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My question is about aircraft load factor and how it relates to turning ability.

I've recently come across this chart for the F-16 that shows the turn rate in degrees per second and the load factor when doing a sustained turn at different speeds. I've always knew the load factor was the sum of the forces that act perpendicular to the aircraft (apart from weight) divided by the weight of the aircraft.
If load factor (when speed is constant) is the only thing related to the centripetal force, then the centripetal acceleration in Gs is equal to the horizontal component of the load factor.

This means that in a sustained turn (where altitude does not change so the vertical component of lift is equal to weight) that horizontal component will be, using pythagorean's theorem, : $ F_c = g \sqrt{(n_z)^2-1}$ , where $n_z$ is the load factor and g is 9.81m/s^2 .

From this we can derive that angular velocity in deg/sec is equal to: $ ω = \dfrac{g \sqrt{(n_z)^2-1}}{V} \dfrac{180}{π}$ .

When looking at the F-16 chart (Ps = 0 line is the one to look at for constant speed) thought, this results do not match.

Using this formula, for example, at mach 0.3 (~97.22 m/s) the F-16 does about 2.95G of load factor, which results in 16.05 deg/sec of turn rate, but the chart shows a little over 17. In general the lower the speed (so higher the angle of attack) the bigger the difference between the formula and the chart.

Any thoughts? Maybe the Load factor in the chart is instead the normal Load factor $ N_z $ instead of $ n_z $?

F-16 e-m diagram

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  • $\begingroup$ The altitude is sea level, it is written in the picture $\endgroup$
    – Giovanex
    Commented Aug 7 at 18:01
  • $\begingroup$ Might be the extra engine thrust at higher angle of attack, the lift vector is smaller. $\endgroup$
    – Koyovis
    Commented Aug 8 at 1:18
  • $\begingroup$ Isn't the sin(AoA) component of thrust already included in the load factor thought? $\endgroup$
    – Giovanex
    Commented Aug 8 at 12:17
  • $\begingroup$ @Giovanex For a standard day the mach number at sea level is $340.3 \frac{m}{s}$, therefore Mach 0.3 is equal to $0.3\cdot 340.3 \frac{m}{s}=102.09$. However, I actually think your formula is wrong... $\endgroup$
    – U_flow
    Commented Aug 8 at 12:34
  • $\begingroup$ But it am not 100% sure where... $\endgroup$
    – U_flow
    Commented Aug 8 at 17:52

1 Answer 1

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The F-16 uses normal load factor (Nz) in the flight control system, and also on HUD to be read by the pilot. When you watch closely to the 1g line of this chart, you can notice that it even got positive turn rate.

Normal load factor Nz is the sum of forces (apart from weight) that act perpendicular to the aircraft body X axis (Xbody), divided by the weight of the aircraft. Those forces are called normal force, and can be calculated by (Total Drag * sinα + Total Lift * cosα). See equation IV-34 below.

This is not to be confused with load factor $n$ which is the sum of forces (apart from weight) that act perpendicular to the relative wind, divided by the weight of the aircraft.

F16 axis system Nz calculation

Images are from DTIC ADA 189675.

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  • $\begingroup$ Thank you very much! Did a quick calculation and, in a steady turn n_z ends up being N_z/cos(AoA) , and it matches perfectly the results of the graph $\endgroup$
    – Giovanex
    Commented Aug 11 at 14:10
  • $\begingroup$ @Giovanex If this answers your question well, please consider accepting this answer. $\endgroup$
    – Ralph J ♦
    Commented Aug 11 at 16:02
  • $\begingroup$ @Giovanex It's not as simple as dividing by cos(AoA). From the equations, load factor = (Total Lift + Thrust * sinα) / Weight, while Nz = (Total Lift * cosα + Total Drag * sinα) / Weight. So they’re not directly related. $\endgroup$
    – LJQCN101
    Commented Aug 13 at 17:20
  • $\begingroup$ @LJQCN101 in a sustained turn at constant speed and altitude: thrust cos(AoA) = drag . By writing drag as Thrust cos(AoA) in the Nz definition you get: Nz = (Lift cos(AoA) + Thrust cos(AoA) sin(AoA) . One can rewrite this as: Nz = (Lift + thrust sin(AoA) ) cos(AoA) and from there Nz = nz cos(AoA) $\endgroup$
    – Giovanex
    Commented Aug 17 at 21:18
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    $\begingroup$ @Giovanex Yes you're correct, it applies when Ps=0. $\endgroup$
    – LJQCN101
    Commented Aug 18 at 12:39

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