Tranquillus, Analysis Data Item Vol. 2 by Sina Golshany

Class II Performance: 

Power Extractions (has not been considered due to inconstancies)

Mechanical Power Extractions: The fuel pump mechanical power requirement is given by:

PmechFP 

0.00014c j TUnInsavail

 fp

Segment 2 TUnIns Avail 44000 lb 0.32 cj

 FP  hp

0.85 0.80

Phydr

5000 psi

Vhyd

74.00 gpm

Pmech

113.28 hp

Pmech fp

2.28 hp

Pmechhyd

277.50 hp

Electrical Power Extraction: The electrical power extracted from the engine is found from: Pelec 

0.00134 Pelec

Pelecreq

175000 VA

 gen

0.85

 gen

Pneumatic Power Extraction: For jets, the pneumatic power extraction requirement is obtained from:

Ppneu

m b TreqU1 m a  550 0.030

m b m a

Total Power Extraction: Flight Segment 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17



Pextr 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34 654.34

Installed Thrust

The available installed thrust from a subsonic jet engine is found from:







Tavail  TUnInsavail 1  0.35 K EngPerf M 1 1  inlinc  550

Pextra M 1a

The engine performance factor is a function of the steady state Mach number and is found from Figure 6.37 of Airplane Design Part VI. The speed of sound is found at the input altitude.

a  RT Flight Segment

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17



M1 0.037 0.258 0.369 0.375 0.674 0.762 0.801 0.532 0.295 0.258 0.037 .295 0.374 0.423 0.374 0.200 0.200

inl

inc

1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

TUnInsavail lb Tavail lb 44000

11173

42941 11476 11476 11682 10853 10943 24253 14000 11867 44000 38000 33198 38000 44000 32000

Thrust – Speed Relationship:

Theory: Based on the three points supplied, a quadratic equation is calculated for the available thrust/power. Available thrust for jet aircraft:



Tavail  AThrustV 2  BThrustV  CThrust Segment 2: lb AThrust 2 kts lb B power 2 kts C power lbf Segment 3: lb AThrust 2 kts



Eqn.(1)

0.002 -4.167 12392.857

0.002

lb kts 2 C power lbf B power

Segment 4: lb AThrust 2 kts lb B power 2 kts C power lbf Segment 5: lb AThrust 2 kts lb B power 2 kts C power lbf Segment 6: lb AThrust 2 kts lb B power 2 kts C power lbf Segment 7: lb AThrust 2 kts lb B power 2 kts C power lbf Segment 8: lb AThrust 2 kts lb B power 2 kts C power lbf

-4.167 12392.857

0.002 -4.167 12392.857

Segment 9: lb AThrust 2 kts lb B power 2 kts C power lbf

0.002 -4.167 12392.857

Segment 10: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 11: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 12: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 13: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 14: 0.002 lb AThrust 2 kts

lb kts 2 C power lbf B power

-4.167 12392.857

Segment 15: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 16: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf Segment 17: 0.002 lb AThrust 2 kts -4.167 lb B power 2 kts 12392.857 C power lbf

Take Off distance: The take-off distance is found from:

S TO

  V3    VS   TO  1   TO .hobs     LOF   

2   W    T  0.72C D 0TO         g      S  TO   W  C Lmax,TO TO   hobs gC LmaxTO 1  1.414 LOF 

1     1.414          

Eqn.(2)

The height of the obstacle and the ratio of take-off thrust (or power) at the current take-off altitude to that at sea-level and ISA depend on the certification of the airplane:

For FAR-25;

h obs  35

Eqn.(3)

 TO  1.15 For jet engines, the mean thrust taken at a speed of 0.707 times the liftoff speed is obtained from: 5  BPR T  0.75Tset Eqn.(4) 4  BPR The flight path angle at liftoff is found from: T  0.3   ARw  W  TO The balanced field length is calculated from:

 LOF  0.9

Eqn.(5)

   W     0 . 863 0 . 694  gC h  Lmax,TO obs     655  1  S  TO     BFL     1  2.3  2   2min 0.694 gC Lmax,TO  T  0.72C D 0,TO    W    g  C   Lmax,TO 





    2.7        

Eqn.(6)

The second segment climb gradient, OEI, is determined from:

 L   T  2         W  TO ,OEI  D  TO 

1

Eqn.(7)

The take-off ground run is found with the following:

VL2,OF S TO ,G 

2g T  W

 0.72C D 0,TO      g  C Lmax,TO  TO 

   

Eqn.(8)

Where:

V LOF  1.2VSTO for light aircrafts C LmaxTO

2.100

C DO ,TO

0.0422

L    D  TO

7.80

 g ,FAR  25 ,Asphalt 0.0200 0.40 a g V3 VS TO

1.30

TSet

42000

BPR T CL ,TO

10.00 0.0 deg. 5.9726 rad-1

CL0 ,TO

0.7403

VSTO

119.22 kts

VLOF

143.06 kts

S TO

6749 ft

S TO ,G

4240 ft



Maximum cruising speed:

Theory: For an aircraft equipped with jet engines, the maximum cruise speed is found when:

Tavail  Treq

Eqn.(1)

The thrust available is found from the Thrust/Speed curve defined as follows:



Tavail  FCr AthrustVCr2 max  BThrustVCrmax  CThrust



Eqn.(2)

The required thrust is found using the following equation:

 CD 0 Clean , M S wVCr2 max Treq    2 cos    T 

   2WCr2 BDPclean   2   S V     cos   w Cr T   max 

Eqn.(3)

An Auto-CAD script used to plot the data available and largest answer has been chosen as the maximum cruise speed: The lift coefficient at the maximum cruise speed is found from:

CD  CD 0 clean , M



CL @ VCr

max

Eqn.(4)

BDPclean

The drag coefficient at the maximum cruise speed is calculated from:

CD 

Treq cos  T 

Eqn.(5)

0.5VCr2 max S w

The angle of attack is found from:



CL @ VCrmax CL

 0

Eqn.(6)

The thrust required for jet driven airplanes is calculated from:

Treq 

550 P Preq

VCruise, Max



Eqn.(7)

VCrmax 663.72 kts.

Range at constant speed:

For an aircraft traveling at constant speed, the range is defined as:

   C   WCr RCrV const  326 P  L  ln  c P  C D   WCr  WFCr The steady state lift coefficient is calculated from

CL 

WCr 

   

Eqn.(1)

WFCr

4500.00 lb

C D 0Clean , M

0.0175

B DPclean

0.0337

0.41

 Treq sin   T  2 0.5U 12 S w

Eqn.(2)

In Cruise segment:

Treq

9226 lb

Tavail

11845 lb

CL RCr ,V CTS

0.37 deg. 0.4979 567.8 nm





Range at constant Altitude:

For aircraft equipped with jet engines:  1.677  1  CL AR   Rh  CTS   WCr  WCr  WFCr    cJ  S w  CD 





Eqn.(1)

The airplane drag coefficient is calculated from: C D  C D 0,Clean  BDPclean C L2AR

Eqn.(2)

The optimum lift coefficient during the cruise for aircraft equipped with jet engine is defined as: CLopt

max R



CD 0,Clean 3BDPClean

Eqn.(3)

The constant flight speed during cruise is calculated from:

U1 

2WCr  WFCr

C L S w

Eqn.(4)

C LAR

0.4150

Treq

9226 lb

Tavail

11845 lb

U1 C Lopt , MaxR

0.37 deg 475.00 kts. 0.4157



RCr ,h CTS Altitude

2849.7 nm 37800 ft.



