Flywheel Launch Calculation

My dad and I worked together to find the velocity and angle for the most ideal launching. If you find an error or have a question, feel free to ask :stuck_out_tongue: (and feel free to answer others’ questions because I might not know the answer) Post your discoveries and stuff if you want to.

There will be the most surface area if a ball enters the net perpendicularly. The net angle is approximately 36 degrees. We also approximated the distance to be 4 meters at 1 meter tall.


The picture above shows a parabola with two known points. The standard form for a parabola is y = ax² + bx +c. The two points are plugged in.

(0,0)    0 = c
(4,1)    1 = 16a + 4b

Next, the slope needs to be calculated.
The derivative of standard form is dy/dx = 2ax + b.

The slope of the red line is
tan(270° + 36°) = -1.376
This is equal to the derivative at (4,1), so
2ax + b = -1.376
8a + b = -1.376

Now, we have two equations, so we can find a and b.
The elimination method results in
3.75276 = 2b
b = 1.876
1 = 16a + 4b
a = -0.4065

The equation for the parabola is
y = -0.4065x² + 1.876x

Next, we will find the maximum point of the parabola.

dy/dx = -0.813x + 1.876 = 0
0.813x = 1.876
x = 2.3075 thus x1 = 2.3075 and y1 = 2.1644
Using the acceleration formula,
y = Gt²/2
y1 = Gt1²/2
2.1644 = 9.8t1²1/2
t1 = 0.6646 (-.6646 can be removed because t cannot be negative)

Now that we know x1, y1, and t1, we can find Vx, which is constant throughout.
Vx = Change in x / Change in t
= x1/t1
= 3.472
The initial y velocity means the launch speed. It will be represented with V0y.
V0y = Gt1 because it will be the same as if it is dropped from the max value of the parabola.
V0y = Gt1 = 6.513

With Vx and V0y, the launch angle can be determined.
tan a = V0y/Vx
a = 61.938°

The initial velocity can be found, too.
V0 = √(Vx² + V0y²) = 7.3806 m/sec


Yay, we’ve made it this far. To summarize, 62° at 7.38 m/sec.


Bonus: Launch RPM Calculation

I need to state the conditions for this calculation. The wheel is 5", and it will be spinning at twice the velocity of the launch (n RPM).

2V0 = n * (1/60) * 2π * radius of wheel * 0.0254 (meter to inches)
n = (120 * V0)/(2πr) RPM
n = 2219.65 RPM

2219.65 RPM on a 5" wheel at 62°

I have made an Excel spreadsheet of gear ratios, so here is a picture of it.
https://cdn.pbrd.co/images/1nIBQ9GT.png
2.jpg
1.jpg
3.png
4.jpg

yay, my team was trying to decide on a good angle, and we were like “60 sounds like a good middle ground, lets do 60” then we calculated the required launch velocity based on that angle, so we got the same answer you did, by random luck, thanks for the info though, good to know we have something right

You might be interesting in reading a thread we had going near the start of the season, with the goal of making some form of calculator to automatically work things out. In this we involved air resistance, not just assuming no air resistance as you have in this post. It’s great to see you actually using maths though, and not just trial and error! :smiley:

Thread: https://vexforum.com/t/launching-speed-calculator/29576/1
Calculator: http://vex.us.nallen.me/extras/nbn_shooting

There was a lot of iterations on the equations we used, and we continually found issues and improved them. At the end there were two real options:

  1. Switch to using an ODE solver (complicated, takes longer to solve) to get correct values for c*v^2 air resistance.
  2. Keep using the “incorrect” equations that give an approximation somewhere between cv and cv^2 air resistance.

I decided to go with Option 2, others (capow08) wrote MATLAB scripts to actually solve the ODEs properly and get more correct values. But for me, the complexity of integrating this as a web service was too much work for the benefit gained.

If you’re really interested in how to do calculations involving air resistance I’d definitely suggest reading through the thread and our findings along the way! :slight_smile:

Only about .3 m/s and a few degrees off! Good enough for today :stuck_out_tongue:

My dad and I might try to work out those air resistance kinks (that’s euphemism) some day. Right now, I’m wondering about the effect of backspin. Do you (or does anyone) know anything about that? Now I need to start working on my history paper :slight_smile:

backspin is good I found for shooting like in basketball :wink: