Can an airfoil fly upside-down ?
A positively cambered airfoil has a zero-lift angle of \(-3.5^{\circ}\). The lift slope for this airfoil is \(0.12\) per degree. Find the lift coefficient for this airfoil at an angle of attack of \(5^{\circ}\). If this airfoil is turned upside-down to the same \(5^{\circ}\) angle of attack, what is its lift coefficient. What will be the angle of attack to generate the same lift as when the airfoil is right-side-up at \(5^{\circ}\) angle of attack.
From the figure, Lift slope with respect to zero lift angle of attack, \( a = \frac{dcl}{d\alpha }\)\[\Rightarrow a = \frac{c_{l} – c_{l,0}}{\alpha – \alpha _{L=0}}\]\[\Rightarrow a = \frac{c_{l} – 0}{\alpha – \alpha _{L=0}}\]\[\Rightarrow a = \frac{c_{l}}{\alpha – \alpha _{L=0}}\]\[c_{l} = a\left ( \alpha – \alpha _{L=0} \right )\]Therefore, at an angle of attack of \(5^{\circ}\), when the airfoil is right-side-up, coefficient of lift is\[c_{l} = 0.1\left \{ 5^{\circ} – \left ( -3.5^{\circ} \right ) \right \}\]\[=0.1\left \{ 8.5^{\circ} \right \}\]\[=0.85\]When, the airfoil is turned upside-down, its zero-lift angle of attack is \(3.5^{\circ}\). Therefore, coefficient of lift in this condition at \(5^{\circ}\) angle of attack is, \[c_{l} = 0.1\left ( 5^{\circ} – 3.5^{\circ} \right )\] \[ = 0.1\left ( 1.5^{\circ} \right )\]\[= 0.15\]At this condition, to generate the same amount of lift as when the airfoil was right-side-up at \(5^{\circ}\) angle of attack, can be found as\[c_{l} = a\left ( \alpha – \alpha _{L = 0} \right )\]\[\Rightarrow 0.85 = 0.1\left ( \alpha – 3.5^{\circ} \right )\]\[\Rightarrow \alpha = 12^{\circ}\]Therefore, an airfoil can fly upside-down. An airplane can also make an inverted flight, provided its angle of attack is more, in order to generate the same amount lift. In order to increase the angle of attack, an airplane’s nose can be pointed more skyward during the inverted flight.