Endurance at Constant Altitude:

For airplanes equipped with jet engines, the constant altitude endurance is calculated from:  1  CL   W  Cr  Eqn.(4) ECrhCTS  60   AE  ln   c C W W  Fl    j  D   Cr For airplanes equipped with jet engines, the airplane lift coefficient for the maximum endurance is found from:

CLE max 

CD 0 clean BDP ,Clean

Eqn.(5)

The steady state flight speeds during the endurance/loiter is calculated from:

U1 

W   2WCr  Fl  Treq sin   T  2   S w C LAE

Eqn.(6)

The airplane angle of attack during the endurance/loiter is calculated from:



C LAE  C L0  C Li , h ih  C L e  cv C L

Eqn.(7)

In general, for any endurance/loiter condition:

Treq

U 12 S w C D  2 cos  T 

WFl

16458.00 lb

Tavail

11845 lbf

Treq

9226 lbf

 U1 C LE  Max .

0.37 deg 475.00 kts 0.7619

ECr ,h CTS

425.6 min≈7h. 6 min.

Eqn.(8)



Turn Radius

The airplane instantaneous turn performances are calculated as follows: The load factor in the turn maneuver is found from: nTurn 

Treq sin   T   0.5 VM2 C Lmax S w WM

Eqn.(1)

The bank angle in the turn maneuver is calculated from:



2   tan 1 nturn 1



Eqn.(2)

The rate of the turn in the turn maneuver is found from:

TurnRate 

g tan  VM

VM Treq

234.47 kts 9226 lb

Tavail

11845 lb



52.4

TurnRate

0.1050 rad

Rturn

3770.45 ft

nturn

1.64 g



Eqn.(3)

sec .

Landing Distance

The stall speed of the airplane in the landing configuration is calculated from:

VSL 

2WL  Tset sin   T  S w C Lmax, L

Eqn.(1)

The angle of attack during the landing is found from:



C Lmax, L  C L0 L C LL

Eqn.(2)

The airplane approach speed at the obstacle height for FAR-25 is calculated from:

V A  1.3VS L

Eqn.(3)

The distance from the obstacle to touchdown is found from:

 1  V A2  V 2 TD    hobs  S air     2g   The intermediate quantity is defined by:

 

0.5 V A2 S w C DA  Tser WL

Eqn.(4)

Eqn.(5)

The drag coefficient in the approach condition is found from: C DA  C D 0, L , fdown  B DPL _ down C L2A

Eqn.(7)

The lift coefficient in the approach configuration for FAR-25 requirements is calculated from:

C LA 

C Lmax, L f

Eqn.(8)

Where: f=1.3 The height of the obstacle is defined as 50 ft by write brothers. The velocity of the airplane at touchdown is found from:

VTD

   2   V A 1      n 

Eqn.(9)

The length of the ground roll is determined by:

S LG 

VTD2 2a

Eqn.(10)

C Lmax, L

2.200

113000.0 lb

C D 0, Ldown

0.0662

B DPL _ down

0.0416

0.40

g n Tset

0.10 (Typical Pilot) 42000 lb



VS L

0.01 105.96 kts

VA S air

137.75 kts 3704 ft

S LG

2096 ft

5800 ft



Stall Speed:

The stall speed performance is evaluated using the following equation:

VS 

2W  Tset sin  current  T  S w C Lmax

Eqn.(1)

The angle of attack is found from:



C Lmax  C L0 C L

VS ,T .O

119.22 kts

VS , Land

105.96 kts



Eqn.(3)

Climb performance:

The climb performance calculation involves an iterative process to determine the angle of attack. The angle of attack can be found from:

 Wcl  Tset sin   T      C L0  C Li .ih  C L   h e 0.5VCl2 SW    C L

Eqn.(1)

The equivalent take-off thrust for engines with propellers is calculated from: Tset 

550 P SHPset VCl

Eqn.(2)

The airplane lift coefficient is calculated from:

C L  C L0  C L   C Lih ih  C L e  e

Eqn.(3)

The rate of climb for the airplane can be determined from:

 T   C  V    V RC  60VCl  set    D  cl .1   Cl  WCl   CL  Wcl    g

 dU      dh 

1

Eqn.(4)

The airplane drag is calculated from:

Drag  0.5 VCl2 SW C D

Eqn.(5)

The airplane drag coefficient is found from: C DA  C D 0, L , fdown  B DPL _ down C L2A

Eqn.(6)

The climb gradient for the airplane can be determined from:

CGR 

RC 60VCl

Eqn.(7)

The specific excess power of the airplane is determined from:

PSpExPwr 

60Tavail  Tset VCl WCl

Eqn.(8)

In Climb Segment: Alt.



R/C CGR

22500 ft -0.28 deg 606.46

ft min

0.02

PSpExPwr

141.550

ft min

Time to climb: h2

∫(1 RC )dh

CGR PSpExPwr t

0.02 141.550 ft/ min 24.50 min

Class II Stability Derivatives: Steady state coefficients:

 Steady state lift Coefficient: Theory: The airplane steady state lift coefficient is given by: C L1 

n.W cos   Tset sin   T   C LN , Prop q1 S w

The steady state dynamic pressure is calculated from:

q

1 U 12 2

Segment 1: (N/A) Segment 2:

Tavail

43864 lb

C L1

5.00 deg 2.0624



Segment 3:

Tavail  C L1

Segment 4:

Tavail

11476 lb

CL1

1.33 deg 0.5536



Segment 6:

Tavail

10853 lb

CL1

-0.28 deg 0.4709



11476 lb 0.00 deg 0.5864 Segment 5:

Tavail  C L1

11682 lb 1.88 deg 0.6502

Segment 7:

Tavail  C L1

11845 lb 0.37 deg 0.4979

Segment 8:

Tavail  C L1

11213 lb -0.19 deg 0.3907

Segment 9:

Tavail  C L1

Segment 10:

Tavail  CL1

11867 lb 7.15 deg 1.2646

Segment 12:

Tavail  CL1

44000 lb 2.02 deg 0.6415

33198 lb 0.00 deg 0.4578

Tavail  C L1



44000 lb 11.83 deg 2.0528

n/a n/a n/a

Segment 13:

Tavail  C L1

38000 lb 3.28 deg 0.5836 Segment 15:

Tavail  C L1

Segment 16:

Tavail  CL1

9.82 deg 1.6282

Segment 11:

Segment 14:

Tavail  C L1

14000 lb

38000 lb 0.16 deg 0.4012

Segment 17:

Tavail  C L1

32000 lb 0.00 deg 0.4074

Steady State Thrust Force Coefficient

Theory: The airplane steady state thrust coefficient is defined as:

C Tx ,1 

Tset cosT    q1 S w

Segment 1: (N/A)

Eqn.(1)

Segment 2:

Segment 3:

CTZ 1

n/a

CTZ 1

0.0142

CTZ 1

0.0000

C T X ,1

n/a

C T X ,1

0.3139

C T X ,1

0.0420

Segment 4:

Segment 5:

Segment 6:

CTZ 1

-0.0010

CTZ 1

-0.0010

CTZ 1

0.0002

C T X ,1

0.0210

C T X ,1

0.0000

C T X ,1

0.0392

Segment 7:

Segment 8:

Segment 9:

CTZ 1

-0.0003

CTZ 1

0.0003

CTZ 1

0.0012

C T X ,1

0.0477

C T X ,1

0.0810

C T X ,1

0.0800

Segment 10:

Segment 11:

Segment 12:

CTZ 1

-0.0184

CTZ 1

n/a

CTZ 1

-0.0089

C T X ,1

0.1467

C T X ,1

n/a

C T X ,1

0.2514

Segment 13:

Segment 14:

Segment 15:

