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%Copyright Andre Masella. All rights reserve. Do not reproduce without express permission.
\title{The Mechanics of Falling Off a Bicycle}
\author{Andre Masella}

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\maketitle

\section{Ejection of the Rider by Torque Anti-parallel to the Direction of Motion}
\subsection{Definitions}
\begin{itemize}
\item  $m_\mathrm{total} = m_\mathrm{bike} + m_\mathrm{rider}$ where $m_\mathrm{bike}$ is the mass of the bike and $m_\mathrm{rider}$ is the mass of the cyclist.
\item $\mu_\mathrm{r-c}$ is the coefficient of static friction between tyre rubber and concrete. 
\item $\mu_\mathrm{r-d}$ is the coefficient of static friction between tyre rubber and dry, packed earth.
\item $\Delta t_\mathrm{stop}$ is the time at which the bicycles comes to rest assuming it left the paved surface at $t=0$.
\item $\Delta r$ is the distance from the edge of the wheel to the point where the tyre would make contact with the curbing. 
\item The positive $x$-axis is the direction in which the bicycle is travelling. The positive $y$-axis is the direction in which the bicycle has travelled off the paved surface. The positive $z$-axis is anti-parallel to gravity while still being orthogonal to the $x$ and $y$ axes. 
\item The moment of inertia of the wheel is calculated about the $y$-axis at the point of attachment of the axle.
\item The moment of inertia of the bike is calculated about the $x$-axis through the centre of gravity.
\end{itemize}

\subsection{Torques About the Front Wheel}

$$\tau _\mathrm{chain} + \tau _\mathrm{pavement} + \tau _\mathrm{curbing} = I_\mathrm{wheel} \alpha _y $$
$$r_\mathrm{sprocket} F_\mathrm{chain} - r_\mathrm{wheel} (m_\mathrm{total} g \mu_\mathrm{r-d}) + ( r_\mathrm{wheel} - \Delta r) F_a = I_\mathrm{wheel} \alpha _y $$
$$r_\mathrm{wheel} \int\!\int_0^{\Delta t_\mathrm{stop}} \alpha _y dt^2 = \Delta d_\mathrm{stop}$$

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\subsection{Force Applied from the Curbing}

$$F_a = m_\mathrm{total} g \sin {\pi \over 6 } - \mu _\mathrm{r-d} \cos {\pi \over 6 }$$

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\subsection{Torque About the Bike}

$$r_\mathrm{wheel} F_a = I_\mathrm{bike} \alpha _x$$

\subsection{Velocity of Impact}
$$ v_\mathrm{final} = g \Delta t _\mathrm{stop} + \int_0^{\Delta t_\mathrm{stop}} r_\mathrm{wheel} \alpha _x dt $$
and by conservation of momentum
$$ \int F dt = { v_\mathrm{final} m_\mathrm{rider} } $$
where $F$ is the force exerted on the cyclist as they hit the pavement.\par

This is full of glaring errors (most of them approximating things to point objects), but it provides a decent approximation of how much force my hip and shoulder took while riding down Keats~Way just before University~Avenue on the south side in Waterloo, Ontario, Canada at approximately 8:20AM on my way to class.\par

\section{Limit of Young's Modulus as Applied to the Scaphoid}

\subsection{Definitions}
\begin{itemize}
\item  $m_\mathrm{rider}$ is the mass of the cyclist. 
\item $h$ is the height, from the ground, where the centre of mass of the rider is located.
\item $g$ is acceleration due to gravity for a body on the Earth.
\end{itemize}
\subsection{Motion of the Grade and Rear Tyre}

At time $t_0$, braking as applied to the rear wheel of the bicycle when moving with approximately the constant velocity $v_0$. An extremely short time later, $t_1$, with a velocity $v_1\sim v_0$, the gravel underneath the rear tyre began to move perpendicular to the direction of motion down a gradient. The friction between the tyre and the loose gravel was quite high, so the bike accelerated  very quickly down the gradient.

\subsection{Inertia and Energies of the Rider}

Since the acceleration of the bike was down and perpendicular to $v$, with a large acceleration, we can assume there was no force imparted to the rider. That is, we can assume the bike ``disappeared'' from under the rider with out affecting his inertia. This means the energy of the rider can be described as:
$$E = \frac{1}{2} m_\mathrm{rider} v^2 + m_\mathrm{rider} g h$$
due to conservation of energy, the energy when the rider reaches the ground can be described as:
$$E = \frac{1}{2} m_\mathrm{rider} 0^2 + m_\mathrm{rider} g 0 + \frac{A \Delta L ^2 \mathcal{E}}{L}$$
Now, some of the force will be absorbed by the soft tissues, so only a certain fraction of the total energy is actually channelled to the bone. Therefore,
$$\frac{1}{2} m_\mathrm{rider} v^2 + m_\mathrm{rider} g h = \frac{A_\mathrm{scaphoid} \Delta L_\mathrm{scaphoid} ^2 \mathcal{E}_\mathrm{scaphoid}}{L_\mathrm{scaphoid}} + \frac{A_\mathrm{flesh} \Delta L_\mathrm{flesh} ^2 \mathcal{E}_\mathrm{flesh}}{L_\mathrm{flesh}}$$
\par

This is precisely what happened to my scaphoid bone on Fischer-Hallman~Road North between the two intersections of Roxton~Drive around 8:30PM Sunday night. Since the scaphoid was fractured, Young's modulus,  $ \mathcal{E}_\mathrm{scaphoid}$, must have exceeded $170 \frac{N}{m^2}$.
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