Is it possible that a shortcut will take LONGER to fly?

My answer was shorter and no less accurate. :)

I think what confuses people is the false, but intuitive fact that flying directly downwind will make up for the lost time of flying directly upwind the same distance. ie flying from A to B in headwind, then B to A downwind. This is of course a closed loop, and it will always be faster with no wind.

What matters here is the distance you have to fly through the air masses. Let’s say the wind moves perpendicular to the line A-B. Flying A-B-A with no wind correction will bring you downwind a distance exactly equal the wind velocity divided by the time you have used, point C. You then have to fly upwind that distance to get to A again, meaning you not only have to fly an extra distance (C-A), but you also have do it at a slower speed over ground.That higher speed over ground flying A-B-C won’t help you one bit.

Peter wrote:

But I have no idea how you would prove it.

With some equations, of course :-)

A simple way of doing it would be: let x be windspeed: f(x) is the function of time to fly the course with some wind speed, and TAS < x < 0. Take a constant C, which is the time taken to fly the course with zero wind. Then prove f(x)-C > 0 for any legal value of x and show the limit of f(x)-C is zero as you approach zero from the right, and you’re done.

Here’s a graph:

https://www.desmos.com/calculator/1fkmqyageq

Without the effort of going to a mathematical proof the graph shows whatever you do with d (wind direction) you always get an increasing function in the range t < x < 0 (the function h(x) here is for a A-B-C-A course as described in Peter’s first post, with A-B = 100 and B-C = 100. t is the true airspeed. The red line in the graph is how much longer in minutes it takes to fly the course with wind versus flying the course with zero wind.

bookworm wrote:

Is there a steady wind condition in which the closed course A → B → C → A is faster than with no wind at all?

No. With a closed course, you’ll always have tailwind on a part, and headwind on another part. Because the time taken has the speed in the denominator of the fraction, headwind (removing speed) has a bigger effect than tailwind (adding speed), because you spend more time in the headwind than in the tailwind. This is the same kind of effect than with Ahm’s law, that programmers may be familiar with.

In other words (v being the speed and w the wind):
1/(v-w)-1/v > 1/v-1/(v+w)
or to make it more readable, using absolute values:
|1/v – 1/(v-w)| > |1/v-1/(v+w)|

This can be seen from the convexity of the function (lambda w.1/(v+w)), that is the fact that the second derivative in w of the expression 1/(v+w), namely 2/(v+w)^3, is positive, while the first derivative is negative. Or equivalently, the third term of the Taylor series development:
1/(v+w) = 1/v – w/v^2 + w^2/v^3 + O(w^4)
more schematically:
the time taken at speed v+w is the time taken at speed v minus a value proportional to the wind plus a time proportional the square of the wind plus a smaller term we will consider negligible. So looking at the time in headwind and in crosswind, 1/(v+w) + 1/(v-w), the second terms will cancel out, but the third term will come twice positive.

The above explains it for a straight path A to B and B to A direct.

Without loss of generality, consider that you do, projected on the direction of the wind, only one round trip. (If you don’t, just consider every round trip separately.)
Consider A your starting point and B the furthest you go in the direction (or against) of the wind (the projection of your path on the line having the wind as direction and A as a point), and d the distance from A to the projection of B. Since whatever path you take from A to B to A (any intermediary points you put between A and B and between B and A), you will always need to go at least distance d against the wind and distance d with the wind, you will always have the above effect of having the w^2 term twice.

As usual, if you “intuitively” believe that the effect of w is monotonic (that is, a bigger wind will only have a bigger effect, not a smaller one), you can see it by going to the limit: as w approaches v, the time in headwind becomes infinite, but the time in tailwind only halves.

Peter wrote:

It depends on the wind direction. But obviously you knew that.

On a closed course it doesn’t.

(Except if of course the course is around the globe as kwlf state before)

bookworm wrote:

Here’a a related one. Is there a steady wind condition in which the closed course A → B → C → A is faster than with no wind at all?

No, over a closed course no wind always results in a shorter time than with any steady wind from any direction. For the same reason actually, with wind, your distance travelled through the air is longer than with no wind.

Other than some sort of cyclone? Intuitively, no (for A → B → A def not)