Shooting Physics

[flamer075]

Hi, my team is trying to work out the physics behind our shooter to see how far we could shoot it with different motors. I am not very advanced in physics, so I was wondering if anyone could give me the equations for the way you would figure out the acceleration of the frisbee as it is exiting a quarter circle shooter and a two wheel linear shooter. Thanks for any help.

[Sparks333]

Hello!

So, did a bit of back-of-the-napkin physics, and came up with a highly simplified model that can be described in terms of a simple equation. Keep in mid this makes a LOT of assumptions - stuff like the edge of the frisbee does not slide, the amount of energy the motor imparts on the system during launch is negligible, etc, but this ought to get you in the ballpark.

The energy of a shooter prior to shooting is:

1/2 * I * omega_start^2

Where I is the inertia of the motor/wheel (with a multiplier effect for gearboxes) and omega_start is the angular velocity of the wheel prior to shooting. The energy of the shooter and the frisbee together after shooting is

(1/2 * I * omega_end^2) + (1/2 * M_frisbee *V_frisbee^2)

where I is the same as before, omega_end is the angular velocity after the frisbee has left the shooter, M_frisbee is the frisbee mass, and V_frisbee is how fast the frisbee has left the shooter.

Still too many unknowns. So, you know that when the frisbee leaves the shooter, one side is going the same speed as the wheel, and the other side is rotating against the side of the shooter, and not moving. Therefore, we know that the velocity of the frisbee is half of the velocity of the edge touching the shooting wheel, or

V_frisbee = omega_end * r_wheel / 2

where r_wheel is the radius of the shooter wheel (all of these are in metric, by the way).

Combining and simplifying the two equations, we get:

I * omega_start^2 = I * omega_end^2 + ((M_frisbee/4) * omega_end^2 * r_wheel^2)

As an experiment, if you take a standard 8" AndyMark pneumatic wheel (r = 0.1016 meters, mass of 0.5126 kg --> moment of inertia calculated to be about 0.00265 kg m^2) and hook it straight onto a CIM and let it spin up to maximum power (277.89 radians/sec) and throw in a 0.2 kg frisbee, this equation predicts that you'll end up with a screamer of a frisbee heading out at 13 meters per second, and have a shooter wheel spinning at about 254 radians/sec. With this equation, the only difference between a curved shooter and a linear shooter is the moments of inertia - in a linear shooter, with multiple wheels, the inertias add linearly, whereas since a curved shooter has one wheel, it gets used as is.

Don't expect to actually get the numbers I've posted here in real life - there are way, waaaay too many simplifications for it to work that well - but this isn't a bad upper limit on performance.

Good luck, have fun!

Sparks

[Ether]

Originally Posted by Sparks333

The energy of the shooter and the frisbee together after shooting is

(1/2 * I * omega_end^2) + (1/2 * M_frisbee *V_frisbee^2)

Nope. The frisbee is translating and rotating. Its kinetic energy is (3/4)MV²

[Sparks333]

Hello!

You caught me Ether - that's why I shouldn't derive physics formulas a 2AM.

For people following at home, Ether modified the equation with the assumption that the frisbee can be treated as a thin disk of homogenous density - not strictly true, but definitely close enough for this sort of equation. The more general form of the equation is

I_wheel * omega_start^2 = (I_wheel * omega_end^2) + I_frisbee * (V_frisbee / r_frisbee)^2 + (M_frisbee * V_frisbee^2)

Due to the frisbee's large diameter compared to most shooting wheels, its moment of inertia is definitely non-negligible, despite having a much lower weight (moments of inertia of disks grow with the square of the radius, and linearly with mass).

Thanks for keeping me honest,

Sparks

[Sparks333]

I realized the equation still isn't in a state that could be considered 'simple' - solvable with the Magic of Algebra, yes, but not simple.

Isolating for V_frisbee:

V_frisbee = (sqrt((4*M_wheel + 3*M_frisbee)*M_wheel) * omega_start * r_wheel)/(4*M_wheel + 3*M_frisbee)

This makes the assumption discussed above that the shooter wheel is a homogenous cylinder with mass M_wheel spinning at omega_start radians per second prior to launch. This equation is only good for single-wheel shooters - otherwise, you'll have to go back and solve and add the individual moments of inertia. As previously discussed, this assumes the frisbee doesn't skid on either the wheel or the outer edge of the shooter, that system is closed (i.e. the motor driving the shooter wheel doesn't add much energy in the time it takes for the frisbee to be shot), and that all 100% of the energy lost by the spinning wheel is transferred to the frisbee.

Luck,

Sparks