# Is the “effective wing span” reduced in a turn?

The Ace the technical pilot interview book says that in a bank the aircraft's lift is reduced due to an effective reduction in wing span. I know in a bank the vertical component of the lift vector is reduced due to the inclination of another component of lift towards the banked side. So the overall lift required to counter the weight is increased. That's why we need to increase the AOA or speed to maintain a level turn. But I am a bit confused about how the effective wing span is reduced, can anyone explain?

"reduced effective wing span" is just an analogy to help some people understand the concept of lift in a turn. The Total Lift is not reduced, it is just applied in a different vector in relation to gravity. The vertical lift component (in relation to gravity) is reduced and you can then imagine that the effective wingspan is also reduced by the same amount.

Banked turn (from Wikipedia)

That's kind of a strange way to view the problem of decreased vertical lift in a turn, but essentially it's true. If you look at an airplane in a bank of angle theta, and imagine the tip-to-tip span of the wings as the hypotenuse of a right triangle, then the base of the triangle, which is the length of the hypotenuse multiplied by the cosine of theta will be shorter than the span is.

But that is an odd way to consider that problem.

Well, in a certain frame that makes sense. If you look at the airplane from the ground, in a bank, you will see a wing span of

$$\text{Wing Span} \times cos(\theta)$$

This is equivalent to the vector force that is applied in the vertical direction*

$$\text{Lift} \times cos(\theta)$$

This is not the same thing, but they are sort of pseudo-related quantities where all of the factors that affect each quantity align very conveniently.

*As an aside, this is where you get those tables for G-load in a turn, which ends up being $${1 \over cos(\theta)}$$