Possible formula to calculate ball speed from wheel speed & radius?

So far, the best formula i found is:

v = radius (in meters) * flywheel rpm * 0.10472

would this be accurate to calculate ball rpm from? assuming losses, the ball would end up slower than you calculated.

Assuming you were using a 3.25in traction wheel, that would convert to 0.08255 meters.

If you had a double fly wheel, both with 0.08255 meter wheels, both spinning 1000rpm for example, would this be the correct to assume that the output speed of the ball would be

0.08255 * 1000 * 0.10472 = 8.6 meters per second?

There will be losses, i doubt the ball would be accelerate from 0 to 8.6 meters per second.

So, assuming the ball speed is the velocity of the ball the moment after the wheels launch it, I can also calculate the ball speed, but in a different way. I will try your way to see if it yields similar results to mine.

Alright, here I go:

The inital velocity can be considered as one component, or two components that combine into the one. Take triangle abc. c is your initial velocity. a and b are the two components that make up this velocity. You also have an angle, <ac. Regardless of the angle, your initial velocity © is the same. when <ac=0, then your x component of your velocity (a) is equal to your initial velocity.

So, launch a ball at a 0-degree angle off of a flat table. Use a stopwatch or the like to find the time it takes from when the ball exits the launcher to when the ball hits the ground. mark where it hits. Find the horizontal distance from the ball’s start point to the place where it hit. you may want to do this a few times to get an average.

Since vertical velocity is independent of the horizontal velocity, you can calculate the ball’s initial horizontal velocity by dividing the average horizontal distance by the average time. Since the ball was launched at 0 degrees, the initial horizontal velocity is equal to the initial velocity. So, now you have the initial velocity.

To go one step further, you can use the below equations with a parametric graphing calculator to draw the trajectory of the ball.

For the below equations: X and Y are graphic coordinates, T is the parameter, Vi is the initial velocity, K is the launch angle, H is the height of the launcher off of the ground, and A is the acceleration due to gravity (remember, it’s negative)
The variables Vi, K, A, and H are constants that you must provide. K is the only variable that you must provide that changes, really.
X(T) = Vi * cos(K) * T
Y(T) = (Vi * sin(K) * T) + (0.5 * A * (T * T)) + H

To graph it on a Y= graph;
Y = (X * tan(K)) + (0.5 * A * ((X /(Vi * cos(K)))^2) + H

To test my method, I had one of the younger kids put together a respectable double-flywheel. After the horizontal launch, I informed said kids that their launcher was about 1.61 meters per second short of being able to shoot the ball into the high goal from the loading zone. I then set up a line on the floor that was where the flywheel had to shoot past in the horizontal to get the ball in the high goal. The boys reworked the mechanism until the ball mostly hit the line. So, they moved the launcher to the starting tile, and were able to then shoot the ball into the high goal. All of this was for naught, as one of the young’uns decided that the double flywheel was “too slow” so now they’re back to square one. Kind of.

Won’t we both have a slight error in ours, ex. motors returning up to speed after the ball comes in contact, slippage between the ball and the flywheels, actual motor speed because resistance?

All of those should be independent of the launch angle, and thus would already be integrated into the launch velocity already. I say should because my calculations were off by at most six centimeters in my tests.