Wind Circle Solution

Wind Circle

lementary time-distance-speed calculations apply: Flying in still-air at an airspeed of 228 knots over a leg between waypoints 562 nautical miles apart would require 2:28 -- not 2:32. Obviously the flight must have been adversely affected by winds aloft. Its ground speed was apparently slowed to 222 knots.

The difference, 6 knots, would seem to be the speed of the wind faced by the flight but not necessarily the solution to the Wind Circle puzzle. For that we must take into account crosswind as well as headwind. As a vector, wind has direction as well as magnitude. The puzzle did not provide enough information to determine wind direction. Or did it?

The return trip the same day took exactly the same time for the same leg.

A headwind facing any given course would of course be a tailwind enjoyed by the reverse course. The return trip over the same leg might be expected to take 8 minutes less time. Winds tend to be steady, changing only gradually. Reversing a wind's direction within a day is unlikely. An explanation that seems reasonable is that the wind aloft at the cruise altitude is perpendicular to the course of the two flights over the common route in both directions.

The aircraft's Ground Speed vG is a vector aligned with the course for the flight, and the Wind Speed vW is a vector perpendicular to the course, together forming a right triangle with the aircraft's Airspeed vA as the hypotenuse. By the Theorem of Pythagoras...

vW² = vA² - vG²

...which, when solved for vW, gives us the solution to the Wind Circle puzzle.

The wind at cruise altitude is 52 knots perpendicular to the course line.

The Airbus must be flying its course directly across a jet stream, which can reach speeds of over 200 knots. We have solved the puzzle but we have not explained the title "Wind Circle."

very pilot knows that flights do not benefit from tailwinds as often as they suffer from headwinds. The puzzle gives something of a hint about that curious asymmetry and might stimulate sophisticated solvers to explore the magnitudes.

This picture shows Airspeed as a vector vA retrospectively superimposed upon the contrails of an aircraft flying at cruise altitude. The red arrow indicates that a wind vector vW in that region of the sky is acting on the flight causing the aircraft to be blown to the right. An observer standing on the ground would see the aircraft following a Ground Track to the right of its heading as represented by the green arrow. In that direction lies the flight's intended destination.

Indeed, the pilot has deliberately 'taken up' a heading to the left of course by a 'Wind Correction Angle' aA. The resultant flight vector is indicated as vG. Unlike the wind vector in the puzzle, the red arrow is not perpendicular to the green arrow. In the depicted case, the wind vector vW is acting at an angle aW to the Ground Track. We see two components: a crosswind and a headwind, such that vG < vA.

Meanwhile, our observer standing on the ground looking up into the sky will see the contrails drifting to the right with the wind.

n flight, as in geometry, the shortest distance between two points is a straight line. In the presence of winds aloft, the pilot must adjust the heading of the aircraft to cancel out crosswind in order to achieve the intended Ground Track to the destination. The pilot has no way to correct for headwind -- and no desire to correct for tailwind.

General 'wind triangle' calculations apply the following equations:

Wind Correction Angle........ aA = arc sin (vW sin aW)
Ground Speed.................... vG = vA cos aA - vW cos aW

...wherein the independent variables are vW and aW (The wind bloweth where it listeth).
Holding wind speed vW constant, one can ascertain the effect of wind direction aW on ground speed vG.

Generalizing still further, sophisticated solvers will probably want to represent windspeed and groundspeed as percentages of Airspeed vA...

Ground Speed: [100%] vG / vA = [100%] cos aA - [100%] (vW / vA) cos aW

Might as well be systematic about the investigation and generate a complete Wind Circle for all possible wind angles, thereby producing a graph like this...

...from which we see the worst-case effect of a direct headwind (aW = 0^o) and best-case effect of a direct tailwind (aW = 180^o). The Green Flight in the puzzle is represented in the green curve, in which vW / vA = 52 / 228 Kts. With the same direct crosswind for either direction of flight, the Ground Speed is slowed to 222 Kts (-2.6%).

We also see from the Wind Circle above that, indeed, "on the average" (with the wind blowing from any arbitrary direction), more often than not, ground speed is less than airspeed. The dashed blue curve shows an extreme case, with vW = 50% of vA, such that headwind components are in effect for 210 degrees -- 58% of the full Wind Circle.

s expected, the impact worsens at higher relative Wind Speeds, which can be seen in the graph on the right. The Green Flight in the puzzle flying in any arbitrary direction at with respect to the jet stream (with vA = 228 Kts and vW = 52 Kts) will on average suffer a 3.6-Kt headwind (-1.6%).

Finally, remembering to keep time in the numerator, the sophisticated solver will find that for a round trip between waypoints, the overall average ground speed will range from 223 Kts down to 216 Kts depending on the wind direction aW.