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Chapter 6Aero Vehicle PerformanceAccelerated Performance

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CONTENTS• PART I: Turning

– Level turn• Minimum turn radius• Maximum turn rate

– Pull-up maneuver– Pull down maneuver– The V-n diagram

• PART II: Energy concepts: Accelerated rate of climb• PART III: Takeoff performance• PART IV: Landing performance

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Chapter 6 Accelerated Performance

(Part I)Turning Performance

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EOM FOR TURNING

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6

EOM for Flight in a Vertical Plane

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TURNING PERFORMANCELEVEL TURN

• H PLANE• Climb angle zero• AC banked•Alt constant

Ф

Ф

WE ARE USING ANGLE Ф

• You will note in the front view of the aircraft that the plane is rolled away from the horizontal and vertical planes by the angle φ discussed earlier. The lift vector L acting on the aircraft is also rolled by that same angle such that it no longer directly opposes the plane's weight. Now observe the following diagram.

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In both cases, the aircraft is rolled to the same bank angle. In the first case, however, the vertical component of lift is less than the weight. Because of this inequality, the greater force imparted by the weight will pull the aircraft downward and it does not maintain the same altitude. The pilot can overcome this behavior by pulling the stick back to increase the lift of the plane and maintain the same altitude. It is for this reason that we refer to the maneuver as a level turn, since the aircraft is banked into a turning motion but maintains the same altitude.

Level turn• AC put in a bank . Finite radius of turn r2 (R) • Curved path is in horizontal plane parallel to ground.• Altitude remains constant.• Climb angle θ is zero. Thus r1 is infinity. No acceleration in

vertical plane• Thrust is parallel with drag thus є is zero• Velocity is constant• Stick is pulled back to maintain constant altitude

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EOM for Flight in a Vertical Plane• Applying Newton’s 2nd law in the flight path

direction, gives

[4.5]

• Similarly, in the direction perpendicular to the flight path, we have

where r1 is the radius of curvature of the flight path

• Again, applying Newton’s 2nd law in the perpendi-cular direction, gives

!__________[4.6]

• EOM in horizontal plane

sincos WDTdt

dVm

coscossincos WTLF

1

2

rVa

coscossincos1

2

WTLr

Vm

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sinsinsincos

2

2

TLr

Vm

13

EOM for level turn• General EOM in flight path direction is

• For level turn with constant velocity and θ & є zero it reduces to T= D

• General EOM in the perpendicular to flight path direction is

• With θ as zero r1 becomes infinity. Thus equation in this case reduces to

L cosΦ = W 6.1 L = W / cosΦ

sincos WDTdt

dVm

coscossincos1

2

WTLr

Vm

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EOM for level turn• EOM in horizontal plane

• Reduces to

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sinsinsincos

2

2

TLr

Vm

2

2

sin (6.2)Vm Lr

2

2

sin (6.3)

with r replaced by R

Vm LR

• Component of lift balances the centrifugal force

15

EOM for level turn

• Three equations reduce to• T=D• L cosΦ = W 6.1

L = W / cosΦ •

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2

2

sin (6.3)

with r replaced by R

Vm LR

16

EOM for level turn

• Look at equation 6.1• L cosΦ = W 6.1

L = W / cosΦ

• Thus in turning lift requirement is increased as bank angle increases and at Φ = 90 degrees , L requirement is infinity.

• Hence sustained 90 degrees turn is not possible

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LEVEL TURN

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• Performance characteristics of interest– Turn radius R to be minimum– Turn rate: which is the angular velocity ω to be maximum

TURN RADIUS

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TURN RADIUS• LOAD FACTOR: It is defined as the lift to drag ratio

– n = L/W• The load factor describes how many g's act on an aircraft in

any given maneuver. For example, a plane with a total lift five times greater than its weight experiences a load factor of 5 g's. In more physical terms, we often refer to the load factor as "apparent weight." In other words, a pilot pulling 5 g's feels like he weighs five times more than normal because of the additional force acting on his body.

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TURN RADIUS• LOAD FACTOR: It is defined as

• n = L/W• n=5 means lift requirement is 5xW

• From equation L cosΦ = W (6.1)

• With• We get• Thus bank angle depends upon load factor for maximum

bank angle

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LnW

1cosArcn

nWLL

W 11cos

TURN RADIUS

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2 2 2

6.7sin sin sin

mV W V VRL L g gn

2

sinVm LR

• In eq 6.3 replace m = W/g and

solve for R

• We know 1cosn

2 2cos sin 1 2

21 sin 1n

22

1 1sin 1 1 6.8nn n

Putting 6.8 in 6.7

TURN RADIUS

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2

2(6.9)

1

VRg n

Minimum Turn Radius

THUS TO HAVE SMALLEST POSSIBLE RADIUS OF TURN FLY AT• Highest possible value of n• Lowest possible velocity at which corresponding n can be sustained

Angular Velocity• We know

• Replacing R from eq 6.9

• To have largest possible turn rate we want– Highest possible load factor– Lowest possible velocity

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d Vdt R

2 1 (6.11)g nV

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Load Factor

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1 2 2, 02 ( )LDD V S C KC

T D1 22 LL nW V SC

2

2L

nWCV S

22

, 0 2

1 22

DnWT V S C K

V S

1/21 2, 01 22

2( / ) /DV T Cn V

K W S W W S

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• It can have values •THUS TO HAVE max LF

