It's an equation to get an approximate value of the total ground take-off distance (including the distance after reaching Vr when it begins to rotate, until all wheels stop touching the ground). This is the equation that is on some lecture notes I found on the Internet:

The subindex "máx." and "referencia" are in spanish, and refer to "maximum" and "reference" respectively

It also specifies that in this equation, drag (D) and Lift (L) are not calculated with the maxmimum lift coefficient with surfaces in take-off configuration, but with a "typical lift coefficient" of 0.1. Drag (D) is calculated through the drag polar.


1 Answer 1


I'm not sure where you got the equation from, but it can relatively easily be deduced from the following, so really anybody with basic aerospace knowledge could have come up with it.

Acceleration: $$ a= \frac{T-D}{m}$$

Time: $$ t = \frac{V_{LOF}}{a}$$

Distance: $$ d = \frac{1}{2}at^2 = \frac{V_{LOF}^2 m}{2(T-D)}$$

And then, for some reason, the original author decided to substitute

$$ m = \frac{W}{g} $$

and $$L = W = \frac{\rho}{2} V_{LOF}^2 S C_{L_{LOF}}$$

at the point of liftoff (weight equals lift).

I simplified $(T-D)$ a bit, but it's a constant anyway, as shown in your equation, evaluated at $0.7V_{LOF}$ in take-off configuration.

The factor $1.44$ comes from $C_{L_{max}}$ being used in the denominator which is $1.2^2$ higher than $C_{L_{TO}}$ because take-off speed $V_{LOF}$ is $1.2$ higher than stall speed in take-off configuration.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Charles
    May 12, 2022 at 17:17

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