Density Altitude and Aircraft Performance
Landing Distance Graphs and Tables
Headwind and Crosswind Component Graph
Computing Weight and Balance Problems
Aircraft performance charts show a pilot what can be expected of an airplane (rate of climb, takeoff roll, etc.) under stipulated conditions. Prediction of performance is based upon a sea level temperature of +15°C (+59°F) and atmospheric pressure of 29.92 "Hg (1013.2 mb). This combination of temperature and pressure is called a standard day. When the air is at a standard density, temperature and/or pressure deviations from standard will change the air density or the density altitude, which affects aircraft performance. Performance charts 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 result in a density altitude higher than the pressure altitude which, in turn, results in reduced aircraft performance.
Problem:
Using the Density Altitude Chart shown in FAA Figure 8, and the following conditions, determine the density altitude.
Conditions:
Altimeter setting: 30.35
Airport temperature: +25°F
Airport elevation: 3,894 feet
Solution:
Determine the applicable altitude correction for the altimeter setting of 30.35 "Hg. See FAA Figure 8. That setting is not shown on the chart, so it is necessary to interpolate between the correction factors shown for 30.30 "Hg and 30.40 "Hg. To interpolate, add the two factors and divide by 2:
-348 + (-440) = -788
-788 ÷ 2 = -394
Since the result is a negative number, subtract that value from the given airport elevation:
|
3,894 |
feet |
|
– 394 |
feet |
|
3,500 |
feet |
Along the bottom of the chart, locate the given OAT (+25°F). From that point, proceed upward until intersecting the pressure altitude line that is equal to the corrected airport elevation (3,500 feet). From that point, proceed to the left and read the density altitude (2,000 feet).
Note that high-density altitude reduces propeller efficiency as well as overall aircraft performance.
The Takeoff Distance Graph, FAA Figure 40, allows the pilot to determine the ground roll required for takeoff under various conditions. It also shows the total distance required for a takeoff and climb to clear a 50-foot obstacle.
Problem:
Using the Takeoff Distance Graph shown in FAA Figure 40, determine the approximate ground roll distance for takeoff and the total distance for a takeoff to clear a 50-foot obstacle under the following conditions:
OAT: 90°F
Pressure altitude: 2,000 feet
Takeoff weight: 2,500 lbs
Headwind component: 20 knots
Solution:
Some landing distance graphs such as the one shown in FAA Figure 37, are used in the same manner as the takeoff distance graph.
Another type of landing distance table is shown in FAA Figure 38.
Problem:
Using the Landing Distance Table shown in FAA Figure 38, determine the total distance required to land over a 50-foot obstacle.
Pressure altitude: 5,000 feet
Headwind: 8 knots
Temperature: 41°F
Runway: Hard surface
Solution:
Enter FAA Figure 38 at 5,000 feet and 41°F to find 1,195 feet is required to clear a 50-foot obstacle. Note #1 states an additional correction is necessary for the headwind, 10% for each 4 knots. This would mean 20% for 8 knots. 20% of 1,195 feet is 239 feet, resulting in 956 feet—total distance required to land:
1,195 × .20 = 239 feet
1,195 – 239 = 956 feet
In general, taking off into a wind improves aircraft performance, and reduces the length of runway 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 aircraft manufacturer determines the safe limit for taking off or landing with a crosswind and establishes the maximum allowable crosswind component. The graph shown in FAA Figure 36 is used to determine what extent a wind of a given direction and speed is felt as a headwind and/or crosswind.
Problem:
The wind is reported to be from 085° at 30 knots, and you plan to land on Runway 11. What will the headwind and crosswind components be?
Solution:
Determine the angular difference between the wind direction and the runway:
|
110° |
runway |
|
– 085° |
wind |
|
025° |
difference |
All aircraft leave two types of wake turbulence: prop or jet blast, and wing-tip vortices.
The prop or jet blast could be hazardous to light aircraft on the ground behind large aircraft which are either taxiing or running-up their engines. In the air, jet or prop blast dissipates rapidly.
Wing-tip vortices are a by-product of lift. When a wing is flown at a positive angle of attack, a pressure differential is created between the upper and lower wing surfaces, and the pressure above the wing will be lower than the pressure below the wing. In attempting to equalize the pressure, air moves outward, upward, and around the wing tip, setting up a vortex which trails behind each wing. See Figure 4-1.

