Aircraft Performance (Aircraft performance)

(Refer to figure 8.) What is the effect of a temperature increase from 35 to 50 °F on the density altitude if the pressure altitude remains at 3,000 feet MSL? View Figure 8

1,000-foot increase. Enter chart at 35°F. Move up to 3,000 Ft. pressure line. Read left the density altitude of 1,900 Ft. Reenter chart at 50 °F. move up to 3,000 Ft. pressure line. Read left, density altitude of 2,900. The difference between the two is a 1,000 foot increase.

(Refer to figure 38.) Determine the total distance required to land over a 50-foot obstacle. View Figure 38 Pressure altitude 7,500 ft Headwind 8 kts Temperature 32 F RunwayHard surface

1,004 feet. Per the chart, the distance to clear a 50' obstacle for 7,500' is 1,255'. Apply note #1 for the 8 knots of headwind and decrease the distance by 20%. 1,255 x .20 = 251. 1,255 - 251 = 1,004.

(Refer to figure 40.) Determine the approximate ground roll distance required for takeoff. View Figure 40 OAT 38 C Pressure altitude 2,000 ft Takeoff weight 2,750 lb Headwind component Calm

1,150 feet. Move up from 38° C to 2,000'. Move right to the ref. line. Move right & down to 2,750. Move right to the next ref. line. Since there is no headwind, move right to the third ref. line. Move right to find a ground roll of 1,150'.

(Refer to figure 8.) What is the effect of a temperature increase from 30 to 50 °F on the density altitude if the pressure altitude remains at 3,000 feet MSL? View Figure 8

1,300-foot increase. Move up from 30° F to a 3,000' pressure altitude and you will find a density altitude of 1,600'. Move up from 50° F to a 3,000' pressure altitude and you will find a density altitude of 2,900'. The difference is an increase of 1,300'.

(Refer to figure 8.) Determine the pressure altitude at an airport that is 1,386 feet MSL with an altimeter setting of 29.97. View Figure 8

1,341 feet MSL. Using the table to the right of the chart, interpolate between the 29.92 and 30.0 correction values to find a value of -45.63'. Decrease the airport elevation by 45.63'. This gives an answer of 1,340.37'. 1,341' is the closest answer.

(Refer to Figure 40.) What is the approximate ground roll distance necessary for takeoff under the following conditions? OAT - 100°F, Pressure altitude - 4,000 ft, Takeoff weight - 2,700 lb, Headwind component - Calm View Figure 40

1,400 feet. Enter the chart at 100° F. and draw a vertical line to the 4,000 pressure altitude line. Draw a horizontal line to the first reference line and proceed proportionally down to the right to intersect the 2,700 pound takeoff weight line. Now draw a horizontal line all the way to the right and read approximately 1,400 feet because the wind component and obstacle height are 0.

(Refer to Figure 40.) What is the necessary distance required for takeoff to clear a 50-foot obstacle? OAT - Std, Pressure altitude - Sea level, Takeoff weight - 2,700 lb, Headwind component - Calm View Figure 40

1,400 feet. Enter the chart at Standard OAT of 15 degrees Celsius. Draw a vertical line up until it intersects the S.L. pressure altitude curve. Draw a horizontal line to the right until it intersects with the weight reference line. Follow the contour of the weight lines until it intersects with the 2,700 lbs. Move horizontally across to the 0 wind component line. Now move horizontally to the right to the 0 obstacle height line and move proportionally up to the right and read just below 1,400'.

Refer to figure 37.) Determine the total distance required to land. View Figure 37 OAT 32 F Pressure altitude 8,000 ft Weight 2,600 lb Headwind component20 kts Obstacle 50 ft

1,400 feet. Move up from 32° F to 8,000'. Move right to the ref. line. Move right & down to 2,600. Move right to the next ref. line. Move right & down to the 20 knot wind. Move right to the third ref. line. Move right & up to the 50' height to find 1,400'.