• Fly at Max trust available• In addition design parameters and altitude matter

1/21 2, 01 22

2( / ) /DV T Cn V

K W S W W S

Max Load Factor

1/21 2, 022

maxmax

1( / ) 2 /

DV T Cn VK W S W W S

max1 n n

Eq 6.17

Eq 6.18

The Max Load Factor• We know

• Max value at a given L/D

• Max of max at max L/D

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L L D L TnW D W D W

maxmax

L TnD W

max max

ML TnD W

Eq 6.21

Low velocity Limit on max value of n

• At stalling speed CL is max

• Max bank angle will be

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212

LL Sn V C

W W

max

2 max( )12 /

L

LCCn VW S

max

1maxcos n

Eq 6.23

Low velocity limit on max n

• At stalling speed CL is max• Stalling speed is given as

• Thus as n ↑ stall speed also ↑• Velocity cant be less than stall velocity

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1 22 LnWL V SC

2L

nWSCV

max

2stall ( )L

W nS CV

21

nVthus stall Eq 6.25

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Max Bank Angle• Max bank angle will be

• Instantaneous 90 degree bank possible• Sustained 90 degree bank not possible as it will require

n = infinity or lift required will be infinity

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max

1maxcos n

nmax Values• We studied three nmax values

– One by equation 6.18

– Two by equation 6.21

– Three by equation 6.23

– And corresponding max bank angle

– This all information is presented graphically on next slide

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1/21 2, 022

maxmax

1( / ) 2 /

DV T Cn VK W S W W S

max max

ML TnD W

max

2 max( )12 /

L

LCCn VW S

max

1maxcos n

Constraints on Max Value of n

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1/21 2, 022

maxmax

1( / ) 2 /

DV T Cn VK W S W W S

max

2stall ( )L

W nS CV

max

1maxcos n

Left of point A eq 6.18 is not applicable rather eq6.23 is applicable

Right of point A max value of n is limited by thrust and on left it is limited by stall

• Point B correspond to n max of max at L/D)max. • on left of B lift dependent drag and on right zero lift drag is dominant

Far right point is limited by excess power to attain max velocity with n=1

This graph applicable for one altitude

We will discuss point C and D with example 6.1

MINIMUM RADIUS OF TURN

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MINIMUM RADIUS OF TURN• We know

• Which can be written as

• For min value differentiate and equate it to zero

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2

2(6.9)

1

VRg n

MINIMUM RADIUS OF TURN

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1/21 2, 01 22

2( / ) /DV T Cn V

K W S W W S

MINIMUM RADIUS OF TURN

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Load factor corresponding to this velocity is obtained by putting eq 6.30 in eq 6.28

MINIMUM RADIUS OF TURN

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EXAMPLE 6.1

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Constraints on Max Value of n

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1/21 2, 022

maxmax

1( / ) 2 /

DV T Cn VK W S W W S

max

2stall ( )L

W nS CV

max

1maxcos n

Left of point A eq 6.18 is not applicable rather eq6.23 is applicable

Right of point A max value of n is limited by thrust and on left it is limited by stall

• Point B correspond to n max of max at L/D)max. • on left of b lift dependent drag is dominant and on right zero lift drag is dominant

Far right point is limited by excess power to attain max velocity with n=1This graph

applicable for one altitude

Point C for theoretical min R is below stall. Point A is min R possible

Point D is theoretical max turn rate point. Max posible turn rate is at point A

EXAMPLE 6.1

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Turn Rate or Angular Velocity• We know

• Replacing R from eq 6.9

• To have largest possible turn rate we want– Highest possible load factor– Lowest possible velocity

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d Vdt R

2 1 (6.11)g nV

MAXIMUM TRURN RATE

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By following the same procedure as for Minimum radius it can be shown that

EXAMPLE 6.2

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EXAMPLE 6.2

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Coordinated level turn

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• This very photogenic biplane was flown by Greg Shelton with Ashley Battles as the wing-walker, going through a pretty hair-raising sequence of climbs, barrel rolls and hammerheads. Apparently Ashley is a pilot herself, maybe that makes it a bit easier to put yourself through maneuvers like this.

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PULL-UP MANEUVER

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PULL-UP MANEUVER

Chapter 6 Figure

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57

EOM for PULL-UP

• General EOM in the perpendicular to flight path direction is

• With Φ and є as zero and r1 replaced by R we have

• Instantaneous pull up is initiated from straight and level flight– L=W with θ = 0 thus we get

coscossincos1

2

WTLr

Vm

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2

cosVm L WR

2Vm L WR

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PULL-UP

Pull Down

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60

EOM for PULL-for pull down

• From the figure

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2Vm L WR

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PULL DOWN

62

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Chapter 5 - PART A 64

65

V-n DIAGRAM• Limitations;

– Aerodynamic– Structural ( For +ive and -ive load factor )

• Limit Load Factor: Permanent deformation of structure• Ultimate Load Factor: Failure/Breakage of strucrure

– Dynamic Pressure Limit: In dive etc. Around 1.2 times Vmax in level flight

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V-nDIAGRAM

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max

2 max( )12 /

L

LCC

n VW S

Dynamic Pressure Limit

Corner velocity

* max

max

2( )L

n WVC S

DESIGN LOAD FACTORS• Excessive load factors must be avoided because of the possibility of exceeding the

structural strength of the aircraft.• Aviation authorities specify the load factor limits within which different classes of aircraft

are required to operate without damage. For example, the US Federal Aviation Regulations prescribe the following limits (for the most restrictive case):

– For commercial transport airplanes, from -1 to +2.5 (or up to +3.8 depending on design takeoff weight)

– For light airplanes, from -1.5 to +3.8 – For aerobatic airplanes, from -3 to +6 – For helicopters, from -1 to +3.5

• However, many aircraft types, in particular aerobatic airplanes, are designed so that they can tolerate load factors much higher than the minimum required. For example, the Sukhoi Su-26 family have load factors limits of -10 to +12.

