better now

    With the 45 degree angle shown (1:30) changes the amount of over balance. If the weight "flips" at 3 o'clock then
    12:00 o'clock to 3 o'clock is balanced by 3 o'clock to 6 o'clock. This means it is the same as 6 o'clock to 12:00 o'clock.
    With over balance happening from 1:30 to 4:30 then that will allow for 90 degrees of rotation of over balance.
    With what I am showing about Mt 20 is how I calculate force and work. The 2nd drawing is what might work. By doing
    things this way it might be easier for everybody to understand. Then if someone wants to try the 2nd drawing they can.
    I do have a design I want to try which is more advanced. And like what is shown some usage of either Conservation of
    Angular Momentum or just requiring less work to reset the levers lifting a weight or possibly even both weights. I am
    kind of hopeful that once what i'm explaining is understood that there might be some questions on people's minds.
    Back to Mt 20. If the weight's fulcrum is 1 meter from the center of the axle and the weight being rotated is 15 cm's
    from the fulcrum it would need to be lifted between 25 & 26 cm's. Why I say lifted is because of the weight moving upwards.
    This means the weight on each opposing lever would need to drop 15 cm's. And this is where we can consider if enough
    over balance exists to reset both opposing levers.
    What does become apparent is that "lifting" a weight 25 cm's to achieve 15 cm's of overbalance is wasting work. And this
    is why shifting a weight is more efficient. what will be realized is that weight's will need to move closer to the axle. This is
    because it's path becomes shorter as it rotates around the wheel. And this can reduce the work needed to move any weight
    on a wheel.



    Jim
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  3. #62
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    This might show things better. Both levers would drop lifting or rotating the top weight. If the center of mass for the top weight
    moves 15 cm's away from the axle then the weights on the 2 levers need to drop at least 7.5 cm's each to perform equal work. With
    this there is a way (hypothetically at this point) that will allow the wheel to rotate with the levers needing less energy to rotate with
    the wheel.


    Jim

    p.s., I'll start showing the math. I'm not sure how familiar everyone is with torque and work and this is a little different
    so will not be trying to offend anyone.


    edited to add; the over balance as shown is 7.5 cm's (3 inches) and 500 grams (20 oz.'s ?). If the work done is calculated then the weight on the long lever by the hub is
    500 g's * (Pi15 cm's = D) = W no. 1

    The over balanced weight is calculated as
    500 g's * (Pi * [1 meter *7.5 cm's] = D) = W no. 2

    Divide no.1 / no. 2 = ratio of work performed

    what that will show is that the over balance doing the work does much more work than what is necessary to reset the 2 long levers. Could / should work as posted.
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    Last edited by James_; 28th March 2017 at 10:00 AM. Reason: add 3rd image and math

  4. #63
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    I did some math and I realized one issue that I've had with the metric system and that is calculating work. What I did was to calculate work in N-m's of work.
    Everyone knows W = M * D (work = mass x distance) so what I did was to multiply N-m's of force by distance. And with what I showed in Mt 20 it does show a net force.
    While Bessler showed his weights rotating and hitting stops, they might have. Witnesses said that they heard 8 knocking sounds per revolution. An easy way to account for this is to have the weights rotate at about (45° ATC). And with 8 weights, one would be placed every 45°.
    With the math, hopefully someone will go over this;

    This is what I keep coming back to;
    (5.635 N-mf - 4.9 N-mf) * (1.15m * Pi) = 2.65 overbalanced weight's net work
    4.9 N-mf * (0.15m * Pi) = 2.31 out of balance weight on a long ever

    The 2 figures are N-m's of work, this is why Pi is a part of the calculation. And 0.34 N-mf does show potential but offers no guarantee.
    What this does take into consideration is the much longer path the overbalanced weight takes.

    4.9 N-mf is a 1/2 Kg weight. What I'm having trouble considering is that a weight rotating 15 cm's from the axle (on the long levers) has the same force as a weight
    that is 15 cm's further from the axle than it's opposing weight (overbalanced weight) yet will travel 1.57 meters more. That could be why I find Mt 20 intriguing. It's not exactly what someone
    would consider thinking rationally.


    Jim
    Last edited by James_; 29th March 2017 at 01:24 AM. Reason: adjust spacing in post and revise math and add more comments

  5. #64
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    Default Test Examples

    The 2 drawings are simple tests that can be done to see if the lever / weight configurations can rotate 180°.
    The placement of the weights are as follows'

    Example 1:
    weight at 12:00 is centered 50 cm's from the center of the axle.
    The weight at 6 o'clock is centered 42.5 cm's from the center of the axle.
    The weight on the long lever's fulcrum at 8 o'clock is 10 cm's to the left of the axle and down 10 cm's.
    If the lever (used as an example and is not needed for this test) is parallel to the floor, then the dimensions mentioned above
    are to go from that plane.
    The long lever on the right has it's weight 10 cm's to the right of the axle.

