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SOLUTION 1 Enter 2300 lbs at the bottom of the graph under Indicated Airspeed and move vertically to the oo flap curve (see symbols at the top of the graph), then move horizontally to the right to read the stall speed of 57.5 kts. EXAMPLE 2 2300 lbs,

oo flaps,

Private Pilot Ucence Revision:

what is the stalling lAS at 45° of bank.

©A vex Air Training 03/2008

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I I I I I I I I I I I I I I I I I I I I

PERFORMANCE GRAPHS Page 3-4

SOLUTION 2 Enter the graph in the same manner as in example 1, but this time at the bank angle reference line (indicated by 0) follow the closest curve, staying parallel to it, until intersecting the 45° bank line. From this point move horizontally to the right to read the stall speed of approximately 67 kts. EXAMPLE 3 2350 lbs, 25° flaps, what is the stalling lAS at 30° of bank SOLUTION 3 Enter the graph in the same manner as the previous example, but select the correct flap setting. Move horizontally to the right to the bank reference line and follow the closest curve until intersecting the 30° bank line. Move horizontally to the right to read the stall speed of approximately 58 kts.

D. TAKE-OFF PERFORMANCE 1. FLAPS UP TAKE-OFF PERFORMANCE (Figure 3-3 on the next page) The purpose of this graph is to calculate the ground roll, i.e. the distance required to become airborne, as well as the take-off distance required from the commencement of the ground roll until the aircraft has reached 50 ft above the ground without flaps. The term barrier, (sometimes referred to as "screen" or "obstacle") does not imply that there is an obstacle to clear. It is merely a safety height. that whilst the runway is obviously required for the ground roll the take-off distance can include the use of a clearway. This graph contains four sections. Starting from the left is temperature and pressure altitude. The temperatures are all in oc, with an ISA or Standard temperature curve provided, and the pressure altitudes in increments of 1000 ft, but note that only the even altitudes are indicated. Next is the weight section which allows the selection of a variety of take-off weights from 1950 lbs up to the gross weight of 2500 lbs; weight increments are in 25 lbs. The next section is wind, which allows for a headwind component of up to 15 kts, and a tailwind of 5 kts. The last section is used to determine the take-off distance. A table at the top left of the graph may be used to determine the take-off speeds, including the lift off speed and the recommended speed at the 50 ft barrier height. There is some similarity between this and the remaining graphs, and it is always important to check the increments in values, whether for pressure altitude, temperature, weight, wind or distance. The graph may be entered from the left or the right depending on the question, or the circumstance. For example, entering from the left and moving continually to the right with the given values will produce the ground roll and take-off distance, but given limited ground roll or take-off distance available, we could enter the graph from the right and determine the maximum take-off weight, or temperature. EXAMPLE 1 Pressure altitude 2000 ft, temperature +25°C, weight 2325 lbs, 5kt headwind. What is the ground roll and the take-off distance to the 50ft barrier height? SOLUTION 1 Enter with the temperature and move vertically to the pressure altitude curve. From this intersection move horizontally to the right to the weight reference line and then follow the curves diagonally down to the right to intersect the weight of 2325 lbs. From this point move horizontally to the right to the wind reference line and follow the curves diagonally down to the right to intersect the headwind of 5 kts. From this point move horizontally to the right to the ground roll reference line. To determine the ground roll continue moving horizontally to the right to read 1100 ft. To find the take-off distance to 50 ft go back to the last point on the ground roll reference line and follow the curves diagonally up to the right to the 50 ft barrier reference line to read approximately 1950 ft. Be aware that very few of the curves are parallel, some converge, some diverge so a sharp eye is required.



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SOLUTION 2 Use the method described in the previous solution, but in this case the aircraft is at gross weight so from the weight reference line just continue horizontally to the right to the wind reference line. From here, that a tailwind will increase the ground roll so when you reach the wind reference line follow the curves up to 5 kts, before moving horizontally to the right read a ground roll of 1900 ft. EXAMPLE 3 Pressure altitude 1500 ft, temperature ISA, weight 2400 lbs, zero wind. What is the take-off distance to 50 ft? SOLUTION 3 You do not need to calculate the temperature, use the STD TEMP curve. Pressure altitudes are in increments of 1000 ft, so find the mid-point between 1000 ft and 2000 ft on the standard temperature curve, proceed as before, moving horizontally from the wind reference line to the ground roll line because there is zero wind. Take-off distance to 50 ft is approximately 1900 ft. 2. 25° FLAPS TAKE-OFF PERFORMANCE (Figure 3-4, on the next page). This graph is used to determine ground roll and/or take-off distance to a barrier height of 50 ft with 25° flaps. The method of using this graph is exactly the same as the previous one. EXAMPLE 1 Airfield elevation 2300 ft, QNH 1023, temperature 30°C, weight 2450 lbs, headwind 10 kts. What is the take-off distance to a 50 ft barrier height? SOLUTION 1 To find the pressure altitude: 1023 - 1013 = 10 hPa x 30 ft/hPa = 300 ft, 2300 ft - 300 ft = pressure altitude 2000 ft. Enter the figures in the same manner as in the previous graph. The take-off distance to 50 ft is approximately 1800 ft. EXAMPLE 2 Airfield elevation 2850 ft, QNH 1008, temperature ISA -9°, weight 2500 lbs, 5 kts tailwind, what is the ground roll? SOLUTION 2 1013 - 1008 = 5 hPa x 30 ft = 150 ft + 2850 ft = pressure altitude 3000 ft. The ISA temperature at 3000 ft is: 3 x -2° + 15° = +9°C - go (deviation) = 0°C. Enter the graph with approximately 1500 ft.


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E. CLIMB PERFORMANCE 1 . CLIMB PERFORMANCE (Figure 3-5)

The purpose of this graph is to determine the rate of climb at gross weight given a temperature and pressure altitude. EXAMPLE

Pressure altitude 7500 ft, temperature 5°C. What is the rate of climb? SOLUTION

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Private Pilot Licence Revision:


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2. TIME, DISTANCE AND FUEL TO CLIMB (Figure 3-6) This graph can be used to determine up to three items as the title suggests; the time taken, the distance covered and the fuel used in a climb. 0

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EXAMPLE An aeroplane climbs from pressure altitude 3500 ft, temperature +20°C to FL085, temperature ISA. Determine the time, distance and fuel for the climb. SOLUTION Two calculations are required. First calculate the time, distance and fuel from zero ft to FL085, (pressure altitude 8500 ft), then from zero ft to 3500 ft and subtract these figures from the first set. From the intersection of the standard temperature curve and 8500 ft move horizontally to the right to each curve in turn, moving vertically downwards to read time, distance and fuel. Then repeat the exercise from 3500 ft:



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SOLUTION Enter temperature and pressure altitude as in the previous graphs. From the intersection move horizontally to the right to cut the 65% curve, then move vertically down to read the RPM setting of approximately 2425 RPM 2. SPEED POWER - PERFORMANCE CRUISE (Figure 3-8) This graph is used to determine the TAS achieved, given a pressure altitude, temperature and percentage power.

Figure 3-8: Speed Power - Performance Cruise

EXAMPLE FL075, temperature +5°C. What is theTAS at 65% power? SOLUTION Enter temperature and pressure altitude as in the previous graphs. From the intersection move horizontally to the right to cut the 65% curve, then move vertically down to read the TAS of approximately 132 kts Private Pilot Ucence Revision:


I



3. SPEED POWER - ECONOMY CRUISE (Figure 3-9) Although similar to the previous graph, (Figure 3-8), the difference between the two graphs is the mixture setting. Figure 3-8 is performance cruise, and Figure 3-9 is economy cruise. Both graphs provide details of the mixture settings required to achieve the stated performance, i.e TAS. It follows that this graph, Figure 3-9, should provide a slightly lower TAS for the same power setting as used in Figure 3-8 because the mixture is leaner. ~~-r~~~~-r~~~~~~~~~~~~~~~~~

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I


PERFORMANCE GRAPHS Page 3-13 4. BEST POWER MIXTURE RANGE (Figure 3-10) This graph is used to determine the range that can be expected at a given pressure altitude, temperature and percentage power setting. The figures are based on maximum usable fuel and two options are available; range with reserve and range without reserve.

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EXAMPLE FL075, temperature +5°C. What is the range with and without reserve at 65% power? SOLUTION Enter temperature and pressure altitude and then move horizontally to the right to cut each 65% power curve in turn, move vertically down from each intersection to read the respective ranges as approximately 590 nm with reserve and approximately 670 nm without reserve.



I


PERFORMANCE GRAPHS Page 3-14 5. BEST ECONOMY MIXTURE RANGE (Figure 3-11)

Although similar to the previous graph, (Figure 3-10), the difference between the two graphs is the mixture setting. Figure 3-10 is power range, and Figure 3-11 is economy range. Both graphs provide details of the mixture settings required to achieve the stated performance, i.e range. It follows that, although the TAS is slightly less (Figures 3-8 and 3-9), this graph, Figure 3-11, should provide an increased range for the same power setting as used in Figure 3-10 because the mixture is leaner.

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FL075, temperature +5°C. What is the range with and without reserve at 65% power? SOLUTION

Enter temperature and pressure altitude and then move horizontally to the right to cut each 65% power curve in turn, move vertically down from each intersection to read the respective ranges as approximately 680 nm with reserve and approximately 790 nm without reserve.



I



6. ENDURANCE (Figure 3-12) This graph is used to determine the endurance that can be expected at a given pressure altitude, temperature and percentage power setting. The figures are based on maximum usable fuel and two options are available; endurance with reserve and endurance without reserve. Note that the endurance is in hours and decimals.

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EXAMPLE FL075, temperature +5°C. What is the range with and without reserve at 65% power? SOLUTION Enter temperature and pressure altitude and then move horizontally to the right to cut each 65% power curve in turn, move vertically down from each intersection to read the respective endurance as approximately 5.35 hours with reserve and approximately 6 hours without reserve.



I



G. DESCENDING 1. TIME, DISTANCE AND FUEL TO DESCEND (Figure 3-13) This graph is similar to Figure 3-6, Time, Distance and Fuel to Climb and, where a descent from one altitude to another is planned, should be used in the same way.

