The question is often asked: Can an airfoil fly upside-down? To answer this, make the following calculation. Consider a positively cambered airfoil with a zero-lift angle of −3◦. The lift slope is 0.1 per degree.
(a) Calculate the lift coefficient at an angle of attack of 5◦. (b) Now imagine the same airfoil turned upside-down, but at the same 5◦ angle of
attack as part (a). Calculate its lift coefficient. (c) At what angle of attack must the upside-down airfoil be set to generate the same lift as that when it is right-side-up at a 5◦ angle of attack?
The Correct Answer and Explanation is:
To answer the question “Can an airfoil fly upside-down?”, we need to explore how lift coefficient changes with angle of attack and how camber affects this behavior.
Given Data:
- Zero-lift angle of attack (α₀): −3°
- Lift slope (a): 0.1 per degree
The lift coefficient CLC_L is calculated using the linear lift slope equation: CL=a(α−α0)C_L = a(\alpha – \alpha_0)
(a) Lift coefficient at 5° (right-side-up)
CL=0.1×(5−(−3))=0.1×8=0.8C_L = 0.1 \times (5 – (-3)) = 0.1 \times 8 = 0.8
✅ Answer (a): CL=0.8C_L = 0.8
(b) Lift coefficient when turned upside-down, still at 5°
When the airfoil is flipped upside-down, the shape is inverted, and so is the camber. The new zero-lift angle becomes +3°, the mirror of −3°. CL=0.1×(5−3)=0.1×2=0.2C_L = 0.1 \times (5 – 3) = 0.1 \times 2 = 0.2
✅ Answer (b): CL=0.2C_L = 0.2
(c) What angle of attack gives same lift (0.8) upside-down?
We now want CL=0.8C_L = 0.8 for the inverted airfoil (zero-lift angle is +3°).
Using: 0.8=0.1(α−3)⇒α−3=8⇒α=11°0.8 = 0.1(\alpha – 3) \Rightarrow \alpha – 3 = 8 \Rightarrow \alpha = 11°
✅ Answer (c): α=11∘\alpha = 11^\circ
✈️ Explanation (~300 words)
An airfoil generates lift largely due to its shape and angle of attack. A cambered airfoil, which is curved more on the top than bottom, can generate lift even when its angle of attack is zero or slightly negative. The angle at which it produces no lift is called the zero-lift angle of attack, which is given as −3° in this case.
In part (a), we calculated the lift coefficient (CLC_L) at 5° angle of attack. Using the formula CL=a(α−α0)C_L = a(\alpha – \alpha_0), we find that a positively cambered airfoil at 5° produces a substantial lift coefficient of 0.8.
In part (b), the airfoil is flipped upside-down but kept at the same angle of attack. Flipping it inverts the camber, making the zero-lift angle +3°. Now, at 5°, the airfoil only sees a 2° effective angle of attack (5 − 3), yielding a much smaller lift coefficient of 0.2.
In part (c), we determine the angle of attack needed for the inverted airfoil to match the original 0.8 lift coefficient. We solve for the angle where the effective angle of attack is 8° (since 0.8=0.1×80.8 = 0.1 \times 8), which requires the inverted airfoil to be set at 11°.
Thus, an airfoil can fly upside-down, but to generate the same lift, it must fly at a higher angle of attack to compensate for the reversed camber. This principle is used in aerobatic aircraft and jet fighters, which often fly inverted.