Headwind and Crosswind Components
Takeoff and Landing Considerations
Fuel Consumption vs. Brake Horsepower
Time, Fuel, and Distance to Climb
Cruise and Range Performance Table
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.
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:
Construct a table, with column headings as follows:
Item | Weight | Arm | Moment |
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 |
Total the weights and moments:
|
Weight |
Moment |
|
|
160 |
7,200 |
|
|
170 |
24,650 |
|
|
+ 105 |
+ 19,425 |
|
|
435 lbs |
51,275 lbs-in |
Determine the CG by dividing the total moment by the total weight:
CG = |
Total moment |
Total weight |
CG = 51,275 ÷ 435 = 117.8 inches
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:
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 |
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 |
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:
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
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
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
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 |
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 |
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:
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 |
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 |
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.
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:
Compute the angle between the wind and the runway:
190° |
wind direction |
–130° |
runway heading |
60° |
wind angle across runway |

Figure 8-3. Wind components
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.
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.
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:
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:
Use the same procedure for landing distance graphs as for the takeoff distance graph.
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:
Find the total endurance possible with 38 gallons at 11.4 GHP.
38 ÷ 11.4 = 3.33 hr (no reserve)
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 | |
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.
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:
Calculate the fuel required to climb:
51 – 12 = 39 lbs
39 × 1.19 = 46.4 lbs
Apply Note #1:
46.4 lbs + 12.0 lbs start and taxi = 58.4 lbs total
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:
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 |
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:
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
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)
Find the flight time with VFR day fuel reserve (30 minutes):
6.50 |
no reserve |
–0.50 |
VFR day reserve |
6.00 |
with reserve |
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:
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.
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
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
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 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:
[10-2024]