    If a disc can rotate 180° then it's weights are in position to both drop again.

    Example 2 shows the levers at a right angle to gravity and are centered on the axle. One weight will be 10 cm's below the plane of it's fulcrum.

    Both tests can use the same disc. For Americans, 50 cm's = 20 inches and 12 oz. weights will work.
    And with the metric weights, 325 g's for all 4 weights.
    And with such a test, if neither configuration works then a grindstone is necessary.



    Jim
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  6. #65
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    With Mt 20 it probably would not work if more weights and levers were used than what is in this drawing. This is because all weights
    would be counterbalanced. Since the 2 opposing long levers are 90° to the over and under balanced weights there is no counterbalancing
    going on. Instead torque should rotate the wheel. This could possibly be the simplest design that might work. And as I mentioned in my
    last post, if the weights are positioned on a piece of plywood as shown on the wheel, can it rotate 180° ? If it does then this design might
    work.
    With the other diagram, if anyone is interested in that then I can discuss it. It's the same as this one but the levers are centered at a right angle to the axle.



    Jim

    p.s., sometimes I will repeat information because this is new to most people.
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    @All,
    Even though I don't really have a place to work I'll see if I can't build a test "wheel". It would be to see if 180° of rotation is possible. It would be
    a + on a stand. I'll go over things this weekend and before I do anything else I will post some specific numbers and from one side to the other it
    would be about 50 cm's across. And as far as this goes I will need to order 4 weights that will be about 340 grams / 12 ounces.
    As for my doing a build, I'll probably need my medical situation resolved.

    Jim

    While it is April 1st in the U.S. and is April Fools Day I will need to skip the month of April.
    Just not sure if the surgeon I'll be seeing is going to do anything and can't rent a shop (storage unit)
    to work in until next month.
    Usually I'd say this is a lot of work but I think I've learned enough to where it's not a lot of work
    anymore.
    Last edited by James_; 2nd April 2017 at 01:41 AM. Reason: add comments

  8. #67
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    All the best for a good result and a speedie recovery Jim....the shed eagerly awaits your return...Cheers, crowie

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    Quote Originally Posted by James_ View Post
    I did some math and I realized one issue that I've had with the metric system and that is calculating work. What I did was to calculate work in N-m's of work.
    Everyone knows W = M * D (work = mass x distance) so what I did was to multiply N-m's of force by distance. Force is just newtons (N). If you multiply Newton-metres by meters you end up with Newton-meters^2. By definition, a N-m is a Joule (J), so the unit you are describing is a Joule-metre (Jm). I may be wrong, but I don't think a Joule-metre is a meaningful measure of anything.

    And with what I showed in Mt 20 it does show a net force.
    While Bessler showed his weights rotating and hitting stops, they might have. Witnesses said that they heard 8 knocking sounds per revolution. An easy way to account for this is to have the weights rotate at about (45° ATC). And with 8 weights, one would be placed every 45°.
    With the math, hopefully someone will go over this;

    This is what I keep coming back to;
    (5.635 N-mf - 4.9 N-mf) * (1.15m * Pi) = 2.65 overbalanced weight's net work
    4.9 N-mf * (0.15m * Pi) = 2.31 out of balance weight on a long ever

    The 2 figures are N-m's of work, this is why Pi is a part of the calculation. And 0.34 N-mf does show potential but offers no guarantee.
    What this does take into consideration is the much longer path the overbalanced weight takes.

    4.9 N-mf is a 1/2 Kg weight. What I'm having trouble considering is that a weight rotating 15 cm's from the axle (on the long levers) has the same force as a weight that is 15 cm's further from the axle than it's opposing weight (overbalanced weight) yet will travel 1.57 meters more. That could be why I find Mt 20 intriguing. It's not exactly what someone would consider thinking rationally.


    Jim
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    regards from Alberta, Canada

    ian

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    Ian,
    I've often found in.lbs. or ft.lbs. of work or torque for reference works quite well. And at the same time I think 96% of the people in the world use the metric system. And for what I am hoping to explain someone might know how a person using the metric system would quantify 375 grams dropping 10 cm's as a value related to work being performed. With in.lbs. It would be about 13 oz.'s * 4 in. = 3.25 in.lbs. And this means that 3.25 lbs. can be lifted or dropped 1 in. or 1 lb. can be lifted or dropped 3.25 in.
    And for everything else I prefer the metric system because it works great with trigonometry and allows for easy scaling and 9.8 m/s I prefer to 33 feet/s.


