# Forces in a Turn ## Forces in a Turn

Turn forces and bank angle

The diagram below shows the relationships between centripetal force, weight, lift and bank angle.

In a level turn, the vertical component of the lift (Lvc) balances the aircraft weight and the horizontal component of lift (Lhc) provides the centripetal force.

(Note: in a right-angle triangle the tangent of an angle is the ratio of the side opposite the angle to that adjacent to the angle. Thus, the tangent of the bank angle is equal to the centripetal force [cf] divided by the weight — or tan ø = cf/W. Or, it can be expressed as tan ø = V²/gr . In the diagram, I have created a parallelogram of forces so that all horizontal lines represent the centripetal force or Lhc and all vertical lines represent the weight or Lvc.)

Let's look at the Jabiru, of mass 400 kg, in a 250 m radius horizontal turn at a constant speed of 97 knots or 50 m/s:

Centripetal acceleration = V² / r = 50 × 50 / 250 = 10 m/s²
Centripetal force required = mass × V² / r = mass × 10 = 400 × 10 = 4000 N

The centripetal force of 4000 N is provided by the horizontal component of the lift force produced by the wings when banked at an angle from the horizontal. The correct bank angle depends on the airspeed and radius; think about a motorbike taking a curve in the road. During the level turn, the lift force must also have a vertical component to balance the aircraft's weight, in this case it is also 4000 N. But the total required force is not the sum of 4000 N + 4000 N = 8000 N; it is less and we have to find the one — and only one — bank angle where Lvc is equal to the weight and Lhc is equal to the required centripetal force.

What then will be the correct bank angle (ø) for a balanced turn? Well, we can calculate it easily if you have access to trigonometrical tables. If you haven't then refer to the abridged version below.

So, in a level turn requiring 4000 N centripetal force with weight 4000 N, the tangent of the bank angle = cf/W = 4000/4000 = 1.0, and thus (from the table) the angle = 45°. Actually, the bank angle would be 45° for any aircraft of any weight moving at 97 knots in a turn radius of 250 metres — provided the aircraft can fly at that speed, of course. (Do the sums with an aircraft of mass 2500 kg, thus weight = 25 000 N.).

Now, what total lift force will the wings need to provide in a level turn if the actual weight component (aircraft plus contents) is 4000 N and the radial component also 4000 N?

Resultant total lift force = actual weight divided by the cosine of the bank angle or L = W / cos ø. Weight is 4000 N, cosine of 45° is 0.707 = 4000/0.707 = 5660 N.

The load on the structure in the turn is 5660/4000 = 1.41 times normal, or 1.41g. Alternatively the 'load factor' = 1/cosine (bank angle); so, cosine 45° is 0.707 = 1/0.707 = 1.41g.

In aviation usage, 'g' denotes the acceleration caused by the force of gravity. When an aircraft is airborne maintaining a constant velocity and altitude — the total lift produced equals the aircraft's weight and that lift is expressed as being equivalent to a '1g' load. Similarly, when the aircraft is parked on the ground, the load on the aircraft wheels (its weight) is a 1g load.

Any time an aircraft's velocity is changed, there are positive or negative acceleration forces applied to the aircraft and felt by its occupants. The resultant manoeuvring 'load factor' is normally expressed in terms of g load, which is the ratio of the forces experienced during the acceleration to the forces existing at the normal 1g flight state.

You will come across terms such as '2g turn' or 'pulling 2g'. What is being implied is that during a particular manoeuvre the lift force is doubled and a radial acceleration is applied to the airframe — for the Jabiru a 2g load = 400 kg × 20 m/s² = 8000 N. The occupants will also feel they weigh twice as much. This is centripetal force and 'radial g'; it applies whether the aircraft is changing direction in the horizontal plane, the vertical plane or anything between.

You may also come across mention of 'negative g'. It is conventional to describe g as positive when the lift produced is in the normal direction relative to the aircraft. When the lift direction is reversed, it is described as negative g. Reduced g and negative g can occur momentarily in turbulence. An aircraft experiencing a sustained 1g negative loading is flying in equilibrium, but upside down. It is also possible for some high-powered aerobatic aircraft to fly an 'outside' loop; i.e. the pilot's head is on the outside of the loop rather than the inside, and the aircraft (and its very uncomfortable occupants), will be experiencing various negative g values all the way around the manoeuvre.

It can be a little misleading when using terms such as 2g. For instance, it was said earlier that a lightly loaded Jabiru has a mass of 340 kg, and if you again do the preceding centripetal force calculation in a 45° banked turn using 340 kg mass you will find that the centripetal acceleration is 10 m/s², centripetal force is 3400 N, weight is 3400 N and total lift = 4800 N. The actual load is 20% less but it is still a 1.41g turn; i.e. the ratio 4800/3400 = 1.41.

Rather than thinking in terms of ratios, it may be more appropriate to consider the actual loads being applied to the aircraft structures. The norm is to use the lift load produced by the wing as the primary structural load reference. In the previous case the load produced is 5660/8 = 707 N/m², compared to the 500 N/m² load in normal cruise.
Increasing the lift force in a turn

You might wonder how does the Jabiru increase the lift if it maintains the same cruise speed in the level turn? Well, the only value in the equation — lift = CL × ½rV² × S — that can then be changed is the lift coefficient. This must be increased by the pilot increasing the angle of attack. (Conversely if CL — the angle of attack — is increased during a constant speed manoeuvre the lift — and consequently the load factor — must increase.) Increasing aoa will also increase induced drag, so that the pilot must also increase thrust to maintain the same airspeed. Thus, the maximum rate of turn for an aircraft will also be limited by the amount of additional power available to overcome induced drag.

For a level turn, the slowest possible speed and the steepest possible bank angle will provide both the smallest radius and the fastest rate of turn; but there are limitations — see this quiz answer. While you are there, also read question 22. The radius of turn = V²/g tan ø metres.

If you consider an aerobatic aircraft weighing 10 000 N and making a turn in the vertical plane —such as a loop — and imagine that the centripetal acceleration is 2g; what will be the load factor at various points of the turn? Actually, the centripetal acceleration varies all the way around because the airspeed and radius must vary. For simplicity we will ignore this and say that it is 2g all around. If the acceleration is 2g then the centripetal force must be 20 000 N all the way around.

A turn in the vertical plane differs from a horizontal turn in that, at both sides of the loop, the wings do not have to provide any lift component to counter weight, only lift for the centripetal force — so the total load at those points is 20 000 N or 2g. At the top, with the aircraft inverted, the weight is directed towards the centre of the turn and provides 10 000 N of the centripetal force while the wings need to provide only 10 000 N. Thus, the total load is only 10 000 N or 1g, whereas at the bottom of a continuing turn the wings provide all the centripetal force plus counter the weight — so the load there is 30 000 N or 3g.

This highlights an important point: when acceleration loads are reinforced by the acceleration of gravity, the total load can be very high.

If you have difficulty in conceiving the centripetal force loading on the wings, think about it in terms of the reaction momentum, centrifugal force which, from within the aircraft, is seen as a force pushing the vehicle and its occupants to the outside of the turn and the lift (centripetal force) is counteracting it. Centrifugal force is always expressed as g multiples.
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