# Why does increasing speed in a turn not change load factor?

From what I've read, load factor (n) is the forces of lift (F) divided by weight (W), so the equation would be n=F/W. In a question about a constant bank angle turn, an increase in airspeed will decrease the rate of turn, which I understand, but will not affect the load factor. If airspeed increases, doesn't lift increase, which should increase* the load factor? I'm missing something here about the calculation of the forces of lift, but I'm not sure what.

*edit

• When you increase airspeed but change nothing else, lift will increase. But if you apply more downward pitch to stay at the same altitude, your lift stays the same. Maybe whatever question you refer to assumes no change in altitude. Jan 16 at 3:49
• If the weight of the aircraft is unchanged, why wouldn't the lift be? Increasing the speed increases the dynamic pressure, but you must reduce the $C_l$ proportionally so that the lift remains unchanged. Jan 16 at 4:33
• Is the flight path constrained to remain horizontal? (See my answer for context) Jan 16 at 21:09

It's just like a plane in straight and level flight. If you increase speed by increasing thrust, Angle of Attack must be reduced to maintain level flight (not climb).

Same thing for 30 degree bank angle turn.

Radius of turn is V$$^2$$ knots/11.26×tan bank angle in feet

For level flight vertical lift component = Weight. Load factor at 30 degrees bank is 1/cosine 30 degrees = 1.15 G.

The $$cosine$$ of the lift component is the vertical lift component. The $$sine$$ of the lift component is the centripetal force into the turn.

Since these do not change, load factor is the same, and the rate of turn decreases.

Rate of turn is 1091×tangent bank angle/V knots in degrees/second

You need to think about what are the constraints and what are the variables. If bank angle is a "given" (i.e. constrained) but turn radius or rate are not, and the airplane is constrained to remain in horizontal flight (not climbing, and especially not "pulling G's" to curve the flight path ever-more-steeply-upward in an approximation of the start of a loop), well, then for any given bank angle, why should increasing the airspeed increase the load factor?

Keep in mind that for any given bank angle, every extra knot of airspeed mandates a decrease in the angle-of-attack of the wing in order that the flight path remain horizontal rather than starting to curve upwards. So why should the load factor increase?

At the end of the day, if the only "givens" are bank angle and the fact that the flight path must remain horizontal, then the flight dynamics don't "care" about airspeed at all as far as load factor is concerned. The load factor will always equal (1/ (cosine (bank angle))).

On the other hand, if turn radius or turn rate is included as one of the constraints, and bank angle is not, then everything changes: an increase in speed will mandate an increase in bank angle and load factor.

If airspeed increases, doesn't lift increase, which should increase the load factor? I'm missing something here about the calculation of the forces of lift, but I'm not sure what.

You are missing that the angle-of-attack is not constrained to be constant, but the flight path is constrained to be horizontal, and that's why increasing the airspeed doesn't increase the lift force.

P.S. your question didn't actually state that the flight path was constrained to remain horizontal. But in the absence of information to the contrary, that's pretty much presumed (though it really should be stated explicitly). If the airplane is allowed to start entering a loop as soon as we increase the power, then situation becomes immensely more complex, and since we aren't given any explicit constraints on angle-of-attack (e.g. a-o-a is presumed to remain constant?), we're basically left with an unsolvable (insufficiently constrained) problem.