In a message dated 25/11/2002, rpratt@............. writes:

> In rolling contact the center of rotation of the ball will be the center of 
> the ball and it will be in translation.  The instantanious center of 
> rotation of the total boom will be the contact point. (wheel and axle)  

Hi Randall,

     Sorry, but I disagree. Assume that the radius of the ball is R. Consider 
the boom rotating by a small angle A. The contact point moves a distance RA 
to one side and the boom load acts through the point of contact. This line 
intersects with the axis at a point R below the surface. This is the centre 
of rotation for small angular movements.

If the compressive force in the boom is axial then any deflection from centre 
will 
> throw the force off axis. There will have to be contact friction sufficient 
> to prevent the rotated boom from sliding on the plate since the boom is no 
> longer perpendicular and this force will be tangent to the ball.  The 
> friction force will be in the direction of pushing the boom toward centre.

       Sure there will be a small frictional force between the ball and the 
surface, but this lies in the plane of the surface and it does not effect the 
rolling motion, to a first order approximation. It is also quite small, it is 
a balanced force and is just sufficient to prevent the contact point from 
sliding. Do not confuse the coefficient of friction with the frictional 
force, or the resistance to rolling movement with static friction.

> As to my earlier thought that the axis will be offset, I calculate that for 
> a 24" boom and .125" motion the pivot will move .0003" left or right using 
> a .125" ball.  With 12" between upper and lower pivot this would be less 
> than 1% of the tilt angle you suggested and may not be a significant 
> factor.

     You need to consider the centre of rotation as a point on the axis R 
below the plane surface. This is one point that you consider when measuring 
the angle of the rotational axis to the vertical axis. 
  
> May I suggest an experiment for the proponents of the ball?  Operate with as 
> little damping as possible and compare to a true sine wave equally damped.

     It may be easier to just measure 1) the period and 2) the damping 
factor. The damping gives an exponential fall off in the amplitude of the 
oscillation. These should be easy to do using your A/D logging program. 

     Regards,

     Chris Chapman
In a message dated 25/11/2002, rpratt@............. writes:



In rolling contact the center of rotation of the ball will be the center of the ball and it will be in translation.  The instantanious center of rotation of the total boom will be the contact point. (wheel and axle)  




Hi Randall,



     Sorry, but I disagree. Assume that the radius of the ball is R. Consider the boom rotating by a small angle A. The contact point moves a distance RA to one side and the boom load acts through the point of contact. This line intersects with the axis at a point R below the surface. This is the centre of rotation for small angular movements.



If the compressive force in the boom is axial then any deflection from centre will 

throw the force off axis. There will have to be contact friction sufficient to prevent the rotated boom from sliding on the plate since the boom is no longer perpendicular and this force will be tangent to the ball.  The friction force will be in the direction of pushing the boom toward centre.




       Sure there will be a small frictional force between the ball and the surface, but this lies in the plane of the surface and it does not effect the rolling motion, to a first order approximation. It is also quite small, it is a balanced force and is just sufficient to prevent the contact point from sliding. Do not confuse the coefficient of friction with the frictional force, or the resistance to rolling movement with static friction.



As to my earlier thought that the axis will be offset, I calculate that for a 24" boom and .125" motion the pivot will move .0003" left or right using a .125" ball.  With 12" between upper and lower pivot this would be less than 1% of the tilt angle you suggested and may not be a significant factor.




     You need to consider the centre of rotation as a point on the axis R below the plane surface. This is one point that you consider when measuring the angle of the rotational axis to the vertical axis. 

  

May I suggest an experiment for the proponents of the ball?  Operate with as little damping as possible and compare to a true sine wave equally damped.




     It may be easier to just measure 1) the period and 2) the damping factor. The damping gives an exponential fall off in the amplitude of the oscillation. These should be easy to do using your A/D logging program. 



     Regards,



     Chris Chapman