Chapter 4
Aircraft Performance

Altitude

Density altitude is the altitude in standard air where the density is the same as the existing density. It is affected by the pressure, temperature, and moisture content of the air. A decrease in pressure and an increase in temperature both decrease the density of the air and increase the density altitude. Since water vapor is less dense than dry air, an increase in the humidity of the air will also increase the density altitude. Density altitude is used for computing aircraft performance, and it may be found on charts or by the use of flight computers.

Problem:

Find the density altitude at an airport elevation of 5,515 feet, an outside air temperature (OAT) of 30°C, and an altimeter setting of 29.40 inches of mercury, using FAA Figure 24.

Solution:

1. Find the pressure altitude of the airport by applying the Pressure Altitude Correction Factor (found to the right of the chart) to the airport elevation. PA Conversion Factor for an altimeter setting of 29.40 is 485 feet. Pressure altitude of the airport is: 5,515 + 485 = 6,000 feet.

In an actual flight situation, pressure altitude is found by setting the barometric scale of the altimeter to 29.92" Hg, and reading the altimeter.

2. Follow the vertical line for 30°C upward until it intersects the diagonal line for 6,000 feet pressure altitude. From this intersection, project a line to the left until it intersects the density altitude scale. This scale is intersected at 9,100 feet.

Density altitude can also be found quickly and accurately by using the CX-3 Flight Computer:

1. Select Altitude from the FLT menu.

2. Enter the elevation of 5,515 feet in the indicated altitude (IAlt) field.

In an actual flight situation, you would find the pressure altitude by setting your altimeter barometric scale to 29.92 inches of mercury and then reading the altimeter.

3. Enter the barometric pressure (Baro) of 29.40 inches of mercury.

4. Enter the temperature (OAT) of 30°C.

5. The density altitude is 9,011 feet. The difference between the answer found using the chart and the answer found using the computer is due to the greater accuracy of the computer.

Takeoff Performance

The upslope or downslope of a runway (runway gradient) is important when runway length and takeoff distance are critical. An uphill slope of a runway produces a retarding force which decreases acceleration and results in a longer ground run needed for takeoff.

The takeoff ground roll and distance needed to clear a 50-foot obstacle can be computed by using the Takeoff Data Chart in FAA Figure 26. Charts of this type are furnished in the POH, and their use requires a great deal of interpolation.

Problem:

Find the takeoff run required for this airplane at a pressure altitude of 2,500 feet, temperature of 24°C, weight of 2,400 pounds, and a 25-knot headwind, refer to FAA Figure 26.

Solution:

1. Find the ground roll required for a 25-knot headwind at 2,200 pounds.

a. Find the ground roll for 2,500 feet, 15-knot headwind at 2,200 pounds: 245 feet.

b. Find the ground roll for 2,500 feet, 30-knot headwind at 2,200 pounds: 120 feet.

c. The difference for a 15-knot headwind increase is 125 feet.

d. Every 5-knots of headwind decreases the ground roll by 125 ÷ 3 = 41.7 feet

e. The ground roll required for a 25-knot headwind at 2,200 pounds is: 245 – (2 × 41.7) = 161.6 feet.

2. Find the ground roll required for a 25-knot headwind at 2,600 pounds.

a. Find the ground roll for 2,500 feet, 15-knot headwind at 2,600 pounds: 370 feet.

b. Find the ground roll for 2,500 feet, 30-knot headwind at 2,600 pounds: 200 feet.

c. The difference for a 15-knot headwind increase is 170 feet.

d. Every 5-knots of headwind decreases the ground roll by 170 ÷ 3 = 56.7 feet

e. The ground roll required for a 25-knot headwind at 2,600 pounds is: 370 – (2 × 56.7) = 256.6 feet.

3. Find the required ground roll required for 2,400 pounds.

a. Ground roll for 2,200 pounds = 161.6 feet.

b. Ground roll for 2,600 pounds = 256.6 feet.

c. Difference = 95 feet

d. The ground roll increases 95 ÷ 4 = 23.75 feet for each 100-pound increase in weight.

e. The ground roll required for 2,400 pounds is: 161.6 + (2 × 23.75) = 209.1 feet

4. Apply correction for nonstandard temperature. The average change in temperature used by this chart is 2°C for each thousand feet.

a. Standard temperature at 2,500 feet is 10°C.

b. Existing temperature is 24°C. This is 24 – 10 = 14°C above standard

c. The ground roll increases 10% for each 14°C above standard temperature.

d. The ground roll is 209.1 + 10% = 230.01 feet.

