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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Sailboat} (Homework January 16, 2013) \\
\end{center}
\setcounter{equation}{0}
\bigskip

One of the major inventions of mankind (not known in the antique)
was the keel (or skeg in dinghies), which allows sailing boats to 
zigzag against the wind. Below you find a sketch for the basic idea, 
where
\begin{equation}
 0<\beta<\alpha<\frac{\pi}{2}
\end{equation} 
holds for the angles. Calculate the forward component $\vec{F}^b$
from the wind force $\vec{F}$ on the boat under the assumption
that friction and perpendicular drift of the boat can be neglected.
\smallskip

\noindent {\bf Solution:} \smallskip

\noindent Let $\vec{F}=-F\,\hat{y}$, $F>0$.  From the figure we find 
for the unit vectors parallel and perpendicular to the sail
\begin{eqnarray} 
  \hat{s}_{||}    &=& \sin(\beta)\,\hat{x} - \cos(\beta)\,\hat{y}\,,\\
  \hat{s}_{\perp} &=& \cos(\beta)\,\hat{x} - \sin(\beta)\,\hat{y}\,.
\end{eqnarray}
The projections of the force on these directions are
\begin{eqnarray} 
  \vec{F}^s_{||}    &=& \left(\hat{s}_{||}\cdot\vec{F}\right)\,
  \hat{s}_{||}\,,\\
  \vec{F}^s_{\perp} &=& \left(\hat{s}_{\perp}\cdot\vec{F}\right)\,
  \hat{s}_{\perp}\ =\ \sin(\beta)\,F\,\hat{s}_{\perp}\,,
\end{eqnarray}
of which the $\vec{F}^s_{||}$ component can be discarded, because we
neglect friction. The unit vector $\hat{s}_{\perp}$ can be expanded
into the unit vector parallel and perpendicular to the boat:
\begin{eqnarray} 
  \hat{s}_{\perp} &=& \sin(\alpha-\beta)\,\hat{b}_{||} 
                   - \cos(\alpha-\beta)\,\hat{b}_{\perp}\,.
\end{eqnarray}
As the drift of the boat is neglected, we are only interested in the
component parallel to the boat, which is our result
\begin{eqnarray} 
  \vec{F}^b_{||} &=& \sin(\alpha-\beta)\,\sin(\beta)\,F\,\hat{b}_{||}\,.
\end{eqnarray}

\end{document}

\eject

\noindent The expansion of $\hat{b}_{||}$ in $x$ and $y$ directions 
reads
\begin{eqnarray} 
  \hat{b}_{||} &=& \sin(\alpha)\,\hat{x} + \cos(\alpha)\,\hat{y}
\end{eqnarray}
and the $y$ coponent of the forward force becomes
\begin{eqnarray} 
  \vec{F}^b_y &=& \sin(\alpha-\beta)\,\sin(\beta)\cos(\alpha)\,F\,
                  \hat{y}\ =\ F^b_y\,\hat{y}_{||}
\end{eqnarray}
with $F^b_y=\sin(\alpha-\beta)\,\sin(\beta)\cos(\alpha)\,F$. At 
the endpoint, $\beta=0$ and $\beta=\alpha$ we see $F^b_y=0$ and 
in-between $F^b_y>0$. The derivative with respect to $\beta$ is
\begin{eqnarray} 
  \frac{dF^b_y}{d\,\beta}\left(\beta\right) &=& \left[\sin(\alpha-\beta)
  \,\cos(\beta)-\cos(\alpha-\beta)\,\sin(\beta)\right]\,
  \cos(\alpha)\,F\,.
\end{eqnarray}
At the endpoints we have
\begin{eqnarray} 
  \frac{dF^b_y}{d\,\beta}\left(0\right) = \sin(\alpha) > 0
  ~~{\rm and}~~
  \frac{dF^b_y}{d\,\beta}\left(\alpha\right) = -\sin(\alpha) < 0\,.
\end{eqnarray}
In-between $dF^b_y/d\,\beta$ is a monotonically decreasing function
of $\beta$. Hence, there is precisely one zero and $F^b_y$ takes its
maximum at this point. Inspecting
\begin{eqnarray} 
  \sin(\alpha-\beta_0)\,\cos(\beta_0) -
  \cos(\alpha-\beta_0)\,\sin(\beta_0)\ =\ 0
\end{eqnarray}
we find that the solution is
\begin{eqnarray} 
 \beta_0\ =\ \alpha/2
\end{eqnarray}
and as function of $\alpha$ the maximum value is
\begin{eqnarray} 
  \vec{F}^m_y(\alpha) &=& \sin^2(\alpha/2)\,\cos(\alpha)\,F\,\hat{y}\ 
  =\ F^m_y(\alpha)\,\hat{y}_{||}
\end{eqnarray}
with $F^m_y(\alpha)=\sin^2(\alpha/2)\,\cos(\alpha)\,F$.

