8: Aircraft Performance

Weight and Balance

Headwind and Crosswind Components

Density Altitude

Takeoff and Landing Considerations

Takeoff and Landing Distance

Fuel Consumption vs. Brake Horsepower

Time, Fuel, and Distance to Climb

Cruise Performance Table

Cruise and Range Performance Table

Maximum Rate of Climb

Glide Distance

Weight and Balance

Even though an aircraft has been certificated for flight at a specified maximum gross weight, it may not be safe to take off with that load under all conditions. High altitude, high temperature, and high humidity are additional factors which may require limiting the load to be carried.

In addition to considering the weight, the pilot must ensure the load is arranged to keep the aircraft in balance. The balance point, or center of gravity (CG), is the point at which all of the weight of the system is considered to be concentrated. For an aircraft to be safe to fly, the center of gravity must fall between specified limits. To keep the CG within safe limits, it may be necessary to move weight toward the nose of the aircraft (forward) which moves the CG forward, or toward the tail (aft) which moves the CG aft. The aircraft and the various compartments of the aircraft are designed for specific maximum weights and critical load factors. Compartments placarded for a given weight may be loaded to the maximum allowable load only if the CG is kept within limits and the aircraft is flown safely within critical load factor limits.

The datum is an imaginary vertical reference line from which locations on/in an aircraft are measured. The datum is established by the manufacturer and may vary in location in different aircraft. See Figure 8-1.

Figure 8-1. Datum lines

The arm (or station) is the horizontal distance measured in inches from the datum line to a point on the aircraft. If measured aft toward the tail, the arm is given a positive value; if measured forward toward the nose, the arm is given a negative value. When all arms are positive, the datum is located at the nose or in front of the airplane. See Figure 8-2.

Figure 8-2. Positive and negative arms

The moment is the product of the weight of an object multiplied by its arm and is expressed in pound-inches (lbs-in): Weight × Arm = Moment. The moment index is a moment divided by a constant such as 100 or 1,000. It is used to simplify computations where heavy items and long arms result in large, unmanageable numbers.

The position of the CG is expressed in inches from datum, and is found by dividing the total moments by the total weight:

CG (inches aft of datum) =

Total moment

Total weight

The specified forward and aft points within which the CG must be located for safe flight are called the center of gravity limits. These limits are specified by the manufacturer. The distance between the forward and aft CG limits is called the center of gravity range. The empty weight includes the airframe, engine(s), all items of equipment, unusable fuel, hydraulic fluid, and undrainable oil, or, in some aircraft, all of the oil. The useful load includes the pilot, passengers, baggage, fuel, and oil. The takeoff weight is the empty weight plus the useful load. The landing weight is the takeoff weight minus any fuel used.

Computing Weight and Balance

Problem:

Calculate the weight and balance and determine where the CG would be located for three weights given the following:

Weight D: 160 lbs at 45 inches aft of datum
Weight E: 170 lbs at 145 inches aft of datum
Weight F: 105 lbs at 185 inches aft of datum

Solution:

  1. Construct a table, with column headings as follows:

    Item

    Weight

    Arm

    Moment

  2. Find the moment for each item by multiplying the weight times the arm:

    Item

    Weight

    Arm

    Moment

    Weight D

    160

    45

    7,200

    Weight E

    170

    145

    24,650

    Weight F

    105

    185

    19,425

  3. Total the weights and moments:

    Weight

    Moment

    160

    7,200

    170

    24,650

    + 105

    + 19,425

    435 lbs

    51,275 lbs-in

  4. Determine the CG by dividing the total moment by the total weight:

    CG =

    Total moment

    Total weight

    CG = 51,275 ÷ 435 = 117.8 inches

Graph Weight and Balance Problems

Airplane manufacturers use one of several available systems to provide loading information. Weight and balance computations are greatly simplified by two graphic aids: the loading graph and the center of gravity moment envelope. The graphs shown in FAA Figure 38 are typical of those found in a POH.

Problem:

Determine if the airplane is loaded within limits, given the following:

Empty weight (oil included): 1,271 lbs
Empty weight moment (lbs-in/1,000): 102.04 lbs-in
Pilot and copilot: 400 lbs
Rear seat passenger: 140 lbs
Cargo: 100 lbs
Fuel: 37 gallons

Solution:

  1. Construct a table with the Item, Weight, and Moment. The moment of each item is found by using the loading graph in FAA Figure 38. Locate the weight of each item on the left side of the chart, and proceed to the right to intersect the correct item line. At point of intersection, proceed downward to find the moment for that item at the specified weight.

