Re: Bessler wheel theory part 1


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Posted by MadMax (81.15.212.3) on September 23, 2003 at 12:40:07:

In Reply to: Bessler wheel theory part 1 posted by Patrick on September 23, 2003 at 10:46:30:

: We know that a symmetrically made wheel will have the same amount of mass descending as it does ascending; that is : at any time the weight on either side (if you split the wheel in half vertically down the center) is equal and thus the wheel will inevitably come to rest at a point of equilibrium regardless of the activity occurring within itself.

: For a wheel to be able to continue in motion, one side of the wheel must somehow continually be heavier(displaying more mass) than the other. (that is : if the wheel is rotating clockwise, the right hand side of the wheel must be heavier that the left and vice versa.) {by left and right , I am referring to viewing the upright wheel face on and splitting it down the middle vertically into left and right sides.)

: Since a wheel is round and each part rotates throughout the 360 degree range, any FIXED imbalance will of necessity reposition itself to the opposite side of the wheel and possibility of a continually heavier side will be lost.

: Thus, to maintain a side that is heavier,
: A) a continual “weight transfer” must occur at either the zenith or nadir of the wheel, (at the very top or the very bottom)
: B) this “weight transfer” or ‘flip’ will allow weight to be redistributed to the right hand to keep the right hand side heavier throughout the rotation
: C) this ‘flip’ or ‘transfer’ must be continual or often enough to redistribute weight from the left to the right at the zenith or nadir to keep from breaking the momentum of the wheel.
: D) a transfer of weight at the zenith should intuitively be a push from one side to the next or if from the bottom , a pull from one side to the next.
: E) The weight transfer “effect” must be greater than the sum of various ‘frictions’ which may be occurring within the wheel.
: --Patrick Doucette




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