at Busch Gardens, there is a ride called Da Vinci's Cradle:
http://www.coasterimage.com/pictures/bu ... ides01.htm
it rotates around that center axis. the cradle itself, of course, stays upright. what I don't understand is, if one side of the cradle weighed more than the other (say the right side in the picture), why wouldn't the CF (I think it would be CF) carry it down with more force than up? since the weight on the downswing would be further from center than on the upswing?
there is probably a well-known reason why this wouldn't work but I am stumped!
Da Vinci's Cradle
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Da Vinci's Cradle
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hansvonlieven
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re: Da Vinci's Cradle
The problem outlined was investigated and explained by Desagulier.
Here is a monograph and a diagram:
A demonstration by Dr. Desaguliers in 1719, in regard to the balance of weights at unequal distances from the center of oscillation, showing that the weight P balances the weight W at any position on the cross arm H, I, on the vertical arm B, E, when pivoted to the double-scale beam A, B, and D, E, in which the resolution of forces is made apparent in a practical form so often over-looked by the inventors of perpetual-motion machines.
The cut representing Desaguliers' balance, with his explanation, goes to show how persistently inventors have ignored the geometrical bearing of this problem for nearly two centuries.
Desaguliers' Demonstration.- A, C, B, E, K, D is a balance in the form of a parallelogram passing through a slit in the upright piece, N, O, standing on the pedestal, M, so as to be movable upon the center pins C and K. To the upright pieces, A, D and B, E, of this balance, are fixed at right angles the horizontal pieces F, G and H, I. That the equal weights, P, W, must keep each other in equilibrium is evident; but it does not at first appear so plainly, that if W be removed to V, being suspended at 6, yet it shall still keep P in equilibrium, though the experiment shows it. Nay, if W be successively moved to any of the points, 1, 2,3, E, 4, 5, or 6, the equilibrium will be continued; or if, W hanging at any of those points, P be successively moved to D, or any of the points of suspension on the crosspiece, F, G, P will at any of those places make an equilibrium with W. Now, when the weights are at P and V, if the least weight that is capable to overcome the friction at the points of suspension C and K be added to V, as w, the weight V will overpower, and that as much at V as if it was at W.
As the lines A, C and K, D, C, B and K, E, always continue of the same length in any position of the machine, the pieces A, D and B, E will always continue parallel to one another and perpendicular to the horizon. However, the whole machine turns upon the points C and K, as appears by bringing the balance to any other position, as a, b, e, d, and, therefore, as the weights applied to any part of the pieces F, G and H, I can only bring down the pieces A, D and B, E perpendicularly, in the same manner as if they were applied to the hooks D and E, or to X and Y, the centers of gravity of A, D and B, E, the force of the weights (if their quantity of matter is equal) will be equal, because their velocities will be their perpendicular ascent or descent, which will always be as the equal lines 4l and 4L, whatever part of the pieces F, G and H, I the weights are applied to. But if to the weight at V be added the little weight, w, those two weights will overpower, because in this case the momentum is made up of the sum of V and w multiplied by the common velocity 4L.
Hence it follows, that it is not the distance, C6, multiplied into the weight, V, which makes its momentum, but its perpendicular velocity, L4, multiplied into its mass.
This is still further evident by taking out the pin at K; for then the weight, P, will overbalance the other weight at V, because then their perpendicular ascent and descent will not be equal.
This "paradox" is illustrated in No. 10, first volume of Mechanical Movements, inviting inquiry by students, a model of which has been exhibited to many doubting amateurs by the author.
Hans von Lieven
Here is a monograph and a diagram:
A demonstration by Dr. Desaguliers in 1719, in regard to the balance of weights at unequal distances from the center of oscillation, showing that the weight P balances the weight W at any position on the cross arm H, I, on the vertical arm B, E, when pivoted to the double-scale beam A, B, and D, E, in which the resolution of forces is made apparent in a practical form so often over-looked by the inventors of perpetual-motion machines.
