I found an ingenious design of a spring ball launcher on grabcad and I'm interested in adapting it to accelerate a $43.7 \mathrm{~mm}$ diameter, $45 \mathrm{~g}$ golf ball enough to reach $65.5 \mathrm{~cm}$. My calculations are shown below together with my questions.

spring ball launcher

First, my project puts a limit on using max $1 \frac{N}{m m} $ springs, so I started by calculating the energy necessary to compress the spring:

(1) $ \begin{array}{l} h_{\max }= 65.5 \mathrm{~cm}\\ E_{h}=m \cdot g \cdot h \\ E_{h_{\max }}=0.289 J \\ \end{array} $

That means I need to compress a $0.99 \frac{N}{m m} $ spring about $24 \mathrm{~mm}$ by applying a force of $24 \mathrm{~N}$. This brings my first question: Is that too much for thin PLA/ABS 3D printed parts to handle?

(2) $ \begin{array}{l} k_{s} =0.99 \frac{N}{m m} \\ E_{s} =\frac{1}{2} k \cdot x^{2}\\ x_{s} =\sqrt{\frac{2}{k_{s}}\cdot E_{h_{\max }} }= 24.165 \mathrm{~mm}\\ F_{s} =\ k \cdot x \\ F_{s} = 23.923 \mathrm{~N} \\ \end{array} $

The best spring I could find was this one, and its small diameter might force me to change the initial design of the launcher.
Does anyone think I can find something better or will this do the job?

RS PRO Spring

Then, I started thinking about electric power and the rack and pinion system. I didn't know exactly where to begin so I assumed that moving the red rack $24 \mathrm{~mm}$ in $3 \mathrm{~s}$ was a reasonable estimate for a DC motor. Is that correct? From where should I start this kind of analysis?

Rack and pinion

I arbitrarily decided that the rack would slide the $24 \mathrm{~mm}$ with 3/4 of the gear full rotation. That brought me to a $10 \mathrm{~mm}$ pitch diameter gear with ten $1 \mathrm{~mm}$ gear module. Is that reasonable? How should I decide these values based on the initial $24 \mathrm{~N}$ estimate?

My previous choices brought me an angular velocity of about $15 \mathrm{~rpm}$ (or $1.571 \frac{\text { rad }}{s}$). Together with the initial required energy of $0.289 J$ and another requirement of using at most $12 V$, I found a current consumption of $38 mAV$.

(3) $ \theta =\frac{3}{4} 2 \cdot \pi \cdot \text { rad } \quad w =\frac{\theta}{3}=1.571 \frac{\text { rad }}{s} \\ $

Here I assumed the torque on the gear to be the same as the work done by compressing the spring. Then I set this to be the electrical power necessary to produce this torque.

(4) $E \cdot w=0.454 W$

$ V=12 V \quad I =\frac{(E \cdot w)}{V}=37.837 \mathrm{~mA}$

Are these last equations correct? If so, how do I go about looking for an adequate DC motor that fits my needs?

Any corrections, comments and/or suggestions are welcome!



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