With much help from various school faculty and mentors, I’ve put together a few equations for the motion of a ball tossed from a launcher at the top hoop, attached, for anyone who might want to take a look. These equations allow you to appropriately set the elevation or power of your launcher.
It may be confusing, but begin at the top:
(Since the field measurements are in imperial units, we might as well use imperial units in the equations.)
The first equation is a basic physics equation of motion given acceleration, time, and initial vertical velocity, describing the vertical height of your ball at a given amount of time after it has left your launcher.
Your robot takes care of initial velocity, the horizontal and vertical components of which may be found by some trig; acceleration is standard acceleration due to gravity on earth; time is distance divided by horizontal velocity.
The second equation is the first equation, but with some algebraic manipulation so that V is on the left side of the equation. All variables in the first equation can, with some math, be converted into variables in the second. Consult the definitions.
As our group’s design had a fixed-angle launcher, we derived the ball’s motion in terms of V, which was the only variable we have control over. Everything in the right half of the equation except for the horizontal distance is fixed - the angle the ball leaves at is fixed, the height of the robot and basket are fixed, and acceleration due to gravity is, of course, fixed.
Thus, if we can find the horizontal distance from the basket, using a rangefinder or field markings, we can calculate the velocity with which the ball needs to leave, which can be directly affected by settings on the launcher.
If your team has a launcher that throws at constant velocity but with variable angle, then simply (heh) substitute things and solve the first equation for theta. If your team can vary both, I don’t know what to tell you.
There are some caveats:
You will have to adjust the horizontal distance depending on whether you are aiming at the backboard or at the center of the hoop. If you are aiming at the backboard, you had better put some backspin on your ball, and be within a certain angle of the basket.
The parabola of a ball’s motion has two solutions on the horizontal line extending from the basket. The first solution must be farther than 18 inches from the wall, or the ball will hit the bottom of the hoop. As for how that is found, well, we haven’t gotten to that yet. As far as I can tell, you’d use the quadratic equation to solve for the smaller solution, but that can only be applied indirectly to the first equation.
We will probably be graphing that.
Effectively, this enforces a minimum distance from the hoop from which one must shoot for a particular angle of the launcher. I’m unsure if this will actually be a problem, given the fender.
There is not a solution for sufficiently far distances for a given launcher angle; this threshold is probably far enough that it will not be a problem, but if you are overcome by madness and decide to shoot from the other side of the field, consider this. Again, I’m unsure if this is actually a problem.
This is probably an inadequate explanation, but I hope it helps somebody.