The cut representing Desaguliers' balance, with his explanation, goes to show how persistently inventors have ignored the geometrical bearing of this problem for nearly two centuries.
Desaguliers' Demonstration.- A, C, B, E, K, D is a balance in the form of a parallelogram passing through a slit in the upright piece, N, O, standing on the pedestal, M, so as to be movable upon the center pins C and K. To the upright pieces, A, D and B, E, of this balance, are fixed at right angles the horizontal pieces F, G and H, I. That the equal weights, P, W, must keep each other in equilibrium is evident; but it does not at first appear so plainly, that if W be removed to V, being suspended at 6, yet it shall still keep P in equilibrium, though the experiment shows it. Nay, if W be successively moved to any of the points, 1, 2,3, E, 4, 5, or 6, the equilibrium will be continued; or if, W hanging at any of those points, P be successively moved to D, or any of the points of suspension on the crosspiece, F, G, P will at any of those places make an equilibrium with W. Now, when the weights are at P and V, if the least weight that is capable to overcome the friction at the points of suspension C and K be added to V, as w, the weight V will overpower, and that as much at V as if it was at W.
As the lines A, C and K, D, C, B and K, E, always continue of the same length in any position of the machine, the pieces A, D and B, E will always continue parallel to one another and perpendicular to the horizon. However, the whole machine turns upon the points C and K, as appears by bringing the balance to any other position, as a, b, e, d, and, therefore, as the weights applied to any part of the pieces F, G and H, I can only bring down the pieces A, D and B, E perpendicularly, in the same manner as if they were applied to the hooks D and E, or to X and Y, the centers of gravity of A, D and B, E, the force of the weights (if their quantity of matter is equal) will be equal, because their velocities will be their perpendicular ascent or descent, which will always be as the equal lines 4l and 4L, whatever part of the pieces F, G and H, I the weights are applied to. But if to the weight at V be added the little weight, w, those two weights will overpower, because in this case the momentum is made up of the sum of V and w multiplied by the common velocity 4L.
Hence it follows, that it is not the distance, C6, multiplied into the weight, V, which makes its momentum, but its perpendicular velocity, L4, multiplied into its mass.
This is still further evident by taking out the pin at K; for then the weight, P, will overbalance the other weight at V, because then their perpendicular ascent and descent will not be equal.
This "paradox" is illustrated in No. 10, first volume of Mechanical Movements, inviting inquiry by students, a model of which has been exhibited to many doubting amateurs by the author.
Hans von Lieven
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When all is said and done, more is said than done . Groucho Marx
Bessler shows a much better apparatus for demonstrating the Roberval balance than Desauguliers' - it's a hand-drawn image on the back of MT136. See the following links where I talked about it previously:
http://www.besslerwheel.com/forum/viewt ... 5909#35909
http://www.besslerwheel.com/forum/viewt ... 5933#35933
The Roberval balance only works if the two fixing points of the parallelogram remain in vertical alignment.
Stewart
http://www.besslerwheel.com/forum/viewt ... 5909#35909
http://www.besslerwheel.com/forum/viewt ... 5933#35933
The Roberval balance only works if the two fixing points of the parallelogram remain in vertical alignment.
No - we're trying to reproduce an overbalanced wheel don't forget, so that's one way of doing so. However, what it does do is demonstrate why certain PM designs are balanced and not overbalanced, such as MT50.Frettsy wrote:thanks Hans. so, does this mean that the idea of moving weights in/out from the center of a wheel is useless?
Stewart
re: Da Vinci's Cradle
Its not the position or situation of the weight but where its being expressed.
The weight may dance around where ever but in the instance of the cradle its always being expressed or sent to the end of the rotating counterbalanced arm.
The weight may dance around where ever but in the instance of the cradle its always being expressed or sent to the end of the rotating counterbalanced arm.
Attempt only the impossible.