[Andrew Blair]

Originally Posted by KenWittlief

if you do that math on this part of it, lets say you fire at 12M/s and you fire all 10 in 5 seconds.

that means each ball is flying at 1/2 S intervals, which will put them 6 meters apart

thats a lot of space between the balls in flight: 10" balls with 226" of empty space between them

the odds of two balls hitting in flight is 10/226 = about 5 out of 100.

The odds are 20:1 in your favor (no collisions in midair)

I would be most worried about the goal being able to keep up. Assume that it takes a two seconds to stop a ball and clear it, out of the box, into the chute. Say also, for simplicity sake, that three balls are capable of doing this at the same time. So say you have, best case scenario, 30 balls in ten seconds.

So, you have ten, 3 balls instances of clearing. At two seconds apart. So, to clear all the balls, it would be 10*2 =20 seconds to clear all the balls completely. So, by the end of the ten seconds, there would be ten seconds of balls to clear, backed up. (10/2)*3 balls~15 balls left

So at the end of the period (through my rough estimation), there are fifteen balls to clear, or, towards the end of the autonomous period, there are 10 or 12 to get in the way.

I wonder if 10 or 12 balls, some still jangling around, could present enough of an obstacle to prevent some incoming balls from coming in? And what about the feeder chute? I wonder if that'll jam? I think, on top of needing a killer alliance to do it, this challenge is going to be harder than the intro suggests...