Figure 5.4 Typical lift curve — Page 83, Lesson 90

Let’s start with Figure 5.4, the typical lift curve. This is the graph that ties together angle of attack, coefficient of lift, and airspeed — the three things that decide whether an aircraft stays in the air. On the vertical axis we have CL, the coefficient of lift. That’s a dimensionless number that tells us how efficiently the wing turns dynamic pressure into lift. On the horizontal axis we have angle of attack in degrees. The curve rises steadily as angle of attack increases, then peaks, then falls off sharply. That peak is the stall — the point where the airflow separates and lift collapses. The curve is labelled with three highlighted values of CL: 0.384, 0.552, 0.863, and at the very top, CLMAX = 1.532. CLMAX is the maximum coefficient of lift the wing can produce, and it occurs right at the stall angle of attack. Beyond that, the curve drops — that’s the stalled region. Now, the questions on this page are practice problems, and I want you to work through them with me. Let me walk you through what each one is really asking. a. How many newtons of lift are required for straight and level flight? In straight and level flight, lift exactly equals weight. So the answer is the aircraft’s weight in newtons — the lift force must balance the weight precisely, no more, no less. b. Calculate the airspeed in knots for each highlighted coefficient of lift. This uses the lift equation: L = ½ ρ V² S CL. If you know the weight (which equals lift), the air density ρ, and the wing area S, you can rearrange to solve for velocity V for each CL value. The lower the CL, the higher the speed needed to generate the same lift. c. What is the lowest speed at which the aircraft can be flown in level flight? That’s the stall speed — the speed at CLMAX. Since CLMAX is the maximum lift coefficient, it gives the minimum speed for level flight. Any slower and the wing cannot produce enough lift. d. What coefficient of lift must be used to fly as slowly as possible? That’s CLMAX = 1.532, because maximum CL gives minimum speed. e. Does each angle of attack require a particular speed? Yes — for a given weight and altitude, each angle of attack corresponds to one specific CL, and therefore one specific airspeed. Change the angle of attack, and you must change speed to maintain level flight. f. As speed is increased, what must be done to the angle of attack to maintain level flight? You must reduce the angle of attack. Higher speed means more lift at the same CL, so to keep lift equal to weight, you lower the angle of attack to reduce CL. g. At higher altitude, air density is lower. If angle of attack is kept constant, what must be done to maintain the required lift force? You must increase the airspeed. Since lift depends on density times velocity squared, lower density means you need higher speed to produce the same lift at the same CL. h. At constant altitude, if speed is halved, what must be done to the angle of attack? You must increase the angle of attack. Halving speed quarters the dynamic pressure, so you need a much higher CL — which means a higher angle of attack — to keep lift equal to weight. Now let’s move to Figure 5.5, which compares different aerofoil sections. We have three curves: a symmetrical section with 6% thickness, a symmetrical section with 12% thickness, and a cambered section with 12% thickness. The vertical axis is section lift coefficient, the horizontal axis is section angle of attack in degrees. The key labels tell the story. Camber gives an increase in CLMAX — the cambered section peaks higher than both symmetrical ones. And greater thickness gives a 70% increase in CLMAX — going from 6% to 12% thickness on the symmetrical section raises its maximum lift coefficient by 70%. Now the questions: a. Why does the cambered aerofoil section have a significantly higher CLMAX? Because camber — the curvature of the mean camber line — generates lift even at zero angle of attack, and it shifts the whole lift curve upward. That gives a higher peak. b. For the same angle of attack, why do the symmetrical sections generate less lift than the cambered one? A symmetrical section has no camber, so at zero angle of attack it produces zero lift. The cambered section produces positive lift at zero angle of attack, so at any given angle of attack, its CL is higher. c. Why does the cambered 12% section generate a small amount of lift at slightly negative angles of attack? Because camber alone produces lift even when the angle of attack is negative. The curvature of the section creates a pressure difference that generates lift without needing a positive angle of attack. d. The symmetrical 6% section generates the smallest lift for a given angle of attack. In what way is that favourable? It means the aerofoil is very predictable and has no built-in pitching moment from camber. That makes it easier to control — especially useful for aerobatic or high-manoeuvrability aircraft where you want symmetric behaviour in upright and inverted flight. e. What are the disadvantages of the symmetrical 6% section? It has the lowest CLMAX, so it produces the least lift at any angle of attack. That means higher stall speeds and the need for higher speeds to generate the same lift — which is a performance penalty. So the big picture: camber raises the lift curve, thickness raises the peak, and symmetry gives control predictability at the cost of lift. These are the trade-offs you balance when choosing an aerofoil. Now, these are the book’s practice questions — let’s try them one at a time.