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I'm looking for a formula to calculate the horizontal distance (guess it is the Ground Distance) passed during the phase of ascent (or descent), having the rate of climb in ft/min and the TAS in knots. I did not found a clear answer about this question. I think it may concern with the Climb Gradient.

Refering to the image below, I'm trying to calculate d, having:

  • h = 11200 ft
  • RoC = 1250 ft/min
  • TAS = 75 kn

Rectangle

ClimbGradient = (RoC / TAS) * 6000 / 6076,12 = (1250 / 75) * 0.9874 = 16.45%

d = h / (ClimbGradient * 100) = 11200 / 1645.78711 = 6.80525 nm

Is this correct calculation? I will update this question with the correct formula in other case.

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    $\begingroup$ Air speed does not relate to ground speed so you can't determine the distance covered over the ground. Think about if the aircraft has a very strong headwind, it will be travelling slower over the ground vs an aircraft with a strong tailwind. The rate of descent gives a time, and true air speed gives speed through the air, but not over the ground. $\endgroup$ – Ron Beyer May 4 '17 at 14:18
  • $\begingroup$ @Ron Beyer is correct, but another significant factor is that winds vary through altitudes. Although they are generally higher at higher altitudes, the wind direction will normally vary also. Therefore, for a reliable figure, you need winds aloft profiles through the climb corridor. $\endgroup$ – mongo May 4 '17 at 14:35
  • $\begingroup$ @Yohji please specify if the calculation takes place at zero wind. $\endgroup$ – Koyovis May 4 '17 at 19:05
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    $\begingroup$ @Koyovis the discussed formula take place in still air, and it would be an approximation by this way. $\endgroup$ – Yohji May 5 '17 at 13:47
  • $\begingroup$ The first bit of your computation is correct: 16.45% = 0.1645. Then if you would divide 11,200 feet by 0.1645 you would get pretty close to 11 nm. $\endgroup$ – Koyovis May 5 '17 at 14:01
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You need ground speed not TAS. GS is affected by the head- or tail-wind components.

Assuming in your example GS is 75 knots, then just check how long it takes to change altitude.

11,200 ft at 1,250 ft/min takes 8.96 minutes, or 0.1493 hours.

At a speed of 75 knots, that's 11.2 NM.

$$\frac{\text{altitude change (ft)}}{\text{rate of climb (ft/min)}\cdot60}\cdot\text{ground speed (knots)}$$


Since the density of air changes during a climb/descent, it affects the rate of climb/descent. Fancy planes have green bananas or similar indications on the navigation display to show where you'll reach the target altitude (it updates in real-time).

enter image description here
The green arc is that green banana (altitude range arc).

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Pythagoras has your answer.

You've been climbing for 11200/1250 = 8.96 minutes

In that time you've traveled 75 * 8.96/60 = 11.2 nautical miles = 68,051 ft along the diagonal D

$D^2$ = $d^2$ + $h^2$ => d = 67,123 ft = 11.05 nm

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  • $\begingroup$ You've travelled that distance through the air not over the ground. $\endgroup$ – Ron Beyer May 4 '17 at 14:52
  • $\begingroup$ You've travelled 11.2 nm through the air. $\endgroup$ – Koyovis May 4 '17 at 18:41
  • $\begingroup$ Right, but what if I told you that if you had a strong enough headwind, you could make that climb in 2 miles of ground distance? The question asks about the distance over the ground traveled in the climb, not the distance in the air. $\endgroup$ – Ron Beyer May 4 '17 at 20:25
  • $\begingroup$ I'm assuming the question would be at zero wind. It would lack an input otherwise. $\endgroup$ – Koyovis May 4 '17 at 20:29
  • $\begingroup$ I'm was thinking too about the Pythagorean theorem. The question is about the distance travelled over the ground (the d in the picture), as if the plane for absurdity never takeoff. $\endgroup$ – Yohji May 5 '17 at 13:54

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