--newton's beads--

    i noticed a video on youtube the other day, and they invited others to speculate on this phenomenon, so, i believe i will "have-a-lash", and "take-a-stab-at-it"...here goes--


   newton's beads--

    a newtonian puzzle, that is fun to watch unfold( very carefully, due to much force being in play)...


    the heart of the matter, seems to be that there is only a few degrees variance, from being "straight-in-line", between each bead...so, each bead's few degrees off center, "add-up", to a total possible apex( curve), for this reason, the chain will not kink...once the beads are pulled over the side, the amount of beads "in-free-fall"( over the side), before they touch the ground, determine the speed of the chain coming out of the cup...and the corresponding height of the beads above the cup( a result of the force pulling the beads out of the cup)...before they arch to the ground...in short, i feel that the weight of the beads from the ground, to the arch above the cup, represents the "total force applied" to the beads resting on top of the pile, within the cup...meaning, the higher the cup is raised, the more "total-weight" of the beads are "in-free-fall", within the chain, and the more forcefully( rapidly), the beads will move( creating a higher arch from inside the cup, to the ground...as mentioned above)--

  the math--

    by using "distance to geo-sync" math( see my "distance to geo-sync" post, or "velocity/straight-drop in-atmo" post), we can grasp the problem numerically, and determine the exact value of the pull of the chain, in "foot-pounds", or mph...here goes--


   the weight of one bead, x the number of beads from the arch above the cup, to the ground...equal the total weight of beads in "free-fall"...weight of beads - distance of 1( foot) = "foot-pounds" of force applied( to the beads below the arch above the cup, from the top of the pile)...( see below)--
 
the numbers--

       ( pounds)                                            
  ( weight of 1 bead)  x  ( # of beads)  =  ( total weight of beads)
          .10                   x          .60         =            6.00


  ( total weight of beads)  -  1   =   ( "foot-pounds" of force applied)--
           6.00                                                 5.99

  more later, out of time at the library--

  best wishes, john kruschke--