    Item

    Weight

    Moment/1,000

    Aircraft empty weight

    1,271

    102.04

    Pilot and copilot

    400

    36.00

    Aft passenger

    140

    18.00

    Cargo

    100

    11.50

    Fuel (37.0 gal × 6 lbs/gal)

    + 222

    + 20.00

    Totals

    2,133

    187.54

  2. Compare the total weight and moment to the limits in the CG envelope in FAA Figure 38. Move up the loaded aircraft weight scale to 2,133, then across to 187.54 on the moment/1,000 scale. The weight and CG are within limits.

Weight Change

There are a number of additional weight and balance formulas that may be used to determine the effect on the CG due to a weight change. Use the following formula to find the change in CG, if weight has been added or subtracted. The amount of weight changed and the new total weight must be known, in addition to the distance between the original CG and the point where the weight is being added or subtracted.

Weight lost or gained

=

Change in CG

New total weight

Distance between original CG and
point of weight removed or added

Problem:

Determine the new CG location after 1 hour 45 minutes of flight time, given the following:

Total weight: 4,037 lbs
CG location: Station 67.8
Fuel consumption: 14.7 GPH
Fuel CG: Station 68.0

Solution:

  1. Find the amount of weight change. The aircraft has consumed 14.7 GPH for 1 hour 45 minutes. The total fuel consumed is:

    14.7 GPH × 1.75 hr = 25.7 gal

    Which weighs:

    25.7 gal × 6 lbs/gal = 154 lbs

  2. Determine the new total weight by subtracting the weight of the fuel consumed (154 lbs) from the total weight (4,037 lbs):

    4,037 – 154 = 3,883 lbs new total weight

  3. Find the distance between the original CG (67.8) and the point weight removed (fuel CG = 68.0):

    68.0 – 67.8 = 0.2 inches

  4. Place the three known values into the formula:

    154 lbs

    =

    Change in CG

    3,883 lbs

    0.2 in

    Change in CG =

    154 lbs × 0.2 in

    = 0.01 in

    3,883 lbs

  5. The CG was found to shift approximately 0.01 in. Since the weight was removed aft (68.0 in) of the CG (67.8 in), the CG shifted forward 0.01 in.

    67.80

    original CG

    – 0.01

    forward shift

    67.79

    new CG

Weight Shift

A shift in weight may have a great affect on the CG location even though the total weight may not have changed. The change in CG caused by a weight shift, or the amount of weight to shift in order to move the CG within limits, may be found by the formula below:

Weight shifted

=

Change in CG

Total weight

Distance shifted

Problem:

Determine the new CG after luggage has been shifted, given the following:

Gross weight: 5,000 lbs (including three pieces of luggage)
CG: 98 inches aft of datum (2 inches aft of limits)
2 pieces of luggage: 100 lbs (weighed together)

The luggage is moved from the rear baggage compartment (145 inches aft of datum) to the front compartment (45 inches aft of datum).

Solution:

  1. Find the distance shifted by taking the difference between the original location (baggage compartment 145 inches) and the new location (front compartment 45 inches):

    145 in

    baggage compartment

    – 45 in

    front compartment

    100 in

    distance shifted

  2. Determine the change in CG by substituting the values into the formula:

    100 lbs

    =

    Change in CG

    5,000 lbs

    100 in

    100 lbs × 100 in

    = Change in CG = 2 inches

    5,000 lbs

  3. The CG is moved 2 inches. Since the weight was shifted from the rear compartment (145 inches) to the front compartment (45 inches), the CG moved forward by 2 inches.

    98 in

    original CG

    – 2 in

    CG shift

    96 in

    new CG

The new CG is located 96 inches aft of datum, which places it on the aft limit.

Headwind and Crosswind Components

In general, taking off into a wind improves aircraft performance, and reduces the runway distance required to become airborne. The stronger the wind, the better the aircraft performs. Crosswinds, however, may make the aircraft difficult or impossible to control. The magnitudes of headwind and crosswind components produced by a wind of a given direction and speed can be determined by using the chart as shown in FAA Figure 31.

