202008190508 Solution to 1973-AL-AMATH-I-1

In calm weather an aeroplane has a speed $v$ and a flying range (out-and-back) $R$ . In a north wind of speed $n$ ( $n<v$ ) and in a line of flight that makes a constant angle $\phi$ with the north, I wish to know its new flying range.

Assuming that the maximum time of flight is independent of wind conditions, and that the effect of turning round at the end of the outgoing journey can be ignored.

Let $\mathbf{v}=v_x\, \hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}}$ denote the velocity of the aeroplane in calm weather (i.e., zero wind speed).

Then $\mathbf{v}=(v_x,\, v_y) = (v\sin\phi , \, v\cos\phi )$ , where $v=|\mathbf{v}|$ is its speed, and $\phi$ measured anticlockwise from the north.

In a north wind of speed $n$ , the outgoing speed $v_{\textrm{out}}$ of the plane will decrease by $(v\sin\phi ,\, v\cos\phi -n)$ , and on its return the speed $v_{\textrm{back}}$ will increase by $(v\sin\phi ,\, v\cos\phi +n)$ .

The new range $R'$ is calculated in two parts:

$R'_{\textrm{total}}=R'_{\textrm{out}}+R'_{\textrm{back}}$ .

$\because \qquad R'_{\textrm{out}}=R'_{\textrm{back}}$

$\therefore\qquad R'_{\textrm{out}}=R'_{\textrm{back}}=\displaystyle{\frac{R'}{2}}$

If the plane is on schedule, the time of flight $t=\displaystyle{\frac{R}{v}}$ should be kept unchanged.

I.e.,

$\begin{aligned} t & = t' \\ t & = t'_{\textrm{out}} + t'_{\textrm{back}} \\ \frac{R}{v} & = \frac{R'_{\textrm{out}}}{v'_{\textrm{out}}} + \frac{R'_{\textrm{back}}}{v'_{\textrm{back}}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi -n)^2}} + \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi +n)^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'}{2}\Bigg\{ \frac{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}\Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}\Bigg\} \\ \end{aligned}$

Let $a=v^2+n^2$ and $b=2nv\cos\phi$ .

$\begin{aligned} R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}}{\sqrt{a+b}-\sqrt{a-b}} \Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}(\sqrt{a+b}+\sqrt{a-b})}{(\sqrt{a+b})^2-(\sqrt{a-b})^2} \Bigg\} \\ R' & = \frac{R}{v}\bigg[\frac{(\sqrt{a-b})(a+b)+(\sqrt{a+b})(a-b)}{b}\bigg] \\ R' & = \frac{R}{v}\bigg[ \sqrt{a-b} \bigg( \frac{a}{b} \bigg) +\sqrt{a-b} + \sqrt{a+b} \bigg( \frac{a}{b} \bigg) - \sqrt{a+b} \bigg] \\ \end{aligned}$

(The Law of Cosines)

I need a break. To be continued.

After dinner, things are much clearer. Go back to the earlier line:

$\displaystyle{\frac{R}{v} = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}}}$ .

Now invoke its physical meaning:

i. Let $\mathbf{n}= -n\,\hat{\mathbf{j}}$ ( $n>0$ ) be the north wind velocity vector pointing in the southerly direction.

ii. Let $\mathbf{v}= -v_x\,\hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}}$ ( $v_x,\, v_y>0$ ) be the velocity vector of the airplane pointing in the northwesterly direction.

iii. Let $\mathbf{s}=\mathbf{n}+\mathbf{v}$ be the side subtending the angle $\phi$ enclosed by the two vector arrows $\mathbf{n}$ and $\mathbf{v}$ . That is, the bottom of vector $\mathbf{s}$ is touching the arrowhead of $\mathbf{v}$ , and the arrowhead of $\mathbf{s}$ is touching the tip of $\mathbf{n}$ .

iv. The sides $\mathbf{n}$ , $\mathbf{v}$ , and $\mathbf{s}$ form a triangle of perimeter given by $n+v+s$ .

v. $s^2=|\mathbf{s}|^2= v^2+n^2-2nv\cos\phi$ .

vi. $2nv\cos\phi =2\mathbf{n}\cdot\mathbf{v}$ .

(continue)

$\begin{aligned} \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \dots & = \frac{R'}{\sqrt{4(v^2 -2nv\cos\phi +n^2)}} + \frac{R'}{\sqrt{4(v^2 +2nv\cos\phi +n^2)}} \\ \dots & = \frac{R'}{\sqrt{4(s^2)}} + \frac{R'}{\sqrt{4(s^2 + 4\mathbf{n}\cdot\mathbf{v})}} \\ \dots & = \bigg( \frac{\sqrt{4(s^2+4\mathbf{n}\cdot \mathbf{v})}+\sqrt{4(s^2)}}{\sqrt{4(s^2)(s^2+4\mathbf{n}\cdot\mathbf{v})}} \bigg) R' \\ \dots & = \bigg( \frac{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s}{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}} \bigg) R' \\ R' & = \bigg( \frac{R}{v} \bigg) \bigg( \frac{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}}{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s} \bigg) \\ \end{aligned}$

Roughwork.

Simplifying first the numerator,

$\begin{aligned} & \sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}} \\ & = \sqrt{(v^4+n^4+2n^2v^2-4nv^3\cos\phi -4n^3v\cos\phi + 4n^2v^2\cos^2\phi )+4(v^2 + n^2 -2nv\cos\phi )(nv\cos\phi )} \\ & = \sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2} \\ \end{aligned}$

then the denominator

$\begin{aligned} & \sqrt{s^2+4\mathbf{n}\cdot\mathbf{v}} + s \\ & = \sqrt{v^2+n^2-2nv\cos\phi+4nv\cos\phi } + \sqrt{v^2 + n^2 -2nv\cos\phi } \\ & = \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} \\ \end{aligned}$

Putting them together,

$\begin{aligned} R' & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{ \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} } \bigg) \\ & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2-n^2)^2+(2nv)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )}} \bigg) \\ \end{aligned}$

Unsatisfactory still.

Trimming the numerator again,

$\begin{aligned} & \sqrt{(v^2-n^2)^2\bigg( 1 + \frac{(2nv)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}\bigg)} \\ & = (v^2-n^2)\sqrt{\frac{(v^2+n^2)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}} \\ \end{aligned}$