## Chapter 14

## Maneuvering Performance

## General Turning Performance

In a turn, the airplane is not in a state of equilibrium, there must be a force to accelerate the turn.

In a turn, not all the lift (called effective lift) is opposite weight.

Some lift is tilted into the horizontal component.

## General Turning Performance

This requires the aircraft generate more total lift to the point at which effective lift will equal aircraft weight.

The force that is generated toward the center of the turn is called centripetal force and the acceleration towards the center is called radial acceleration.

## General Turning Performance

The force that is generated outward from the center of the turn is called centrifugal force.

Total lift then is offset by the G’s times the weight.

## Load Factor

Load factor or G is defined as lift/weight or the combined force of centrifugal force and weight.

Bank angle alone is the determining component for load factor.

## Load Factor

Things that do not affect load factor are:

airspeed (as long as the aircraft does not stall)

aircraft type

power setting.

The load factor formula to figure for any angle is:

## Load Factor

If we were to work this formula out to a 90 degree angle we would see that lift approaches infinity.

So theoretically, flight with a 90 degree angle is impossible.

The trick is the fuselage, tail and the vertical component of thrust make up the necessary lift requirement.

## Load factor and stall speed

Stall speed depends on the square root of the G loading or put another way, stall speed increases proportional to the square root of the load factor.

## Load factor and stall speed

So to figure stall speed at a specific bank angle, simply first compute the load factor for that particular bank, take the square root of the load factor then take that times the stall speed in unaccelerated level flight and presto, you’ve got the new stall speed at that bank angle.

What?

## Load factor and stall speed

For example, if the plane stalls at 50 kts, at 60 degrees the load factor is 2. the square root of 2 is 1.4 so 1.4 x 50 = 70 kts.

It is this principle which limits how we maneuver our airplanes.

Consider what a 40 degree bank would do to a plane with a stall speed of 50 in the traffic pattern say turning from upwind to crosswind while doing a Vx climb.

## Load factor and stall speed

The root of the LF is about 1.3 x 50 = 65 kts Poof, stall equals climb speed.

Stall spin die.

## The Vg Diagram, Vn diagram, maneuver envelope or flight envelope

The Vg diagram shows graphical how stall speed increases as the G’s go up.

The curved portion shows the aerodynamic limits of the plane.

The straight line is basically the stall speed at the limit load factor.

For the normal category airplane the load factors are +3.8 and -1.52

## The Vg Diagram

These structural limits are required to be imposed by FAR part 23 and assume:

gross weight

symmetrical loads (no rolling pull outs)

no corrosion metal fatigue or other structural damage

smooth air no gusts.

## The Vg Diagram

Usually a separate gust penetration chart was provided by the manufacture in the past for certain airplanes for calculation of such.

The stress imposed is a function of the lift to weight ratio.

Remember that the amount of lift available to be generated varies with the square of the airspeed.

## How to Break Your Airplane

Let’s say that a plane has a Va of 116kts.

This means that if we were to increase the AOA rapidly to the stall AOA while doing 116kts our 2150lb plane would generate 8170lbs of lift before stalling at 3.8 g.

This is the Va for the plane at 2150 but what about at other weights? Va must decrease with the square root of the Weight decrease.

## How to Break Your Airplane

Lets say the same plane weighs 1500lbs and is again stalled at 116kts.

The same amount of lift of 8170lbs is going to be generated by the wings as in the plane at gross weight.

The positive load factor now is 8170/1500 or 5.44 g.

## How to Break Your Airplane

So the argument goes so what the wings are still generating the same amount of lift of 8170lbs so no problem right?

Wrong! You sleepy student pilots!

The problem is with fixed weight components such as the engine, baggage retractable landing gear etc.

## How to Break Your Airplane

Because of the lighter overall weight of the plane the fixed weight components are subjected to higher acceleration forces.

The engine mounts may not be stressed to handle the weight of the engine at 5.4 gs.

A baggage compartment that can hold 200lbs at 3.8 gs could keep it together with 760lbs of force but at 5.4 gs that same baggage weighs in at 1080lbs.

## How to Break Your Airplane

Likewise if you put 400lbs in the baggage compartment and pulled 3.8 gs a total of 1520lbs is put on a baggage compartment floor stressed to handle 760lbs.

You will now have a nice hole to take beautiful pictures from.

## Radius of Turn

The radius of turn goes up as you go faster at a given bank angle.

In fact, the radius goes up as the V squared goes up.

The radius of turn at 300kts is 4 times bigger as at 150kts.

The formula is:

## Rate of Turn

The rate of turn is based on bank angle.

The higher the bank angle the quicker the rate of turn.

The primary rate of turn we shoot for is 3 degrees/sec or 2min for a 360 turn also known as standard rate.

## Rate of Turn

A easy formula for determining the standard rate bank angle is to take the airspeed divide by 10, half that number and add to the original.

Or you can use Doles:

There is a relationship between bank angle, velocity, rate and radius of turn. pg 217 shows that relationship

## Limits by aerodynamic and structural

The turn may be limited by stall or stress related relationships.

Because of the induced drag generated by the needed increase lift in the turn you may not have enough thrust to overcome the drag.

In this situation, the plane may not reach its structural limits.