(Refer to Figure 37.) What is your total distance to land using the data below? OAT - 90°F, Pressure altitude - 6,000 ft, Weight - 2,700 lb, Headwind component - 20 kts, Obstacle - 50 ft View Figure 37

1,550 feet. Enter the chart at 90° F. and draw a vertical line to the 6,000 pressure altitude line. Draw a horizontal line to the first reference line and proceed proportionally down to the right to intersect the 2,700 pound takeoff weight line. Proceed to the right to the second reference line and move down to the right to intersect the 20 kts. headwind component. Read straight to the right to the third reference line and move proportionately up and read approximately 1,550 feet.

(Refer to figure 8.) What is the effect of a temperature increase from 25 to 50 °F on the density altitude if the pressure altitude remains at 5,000 feet?

1,650-foot increase. Move up from 25° F to a 5,000' pressure altitude and you will find a density altitude of 3,850'. Move up from 50° F to a 5,000' pressure altitude and you will find a density altitude of 5,500'. The difference is 1,650'.

(Refer to figure 8.) What is the effect of a temperature decrease and a pressure altitude increase on the density altitude from 90 °F and 1,250 feet pressure altitude to 55 °F and 1,750 feet pressure altitude? View Figure 8

1,700-foot decrease. Move up from 90° F to a pressure altitude (PA) of 1,250' to find a density altitude (DA) of 3,500'. Move up from 55° F to a PA of 1,750' to find a DA of 1,800'. A decrease of 1,700'.

(Refer to figure 40.) Determine the total distance required for takeoff to clear a 50-foot obstacle. View Figure 40 OAT Std Pressure altitude 4,000 ft Takeoff weight 2,800 lb Headwind component Calm

1,750 feet. Move up the ISA line to 4,000'. Move right to the ref. line. Move right & down to 2,800. Move right to the next ref. line. Since there is no headwind, move right to the third ref. line. Move right & up to the 50' height to find 1,750'.

(Refer to figure 37.) Determine the approximate total distance required to land over a 50-foot obstacle. View Figure 37 OAT 90 F Pressure altitude 4,000 ft Weight 2,800 lb Headwind component 10 kts

1,775 feet. Move up from 90° F to 4,000'. Move right to the ref. line. Move right & down to 2,800. Move right to the next ref. line. Move right & down to the 10 knot wind. Move right to the third ref. line. Move right & up to the 50' height to find 1,775'.

(Refer to Figure 37.) What is your total distance to land using the data below? OAT - Std, Pressure altitude - 6,000 ft, Weight - 2,700 lb, Wind component - Calm, Obstacle - 50 ft.

1,900 feet. Find where the pressure altitude of 6,000 feet intersects the ISA curve. Draw a line from there to the right until it intersects the first reference line and proceed proportionately down and to the right until it intersects the 2,700 lb. line. From there move to the right all the way to the third reference line since the wind component is 0. Move proportionately up the line to the right to account for the 50 ft. obstacle and read approximately 1,900 feet.

Refer to figure 35.) What fuel flow should a pilot expect at 11,000 feet on a standard day with 65 percent maximum continuous power? View Figure 35

11.2 gallons per hour. The entire table is representative of 65% power. Interpolating between a 10,000 foot flow of 11.5 and a 12,000 foot flow of 10.9 results in a value of 11.2 gph.

How far will an aircraft travel in 7.5 minutes with a ground speed of 114 knots?

14.25 NM. Use the formula D = GS x T (distance = groundspeed times time). Divide 7.5 by 60 to convert to hours, or 0.125 hours. Then multiply 114 x 0.125 = 14.25.

(Refer to figure 35.) Approximately what true airspeed should a pilot expect with 65 percent maximum continuous power at 9,500 feet with a temperature of 36 °F below standard? View Figure 35

183 MPH. Use -36° F columns. Interpolate between 8,000 and 10,000 to find the value for 9,500. For 8,000 the TAS is181 MPH, for 10,000 the TAS is 184 MPH. 181 + 184 / 2 = 182.5 (this is the TAS for 9,000). 182.5 + 184 / 2 = 183.25 (this is the TAS for 9,500).

