# I got this wrong when someone asked me. Can you get it right?

**If you take your plane on a round trip flight, will it be faster if there is no wind, or if there is steady wind from one direction? **Don't worry, you won't find this as one of the hardest questions on the Private Pilot Knowledge __Test__ or __Checkride__, it's just a fun question.

Here's the scenario: You're flying from Jena Airport (1R1) to De Quincy Airpark (5R8) in Louisiana, a distance of exactly 100 NM and a course of roughly 220°. Your average true air speed for the flight will be 100 knots. You're going to land at De Quincy, drop a passenger, and head straight back to Jena.

Now, you have a choice of taking off now, when the winds are predicted to be calm for the full journey, or waiting until tomorrow, when the predicted winds will be a steady 20 knots out of 220°. You're trying to minimize your flight time and fuel burn. Which day do you fly? Does it matter?

I was sitting in a crowded pilot's lounge on a rainy day when someone asked us this same question. I thought about it for a second or two, and said it didn't matter whether you flew with no wind or the steady wind. Since it's a round trip, doesn't the headwind get cancelled out by the tailwind on the way back?

Was I right? Turns out I was dead wrong. And I had to do the math to believe it. Here it is:

**With no wind**, the calculations are easy. A 100 NM flight at an average true airspeed of 100 knots will take exactly one hour (easy, right?) to go the first leg, and the same hour coming back. A total flight time of **2 hours.**

**With steady wind of 20 knots from 220° **though, the trip out to De Quincy will be at a ground speed of 100 KTAS minus 20 knots wind = 80 knots ground speed. A glance at my __E6-B__ tells me that a 100 NM trip at 80 knots ground speed will take me 1 hour and 15 minutes. For the return to Jena, my ground speed will be 100 KTAS + 20 knots wind = 120. The return flight will take me 50 minutes. So adding the two legs gives me 1 hour and 15 minutes + 50 minutes = **2 hours and 5 minutes. **I'll lose time flying with this wind versus flying on the calm day!

How can this be?! I was still sure that the headwind and tailwind should cancel out, yet I couldn't argue with the math. I thought of an even more extreme example:

What if instead of a 20 knot wind, it was a 100 knot wind?

Now on the flight out, my ground speed will be 100 KTAS minus 100 knots wind equals....*zero?* I'll be hovering over the ground! Unless I speed up or find calmer air, I'll never reach De Quincy!

It doesn't matter that the flight back would take just 30 minutes (100 NM divided by 200 knots ground speed = 30 minutes). Adding 30 minutes to my infinity ETE on the way out still gives me infinity!

So instead of thinking of the headwind and tailwind as cancelling each other out, what is another way to think about the effects of wind on a round trip flight?

It turns out that because you're exposed to the headwind on the way out for a longer period of time than you experience the tailwind on the return, your* initial penalty* is greater than your *later bonus*. The ideal scenario is no wind, and then as the wind gets stronger out of a constant direction, your overall time enroute gets longer.

Why is this important? Because if you're like I was, you could incorrectly assume that the amount of fuel you've planned for a no wind round trip flight will be just as good on a flight with the winds we've talked about. Either that, or someone in your pilot lounge can make a bet and take you for a quick ten bucks!

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