CTZ 1

-0.0088

CTZ 1

0.0000

CTZ 1

-0.0004

C T X ,1

0.1540

C T X ,1

0.4578

C T X ,1

0.1542

Segment 16:

Segment 17:

CTZ 1

-0.0795

CTZ 1

0.0000

C T X ,1

0.5496

C T X ,1

0.1196



Steady State Thrust Pitching Moment Coefficient

Theory: The airplane steady state thrust pitching moment coefficient for a jet airplane is given by:

 Tavail dT q1S wcw The wing mean geometric chord is given by: CmT1 

cw 

4 1   w  2w 3 1   w 2

Eqn.(2)

Sw ARw

Eqn.(3)

The perpendicular distance from the thrust line to the airplane center of gravity is found from:

d T  Z T  Z cg cos T  X T  X cg sin T

Eqn.(4)

Aircraft assumed to be in trim condition: C mT  C m1

Eqn.(5)

Segment 1:

Segment 2:

Segment 3:

13.25 ft

dT dN

12.63 ft -31.30 ft

dT dN

12.66 ft -30.51 ft

dT dN

12.45 ft -31.07 ft

C m,T

n/a

C m,T

-0.3002

C m,T

-0.0395

Segment 4:

Segment 5:

Segment 6:

13.25 ft

dT dN

12.48 ft -30.22 ft

dT dN

12.59 ft -30.54 ft

dT dN

12.47 ft -31.07 ft

C m,T

0.0731

C m,T

-0.0556

C m,T

-0.0369

Segment 7:

Segment 8:

Segment 9:

13.25 ft

dT dN

12.52 ft -31.06 ft

dT dN

12.31 ft -31.94 ft

dT dN

12.26 ft -31.94 ft

C m,T

-0.0451

C m,T

-0.0752

C m,T

-0.0740

Segment 10:

Segment 11:

Segment 12:

13.25 ft

dT dN

12.29 ft -31.95 ft

dT dN

12.29 ft -31.03 ft

dT dN

12.23 ft -31.44 ft

C m,T

-0.1371

C m,T

n/a

C m,T

-0.2321

Segment 13:

Segment 14:

Segment 15:

13.25 ft

dT dN

12.24 ft -31.07 ft

dT dN

12.24 ft -32.08 ft

dT dN

12.04 ft -31.04 ft

C m,T

-0.1425

C m,T

-0.1246

C m,T

-0.1401

Segment 16:

Segment 17:

13.25 ft

dT dN

11.97 ft -32.01 ft

dT dN

12.14 ft -34.63 ft

C m,T

-0.5041

C m,T

-0.1095



Speed related derivatives:

The airplane drag-coefficient-due-to-speed derivative may be determined in all speed regimes from: C

M

C D 1

Eqn.(1)

C M

The derivative of the airplane drag coefficient with respect to Mach number can be found from the airplane drag polar using the method shown in Figure 10.3 in Airplane Design Part VI: C D

 tan 

C M

Segment 1: C D M C DU

n/a n/a

Segment 4: C D M C DU

0.100 0.0375

Segment 7: C D M C DU

0.300 0.2403

Segment 10: C D M C DU

Eqn.(2)

M CD Diagram 

0.300 0.0598

Segment 2: C D M C DU

0.000 0.0000

Segment 5: C D M C DU

0.200 0.1349

Segment 8: C D M C DU

0.100 0.0532

Segment 11: C D M C DU

n/a n/a

Segment 3: C D M C DU

0.200 0.0739

Segment 6: C D M C DU

0.100 0.0556

Segment 9: C D M C DU

0.200 0.0591

Segment 12: C D M C DU

0.320 0.0945

Segment 13:

Segment 14:

0.000

C D M C DU

0.0000

Segment 16:

C D M C DU

0.0000



0.0000

C D M C DU

n/a 0.0000

Segment 17:

n/a

C D M C DU

n/a

Segment 15:

n/a 0.0000

Lift Coefficient due to Speed Derivative

The airplane lift-coefficient-due-to-speed derivative is defined as: 2 2 M 1 Cos  C C



q



2 2 1 M 1 Cos  C

CL 1

Eqn.(1)

q SW 1 2

U 2 1

Segment: 1 2 3 4 5 6 7 n/a 0.0537 0.0698 0.0683 0.3573 0.1495 0.4980 C LU Segment: 8 9 10 11 12 13 14 0.1088 0.0475 0.0404 n/a 0.0469 0.0712 0.4578 C LU Segment: 15 16 17 0.0490 0.0461 0.0487 C LU



Pitching Moment Coefficient due to Speed Derivative:

The airplane pitching-moment-coefficient-due-to-speed derivative is found from: C

 C

X AC M L1 1 M

Eqn.(1)

The derivative of the airplane aerodynamic center is determined with respect to the Mach number by estimating the airplane aerodynamic center at 2 points close to the specified Mach number. The slope of the line through the 2 points is used to estimate this derivative. Segment 1:

Segment 2:

Segment 3:

x ac

n/a

x ac

-0.0525

x ac

-0.0776

M C mU

n/a

M C mU

0.0133

M C mU

0.0168

Segment 4:

Segment 5:

Segment 6:

x ac

-0.0638

x ac

-0.1957

x ac

-0.1364

M C mU

0.0133

M C mU

0.0858

M C mU

0.0357

Segment 7:

Segment 8:

Segment 9:

x ac

-0.1431

x ac

-0.1247

x ac

-0.0602

M C mU

0.0571

M C mU

0.0255

M C mU

0.0116

Segment 10:

Segment 11:

Segment 12:

x ac

-0.0410

x ac

-0.0115

x ac

-0.0807

M C mU

0.0103

M C mU

n/a

M C mU

0.0176

Segment 13:

Segment 14:

x ac

-0.1257

x ac

-0.1543

x ac

-0.1257

M C mU

0.0269

M C mU

0.0566

M C mU

0.0188

Segment 16:

Segment 17:

x ac

-0.0399

x ac

-0.0822

M C mU

0.0114

M C mU

0.0124



Segment 15:

Thrust Coefficient due to Speed Derivative:

The airplane thrust-coefficient-due-to-speed derivative is defined as:

C Tx , u 

CTx

 u     U1  For jet driven airplanes:  M  T  CTxU   1    2CTx1  q1S w  u 

Eqn.(1)

Eqn.(2)

The installed power can be expressed as:

T  AThrustU12  BThrustU1  CThrust

Eqn.(3)

The thrust due to speed derivative for a jet driven airplane can be further expressed as:  U  CTx ,u   1 .2 AThrustU1  BThrust   2CTx1  q1S w 

Eqn.(4)

Segment: 1 2 3 4 5 6 7 n/a -0.6321 -0.0868 -0.1003 -0.0051 -0.0816 -0.0991 CTx ,U Segment: 8 9 10 11 12 13 14 -0.1649 -0.1637 -0.2993 n/a -0.5065 -0.3110 -0.2735 CTx ,U Segment: 15 16 17 -0.3115 -1.1053 -0.2420 CTx ,U



Thrust Pitching Moment Coefficient due to Speed Derivative

C mT

C mTU 

Eqn.(1)

 u     U1 

The airplane thrust-pitching-moment-coefficient-due-to-speed derivative is computed from: d  C mT ,U   T CTx ,U  cw 

Eqn.(2)

Segment: 1 n/a C mT ,U

2 0.6039

3 0.0815

4 5 6 7 0.0944 0.1161 0.0768 0.0936

Segment: 8 0.1532 C mT ,U

9 0.1514

10 0.2776

11 n/a

Segment: 15 0.2829 C mT ,U

16 0.9985

17 0.2216

12 0.4672

13 0.2873

14 0.2527

-Angle of attack related derivatives: Theory:



