Editorial Note: This is a very math heavy post. They won't all be this way, although it is a very engineering heavy project. Once I get to the prototyping phase, things should settle down a bit. Also, if the math fonts look bad, right click and set the math renderer to SVG.

## Preliminary Flywheel Sizing

Before I get too far into the design of my Dragon*Con costume, I need to do a bit of a study in feasibility. The biggest unknown for my costume is: How big of a flywheel will I need to power my contraptions? In order to start answering this, I need an approximation. Since I want to be able to charge my cell phone using the energy stored in the flywheel, I figured that would be a good place to start. The charger explicitly states the output and barring any gross inefficiencies in any of the electrical circuits, I have calculated the information below.

## Energy Needs

At the peak, my cell phone charger's output is 5VDC @ 1A. Therefore the energy required to run the charger for 1hr is:

$$\begin{aligned}

E_{req.} &= 5VDC \cdot 1A \cdot 1hr\\

E_{req.} &= 5watt \mbox{-} hours

\end{aligned}$$

E_{req.} &= 5VDC \cdot 1A \cdot 1hr\\

E_{req.} &= 5watt \mbox{-} hours

\end{aligned}$$

5watt-hours does not seem like much.

## Flywheel Energy Storage Equations

The rotational energy stored in a flywheel is calculated using the following equation:
\[ E_{k} = \frac{1}{2} I \omega^{2} \]

Where:

- \(E_{k}\) is the rotational energy, in Joules.
- \(I\) is the mass moment of inertia, in \(kg \cdot m^{2}\).
- \(\omega\) is the angular velocity, in radians per second.

Since these calculations are primarily to determine viability of having a spinning ~5lb mass of steel on my back, I decided to assume that the mass moment of inertia could be approximated using the equation for a thick-walled empty cylinder:

Where:

- \(m\) is the mass, in kilograms.
- The \(r\)'s are radii, in meters.

## The Exciting Conclusion

Since I have an idea of how large of a diameter I want my flywheel to be, then my unknown variable becomes \(\omega\). In order to find \(\omega\), I started by defining my variables:

\[\begin{aligned}

r_{internal} &= 0.1143 m ( 4.5in )\\

r_{external} &= 0.1270 m ( 5.0in )\\

M &= 2.268kg( 5lb )\\

E_{k} &= 1800J( 5watt \mbox{-} hours )\\

\end{aligned}\]

r_{internal} &= 0.1143 m ( 4.5in )\\

r_{external} &= 0.1270 m ( 5.0in )\\

M &= 2.268kg( 5lb )\\

E_{k} &= 1800J( 5watt \mbox{-} hours )\\

\end{aligned}\]

Then, substituting the equation for the mass moment of inertia into the rotational energy equation, we get:

\[\begin{aligned}

E_{k} &= \frac{1}{4} m ( r_{external}^{2} + r_{internal}^{2} ) \omega^{2}\\

18000J &= \frac{1}{4} \cdot 2.268kg \cdot ( 0.1270^{2}m + 0.1143^{2}m ) \cdot \omega^{2}\\

\end{aligned}\]

E_{k} &= \frac{1}{4} m ( r_{external}^{2} + r_{internal}^{2} ) \omega^{2}\\

18000J &= \frac{1}{4} \cdot 2.268kg \cdot ( 0.1270^{2}m + 0.1143^{2}m ) \cdot \omega^{2}\\

\end{aligned}\]

Then, we need to solve for \(\omega\):

\[\begin{aligned}

\omega^{2} &= \frac{18000J}{\frac{1}{4} \cdot 2.268kg \cdot ( 0.1270^{2} + 0.1143^{2} )}\\

\therefore \omega &= 1042.8 \frac{radians}{second}\Rightarrow 9958.02RPM\\

\end{aligned}\]

\omega^{2} &= \frac{18000J}{\frac{1}{4} \cdot 2.268kg \cdot ( 0.1270^{2} + 0.1143^{2} )}\\

\therefore \omega &= 1042.8 \frac{radians}{second}\Rightarrow 9958.02RPM\\

\end{aligned}\]

From these calculations, I have estimated that in order to charge my cell phone for 1 hour, I need to spin a 5lb flywheel at ~10,000 RPM. Well, that does not sound too bad at all! Of course, the energy storage requirement will be dependent on number of contraptions that I will be engaging at once, and how much energy they will require to perform their tasks. These calculations merely show that the idea is achievable.

Boy, getting a gear ratio that high is going to be another interesting problem.

Boy, getting a gear ratio that high is going to be another interesting problem.

## Extra Credit

I have created a crude SolidWorks model in order to double-check my mass moment of inertia calculations. Since they seemed to be correct, I decided to play with material types for the same shape. It turns out that for the same shape, a flywheel made of 6063 aluminum would only weigh about 1.79lb. That sounds like it would be much nicer to carry around a convention all day. So, I ran the calculations again with the new material in mind and found out that I would have to spin the flywheel at ~18,000RPM. My preliminary finding on bearings seem to indicate that 20,000RPM is going to be my max, so aluminum might be an option!

After looking at bearings, I also reran the calculations for the steel flywheel spinning at ~20,000 RPM and determined that it would be storing ~20watt-hours. This

*really*demonstrates the power of the squared angular velocity term in the rotational energy equation.
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