Another type of takeoff performance chart is shown in the Maximum Performance Takeoff Distance chart in FAA Figure 28.

Problem:

Determine the approximate total distance required to clear a 50-foot obstacle.

Solution:

1. Find the maximum performance takeoff distance required to takeoff and clear a 50-foot obstacle at 1,000 feet pressure altitude and a gross weight of 5,300 pounds, with an outside air temperature of 20°C.

a. Find the distance required to clear a 50-foot obstacle for 5,500 pounds at 1,000 feet PA and an OAT of 20°C: 1,950 feet.

b. Find the distance required to clear a 50-foot obstacle for 5,100 pounds at 1,000 feet PA and an OAT of 20°C: 1,620 feet.

c. The difference in distance for a 400-pound increase in weight is 330 feet.

d. Each 100-pound increase in weight increases the distance required to clear a 50-foot obstacle by 82.5 feet.

e. The distance required to clear a 50-foot obstacle at 5,300-pound weight is: 1620 + (2 × 82.5) = 1,785 feet.

2. Since the takeoff is made from a sod surface, this distance must be increased by 8%:

1,785 + 8% = 1,927.8 feet.

3. The takeoff is made with a 15-knot headwind, and the takeoff run is reduced by 7% for each 10 knots of headwind:

1,927.8 – 10.5% = 1,725.4 feet

Climb Performance

An airplane’s climb performance can be calculated by using a chart such as the one in FAA Figure 27, which is included in many POHs.

Problem:

Find the indicated airspeed that will give the greatest gain in altitude in a unit of time at 3,200 feet.

Solution:

Follow the line that represents 3,200-foot altitude to the right until it intersects the line for the best rate of climb. From this intersection, follow a line down until it intersects the indicated airspeed scale. An indicated airspeed of 113 KIAS will give the greatest gain in altitude in a unit of time at 3,200 feet.

Range Performance

The maximum range of a propeller-driven airplane occurs when the airplane is operating at the angle of attack that produces the maximum lift over drag (L/D) ratio, L/D_MAX. At the angle of attack that produces L/D_MAX, the induced drag and the parasite drag are equal, and the total drag is minimum.

Glide Distance

Glide distance can be found by using a chart such as FAA Figure 29.

Problem:

Find the distance you can glide from 5,500 feet above the terrain with a 10-knot tailwind.

Solution:

1. Follow a line to the right from the height above terrain of 5,500 feet until it intersects the slanted line.

2. From this intersection, draw a line downward until it intersects the glide distance. This intersection is at 12 statute miles.

3. According to Note 1, increase this distance by 10% for each 10-knot tailwind: 10% × 12 = 1.2 miles.

4. The airplane for which this curve was prepared will glide 12 + 1.2 = 13.2 miles from a height of 5,500 feet above the terrain with the help of a 10-knot tailwind.

Stall Speed Performance

The stall speed of an airplane under the various configurations of flaps and gear, and angles of bank can be found by referring to a Stall Speed Chart such as the one in FAA Figure 25.

Problem:

Find the indicated stall speed for this airplane with the gear down and the flaps at 15° and the bank angle at 30°, referring to FAA Figure 25.

Solution:

1. Find the calibrated airspeed for stall at 20° bank angle with gear down and flaps at 15°: 83 KCAS.

2. Find the calibrated airspeed for stall at 40° bank angle with gear down and flaps at 15°: 92 KCAS.

3. Interpolate to find calibrated airspeed for stall at 30° bank angle with gear down and flaps at 15°:

92 – 83 = 9

9 ÷ 2 = 4.5

83 + 4.5 = 87.5 KCAS

4. Use the Airspeed Calibration table, in the flaps 15° columns to convert 87.5 KCAS to KIAS:

94 – 86 = 8

87.5 is 1.5/8 of the way (that is, 1.5 ÷ 8) between these two speeds; 18.75% of the way between 80 and 90 is:

80 + (10 × .1875) = 81.875 KIAS.

