Inertial reaction

MARC OLHOEFT ( seapower@Alaska.NET )
Wed, 11 Mar 1998 09:57:39 -0900

Bill,I JUST COULD NOT RESIST WHAT AT FIRST SEEMS A VERY BASIC PHYSICS
MODEL IS MUCH MORE FACINATING
1 If the forces are equal and VECTORS parallel and strike the beam
perpendicular to the beam.
2 the arms of the beam are equidistant from pivot.
3 the spheres are not spinning when dropped.
4 no absorption losses occur in the impact.
5 and if the beam is infinitly rigid.
Equal energy will be transmitted to the pivot .
Total energy transferred to the pivot will be the combined kinetic
energy of the spheres/the potential energy(mass) of the beam.

NO NET WORK WILL BE PERFORMED AT THE END OF THE ARM.
AS EQUAL AND OPPOSING FORCES WILL CANCEL EACH OTHER.

HOWEVER VERY INTERESTING THINGS HAPPEN WHEN ANY OF THE ABOVE CRITERIA IS
NOT MET AND MODELING BECOMES MUCH MORE COMPLEX.

1 harmonics will be a factor as in the real world there are no perfectly
rigid beams(varing absorption losses), arm length, varying arm mass
would cause vibratory nodes to reach the pivot at varying times changing
the momentary energy transferand the total transfer due to losses.
2 sphere rotation will be a factor as any sphere rotation will change
total kinetic energy for that sphere ,and how that additional energy is
transmitted to the beam,(friction) at impact ,duration of contact
(absorption losses)vector of net energy gain to beam.
(SHADES OF BESSLERS WHEEL?)
3 ANGLE OF IMPACT (IF IN FACT GRAVITY SUCKS)THEN THE SPHERES WILL BE
SLIGHTLY CLOSER TOGETHER AT IMPACT THAN WHEN THEY WERE DROPPED THEN
THE IMPACT ANGLE TO THE BEAM WILL NOT BE PERPINDICULAR.
4 Resistance to deformity at impact(superball,vrs steel)brings up
complex material relationship impact losses and material ability to
and speed of recovery to deformity.

AS MY OLD CALC PROF SAID "VERY INTERESTING BUT IT IS ALL CRAP"

HOPE ALL OF THIS HELPS PROBABLY JUST MUDDIED THE WATERS