How does descoring work?

[GBIT]

given a trackball center at (0,0)
given a trackball contact point with a rail at (16,y) (the rails are 32" apart)
given the trackball is 20" and assuming it remains approximately spherical, then...

x^2 + y^2 = r^2
16^2 + y^2 = 400
y^2 = 400 - 256
y = 12

So the contact point for the rails will be 12" below the center of the ball. This means that between the rails, the bottom of the ball should be 8" below them, and thus 2" below the maximum allowed height of a robot in the opponent zone. However, this is in an ideal world where the rails are infinitely thin and the ball is perfectly sperical. Since the rails are 1.5" wide and the ball will deform some, the ball may be slightly higher than where I computed it to be, and the contact space for a robot will be very, very small.

Originally Posted by Bongle

Posted this in a couple other threads today.


Glad to hear someone did the actual measurement and it was close

Here's a diagram:
http://www.chiefdelphi.com/forums/at...4&d=1199558995

Umm. Hope not too many people used this number because our team found it to be wrong. We noticed the the equation above used numbers assuming the ball was on center with the pipes. this is not true seeing that the ball rests on the top of the pipes. We found the hang to be (assuming the the 6.5' overpass is on center with the pipes) it would hang only 1.5 inches below the 6' legal limit. If the 6.5' is at the bottom of the pipes it would hang only .67 inches from the 6' limit. this is assuming the the ball in of 40" diameter.



here is the diagram:
http://www.chiefdelphi.com/forums/at...6&d=1199655203