5. 87.5 KCAS is equivalent to 81.875 KIAS.

Landing Performance

The landing ground roll and landing distance needed over a 50-foot obstacle can be computed by using the Normal Landing Chart in FAA Figure 31. Charts of this type are furnished in the POH.

Problem:

Find the total landing distance required for this airplane over a 50-foot obstacle at a pressure altitude of 4,000 feet, temperature of 15°C, weight of 3,000 pounds and a 22-knot headwind.

Solution:

1. Follow the 4,000-foot pressure altitude curve to the right until it intersects the vertical line for 15°C. From this intersection, draw a line to the right until it intersects the left reference line.

2. From this point, draw a line diagonally upward to the right, parallel to the lines that are already there. Follow this line up until it intersects the vertical line for 3,000-pound weight.

3. Draw a line to the right until it intersects the right reference line.

4. From this point, draw a line downward to the right paralleling the existing lines.

5. Draw a line upward from the 22-knot headwind index until it intersects the diagonal line you have just drawn.

6. From this point, draw a line to the right until it intersects the index for the total landing distance over a 50-foot obstacle. This is at 1,175 feet.

Multi-Engine Performance

The single-engine service ceiling of a twin-engine airplane is the maximum density altitude at which the single-engine best rate-of-climb speed (V_YSE) will produce a 50-foot-per-minute rate of climb.

When one engine fails on a conventional, light, twin-engine airplane, performance is not halved, but is actually reduced by 80% or more. When one engine fails, the airplane not only loses power, but the drag increases considerably because of the asymmetric thrust produced as one engine carries all of the load.

The airplane should be capable of allowing a pilot to maintain heading when flying at minimum control speed with the critical engine inoperative (V_MC), but it may not allow altitude to be gained, or even held. Since the power of an unsupercharged engine decreases with altitude, V_MC decreases as altitude is increased. This makes it possible to maintain directional control at a lower airspeed at altitude than at sea level. The thrust moment of the operating engine becomes less at altitude, decreasing the need for the rudder’s yawing force. Since V_MC is a function of engine power, it is possible for the airplane at altitude to reach its stalling speed prior to losing directional control.

When one engine fails on a light, conventional twin-engine airplane, and full power is applied to the operative engine, the airplane will try to roll as well as yaw into the inoperative engine as the airspeed drops below V_MC. This tendency becomes greater as the airspeed is further reduced. Banking toward the inoperative engine increases V_MC at the rate of about 3 knots per degree of bank, while banking toward the operative engine reduces V_MC.

V_MC is highest when the center of gravity is at the rear-most allowable position. The airplane rotates about its center of gravity, and the moments, when considering V_MC, are measured using the CG as the reference. A rearward CG shortens the arm between the CG and the center of the rudder’s horizontal lift. This means that a higher force (higher airspeed) is needed to counteract the engine-out yaw when the CG is aft, than when it is further forward.

Wind Components

We can find the crosswind, headwind, or tail wind component for flight or for takeoff or landing by using a Wind Component Chart, such as that in FAA Figure 30, or by using an electronic flight computer. Charts such as that in FAA Figure 30 are printed on the card of some E6-B flight computers.

Problem:

Find the crosswind component a pilot would have when landing on runway 30 with a wind from 020° at 15 knots.

Solution:

1. The wind angle is the difference between the aircraft heading and the direction from which the wind is blowing. The wind is blowing from 020°, and the aircraft is landing on runway 30. Therefore, the wind angle is (360° – 300°) + 20 = 80°.

2. Follow the arc for 15 knots until it intersects the 80° radial for the wind angle.

3. From the intersection of the 15-knot arc and the radial for the 80° wind angle, draw a line downward until it intersects the crosswind component: 15 knots.