• The maximum load factors, both positive and negative, applicable to an aircraft are usually specified in the pilot's operating handbook.

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SU 26, 29, 31

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SU 31

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31

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• The main features of the new aeroplane were: – high power-to-weight ratio achieved through a rugged design, the weight of the airframe

being kept down to a minimum; – special aerodynamic symmetric-profile wing configuration to enable superior piloting

performance; – a robust airframe structure capable of sustaining +12/-10 G loads; – extensive use of composite materials in the design; – ergonomic cockpit configuration.

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SU 26 VS 31• SU 26 General characteristics• Crew: 1 • Length: 6.83 m (22 ft 5 in) • Wingspan: 7.80 m (25 ft 7 in) • Height: 2.89 m (9 ft 6 in) • Wing area: 11.83 m² (127 ft²) • Empty weight: 736 kg (1,619 lb) • Max. takeoff weight: 1,206 kg (2,653 lb) • Powerplant: 1 × Vedeneyev M-14P radial engine, 270

kW (360 hp) • Performance• Maximum speed: 450 km/h (281 mph) • Cruise speed: 310 km/h (193 mph) • Range: 800 km (500 mi) • Service ceiling: 4,000 m (12,120 ft) • Rate of climb: 18 m/s (3,543 ft/min)• Load factor: -10g to 12g

• SU 31 General characteristics• Crew: 1 • Length: 6.83 m (22.41 ft) • Wingspan: 7.80 m (25.59 ft) • Height: 2.76 m (9.06 ft) • Wing area: 11.83 m² (127.34 sq ft) • Empty weight: 700 kg (1,543 lb) • Max. takeoff weight: 1,050 kg (2,315 lb) • Powerplant: 1 × Vedeneyev M-14PF, 294 kW (400

hp) • Performance• Never exceed speed: 450 km/h (243 knots, 280 mph) • Maximum speed: 331 km/h (178 knots, 205 mph) • Stall speed: 106 km/h (57 knots, 66 mph) • Range: 1,100 km (594 nmi, 684 mi) • Rate of climb: 24 m/s (4,724 ft/min)

• Load factor: -10g to 12g • Roll rate: 7 rad/sec (401 deg/sec

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SU 26• General characteristics• Crew: 1 • Length: 6.83 m (22 ft 5 in) • Wingspan: 7.80 m (25 ft 7 in) • Height: 2.89 m (9 ft 6 in) • Wing area: 11.83 m² (127 ft²) • Empty weight: 736 kg (1,619 lb) • Max. takeoff weight: 1,206 kg (2,653 lb) • Powerplant: 1 × Vedeneyev M-14P radial engine,

270 kW (360 hp) • Performance• Maximum speed: 450 km/h (281 mph) • Cruise speed: 310 km/h (193 mph) • Range: 800 km (500 mi) • Service ceiling: 4,000 m (12,120 ft) • Rate of climb: 18 m/s (3,543 ft/min)• Load factor: -10g to 12g

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SHERDILS

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SHERDILS

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Sherdils: History• Sherdils is the aerobatics display team of the Pakistan Air Force Academy based at

the PAF Academy, Risalpur, Pakistan. The team flies K-8 Karakorum trainer aircraft and used to fly the T-37 "Tweety Bird". Its pilots are instructors from the PAF Academy's Basic Flying Training (BFT) Wing.

• The team was officially formed on 17 August 1972 as a result of efforts by an Academy instructor, Sqn Ldr Bahar-ul-haq.

• The initial performances of the team were highly successful. The "Tweety birds" performed at air shows for foreign dignitaries, including heads of state and military officers.

• Initially, the team had no name. It flew as ‘Sherdils’ for the first time on 19 September 1974

• The Sherdils have transitioned from the T-37 to the newer and more modern Hongdu K-8 "Karakorum" trainer jointly developed with China.

• The type of formation includes – line astern to diamond formation during a loop, then clover-leaf, steep turn, barrel roll and finally, the breath-taking bomb-burst.

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Cuban 8

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Chapter 6 (Part III)

Accelerated PerformanceENERY CONCEPTS: ACCELERATED RATE OF CLIMB

and Specific Excess Power (PS) Plots

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CONTENTS• PART I: Turning

– Level turn• Minimum turn radius• Maximum turn rate

– Pull-up maneuver– Pull down maneuver– The V-n diagram

• PART II: Energy concepts: Accelerated rate of climb• PART III: Takeoff performance• PART IV: Landing performance

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ENERGY HEIGHT• Consider an airplane with

– Mass m– Height h– Velocity V

• Potential energy is (mgh)

• Kinetic energy is ½ mV2

• Total aircraft energy = mgh + ½ mV2

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ENERGY HEIGHT• Specific Energy (He): Total aircraft energy per unit

weight

• Specific energy has units of height and is called energy height

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2 2

2

1/ 2 1/ 2

2

e

e

mgh mV mgh mVHW mg

VH hg

EXAMPLE 6.4

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Contours of Constant Energy Height

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• AC at Point A, B and C • AC at point D

• Compare the two AC

gV

e hH 2

2

SPECIFIC EXCESS POWER (Ps)• Recall equation EOM 4.5 in vertical plane

• For pure climb case with Ф and є as zero

• The equation for accelerated climb reduces to

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sincos WDTdt

dVm

sin

sin 6.57

dVm T D Wdt

dVT D W mdt

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• In equation 6.61– if acceleration is zero

