## Commercial Ground School AVF 221

## Operational Limits

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 is opposite weight

Effective lift or the vertical component of lift is opposite to 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

The horizontal component of lift is what generates this force

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.

Or G’s = lift divided by 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.

To solve for load factor for any bank angle we can use the formula:

F=phi (pronounced phee)

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

Things that do not affect load factor are:

Airspeed (as long as the aircraft does not stall)

Aircraft type

Power setting

The equal and opposite reaction force to centripetal force is centrifugal force

The plane’s inertia, is what causes this force

So inertia opposes the horizontal component of lift

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, 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

the new stall speed at that bank angle.

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.

Load factor and stall speed

Consider what a 48 degree bank would do to the Bonanza with a stall speed of 64 in the traffic pattern say turning from upwind to crosswind while doing a Vx climb at 77 kts.

The LF at 48 degrees bank is about 1.49

The root of the LF of 1.49 is 1.22

So take 1.22 x 64 = 78 kts

Poof, stall speed is 1 kt higher than climb speed of 77.

Stall spin die.

Load factor and stall speed

Intentionally stalling above Va tremendously effects LF

If an aircraft stalls at 50 kts, it would take 4 Gs to stall at 100kts

If you could pull 9 Gs with this plane it would stall at 150kts

Wouldn’t it be cool if we had a graph that would indicate this?

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

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

The curved portion shows the aerodynamic limits of the plane.

The straight lines are basically the stall speed at the limit load factor.

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

Vne

Vg Diagram

Structural Load Limits:

Normal Category +3.8, -1.52

Utility Category +4.4, -1.76

Acrobatic Category +6.0, -3.0

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

A factor of 50% is assumed

Vg Diagram

The lines of maximum lift capability (curved lines) are the first items of importance on the Vg diagram

The aircraft is capable of developing no more than +1 G at 62 mph, the wings level stall speed of the aircraft

Since the maximum load factor varies with the square of the airspeed, the maximum positive lift capability of this aircraft is 2 G at 92 mph, 3 G at 112 mph, 4.4 G at 137 mph

Any load factor above this line is unavailable aerodynamically

The same situation exists for negative lift flight with the exception that the speed necessary to produce a given negative load factor is higher than that to produce the same positive load factor.

If the aircraft is flown at a positive load factor greater than the positive limit load factor of 4.4, structural damage is possible.

Vg Diagram

2 important points:

The intersection of the positive limit load factor and the line of maximum positive lift capability (Va 134kts)

The airspeed at this point is the minimum airspeed at which the limit load of 4.4 Gs can be developed aerodynamically

Any airspeed greater than this provides a positive lift capability sufficient to damage the aircraft.

Conversely, any airspeed less than this does not provide positive lift capability sufficient to cause damage from excessive flight loads

At Va minimum usable turn radius or maneuverability occurs at this condition

The maneuver speed is a valuable reference point, since an aircraft operating below this point most likely will not produce a damaging positive flight load

Any combination of maneuver and gust most likely will not create damage due to excess airload when the aircraft is below the maneuver speed

Vg Diagram

The other point of importance on the Vg diagram is the intersection of the negative limit load factor and line of maximum negative lift capability (100kts)

Any airspeed greater than this provides a negative lift capability sufficient to damage the aircraft

Any airspeed less than this does not provide negative lift capability sufficient to damage the aircraft from excessive flight loads

Vg Diagram

Usually a separate gust penetration chart was provided by the manufacture in the past for certain airplanes for calculation of such. (see Kershner’s book for an example)

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.

Va

The Bonanza is a Utility Category airplane and has a Va of 134kts and a gross weight of 3400lbs.

This means that if we were to increase the AOA rapidly to the stall AOA while doing 134kts, our 3400lb plane would generate 14,960lbs of lift before stalling at 4.4 g.

This is the Va for the plane at 3400 but what about at other weights?

Va must decrease with the square root of the Weight decrease.

Va

Lets say the Bonanza weighs 2,600lbs and is again stalled at 134kts.

The same amount of lift of 14,960lbs is going to be generated by the wings as in the plane at gross weight.

Since LF=lift/weight:

The positive load factor now is 14,960/2,600 or 5.8Gs.

This is 2Gs above the limit of 3.8

Va

So the argument goes, so what, the wings are still generating the same amount of lift of 14,960lbs so no problem right?

Wrong!

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

A

ll these components are now acting on the fuselage at the rate of 5.8 Gs

This places a lot of a stress on the spar carry through which is the piece that bolts your wing spars together

Note the red dye penetrant indicating a crack in this Beech Spar Carry Through

The mounts for these components, the baggage compartment floor ect are not stressed to handle that much weight

Va

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 5.8Gs.

A baggage compartment that can hold 270lbs at 3.8 gs could keep it together with 1026lbs of force but at 5.8Gs that same baggage weighs in at 1566lbs.

Plus you could break your tail off

Va

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.

Do not overload your airplane

Va

So what’s the solution?

Lower the Va airspeed as gross weight decreases

Let’s stick with our example Bonanza at 2600lbs

Our new Va for this weight is now 117kts

This means we can successfully stall with no structural damage at 4.4Gs at 117kts

If we are below 117kts we will stall at less than 4.4Gs

Our 1G stall happens at Vs (top of the CL curve) at 64kts

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 V2 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 highest bank angle we can hold is found at Va

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:

There is a relationship between bank angle, velocity, rate and radius of turn.

At Va, a plane will make the highest rate of turn, the smallest radius of turn and will not be overstressed.

Rate and Radius of Turn

Lets take the SR-71 through the equations

Top speed is Mach 3.5 that’s 2314kts

What is the rate of turn at 80° bank?

2.67 degrees/sec

So almost standard rate

Radius of turn with a 30° bank

824,162 feet or 135 NM

This means that the pilot would have to start his turn over MWH to be able to roll out south over the pacific coast

The Mighty Bonanza

That means to make the smallest, tightest turn in the Bonanza we have to be at 134kts, 76.86 degrees and 4.4 Gs

Using the formula, that puts our radius of turn at 2764 feet

That puts us 5528 foot diameter circle

Muni is 2513 feet

Limits by aerodynamic and structure

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.

Home Work Extra Credit: Draw a Vg Diagram for the Mighty B19.

I will replace your lowest homework score with the score on the Vg Diagram.