    Jim
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    Default Balance(d)

    I'll go over the math behind balanced and out of balance. With Bessler, in his drawings he shows an out of balance lever.
    This is when a lever is not 90° to the axle. Why this matters is that when a lever is at a different angle then it is out of balance.
    The 3rd image shows this. When 2 opposing weights are perpendicular to gravity, 1 weight is at 6 o'clock. Those 3 weights are
    considered in balance.
    The weight that is out of balance is to the left of 12 o'clock. It is this imbalance that needs to be accounted for. And there are 2
    basic options to account for this without using a grind stone. When both levers drop they lift the weights that shift the balance of
    the wheel. The work performed to create an overbalance will always equal the imbalance created by a lever not being 90° to the
    axle. This is at 100% efficiency.
    I'll give everybody a couple of days to consider this. I'll give a hint as to the answer; I'll show 180° of rotation which means
    12 o'clock becomes 6 o'clock and 6 o'clock becomes 12 o'clock and this will be for the 2 weights being lifted.


    Jim

    for this the levered weight to the left of 12 o'clock would counter balance the over balanced weight at 3 o'clock, they would move
    the same distance. This means if the levered weights drop 10 cm's then the weight at 3 o'clock would be 10 cm's further from
    the axle than the weight at 9 o'clock. In the U.S., 10 cm's = 4 in.
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    Last edited by James_; 8th April 2017 at 05:47 AM. Reason: add a comment

  12. #71
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    These 3 images show what I am going to start on after the 1st. That's when I'll be able to have a place to work. It is kind of a simple build because it uses basic leverage/levers with pulleys. The weights that shift for the over balance travel further than weights near the axle. This means less force is needed to do more work.


    Jim

    p.s., 3 times the radius means if the shift is multiplied by 3, then is it more than what the interior weights shift ? If so then a usable over balance should be realized.
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    Last edited by James_; 15th April 2017 at 06:35 AM. Reason: add p.s.

  13. #72
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    @All,
    It's actually a pretty neat trick in leveraging. The fulcrum moves more than the weight does. And this is where mass times distance = work.
    If the fulcrum moves 3 times more than the weight then it needs 1/3 the force.



    Jim
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    Some people might be shocked to realize this has been over looked. The image on the left is what Bessler shows in his Mt 20 while the image on the right is an obvious 3:1 ratio which is acceptable balance because of mass and leverage. While they are the same thing no one has considered a weight being lifted from the rim of a wheel and could only consider the axis as a fulcrum for leveraging.
    And at the same time 2 weights on long levers can lift more than 1/3 their own mass. This is how the over balance/net force is realized.


    Jim

    edited to add; if the weight on the left is on the long lever then it can lift more than 1/3 it's own weight. The weight on the right represents net force or over balance for comparison.
    As I have shown in a previous post the weights on the long levers can be pretty much in line with the 2 weights that shift to create an over balance. It does take time to consider
    something that is not in the usual way or fashion of having been done.

    edited to add; the last 2 images show 2 - 500 gram weights dropping 10 cm's lifting 2 - 250 gram weights 10 cm's.
    obviously 500 grams is twice as much as 250 grams. What everyone is missing is that a net force of 250 grams * 10 cm's * 3 is more than 500 grams * 10 cm's * 1.
    If everyone notices that the fulcrum for the long levers is the same distance from the axle as the center of mass for the "under balanced" weight.

    edited to add; will start on actual build after the 1st. Will take it easy until then.
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    Last edited by James_; 18th April 2017 at 04:32 AM. Reason: to clarify leverage, add 2 more images

  15. #74
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    This is a simple leverage test anyone can try. They will find that the force the weight exerts on the end of the cross beam is 1/3 that of the weight. This is why I placed the fulcrum for the long levers at the outside of the wheel which is what Bessler shows in his Mt 20. And with the 2 weights shifting to create the over balance being as far from center as the long levers fulcrums, would need to get into physics to explain why this is necessary. And at the end of the day it will probably take demonstrating this for people to accept it.
    I think the simplest way that I can explain it is that the heavier weights are rotated/lifted from the rim and not from the axle.

    Jim
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  16. #75
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    This is something I'll start working on after the 1st of the month. It will be the easiest way to test conservation of angular momentum. This would be when a line starts wrapping around the grind stone. This is because as the path the swinging weight is taken has a radius that keeps becoming smaller.
    The link is to a web page that allows for some basic calculations to be performed. And once the bob reaches 6 o'clock (BC or Bottom Center) the grind stone will slowly retract the bob. This should allow the bob to swing higher because conservation of angular momentum dictates that the velocity of the bob will increase.
    The weight will be between 2 solid beams with tracks the weight can ride on.
    This way once the weight reaches a height greater than what it started out at in then can be released so it can move outward. And if successful then this might help to explain why Bessler referenced grind stones in his wheels.
    What would need to be looked for in this is the angle the bob started swinging down from is eclipsed on the ascending side. And with this, 2 weights might be needed. With 2 weights then conservation of momentum and angular momentum might be realized. I think this might be the most basic way that can be tested.

    Pendulum

    @All, with a pendulum I should be able to see what works before committing to a full build. This will allow me to test 2 different methods to conserve momentum.


    Jim
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