4. An aircraft landing on runway 30 with a wind from 020° at 15 knots will have a 15-knot crosswind.

This same problem can be quickly and accurately solved on the CX-3 Flight Computer:

1. Select Wind Component from the FLT menu.

2. Enter the wind direction (WDir) of 020 degrees.

3. Enter the wind speed (WSpd) of 15 knots.

4. Enter the Runway number, 30.

5. Read the crosswind component (X Wnd) of 15 knots and the headwind component (H Wnd) of 3 knots.

Helicopter Performance

The most favorable conditions for rotorcraft performance are the combination of low density altitude, light gross weight, and moderate to strong wind. Low density altitude increases both the performance of the engine and the lift produced by the rotors. The low gross weight allows the helicopter to operate with less than maximum performance, and a moderate to strong wind allows the helicopter to obtain translational lift with a minimum ground speed.

Water vapor is only about 5/8 as heavy as dry air, and when the relative humidity is high, the air is less dense than standard air. Because of this less dense air, the engine develops less power and the rotors produce less lift than they would in standard conditions. If all other factors remain the same, an increase in relative humidity will lower the hover ceiling of a helicopter.

Principles of Weight and Balance

It is important, when loading an airplane for a flight, that the maximum allowable gross weight is not exceeded and that the center of gravity is within the allowable limits. In calculating weight and balance problems, these factors are important:

• The center of gravity of an airplane is measured along its longitudinal axis in inches from a reference called the datum.

• The arm of a weight is the distance in inches from the datum. If the weight is ahead of, or to the left of the datum, the arm is negative, and if it is behind, or to the right of the datum, the arm is positive.

• A moment is a force that tends to cause rotation about a point, and is measured in pound-inches. A moment is the product of the weight of the item, in pounds, and the distance from the datum in inches.

• The center of gravity of an airplane is found by dividing the total moment by the total weight.

Excessive weight reduces the flight performance of an aircraft by:

1. Requiring a higher takeoff speed.

2. Requiring a longer takeoff.

3. Reducing the rate and angle of climb.

4. Lowering the maximum altitude.

5. Giving the aircraft a shorter range.

6. Reducing the cruising speed.

7. Reducing maneuverability.

8. Giving the aircraft a higher stalling speed.

9. Requiring a higher landing speed.

10. Requiring a longer landing roll.

11. Putting an excessive weight on the nosewheel.

With excessive weight, the aircraft will accelerate more slowly with the same power output and a higher airspeed is needed to generate enough lift for takeoff.

Loading an aircraft so its CG is behind the aft limit causes the aircraft to become less longitudinally stable and less controllable. When the CG is at or near its aft limit, the tail surfaces exert less down-load and there is less stabilizing down-pitching moment. This can reduce the aircraft’s ability to recover from stalls and spins. If the CG is too far aft, the elevator may not have enough power to get the nose down to bring the aircraft out of its stalled condition. The aircraft will enter a spin, and the aft CG will cause the spin to flatten out, making recovery very difficult or impossible.

When an aircraft is loaded with its CG at its aft limit, the horizontal tail surfaces will be required to exert the least amount of down-load. This relieves the wing of part of its load and allows it to maintain altitude with a lower angle of attack. The drag is less, and the aircraft has a faster cruising speed.

The disadvantage of having the CG at its aft limit is the decrease in longitudinal stability. A neutral load on the tail surfaces in cruising flight produces the most efficient overall performance and fastest cruising speed, but it also results in longitudinal instability. The problem caused by this instability is overcome by sophisticated automatic flight control systems in many of the modern high-performance aircraft.

When an aircraft is loaded at its most forward CG location, nose-up trim is needed to maintain level cruising flight. Nose-up trim causes the tail surfaces to produce a greater down-load on the aft portion of the fuselage, and this adds to the wing loading and increases the total lift the wing must produce to maintain altitude. When the wing flies at a higher angle of attack to produce the extra lift, it produces more induced drag.

The indicated airspeed produced by a constant power at a constant altitude will be lower when the CG is at the forward limits than it is when the CG is further aft.

Moving the CG from the aft limit to beyond the forward limit will cause the aircraft to stall at a higher speed. The additional downward force produced by the horizontal stabilizer to overcome the nose-down pitching tendency increases the wing loading and requires a higher speed before reaching its stalling angle of attack. The aircraft will also cruise slower because the additional downward tail load increases the wing loading and requires a higher angle of attack.