Drag Coefficient due to Angle of Attack Derivative

The airplane drag-coefficient-due-to-angle-of-attack derivative can be found from:

C D 

2C L1 C L

ARw e

 C D , Power

Eqn.(1)

The Oswald efficiency factor is estimated from Figure 3.2 in Methods for Estimating Stability and Control Derivatives of Conventional Subsonic Airplanes (J.Roskam, Sec. 3.1 PP. 3.1-3.17) and is a function of the wing aspect ratio and the wing taper ratio:

e  f  ARw ,  w  Segment: C L rad 1

1 2 3 4 5 6 7 n/a 5.5737 5.7127 5.6049 6.3452 5.9379 6.8396

1

n/a 0.2001 0.5206 0.1151 0.7721 0.4440 0.1853

C D

  rad 

Segment: C L rad 1

C D

1

0.3833 0.5790 7.0785 n/a

Segment: C L rad 1

15 16 17 6.0697 5.5586 5.5983

1

0.1200 0.1000 0.1000

C D 

  rad 

8 9 10 11 12 13 14 5.8816 5.5585 5.5547 5.8526 5.6320 5.6003 6.8340

  rad 

0.5672 0.1000 0.0000

Pitching Moment Coefficient due to Angle of Attack:

Theory: The airplane pitching-moment-coefficient-due-to-angle-of-attack derivative is found from: C m  C m , P .OFF  C m , Power

Eqn.(1)

The airplane pitching-moment-coefficient-due-to-angle-of-attack derivative, without power effects is determined from:





C m , P.OFF  xcg  x acP .OFF C L , P .OFF

Eqn.(2)

The current static margin of the airplane is found from:

S .M  100x ac  xcg  Segment: C m , rad 1



S.M (%)



Eqn.(3)

1 2 3 4 5 6 7 n/a -0.7425 -0.9540 -0.2341 -1.5793 -0.5780 -0.3400 n/a 13.32

16.70

4.18

9.13

9.73

4.97

Segment: C m , rad 1





S.M (%)

16.58

Segment: C m , rad 1

15 16 17 -1.7187 -1.3748 -2.1622





S.M (%)



8 9 10 11 12 13 14 -0.9747 -1.0255 -1.0331 -1.7868 -1.1245 -0.9763 -0.8473

28.32

18.45

24.73

18.60

30.53

19.97

17.43

12.4

38.62

The airplane thrust-pitching-moment-coefficient-due-to-angle-of-attack derivative is defined as:

C mT 

C mT

Eqn.(1)



The airplane thrust-pitching-moment-coefficient-due-to-angle-of-attack computed from:

  dC   C mT   m  C L   dC L  T 

derivative

Eqn.(2)

The perpendicular distance between the thrust line and the airplane center of gravity is given by:

d T  X T  X cg sin  T  Z T  Z cg cos  T

Segment: dC m dC L TL

1 2 0.0000 0.0000

dC m dC L

n/a

-0.0367 -0.0361 0.0000 0.0000 -0.0338 -0.0265

n/a

-0.0367 -0.0361 0.0000 0.0000 -0.0338 -0.0265

n/a

-0.2043 -0.2065 0.0000 0.0000 -0.2005 -0.1815

dC m dC L

3 0.0000

Eqn.(3)

4 5 6 0.0000 0.0000 0.0000

7 0.0000



C mT , rad 1



Segment: dC m dC L TL

8 0.0000

9 0.0000

10 0.0000

dC m dC L

-0.0352 -0.0379 -0.0391 n/a

-0.0375 -0.0369 -0.0275

13 0.0000

14 0.0000

-0.0352 -0.0379 -0.0391 n/a

-0.0375 -0.0369 -0.0275

-0.2069 -0.2125 -0.2174 n/a

-0.2111 -0.2065 -0.1882

dC m dC L

C mT ,

rad  1

Segment: dC m dC L TL

15 0.0000

dC m dC L

-0.0346 -0.0389 -0.0418

dC m dC L

11 12 0.0000 0.0000

16 0.0000

17 0.0000

-0.0346 -0.0389 -0.0418 T



C mT , rad 1



-0.2102 -0.2164 -0.2340

-Rate of Angle of attack related derivatives:  Drag Coefficient due to Angle of Attack Rate Derivative The airplane drag-coefficient-due-to-angle-of-attack-rate derivative is normally neglected:

C D  0  Lift Coefficient due to Angle of Attack Rate Derivative Theory: The airplane lift-coefficient-due-angle-of-attack-rate derivative is determined from: C

L

 2C

L

 V h

h h

 h

Eqn.(1)



The horizontal tail volume coefficient is solved from: V  h





X AC  X CG S h . h CW SW

Eqn.(2)

Segment: C L rad 1 h



C L rad 1



Segment: C L rad 1 h



C L rad 1



Segment: C L rad 1 h



C L rad 1





1 2 3 4 5 6 7 n/a 1.1738 1.2286 1.1986 1.4365 1.3330 1.6235 n/a 1.1738 1.2286 1.1986 1.4365 1.3330 1.6235



8 9 10 11 12 13 14 1.3499 1.2244 1.1901 1.1460 1.2107 1.2303 1.6387 1.3499 1.2244 1.1901 1.1460 1.2107 1.2303 1.6387



15 16 17 1.2210 1.2002 1.3200 1.2210 1.2002 1.3200

Pitching Moment Coefficient due to Angle of Attack Rate Derivative

Theory: The airplane pitching-moment-coefficient-due-to-angle-of-attack-rate derivative is calculated from:

m

V  h

 2C



Segment: C m rad 1



Segment: C m rad 1



X AC  X CG d h h

h h

X AC  X CG . S h

Segment: C m rad 1



L

 V

d

Eqn.(1)



1 2 3 4 5 6 7 n/a -3.8146 -4.0414 -3.8391 -4.8517 -4.3779 -5.3480



8 9 10 11 12 13 14 -4.4914 -4.1102 -3.9978 -3.7721 -4.0184 -4.0470 -5.5243



15 16 17 -4.0138 -4.0376 -4.6967

-Pitch rate related derivatives:



Drag Coefficient due to Pitch Rate Derivative:

CD  0 q



Lift Coefficient due to Pitch Rate Derivative:

Theory: The airplane lift-coefficient-due-to-pitch-rate derivative: is estimated from C

C

Lq W

C

Eqn.(1) Lq h

The wing contribution to the airplane lift-coefficient-due-to-pitch-rate derivative is found from: ARW  2Cos C C

Lq W



ARW B  2Cos C

 

CL qW |M  0

Eqn.(2)

The compressible sweep correction factor is solved from:

B  1 M

  Cos  C   4W 

Eqn.(3)

The wing contribution of this derivative at Mach equal to zero is determined from: C Lq ,W

M 0



 C L ,W

Clean

 1 2X ac  X cg   C L ,Wf    cw 2 



Eqn.(4)

The horizontal tail contribution to the airplane lift-coefficient-due-to-pitch-rate derivative is found from: C