(Refer to figure 36.) What is the crosswind component for a landing on Runway 18 if the tower reports the wind as 220° at 30 knots? View Figure 36

19 knots. The direction for runway 18 is 180°. The wind is from 220° at 30 knots. 220 - 180 = 40° crosswind. Move up the 40° line until it intersects the 30 wind velocity arc. Move straight down to find a 19 knot crosswind component.

(Refer to Figure 40.) Determine the total distance required for takeoff to clear a 50-foot obstacle. OAT - Std, Pressure altitude - 6,000 ft, Takeoff weight - 2,800 lb, Headwind component - Calm

2,000 feet. Find where the pressure altitude of 6,000 feet intersects the ISA curve. Draw a horizontal line to the right until it intersects with the weight reference line. Follow the contour of the weight lines until it intersects with the 2,800 lbs. From the intersection of 2,800 lbs., draw a horizontal line to the right until it intersects with the wind reference line. Since the winds are calm, there is no adjustment to be made for winds, so continue the horizontal line to the obstacle reference line. Follow the contour of the obstacle lines until it intersects with the 50' obstacle line and read approximately 2,000 feet.

(Refer to Figure 8.) Determine the density altitude for these conditions: Altimeter setting - 30.30, Runway temperature - +25°F, Airport elevation - 3,894 it MSL View Figure 8

2,200 feet MSL. On the Altimeter setting chart find 30.3. Its conversion factor is -348. Subtract this from the airport elevation of 3,894. (3,894-348=3546) Enter the Density Altitude chart at +25°F and go up to the 3,546pressure altitude line and read left to find density altitude of approximately 2,200feet.

(Refer to figure 8.) Determine the pressure altitude with an indicated altitude of 1,380 feet MSL with an altimeter setting of 28.22 at standard temperature. View Figure 8

2,991 feet MSL. Use the table to the right of the chart. Interpolate between settings of 28.2 and 28.3 to find an adjustment of 1,610.6' for 28.22. Round this to 1,611 and add it to the 1,380 foot value to find the answer.

(Refer to figure 35.) Determine the approximate manifold pressure setting with 2,450 RPM to achieve 65 percent maximum continuous power at 6,500 feet with a temperature of 36 °F higher than standard. View Figure 35

21.0 inches Hg. Using the columns for +36° F, interpolate between the 6,000' setting of 21.0 and the 8,000' setting of 20.8 to find a value of 20.95 for 6,500'. Rounded to the nearest tenth, this would be a value of 21.0" Hg.

(Refer to figure 36.) What is the headwind component for a landing on Runway 18 if the tower reports the wind as 220° at 30 knots? View Figure 36

23 knots. The direction for runway 18 is 180°. The wind is from 220° at 30 knots. 220 - 180 = 40° crosswind. Move up the 40° line until it intersects the 30 wind velocity arc. Move straight left to find a 23 knot headwind component.

(Refer to figure 36.) What is the maximum wind velocity for a 30° crosswind if the maximum crosswind component for the airplane is 12 knots?

24 knots. Enter the chart at the 12 knot crosswind component line and move up vertically until it intersects the 30° line. The intersection is just below the midway point between the 20 and 30 wind velocity arcs or about 24 knots.

(Refer to figure 8.) Determine the pressure altitude at an airport that is 3,563 feet MSL with an altimeter setting of 29.96. View Figure 8

3,527 feet MSL. Using the table to the right of the chart, interpolate between the 29.92 and 30.0 correction values to find a value of -36.5'. Decrease the airport elevation by 36.5'. This gives an answer of 3,526.5'. Round this to 3,527'.

(Refer to Figure 35.) Determine your fuel consumption for a 500 nautical mile flight using the following conditions? Pressure altitude - 6,000 ft, Temperature - +26°C, Manifold pressure - 21"" Hg, Wind - Calm View Figure 35

35.6 gallons. Enter the chart at a pressure altitude of 6,000 feet. Find the part of the chart that indicates +26°C—That is on the ISA chart of +20°C. Read 11.5 gallons per hour and a true airspeed of 161 knots. With your flight computer figure your time en route of 3 hrs. and 6 min. Now calculate the fuel consumption as 35.6 gallons.

