= Aerodynamic load factor
{title2=$n=L/W$}
The \b[aerodynamic load factor] $n=L/W$ compares <lift> with <gravitational weight>. The stated convention takes $L$ along the chosen positive lift direction, so $n$ is a signed force ratio, not simply the magnitude of total <acceleration> divided by $g$. A steady level condition can have $n=1$ with zero inertial <acceleration>.<Introduction to Aircraft Performance and Static Stability by Earll Murman>[PDF pp. 14–15, Horizontal Turn and Pull Up/Pull Over/Vertical]{ref}
A focused example is a wings-level pull-up at the bottom of a vertical curved path. Neglect other vertical <forces> and take upward normal acceleration as positive. The <force balance> gives
$$
L-W=\frac{mV^2}{R},\qquad
\boxed{n=1+\frac{V^2}{gR}.}
$$
For stipulated <true airspeed> $V=50\,\mathrm{m/s}$, curvature radius $R=500\,\mathrm m$ and $g=9.81\,\mathrm{m/s^2}$, the upward <centripetal acceleration> is $5\,\mathrm{m/s^2}$ and \b[$n\simeq1.510$]. With $m=600\,\mathrm{kg}$, <lift> is $8886\,\mathrm N$, while the net upward force is $3000\,\mathrm N$. Neither is the other.
In the steady horizontal-turn model, the condition instead is $n=1/\cos\phi$; <bank angle> derives that balance. It does not apply unchanged to the vertical pull-up. At fixed air <mass density>, <wing area> and usable maximum <lift coefficient>, the relation $L=\rho V^2SC_L/2$ gives limiting speed proportional to $\sqrt n$ for positive $n$. For this example, the factor is about $1.229$ relative to the corresponding one-load condition.
A force balance measures aerodynamic load; a trajectory measurement estimates inertial <acceleration>. Comparing them requires the <gravitational weight> term and correct directions. A reported load factor also needs its force-axis convention: thrust, side forces and coupled motion can matter to what an instrument actually records. The definition alone establishes neither structural capability nor available <lift> before <aerodynamic stall>.
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