Lq h

 2C

L

 V h

Eqn.(5)

h h

Segment: C Lq C Lq C Lq

rad  rad  rad 

C Lq C Lq

1

C Lq C Lq

4.056

4.1386 4.0593 4.2029 4.2175 4.3506

1

1.2018 1.2061 1.6503 0.5515 1.0693

1.293

1.7212

5.898

6.4312 5.2433 5.9084 0.1528

6.714

n/a

1

rad  rad  rad  1

4.2891 4.1996 4.1801 4.0784 4.1524 4.1397 4.3915

1

2.4518 2.1933 1.6463 1.0228 1.8538 1.6522 2.6013

1

Segment: C Lq

n/a

Segment: C Lq

rad  rad  rad  1

7.3931 7.0451 6.2488 5.7431 6.6527 6.4341 7.6466

4.1372 4.1861 4.4768

1.13

1

2.1892

3.593

1



5.9091 7.0284 8.7529

Pitching Moment Coefficient due to Pitch Rate Derivative

Theory: The airplane pitching-moment-coefficient-due-to-pitch-rate derivative, also known as the pitch damping derivative, is calculated from:

C mq  C mqW  C mqh

Eqn.(1)

The wing contribution of this derivative is given from:

mq W

  ARW tan 2  C 3 4   W   ARW B  6Cos C B  4   3 2  M  0  ARW tan  C   4W 3   AR  6Cos C   W 4W

 

 Cm qW

Eqn.(2)

The wing contribution at Mach = 0 is estimated from:

   X  2 1 X    ARW 2 W   W      CW  2 CW    K C X Cos  W L C |M  0 ARW  2Cos C W 4W   4W    

  Cm q

Eqn.(3)

The intermediate calculation parameter, X, is given by: 2     1         2 AR x x x x ARW tan  C w   ac cg  cg  2  acw  w    1 4W    X  AR  2 cos    8 w c 24 ARW  6Cos C 4w  4W   2

Eqn.(4)

The correction constant for the wing contribution to pitch damping is obtained from Figure 10.40 in Airplane Design Part VI and is a function of the wing aspect ratio:



K  f ARW W



The compressible sweep correction factor is given by:

B  1  M 12 cos 2  c

Eqn.(5) 4 w

The horizontal tail contribution to the pitch damping derivative is given by: C

mq h

 2C

L

Segment: C mq

C mq

 V h

rad  rad  rad 



X AC  X CG h

h h



Eqn.(6)

1

n/a -13.1808 -13.614 0.098 -13.6291 -13.8157 -14.3316

1

n/a

1

n/a

-3.899

n/a

-4.0545

-13.181 -13.614 0.098 -22.0324 -13.8157 -14.7973

Segment:

rad  rad  rad  1

C mq

1 1

C mq

Segment: h

C mq

rad  rad  rad  1

C mq

1 1

-14.3711 -14.0975 -14.0422 -13.4242 -13.7825 -13.6168 -14.8481 n/a

n/a

-14.3711 -14.0975 -14.0422 -13.4242 -13.7825 -13.6168 -14.8481

-13.5999 -14.0822 -15.9289 n/a

n/a

-13.5999 -14.0822 -15.9289

Detailed Static Margin: Theory: Static Margin stick fixed, known as simply Static Margin, is the non-dimensional distance (in fractions of mean geometric chord) from the aerodynamic center (a.c.) to the center of gravity (c.g.). Cm dC m     xcg  x ac  dC L C L When Static Margin is greater than zero, the aircraft is stable. SM  

Eqn.(1)

Static Margin Stick Free implies the pilot is not holding the control column.

SM free  NPfree  xcg

Eqn.(2)

Static Margin Stick Fixed implies the pilot is holding the control column.

SM fix  NPfix  xcg  x ac  xcg

Eqn.(3)

The neutral point, stick free is that c.g. location for which, C m , stick , free  0

Eqn.(4)

Stick Free conditions apply to reversible flight control systems, where the control surface 'floats'. If a gust hits the horizontal tail while the pilot is not holding the controls, the tail angle of attack will change; that causes the elevator to 'float' to a new angle which is defined by the flotation condition:

C h  0  C hO  C h  h  C h e  e

Eqn.(5)

Differentiating this equation with respect to the elevator angle results in:

C h  h Ch   e d  Eqn.(6)    1     C h e  C h e  d    This change in elevator angle causes a change in static longitudinal stability. For stick-free, C m can be re-written as follows:





C m  C L , wf xcg  x acwf  C L  h h



Sh x ach  x cg S

1  dd    



 e   

Eqn.(7)

Therefore, Sh d  C h  e   x ach 1   1 C L wf S C h w   d   Eqn.(8) xcg  NPfree    C L h  C Sh d  h e  h 1 x ac ,h 1   1    C L S C h   d  wf w  The neutral-point-stick-free is forward of the neutral-point-stick-fixed because, as a general rule: x acwf 

 Ch  e 1    C h e 

C L h

h

   1.0  

Eqn.(9)

This places an additional restriction on the allowable most aft c.g. location of an airplane. Segment: 1 2 3 4 5 6 0.4228 0.4826 0.3837 0.5157 0.4222 0.4403 xcg n/a 0.6158 0.5507 0.5575 0.5135 0.5376 x ac

7 0.4411 0.4908

NPfree

n/a

0.3840 0.3832 0.5735 0.5100 0.4690

0.2144

SM % SM free %

n/a n/a

13.13 -9.86

4.97 -17.00

16.70 -0.06

4.18 6.87

9.13 8.78

9.73 2.87

Segment: 8 9 10 11 12 13 0.3747 0.3747 0.3747 0.4430 0.3558 0.3839 xcg 0.5408 0.3825 0.4123 0.7483 0.5555 0.5582 x ac

14 0.3641 0.4881

NPfree

0.3797 0.5591 0.5599 18.45 0.78

18.60 3.84

n/a

SM % SM free %

16.58 0.50

30.53 n/a

Segment: xcg

15 0.4425

16 0.3692

17 0.1717

x ac

0.7256

0.6165

0.5579

NPfree

0.5738

0.3859

0.3835

SM % SM free %

28.32 13.13

24.73 1.67

38.62 21.18

0.3848 0.3873

0.2148

19.97 2.90

12.40 -14.93

17.43 0.34

Lateral-Directional Stability: -Sideforce Coefficient due to Sideslip Derivative



Sideforce Coefficient due to Sideslip Derivative

The airplane sideforce-coefficient-due-to-sideslip derivative is found from: C C C Y Y Y

Eqn.(1)

C Y f V

The wing contribution to this derivative is given by: C Y

Eqn.(2)

 0.00573  w

The fuselage contribution to the airplane sideforce-coefficient-due-to-sideslip derivative is given by: C Y

 2 K f

S0 J

Eqn.(3)

The wing-fuselage interference factor is found from Figure 10.8 in Airplane Design Part VI and is a function of the Z-location from fuselage centerline to exposed wing root quarter chord point and the fuselage height at the wing root chord:



K  f Z w ,d f J



Eqn.(4)

The vertical tail contribution is given for single and twin vertical tails: For single vertical tails:

 K C Y V L V V

   1 V   

Eqn.(5)

The empirical factor for estimating airplane sideforce-coefficient-due-to-sideslip derivative is found from Figure 10.12 in Airplane Design Part VI and is a function of the vertical tail span and the height of the fuselage at the quarter chord point of the vertical tail section:



K v  f bv , h f v



The intermediate calculation parameter is given by:

  d   3.06  v  0.724   1   d v  1  cos C 4 

 Z fcw  Z Cr  Sv  4     0.40 zf  Sw  w 

 0.009 ARw

Eqn.(6)

For twin vertical tails:

C y v  2C yv

C y v  wfh  S v . C y veff S w

Eqn.(7)