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2. GLIDE RANGE (Figure 3-14) This graph is used to determine the distance covered during a power off glide. It should be used in the same manner as the previous graph. (/)

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Figure 3-14: Glide Range

EXAMPLE Cruising level FL085, temperature ISA + 2°C, terrain elevation 2000 ft, temperature +25°C. What is the glide range? SOLUTION First calculate the temperature at FL085. 8.5 x -2°C = -1rc + 15°C = -2°. The ISA deviation is +2°C, meaning that the temperature at FL085 is 2°C warmer than ISA, therefore -2°C + 2°C = oac. Enter the graph to determine the glide range from FL085 to zero ft and 2000 ft to zero ft. Subtract the one from the other to determine the glide range. FL085 2000ft

Distance 16 nm

fi.J1rrL 10 nm



I



H. LANDING PERFORMANCE LANDING PERFORMANCE (Figure 3-15, on the next page) This graph allows you to determine both the landing ground roll and the landing distance over a 50 ft barrier. Again, the term barrier does not imply that there is a barrier at 50 ft over the runway threshold. It simply means that the aircraft crosses the threshold at 50 ft. Enter the graph in the same manner as the take-off graphs. EXAMPLE 1 Airfield elevation 1800 ft, QNH 1023, temperature +20° C, weight 2250 lbs, 5 kt headwind, What is the landing distance over a 50ft barrier? SOLUTION 1 Pressure altitude = 1023 - 1013 - 10 hPa 30 ft :::;: 300 ft. 1800 - 300 = 1500 ft. Enter all the values in the normal way to read a landing distance of approximately 1175 ft. EXAMPLE 2 Airfield pressure altitude 3500 ft, temperature ISA, weight 2325 lbs, runway 03 in use, surface wind 070/13. What is the landing ground roll? SOLUTION 2 Calculate the headwind component before you start. 070/13, runway 03

= headwind

10 kts.

The ground roll is approximately 725 ft.



I



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I



I. PERFORMANCE TABLES 1. AIRSPEED CORRECTION TABLE (Figure 3-16) This table is used to convert Indicated Airspeed (lAS) to Calibrated, or Rectified, Airspeed (CAS/RAS). Speeds can be given in Miles Per Hour (MPH) or Knots (Kts) usually depending on the age of the aeroplane. In this table, the Indicated airspeed (KIAS) starts at 50 kts, with increments of 10 kts up until 150 KIAS. Interpolation for other speeds will have to be made. Three flap settings are given: 0°, 1oo and 40°.

AIRSPEED CORRECTION TABLE KIAS

50

60

70

80

90 100

110

120

130

140

150

FLAPS UP

KCAS

62

66

75

83

92 102

108

118

129

139

148

FLAPS 10°

KCAS 58

64

73

81

91

102

107

118

FLAPS 40°

KCAS 56

62

71

79

89 103

111

--- - - - - - - - - -- - - - 129

Figure 3-16: Airspeed Correction Table EXAMPLE 1 Indicated Airspeed 70 kts. What is the Calibrated Airspeed with 1oo flaps? SOLUTION 1 Very simple in this case. Locate 70 KIAS and move vertically down to the 1oo Flaps line to read 73 KCAS. EXAMPLE 2 Indicated airspeed 95 KIAS. What is the Calibrated Airspeed with flaps up? SOLUTION 2 Figures are given for 90 KIAS and 100 KIAS, with 92 KCAS and 102 KCAS for Flaps up. 95 KIAS is midway between 90 and 100 KIAS, the mid-point Calibrated Airspeed between 92 and 102 is 97 KCAS. EXAMPLE 3 Indicated Airspeed 78 kts. What is the Calibrated Airspeed with 40° flaps? SOLUTION 3 KIAS 70 KCAS 71

80 79

The difference between 70 and 80 KIAS is 10, whereas the equivalent difference in Calibrated Airspeed is 8, which means that 1 kt of CAS is less than 1 kt of lAS. To determine its value, simply divide 8 by 10. 1 kt of lAS is equal to 0.8 kt CAS. The KIAS figure given was 78 kts, therefore we need to know the equivalent in CAS. Multiply 8 (lAS) by 0.8 (CAS), which gives 6.4. This figure can be rounded down to 6 then added to 71 to provide the answer of 77 kts CAS. EXAMPLE 4 Indicated Airspeed 53 kts. What is the Calibrated Airspeed with 40° flaps? SOLUTION 4 KIAS 50 KCAS 56

60 62

Therefore 6 + 10 x 3


= 1.8.

KCAS 56

+ 2 (1.8 rounded up) = 58 kts

© A vex Air Training 03/2008

I



2. MAXIMUM RATE OF CLIMB TABLE (Figure 3-17) This table is used to determine the maximum rate of climb at a given range of pressure altitudes (from sea level with increments of 2000 ft) and temperatures (from -20°C with increments of 1oo up to +40°C) at a gross weight of 2300 lbs. A range of climb speeds at different pressure altitudes is also provided. For rates of climb at other altitudes and/or temperatures interpolation will be necessary.

MAXIMUM RATE OF CLIMB FLAPS UP, FULL THROTTLE, MIXTURE LEANED FOR MAXIMUM RPM DURING CLIMB WEIGHT PRESS ALT (LBS) (FT) 2300

CLIMB SPEED KIAS

-20°C

-10°C

ooc

10°C

20°C

30°C

40°C

70 70 69 69 68 67 66

650 630 590 550 510 460 390

630 610 560 520 480 410 340

610 580 530 490 450 360 280

570 530 490 450 410 300 220

520 480 440 400 350 250

480 420 390 350 310

430 360 320 300 270

-----

-----

SL 2000 4000 6000 8000 10,000 12,000

RATE OF CLIMB - FPM

---

Figure 3-17: Maximum Rate of Climb Table

EXAMPLE 1 Pressure altitude 4000 ft, temperature + 10°. What is the rate of climb? SOLUTION 1 Enter the table at 4000 ft and move horizontally to the right to read 490 fpm under 10°. EXAMPLE 2 Pressure altitude 5000 ft, temperature +10°. What is the rate of climb? SOLUTION 2 Only figures for 4000 ft and 6000 ft are given: 4000 ft 6000 ft

=

490 fpm fpm

= 450

5000 ft is midway between the two so 490 - 450 = 40 + 2 490 fpm - 20 fpm = 470 fpm at 5000 ft and 10° C.

= 20

fpm.

EXAMPLE 3 Pressure altitude 4000 ft, temperature +5°. What is the rate of climb? SOLUTION 3 Only temperatures of

oo and 1oo are provided,

so:

4000 ft at oo = 530 fpm 4000 ft at 1oo = 490 fpm 530 fpm - 490 fpm = 40 fpm + 2 = 20 fpm 530 fpm - 20 fpm = 510 fpm at +5°C



I



EXAMPLE 4 Pressure altitude 9000 ft, temperature -5°C. What is the rate of climb? SOLUTION 4 Calculate the rate of climb at -5°C for 8000 ft and 10 000 ft. Pressure altitude 9000 ft, temperature -5°C. 8000 10 000

0° 450 360

Therefore 9000 ft/-5° 465 - 40 = 425 fpm

-5° 465 385

-10° 480 410

= 465

- 385

= 80 fpm

+ 2000 x 1000

= 40 fpm

3. STALL SPEEDS (Figure 3-18) This table is used to calculate the stall speeds at various bank angles and flap settings at a gross weight of 2300 lbs. STALL SPEEDS- KNOTS (CAS) WEIGHT

ANGLE OF BANK

CONDITION

oo

20°

40°

60°

FLAPS UP

56

60

66

80

FLAPS 10°

54

58

64

78

FLAPS 40°

50

54

58

70

2300 (LBS) GROSS WEIGHT

Figure 3-18: Stall Speeds

EXAMPLE 1 What is the stall speed with 1oo flap at 20° bank? SOLUTION 1 Very simple. Enter the Flaps 1oo line and read the stall speed under the 20° column, 58 kts. EXAMPLE 2 What is the stall speed, flaps up, at 30° bank? SOLUTION 2 30° is midway between 20° and 30° and the difference in stall speeds is 6 kts. Therefore the stall speed is 63 kts at 30° bank. EXAMPLE 3 What is the stall speed with 40° flap at 45° bank? SOLUTION 3 60° bank 40° bank

= 70 = 58

kts kts

Therefore 20° of bank

= 12 kts.

12 kts + 20° = 0.6 kts per degree. Therefore 5° = 5 x 0.6


=3

kts

+ 58 kts = 61 kts at 45°.


I


PERFORMANCE GRAPHS Page 3-23 4. TAKE-OFF DISTANCE (Figure 3-19) This table is used to calculate both the take-off ground run and the take-off distance to 50 ft obstacle height at a range of pressure altitudes and temperatures and at a ~iven weight, 2300 lbs. Always pay attention to the notes on any graph or table. In this case correcttons must be applied depending on the type of surface and the head or tailwind component.

TAKE-OFF DISTANCE PAVED, LEVEL, DRY RUNWAY, ZERO WIND, FLAPS UP

ooc

TAKE-OFF 20°C 30°C 10°C SPEED PRESSURE WEIGHT (KIAS) TOTAL TOTAL TOTAL TOTAL ALTITUDE (LBS) TO CLEAR GND GND GND GND TO CLEAR TO CLEAR TO CLEAR (FT) LIFT AT RUN RUN 50FT RUN 50FT RUN 50FT 50FT OFF 50FT 60

2300

65

SEALEVEL 1000 2000 3000 4000 5000 6000 7000 8000

860 920 1010 1120 1280 1420 1580 1760 1980

1500 1580 1700 1840 1980 2200 2420 2680 3020

900 970 1070 1190 1360 1510 1680 1870 2100

1580 1670 1800 1950 2100 2330 2560 2830 3180

960 1030 1140 1270 1450 1610 1790 1990 2230

1670 1750 1940 2040 2230 2470 2710 2990 3350

1040 1100 1220 1360 1550 1770 1910 2120

1770 1860 2060 2270 2490 2720 2960 3210

----

----

NOTES: 1. For operation on a dry, grass runway, increase both distances, (i.e. "ground run" and ''total to clear 50ft obstacle") by 5% of the "total to clear 50ft obstacle figure". 2. Reduce take-off distances by 10% for each 10 kts headwind. 3. Increase take-off distances by 10% for each 5 kts tailwind, maximum tailwind component 10 kts.

Figure 3-19: Take-off Distance EXAMPLE 1 Pressure altitude 2000 ft, temperature + 15° C. What is the take-off ground run? SOLUTION 1 10°C ground run 20°C ground run

= =

1070 ft 1140 ft

15° is midway between 10° and 20°, therefore 1140 - 1070 1070 + 35 ft

= 11 05 ft take-off

=

70 ft + 2

= 35

ft

ground roll at 15° C.

EXAMPLE 2 Pressure altitude 3500 ft, temperature +20° C. What is the take-off distance to 50 ft? SOLUTION 2 3000 4000 2230

= 2040 ft = 2230 ft - 2040 = 190 ft


+ 2

= 95

ft + 2040 ft

=

2135 ft at 3500 ft pressure altitude.