Problem:

The wind is reported to be from 190° at 15 knots, and you plan to land on runway 13. What will the headwind and crosswind components be?

Solution:

  1. Compute the angle between the wind and the runway:

    190°

    wind direction

    –130°

    runway heading

    60°

    wind angle across runway

  2. Find the intersection of the 60° angle radial line and the 15-knot wind velocity arc on the graph. From the intersection move downward and read the crosswind component of 13 knots. From the point of intersection move to the left and read the headwind component of 7 knots. See Figure 8-3.

Figure 8-3. Wind components

Density Altitude

Aircraft performance charts indicate what performance (rate of climb, takeoff roll, etc.) can be expected of an aircraft under stipulated conditions. Prediction of performance is based upon standard atmospheric conditions, or the International Standard Atmosphere, which at sea level is a temperature of +15°C (+59°F) and an atmospheric pressure of 29.92 "Hg (1013.2 hectopascals).

In a standard atmosphere, temperature changes at a rate of 2°C (3.5°F) per 1,000 feet, and pressure changes approximately 1 "Hg per 1,000 feet. Temperature and/or pressure deviations from standard will change the air density. The result is a value for density altitude, which affects aircraft performance. Performance charts based on density altitude allow the pilot to predict how an aircraft will perform.

Relative humidity also affects density altitude but is not considered when the performance charts are formulated. A combination of high temperature, high humidity, and high altitude will result in a density altitude higher than the pressure altitude, which results in reduced aircraft performance.

Takeoff and Landing Considerations

During takeoff with a crosswind, rudder is used to maintain directional control and aileron pressure to counter the wind. For both conventional and nosewheel-type airplanes, a higher than normal liftoff airspeed is used. Landing under crosswind conditions requires the direction of motion of the airplane and its longitudinal axis to be parallel to the runway at the moment of touchdown. During gusty wind conditions, both the approach and landing should be with power on.

Operating with an uphill runway slope has no effect on the takeoff speed but will increase the take-off distance.

In the event of an actual engine failure immediately after takeoff and before a safe maneuvering altitude is attained, it is not recommended to attempt to turn back to the airport. Instead, it is generally safer to maintain a safe airspeed, and select a field directly ahead or slightly to either side of the takeoff path.

While diverting to an alternate airport due to an emergency, the heading should be changed to establish the new course immediately. Apply rule-of-thumb computations, estimates, and appropriate shortcuts to calculate wind correction, actual distance, and estimated time and fuel required.

When takeoff power is applied, it is usually necessary to hold considerable pressure on the controls to maintain straight flight and a safe climb attitude. Since the airplane is trimmed for the approach (a low power and low airspeed condition), application of maximum allowable power requires considerable control pressure to maintain a climb pitch attitude. The addition of power tends to raise the airplane’s nose suddenly and it veers to the left.

Takeoff and Landing Distance

The takeoff distance graph such as the graph shown in FAA Figure 32, allows the pilot to determine the ground roll and takeoff distance over a 50-foot obstacle.

Problem:

Using the Obstacle Takeoff Graph (FAA Figure 32), determine the total takeoff distance over a 50-foot obstacle under the following conditions:

Temperature: 30°F
Pressure Altitude: 6,000 feet
Weight: 3,300 lbs
Headwind component: 20 knots

Solution:

  1. Enter the chart at 30°F. Proceed upward until intersecting the 6,000-foot pressure altitude. From this point, proceed to the right until intersecting the first reference line.
  2. From this point on the reference line, proceed diagonally upward to the right (on a line spaced proportionally between the existing guide lines) to the vertical line representing 3,300 pounds. From there, proceed to the right until intersecting the second reference line.
  3. From this point, proceed diagonally downward and to the right (remaining proportionally between the existing guide lines) to the vertical line representing the 20-knot headwind component. From there, proceed to the far right scale and read the total takeoff distance over a 50-foot obstacle of 1,500 feet.

The takeoff distance graph is often used to find the required takeoff distance based on temperature, pressure altitude, weight and wind. The graph may also be used however, to find the maximum weight allowable under certain conditions for a particular takeoff distance.

