Answers from page 77 — Page 98, Lesson 116

Let’s work through these answers together. This is the worked solution to the lift questions from page 77, and it ties directly into the lift curve on Figure 5.27. First, the setup. We’re in straight and level flight, which means lift exactly equals weight. The aircraft’s weight is given as 588 600 newtons. So the answer to (a) is simply that 588 600 newtons of lift are required — because in straight and level flight, lift must balance weight exactly, no more, no less. Now look at the curve. The vertical axis is the coefficient of lift, CL, and the horizontal axis is angle of attack in degrees. The curve rises with angle of attack until it peaks at CLMAX, then it falls away — that peak is the stall. The highlighted CL values on the curve are 1.532, 0.863, 0.552, and 0.384. Each of those corresponds to a particular airspeed at which the aircraft can generate the required 588 600 newtons of lift. The speeds are 150 kt, 200 kt, 250 kt, and 300 kt respectively. So for (b), the airspeed for each highlighted coefficient of lift is as above — the highest CL of 1.532 corresponds to the slowest speed, 150 kt, and the lowest CL of 0.384 corresponds to the fastest speed, 300 kt. That’s the inverse relationship: high CL means slow flight, low CL means fast flight. For (c), the lowest speed at which the aircraft can be flown in level flight is 150 kt. Why? Because that speed corresponds to CLMAX — the maximum coefficient of lift the wing can produce. You cannot fly slower in level flight, because you’d need more lift than the wing can generate, and the wing would stall. That leads directly to (d). The coefficient of lift that must be used to fly as slowly as possible in level flight is CLMAX. That’s the peak of the curve, the maximum lift coefficient, and it defines the stall boundary — the slowest possible speed. For (e), does each angle of attack require a particular speed? Yes. Because for a given weight, lift depends on both angle of attack and speed. To hold lift constant at 588 600 newtons, a specific angle of attack pairs with a specific speed. Change one, and you must change the other. For (f), as speed is increased, what must be done to the angle of attack to maintain level flight? The angle of attack must be decreased. Think of it this way: lift increases with speed squared, so if you speed up and keep the same angle of attack, lift would grow and you’d climb. To hold lift constant, you reduce the angle of attack, which lowers CL and brings lift back down. For (g), at higher altitude, air density is lower. To maintain the required lift force, you must increase the True Airspeed, or TAS. Because lift depends on air density times speed squared — lower density means you need more speed to produce the same lift. Note it’s True Airspeed, not indicated airspeed, because we’re talking about the actual physical speed through the air. Finally, (h). At a constant altitude, if speed is halved, what must be done to the angle of attack to maintain level flight? The angle of attack must be increased so that CL is four times greater. Why four times? Because lift is proportional to the square of the speed. Halve the speed, and speed squared drops to one quarter. To keep lift constant, you must multiply CL by four to compensate. That’s why the answer says CL must be four times greater. So the whole set hangs together on one idea: in straight and level flight, lift equals weight, and you balance the lift equation by trading speed against coefficient of lift — and the angle of attack is your control over CL, up to the limit of CLMAX at the stall.