1 A320 Takeoff Performance-1

(1)

Takeoff Performance

(2)

Takeoff Performance

MTOW

Flexible Takeoff

1 Contaminated Takeoff

2

(3)

RTOW Charts

RTOW charts enable to:

Determine a MTOW and its associated takeoff speeds

Determine a flexible temperature

MTOW

V

₁

, V

_R

, V

₂

T

_flex

MTOW

V

₁

, V

_R

, V

₂

T

_flex

(4)

(5)

(6)

(7)

(8)

(9)

Example 1: MTOW

OAT

= 44°C

Wind

= Calm

Air Conditioning = ON

QNH

= 1003 hPa

Rwy Condition = DRY

(10)

Solution

CONF 1+F, 0 kt, 44 °C

MTOW = 78.2 T

CONF 2, 0 kt, 44 °C

(11)

(12)

Solution

Conf 1+F

Conf 2

CHART READING

78.2

78.7 A/C ON

-1.8

Intermediate Values

76.4 No speed correction

76.9 No speed correction

QNH

Final Values

(13)

(14)

Solution

Conf 1+F

Conf 2

CHART READING

78.2

78.7 A/C ON

-1.8

Intermediate Values

76.4 153/53/55

76.9 150/51/55

QNH

Final Values

(15)

OAT ≤ T

_VMC

44°C ≤ 61°C

(16)

Solution

Conf 1+F

Conf 2

CHART READING

78.2

78.7 A/C ON

-1.8

Intermediate Values

76.4 153/53/55

76.9 150/51/55

QNH

-0.7/ -2

-1/ -1/ -1

-0.7/ -2

0/ 0/ -1

Final Values

75.7 152/52/54

76.2 150/51/54

(17)

Determining Airport Pressure Altitude

Airport elevation

= 489 ft

QNH

= 1003 hPa

Pressure altitude correction =

+300 ft

Airport pressure altitude

=

789 ft

(18)

Check V2 & VMU

Conf 1+F

75.7 152/52/54

Conf 2

76.2 150/51/54

(19)

Result

Conf 1+F

Conf 2

Final Values

75.7 152/52/54

76.2 150/51/54

Check Min V1/VR/V2

Check V2 limited by VMU

MTOW = 76.2 T

150/51/54

CONF 2

(20)

(21)

(22)

Example 2: MTOW

OAT

= 25°C

Wind

= 10 kt

Air Conditioning = ON

QNH

= 998 hPa

Rwy Condition = Wet

(23)

Solution

CONF 1+F, 10 kt, 25 °C

MTOW = 80.6+

0.46 x1+

0.06 x19 = 82.2 T

CONF 2, 10 kt, 25 °C

(24)

(25)

Solution

Conf 1+F

Conf 2

CHART READING

82.2

82.1 A/C ON

-1.8

Intermediate Values

80.4 155/56/58

80.3 152/53/58

WET

Intermediate Values

QNH

Final Values

(26)

Solution: WET Correction

(27)

Solution

Conf 1+F

Conf 2

CHART READING

82.2

82.1 A/C ON

-1.8

Intermediate Values

80.4 155/56/58

80.3 152/53/58

WET

-0.7/ -2

-8/ -2/ -2

-0.2/ -1

-5/ 0/ 0

Intermediate Values

79.7 147/54/56

80.1 147/53/58

QNH

Final Values

(28)

Solution: QNH Correction

(29)

Solution

Conf 1+F

Conf 2

CHART READING

82.2

82.1 A/C ON

-1.8

Intermediate Values

80.4 155/56/58

80.3 152/53/58

WET

-0.7/ -2

-8/ -2/ -2

-0.2/ -1

-5/ 0/ 0

Intermediate Values

79.7 147/54/56

80.1 147/53/58

QNH

-0.7

*1.5

/ -2

-1

*1.5

/ -1

*1.5

/ -1

*1.5

-0.7

*1.5

/ -2

0 *1.5

/ 0

*1.5

/ -1

*1.5

Final Values

78.6 145/53/55

79 147/53/57

(30)

Determining Airport Pressure Altitude

Airport elevation

= 489 ft

QNH

= 998 hPa

Pressure altitude correction =

+400 ft

(31)

Check V2 & VMU

Conf 1+F

78.6 145/53/55

Conf 2

79 147/53/57

(32)

Result

Conf 1+F

Conf 2

Final Values

78.6 145/53/55

79 147/53/57

Check Min V1/VR/V2

Check V2 limited by VMU

MTOW = 79 T

147/53/57

CONF 2

(33)

Example 3: MTOW

OAT

= 55°C

Wind

= -10 kt

Air Conditioning

= ON

QNH

= 1023 hPa

Runway Condition

= DRY

(34)

OAT > T

_MAX

55°C > 54°C

NO TAKEOFF!!!

(35)

(36)

(37)

Max TOW

Actual TOW

Available

Thrust

Required

Thrust

Flat rated Thrust

Tref

Weight

Thrust

EGT Limit

T

Flex Temp

OAT

Thrust-Temperature

AMJ 25-13 ; AC 25-13

Flex Takeoff: what for?