Problem:

Using the information shown in FAA Figure 32, determine the maximum weight allowable to take off over a 50-foot obstacle in 1,500 feet under the following conditions:

Temperature: 30°F
Pressure Altitude: 6,000 feet
Headwind component: 20 knots

Solution:

  1. Enter the chart at 30°F. Proceed upward until intersecting the 6,000-foot pressure altitude line. From this point, proceed to the right until intersecting the first reference line.
  2. From this point, draw a line diagonally upward and to the right (remaining proportionally spaced between the existing guide lines) until intersecting the second reference line.
  3. Re-enter the graph from the right at the 1,500-foot takeoff distance over a 50-foot obstacle. Proceed to the left until intersecting the vertical line representing a 20-knot headwind component. From there, proceed diagonally upward and to the left (remaining proportionally spaced between the existing guide lines) until intersecting the second reference line.
  4. From this point, proceed to the left until intersecting the previously drawn diagonal line. From there, proceed vertically downward and read the maximum allowable takeoff weight of 3,300 pounds.

Use the same procedure for landing distance graphs as for the takeoff distance graph.

Fuel Consumption vs. Brake Horsepower

The fuel consumption vs. brake horsepower graph allows the pilot to determine the gallons per hour used, based on percent of power and mixture settings. See FAA Figure 8.

Problem:

How much flight time is available with a 45-minute fuel reserve under the following conditions?

Fuel on board: 38 gallons
Mixture: Cruise (lean)
Percent power: 55%

Solution:

  1. Determine the fuel consumption (GPH) based on a cruise (lean) setting of 55% power. Locate the point where the diagonal cruise (lean) line and the vertical 55% power line intersect. From the point of intersection, proceed to the left and read the fuel consumption: 11.4 GHP.
  2. Find the total endurance possible with 38 gallons at 11.4 GHP.

    38 ÷ 11.4 = 3.33 hr (no reserve)

  3. Determine the flight time available with a 45-minute (0.75 hours) fuel reserve remaining:

    3.33 hr

    no reserve

    –0.75 hr

    45-minute reserve

    2.58 hr

    with 45-minute reserve

    2.58 hr = 2 hours, 34 minutes

Time, Fuel, and Distance to Climb

The time, fuel and distance to climb tables and graphs allow the pilot to calculate the time required, fuel used and distance covered during a climb to a specified altitude.

Table Method

Problem:

Referring to the time, fuel and distance to climb table (FAA Figure 9), determine the amount of fuel used from engine start to a pressure altitude (PA) of 12,000 feet using a normal rate of climb under the following conditions:

Weight: 3,800 lbs
Airport Pressure altitude: 4,000 feet
Temperature at 4,000 feet: 26°C

Solution:

  1. Locate the section appropriate to 3,800 pounds weight. Read across the 4,000-foot pressure altitude line to the entry under fuel used: 12 pounds.
  2. Read across the 12,000-foot pressure altitude line to the entry under fuel used: 51 pounds.
  3. Calculate the fuel required to climb:

    51 – 12 = 39 lbs

  4. Apply Note #2. (A temperature of 26°C is +19°C, with respect to the standard atmosphere at 4,000 feet.)

    39 × 1.19 = 46.4 lbs

  5. Apply Note #1:

    46.4 lbs + 12.0 lbs start and taxi = 58.4 lbs total

Graph Method

Problem:

Using the time, fuel, and distance to climb graph shown in FAA Figure 15, determine the fuel, time, and distance required to climb to cruise altitude under the following conditions:

Airport pressure altitude: 2,000 feet
Airport temperature: 20°C
Cruise pressure altitude: 10,000 feet
Cruise temperature: 0°C

Solution:

  1. Determine the fuel, time and distance to climb to 10,000 feet at 0°C. Locate the temperature at 0°C on the bottom left side of the graph. Proceed upward to the diagonal line representing the pressure altitude of 10,000 feet.
  2. From this point proceed to the right and stop at each of the lines, forming three points of intersection.
  3. From each of the three points, draw a line downward to read the fuel used (6 gals), time (11 min) and distance (16 NM). These are the values for a climb from sea level to 10,000 feet.
  4. Repeat the steps to find the values for a climb from sea level to 2,000 feet. The fuel used is 1 gallon, time is 2 minutes, and distance is 3 NM.
  5. Find the fuel, time, and distance values for a climb from 2,000 feet to 10,000 feet, by taking the differences for the two altitudes:

    Fuel

    6 (for 10,000 feet) – 1 (for 2,000 feet) = 5 gallons

    Time

    11 (for 10,000 feet) – 2 (for 2,000 feet) = 9 minutes

    Distance

    16 (for 2,000 feet) – 3 (for 2,000 feet) = 13 NM

Cruise Performance Table

The cruise performance table may be used to determine the expected percent power, true airspeed and fuel flow for a particular altitude and power setting. Based on this information, the pilot can find the estimated time en route and fuel required.

