Accelerations

Now the accelerations of an aeroplane along its line of flight are comparatively unimportant. They are probably greatest during the take-off, or, in the nega­tive sense, during the pull-up after landing. But the accelerations due to change in direction of flight are of tremendous importance.

Fig8A Manoeuvres

A clipped-wing Spitfire in mock combat manoeuvres with a Chance Vought Corsair.

As we have already discovered, when a body is compelled to move on a curved path, it is necessary to supply a force towards the centre, this force being directly proportional to the acceleration required. Such a force is called the centripetal force. The body will cause a reaction, that is to say an outward force, on whatever makes it travel on a curved path. This reaction is called by some people the centrifugal force.

If an aeroplane is travelling at a velocity of У metres per second on the cir­cumference of a circle of radius r metres, then the acceleration towards the centre of the circle is Y/r metres per second per second.

Therefore the centripetal (or centrifugal) force is m X acceleration, where m is the mass of the aeroplane in kilograms,

= mY/r newtons

In practice aeroplanes very rarely travel for any length of time on the arc of a circle; but that does not alter the principle, since any small arc of a curve is, for all practical purposes, an arc of some circle with some radius, so all it means is that the centre and the radius of the circle keep changing as the aero­plane manoeuvres.

The acceleration being Y/r shows that the two factors which decide the acceleration, and therefore the necessary force, are velocity and radius, the vel­ocity being squared having the greatest effect. Thus curves at high speed, tight turns at small radius, need large forces towards the centre of the curve.

We can easily work out the acceleration, Y/r. For instance, for an aeroplane travelling at 82m/s on a radius of 200 metres, the acceleration is (82 X 82)/200 = 34m/s2, which is a little less than 4 X the acceleration due to gravity. For convenience, this is often written as 4g. However, before we go any further, we need to try to clear up some misunderstanding about the use of the symbol g. This letter is unfortunately used to represent several different things. Firstly, it represents the gravitational constant. By multiplying the mass by the gravitational constant g, we obtain the weight (written in mathematical terms W=m g). The gravitational constant g has a value of 9.81 m/s2 on the earth’s surface, and this makes it look like an acceleration. In this context it is not an acceleration. As we noted in Chapter 1, a book resting on a fixed table has no acceleration, but it does have a weight equal to m X g newtons. The trouble is that g is also used to represent the acceleration due to gravity. If the book falls off the table, it will accelerate at 9.81 m/s2; the acceleration will then be equal to g (9.81 m/s2). Fortunately, although g is commonly used to repre­sent two different things, it always has a value of 9.81 m/s2, so we can still use it for both purposes, as long as we make quite clear whether we are talking about an acceleration or a weight. Finally, to make things even more con­fusing, pilots (and racing car drivers) often talk about pulling a certain number of ‘g’s. Unfortunately, this usage is so common that we cannot ignore it. This quantity is not an acceleration, it is just a number. It has no units, and it simply represents a factor which, when multiplied by the weight, gives the total force that must be applied to a body to balance the combined effects of gravity and centripetal acceleration. It is really a load factor, because it tells us how the loads and stresses in the airframe increase during a manoeuvre. In this book we will try to make things easier by using a bold letter ‘g’ in inverted commas for this quantity.

Let us try to show how the difference in usage works, by taking a simple example of an aircraft with a mass of 1000 kg, which therefore has a weight of lOOOg newtons, (since W=m g). We will take as an example, the case of this aircraft pulling out of a dive, where it is subjected to a centripetal acceleration of 3g at the bottom of the manoeuvre. The centripetal force required to provide this acceleration will be 1000 X 3g newtons (since force = mass X acceleration). Now 1000 X 3g newtons is a force equal to 3 times the weight. This centripetal force will be provided by the lift from the wings. However, the wing has to support the weight of the aircraft (1000 X g newtons) as well as providing the centripetal acceleration, so the total lift force must be 1000 X (3g +lg) newtons, which is 4 X the weight. The pilot therefore refers to this as a 4 ‘g’ manoeuvre. Not only will the lift and the stresses on the airframe by 4 times their normal level flight value, but the pilot will feel as though he weighs 4 times as much as usual. From the above, you can begin to appreciate the problem. A 4 ‘g’ pullout involves a 3g centripetal acceleration!

Now consider what happens if the aircraft is at the top of a loop at the same speed, and with the same radius of curvature. At the top of the loop, the cen­tripetal acceleration required will still be 3g, and the total force required to produce this will still be 1000 X 3g newtons. However the weight of the air­craft will provide part the centripetal force required (1000 X g newtons), leaving the wing lift (now pulling downward) to provide the extra 1000 X 2g newtons. Notice that although the centripetal acceleration is 3g, the wings only have to provide twice as much lift force as in level flight: that is 2 X the weight. Also, the pilot will not be squashed so firmly into his seat as at the bottom of the pull-out manoeuvre, as his weight is trying to pull him out of the seat. Because the lift required is only 2 X the weight, the load factor is only 2, and the pilot would call this a 2 ‘g’ manouevre.

If the loop were performed at the same speed, but with three times the radius, then the centripetal acceleration would be lg, so the required cen­tripetal force would be 1 X the weight, and the weight alone could provide all the centripetal force required. The wings would need to produce no lift at all. This is therefore called a zero ‘g’ manoeuvre. The pilot would temporarily feel weightless. Indeed, performing an outside loop or ‘bunt’, that is with the air­craft the right way up (see later in this chapter), with a lg centripetal acceleration is used as a way of providing trainee astronauts with an experi­ence of the weightless conditions that they will encounter in space flight.