Figure 4-1. Wing-tip vortices
The strength of a vortex is governed by the weight, speed, and the shape of the wing of the generating aircraft. Maximum vortex strength occurs when the generating aircraft is heavy, clean, and slow.
Vortices generated by large aircraft in flight tend to sink below the flight path of the generating aircraft. A pilot should fly at or above the larger aircraft’s flight path in order to avoid the wake turbulence created by the wing-tip vortices. See Figure 4-2.

Figure 4-2. Vortices in cruise flight
Close to the ground, vortices tend to move laterally. A crosswind will tend to hold the upwind vortex over the landing runway, while a tailwind may move the vortices of a preceding aircraft forward into the touchdown zone.
To avoid wake turbulence when landing, a pilot should note the point where a preceding large aircraft touched down and then land past that point. See Figure 4-3.

Figure 4-3. Touchdown and wake end
On takeoff, lift off should be accomplished prior to reaching the rotation point of a preceding large aircraft; the flight path should then remain upwind and above the preceding aircraft’s flight path. See Figure 4-4.

Figure 4-4. Rotation and wake beginning
Ground effect occurs when flying within one wingspan or less above the surface. The airflow around the wing and wing tips is modified and the resulting pattern reduces the downwash and the induced drag. These changes can result in an aircraft becoming airborne before reaching recommended takeoff speed or floating during an approach to land. See Figure 4-5.

Figure 4-5. Ground effect phenomenon
An airplane leaving ground effect after takeoff will require an increase in angle of attack to maintain the same lift coefficient, which in turn will cause an increase in induced drag and, therefore, require increased thrust.
The Cruise Power Setting Table (FAA Figure 35) may be used to forecast fuel flow and true airspeed and, therefore allow a pilot to determine the amount of fuel required and the estimated time en route.
Problem:
Using the Cruise Power Setting Table shown in FAA Figure 35, determine the expected fuel consumption for a 1,000-statute-mile flight under the following conditions:
Pressure altitude: 8,000 feet
Temperature: -19°C
Manifold pressure: 19.5 "Hg
Wind: Calm
Solution:
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 some weight less than the maximum allowable.
Some of the problems caused by overloading an aircraft are:
The empty weight is obtained from appropriate charts. It includes the airframe, engine, all fixed equipment, and unusable fuel and oil. Some aircraft include all oil in the aircraft empty weight. 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.
The arm 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. See Figure 4-6.

