Cannon Design!!!!!! >_<;

[EricH]

You may want a belt system, but if you need to go with a launcher, you want to be able to go horizontally after the vertical launch.

Let's say you need to go 3 ft. horizontally from the top of the robot. You need enough speed to go that far left over from the upwards launch.

Let's start with the horizontal portion. Launching at 4 feet, you need to clear 3 feet. So you can't go more than 1 foot downwards, and you need to go 3 feet horizontally, reliably. You'll have an upwards trajectory from the launch, but we'll ignore that for now. We need to go horizontally and vertically, so for vertical motion, s(t) = -1/2g*t^2 + v*t + hi, where g is gravity, hi is initial height, and v is the initial velocity. hi is 4 ft and v starts at zero. s(t) ends up at -1. So... -1 = -1/2*32*t^2 + 0 + 4, -5 = -16*t^2, 5/16 = t^2, t = 0.559 seconds.

Now, how fast does the ball need to go to go 3 feet in that time?
s(t) = v * t for horizontal motion. s(t) needs to end up being 3 ft, and we have t from above = 0.559 s.

3 = v*.559, v = 3/.559, v = 5.3 f/s coming off the top of the chute.

So, if it has to go 5.3f/s coming out, and it has to climb 3 feet, let's look at that equation for vertical motion again. s(t) = -1/2g*t^2 + v*t +hi. We'll assume you want to lift the ball 4 ft.

But we're interested in v(t), so we take the derivative of s(t). v(t) = -g*t+v.

5.3 = -32*t+v; t = (5.3 - v)/-32.

Substitute into the s(t) and you get 4 =-1/2*32*((5.3-v)/-32)^2 + v(5.3-v)/-32 + 1; 3 = -16(28.09 -10.6v+v^2)/1024 -0.165v+(v^2)/32

3 = -0.438 + 0.165v +.0156v^2 - .0165v + 0.0313v^2
3.438 = 0.0469v^2
v = 8.56 f/s launch speed minimum. 10 f/s would be better to start the balls at.

I'm sure that you can run the numbers to figure out what speed the wheels need to spin to impart that speed; wider wheels may help as well to avoid digging into the holes in the ball.