The wing-fuselage-horizontal-tail interference on airplane sideforce-coefficient-due-tosideslip derivative is found from Figure 10.17 in Airplane Design Part VI and is a function of the vertical tail span, the depth of the fuselage at the region of the vertical tail, the distance between the vertical tail panels in the Y-direction, and the length of the fuselage: C y , V  wfh  C y , V eff 

 f bv ,2r1 , bh , l f



Segment: C y ,W

1 -0.0458

2 -0.0115

3 -0.0115

4 -0.0458

5 -0.0458

6 -0.0115

7 -0.0115

C y , f

-0.1081

-0.1065

-0.1081

-0.1065

C y  ,V

n/a

-0.7672

-0.7788

-0.9733

-1.0568

-1.0117

-1.1118

C y

n/a

-0.8852

-0.8968

-0.7164

-0.7587

-1.1297

-1.2298

Segment: C y ,W

8 -0.0115

9 -0.0115

10 -0.0115

11 -0.0458

12 -0.0115

13 -0.0115

14 -0.0458

C y , f

-0.1065

-0.1081

-0.1065

C y  ,V

-1.0054

-0.9532

-0.9526

n/a

-0.7706

-0.7793

-1.118

C y

-0.9232

-0.8886

-1.0706

n/a

-0.8886

-0.8973

-1.2298

Segment: C y ,W

15 -0.0458

16 -0.0115

17 -0.0115

C y , f

-0.1081

-0.1065

C y  ,V

-0.5623

-0.9526

-0.7789

C y

-0.7162

-0.8809

-0.8969



Rolling Moment Coefficient due to Sideslip Derivative:

Theory: The airplane rolling-moment-coefficient-due-to-sideslip derivative, also known as the dihedral effect, is given by: Eqn.(1)

C C C C l l l l Wf h V

The horizontal tail contribution to the dihedral effect is given by: S h bh S w bw

C l , h  C l h , f

Eqn.(2)

The wing-fuselage, horizontal tail-fuselage and/or canard-fuselage contributions to the dihedral effect are found from:



C  57.3 X  Y  Z l Wf



Eqn.(2)

The first intermediate calculation parameter, X, is calculated from:

  Cl  X C LWf   C L 

    C

 Cl  K M K f    CL  4W

      A 

Eqn.(3)

The sweep contribution is found from Figure 10.20 in Airplane Design Part VI and is a function of the lifting surface half chord sweep angle, the lifting surface aspect ratio, and the lifting surface taper ratio:

 Cl    CL 

    C



 f  C , ARW , W 2



The compressibility correction to the sweep contribution is found from Figure 10.21 in Airplane Design Part VI and is a function of the lifting surface half chord sweep angle, the lifting surface aspect ratio and the steady state flight Mach number: K

M



 f  C , ARW , W 2



The fuselage correction factor is obtained from Figure 10.22 in Airplane Design Part VI and is a function of the lifting surface aspect ratio, lifting surface mid-chord sweep angle, the length of the fuselage and the lifting surface span: K

 

 f  ARW ,  C

  

, l f ,bW

The X-distance between the fuselage nose and the wing tip-mid-chord point is given by: c r ( l .S )  bl .s   l f   X apex ,l .S  tan  C ,l .s   X apex f  2 4  2  

Eqn.(4)

The lifting surface root chord length is calculated from: c r (l .S ) 

l .S  1

S l .S . ARl .S .

Eqn.(5)

The aspect ratio contribution is obtained from Figure 10.23 in Airplane Design Part VI and is a function of the lifting surface aspect ratio and the lifting surface taper ratio:

 C l  C  L

   f  AR,    

Eqn.(6)

The second intermediate calculation parameter, Y, is given by:

  Cl Y    K M    C l      





Z ,l . s

Eqn.(7)

The lifting surface dihedral effect is found from Figure 10.24 in Airplane Design Part VI and is a function of the lifting surface aspect ratio, the lifting surface taper ratio, and lifting surface half chord sweep angle:

C l

 f  AR,  ,  c  2   

The compressibility correction to the lifting surface dihedral effect can be found from Figure 10.25 in Airplane Design Part VI and is a function of the steady state flight Mach number, the lifting surface aspect ratio, and the lifting surface half chord sweep angle: K

M

 

 f  M , ARW ,  C

  

The fuselage induced effect on the lifting surface is solved from the following set of equations. z w  z fcl . s  z cr

4 l .S For a wing placement inside of the fuselage diameter:

z w  0.7 D fl .s tan l .S 

Eqn.(8)

D fls 2

The fuselage induced effect on the lifting surface for a low wing is found from:

C l , fus

 2  1.2 ARl .S  z w  D fl .s   1.4 tan l .s  2  D 57.3 b  l .s  fl .s 2

Eqn.(9)

For a wing placement outside of the fuselage diameter:

z w  0.7 D f ,l .s tan l .s 

D fl .s

Eqn.(10)

The fuselage induced effect on the lifting surface for a high wing is found from: C l , fus

 2 D fl .s D 2fl , s 1.2 ARl .s     57.3  z w  0.7 D f ,l .s tan l .s z w  0.7 D f ,l .s tan l .s  bl2.s  

Eqn.(11)

The third intermediate calculation parameter, Z, is written as:

 C l   Z   tan  t C  4 h   t tan  C 4W 

   

Eqn.(12)

The lifting surface twist correction factor is obtained from Figure 10.26 in Airplane Design Part VI and is a function of the lifting surface aspect ratio and the lifting surface taper ratio:

C l

 t ,l .s tan  c

 f  ARl .s , l .s 

Eqn.(13)

4l .s

The contribution to the dihedral effect from the vertical tail is found from:









 Z acV  Z cg cos   x acv  xcg sin    C lV  C y v   bw  

Eqn.(14)

Segment: C lW rad 1

1 n/a

2 -0.0265

3 -0.0268

4 -0.0268

5 -0.0304

6 -0.1298

7 -0.0284

1

n/a

0.0097

0.0010

-0.0001

-0.0041

0.0012

n/a

-0.1055

-0.0905

-0.1117

-0.1174

-0.0738

-0.1285

n/a

-0.1357

-0.1410

-0.1594

-0.1929

-0.2064

-0.1754

8 -0.0334

9 -0.0265

10 -0.0265

11 n/a

12 -0.0266

13 -0.0268

14 -0.0334

0.0019

0.0058

0.0047

n/a

0.0012

-.0021

0.0050

-0.1361

-0.0418

-0.0627

n/a

-0.0761

-0.0696

-0.1367

-0.2059

-0.1773

-0.1670

n/a

-0.1554

-0.1393

-0.1970

15 -0.1072

16 -0.0265

17 -0.0265

0.0031

0.0157

-0.0566

-0.0238

-0.1743

-0.1524

 rad rad rad

C l h C l V C l

1 1

   

Segment: C lW rad 1

 rad rad rad

C l h C l V C l

1 1 1

   

Segment: C lW rad 1

 rad rad rad

C l h C l V C l



1 1 1

   

Yawing Moment Coefficient due to Sideslip:

Theory: The airplane yawing-moment-coefficient-due-to-sideslip derivative, or static directional stability, is determined from:

C

n

n W

C

n

C f

Eqn.(1)

n V

The wing contribution is only important at high angles of attack. For preliminary design purposes: C

n W

Eqn.(2)

0

The contribution of the fuselage to the static directional stability is found from:

C n , f  57.3K N K Rl

S B,s L f

Eqn.(3)

S w bw The empirical factor for wing-fuselage interference is obtained from Figure 10.28 in Airplane Design Part VI and is a function of the airplane center of gravity location, fuselage length, fuselage side projected area, fuselage height at the quarter and three-quarter length, maximum fuselage height, and maximum fuselage width: K



 f X CG , l f , S B , h1 , h2 , hMax ,W f s



The effect of fuselage Reynold's number on wing-fuselage directional stability is found from Figure 10.29 in Airplane Design Part VI and is a function of the fuselage Reynold's number: K