(Refer to figure 38.) Determine the approximate landing ground roll distance. View Figure 38 Pressure altitude 1,250 ft Headwind 8 kts Temperature Std

366 feet. 1,250' is between sea level and 2,500'. Interpolate: 445 + 470 / 2 = 457.5. Apply note #1 for the 8 knots of headwind and decrease the distance by 20%. 457.5 x .20 = 91.5'. 457.5 - 91.5 = 366'.

(Refer to figure 38.) Determine the approximate landing ground roll distance. View Figure 38 Pressure altitude Sea level Headwind 4 kts Temperature Std

401 feet. From the chart, the ground roll at sea level is found to be 445'. Apply note #1 for the 4 knots of headwind and decrease the distance by 10%. 445 x .10 = 44.5'. 445 - 44.5 = 400.5. Round this value to 401'.

(Refer to figure 38.) Determine the approximate landing ground roll distance. View Figure 38 Pressure altitude 5,000 ft Headwind Calm Temperature 101 F

545 feet. Per the chart, the ground roll for 5,000' is 495'. Apply note #2 for the non standard temperature and increase the distance by 10%. 495 x .10 = 49.5. 495 + 49.5 = 544.5.

(Refer to figure 40.) Determine the approximate ground roll distance required for takeoff. View Figure 40 OAT 32 C Pressure altitude 2,000 ft Takeoff weight 2,500 lb Headwind component 20 kts

650 feet. Move up from 32° C to 2,000'. Move right to the ref. line. Move right & down to 2,500. Move right to the next ref. line. Move right & down to the 20 knot wind. Move right to find a ground roll of 650'.

(Refer to figure 35.) What is the expected fuel consumption for a 1,000-nautical mile flight under the following conditions? View Figure 35 Pressure altitude 8,000 ft Temperature 22° C Manifold pressure 20.8 inches Hg Wind Calm

70.1 gallons. Per the chart, fuel flow is 11.5 GPH at 164 KTS TAS at 8,000' PA and 22° C IOAT. With a flight computer, calculate the no wind time en route for the 1,000 miles as 6 h 6 m and the fuel consumption as 70.1 gallons.

(Refer to figure 8.) Determine the density altitude for these conditions: View Figure 8 Altimeter setting 29.25 Runway temperature +81 F Airport elevation 5,250 ft MSL

8,500 feet MSL. Determine the pressure altitude (PA) correction for the 29.25 setting by interpolating between 29.2 and 29.3 to find a value of 626'. Add this to the 5,250' value for a PA of 5,876'. Move up from 81° F to the PA and read the density altitude at left.

(Refer to Figure 40.) What is the approximate ground roll distance necessary for takeoff under the following conditions? OAT - 90°F, Pressure altitude - 4,000 ft, Takeoff weight - 2,600 lb, Headwind component - 20 kts View Figure 40

800 feet. Enter the chart at 90° F. and draw a vertical line to the 4,000 pressure altitude line. Draw a horizontal line to the first reference line and proceed proportionally down to the right to intersect the 2,600 pound takeoff weight line. Now draw a horizontal line to the right to the 0 wind component line and proceed proportionately down to the right to the 20 knot wind component line. From there move horizontally to the right and read approximately 800 feet.

(Refer to figure 38.) Determine the total distance required to land over a 50-foot obstacle. View Figure 38 Pressure altitude 3,750 ft Headwind 12 kts Temperature Std

816 feet. 3,750' is between 2,500' and 5,000'. Interpolate: 1,135 + 1,195 / 2 = 1,165. Apply note #1 for the 12 knots of headwind and decrease the distance by 30%. 1,165 x .30 = 349.5. 1,165 - 349.5 = 815.5. Round this value to 816'.

