7 Stalling

Composition and Resolution of Forces

A force is a vector quantity. It has magnitude and direction, and it can be represented by a straight line passing through the point at which it is applied, its length representing the magnitude of the force, and its direction corresponding to that in which the force is acting.

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FORCE		FORCE
VECTOR	FORCE	VECTOR

FORCE FORCE

VECTOR

FORCE

VECTOR

Figure 7.21 The resolution of a force into two vectors and the addition of vectors to form a resultant

As vector quantities, forces can be added or subtracted to form a resultant force, or they can be resolved - split into two or more component parts by the simple process of drawing the vectors to represent them. Figure 7.21.

Using Trigonometry to Resolve Forces

If one of the angles and the length of one of the sides of a right angled triangle are known, it is possible to calculate the length of the other sides using trigonometry. This technique is used when resolving a force into its horizontal and vertical components.

Hypotenuse

Opposite

Ad j acen t

TAN		=	Opp	SIN		=	Opp	COS		=	Ad j
			Ad j				Hyp				Hyp

Figure 7.22

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Lift Increase in a Level Turn

45º	LIFT INCREASE
	REQUIRED
ADJACENT 1	L HYPOTENUSE

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W

Figure 7.23

Figure 7.23 shows an aircraft in a level 45° bank turn. Weight always acts vertically downwards. For the aircraft to maintain altitude, the UP force must be the same as the DOWN force. Lift is inclined from the horizontal by the bank angle of 45° and can be resolved into two components, or vectors; one vertical and one horizontal. It can be SEEN from the illustration that in a level turn, lift must be increased in order to produce an upwards force vector equal to weight. We know the vertical force must be equal to the weight, so the vertical force can be represented by 1. The relationship between the vertical force and lift can be found using trigonometry, where φ (phi) is the bank angle:

	cos φ	=	ADJ (1)	transposing this formula gives, L =	1
			HYP (L)		cos φ

In this case φ = 45 degrees

1

L = 0.707 = 1.41

This shows that:

In a 45° bank, LIFT must be greater than weight by a factor of 1.41

Another way of saying the same thing: in a level 45° bank turn, lift must be increased by 41%.

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7 Stalling

Effect of Load Factor on Stall Speed

It has been demonstrated that to bank an aircraft and maintain altitude, lift has to be greater than weight. And that additional lift in a turn is obtained by increasing the angle of attack. To consider the relationship between lift and weight we use Load Factor.

LOAD FACTOR (n) or ‘g’ =	LIFT
	WEIGHT

(a)Increasing lift in a turn, increases the load factor.

(b)As bank angle increases, load factor increases.

In straight and level flight at CLMAX it would be impossible to turn AND maintain altitude. Trying to increase lift would stall the aircraft. If a turn was started at an IAS above the stall speed, at some bank angle CL would reach its maximum and the aircraft would stall at a speed higher than the 1g stall speed.

The increase of lift in a level turn is a function of the bank angle only. Using the following formula, it is possible to calculate stall speed as a function of bank angle or load factor. VSt is the stall speed in a turn

							Load factor does not
VSt	= VS	√		1
				cos φ			affect stall angle

Using our example aeroplane: the 1g stall speed is 150 knots CAS, so what will be the stall speed in a 45° bank?

VSt = 150 √	1	= 178 knots CAS
	0.707

In a 60° bank the stall speed will be:

VSt = 150 √	1	= 212 knots CAS
	0.5

Stall speed in a 45° bank is 19% greater than VS1g and in a 60° bank the stall speed is 41% greater than VS1g, and since these are ratios, this will be true for any aircraft.

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As bank angle is increased, stall speed will increase at an increasing rate. While operating at high CL, during take-off and landing in particular, only moderate bank angles should be used to manoeuvre the aircraft. For a modern high speed jet transport aircraft, the absolute maximum bank angle which should be used in service is 30° (excluding emergency manoeuvres). The normal maximum would be 25°, but at higher altitude the normal maximum is 10° to 15°.

If the 1g stall speed is 150 kt, calculate the stall speed in a 25° and a 30° bank turn. (Answers on page 191).

If the stall speed in a 15° bank turn is 153 kt CAS and it is necessary to calculate the stall speed in a 45° bank turn, you would need to calculate the 1g stall speed first, as follows:

VSt

VSt

=

VS1g

√

1

transposition gives VS1g

=

cos 15°

√

1

cos 15°

VS1g

=

153

=

150 kt CAS

1.02

Effect of High Lift Devices on Stall Speed

Modern high speed jet transport aircraft have swept wings with relatively low thickness/chord

ratios (e.g. 12% for an A310). The overall value of CLMAX for these wings is fairly low and the clean stalling speed correspondingly high. In order to reduce the landing and take-off speeds,

various devices are used to increase the usable value of CLMAX. In addition to decreasing the stall speed, these high lift devices will usually alter the stalling characteristics. The devices include:

a)leading edge flaps and slats

b)trailing edge flaps

			√
From the 1g stall formula:	VS1g	=				L
					½	ρ CLMAX S

it can be seen that an increase in CLMAX will reduce the stall speed. It is possible, with the most modern high lift devices, to increase CLMAX by as much as 100%. High lift devices will be fully described in Chapter 8. High lift devices decrease stall speed, hence minimum flight speed, so

provide a shorter take-off and landing run - this is their sole purpose.

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