To enable takeoff without engines at full rate reduces: • the probability of a failure (safety aspect)

• the engine deterioration rate and associated maintenance costs (economic aspect)

(38)

(39)

(40)

(41)

(42)

(43)

(44)

Example 4: Flexible Temperature

Actual Takeoff Weight = 68 000 kg

OAT

= 25°C

Head Wind

= 10 kt

Air Conditioning

= ON

QNH

= 998 hPa

(45)

Solution

Conf 1+F

Conf 2

CHART READING

82.2

82.1 A/C ON

-1.8

Intermediate Values

80.4 155/56/58

80.3 152/53/58

WET

-0.7/ -2

-8/ -2/ -2

-0.2/ -1

-5/ 0/ 0

Intermediate Values

79.7 147/54/56

80.1 147/53/58

QNH

-0.7

*1.5

/ -2

-1

*1.5

/ -1

*1.5

/ -1

*1.5

-0.7

*1.5

/ -2

0 *1.5

/ 0

*1.5

/ -1

*1.5

Final Values

78.6 145/53/55

79 147/53/57

(46)

Solution: ATOW & MTOW

Check that the speeds are higher than minimum speeds

from the chart and from VMU table. (It is reminded that if the

speed checks are not fulfilled, the corrections must be

recalculated using those provided on lines 3 and 4).

Since the speed check is fulfilled:

Conf 1+F

Conf 2

Final Values

78.6 145/53/55

79 147/53/57

(47)

Solution

Conf 1+F

Conf 2

Flexible Temperature

67°C

66°C

(48)

Solution

Conf 1+F

Conf 2

FLEX TEMPERATURE

(chart reading)

67°C

153/53/54

66°C

149/49/53

A/C ON

Intermediate Values

WET

Intermediate Values

QNH

(49)

(50)

Solution

Conf 1+F

Conf 2

FLEX TEMPERATURE

(chart reading)

67°C

153/53/54

66°C

149/49/53

A/C ON

-3°C

No speed correction

-3°C

No speed correction

Intermediate Values

64°C

153/53/54

63°C

149/49/53

WET

Intermediate Values

QNH

(51)

Solution: WET Correction

Conf 1+F

T

_flex

≤ T

_VMC

64°C ≤ 69°C

Conf 2

T

_flex

≤ T

_VMC

63°C ≤ 69°C

(52)

Solution

Conf 1+F

Conf 2

FLEX TEMPERATURE

(chart reading)

67°C

153/53/54

66°C

149/49/53

A/C ON

-3°C

No speed correction

-3°C

No speed correction

Intermediate Values

64°C

153/53/54

63°C

149/49/53

WET

-0.7/ -2

-8/ -2/ -2

-0.2/ -1

-5/ 0/ 0

Intermediate Values

62°C

145/51/52

62°C

144/49/53

QNH

(53)

Solution: QNH Correction

Conf 1+F

T

_flex

> T

_VMC

62°C > 61°C

Conf 2

T

_flex

> T

_VMC

62°C > 61°C

(54)

Solution

Conf 1+F

Conf 2

FLEX TEMPERATURE

(chart reading)

67°C

153/53/54

66°C

149/49/53

A/C ON

-3°C

No speed correction

-3°C

No speed correction

Intermediate Values

64°C

153/53/54

63°C

149/49/53

WET

-0.7/ -2

-8/ -2/ -2

-0.2/ -1

-5/ 0/ 0

Intermediate Values

62°C

145/51/52

62°C

144/49/53

QNH

-0.7/ -2

*1.5

No speed correction

-0.7/ -2

*1.5

No speed correction

(55)

Determining Airport Pressure Altitude

Airport elevation

= 489 ft

QNH

= 998 hPa

Pressure altitude correction =

+400 ft

Airport pressure altitude

=

889 ft

(56)

Check V2 & VMU

Conf 1+F

59°C

145/51/52

Conf 2

59°C

144/49/53

(57)

Result

Conf 1+F

Conf 2

Final Values

59°C

145/51/52

59°C

144/49/53

Check Min. speeds

Check V2 limited by VMU

Check that:

OAT ≤ T

_max

T

_flex

≤ T

_flex,max

T

_flex

> OAT

T

_flex

> T

_ref

Retain CONF 2

when flexible

temperature

difference between

two flap settings is

low, use the highest

flap setting.

(58)

(59)

Takeoff Performance

Contaminated Takeoff

2 MTOW

Flexible Takeoff

1

(60)

(61)

Definitions

Dry Snow

Wet Snow

Slush

Standing

Compacted

Wet runway

corrections

any depth

3 mm

6.3 mm

6.3mm

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

3 mm

(62)

Definitions

any depth

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

Contaminated

runway corrections

B. 12.7mm (1/2 inch) water

A. 6.3 mm (1/4 inch) water

(63)

Definitions

Dry Snow

Wet Snow

Slush

Standing

15 mm

Compacted

4 mm

any depth

3 mm

6.3 mm

6.3mm

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

3 mm

_{3 mm}

Contaminated

runway corrections

C. 6.3 mm (1/4 inch) slush

(64)

Definitions

any depth

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

Contaminated

runway corrections

D. 12.7mm (1/2 inch) slush

(65)

Definitions

Dry Snow

Wet Snow

Slush

Standing

15 mm

Compacted

4 mm

any depth

3 mm

6.3 mm

6.3mm

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

3 mm

_{3 mm}

Contaminated

runway corrections

E. Compacted snow

(66)

Definitions

any depth

12.7 mm

12.7mm

12.7 mm

25.4 mm

50.8 mm

101.6 mm

(67)

Example 5: Contaminated TOW

Runway length = 3 000 m (no clearway)

OAT

= 5°C

Wind

= 5 kt Tail Wind

Runways Status = 8 mm slush

CONF 1+F

TOW

V1/VR/V2

TOW

(68)

Solution: MTOW

(69)

Solution

Corrected weight = 80 200 – 14 700 =

65 500 kg

Determine maximum takeoff weight and associated

speeds:

(70)

Result

MTOW =

65 500 kg

V1 =

141 kt

, VR =

147 kt

, V2 =

148 kt

MTOW =

65 500 kg

(71)