Figure 4-6. Positive and negative arm
The moment is the product of the weight of an object multiplied by its arm and is expressed in pound-inches (lbs-in). A formula that is used to find moment is usually expressed as follows:
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. This is usually expressed as MOM/100 or MOM/1000, etc.
The center of gravity (CG) is the point about which an aircraft will balance, and it is expressed in inches from datum. The center of gravity is found by dividing the total moment by the total weight, and the formula is usually expressed as follows:
Total moment |
= CG (inches aft of datum) |
Total weight |
Standard weights have been established for numerous items involved in weight and balance computations. These weights should not be used if actual weights are available. Some of the standard weights are:
|
Item |
Weight |
|
General aviation crew and passenger |
190 lbs each |
|
Gasoline |
6 lbs/U.S. gallon |
|
Oil |
7.5 lbs/U.S. gallon |
|
Water |
8.35 lbs/U.S. gallon |
In addition to considering the weight to be carried, the pilot must ensure that 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 airplane is considered to be concentrated. For an aircraft to be safe to fly, the center of gravity must fall between specified limits. All aircraft categories have unique balance requirements; some need to be balanced precisely while others are less sensitive to fore and aft weight distribution. The specific balance requirements for each aircraft category are covered in the pilot’s operating handbook (POH) and will be tested on during the practical (checkride). 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 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.
Weight and loading is typically used for weight-shift control. The cart normally does not have balance requirements, but the attachment point to the cart is specified in the POH.
Weight and loading is used for powered parachutes. The POH procedure for the fore and aft attachment points of the wing to the cart must be followed so that the cart will obtain the proper deck angle (the angle of the cart bottom with the horizontal plane).
Problem:
Determine if the airplane weight and balance is within limits.
Front seat occupants: 340 lbs
Rear seat occupants: 295 lbs
Fuel (main wing tanks): 44 gal
Baggage: 56 lbs
Solution:
Determine the total weight. Using the information provided in the question, enter the items and the weights:
|
Item |
Weight (lbs) |
Arm (in) |
Moment (lbs-in) |
|
Front seat occupants |
340 |
||
|
Rear seat occupants |
295 |
||
|
Fuel (main wing tanks) |
264* |
||
|
Baggage |
56 |
||
|
Empty weight |
+ 2,015** |
||
|
2,970 |
|||
|
* 44 gal × 6 lbs/gal |
|||
Fill in the arms (found in FAA Figure 32) and calculate the moments. Moments are determined by multiplying in each row: weight × arm. Then divide that moment by 100 to keep the number a manageable size, and finally, total all moments.
|
Item |
Weight (lbs) |
Arm (in) |
Moment/100 (lbs-in) |
|
Front seat occupants |
340 |
85 |
289.0 |
|
Rear seat occupants |
295 |
121 |
357.0 |
|
Fuel (main wing tanks) |
264 |
75 |
198.0 |
|
Baggage |
56 |
140 |
78.4 |
|
Empty weight |
+ 2,015 |
1,554.0 |
|
|
2,970 |
2,476.4 |
Problem:
Referring to FAA Figure 34, determine the maximum amount of baggage that may be loaded aboard the airplane for the CG to remain within the loading envelope.
|
Item |
Weight (lbs) |
Moment/1,000 (lbs-in) |
|
Empty weight |
1,350 |
51.5 |
|
Pilot, front passenger |
250 |
— |
|
Rear passengers |
400 |
— |
|
Baggage |
— |
— |
|
Fuel 30 gal. |
— |
— |
|
Oil 8 qt. |
— |
— |
Solution:
Determine the total weight. Using the information provided in the question, enter the items and the weights:
|
Item |
Weight (lbs) |
Moment/1,000 (lbs-in) |
|
Pilot, front passenger |
250 |
|
|
Rear passengers |
400 |
|
|
Fuel |
180* |
|
|
Oil |
15** |
|
|
Baggage |
— |
|
|
Empty weight |
+ 1,350† |
|
|
2,195 |
||
|
*30 gal × 6 lbs/gal |
||
Determine the moments. Using FAA Figure 34, locate the moments for all the given weights:
|
Item |
Weight (lbs) |
Moment/1,000 (lbs-in) |
||
|
Pilot, front passenger |
250 |
9.3 |
||
|
Rear passengers |
400 |
29.3 |
||
|
Fuel |
180 |
8.7 |
||
|
Oil |
15 |
-0.2* |
||
|
Baggage |
— |
— |
||
|
Empty weight |
+ 1,350 |
51.5 |
||
|
2,195 |
98.6 |
|||
|
*from note in center of FAA Figure 34 |
||||
From the Center of Gravity Moment Envelope, at the bottom of FAA Figure 34, determine the maximum allowable gross weight of the aircraft, as indicated by the top of the “Normal Category” envelope (2,300 pounds). Comparing the total weight of this problem with the maximum allowable weight shows that 105 pounds of baggage could be carried:
|
2,300 |
|
– 2,195 |
|
105 |
From the Loading Graph, determine the MOM/1000 of 105 pounds of baggage. Add that amount to the previous total MOM/1000:
|
Item |
Weight (lbs) |
Moment/1,000 (lbs-in) |
|
|
Totals |
2,195 |
98.6 |
|
|
Baggage |
+ 105 |
+ 10.0 |
|
|
New Totals |
2,300 |
108.6 |
[10-2024]