• Which is the rate of climb– If height is constant

• Which is acceleration of the AC

• Thus Ps is a measure of climbing and accelerated performance of any aircraft

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dtdV

gV

dtdh

sP

ROCP dtdh

gV

dtdh

s )0(

dtdV

gV

dtdV

gV

sP 0

SPECIFIC EXCESS POWER (Ps)

RELATION BETWEEN ENERGY HEIGHT (He) and Ps

• We have seen

• And

• Differentiate first equation wrt time

• ComparingPs = d He /dt

Thus Ps is a measure of how quickly an aircraft can change its energy height from one level to another one

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2

2eVH h

g

dtdV

gV

dtdh

sP

HOW TO DRAW Ps Plots• We have seen

• At a given altitude h1 find excess power for a range of velocities/mach numbers as in example 5.1

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EXAMPLE 5.13

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HOW TO DRAW Ps Plots• We have seen

• At a given altitude h1 find excess power for a range of velocities/mach numbers as in example 5.13

• Plot these Ps values vs mach no for height h1 as shown

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HOW TO DRAW Ps Plots• We have seen

• At a given altitude h1 find excess power for a range of velocities/mach numbers as in example 5.13

• Plot these Ps values vs mach no for height h1 as shown

• Repeat the procedure for another altitude h2• Plot these Ps values vs mach no for height h2• Repeat for other height h3 and so on

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HOW TO DRAW Ps Plots• We have seen

• At a given altitude h1 find excess power for a range of velocities/mach numbers as in example 5.13

• Plot these Ps values vs mach no for height h1 as shown

• Repeat the procedure for another altitude h2• Plot these Ps values vs mach no for height h2• Repeat for other height h3 and so on• Now from this plot find values of constant Ps at various

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04/28/23 Chapter 6 – PART 2 97

sftP

sftP

sftP

sftP

s

s

s

s

0

100

200

300

HOW TO DRAW Ps Plots• We have seen

• At a given altitude h1 find excess power for a range of velocities/mach numbers as in example 5.13

• Plot these Ps values vs mach no for height h1 as shown

• Repeat the procedure for another altitude h2• Plot these Ps values vs mach no for height h2• Repeat for other height h3 and so on• Now from this plot find values of constant Ps at various altitudes• Now plot these constant Ps plots

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04/28/23 Chapter 6 – PART 2 99

sftP

sftP

sftP

sftP

s

s

s

s

0

100

200

300

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From USAF Academy Book

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Subsonic and Supersonic plots

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Subsonic Vs supersonic plot

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Supersonic AC Plots

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Dynamic Pressure Limit

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Minimum Time or Optimum Energy Trajectory

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Minimum Time or Optimum Energy Trajectory

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Comparison of Actual and Theoretical Climb

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Time to Climb

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Time to Climb

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Minimum Fuel to Climb Trajectory

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0T

With edhdt

Eq 5.13

HOW TO DRAW Fs Plots• We have seen

• At a given altitude h1 find above value (Fs )for a range of velocities/mach numbers

• Plot these values of Fs

• Repeat for another altitude h2• Plot the values of Fs vs mach no• Now from first this plot find values of constant Fs at various

altitudes• Now plot these constant Fs plots

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Fuel to Climb

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Comparison of AC on Specific Excess Power Plots

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126

Chapter 6 (PART III)Airplane Performance

Accelerated PerformanceTAKEOFF (TO)

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CONTENTS• PART I: Turning

– Level turn• Minimum turn radius• Maximum turn rate

– Pull-up maneuver– Pull down maneuver– The V-n diagram

• PART II: Energy concepts: Accelerated rate of climb• PART III: Takeoff performance• PART IV: Landing performance

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TAKEOFF (TO)

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DIFFERENT VELOCITIES DURING TO

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Takeoff: Important Speeds• The following speeds are of importance in the take-off field length calculation:

– VS Stall speed

– Vmu Minimum Unstick Speed. Minimum airspeed at which airplane can safely lift off ground and continue take-off.

– Vmcg Minimum control speed on the ground. At this speed the aircraft must be able to continue a straight path down the runway with a failed engine, without relying on nose gear reactions.

– Vmc Minimum Control Speed. Minimum airspeed at which when critical engine is made inoperative, it is still possible to recover control of the airplane and maintain straight flight.

– V1 Decision speed, a short time after critical engine failure speed. Above this speed, aerodynamic controls alone must be adequate to proceed safely with takeoff.

– VR Rotation Speed. Must be greater than V1 and greater than 1.05 Vmc

– Vlo Lift-off Speed. Must be greater than 1.1 Vmu with all engines, or 1.05 Vmu with engine out.

– V2 Take-off climb speed is the demonstrated airspeed at the 35 ft height. Must be greater than 1.1 V mc and 1.2 Vs, the stalling speed in the take-off configuration.