I



EXAMPLE 3 Pressure altitude 5000 ft, temperature +25°C, 10 kt headwind, dry grass runway. What is the take-off ground run? SOLUTION 3 20°C ground run 30°C ground run

= 1610 ft = 1770 ft

25° is midway between 20° and 30°, therefore 1770 - 1610

= 160 ft

+ 2

= 80 ft

1610 + 80 ft = 1690 ft take-off ground roll at 25° C. But, do not forget the corrections for surface and headwind! Note 1 requires an increase in the ground run of 5% of the total to clear 50 ft obstacle distance, not the ground run distance. So:

= 2470 ft = 2720 ft 2720 - 2470 = 250 ft + 2 = 125 + 2470 = 2595 ft distance to 50 ft obstacle. 5% of 2595 = 2595 + 100 x 5 = 129.75 Ground roll correction for surface = 1690 ft + 130 (129. 75 rounded up) = 1820 ft. 20°C total to 50 ft distance 30°C total to 50 ft distance

This distance must now be corrected for the 10 kt headwind, i.e. a reduction of 10%. 1820 - 182

=

1638 ft.

5. CRUISE PERFORMANCE (Figure 3-20) The purpose of this table is to determine the percentage power, TAS and fuel consumption (GPH) given a pressure altitude, temperature and power setting (RPM). Some tables also include the endurance and range. Be aware that the temperatures are ISA values.

CRUISE PERFORMANCE GROSS WEIGHT 2300 LBS, MIXTURE LEANED FOR CRUISE, ZERO WIND PRESSURE ALTITUDE

TEMPERATURE oc

RPM %BHP 87 82 76 69 60

ISA- 20° KTAS 120 117 113 108 102

GPH 9.8 9.2 8.6 8.0 7.4

%BHP 82 77 70 64 58

ISA KTAS 118 117 113 107 100

GPH 9.5 9.1 8.4 7.8 7.2

%BHP 78 74 68 61 55

9.8 8.9 8.3 7.7 7.2

82 75 68 62 56

118 113 108 102 95

9.5 8.7 8.1 7.6 7.0

78 71 65 61 54

118 112 106 100 93

9.2 8.6 8.0 7.5 6.9

78 71 65 59 53

116 110 104 100 94

9.1 8.4 7.9 7.5 7.1

75 69 62 56 51

116 110 103 99 93

8.9 8.4 7.8 7.4 7.0

2000

2600 2500 2400 2300 2200

4000

2600 2500 2400 2300 2200

85 80 73 66 58

120 116 112 106 100

6000

2600 2500 2400 2300 2200

81 74 67 61 55

118 113 108 102 96

9.3 8.7 8.1 7.6 7.2

ISA +20° KTAS GPH 118 9.2 116 8.9 111 8.2 7.6 104 7.0 98

Figure 3-20: Cruise Performance



I


PERFORMANCE GRAPHS Page 3-25 As with previous tables, interpolation will be necessary for pressure altitudes or temperatures other than those listed.

EXAMPLE 1 Pressure altitude 2000 ft, 2300 RPM, temperature ISA. What is theTAS and fuel flow? SOLUTION 1 Simply enter the table at 2000 ft and 2300 RPM. move horizontally to the right and in the ISA column read KTAS 107, (TAS 107 kts) and fuel flow 7.8 GPH. EXAMPLE 2 Pressure altitude 4000 ft, 2400 RPM, temperature ISA -10°C. What is the TAS and fuel flow. SOLUTION 2 The temperature columns only allow for ISA and ISA -20°C, so in this case interpolation will be necessary, which is fairly simple because the temperature given in the question is exactly halfway between the two. TAS ISA -20°C 112 kts ISA 108 kts ISA -10°C will be 110 KTAS. GPH ISA -20°C 8.3 GPH ISA 8.1 GPH ISA -10°C will be 8.2 GPH. EXAMPLE 3 Pressure altitude 3000 ft, 2500 RPM, temperature ISA. What is the TAS? SOLUTION 3 In this case interpolation between pressure altitudes is necessary, which is also fairly simple because the altitude in the question is halfway between those provided in the table. 2000 ft 4000 ft

ISA ISA

117 KTAS 113 KTAS

3000 ft will therefore be 115 KTAS



I


PERFORMANCE GRAPHS Page 3-26 6. LANDING DISTANCE (Figure 3-21) This table is similar to the Take-off Distance Table (Figure 3-19) and is used to determine the landing distance over a 50ft obstacle, or the landing ground roll, (i.e. the distance covered from touch down to a full stop). Temperatures are given, from oo to 30°, in increments of 10°. Interpolation for temperatures in between, as well as pressure altitudes in between those provided, will have to be made. Be aware of the notes, 1 to 3, regarding the correction for head or tail winds and dry, grass runway.

LANDING DISTANCE TABLE PAVED, LEVEL, DRY RUNWAY, ZERO WIND, 40° FLAP, POWER OFF PRESSURE GROSS KIASAT ALTITUDE WEIGHT 50FT GND FEET (LBS) ROLL 2300

60

SEA LEVEL 1000 2000 3000 4000 5000 6000 7000 8000

620 650 680 710 740 780 820 865 915

ooc

20°C

10°C

30°C

TOTAL OVER 50FT

GND ROLL

TOTAL OVER 50FT

GND ROLL

TOTAL OVER 50FT

GND ROLL

TOTAL OVER 50FT

1140 1170 1200 1240 1290 1340 1400 1460 1530

650 680 710 740 780 820 870 920 990

1200 1230 1270 1310 1350 1400 1470 1530 1600

680 720 760 810 860 920 980 1050 1130

1270 1310 1360 1410 1470 1530 1600 1680 1770

720 770 820 880 940 1000 1070 1150 1230

1350 1400 1450 1510 1570 1630 1700 1780 1870

NOTES: 1. Reduce landing distance by 10% for each 5 knots of headwind. 2. Increase landing distance by 10% for each 2 knots of tailwind. 3. For operation on a dry, grass runway, increase both distances, (i.e. "ground run" and "total to clear 50ft obstacle") by 15% of the "total to clear 50ft obstacle figure". Figure 3-21: Landing Distance

EXAMPLE 1 Pressure altitude 2500 ft, temperature + 1oo C, 5 kts headwind. What is the landing ground roll? SOLUTION 1 Interpolate for 2500 ft. 3000 2000 740- 710

740 ft 710ft

= 30ft

+ 2

=

15ft + 710ft

= 725ft

ground roll.

Now apply the correction for 5 kts headwind: 725 + 100 x 10

= 73ft

(72.5 ft rounded up). 725 - 73

= 652ft.

EXAMPLE 2 Airfield elevation 2850 ft, QNH 1008, temperature + 15°C, dry grass runway. What is the landing distance over a 50 ft obstacle? SOLUTION 2 Pressure altitude

=

1013 hPa - 1008 hPa

=5

hPa x 30 ft

= 150 ft

+ 2850 ft

= 3000 ft.

Interpolate for temperature. +20° +10°


1410 ft 1310 ft


I



1410- 1310

= 100ft

-;.- 2

= 50ft +

1310 ft

= 1360 ft

Now correct for surface. Note 3 requires that ground roll and total to clear 50 ft obstacle distances must be increased by 15% of the total to clear 50 ft obstacle distance, so: 1360 ft

-7-

100

X

15

= 204ft +

1360

=

1564 ft.

EXAMPLE 3

Pressure altitude 5000 ft, temperature 30°C, 4 kts tailwind, dry, grass runway. What is the landing distance over a 50 ft obstacle? SOLUTION 3

Landing distance at 5000 ft, 30°C

=

1630 ft from the table.

Correction for tailwind is 10% for each 2 kts of tailwind, so: 1630

-7-

100 X 20

= 326 ft +

1630 ft

= 1956 ft.

Now correct for surface. 1956 -;.- 100 x 15


= 293 ft

(293.4 rounded down)

+ 1956 ft = 2249 ft.


I


QUESTIONS Page 3-01

QUESTIONS 1. With reference to Figure 3-3. Airfield elevation 3850 ft, QNH 1008 hPa, temperature ISA, aircraft weight 2400 lbs, zero wind. The take-off distance to a 50ft barrier is approximately:-

(a) (b) (c) (d)

2200 2300 2375 2400

ft, ft, ft, ft

2. With reference to Figure 3-5. FL065, temperature ISA +3°C, the rate of climb achieved is approximately:(a) (b) (c) (d)

425 400 450 410

fpm, fpm, fpm, fpm.

3. With reference to Figure 3-7. FL095, temperature ISA, 70% power. The RPM required is approximately:(a) (b) (c) (d)

2450 2475 2510 2420

RPM, RPM, RPM, RPM.

4. With reference to Figure 3-8. FL095, temperature ISA, 55% power. TheTAS is approximately:(a) (b) (c) (d)

112 122 108 118

kts, kts, kts, kts

5. With reference to Figure 3-10. FL095, temperature ISA, 55% power. The range without reserve is approximately:-

(a) (b) (c) (d)

620 700 670 650

nm, nm, nm, nm.

6. With reference to Figure 3-12. FL095, temperature ISA, 55% power. The endurance with reserve is approximately:(a) (b) (c) (d)

6.25 7.15 6.45 6.05

hours, hours, hours, hours.

7. With reference to Figure 3-13. FL105, temperature ISA, the distance covered during a descent to sea level is approximately:-

(a) 45.5 nm, (b) 52 nm, (c) 47.5 nm., (d) 42.5 nm.



I


QUESTIONS Page 3-02 8. With reference to Figure 3-15. Pressure altitude 5000 ft, temperature +25°C, weight 2200 lbs, 15 kts headwind. The landing distance required over a 50 ft barrier height is approximately:-

(a) (b) (c) (d)

1130 ft, 1100 ft. 1250 ft, 1190 ft

9. With reference to Figure 3-15. Airiield elevation 2120 ft, QNH 1017, temperature +15°C, weight 2425 lbs, 5 kts tailwind. The landing ground roll is approximately:(a) (b) (c) (d)

1000 ft, 1070 ft, 1150 ft, 1030 ft.

10. With reference to Figure 3-16. KIAS 58 kts, the KCAS with 10° flaps is:(a) (b) (c) (d)

58 60 59 63

kts, kts, kts, kts

11. With reference to Figure 3-17. Pressure altitude 7000 ft, temperature -5°C. The rate of climb is:(a) (b) (c) (d)

505 495 485 475

fpm, fpm, fpm, fpm

12. With reference to Figure 3-19. Airfield elevation 1710 ft, QNH 1020, temperature +20°C, dry grass runway, 10 kt headwind. The take-off ground roll is:(a) (b) (c) (d)

1059 ft, 1743 ft, 1154 ft, 1023 ft.

13. With reference to Figure 3-20. Pressure altitude 5000 ft, OAT +5°C, 2600 RPM. The KTAS is:(a) (b) (c) (d)

118 116 115 117

kts, kts, kts, kts.