\eject

We want to find the maximum of the function
\begin{eqnarray} 
  F^m_y(\alpha)=\sin^2(\alpha/2)\,\cos(\alpha)\,F
\end{eqnarray}
in the range $0\le\alpha\le\pi/2$. By inspection we find for the 
endpoint $\alpha=0$ and $\alpha=\pi/2$ we find $F^m_y=0$ and
in-between $F^m_y>0$. The derivative is 
\begin{eqnarray} 
  \frac{dF^m_y}{d\,\alpha} &=& \left[ \sin(\alpha/2)\,\cos(\alpha/2)\,
  \cos(\alpha)- \sin^2(\alpha/2)\,\sin(\alpha)\right]\,F\\ &=& 
  \left[\sin(\alpha/2)\,\cos^3(\alpha/2)-\sin^3(\alpha/2)\,\cos(\alpha/2)
  - 2\,\sin^3(\alpha/2)\,\cos(\alpha/2)\right]\,F \nonumber \\ &=&
  \left[\sin(\alpha/2)\,\cos^3(\alpha/2) -
  3\,\sin^3(\alpha/2)\,\cos(\alpha/2)\right]\,F \nonumber \\ &=&
  \left[\cos^3(\alpha/2) - 3\,\sin^2(\alpha/2)\,\cos(\alpha/2)\right]
  \,\sin(\alpha/2)\,F \nonumber \\ &=& \left[4\,\cos^3(\alpha/2) - 
  3\,\cos(\alpha/2)\right]\,\sin(\alpha/2)\,F\,.
\end{eqnarray}
Substituting $x=\cos(\alpha/2)$ we have to find solutions of
\begin{eqnarray} 
  0 &=& 4 x^3-3 x\ =\ (4 x^2-3)\,x
\end{eqnarray}
of which 
\begin{eqnarray} 
  x_0 &=& \sqrt{3}/2
\end{eqnarray}
corresponds to the maximum of the function $F^m_y(\alpha)$ in the
given range. The corresponding angle is in degree $\alpha_0=60^o$.
The numerical value for $F^{\max}_y=F^m_y(\alpha_0)$ follows from 
\begin{eqnarray} \nonumber
  F^m_y(\alpha_0) &=& \sin^2(\alpha_0/2)\,\cos(\alpha_0)\,F\ =\
  \left[\sin^2(\alpha_0/2)\,\cos^2(\alpha_0/2) -
  \sin^4(\alpha_0/2)\right]\,F
\end{eqnarray}
and $\sin^2(\alpha_0/2)=1-\cos^2(\alpha_0/2)=1-3/4=1/4$ to be
\begin{eqnarray} \nonumber
  F^{\max}_y&=&\left[\left(\frac{1}{4}\right)\,\left(\frac{3}{4}\right) 
  -\left(\frac{1}{4}\right)^2\right]\,F\ =\ \left(\frac{1}{8}\right)\,F\,.
\end{eqnarray}

\end{document}

