Is there a formula to calculate ground distance traveled given rate of climb and true airspeed? 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



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.
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).

The green arc is that green banana (altitude range arc).