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CALCULATIONS OF GROUND ROLL

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F = T

CALCULATIONS OF GROUND ROLL

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CALCULATIONS OF GROUND ROLL

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CALCULATIONS OF GROUND ROLL

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NET FORWARD FORCE IS ALMOST CONSTANT, THUST CONSTANT ACCELERATION

CALCULATIONS OF GROUND ROLL

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Use text book value 1.1 times

CALCULATIONS OF GROUND ROLL

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From Anderson book

CALCULATIONS OF GROUND ROLL• With

• With Take off velocity = 1.1 Stall velocity

• Thus

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MINIMIZING THE RESISTANCE

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FACTORS EFFECTING T.O. DISTANCE• In addition to pilot factor, AC configuration, throttle

setting and thrust augmentation following factors may cause variation from a standard day sea level take off performance :– WEIGHT– WIND– RUNWAY SLOPE– PRESSURE ALTITUDE– TEMPRATURE

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WEIGHT EFFECT

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WIND EFFECT

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S

RUNWAY SLOPE

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Out of headwind and slope both normally choice is Takeoff into head wind.

TEMPRATURE AND DENSITY ALTITUDE

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VARIATION OF FORCES DURING TO

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CALCULATION OF ARIAL DISTANCE TO CLEAR OBSTACLE

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CALCULATION OF ARIAL DISTANCE TO CLEAR OBSTACLE

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ARIAL DISTANCE

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FACTORS EFFECTING DISTANCEInfluencing factor Increase in Take-off Distance Increase in Landing Distance

10% increase in weight 20% 10%

Increase of 1000' in runway altitude (1" MAP)

10% 5%

Increase in temperature of 10°C above ISA

10% 5%

Dry short grass (under 5") 20% 20%

Dry long grass (5 - 10") 25% 30%

Wet short / long grass 25% / 30% 30% / 40%

2% uphill slope 10% -10%

1% downhill slope -5% 5%

Tailwind component of 10% of liftoff speed

20% 20%

Soft ground or snow at least 25% at least 25%

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162

Chapter 6 (PART IV)Airplane Performance

Accelerated PerformanceLANDING

IZHAR KAZMI

CONTENTS• PART I: Turning

– Level turn• Minimum turn radius• Maximum turn rate

– Pull-up maneuver– Pull down maneuver– The V-n diagram

• PART II: Energy concepts: Accelerated rate of climb• PART III: Takeoff performance• PART IV: Landing performance

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LANDING

04/28/23 Chapter 6 – PART 2 165

THE LANDING PATH AND LANDING DISTANCE

LANDING PERFORMANCE.• VERY CRUCIAL PHASE OF FLYING• PERFORMANCE ANNALYSIS VERY SIMILAR TO TAKE OFF• THREE STAGES

– Aerial stage• Descent from50 feet obstacle to flare out• Velocity 1.3 or 1.2 Times of stall speed for Civil and mil Ac respectively

– Flare out Stage• Transition from straight descending approach path to horizontal parallel with runway path• Distance from 50 feet to start of fare out is called approach distance

– Touchdown/ground roll stage• Touchdown starts when wheels touch the ground• Velocity is 1.15 or 1.1 times of stall Velocity for Civil and Mil AC Respectively• Distance from start of flare to touch down is called flare distance• Free roll for some time• Distance from TD to zero velocity is called the ground roll distance

• Thus total landing distance = Approach distance + Flare distance + Ground roll distance

04/28/23 Chapter 5 - PART A 166

LANDING DISTANCE CALCULATIONS

• Total landing distance = Approach distance + Flare distance + Ground roll distance

• We will proceed to calculate:– Approach distance– Flare distance– Ground roll distance

04/28/23 Chapter 5 - PART A 167

04/28/23 Chapter 6 – PART 2 168

CALCULATION OF APPROACH DISTANCE

IZHAR KAZMI 169

CALCULATION OF APPROACH DISTANCE

IZHAR KAZMI 170

aaf uu

50 fa

a

hTan

s

a

fa

hS

tan50

CALCULATION OF APPROACH DISTANCE

04/28/23 Chapter 6 – PART 2 171

50 fa

a

hTan

s

aaf uu

a

fa

hS

tan50

CALCULATION OF APPROACH DISTANCE

CALCULATION OF APPROACH DISTANCE• Total landing distance = Approach distance + Flare distance + Ground roll

distance• Approach distance from the geometry is given as

• In order to calculate approach distance we need to know– Flare height– And angle

• Now we will find the above mentioned two unknowns and

IZHAR KAZMI 172

a

fa

hS

tan50

afh

fh a

04/28/23 Chapter 6 – PART 2 173

GEOMETRY OF THE LANDING FLARE

04/28/23 Chapter 6 – PART 2 174

GEOMETRY OF THE LANDING FLARE

afu

)cos1(

cos

af

af

Rh

RRh

To know flare height we need to know R and θ

aR cos

fh

04/28/23 Chapter 6 – PART 2 175

CALCULATION OF APPROACH DISTANCE

aW cos

aW sin

a

a

WLWTD

cos

sin

EOMs in FP and Perpendicular to FP directions are:

04/28/23 Chapter 6 – PART 2 176

CALCULATION OF APPROACH DISTANCE

04/28/23 Chapter 5 - PART A 177

04/28/23 Chapter 5 - PART A 178

2

26.41

1

VRg n

6.107

CALCULATION OF APPROACH DISTANCE• Total landing distance = Approach distance + Flare distance + Ground roll

distance• Approach distance from the geometry is given as

• In order to calculate approach distance we needed to know– Flare height– And angle

• Now we have found the two unknowns and as

IZHAR KAZMI 179

a

fa

hS

tan50

afh

fh a

CLCULATION OF FLARE DISTANCE

• Total landing distance = Approach distance + Flare distance + Ground roll distance

• We have seen how to find approach distance• Now we will proceed to calculate:

– Flare distance

04/28/23 Chapter 5 - PART A 180

04/28/23 Chapter 6 – PART 2 181

GEOMETRY OF THE LANDING FLARE AND CLCULATION OF FLARE DISTANCE

04/28/23 Chapter 6 – PART 2 182

GEOMETRY OF THE LANDING FLARE

af

aaf

RS

uu

sin

af RS sin

04/28/23 Chapter 6 – PART 2 183

CALCULATION OF FLARE DISTANCE

Where

CALCULATION OF LANDING GROUND ROLL

• Total landing distance = Approach distance + Flare distance + Ground roll distance

• We have seen how to find – Approach distance– Flare distance

• Now we will proceed to calculate:– Ground roll distance

• But before that we will have a look how drag is effected during landing and takeoff

04/28/23 Chapter 5 - PART A 184

LIFT AND DRAG FORCES DURING LANDING AND TAKEOFF GROUND ROLL

• During Landing and Takeoff drag is effected due to – Landing gears additional drag – Reduction in induced drag due to ground effect by factor G

• Total drag = parasite drag + wave drag + induced drag• In coefficient form, this equation is

• Original drag polar of AC is given as• With k1+ k2+ k3 = K

04/28/23 Chapter 5 - PART A 185

AReC

CCC LwDeDD

2

,,

20, LDD KCCC

0DC

IZHAR KAZMI 186

LIFT AND DRAG FORCES DURING LANDING AND TAKEOFF GROUND ROLL

04/28/23 Chapter 5 - PART A 187

CALCULATION OF LANDING GROUND ROLL

• Total landing distance = Approach distance + Flare distance + Ground roll distance

• We have seen how to find – Approach distance– Flare distance

• Now we will proceed to calculate:– Ground roll distance

04/28/23 Chapter 5 - PART A 188

04/28/23 Chapter 6 – PART 2 189

GEOMETRY OF LANDING

•Spoiler•Airbrakes•Drogue chute•Wheel brakes

Lift Spoilers

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Air Brakes

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Drogue Chute

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04/28/23 Chapter 6 – PART 2 193

CALCULATION OF GROUND ROLL

IZHAR KAZMI 194

F = T during landingT=0 orT= -T (If thrust reversal mechanism installed)

Analysis similar to takeoff. Consider TO fig below

04/28/23 Chapter 5 - PART A 195

We will use two approaches

04/28/23 Chapter 6 – PART 2 196

CALCULATION OF GROUND ROLL FIRST APPROACH

04/28/23 Chapter 6 – PART 2 197

CALCULATION OF GROUND ROLL

LANDING GROUND ROLL• We know

04/28/23 Chapter 5 - PART A 198

2

2 2

( )

2( )

tan

2 2 ave

V dV dVds dtds dt V dt V dV dV dVdt dVdt dt

If acceleration is cons t Integrating

V Vs dV adt

04/28/23 Chapter 6 – PART 2 199

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 200

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 201

CALCULATION OF GROUND ROLL

LANDING GROUND ROLL2ND APPROACH

• We know

• We also saw in Takeoff

04/28/23 Chapter 5 - PART A 202

2

2 2

( )

2( )

tan

2 2 ave

V dV dVds dtds dt V dt V dV dV dVdt dVdt dt

If acceleration is cons t Integrating

V Vs dV adt

• We know

• Using equation for landing

• Average acceleration is given as

• We have already seen

• Using touch down velocity and weight

04/28/23 Chapter 5 - PART A 203

Fg FgFa m mg W

0.7( )

TDave r V

gFga D u W LW W

2 2

2 2 ave

V Vs dV adt

04/28/23 Chapter 5 - PART A 204

LANDING GROUND ROLL WITH THRUST REVERSAL

• Where lift and drag forces are given as

• Where G is the ground effect factor

04/28/23 Chapter 5 - PART A 205

LANDING ROLL WITH THRUST REVERSAL AND FREE ROLL CONSIDERATIONS

04/28/23 Chapter 5 - PART A 206

FACTORS EFFECTING LANDING DISTANCE• In addition to pilot factor and AC configuration

following factors may cause variation from a standard day sea level landing performance :– WEIGHT– WIND– RUNWAY SLOPE– PRESSURE ALTITUDE– TEMPRATURE– TRUST REVESAL ( % APPLIED)

IZHAR KAZMI 207

FACTORS EFFECTING DISTANCESInfluencing factor Increase in Take-off Distance Increase in Landing Distance

10% increase in weight 20% 10%

Increase of 1000' in runway altitude (1" MAP)

10% 5%

Increase in temperature of 10°C above ISA

10% 5%

Dry short grass (under 5") 20% 20%

Dry long grass (5 - 10") 25% 30%

Wet short / long grass 25% / 30% 30% / 40%

2% uphill slope 10% -10%

1% downhill slope -5% 5%

Tailwind component of 10% of liftoff speed

20% 20%

Soft ground or snow at least 25% at least 25%

IZHAR KAZMI 208

04/28/23 Chapter 6 – PART 2 209

EXAMPLE 6.7

04/28/23 Chapter 6 – PART 2 210

EXAMPLE 6.7

04/28/23 Chapter 6 – PART 2 211

Max Lift Coefficient Values (TABLE )

LIFT AND DRAG FORCES DURING LANDING AND TAKEOFF GROUND ROLL

04/28/23 Chapter 5 - PART A 212

04/28/23 Chapter 6 – PART 2 213

EXAMPLE 6.7

IZHAR KAZMI 214

END OF COURSE

Course Objectives• By the end of the course students will be able to carry out

– Unaccelerated performance analysis of aircraft including• Range• Endurance• Steady climb• Gliding

– Accelerated performance analysis of aircraft including:• Turning• Energy concepts: Accelerated climb• Takeoff performance• Landing performance

IZHAR KAZMI 215

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IZHAR KAZMI 217

IZHAR KAZMI 218

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JF 17 Thunder

IZHAR KAZMI 220

• JF-17 Thunder is an advanced, light-weight, all weather, day / night multi-role fighter aircraft

• developed as a joint venture between Pakistan Aeronautical Complex (PAC), Kamra and Chengdu Aircraft Industry Corporation (CAC) of China.