Problem:

Using the Cruise Performance Table (FAA Figure 12), find the approximate flight time available under the following conditions allowing for VFR day fuel reserve:

Pressure Altitude: 18,000 feet
Temperature: -1°C
Power (Best fuel economy): 22 RPM — 20 in. manifold pressure
Usable fuel: 344 lbs

Solution:

  1. Determine the flight time available by finding the rate of fuel consumption. Enter the table on the left column (RPM) and read down to 2,200 RPM.
  2. Find line for 20 in. MP under the second column (MP). Follow to the right, to the column that represents a temperature of -1°C (20°C above standard temperature).
  3. Continue to the right, noting 43% BHP, and in the next column, read the true airspeed of 124 knots, and a fuel flow of 59 pounds per hour (pph).
  4. Find the fuel rate for best fuel economy. The chart notes in the upper right-hand corner state for best fuel economy at 70% power or less, operate at 6 pph leaner than shown in chart. For the given conditions, the engine is operating at 43% BHP, so the note does apply. The fuel burn for best fuel economy is brought down to:

    59 – 6 = 53 pph

  5. Determine the approximate flight time based on 344 lbs of usable fuel and a fuel rate of 53 pph:

    344 ÷ 53 = 6.50 hours (no reserve)

  6. Find the flight time with VFR day fuel reserve (30 minutes):

    6.50

    no reserve

    –0.50

    VFR day reserve

    6.00

    with reserve

Cruise and Range Performance Table

Referring to the cruise and range performance table, a pilot can determine true airspeed, fuel consumption, endurance, and range based on altitude and power setting.

Problem:

Using FAA Figure 11, find the approximate true airspeed and fuel consumption per hour at an altitude of 7,500 feet, 52 percent power.

Solution:

  1. Enter the chart at the altitude to be flown (7,500 feet) on the left.
  2. Under the percent brake horsepower (BHP) column, refer to the line representing 52 percent power. Read to the right and find the true airspeed (105 mph) and fuel consumption (6.2 GPH).

Maximum Rate of Climb

The maximum rate of climb table illustrates the expected climb performance based on pressure altitude and temperature.

Problem:

Determine the maximum rate of climb under the given conditions:

Weight: 3,400 lbs
Pressure altitude: 7,000 feet
Temperature: +15°C

Solution:

To find the maximum rate of climb for 7,000 feet at 15°C, interpolation is necessary for both altitude and temperature. The interpolation may be accomplished in any order.

  1. To find the value for 15°C, interpolation is used between 0° and 20°C. The difference between the two temperatures at 4,000 feet is 155 fpm (1,220 fpm – 1,065 fpm). The temperature of 15°C is 3/4 of the way between the 0°C and 20°C value. At 15°C for 4,000 feet, the rate of climb is 1,104 fpm.

    1,220 – 116-3/4 = 1,104

    or 1,065 + 39-1/4 = 1,104

  2. The same method is applied to 8,000 feet to find the value for 15°C. The difference between 0°C and 20°C is 155 fpm (1,110 fpm – 955 fpm). Taking 3/4 of 155 fpm also gives 116 fpm. The rate of climb at 8,000 feet is 994 fpm.

    1,110 – 116 = 994 or 955 + 39 = 994

  3. Interpolate for an altitude of 7,000 feet. The altitude of 7,000 feet is 3/4 of the way between 4,000 feet and 8,000 feet. The overall difference is 110 fpm, 3/4 of which would be 82.5 fpm. The rate for 7,000 feet is 1,021 fpm.

    1,114 – 82.5 = 1,031

    or 994 + 27.5 = 1,021

Glide Distance

Glide distance can be found by using a chart such as FAA Figure 3A.

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 8 nautical miles.

[10-2024]