How can I calculate the distance on the ground during the climb?

Assuming the climb power is 700ft/min, the climb speed is 65 kts, the wind 260/10kts on runway 26. Start: 700ft Top of climb: 3500ft

How far is the distance flown on the ground (in e.g. ft) and the GS?

What I understood: The difference in altitude is 2800ft and therefore 2800ft / 700ft min = 4 minutes. I can say that I fly 4 minutes long 65 knots: 65kts -> 6582 ft * 4 = 26 328 ft but I don't use the wind in this calculation. Can someone please help me?


  • 1
    $\begingroup$ How much did the air mass (wind) travel in four minutes? $\endgroup$
    – Sanchises
    Commented Dec 2, 2019 at 19:31

1 Answer 1


That 65kts "climb speed" -- is that ground speed or air speed?

If it's ground speed, then you already have your answer. Since the question is about ground speed and distance travelled, then the wind doesn't factor in to it.

If it's air speed (which is far, far more likely), then you need to figure out your ground speed first. This is where the wind comes in. Now, you didn't specify which direction you're flying, but we can assume from the mention of "runway 26" that you're flying 260 degrees. That means you're flying into a direct headwind, which means in turn that your ground speed is simply your airspeed minus the wind speed. Then use that value in your distance calculation.

  • $\begingroup$ Thanks for your help! Correct, 65 knots is airspeed. But I don‘t understand why the GS is just airspeed minus wind? If I climb with an angle of (almost) 89 degrees (or 100.000ft/min) than the GS can never be 65-10kts because it is something else as to climb with an angle of 1 degree (or about 10ft/min) isn‘t it? $\endgroup$
    – epus
    Commented Dec 3, 2019 at 5:37
  • $\begingroup$ 65kts with a 700ft/min climb gives a climb angle of 6.7 degrees, so that the horizontal component of the speed is 65*cos(6.7)=64.6 so you typically neglect those 0.4 knots. Other errors as exact wind speed will be bigger. $\endgroup$
    – iamc
    Commented Dec 3, 2019 at 9:04
  • $\begingroup$ Thanks, good argument :) That's it! $\endgroup$
    – epus
    Commented Dec 3, 2019 at 9:47

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