One advantage of a forward CG location is that an aircraft becomes more longitudinally stable as the CG is moved forward. The forward CG increases the tendency for the nose to drop as the airspeed decreases.

Forward CG is most critical during landing. If it is too far forward, the elevators may not have enough power to get the tail down enough for a proper approach and flare.

A high gross weight and a forward CG will increase the stall speed of an aircraft. Both the forward CG and the high gross weight require the aircraft to be flown at a higher speed before it reaches its stalling angle of attack.

The Law of the Lever

Weight and balance problems are based on the physical law of the lever. This law states that a lever is balanced when the weight on one side of the fulcrum multiplied by its arm is equal to the weight on the opposite side multiplied by its arm. In other words, the lever is balanced when the algebraic sum of the moments about the fulcrum is zero. This is the condition in which the positive moments (those that try to rotate the lever clockwise) are equal to the negative moments (those that try to rotate it counter-clockwise).

Weight A × Arm A = Weight B × Arm B

Problem:

Item	Weight (lbs)	Arm (in)	Moment (lb-in)
Weight A	90	-50	-4,500
Weight B	?	-25	+4,500
			0

Solution:

Weight B × 25 = 4,500

4,500 ÷ 25 = Weight B

Weight B = 180 lbs

A common weight and balance problem involves shifting the CG by moving around passengers or cargo. This can be solved by a simple math equation:

Finding the Center of Gravity

The datum is a location specified by the manufacturer of an aircraft from which all measurements are made to determine the weight and balance of an aircraft. The datum may be located anywhere. Commonly used datum locations are:

• Leading edge of the wing

• Center line of the main spar

• A given distance ahead of the nose of the aircraft

Locating the datum ahead of the nose of the aircraft makes all moments positive and prevents many arithmetic errors.

In weight and balance problems, weight is specified in pounds, and the arm is specified in inches. The moment is the product of the distance between the item and the datum multiplied by the weight of the item. It is expressed in pound-inches. The arm of each item is its distance from the datum in inches. Find the total weight and the total moment and divide the total moment by the total weight to find the distance from the datum to the center of gravity.

Problem:

Find the center of gravity when this information is known.

Weight A	120 lbs	at 15" aft of datum
Weight B	200 lbs	at 117" aft of datum
Weight C	75 lbs	at 195" aft of datum

Solution:

Make a table:

Item	Weight	Arm	Moment
Weight A	120	15	1,800
Weight B	200	117	23,400
Weight C	+75	195	+14,625
Totals	395		39,825

CG = Total Moment ÷ Total Weight

39,825 ÷ 395 = 100.8 inches from the datum

This problem can also be quickly and accurately worked on the CX-3 Flight Computer:

1. Select Weight and Balance from the W/B menu.

2. Set a reduction factor of 1.

3. For item 1, enter Weight A of 120 lbs and Arm of 15 inches.

4. For item 2, enter Weight B of 200 lbs and Arm of 117 inches.

5. For item 3, enter Weight C of 75 lbs and Arm of 195 inches.

Read a total weight of 395 pounds and CG of 100.8 under the Totals section.

To find the CG of an airplane using the charts of FAA Figure 36, compile the same “Weight × Arm = Moment” chart previously used. To find the moments not given in the charts, multiply the weight by the arms that are given and divide by 100:

Item	Weight		Moment
Airplane	2,110		1,652
Pilot & front seat	375	× 85/100	319
Rear passengers (aft)	245	× 136/100	333
Baggage	65	× 150/100	98
Fuel (70 gal.)	+420		+315
Totals	3,215		2,717

The allowable gross weight of the airplane as shown in the Gross Weight Moment Limits Chart is 3,400 pounds. The airplane, as loaded, weighs 3,215 pounds, which is 185 pounds under the allowable gross weight.

In the Gross Weight Moment Limits Chart, follow a line representing 3,215 pounds to the left until it intersects the diagonal line representing the 2,717 inch-pound moment/100. This intersection is within the envelope, which means that the CG is located within the allowable limits.