14. With reference to Figure 3-21. Airiield elevation 2590 ft, QNH 1016, temperature + 1ooc, 10 kts headwind. The landing ground roll is:(a) 580ft, (b) 725ft, (c) 635 ft, (d) 555 ft.



I


ANSWERS Page 3-A1

ANSWERS 1. (b) The graph must be entered with pressure altitude. 1013 - 1008 = 5 hPa x 30 ft = 150 ft. Add this to airfield elevation because the QNH is lower. Pressure altitude = 4000 ft. Use the intersection of the pressure altitude and the ISA curve and then follow the procedure explained in the various solutions to examples dealing with Figure 3-3.

2. (a) The graph must be entered with outside air temperature (OAn, so calculate the OAT at FL065 from the information given: 6500 x -2°/1000 ft = -13° + 15° = +2° + 3° (ISA +3°C) = +5°C. Now enter the graph in the manner explained in the various solutions to the examples dealing with Figure 3-5.

3. (c) Enter the graph with 9500 feet (FL095) and the intersection of the ISA temperature curve. Follow the method explained in the solutions to the various examples. 4. (d) Enter the graph with 9500 feet (FL095) and the intersection of the ISA temperature curve. Follow the method explained in the solutions to the various examples. 5. (b) Enter the graph with 9500 feet (FL095) and the intersection of the ISA temperature curve. Follow the method explained in the solutions to the various examples.

6. (a) Enter the graph with 9500 feet (FL095) and the intersection of the ISA temperature curve. Follow the method explained in the solutions to the various examples.

7. (c) Enter the graph with 10500 feet (FL 105) and the intersection of the ISA temperature curve. Follow the method explained in the solutions to the various examples. 8. (d) Follow the method explained in the solutions to the various examples. 9. (b) The graph must be entered with pressure altitude. 1017 - 1013 = 4 hPa x 30 ft = 120 ft. Subtract this from airfield elevation because the QNH is higher. Pressure altitude ples.

=

2000 ft, now follow the method explained in the solutions to the various exam-

10. (d) KIAS KCAS 64 - 58

50 58

=6


60 64

kts + 10 x 8

= 4.8

(5 rounded up) + 58

= 63

KCAS


I


ANSWERS Page 3-A2

11. (c)

oo 6000 8000

-50 505 465

490 450

505 - 465

= 40

+ 2

= 20 fpm.

-10° 520 480 505 - 20 - 485 fpm.

12. (a) First calculate the pressure altitude. 1020 - 1013

=7

hPa x 30 ft

= 210 ft.

Airfield elevation 1710 ft - 210 ft

+20° 1140 1030 11 0 ft + 2 = 55 ft + 1030 ft requires 5% of take-oft distance.

2000 1000

=

=

1500 ft.

1085 ft ground run. BUT, dry grass runway

+20° 1940 1750 190 + 2 = 95 ft + 1750 = 1845 take-oft distance at 1500 ft. 1845 + 100 x 5 = 92.25 (92 rounded down) ground run 1085 + 92 ft = 1177 ft. BUT, 10 kt headwind requires 10% reduction. 1177 + 100 x 10 = 117.7 (118 rounded up). 1177- 118 = 1059 ft. 2000 1000

13. (d) In this case the OAT is given and not an ISA deviation. Therefore the ISA temperature at 5000 ft must be calculated. ISA temperature at 5000 x -2°/1000 ft temperature is in fact ISA.

=

-10° + 15°

=

+5°C. Since the OAT is also +5°C then the

The KTAS at 4000 ft is 118 kts and at 6000 ft it is 116 kts. 5000 ft is mid-way between the two, so the TAS is 117 kts. 14. (a) Pressure altitude 3000 2000

= 1016 -

1013

=3

hPa x 30 ft

740 710 30 + 2 :::: 15 ft + 71 0

= 90 ft.

2590 - 90 ft

= 2500 ft.

= 725 ft.

Headwind correction is a reduction of 10% for each 5 kts. The headwind is 10 kts so a reduction of 20% is required. 725 + 100

X


20

= 145ft.

725- 145

= 580ft.


I

FUEL WEIGHT AND PERFORMANCE Page 4-1

CHAPTER 4

I I I I I

FUEL WEIGHT AND PERFORMANCE A. INTRODUCTION The calculation of fuel requirements is one of the more important aspects of good pre-flight planning for two reasons: (i) The weight of the fuel required will have a significant influence on weight and balance calculations.

(ii) Having loaded the fuel, what performance can be obtained from it.

B. FUEL WEIGHT CALCULATIONS Two methods may be used to calculate fuel weight. (i) SPECIFIC WEIGHT (SW)

I I I I I I I I I I I I I I I

This is merely a statement in words, ie 1 US gallon

=6

Lbs.

Multiply the number of USG by 6 to determine the weight of fuel in pounds. This is probably the most common and certainly the easiest method, however we must that whilst American aircraft work with US gallons, in South Africa we purchase fuel in litres. The following conversion factors will help with fuel/weight calculations: 1 US Gallon 1 Kilogram

=

3.785 litres Lbs.

= 2.205

(ii) SPECIFIC GRAVITY (SG) This is a far more accurate method, but it does require a little more effort and an accurate SG of the fuel to be loaded. Very simply, Specific Gravity (SG) is a ratio. It compares the weight of a volume of liquid with an equal volume of water. It follows, therefore, that the SG of water is 1 and one litre of water weighs one kilogram. Aviation fuel weighs less than water and the SG may be expressed as, for example 0.74, which means that one litre of Avgas weighs 0.74 of a kilogram. Again, we have the problem of US gallons and pounds. To find the weight of 1 USG of Avgas, it must first be converted into litres. The equivalent weight will be kilograms so if we need the weight in pounds, use the applicable conversion factor. When working with SG the formula: Volume

x SG = Weight

If the volume is required, then: Volume = Weight + SG

EXAMPLE 1 Determine the weight in pounds of 45 USG, SG 0.74 SOLUTION 1 45 X 3.785

=

170.325 litres x 0.74

=

126 kg x 2.205

=

277.8 lbs.

EXAMPLE 2 After loading an aircraft with engers and baggage the pilot has 250 lbs available for fuel. If the SG is 0.73, how many USG can be loaded? SOLUTION 2 Convert 250 lbs to kg Private Pilot Ucence Revision:


I


FUEL WEIGHT AND PERFORMANCE Page 4-2

= 113.38 kg Volume = weight + 113.38 + 0. 73 = 155.3 litres +

250 -;- 2.205

SG, so: 3. 785

= 41

USG.

The SG of aviation fuel is not constant, it varies with both temperature and octane.

C. FUEL CONSUMPTION AND PERFORMANCE 1. FUEL CONSUMPTION CONSIDERATIONS In most light aircraft fuel consumption is expressed in USG per hour and aircraft performance graphs allow us to calculate the expected fuel flow at a particular percenta~e power setting. With this information, we can calculate the endurance expected from a given quantity of fuel. This calculation is accomplished using the navigation computer. When considering fuel requirements for a flight always the reserves. For a long flight with possible changes in weather, either load extra fuel or plan for possible en route refuelling stops. Fuel consumption does, of course, vary with percentage power setting. The higher the power, the greater the fuel flow. During the course of a practical navigation exercise, three power settings may be used: 1. Take-off and climb; 2. Cruise; 3. Descent. For the average four seater light aircraft the change in fuel flow in each case may only be two or three gallons per hour. But for a large aircraft there is a significant change in fuel consumption. Some aircraft flight manuals contain graphs enabling the pilot to calculate the fuel required, time taken and distance covered during the climb and the descent. This aspect of flight planning was covered in Chapter 3. Fuel planning is an important part of flight planning. It affects operational ability as well as weight and balance aspects. Fuel should never be sacrificed for extra baggage or engers. See Chapter 5, Mass and Balance. Despite the most meticulous preparation for navigation flights, things can go wrong, weather conditions may deteriorate, winds may be stronger than forecast resulting in lower ground speeds and increased flight times. The pilot must always be confident that there is sufficient fuel to cope with changing circumstances, or that an aerodrome with fuelling facilities is within reach. The aircraft manufacturer goes to a great deal of trouble to provide performance graphs or tables to enable the pilot to accurately calculate the potential of the aircraft given certain conditions. The pilot should always pay attention to correct fuel calculations, mixture leaning procedures and expected fuel consumption at the selected power settings. 2. FUEL CALCULATIONS Several factors are involved in determining the amount of fuel to be carried for a flight. The fuel flow at a particular power setting can be extracted from an aircraft performance graph, another graph will be used to determine the TAS expected at the same power setting. The TAS is used to calculate the groundspeed. So, if we calculate the flying time for the flight, using groundspeed and distance, and we know the fuel consumption, we can calculate the fuel burn off as well as the reserve fuel requirement. The slide rule side of the computer is also used for fuel calculations. (a) FUEL REQUIRED Fuel consumption can be expressed in US gallons per hour (US Gph), litres per hour (LPH), pounds per hour (Lbs/hr) or kilograms per hour (Kg/hr). In American light aircraft US Gph is the common method, so to find the fuel required we need the following: Groundspeed, distance (or simply the flying time for the flight), reserve requirements and the fuel flow (F/F). EXAMPLE 1 Distance A to 8, 205 nm, GS 122, F/F 8.0 US Gph, reserve 45 minutes. Determine the amount of fuel to be loaded. SOLUTION 1 Calculate the flying time using the navigation computer.



I

I I I I I I I I I I I I I I I I

I I I I

FUEL WEIGHT AND PERFORMANCE Page 4-3 Flying time A - 8 = 1 hr 41 minutes. Since the total fuel required for the flight comprises burn-off and reserve, it is easier to add the two together and do one calculation: 1:41 minutes

+ 45 minutes reserve = 2:26 or 146 minutes.

Align the F/F (8.0) on the outer scale with the 60 minutes mark on the inner scale. Locate 146 minutes on the inner scale and immediately opposite on the outer scale read the fuel required in USG (19.5 USG).

Figure 4-1: Calculating Fuel Required

(b) ENDURANCE The term endurance in the context of fuel calculations refers to the flying time that can be obtained from a given quantity of fuel. It follows, then, that if the amount of fuel on board the aircraft is known and the F/F is also known, the endurance can be calculated. EXAMPLE 1 Fuel on board is 40 USG, F/F 8.6 Gph. Determine the endurance. SOLUTION 1 Rotate the scale and place the F/F (8.6) on the outer scale over the 60 minutes mark on the inner scale. Locate the total fuel on board (40 USG) on the outer scale and immediately opposite on the inside scale read the time (endurance) in minutes (280 minutes) or 4 hours 40 minutes.

Figure 4-2: Calculating Endurance

The endurance can also be calculated if the total weight of the fuel on board is given and the fuel flow also in weight. EXAMPLE 2 Usable fuel on board is 160 lbs. F/F 45 lbs/hr. Determine the endurance. SOLUTION 2 Use Figure 4-2 as a guide but substitute weight for volume. Answer minutes).