• It possesses excellent air-to-air and air-to-surface combat capabilities. • The state-of-the art avionics, optimally integrated sub-systems, computerized flight

controls and capability to employ latest weapons provides decisive advantage to JF-17 over adversaries of same class.

• This, all weather, multi-role light combat fighter has remarkable high combat maneuverability at medium and low altitude.

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Physical Parameters

Length 49 ft

Height 15.5 ft

Wingspan 31 ft

Empty Weight 14,520 lb

Performance Parameters

Maximum Take Off Weight 27,300 lb

Max Mach No 1.6

Maximum Speed 700 Knots IAS

Service Ceiling 55,500 ft

Thrust to Weight Ratio 0.95

Maximum Engine Thrust 19,000 lb

G Limit +8,-3

Ferry Range 1,880 NM

Armament

No of Stations 7

Total Load Capacity 8000 lb

223

General characteristics

• Crew: 1• Length: 14.93 m (49 ft[139])• Wingspan: 9.45 m (31 ft, including 2 wingtip missiles.[139])• Height: 4.72 m (15 ft 6 in[139])• Wing area: 24.4 m²[140] (263 ft²)• Empty weight: 6,586 kg (14,520 lb[139])• Loaded weight: 9,100 kg (20,062 lb)• Useful load: 3000 kg (6600 lb)• Max. takeoff weight: 12,383 kg (27,300 lb[139])• Powerplant: 1 × Klimov RD-93

– Dry thrust: 49.4 kN[16] / 51.2 kN (11,106 lbf / 11,510 lbf)– Thrust with afterburner: 84.5 kN (19,000 lbf[139])

• G-limit: +8 g / -3 g[139]

• Internal Fuel Capacity: 2,300 kg (5,130 lb[139])

IZHAR KAZMI 224

Performance

• Maximum speed: Mach 1.6[141]

• Combat radius: 1,352 km[16] (840 mi)• Ferry range: 3,482 km(1,880 NM[139])• Service ceiling: 16,920 m (55,500 ft[139])• Thrust/weight: 0.95 [139]

IZHAR KAZMI 225

Armament• Guns: 1× 23 mm GSh-23-2 twin-barrel cannon (can be replaced with 30 mm GSh-30-2)• Hardpoints: 7 in total (4× under-wing, 2× wing-tip, 1× under-fuselage; pylon stations number

3, 4 and 5 are wet-plumb capable) with a capacity of 3,629 kg (8,001 lb) for external fuel and ordnance

• Missiles: ** Air-to-air missiles: • MAA-1 Piranha (Short-range)[143]

• AIM-9L/M (Short-range)• PL-5EII (Short-range)[144]

• PL-9C (Short-range)• PL-12 / SD-10 (Beyond visual range)[144]

– Air-to-surface missiles: • MAR-1 (Anti-radiation missile) - Planned.[92][145]

• C-802A (Anti-ship missiles)[72][144]

• C-803 (Anti-ship missiles)• CM-400AKG (Anti-ship missiles)[146]

• Ra'ad ALCM (Nuclear capable Stealth Cruise missile)[147]

IZHAR KAZMI 226

BOMBS– Un guided

• Mk-82 (general purpose bomb)• Mk-84 (general purpose bomb)• Matra Durandal (anti-runway bomb)• CBU-100/Mk-20 Rockeye (anti-armour cluster bomb)

– Precision guided munitions (PGM): • GBU-10 (Laser-guided)• GBU-12 (Laser-guided)• LT-2 (Laser-guided)• H-2 (electro-optically guided)• H-4 (electro-optically guided)[3]

• LS-6[disambiguation needed] (satellite-guided glide bombs)[142]

• Satellite-guided bombs[3]

IZHAR KAZMI 227

IZHAR KAZMI 228

– Air-to-surface missiles: • MAR-1 (Anti-radiation missile) - Planned.[92][145]

• C-802A (Anti-ship missiles)[72][144]

• C-803 (Anti-ship missiles)• CM-400AKG (Anti-ship missiles)[146]

• Ra'ad ALCM (Nuclear capable Stealth Cruise missile)[147]

• Bombs: ** Unguided bombs:

IZHAR KAZMI 229

Systems

IZHAR KAZMI 230

Aerodynamic Configuration• Bifurcated side air inlet with incorporation of latest technology for improved

performance • Leading edge maneuvering flaps • Trailing edge flaps • Twin Ventral Fins

IZHAR KAZMI 231

Landing Gear• Nose gear with steering • Main gear with paddle controlled hydraulic brakes and nti-skid braking system

Comfortable Cockpit Escape and Egress System• NVG compatible cockpit conforming to US MIL Standard, suitable for 3% to 98%

percentile range of pilots. • Single Piece Stretch Acrylic Transparent Canopy providing a good all around Field

of View • Ejection Seat

– Latest Martin Baker high performance ejection seat – Canopy severance system for additional safety – French oxygen regulation system – Passive Leg Restraint System

