murilo wrote:The modeling I ask for is a model where one can make N changes and tries,
and the computer will calculate on line to all weight resultants, losses and 'g' playing.
That is what I did. Below is just a small sample of the results.
murilo wrote:Jim, you always insist in 2 stuffs: you always draw that narrow rail who confines the inside rods, that will not fall free due opposites frictions at 180º of each rod.
This rail you insist is supposed to conduce the chain that of sure will brake, thanks to your version. (the force transference is also very poor.)
The rails can be turned on or off in the computer. (See the right side shows the rails turned off.) The rails are NOT used in any calculation. They are there only for looks.
murilo wrote:Jim also insist in avoid the use of the wheel with hooks, which transform the novel in something absolutely different than the above narrow rails. The hooks wheel receives a punctual mass and also the counter opposite mass, all confronted over wheels axle.
The hooks are drawn as simple lines because it takes a lot of time and extra programming to make fancy hooks. The formulas assume hooks.
Each weight either exerts its 'g' force at the inner radius (shown as blue) or it exerts its 'g' force at the outer radius (shown as red). Because the top and bottom are symmetrical everything above the top axle and below the bottom axle can be ignored since they will balance.
Therefore we only need to look at what happens in between the axles.
All the weight on the left side between the axles rests on the inner blue radius.
All the weight on the right side between the axles rests on the outer red radius.
The ratio of the number of blue weights to the number of red weights is inverse to the ratio of blue radius to the red radius. Thus the total of the blue weights times the blue radius always equals the total of the red weights times the red radius. And thus the mechanism balances. This is a mathematical truth.
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