213 minutes (3hrs 33

EXAMPLE 3 An aircraft flies from A to 8, distance 245 nm, TAS 110 kts, TR 230° (T), forecast WN 180°/15, cruise fuel flow 9.5 USG/hr. Calculate the fuel required for the flight plus 45 minutes reserve. SOLUTION 3 In order to calculate the fuel required, we must first calculate the expected flying time. This is a function of groundspeed and distance. Using the navigation computer, calculate the groundspeed. Groundspeed = 100 kts. Flying time = 245 nm @ 100 kts = 2 hours 27 minutes. Total fuel required will be for the flight from A to 8 plus the reserve, so:



I


FUEL WEIGHT AND PERFORMANCE Page 4-4 2 hours 27 minutes + 45 minutes = 3 hours 12 minutes @ 9.5 USG/hr = 30.4 USG.

Taking this one step further let us assume that the pilot loads extra fuel and has 45 USG of usable fuel on board. On reaching B, the weather has deteriorated and the pilot cannot land for the time being. Since he has extra fuel on board the pilot decides to delay divertin9 to the alternate and instead remain in the vicinity of the airport to see if the weather w111 clear. Assuming the same power setting is used, and therefore the same fuel flow, how long can the pilot hold without infringing the reserve?

45 USG - 30.4 USG computer:

=

14.6 USG available to hold @ 9.5 USG/hr. Using the navigation

The pilot can hold for 92 minutes or 1 hour 32 minutes. EXAMPLE 4 An aircraft departs A for B with 34 USG of fuel on board, which includes 45 minutes reserve at cruise fuel flow. Cruise TAS 125 kts, headwind component 10 kts, cruise fuel flow 10.6 USG/hr, distance A to B 180 nm. Allow 3.3 USG for taxi, take-off and climb. Without infringing the reserve fuel, how long can the aircraft hold over B? SOLUTION 4 Again, fuel required is a function of flying time. In this example a TAS is given and instead of the WN and track a headwind component is provided. If the TAS is 125 kts and there is a headwind component of 10 kts, then the groundspeed is: 125 - 10 = 115 kts. Flying time

= 180

nm @ GS 115

=

1 hr 34 minutes.

Fuel required = 1 hours 34 minutes plus 45 minutes reserve = 2 hours 19 minutes @ 10.6 USG/hr = 24.6 USG. now to add the 3.3 USG used for taxi, take-off and climb and we will have the total fuel required: 24.6 + 3.3 = 27.9 USG. Fuel remaining at B

= 34 - 27.9 =

6.1 USG @ 10.6 GPH

= 35

minutes.

3. FUEL PERFORMANCE Fuel performance may be expressed as nm/litre, nm/Kg, nm/lb or nm/USG. If the aircraft's speed is included, then this may be further expressed as Air Nautical Miles (TAS) or Ground Nautical Miles (GIS). EXAMPLE 1 An aircraft with a ground speed of 100 Kts has a fuel consumption of 7.6 USG per hour, what is the fuel performance in ground nautical miles per USG? Both fuel flow and speed are expressed per hour, so:

100 7.6

= 13.16 GNM/USG

EXAMPLE 2 An aircraft with a ground speed of 95 kts, has a fuel consumption of 8 USG/hr. If the SG is 0. 71 what is the fuel performance in ground nautical miles per Kg.

8

X

3.785

_9.5_

21.5

=

X

0.71

= 21 .5 Kg.

4.4 GNM/Kg



I


QUESTIONS Page 4-01

QUESTIONS 1. 42 USG of fuel, SG 0.71. The weight of the fuel in lbs is:(a) {b) (c) (d)

248.9 223.9 262.9 238.9

lbs, lbs, lbs, lbs.

2. 155 litres of fuel, SG 0.70. The weight of the fuel in lbs is:(a) (b) (c) (d)

229 248 223 239

lbs, lbs, lbs, lbs.

3. 120 kg of fuel, SG 0.71. The volume of the fuel in USG is:(a) (b) (c) (d)

41.55 44.65 47.45 46.45

USG, USG USG. USG.

4. The usable fuel on board is 46 USG, F/F 7.8 Gph. The endurance is:(a) (b) (c) (d)

5 5 5 6

hours hours hours hours

54 42 33 12

minutes, minutes, minutes, minutes.

5. The trip fuel is 34 USG and the endurance is 3 hrs 35 minutes. The fuel flow is:(a) (b) (c) (d)

7.8 8.4 9.5 8.8

USG/hr, USG/hr. USG/hr., USG/hr.

6. An aircraft with a ground speed of 108 Kts has a fuel consumption of 6.8 USG per hour. The fuel performance in ground nautical miles per USG is:(a) (b) (c) (d)

14.78 15.24 15.88 16.25

GNM/USG, GNM/USG, GNM/USG., GNM/USG

7. If 30 USG of fuel weighs 183 lbs, the SG is:(a) 0.70, (b) 0.71' (c) 0.72, (d) 0.73. 8. The usable fuel on board is 38 USG, F/F 7.2 Gph. The endurance is:(a) (b) (c) (d)

5 5 5 5

hours hours hours hours

24 17 03 10


minutes, minutes, minutes, minutes.


I


QUESTIONS Page 4-02

9. The trip fuel is 28 USG and the endurance is 2 hrs 45 minutes. The fuel flow is:(a) {b) (c) {d)

10.8 USG/hr, 9.2 USG/hr. 9.8 USG/hr., 10.2 USG/hr.

10. An aircraft with a TAS of 100 Kts has a fuel consumption of 7.2 USG per hour. The fuel performance in air nautical miles per USG is:(a) {b) {c) (d)

13.38 GNM/USG, 13.88 GNM/USG, 14.88 GNM/USG., 14.28 GNM/USG

11. If 40 USG of fuel weighs 244 lbs, the SG is:(a) {b) (c) (d)

0.73, 0.71' 0.72, 0.70.



I


ANSWERS Page 4-A1

ANSWERS 1. (a) 42 USG x 3.785 = 158.97 litres x SG 0.71 = 112 .87 kg x 2.206

= 248.9

lbs

2. (d) 155 x SG 0.70

=

108.5 kg x 2.205

= 239

lbs

3. (b) 120

0.71

7

= 169

litres

7

3.785

= 44.65

USG

4. (a) Use the navigation computer. Place 7.8 on the outside scale over 60 on the inside scale. Opposite 46 on the outside scale read 354 minutes on the inside scale. 354 minutes = 5 hours and 54 minutes.

5. (c) Use the navigation computer. Place 34 on the outside scale over 215 minutes (3 hrs 35 mins) on the inside scale. Read the fuel flow on the outside scale over the 60 on the inside scale. 9.5 USG per hour.

6. (c) Both figures are per hour; in one hour the aircraft will travel 108 nm and in one hour it will consume 6.8 USG, so: 108 kts

7

6.8 GPH

= 15.88 GNM/USG

7. (d) Convert 30 USG into litres (113.55) and 183 lbs into Kg (82.99) If Volume x SG = weight, then Weight 7 SG = Volume, and Weight 7 Volume = SG. 82.99

7

113.55

= 0. 73.

8. (b) Use the navigation computer. Place 7.2 on the outside scale over 60 on the inside scale. Opposite 38 on the outside scale read 317 minutes on the inside scale. 317 minutes = 5 hours and 17 minutes. 9. (d) Use the navigation computer. Place 28 on the outside scale over 165 minutes (2 hrs 45 mins) on the inside scale. Read the fuel flow on the outside scale over the 60 on the inside scale. 10.2 USG per hour.



I


ANSWERS Page 4-A2

10. (b) Both figures are per hour; in one hour the aircraft will travel 100 nm and in one hour it will consume 7.2 USG, so: 100 kts + 7.2 GPH

= 13.88 GNM/USG

11. (a) Convert 40 USG into litres (151.4) and 245 lbs into Kg (110.65) Weight + Volume 110.65 + 151.4


= SG.

= 0.73.


I


MASS AND BALANCE Page 5-1

CHAPTER 5

MASS AND BALANCE A. INTRODUCTION Traditionally, the main concern when loading an aircraft is that the weight of the loaded aircraft should not exceed the Maximum All Up Weight (MAUW) as determined by the manufacturer. This is certainly important, but equally so is the balance of the aircraft. In this regard we are referring to the final location of the Centre of Gravity (CG). In order to determine this, a loading form is provided by the manufacturer which, in conjunction with the CG envelope, will determine whether or not the aircraft is within the specified CG limits. 1. WEIGHT The effect of weight plays a significant role in of flight performance. Excessive weight may lead to the following: (a) a higher take-off speed, (b) a longer take-off run, (c) a reduction in the rate of climb, (d) greater fuel consumption, (e) reduced range, (f) a higher stalling speed, (g) a higher landing speed, (h) a longer landing roll. The maximum weight that an aircraft can carry is dependant upon a combination of structural strength and the ability of the wing to generate sufficient lift to the weight. The maximum operating weight of the aircraft is determined by the manufacturer and will be published in the Aircraft Flight Manual. 2. BALANCE Balance refers to the final location of the Centre of Gravity (CG) of the aircraft and this plays an important role in the ability of the pilot to control the aircraft in flight. If the CG is too far forward the aircraft will become nose heavy, making if difficult for the pilot to rotate. An aft CG plays a similar role in reducing elevator effectiveness; the aircraft will become tail heavy and may rotate prematurely during take-off and present difficulties during the landing. The manufacturer will specify the maximum fore and aft CG limits in the form of a CG envelope, or graph, which will also be published in the Aircraft Flight Manual.

B. DEFINITIONS 1. AIRCRAFT WEIGHT (a) BASIC EMPTY WEIGHT (SEW) The Basic Empty Weight is the Empty Weight plus full oil and unusable fuel. (b) OPERATING EMPTY WEIGHT (OEW) The Operating Empty Weight is the Basic Empty Weight plus crew and baggage. (c) ZERO FUEL WEIGHT (ZFW) The Zero Fuel Weight is the Operating Empty Weight plus engers, baggage and cargo (collectively called payload). Usable fuel is not included. (d) RAMP WEIGHT The maximum ramp weight is the Zero Fuel Weight plus usable fuel. (e) TAKE-OFF WEIGHT The Take-off Weight is the weight of the aircraft at lift-off from the runway.