(Refer to figure 38.) Determine the total distance required to land over a 50-foot obstacle. View Figure 38 Pressure altitude 5,000 ft Headwind 8 kts Temperature 41 F Runway Hard surface

956 feet. Per the chart, the distance to clear a 50' obstacle for 5,000' is 1,195'. Apply note #1 for the 8 knots of headwind and decrease the distance by 20%. 1,195 x .20 = 239. 1,195 - 239 = 956'.

Ground effect is most likely to result in which problem?

Becoming airborne before reaching recommended takeoff speed. When flown near the ground, the vertical component of airflow is restricted and modified. A wing operating in ground effect produces less drag and is more efficient. This can allow takeoff at slower than recommended speeds.

What effect does high density altitude, as compared to low density altitude, have on propeller efficiency and why?

Efficiency is reduced because the propeller exerts less force at high density altitudes than at low density altitudes. High density altitude conditions reduce prop efficiency due to a reduction of air molecules for a given volume of air. Therefore, the higher the density altitude the less efficient the propeller becomes.

Which combination of atmospheric conditions will reduce aircraft takeoff and climb performance?

High temperature, high relative humidity, and high density altitude. High temperature and high relative humidity contribute to lower air density which is by definition a higher density altitude. High density altitude decreases the power output of the engine and the efficiency of airfoils, decreasing performance.

What must a pilot be aware of as a result of ground effect?

Induced drag decreases; therefore, any excess speed at the point of flare may cause considerable floating. When the wing is under the influence of ground effect, there is a reduction of upwash, downwash, and tip vortices. As a result of the reduced tip vortices, induced drag is reduced. The reduced drag can cause floating on landing.

What effect, if any, does high humidity have on aircraft performance?

It decreases performance. A molecule of water vapor weighs less than a molecule of dry air. For a given volume of air, moist air weighs less, is less dense, than dry air. Lower air density decreases performance.

What effect does high density altitude have on aircraft performance?

It reduces climb performance. High density altitude is equivalent to lower air density. Lower air density decreases the power output of the engine and the efficiency of the airfoils, decreasing climb performance.

Maneuvering speed is:

Not marked on the airspeed indicator. Design maneuvering speed (VA)—the maximum speed at which the structural design's limit load can be imposed (either by gusts or full deflection of the control surfaces) without causing structural damage. Maneuvering speed varies depending on the weight of the airplane. It should be determined from the particular airplane's operating limitations provided by the manufacturer.

(Refer to figure 36.) With a reported wind of south at 20 knots, which runway (10, 14, or 24) is appropriate for an airplane with a 13-knot maximum crosswind component? View Figure 36

Runway 14. Runway 14 is the closest to the wind direction of 180. Move up the 40° line to the 20 knot arc. Move straight down to find a crosswind component just under 13 knots. Therefore, the aircraft can safely land on runway 14 under the given conditions.

(Refer to figure 36.) With a reported wind of north at 20 knots, which runway (6, 29, or 32) is acceptable for use for an airplane with a 13-knot maximum crosswind component?

Runway 32. Runway 32 is the closest to the wind direction of 360. Move up the 40° line to the 20 knot arc. Move straight down to find a crosswind component just under 13 knots. Therefore, the aircraft can safely land on runway 32 under the given conditions.

(Refer to figure 36.) Determine the maximum wind velocity for a 45° crosswind if the maximum crosswind component for the airplane is 25 knots. View Figure 36

Starting on the horizontal axis at a crosswind component of 25 knots, follow the line up to the 45° line. Follow the arc around from that point to read a total wind velocity of 35 knots.

What is ground effect?

The result of the interference of the surface of the Earth with the airflow patterns about an airplane. The surface interferes with the airflow patterns and changes the vertical airflow pattern, decreasing the angle of attack of the wing and thus decreasing the induced drag.

Floating caused by the phenomenon of ground effect will be most realized during an approach to land when at

less than the length of the wingspan above the surface. Ground effect is only apparent within one wing span or less from the ground.