IZHAR KAZMI 232

Environment Control System and Oxygen System• Effective control of cockpit pressure and temperature • Effective temperature and humidity control of cockpit and avionics for optimum

performance • Efficient Anti G system for Pilot • Oxygen supply duration three hours Flight Control System• Composite Flight Control System comprising conventional controls with stability

augmentation in roll and yaw axis and fly by wire in pitch axis • Quad-redundancy in Fly By Wire System • Autopilot with Altitude hold and Attitude hold modes

IZHAR KAZMI 233

Fuel System• Total internal fuel 5130 lb (3000 liters) • Single point pressure refueling system • External Fuel

– One center line drop tank 800 liters • Two under wing drop tanks 800/1100 liters

IZHAR KAZMI 234

Avionics

IZHAR KAZMI 235

Weapons• The aircraft is fitted with modern Stores Management System incorporating

accurate weapon delivery modes and solutions involving minimum pilot workload. The aircraft is capable of carrying some of the most modern as well as conventional weapons, including:

– 70-100 Km range beyond visual range active missiles – Highly agile Imaging infra red short range missiles – Air to sea missiles – Anti radiation missiles – Laser guided weapons – Programmable cluster bombs – Runway penetration bombs – General purpose bombs – Training bombs – 23 mm double barrel gun

IZHAR KAZMI 236

AIR to Air Configuration

IZHAR KAZMI 237

Air to Ground Configuration

IZHAR KAZMI 238

JF 17 Vs F 16

IZHAR KAZMI 239

IZHAR KAZMI 240

END OF COURSE

THANK YOU

» EXTRA SLIDES

IZHAR KAZMI 241

Approach Distance

04/28/23 Chapter 5 - PART A 242

50 fa

a

hTan

s

50 fa

a

hs

Tan

From Fig 6.17

506.108f

aa

hs

Tan

LANDING GROUND ROLL• We know

• We also saw in Takeoff

04/28/23 Chapter 5 - PART A 243

2

2 2

( )

2( )

tan

2 2 ave

V dV dVds dtds dt V dt V dV dV dVdt dVdt dt

If acceleration is cons t Integrating

V Vs dV adt

IZHAR KAZMI 244

F = T=0 during landing

04/28/23 Chapter 5 - PART A 245

• We know

• Using equation for landing

• Average acceleration is given as

• We have already seen

• Using touch down velocity and weight

04/28/23 Chapter 5 - PART A 246

Fg FgFa m mg W

0.7( )

TDave r V

gFga D u W LW W

2 2

2 2 ave

V Vs dV adt

04/28/23 Chapter 5 - PART A 247

LANDING GROUND ROLL WITH THRUST REVERSAL

• Where lift and drag forces are given as

• Where G is the ground effect factor

04/28/23 Chapter 5 - PART A 248

LANDING ROLL WITH THRUST REVERSAL AND FREE ROLL CONSIDERATIONS

04/28/23 Chapter 5 - PART A 249

04/28/23 Chapter 6 – PART 2 250

Max Lift Coefficient Values (TABLE )

04/28/23 Chapter 6 – PART 2 251

EXAMPLE 6.7

04/28/23 Chapter 6 – PART 2 252

EXAMPLE 6.7

04/28/23 Chapter 6 – PART 2 253

EXAMPLE 6.7

FACTORS EFFECTING DISTANCEInfluencing factor Increase in Take-off Distance Increase in Landing Distance

10% increase in weight 20% 10%

Increase of 1000' in runway altitude (1" MAP)

10% 5%

Increase in temperature of 10°C above ISA

10% 5%

Dry short grass (under 5") 20% 20%

Dry long grass (5 - 10") 25% 30%

Wet short / long grass 25% / 30% 30% / 40%

2% uphill slope 10% -10%

1% downhill slope -5% 5%

Tailwind component of 10% of liftoff speed

20% 20%

Soft ground or snow at least 25% at least 25%

IZHAR KAZMI 254

04/28/23 Chapter 6 – PART 2 255

ALTERNATE APPROACHCALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 256

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 257

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 258

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 259

CALCULATION OF GROUND ROLL

04/28/23 Chapter 6 – PART 2 260

CALCULATION OF GROUND ROLL

LANDING GROUND ROLL• We know

• We also saw in Takeoff

04/28/23 Chapter 5 - PART A 261

2

2 2

( )

2( )

tan

2 2 ave

V dV dVds dtds dt V dt V dV dV dVdt dVdt dt

If acceleration is cons t Integrating

V Vs dV adt

• Fits have been made to the FAR field length requirements of 2,3,and 4 engine jet aircraft vs. the parameter:

W is the take-off gross weight (lbs).

Sref is the reference wing area (sq ft).

s is the ratio of air density under the conditions of interest which might well be a hot day in Denver or another high altitude airport.

CLmax is the aircraft maximum lift coefficient in the take-off configuration.

T is the total installed thrust (all engines running). It varies with speed and must be evaluated at 70% of the lift-off speed which we take as 1.2 Vs. The variation of thrust with speed shown here may be used for this calculation if detailed engine data is not available.

IZHAR KAZMI 262

• For 2 engine aircraft: TOFL = 857.4 + 28.43 Index + .0185 Index2

For 3 engine aircraft: TOFL = 667.9 + 26.91 Index + .0123 Index2

For 4 engine aircraft: TOFL = 486.7 + 26.20 Index + .0093 Index2

Since for four engine aircraft, the all-engines operating (with 15% pad) case is critical, one may use this fit for the all-engines operating case with 2 or 3 engines as well. Note that the 15% markup is already included.

IZHAR KAZMI 263

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