I


MASS AND BALANCE Page 5-2 (f) LANDING WEIGHT The Landing Weight is the weight of the aircraft at touchdown. 2. STRUCTURAL CONSIDERATIONS

(a) Maximum Take-off Weight (MTOW) The MTOW is the maximum allowable weight permitted at take-off. (b) Maximum Zero Fuel Weight (MZFW) The MZFW is the maximum allowable weight of the aircraft before usable fuel is added. (c) Maximum Ramp Weight (MAW) The MAW is the maximum weight permitted prior to taxying. In large aircraft it may exceed the MTOW by the taxi fuel allowance. (d) Maximum Landing Weight (MLW) The MLW is the maximum allowable weight for landing. For many light aircraft the MLW equals the MTOW. For other aircraft the MLW is less than the MTOW. In this case the fuel burn-off must be sufficient to reduce the take-off weight to below the MLW. In an emergency fuel may have to be dumped prior to landing.

C. BALANCE TERMINOLOGY (a) CENTRE OF GRAVITY (CG) This is the point about which the aircraft would balance if suspended at that point, or the theoretical point at which the total weight of the aircraft is assumed to be concentrated.

CG Figure 5-1: Centre of Gravity

(b) DATUM OR REFERENCE POINT This is a reference from which all measurements regarding balance are taken. Its location is determined by the aircraft manufacturer and is may be expressed either as a distance in inches forward of the leading edge of the wing, for example 79.6 inches as indicated in Figure 5-2, or another part of the aircraft such as the firewall. Datum

Figure 5-2: Datum Private Pilot Ucence Revision:


I



(c) ARM This is the horizontal distance, usually expressed in inches, from the Datum to each position where weight is to be loaded, such as the bag!;lage compartment, fuel tanks or pilot and enger seats. The arm is referred to in some atrcraft manuals as a Flight Station (FS), for example an arm of 35 inches is the same as FS 35. (d) MOMENT A moment is the product of the weight of an item multiplied by its arm. It is the effect of the gravitational force that the weight of an item would have as a result of its distance from the datum. The moment is calculated by the formula: Weight x Arm

=

Moment

Where, in American aircraft, the weight is in pounds (lb) and the arm is in inches ("). The resultant moment is expressed in Inch/pounds. For example: 340 lbs {W) x 94.6 inches (A)

= 32164

(M)

(e) REDUCTION FACTOR (RF) With large aircraft the determined moments are often very large and cumbersome to work with in calculations. For this reason, a suitable reduction factor (RF) is used, for example a RF of 100. The moment is now divided by the RF to make it more workable. 32164 100

=

321.64

The same RF must be used throughout the calculations. Once a moment has been divided by an RF, it now becomes known as an index. It is usually written as, for example, Moment/1 00 or Mom/100. (f) PRINCIPLES OF BALANCE

The principle of balance is easily explained. Figure 5-3 shows the effect a weight of 100 lbs would have if positioned 20 inches from a fulcrum or datum. Datum

n

100 lbs 20inches

~~----------------+•

Moment: 20" x 100 lbs = 2000 inch/pounds

Fulcrum Figure 5-3: Calculating a Moment

The result is a force of 2000 inch-pounds. To balance this, a similar force must be positioned on the opposite side of the fulcrum. Datum

n

n

100 lbs

100 lbs 20 inches

20 inches

.......- - - - - + • .......- - - - - + •

A Moment: 20" x 100 lbs = 2000 inch/pounds

B Moment: 20" x 100 lbs = 2000 inch/pounds

Fulcrum

Figure 5-4: Establishing Balance



I


MASS AND BALANCE Page 5-4 Balance is achieved in this case by placing a similar weight at a similar distance from the fulcrum. However, it is the moments that must balance. We could place a heavier weight at a shorter distance from the fulcrum and still achieve balance. See Figure 5-5. Datum 200 lbs

n

j. ----+li+--• ----+:n

100 lbs

10 inches

20 inches

+---111

A

B

Moment: 10"x 200/bs = 2000 inch/pounds

Fulcrum

Moment: 20"x 100 lbs = 2000 inch/pounds

Figure 5-5: Another Method of Achieving Balance

In this example a weight of 200 lbs is placed 10 inches from the achieved is also 2000 inch-pounds and balance is also achieved.

fulcrum, the moment thus

D. BASIC PRINCIPLES OF MASS AND BALANCE Ultimately it is the responsibility of the pilot to ensure that the aircraft is correctly loaded. In this regard it is a question of finding a balance between operational ability and safety, and a sound knowledge of the principles of weight and balance is vital. To assist the pilot in loading the aircraft correctly, the manufacturer produces a loading form indicating the arm of each position where weight will be loaded. On completion of the loading form the final CG and total weight of the loaded aircraft are entered into a CG Range envelope to ensure that both items are within the limits specified by the manufacturer. The Loading Form and CG Envelope varies from manufacturer to manufacturer, two different systems are explained in this section. 1. LOADING FORM 1 An example of a loading form appears in Figure 5-6. Weight (lbs) Basic Empty Weight Pilot and Front enger engers (Aft) Fuel Baggage

1530

Arm Aft Datum (inches) 86.8

Moment (inch-pounds) 132804

78.5 115.3 97.0 144.5

Total Loaded Aeroplane Figure 5-6: Loading Form

Completion of the form is straight forward once the pilot has determined the weight of the items he or she intends loading. Notice that the Weight column is in pounds which means that all items, fuel, pilot, engers and baggage must be converted into the same units. EXAMPLE Pilot 75 kg, one enger 80 kg, one enger 50 kg, 40 USG of fuel at SW 6 lb per USG, baggage 15 kg. Each item will be converted to pounds and then entered on the loading form. In this example, the 80 kg enger will be located alongside the pilot. For ease of calculation, the weights will be rounded up.



I

I I I

I I I I I I I I I I I I I I I I I


METHOD (a) enter the weight of each item in the relevant column; (b) total the weight column to ensure that the MTOW has not been exceeded (in this case the MTOW is 2500 lbs); (c) cross multiply the weight of each item by its arm and enter the result in the moment column; (d) total the moment column and divide it by the total weight to determine the final CG. : if Weight x Arm

= Moment,

Moment + Weight

then:

= Arm Weight (lbs)

Arm Aft Datum (inches)

Moment (inch-pounds)

1530

86.8

132804

Pilot and Front enger

342

78.5

26847

engers (Aft)

110

115.3

12683

Fuel

240

97.0

23280

33

144.5

4768

Basic Empty Weight

Baggage Total Loaded Aeroplane

2255

200382

88.86

Figure 5-7: Completed Loading Form

The total weight of our example aircraft is 2255 lb, which is less than the MTOW, and so in this respect it is legal for flight. Confirmation of this can be achieved by locating both the total weight and loaded CG on the CG Envelope. 2. CG ENVELOPE 1 Enter with the weight from the right and move horizontally left to intersect the CG vertically from the top.



I


I I

88.86 inches

2600

/N~MAL CATECjORY 2200

2000

I I I

/

(f)

v

./

v

v

2255/bs

"

Q

z

::::l

0

1800

a..

UTIUTY CATEGORY

~ 1--

-

<.? 1600

f--

I

w

~

,....---

1400

f--

1200 -

1000

I I I I

-

1'

I

80

82

86

84

88

90

92

94

INCHES AFT OF DATUM

Figure 5-8: CG Envelope

The combination of weight and CG is comfortably located within all the limits of the CG Envelope. In the event that the CG was too far forward or aft, weight would have to be moved. If the total weight of the aircraft exceeds the MTOW, then weight will have to be removed from the aircraft. This is not an easy choice, but beware of sacrificing fuel for baggage or engers! 3. LOADING FORM 2 This type of loading form requires the use of a loading graph to determine the moment directly from the weight to be loaded at a particular position. Even though this is a light aircraft, the form, and the CG envelope, make use of a reduction factor of 1000 (MOM/1000) . WEIGHT (lbs)

MOMENT (lb-ins/1 000)

1400

51 .5

Oil (8 qts)

15

-0.2

Fuel (45 USG@ 61bs/Gal)

270

13.0

Pilot and Front enger

320

12.0

Rear engers

200

14.5

30

3.0

2235

93 .8

LOADING FORM Licenced Empty Weight

I I I

Baggage TOTAL WEIGHT AND MOMENT

Figure 5-9: Loading Form 2 Private Pilot Licence Revision:


I


LOADING GRAPH 400 350 (j) Cl

300

z

:J

0

250

1I

200

~

I I I

(!)

iii ~

150

Cl

<:

9

100

~~Gl

J>.<">

50

00

5

10

15

20

25

30

LOAD MOMENT/1 000 (POUNDS - INCHES) Figure 5:10 Loading Graph

I I I I I

Once the weights of the different items have been established the loading graph is used to determine the moment of each weight at its location. In this case no cross multiplying is required, ie Weight x Arm = Moment. The fuel weight is calculated by converting the volume in USG into pounds using a specific weight (SW) of 6 lb/USG. The weight of 270 lbs is now entered horizontally from the left until the fuel reference line is reached. Now move vertically down to establish a moment of 13.0. This figure is entered in the loading form, along with the other weights and moments. Finally total the weight and moment columns. It is not necessary to establish the CG. Simply enter the total weight and the total moment in the CG envelope and ensure that the intersection of the two figures lies within the CG envelope. 93.8 inches

2300 (j)

~ 2200 :J

0

~ 1I (.')

iii

OR MAL

2100

C T G RY

2000

~

ti:

1900

~

()

a: 1800

<( Cl

w 1700

Cl

<:

9

I I I I

1600 1500 45

50

55 60 65 70 75 80 85 90 95 100 105 LOADED AIRCRAFT MOMENT/1000 (POUNDS -INCHES)

110

Figure 5-11: CG Envelope 2 Private Pilot Ucence Revision:


I



3. NORMAL AND UTILITY CATEGORIES It will have been noticed that both of the CG envelopes used in the previous examples include shaded areas marked "Utility Category", whilst the rest of the graph is labelled "Normal Category". These two refer to the category of operation of the aircraft and the location of the loaded CG is extremely important in this regard. NORMAL CATEGORY As the name implies this category of operation includes the manoeuvres carried out in "normal" flight including, from the training perspective, elements such as stalls and steep turns at bank angles less than 60°. UTILITY CATEGORY In this case the manoeuvres include those that can be conducted in the normal category as well as spins and steep turns at bank angles greater than 60°. The manoeuvres that can be conducted are limited by both weight and location of the CG. In the case of the utility category it can be seen that the CG aft limit is much further forward than that of the normal category. An aft CG in an aircraft which is being used for spin training could delay spin recovery, and the lower all up weight reduces the possibility of exceeding the specified load factors during the various manoeuvres. 4. CHANGE OF WEIGHT Two situations can arise here, both of which will require a recalculation of the weight and balance: (a) WEIGHT ADDITION OR REMOVAL Our example aircraft in loading form 1 came in significantly under the MTOW with 245 lb of available payload remaining. Let us assume the pilot loads a further 10 USG of fuel (60 lbs) and another 50 lbs of baggage. This means an additional 11 0 lb, which still leaves the aircraft under the MTOW, but does result in a movement of the CG, perhaps very little, but perhaps not. We must re-calculate the CG. This can be done quite easily without the use of the loading form: Weight 2255 + 60 + 50 2365

x X X X

Arm 88.86 97.0 144.5

= = = =

Moment 200382 5820 7225 213427

Now, divide the new total moment by the new total weight: 213427 + 2365

= new

CG 90.24

The old CG was at 88.86, so the new CG has moved 1.38" further aft. A quick check of the CG envelope tells us that we are still within limits. (b) Weight shifting. If weight is shifted from one location to another in the aircraft the CG will also be affected. Using our example aircraft in loading form 1, let us move the front enger to the rear seat and the rear enger to the front seat. The front enger weighs 80 kg (176 lbs) and was located at 78.5", the rear enger 50 kg (110 lbs) was located at 115.3". The calculation is as follows: Weight 2255 - 176 +176 - 110 ±.11.Q. 2255

X X X X X X

Arm 88.86 78.5 115.3 115.3 78.5

Therefore the new CG still within limits.


