Wind doesn’t “cancel itself out” — here comes the math

I recently had an interesting conversation with a CFI. We were discussing trip planning, specifically a hypothetical medium-distance out-and-back flight. Ever the teacher, he mischievously asked, “What about winds? If you had a significant headwind heading out, would it matter much with respect to your trip time?” Without giving it much thought, I replied that it would not — that the headwind heading east would become a tailwind heading west, and that the effects would therefore be canceled.

I was wrong.

Let’s illustrate using a simple example: a 100 knot plane traveling 50 miles east, then 50 miles west. In a no-wind situation our groundspeed would equal our airspeed, so the trip would take 30 minutes out, then 30 minutes back for a total of one hour. That’s straightforward.

Adding wind to the scenario, let’s get extreme: say there was a 50 knot wind from the east. Groundspeed heading east would be 100 – 50, or 50 knots. This is your first clue that the winds wouldn’t cancel out, for at 50 knots groundspeed, the eastbound leg by itself would take an hour. That’s the same as the entire no-wind trip, and we would still have the return leg to fly! (The return leg would take 50/150, or 0.33 hours, for a total trip time of 1.33 hours.)

That is, of course, an unlikely scenario and is purely to illustrate the point. What about the real world? What does the picture look like for typical GA piston-engine aircraft? The situation is fortunately not as dire. Here’s a chart for a 120 knot aircraft covering a 200 mile round trip, accompanied by the raw data:

As you can see, for more realistic wind values (25 knots or less) the total time penalty doesn’t exceed 5%. And note that if you change the 200 mile trip to a 2,000 mile trip, the percentages do not change.

If you change airspeed, the percentages do change dramatically. If you’re lucky enough to cruise at 160 knots, for example, you can endure a 35 knot wind with only a 5% total time penalty:

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