<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
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{\bf Question}

An aircraft has speed $v$ and a flying range (out and back) of $R$
in calm weather.  Prove that in a north wind (i.e. from the north)
of speed $n$ ($n&lt;v$) its range is $$ \frac{R(v^2 - n^2)}{v(v^2 -
n^2 \sin^2 \phi)^{\frac{1}{2}}}$$ in a direction whose true
bearing from north is $\phi$.

What is the maximum value of this range and it what directions may
it be attained?

\vspace{.25in}

{\bf Answer}

The range of the plane is determined by its flying time.  In still
air flying time is $2RV^{-1}$.   Hence when there is a wind the
total time is   $2RV^{-1}$.

I.e. $T_{\rm out} + T_{\rm in} = 2RV^{-1}$

where  $T_{\rm out}$ and $T_{\rm in}$ are the times for the
outward and inward legs.

Now if the range in the wind is $R_\phi$ then $T_{\rm out} =
\frac{R}{v_{\rm out}}$ and $T_{\rm in} = \frac{R}{v_{\rm in}}$
where $v_{\rm out}$ and $v_{\rm in}$ are outward and inward speeds
relative to the ground.

\bigskip

\underline {Outward trip}

$\ds {\bf v}_{\rm out} = {\bf v}_{AW} + {\bf v}_W$

where ${\bf v}_{AW}$ is the velocity of the aeroplane relative to
the wind

and ${\bf v}_W$ is the velocity of the wind relative to the
ground.


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\put(3,0){\vector(-1,2){1}}

\put(2.2,2){\makebox(0,0){$v$}}

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\put(3,0){\vector(-2,1){1}}

\put(1.5,0.25){\makebox(0,0)[l]{${\bf v}_{\rm out}$}}

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\put(0.8,2.5){\makebox(0,0){$n$}}

\put(2.85,.5){\makebox(0,0){$\phi$}}

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\put(1.2,1.1){\makebox(0,0)[l]{$\pi - \phi$}}

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\begin{eqnarray*}{\rm cosine\ rule\ } v^2 &amp; = &amp; v_{\rm out}^2 + n^2 -
2n v_{\rm out} \cos(\pi - \phi) \\ \Rightarrow 0 &amp; = &amp; v_{\rm
out}^2 + 2n\cos \phi v_{\rm out} - v^2 \\ \Rightarrow v_{\rm out}
&amp; = &amp; \frac{-2n\cos \phi \pm \sqrt{4n^2 \cos^2 \phi - 4n^2 +
4v^2}}{2}
\\ \Rightarrow v_{\rm out} &amp; = &amp; -n\cos \phi + \sqrt{n^2 \cos^2 \phi - n^2
+ v^2}
\end{eqnarray*}

A similar calculation for outward trip gives $$v_{\rm in} = n\cos
\phi + \sqrt{v^2 - n^2 \sin^2 \phi}$$

$\ds\Rightarrow \frac{R_{\phi}}{v_{\rm in}} + \frac{R_\phi}{v_{\rm
out}}$

$\ds  =  \left[ \frac{1}{-n\cos \phi + \sqrt{v^2 - n^2 \sin^2
\phi}} + \frac{1}{n\cos \phi + \sqrt{v^2 - n^2 \sin^2 \phi}}
\right] R_\phi  =  \frac{2R}{v}$

$\ds \Rightarrow R_\phi  =  \frac{R(v^2 - n^2)}{v \sqrt{v^2 - n^2
\sin^2 \phi}}$

\bigskip

By inspection $R_\phi$ maximum when the denominator in minimum at
$\phi = \pm \frac{\pi}{2}$ in which case the maximum range is $\ds
R\left( 1 - \frac{n^2}{v^2}\right)$ in headings due east or west.


\end{document}
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