= 202811

= = = = = =

Moment 200382 13816 20293 12683 62~5

202811

+ 2255

= 89.94,

which means it has moved aft by 1.08" and is


I


MASS AND BALANCE Page 5-9 (c) Fuel Burn Off Our example aircraft in loading form 1 carried 40 USG (240 lbs) of fuel. As the fuel is consumed during the flight the CG will also move. If we assume trip fuel as 32 USG (192 lbs), the landing CG can be calculated as follows: Weight 2255 192 2063

Arm 88.86 97.0

X X X

Therefore new CG

=

= = =

Moment 200382 18624 181758

181758 + 2063

= 88.1 0"

and is still within the CG limits.

E. MAXIMUM FLOOR LOAD The maximum floor load represents the weight bearing strength of the aircraft's floor in the baggage or cargo area. It indicates the maximum weight that can be accommodated per surface area. It should be ed that as this is weight per area, the height of an object to be carried does not normally need to be considered. To calculate the area of a square or rectangular object, use the following formula: Area = Length x Width

To calculate the area of a circle, use the following formula: Jtr2 where r = radius and n = 3.142 To solve the floor load calculation, use the following formula: Maximum Weight = Area of Object x Maximum Floor Load

EXAMPLE 1 A box is 4 feet long, 2 feet wide and weights 8 lbs. The maximum floor load is 60 lbs per square foot. What is the maximum weight that could be loaded into the box? SOLUTION 1 4 x 2 x 60

= 480

lbs minus the weight of the box (8 lbs)

= 472

lbs.

For the purpose of calculation it is assumed that the weight will be evenly distributed throughout the box. EXAMPLE 2 If the maximum floor load is 30 Kg/sq m, what is the maximum weight that can be loaded into a box measuring 1.5 m x 0.5 m and weighing 3 kg? SOLUTION 2 1.5 x 0.5

= 0.75

sq m x 30 kg/sq m

= 22.5

kg- 3 kg= 19.5 kg

EXAMPLE 3 If the maximum floor load is 40 kg/sq m, what is the maximum weight that can be loaded into a barrel, the radius of which is .75 m and which weighs 4 kg SOLUTION 3 3.142 X .75 X .75 X 35 = 61.86 kg- 4 kg


= 57.86

kg


I


QUESTIONS Page 5-01

QUESTIONS 1. A loaded aircraft weighs 2300 lbs, with a CG of 90.12 inches. A enger weighing 165 lbs moves from FS 78 to FS 142. The new CG is:(a) 92.8,

(b) 94.7, (c) 94.1, (d) 93.1

2. A loaded aircraft weighs 2440 lbs, CG 88.9 inches. Trip fuel is 150 lbs. If the fuel tank arm is 97 inches, the CG on landing will be:-

(a) 88.37, (b) 87.56, (c) 89.35, (d) 88.97.

3. A loaded aircraft weighs 2200 lbs, with a CG of 89.4 inches. A enger weighing 185 lbs moves from FS 132 to FS 81. the new CG is:(a) (b) (c) (d)

88.84, 87.66, 85.77, 85.11.

4. A loaded aircraft weighs 2350 lbs, CG 90.1 inches. Trip fuel is 185 lbs. If the fuel tank arm is 96 inches, the CG on landing will be:(a) (b) (c) (d)

89.17, 88.98, 89.59, 89.98.

5. A loaded aircraft weighs 2500 lbs, CG 91 .5 inches. A enger, weighing 100 lbs, seated at FS 115.3 changes place with a enger, weighing 164 lbs, seated at FS 78.5. The new CG is:(a) (b) (c) (d)

91.5, 91.85, 91.11, 92.44.

6. A box measures 2 ft x 2 ft and weighs 5 lbs. If the aircraft's maximum floor load is 60 lbs/sq ft, the maximum weight that can be loaded in the box is:(a) (b) (c) (d)

235 245 240 245

lbs, lbs, lbs, lbs.

7. A box measures 1 metre x 1.5 metres and weighs 4 kg. If the aircraft's maximum floor load is 20 kg/sq metres, the maximum weight that can be loaded in the box is:(a) 30 kg, (b) 26 kg, (c) 36 kg,

(d) 34 kg.



I


QUESTIONS Page 5-02

8. With regard to CG calculations:moment = weight + arm, weight + moment = arm arm + moment = weight, moment = weight x arm.

(a) (b) (c) (d)

9. With regard to CG calculations:moment + weight = arm, weight + moment = arm arm + moment = weight, moment x weight = arm.

(a) (b) (c) (d)

10. With regard to CG calculations:(a) (b) (c) (d)

moment = weight + arm, weight + moment = arm moment + arm = weight, moment x arm = weight.

11. Aircraft landing weight 2100 lbs, CG 87.3. 148 lbs of fuel from FS 97 was used for the flight. The aircraft's CG on take-off was:(a) (b) (c) (d)

88.3, 87.9, 87.4, 88.1.

12. Aircraft landing weight 2300 lbs, CG 88.8. 162 lbs of fuel from FS 97 was used for the flight. The aircraft's CG on take-off was:(a) (b) (c) (d)

88.5, 89.0, 89.6, 89.3.

13. The following figures appear on a loadsheet Empty aircraft Pilot and front enger Rear engers Fuel Baggage

WEIGHT 1530 340 220 230 40

ARM 86.8 78.5 115.3 97.0 144.5

MOMENT 132804

The CG of the loaded aircraft is:(a) (b) (c) (d)

90.2, 90.5, 90.9, 89.9.



I


QUESTIONS Page 5-03

14. The following figures appear on a loadsheet:

Empty aircraft Pilot and front enger Rear engers Fuel Baggage

WEIGHT 1530 360 140 300 60

ARM 86.8 78.5 115.3 97.0 144.5

MOMENT 132804

The CG of the loaded aircraft is:(a) 90.2,

(b) 89.2, (c) 89.9, (d) 89.5.



I


ANSWERS Page 5-A1

ANSWERS 1. (b) 2300 -165 +165 2300 217836

X X X

90.12 78.0 142.0

207276 12870 23430 217836

= :=;:

+ 2300 = 94.7

2. (a) 2440 -150 2290

X X

88.9 97.0

= =

216916 14550 202366

= = =

196680 24420 14985 187245

202366 + 2290 = 88.37 3. (d) 2200 -185 +185 2200 187245

X X X

89.4 132.0 81.0

+ 2200 = 85.11

4. (c) 2350 -185 2165

X X

211735 17760 193975

90.1 96.0

193975 + 2165 = 89.59 5. (d) 2500 X -100 X X +100 -164 X X +164 2500 231105.2 +

91.5 115.3 78.5 78.5 115.3 2500

228750 11530 7850 12874 18909.2 231105.2

= = =

= 92.44

6. (a) 2 ft x 2 ft

=4

sq ft x 60 lbs/sq ft

= 240

lbs - 5 lbs

= 235

lbs.

30 kg - 4 kg

= 26

kg.

7. (b) 1 x 1.5 - 1.5 sq m x 20 kg/sq metre

=

8 (d) moment == weight x arm. See page 5-3. 9. (a) moment + weight


= arm.

See page 5-5.


I

I I I I 1 I I I I I I I I I I I I I I I

ANSWERS

Page 5-A2

10. (c) If weight x arm = moment, and moment + weight = arm, then moment + arm = weight.

11. (b) 2100 148

87.3 97

183330 14356 197686

2248 197686 + 2248

= 87.9

12. (d) 2300 162 2462

88.8 97

204240 15714 219954

219954 + 2462 = 89.3 13. (a) Empty aircraft Pilot and front enger Rear engers Fuel

Baggage

WEIGHT

ARM

MOMENT

1530 340 220 230 40 2360

86.8 78.5 115.3 97.0 144.5 90.2

132804 26690 25366 22310 5780 212950

WEIGHT

ARM

MOMENT

1530 360 140 300 60

86.8 78.5 115.3 97.0 144.5

132804 28260 16142 29100 8670

2390

89.9

214976

14. (c) Empty aircraft Pilot and front enger Rear engers Fuel

Baggage



I

AIRCRAFT PERFORMANCE Page 6-1

CHAPTER 6


AIRCRAFT PERFORMANCE A. AQUAPLANING or HYDROPLANING In order for an aeroplane to maintain directional control and effective braking on the ground, sufficient friction between the tyre and the surface is required. A surface which is covered by water, whether lightly or heavily is particularly dangerous because it can cause the tyre to be lifted from the surface completely with the consequent loss of traction and, therefore, control. Such a condition is known as aquaplaning or hydroplaning. As we saw earlier in this manual, the take-off and Iandin!;] performance charts normally only provide runway lengths for operations on a dry surface. A contaminated runway will make field length figures for dry runways invalid and may even prohibit take-off or landing. 1. LANDING ON SLIPPERY RUNWAYS (a) SLIPPERY RUNWAY TRACTION The presence water on a runwaf prevents between the tyre and the runway surface necessary for the development o maximum friction forces, thus reducing the availability of the friction required for braking and directional control. In some cases the possibility exists that wheel spin-up, (i.e. going from zero rotation during the approach to the full rotational speed required by the landing speed of the aeroplane), on touchdown may be inhibited, which reduces braking effectiveness and braking techniques. A tyre rolling along on a wet runway is constantly squeezing water from under its tread. The water pressure involved in this squeezing action can, at high speeds, keep the tyre, or portions of it, off the runway thereby reducing tyre-to-pavement friction. (b) VISCOUS HYDROPLANING This can occur on wet runways whether that are just damp or are coated by a very thin layer of water. It generally occurs at low groundspeeds and is normally of a very short duration. Typically, on touchdown, the damp surface causes the tyres to slide or skid until they spin up to their normal rotation speed. During this time braking action is reduced and the landing roll will increase. A rough runway surface will assist in breaking up the layer of water, but deposits of rubber from constant aeroplane landings in the same area can prevent the removal of water and increase the chances of hydroplaning. (c) DYNAMIC HYDROPLANING This occurs when a tyre is actually lifted off the runway surface and rides on a "wedge" of water like a water ski. The conditions required to cause this are standing water and high speeds and the hydroplaning will continue until the either speed has decreased or the depth of water has reduced. The standing water required to produce true dynamic hydroplaning would have to be deeper than the groove depth of a tyre tread - a condition that might only be obtained by a combination of worn tyres and heavy rainfall. (d) REVERTED RUBBER HYDROPLANING This occurs when the lack of friction does not allow the wheel to rotate and it skids on a smooth or wet surface. The heat created by friction can melt the rubber and turn the water on the runway to superheated steam. The reverted rubber can, in effect, seal the steam pressure under the tyre and hold it off the runway. Reverted rubber hydroplaning most typically occurs at lower speeds and can best be prevented by avoiding the need for hard braking on the last part of the runway. Clearly, then, the possibility of aquaplaning will increase as the depth of the tyre tread reduces and it is important to check the tyre tread carefully during pre-flight inspections. 2. FLARE AND TOUCHDOWN Trying to land as gently as possible on a slippery runway is not recommended. Too often this tends to extend the flare, u precious runway. Making a moderately firm touchdown will prevent extended flare, helps to break through the water and contributes to faster wheel spinup. A rough formula for calculating the aquaplaning groundspeed of an aeroplane is to multiply the square root



