3 robots designs I hope to see this season

[JesseK]

Assuming a 60 degree arc length of contact on a 8" 1-wheel shooter at ~2400 RPM, the ball needs a minimum of 500 watts transferred to it after inefficiencies in order to go 54 feet (~42 ft/s for middle goal). This is doable with a heavy flywheel that has plenty of spin-up time and 4 Banebots RS-550 motors connected to the same wheel. At 63% overall efficiency, that's ~67 amps of current while a ball is in the shooter. If you sit in one spot, that's chump change.

The kinematic equations are your friends this year.
Vf^2 = Vi^2 + 2 * a * d
You have D, Vf, and Vi. You can find a, which then can give you required force via F = m * a, and then a torque via Tau = Force * radius of the wheel. Then you can multiply Torque (N*m) * Wheel speed (rotations / second) and you get power required.

Of course, that assumes a perfect flywheel effect and a perfect contact patch between ball and wheel, so YMMV. Prototype it.

It also assumes that a 84" tall robot will never block the shot, heh. We aren't even attempting it because the trajectory length is so long that a variance of ~0.7 degrees in the shooter causes the shot to miss. That's not worth putting 4 motors on the shooter.

[JamesCH95]

Originally Posted by JesseK

Assuming a 60 degree arc length of contact on a 8" 1-wheel shooter at ~2400 RPM, the ball needs a minimum of 500 watts transferred to it after inefficiencies in order to go 54 feet (~42 ft/s for middle goal). This is doable with a heavy flywheel that has plenty of spin-up time and 4 Banebots RS-550 motors connected to the same wheel. At 63% overall efficiency, that's ~67 amps of current while a ball is in the shooter. If you sit in one spot, that's chump change.

The kinematic equations are your friends this year.
Vf^2 = Vi^2 + 2 * a * d
You have D, Vf, and Vi. You can find a, which then can give you required force via F = m * a, and then a torque via Tau = Force * radius of the wheel. Then you can multiply Torque (N*m) * Wheel speed (rotations / second) and you get power required.

Of course, that assumes a perfect flywheel effect and a perfect contact patch between ball and wheel, so YMMV. Prototype it.

It also assumes that a 84" tall robot will never block the shot, heh. We aren't even attempting it because the trajectory length is so long that a variance of ~0.7 degrees in the shooter causes the shot to miss. That's not worth putting 4 motors on the shooter.

500 watts for what length of time?

[JesseK]

Originally Posted by JamesCH95

500 watts for what length of time?

Well, since I know the watts required, and the distance of the arc length in the shooter wheel, I suppose we can calculate time since [Power = Force * distance / time]. Yet the rotational power required was calculate via [Power = Torque * Rotation Speed]. Thus we didn't need time. It does make a couple of assumptions, such as Power In = Power Out (power required by the ball is available in the shooter wheel).

The whole point of the exercise is to figure out what motors to use and what shooter wheel radius to use. It's "doable", but personally I wouldn't bet my season on it being successful.

[JamesCH95]

Originally Posted by JesseK

Well, since I know the watts required, and the distance of the arc length in the shooter wheel, I suppose we can calculate time since [Power = Force * distance / time]. Yet the rotational power required was calculate via [Power = Torque * Rotation Speed]. Thus we didn't need time. It does make a couple of assumptions, such as Power In = Power Out (power required by the ball is available in the shooter wheel).

The whole point of the exercise is to figure out what motors to use and what shooter wheel radius to use. It's "doable", but personally I wouldn't bet my season on it being successful.

Okay, I didn't see anywhere that you assumed length of time/distance over which the ball would be in contact with the wheel and was curious. I also wanted to make sure you didn't have a units jumble somewhere.

[JesseK]

Originally Posted by JamesCH95

Okay, I didn't see anywhere that you assumed length of time/distance over which the ball would be in contact with the wheel and was curious. I also wanted to make sure you didn't have a units jumble somewhere.

Heh, distance was in the first few words of the post, but perhaps it should be "8-inch diameter 1-wheel shooter". And personally before I run calculations I convert everything to metric since most of the motors specs I have are metric (Newton-meter for torque). The units check out.