[CRUSH-build] ball throwing mechanisms

[Bill Bennett]

Here's a short video of the Team 246 ball shooter mock up. Nothing really built, just testing the wheel shooter concept with a motor, wheel and board. Looks like it worked pretty good.

http://crush1011.org/temp/Team246BallLauncher.mpg 

The chiefdelphi thread about it is here: http://www.chiefdelphi.com/forums/showpost.php?p=430096&postcount=1

A second video from team 1251 is here: http://www.chiefdelphi.com/forums/showpost.php?p=430121&postcount=2 and their video is here:
http://crush1011.org/temp/Team1251BallLauncher.mpg 


The new motor curve is posted by FIRST as part of the kit descriptions. It's here: http://www2.usfirst.org/2006comp/Specs/CIM_Motor_FR801-005.pdf

It's a little slower and a little stronger than the smaller CIM. Since the smaller motors fit the drive train gearboxes, I suggest using the bigger motors with the thrower and collector. 

Here's a quick and rough analysis: 

We want the ball to leave the robot at 12 m/s. Since it is thrown from about 4 feet below the exit point, it will decelerate due to gravity plus friction. If we accomplish a smooth track with large enough radii, as we discussed in the meeting today, the loss due to friction should be small. The loss due to gravity is easy to estimate. The ball will travel upwards about 4 feet = 1.2 meter. It's going at something over 12 m/s average through this distance. So it will take about 1.2/12 = 0.1 seconds to ascend. During that time gravity is pulling it down with an acceleration of 9.8 m/s^2, so it decelerates by 9.8*0.1 = 0.98. So maybe gravity+friction will slow the ball down by around 1 m/s, so we'll try to have it leave the spinner at 13 m/s.

The motor runs with no load at about 2500 rpm. After the mechanism spins up to speed, there is no load except for mechanism friction, and since this will probably be a direct drive wheel, it should be minimal. I'll guess that it will spin at about 2400 rpm or 40 r/s. The linear speed of the circumference of a spinning wheel is pi * d * angular velocity in rev/sec. We want this to be 13 m/s. So

13 = pi * d * 40
solve for d
d = 13 / (pi * 40)

d = 0.10 meters = 3.9 inches

A four inch wheel should be about right.

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