To solve this type of problem on the CX-3 Flight Computer:

1. Multiply the empty weight moment of 1,652 by the reduction factor of 100 to get the moment of 165,200. Divide this by the empty weight of 2,110 to get the empty weight center of gravity (EWCG) of 78.29 inches.

2. Press the “W/B” button and select “Weight and Balance” and set RF to 1.

3. For Item 1, enter the airplane empty weight of 2,110 lbs and EWCG (Arm) of 78.29 inches.

4. For Item 2, select Add and enter the weight of the pilot and front passenger, 375 lbs, and front seat arm of 85 inches.

5. For Item 3, select Add and enter weight of rear passengers, 245 lbs, and arm of 136 inches.

6. For Item 4, select Add and enter baggage weight of 65 lbs and arm of 150 inches.

7. For Item 5, select Add and enter fuel weight of 420 lbs and arm of 70 inches.

8. Under TOTALS, read a gross weight of 3,215 lbs and a CG of 84.5. The allowable gross weight is 3,400 lbs, so the airplane is 185 pounds under the allowable gross weight. Using FAA Figure 36, determine that the aircraft’s CG is within limits.

Weight to be Added or Removed

The amount of weight needed to be added or removed at a particular station to move the CG to a specific location can be found by using these steps.

Problem:

The aircraft weighs 2,900 pounds. CG location is at station 115.0. Aft CG limit is at station 116.0. What is the maximum weight which can be added at station 130.0 without exceeding the aft CG limit?

Solution:

The formula to use for this problem is:

Rearrange the formula:

Problem:

Find the new location of the CG after a given weight is added or removed:

1. Find the new location of the CG after 90 pounds are removed from Station 140.

2. Aircraft weight is 6,230 pounds.

3. CG location is Station 79.

Solution:

1. Make a chart similar to this:

	Item	Weight	Arm	Moment
	Airplane	6,230	79.0	492,170
	Weight	–90	140.0	–12,600
	New Totals	6,140		479,570

2. To find the new CG, divide the new total moment by the new total weight:

CG = Moment ÷ Weight

479,570 ÷ 6,140 = 78.1

The CG after the removal of the weight is located at station 78.1.

This problem can be solved quickly and accurately on the CX-3 Flight Computer:

1. Press the “W/B” button and select “Weight and Balance” and set RF to 1.

2. For Item 1, enter the aircraft weight of 6,320 lbs and Arm of 79 inches.

3. For Item 2, enter the weight to be removed, -90 lbs, and Arm of 140 inches.

4. Under TOTALS, read the aircraft weight of 6,140 lbs and new CG of 78.11 inches.

Weight to be Shifted

The amount of weight that must be shifted to move the CG to a desired location may be found by using this formula:

Rearrange the formula:

This formula can also be arranged to find the new CG after a weight is shifted by rearranging it as:

To solve this type of problem on the CX-3 Flight Computer:

1. Press the “W/B” button and select “Weight Shift Formula.”

2. Enter the Item Weight to be shifted (Item Wt), 100 lbs.

3. Enter the aircraft weight (Total Wt), 2,800 lbs.

4. Since the 100 lbs is being shifted from 130 inches to 30 inches, this is a change in Arm of -100. Enter a change in Arm (ΔArm) of -100 inches.

5. Press enter to read a change of center of gravity (∆CG) of -3.57 inches.

6. Subtract 3.57 from the original CG of 120 to get a new CG of 116.4 inches, which is forward of the allowable CG limit of 117 inches.

Helicopter Weight and Balance

If a helicopter is loaded with its CG aft of the allowable limits, it will have a tail-low attitude, and will require an excessive forward displacement of the cyclic stick to maintain the desired position over the ground while hovering in a no-wind condition. It may even be impossible to hover. This situation will be aggravated if the fuel tanks are located forward of the CG, because, as fuel is used from these tanks, the CG will shift further aft.

With the CG aft of the limit, it may not be possible to fly in the upper allowable airspeed range, because there is not enough forward cyclic displacement to maintain the necessary nose-low attitude. The helicopter will be dangerous to fly in gusty or rough air. If it should pitch up because of gusty winds during high-speed flight, the nose will start to rise, and full forward cyclic stick may not be sufficient to hold it down, or to lower it once it rises.