Stalling — Page 179, Lesson 209

Let’s pick up with the bank-angle effect on stall speed, because that’s the first thing in front of us. As you increase bank angle, the stall speed rises — and it rises at an increasing rate. That means the relationship isn’t linear; each extra degree of bank costs you more than the last one. The reason is that in a banked turn, the wing has to produce more lift to support the aircraft’s weight plus the centrifugal force of the turn. That extra lift demand pushes you closer to the stall. Now, because of this, when you’re operating at high lift coefficient — and that’s exactly what happens during take-off and landing — you should only use moderate bank angles to manoeuvre. For a modern high-speed jet transport, the absolute maximum bank angle you should use in service is 30°, and that excludes emergency manoeuvres. The normal maximum is 25°. But at higher altitude, the normal maximum drops to 10° to 15°. So altitude itself restricts your bank angle, because the air is thinner and the margins are tighter. Let me give you the formula that governs all of this. The stall speed in a turn, which we write as VSt, equals the 1g stall speed, VS1g, divided by the square root of the cosine of the bank angle. In symbols: VSt = VS1g divided by the square root of (1 over cos of the bank angle). Wait — let me be precise. The formula is VSt = VS1g × the square root of (1 / cos φ), where φ is the bank angle. So the stall speed in the turn is the 1g stall speed multiplied by the square root of the reciprocal of the cosine of the bank angle. Let’s work the numbers. Suppose the 1g stall speed is 150 knots. In a 25° bank turn, you’d calculate the stall speed using that formula. In a 30° bank turn, you’d do the same. The answers are on page 191, but the point is you can see how the stall speed climbs as the bank increases. Now here’s a practical twist. Suppose you know the stall speed in a 15° bank turn is 153 knots CAS — that’s calibrated airspeed — and you need the stall speed in a 45° bank turn. You can’t just scale directly, because you don’t know the 1g stall speed. So you have to work backwards first. You take the formula VSt = VS1g × the square root of (1 / cos 15°), and you transpose it to solve for VS1g. That gives you VS1g = VSt divided by the square root of (1 / cos 15°). Plug in 153 knots and the square root of 1.02, and you get VS1g = 150 knots CAS. So the 1g stall speed is 150 knots. Once you have that, you can compute the stall speed in the 45° bank turn using the same formula. Now let’s move to the effect of high-lift devices on stall speed. Modern high-speed jet transports have swept wings with relatively low thickness-to-chord ratios — for example, 12% for an A310. That means the wing is thin relative to its chord, and the overall value of CLMAX — the maximum lift coefficient — is fairly low. Consequently, the clean stalling speed is correspondingly high. “Clean” means with no flaps or slats deployed. To reduce landing and take-off speeds, we use various devices to increase the usable value of CLMAX. And here’s the key point: in addition to decreasing the stall speed, these high-lift devices will usually alter the stalling characteristics themselves. The devices include leading-edge flaps and slats, and trailing-edge flaps. Let’s look at the 1g stall formula to see why this works. The formula is VS1g = the square root of (L / (½ ρ CLMAX S)). Here, L is the lift force, ρ is the air density, CLMAX is the maximum lift coefficient, and S is the wing area. From this formula, you can see that an increase in CLMAX will reduce the stall speed — because CLMAX is in the denominator under the square root. So if you raise CLMAX, the stall speed drops. With the most modern high-lift devices, you can increase CLMAX by as much as 100% — that is, you can double it. High-lift devices are fully described in Chapter 8, but for now, understand their sole purpose: they decrease stall speed, hence minimum flight speed, and so they provide a shorter take-off and landing run. That’s it — that’s their only purpose. Now let’s talk about the effect of centre of gravity position on stall speed. There’s a regulatory requirement here: CS-25.103(b) states that VCLMAX — that’s the reference stall speed — is determined with the CG position that results in the highest value of reference stall speed. So the certification process deliberately picks the worst-case CG. Let me explain the physics. Look at Figure 7.24. You have the weight W acting downward, the lift L acting upward, and the centre of pressure CP. If the CG is in front of the CP, you get a nose-down pitching moment. If there’s no thrust or drag moment to oppose it, the tailplane must provide a down load to maintain equilibrium. So the tail pushes down. Now, because the tail is pushing down, the wing’s lift must be increased to maintain an upwards force equal to the increased downwards force. In other words, the total downward force is now weight plus tail down load, so lift has to match that. From the 1g stall formula, you can see that CLMAX will be divisible into the increased lift force more times — meaning the stall speed goes up. So the bottom line: forward movement of the CG increases stall speed. That’s the key takeaway — a forward CG makes the aircraft stall at a higher speed, because the tail has to work harder to hold the nose up, and the wing has to produce more lift to compensate. Let me make sure you’ve got the whole picture. Bank angle raises stall speed, and the effect accelerates as bank increases. High-lift devices raise CLMAX and lower stall speed, giving shorter take-off and landing runs. And a forward CG raises stall speed because of the tail down load. All of these tie back to that one formula: VS1g = the square root of (L / (½ ρ CLMAX S)).