Above, $O$ is the centre of the sphere, $x$ is its radius. Writing the area of trapezoid $ABCD$ as a sum of three triangles, we can easily solve for $x$ to get:
$$
x=\frac{1200\sqrt{3}}{54}=38.490017\ldots
$$
No trig or calculus needed, only geometry.
Note that $\theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $\theta_{out}=\theta_{in}$
The case of very long legs is similar to this: $\theta_{out}=\theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $\theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $\theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$\theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=\sin \theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (\theta_{in} + \theta_{out}) / 180$ and
$t_{Tin} = 1 + (\theta_{in} + \theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(\cos \theta_{in}, \sin \theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + \cos \theta_{in}, 3w_2 - \sin \theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
\alpha_{out} = \arctan \frac{ 3w_2 - \sin \theta_{in} }{ 3w_1 + \cos \theta_{in} }
$$
Note this is not $\theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $\alpha_{out}$,
and he is able to determine corresponding $\theta_{out}$. I am lazy
to do the calculation of the corresponding $\theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - \cos \theta_{out} + w_1, - \sin \theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = \frac { 3w_1 + \cos \theta_{in} }{ - \cos \theta_{out} + w_1 }
= - \frac { 3w_2 - \sin \theta_{in} }{ - \sin \theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $\theta_{in}=w_2$ and
$
\theta_{out} = \alpha_{out} = 2 \theta_{in}
$.
Best Answer
Here is my solution. Here is a picture.
$AE', AG, EF, DD'$ are horizontal, $CD', CG, DF, EE'$ are vertical, $AB$ is at angle $\pi/4 = 45^\circ$, $\angle ABC$ is a right angle. I will solve for angle $\theta=\angle DEF = \angle CAG$. Then the angle we want is $3\pi/4-\theta$.
In the original problem, we are given lengths $AB=18, DD'=1,AE'=1, DF=10$. Using these, we get:
$\angle ACB = \theta+\pi/4$, so $AC = 18 \csc(\theta+\pi/4)$ and $CG = 18 \sin\theta\csc(\theta+\pi/4)$.
In right triangle $\triangle D D' C$ we have $DD'=1$ so $CD' = \tan(\theta)$; similarly $EE'=\tan(\theta)$.
Subtracting, $DF = CG - CD' - EE' = 18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta)$
The problem to solve is $$ 18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta) = 10 $$
This has two solutions,
The one near $\pi/4$ is $$ \theta = -\arctan \frac{\left( 3\,\sqrt {2}-9+\sqrt {89-54\,\sqrt {2}} \right) \sqrt {2}}{2} $$ (We have to solve a quadratic equation for $\tan\theta$.)
We get $\theta \approx 0.7049$ so the angle we want is $3\pi/4 - \theta \approx 1.6513 \approx 94.6106^\circ$.
In the other example, things are the same except $DF=13$ by subtracting. We must solve $$ 18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta) = 13 $$ The solution near $\pi/4$ is $$ -\arctan \frac{\left( 15\,\sqrt {2}-36+\sqrt {1538-1080\,\sqrt {2}} \right) \sqrt {2}}{8} $$
We get $\theta \approx 1.1145$, so $3\pi/4-\theta \approx 1.2417 \approx 71.146^\circ$