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Problem: A pilot wishes to fly from Bayfield to London, a distance of 85 km on a bearing of 160°. The speed of the plane in still air is 250 km/h. A 20 km/h wind is blowing on a bearing of 030°. Remembering that she must fly on a bearing of 160° relative to the ground (i.e the resultant must be on that bearing),

- find the heading she should take to reach her destination.
- how long the trip will take.

I’ve drawn the diagram for this problem numerous times and I cannot seem to figure out where to start… I know the answers to the questions because of other websites, but the answers they give don’t make sense to me as they aren’t very explanatory (it just shows the equations and how to get numbers)… I’m looking for a more thorough explanation for an answer as I’m trying to understand the problem (not just find the answer).

Please help!

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To get the concepts going:

- Draw a line 20km long on a bearing of 30°. This is where the plane would have got to after an hour if it had been hovering stationary in the moving air.
- Pick a bearing – any bearing, and, starting at the end of the previous line, draw a line 85km long in that direction.

The end of the second line shows where the plane will be after an hour’s flight along that heading. *The “do all the wind and then do all the flying” approach does not introduce errors. To see this, draw two lines half the length, representing half an hour’s flight, then do it again, representing the second half-hour. You will end up in the same place as in the original “1-hour” exercise. If you have the patience, you can repeat the exercise with 60 minute-long segments, or 3600 second-long ones,…*

You want to be on a bearing of 160° from your starting point. So draw an infinitely long line on that bearing.

Now the heading you should actually be flying on is the heading which, after step 2, intersects that infinite line. In your place, I’d do it like this:

- Draw the 20km line on a bearing of 30°.
- Draw the infinite 160° line.
- Set your compasses or dividers to straddle a distance of 85km. Put one leg on the end of the 20km line and the other leg on the infinite 160° line. For convenience of explanation, draw a line between the legs.

The direction of the line you have just drawn is the heading to be flown.

The distance between the start of the 20km line and the end of the line you have just drawn is the ground distance flown in an hour. Having a ground speed in km/h you should easily be able to work out how the trip will take.

There are lots of other methods you might use to get the answer (for instance, fly first and get blown after), but I hope that this approach will get you to understand what is really going on.

And what happened to nautical miles, anyway?

Here’s a translation of Martin Kochanski’s excellent explanation of the concepts behind the solution into mathematical manipulations.

The solution amounts to finding the intersection between the line to the destination and a circle with radius equal to the airspeed and center that’s been offset by the wind speed and direction. A brute-force approach is to set up the equations of this circle and line, solve them simultaneously, select which of the possible two solutions is the correct one, then find the slope of the resulting line and find its arctangent (taking care to get it in the correct quadrant) to get the required heading. If there’s no solution, you can’t get there from here—the plane can’t fly fast enough to overcome the wind. This is a tedious computation. You can do something simpler by taking advantage of the geometry.

Referring to the diagram above, the line $OA$ is the “infinite line” to the destination with bearing $\beta$ and $OW$ is the wind vector with direction $\alpha$ and distance equal to the wind speed $w$. The circle is centered at $W$, while $WA$ is a radius with length equal to the airspeed $a$. The heading that we need is in the direction of $WA$, which you can see is $\beta+\phi$ by drawing a line parallel to $OA$ through $W$. By the Law of Sines, we have $${\sin(\beta-\alpha)\over WA}={\sin\phi\over WO}$$ which becomes $$\sin\phi=\frac wa\sin(\beta-\alpha)$$ after substituting and rearranging. If $\sin(\beta-\alpha)$ is negative, we’ll end up with a negative value for $\phi$, but that’s as it should be: in this case the wind is blowing to the right relative to the desired bearing, so the plane’s heading needs to be adjusted counterclockwise.

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