I


AIRCRAFT PERFORMANCE Page 6-2 of the tyre pressure in pounds per square inch (psi) by 9. Thus an aeroplane with a tyre pressure of 30 psi would have an aquaplaning ground speed of approximately 49 knots. 3. THE ROLLOUT

Once the aircraft has touched down the minimum stopping distance is achieved by making optimum use of the aircraft braking system. Lowering the nose promptly, as soon as the main wheels are on the runway, helps place the weight of the aircraft on the tyres by decreasing lift. Whereas the weight on the main gear tyres can be negligible at the point of touchdown, the decrease in lift produced by lowering the nose quickly transfers weight to the tyres. Directional control stabilizes and wheel spinup will be well established.

B.WAKETURBULENCE Wake turbulence vortices are present behind any aircraft and are particularly severe when generated by large and wide-bodied aircraft. These vortices comprise two counter-rotating cylindrical air masses, trailing aft of aircraft. In level flight, in stable air conditions, these vortices descend slowly at sink rates of 400 - 500 fpm to stabilise approximately 1000 ft below the aircraft's flight path rema1ning there for up to 4 minutes. If the air is unstable the vortices disperse fairly quickly. Should there be any wind, the effect of such must be borne in mind as the vortices drift with the prevailing wind. In order to avoid the turbulence, a following aircraft should always fly above the flight path of the aircraft. The vortices are most dangerous to following aircraft during the take-off, initial climb, final approach and landing phases of flight. When close to the ground they drift sideways from the track of the generating aircraft. If there is a prevailing crosswind the downwind vortices could affect a possible second, parallel runway. Various measures are used to try to protect aircraft from the effects of wake turbulence. The most commonly known method is that of the aircraft wake turbulence categories used on ATC flight plans: Light (L) Medium (M) Heavy (H)

- at or below 7000 kg - between 7000 and 136 000 kg - greater than 136 000 kg.

SEPARATION Although wake turbulence is generated by all aircraft in varying amounts, the greatest threat would be to a light aircraft following a heavy aircraft and therefore the following separation is generally applied: (a) RADAR SEPARATION Heavy followed by heavy or medium - 5 nm Heavy followed by light - 6 nm Medium followed by heavy, medium or light - 5 nm Light followed by light - 5 nm (b) TIME SEPARATION (i) TAKE-OFF When taking off behind a heavy aircraft, light or medium aircraft are separated by 2 minutes if: -

using the same runway crossing in a runway where the flight paths cross using a parallel runway within 760 metres using a parallel runway more that 760 metres away if the flight paths cross

The separation is increased to 3 minutes when: - taking off from an intersection - taking off from the intersection of a parallel runway within 760 metres if the flight paths cross. Also: When taking off after a heavy aircraft, get airborne before its lift off point, try to remain upwind and above its flight path. When taking off after a heavy aircraft has landed get airborne after its touchdown point.



I I I

AIRCRAFT PERFORMANCE Page 6-3

(ii) ARRIVING Arriving aircraft are separated by 2 minutes in the case of a medium aircraft following a heavy aircraft, and 3 minutes in the case of a light aircraft following a heavy aircraft. Also : When landing after a heavy aircraft has landed, land after its touchdown point. When landing after a heavy aircraft has taken off, touch down before its lift off point. If landing on a cross runway, try and stop before the intersection.

C. WIND SHEAR

I I I

Low level wind shear as an operational problem constitutes a significant hazard to aircraft during the landing and take-off phase. Wind shear is defined as a change in wind speed and/or direction occurring in a relatively short distance. Such changes may occur with height (vertical wind sheer) or lateral distance (horizontal wind shear) . These conditions cause changes in airspeed, and the flight path is adversely affected. When an aircraft is in the cruise configuration , any wind induced airspeed changes tend to be balanced after a short period by a corresponding aircraft inertial speed change, ie drag/thrust imbalance is regained. However, if the wind change is rapid enough to exceed the aircraft's acceleration/ deceleration capacity and large enough to match its airspeed margin over the minimum approach or climb speed for a given configuration , then a potential hazard exists. Aircraft taking off tends to overshoot

Aircraft landing tends to undershoot

Lower power setting required than before wind shear encountered

I I I

Figure 6-1: Effect of Reduction in Headwind

With reference to Figure 6-1, when an aircraft which is making a stable approach encounters a rapid reduction in the headwind component, the initial acceleration capabilities of the aircraft are exceeded and a rapid reduction in airspeed results to below the minimum approach speed. A loss of lift and a consequent undershoot of the glide path follows, which requ ires that the pilot either increase the power or lower the nose in order to regain airspeed . Such an occurrence may be hazardous where terrain obstacle clearance limits are marginal. However, once re-established on the glide path, the power setting will be lower than the original setting due to the reduced wind speed . Similarly, when an aircraft in the climb-out phase encounters a rapid increase in the headwind component, the initial deceleration capabilities of the aircraft are exceeded and a rapid increase in airspeed above the recommended climb speed occurs, resulting in increased lift and increased climb performance with a subsequent deviation above the flight path (overshoot) .

I I I I

When an aircraft, making a stable approach, encounters a rapid increase in the headwind component, the initial deceleration capabilities of the aircraft are exceeded and a rapid increase of airspeed results above the minimum approach speed, creating in an increase of lift and a consequent overshoot of the glide path, which requires a decrease in power to regain the approach speed. Once re-established on the glide path , the power setting will be higher that the original setting due to the increased wind speed. See Figure 6-2 .



I

I I

AIRCRAFT PERFORMANCE Page 6-4 Aircraft landing tends to overshoot

Aircraft taking off tends to undershoot

Aircraft flight path

\ .., ...... ~ Required glide path

Higher power setting required than before wind shear encountered

I I

Figure 6-2: Effect of Increase in Headwind

Similarly, when an aircraft in the climb-out phase encounters a rapid decrease in the headwind component, the initial acceleration capabilities of the aircraft are exceeded resulting in a rapid decrease in the recommended climb airspeed, a reduction of lift and a reduced climb performance with a subsequent deviation below the flight path (undershoot). This requires a power increase to regain airspeed (if extra power is available). Once again , this is a potentially hazardous situation where terrain obstacle clearance limits are marginal.

D. ICING Although VFR pilots should not intentionally fly into known icing conditions it is important to understand the implications of ice accretion. Airframe icing affects an aeroplane's performance in two ways. (a) WEIGHT INCREASE

I I I I I I I I I

We have already seen how weight can influence an aircraft's performance in of take-off ground roll , take-off distance and rate of climb, and how important it is to remain within the limits specified by the manufacturer. Any build up of ice on the airframe can considerably increase the aeroplane's weight, with the resultant decrease in performance. that one litre of water weighs one kilogram and even a thin layer of ice over the airframe will weigh many kilograms. If there is ice on the airframe before flight remove it before going anywhere. (b) EFFECTS ON COEFFICIENT OF LIFT (CL) The other problem associated with ice on the airframe is its effect on the production of lift. Ice, even frost which may have formed overnight for that matter, over the upper surface of the wings will distort the shape of the wing, decrease its lift potential and increase the drag. Similarly ice on the propeller will reduce its thrust potential. Again, if there is ice or even a light coating of frost on the aeroplane before flight, remove it.



I


QUESTIONS Page 6-Q1

QUESTIONS 1. When landing behind a heavier aircraft the most appropriate action for a light aircraft would be to:(a) land before the point of touchdown of the heavier aircraft, (b) land at the same touchdown point of the heavier aircraft, (c) land beyond the touchdown point of the heavier aircraft. 2. When landing behind a heavier aircraft which has just taken off, the most appropriate action for a light aircraft would be to:(a) land before the lift-off point of the heavier aircraft, (b) land at the same lift-off point of the heavier aircraft, (c) land beyond the lift-off point of the heavier aircraft. 3. When taking off behind a heavier aircraft which has just taken off, the most appropriate action for a light aircraft would be to:(a) become airborne after the lift-off point of the heavier aircraft, (b) become airborne before the lift-off point of the heavier aircraft, (c) become airborne at the same lift-off point of the heavier aircraft,. 4. With regard to aquaplaning when landing on a wet runway:(a) (b) (c) (d)

aquaplaning should only be expected at low groundspeeds, the most appropriate action is to land as softly as possible, over-inflated tyres are more prone to aquaplaning than under-inflated tyres, a firm touchdown, without bouncing, should be carried out.

5. With regard to windshear on final approach to land:(a) (b) (c) (d)

a decrease in the headwind will cause the aircraft to overshoot, a decrease in the headwind will cause the aircraft to undershoot, an increase in the headwind will cause the aircraft to undershoot, an increase in the headwind will have no effect on the flight path.

6. With regard to windshear during the climb out after take-off:(a) (b) (c) (d)

a decrease in the headwind will cause the aircraft to overshoot, an increase in the headwind will have no effect on the flight path, an increase in the headwind will cause the aircraft to undershoot, an increase in the headwind will cause the aircraft to overshoot.



I


ANSWERS Page 6-A1

ANSWERS 1. (c) See page 6-3, (ii) ARRIVING. 2. (a) See page 6-3, (ii) ARRIVING. 3. (b) See page 6-2, (i) TAKE-OFF

4. (d) See page 6-1, 2. FLARE AND TOUCHDOWN 5. (b) See page 6-3, C. WIND SHEAR 6. (d) See page 6-3, C. WIND SHEAR



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