 

 f R Nf

The vertical tail contribution to this derivative is solved from: C

n V

lV Cos  ZV Sin  C Y bV V

Segment: C n , f rad 1

C n ,v C n

  rad  rad 

1 2 3 4 5 6 7 -0.0134 -0.0651 -0.0662 -0.0674 -0.0622 -0.0649 -0.0629

1

n/a

0.3270

0.3402

0.4289

0.4787

0.4512

0.4988

1

n/a

0.2618

0.2740

0.3615

0.4165

0.3863

0.4359

Segment: C n , f rad 1

C n ,v

Eqn.(1)

  rad  1

8 9 10 11 12 13 14 -0.0637 -0.0610 -0.0608 -0.0136 -0.0641 -0.0657 -0.0610 0.3576

0.4461

0.4435

n/a

0.3419

0.3449

0.5076

rad  1

C n

Segment: C n , f rad 1

  rad  rad 

C n ,v C n 

0.2938

0.3851

0.3827

n/a

0.2778

0.2793

0.4466

15 16 17 -0.0172 -0.0606 -0.0592

1

0.2387

0.4471

0.3638

1

0.2215

0.3865

0.3046

Thrust Sideforce Coefficient due to Sideslip Derivative

The airplane thrust sideforce-coefficient-due-to-sideslip-derivative is neglected for jet powered airplanes.



Thrust Yawing Moment Coefficient due to Sideslip Derivative

The airplane thrust yawing-moment-coefficient-due-to-sideslip-derivative is neglected for jet powered airplanes.

-Sideslip related derivatives:



Sideforce Coefficient due to Sideslip Rate Derivative

The airplane sideforce-coefficient-due-to-sideslip-rate derivative is estimated by:  SV lV Cos  Z V Sin  2C C Y L  bW V  SW

Eqn.(1)

Where:

l P  X acV  X acW And z P  z acv  z acw

Eqn.(2) Eqn.(3)

The change in sidewash angle due to the change in sideslip is determined from:  











57.3



  e t Wf

Eqn.(4)

The angle-of-attack contribution to sidewash is found from Figure 10.30 in Airplane Design Part VI and is a function of wing aspect ratio, steady state flight Mach number, wing taper ratio, wing leading edge sweep angle, wing span, and the relative distances between the wing and vertical tail aerodynamic centers: 





 f ARW , M , W ,  LE , bW , ZV W



The wing dihedral contribution to sidewash is obtained from Figure 10.31 in Airplane Design Part VI and is a function of wing aspect ratio, steady state flight Mach number, wing taper ratio, wing leading edge sweep angle, wing span, and the vertical distance between the wing and vertical tail aerodynamic centers: 





 f ARW , M , W ,  LE ,bW , ZV W



The wing twist contribution to sidewash is defined from Figure 10.32 in Airplane Design Part VI and is a function of wing aspect ratio, Mach number, wing taper ratio, wing leading edge sweep angle, wing span, and relative distances between the wing and vertical tail aerodynamic centers: 





 f ARW , M , W ,  LE ,bW , ZV W



The fuselage contribution to sidewash is found from Figure 10.33 in Airplane Design Part VI and is a function of wing aspect ratio, steady state flight Mach number, wing taper ratio, wing leading edge sweep angle, wing span, relative distance between the wing and the vertical tail aerodynamic centers, and maximum fuselage diameter. The wing position on the body has an effect on this term too. It has a positive value for a low-wing configuration and changes sign for a high-wing Configuration. 





 f ARW , M , W ,  LE , bW , ZV W



Segment: C y

rad 

1 2 3 4 5 6 7 n/a 0.0152 0.0119 -0.0226 -0.0261 0.0102 0.0071

Segment: C y

rad 

8 9 10 11 12 13 14 0.0090 0.0146 -0.0331 n/a 0.0010 -0.0034 0.0093

Segment: C y

rad 

15 16 17 -0.0173 -0.0592 -0.0083

1



Rolling Moment Coefficient due to Sideslip Rate Derivative

The airplane rolling-moment-coefficient-due-to-sideslip-rate derivative is calculated from: Cl  C y 



ZV Cos lV Sin bW

Segment: C l rad 1



Segment: C l rad 1



Segment: C l rad 1





Eqn.(1)



1 2 3 4 5 6 7 n/a 0.0029 0.0018 -0.0029 -0.0034 0.0018 0.0013



8 9 10 11 12 13 14 0.0015 0.0023 -0.0040 n/a 0.0002 -0.0004 0.0017



15 16 17 -0.0024 -0.0054 0.0014

Yawing Moment Coefficient due to Sideslip Rate Derivative

The airplane yawing-moment-coefficient-due-to-sideslip-rate derivative is found from: C

n 

lV Cos  ZV Sin C Y  bW

Eqn.(1)

Segment: Cn

rad 

1 2 3 4 5 6 7 n/a 0.0066 0.0053 -0.0100 -0.0115 0.0047 0.0033

Segment: Cn

rad 

8 9 10 11 12 13 14 0.0040 0.0065 -0.0159 n/a 0.0005 -0.0016 0.0043

Segment: Cn

rad 

15 16 17 -0.0076 -0.0277 0.0038



1

Sideforce Coefficient due to Roll Rate Derivative:

The airplane sideforce-coefficient-due-to-roll-rate derivative is primarily influenced by the vertical tail and may be determined from:  2C C Y YP V

ZV Cos  lV Sin  ZV bW



 

 3Sin 1 4 ZSinW Cl |  0 W P |C L  0

Eqn.(1)

The roll damping derivative of the wing without dihedral and at zero lift is found from:

  Cl P   K   

ClP 0  K  C L 0

Eqn.(2)

The roll damping parameter at zero lift is found from Figure 10.35 in Airplane Design Part VI and is a function of the wing aspect ratio, the Prandtl-Glauert transformation factor, the sectional lift curve slope obtained through the Prandtl-Glauert transformation factor, the wing quarter chord sweep angle, and the taper ratio:

  Cl P   K 

     f  ARW ,  , K ,  C , W   4W  |CL 0 

The ratio of incompressible sectional lift curve slope with 2p is given by:

f gap , wo cl ,W M 0

k

Eqn.(2)

2

Segment: C yp

rad 

1 2 3 4 5 6 7 n/a -0.0781 -0.0491 -0.1808 -0.1755 -0.0560 -0.0528

Segment: C yp

rad 

8 9 10 11 12 13 14 -0.0535 -0.0585 0.0615 n/a -0.0245 -0.0098 -0.0592

Segment: C yp

rad 

15 16 17 -0.1890 0.0717 -0.0490



1

Rolling Moment Coefficient due to Roll Rate Derivative

The airplane rolling-moment-coefficient-due-to-roll-rate derivative, also known as the roll damping derivative, is estimated from:

C l p  C l p , w  C lPh  C lP ,V

Eqn.(1)

The contribution of the horizontal tail given by:

 bh   Cl    P h 2 SW  bW  1

C lP h

Eqn.(2)

The intermediate calculation parameter can be calculated from:    ClP    k 

C 

l P l .s

 k      C L 0  

C  L ,l . S   C L ,l . S

CL ,l . s C L , 0

 CL P   C  LP   0

   C lP   







drag l . s

Eqn.(3)

The roll damping parameter at zero lift for the lifting surface is obtained from Figures 10.35 in Airplane Design Part VI and is a function of the lifting surface aspect ratio, the PrandtlGlauert transformation factor, the sectional lift curve slope of the lifting surface, the lifting surface quarter chord sweep angle, and the lifting surface taper ratio:

 C lP    ,  l.S   f  ARl.S ,  ,  C   4 l.S   K C 0  L The ratio of the incompressible sectional lift curve slope of the lifting surface to 2p is defined as:

 C   C  l  X  M   l X  M 0 K 2

Eqn.(4)

2



The dihedral effect parameter is found from:

   

4Z  ZW P   1  W Sin  12 Cl bW  bW P  0 Cl

  

Eqn.(5)

Sin 2

The drag contribution is determined from:

C





l P drag w



C 

lP C D,L

2 LW

C

 C L , f





 0.125 C D 0 w  C D 0 flap



Eqn.(6)

C





l P drag w



C 

lP C D,L

C L2

C L2h  0.125C D 0h

Eqn.(7)

The drag-due-to-lift roll damping parameter is found from Figure 10.36 in Airplane Design Part VI and is a function of the lifting surface aspect ratio and the wing quarter-chord sweep angle:

  Cl

P C DL   f  ARW ,  C 2 4W  CL W

  

The vertical tail also contributes to the roll damping derivative by:

C lP ,V 





2 z v cos   lv sin   z v cos   lv sin    Z ac,V  Z cg  C y ,V bw2

Eqn.(8)

Segment: C lP ,W rad 1

1 n/a

2 -0.4659

3 -0.4748

4 -0.4633

5 -0.5210

6 -0.4962

7 -0.5704

1

n/a

-0.0012

1

n/a

-0.0041

0.0000

-0.0010

-0.0028

-0.0006

-0.0008

n/a

-0.4712

-0.4760

-0.4654

-0.5249

-0.4980

-0.5724

Segment: C lP ,W rad 1

8 -0.4921

9 -0.4731

10 -0.4866

11 n/a

12 -0.4791

13 -0.4783

14 -0.5696

1

-0.0012

n/a

-0.0012

1

-0.0003

-0.0012

-0.0072

n/a

-0.0024

-0.0035

0.0000

-0.4936

-0.4755

-0.4950

n/a

-0.4826

-0.4829

-0.5708

Segment: C lP ,W rad 1

15 -0.4609

16 -0.4539

17 -0.4724

1

-0.0012

1

-0.0001

-0.0040

0.0000

-0.4622

-0.5005

-0.4735

  rad  rad  rad 

ClP ,h ClP ,v Cl p

1

  rad  rad  rad 

ClP ,h ClP ,v Cl p

1

  rad  rad  rad 

ClP ,h ClP ,v Cl p

1



Yawing Moment Coefficient due to Roll Rate Derivative

The airplane yawing-moment-coefficient-due-to-roll-rate derivative is determined from: CnP  CnP ,W  CnP ,V

Eqn.(1)

The wing contribution to this derivative is given by:

 C nP   C nP  C C    nP L t C L C 0 W L    t |M

 CnP     f  f

  f f

Eqn.(2)

The airplane zero-lift contribution is calculated from: ARW  4Cos C  C nP  4W     C L C 0 AR B  4Cos W C   L 4W

 CnP  CL 

X

  C 0  L

Eqn.(3)

M 0

The lift coefficient contribution at zero lift and at zero Mach is solved from:

 C n, P   CL

 1   6  C L 0 , M 0

 ARw  6 ARw  cos  C 4  ARw  4 cos c

 Y 

Eqn.(4)

4 w

The variable Y is:

  tan    C 4 Y   S .M    ARw    

2   tan       c 4  w     12     

Eqn.(5)

The compressible sweep correction factor B is obtained from:

B  1  M  cos  C  4 w  2 1

Eqn.(6)

The intermediate calculation parameter X is: 1   2 ARW B   ARW B  Cos C  tan  C  2 4W 4W  X   ARW  1  ARW  Cos C  tan 2  C  2 4W 4W 

    

Eqn.(7)

The wing twist contribution is obtained from Figure 10.37 in Airplane Design Part VI and is a function of wing aspect ratio and wing taper ratio:







 f ARW , W

t

The contribution due to symmetrical flap deflection is determined from Figure 10.38 in Airplane Design Part VI and is a function of wing aspect ratio, wing taper ratio, flaps inboard and outboard stations in terms of wing half span, and wing span:



C n

P  f AR ,  , , W W i f O f ,bW   f f



The product of the derivative of the angle of attack with respect to flap deflection and the flap deflection angle is found from: 



f

Cl

Eqn.(8)

Cl  f 

The vertical tail contribution to the yawing-moment-due-to-roll-rate derivative is solved by: C

nP V





l Cos  ZV Sin 2 V

ZV Cos lV Sin  ZV .C y 

Eqn.(9) v

Segment: C nP , f rad 1

1 n/a

2 0.0001

3 0.0001

4 0.0000

5 0.0000

6 0.0000

7 0.0000

1

n/a

-0.0380

-0.0717

-0.0820

-0.1266

-0.0700

-0.0683

1

n/a

0.0127

0.0000

-0.0042

-0.0153

0.0020

–0.0029

n/a

-0.0253

-0.0717

-0.0862

-0.1419

-0.0680

-0.0711

  rad  rad  rad 

C nP , w C nP , v

C nP

1

Segment: C nP , f rad 1

9 0.0002

10 0.0001

11 n/a

12 0.0002

13 0.0000

14 0.0000

1

-0.0732

-0.0614

-0.1587

n/a

-0.0967

-0.1133

-0.0644

1

0.0011

0.0044

-0.0510

n/a

-0.0107

-0.0174

0.000

-0.0721

-0.0570

-0.2096

n/a

-0.1074

-0.1307

-0.0644

15 0.0000

16 0.0001

17 0.0000

1

-0.0762

-0.1964

-0.0762

1

-0.0006

-0.0548

0.0000

-0.0768

-0.2512

-0.0762

  rad  rad  rad 

C nP , w C nP , v

8 0.0000

1

C nP

Segment: C nP , f rad 1

  rad  rad  rad 

C nP , w C nP , v

1

C nP

-Yaw rate related derivatives:



Sideforce Coefficient due to Yaw Rate Derivative

The airplane sideforce-coefficient-due-to-yaw-rate derivative is primarily influenced by the vertical tail and may be determined from: C

 2C

lV Cos  Z z Sin y



Segment: C yr rad 1



Segment: C yr rad 1





Segment: rad 1 C yr

Eqn.(1)



1 2 3 4 5 6 7 n/a 0.6539 0.6803 0.4717 0.5163 0.9025 0.9976



8 9 10 11 12 13 14 0.7151 0.6818 0.8869 n/a 0.6837 0.6899 1.0151



15 16 17 0.4774 0.6980 0.7276

Rolling Moment Coefficient due to Yaw Rate Derivative

The airplane rolling-moment-coefficient-due-to-yaw-rate derivative is given by:

Eqn.(2)

C C C lr lr lr V W

The wing contribution is found from:

C

lr w

 Cl r  Lw   CL

Cl  r   C  0   L



Cl

t

r   t

Cl

  f f



f



Eqn.(3) f

The slope of the rolling-moment-due-to-roll-rate at zero-lift is solved from: 1

 Cl r   CL 

   C  0  L

  2 B AR w  2Cos C  4w  

AR w  2Cos C 1

 

AR w 1 B

AR w  4Cos C

tan 4w

X

C

 Cl r   CL 

  C  0  L

Eqn.(4)

M 0

8 4w

The variable X can be expanded into the following: AR w B  2Cos C X  AR w B  4Cos C

2 tan  C

Eqn.(5)

8 4w

The variable B is defined from: B  1 M

2

  Cos C  4w  

Eqn.(6)

The slope of the low-speed rolling-moment-coefficient-due-to-yaw-rate at zero lift is found from Figure 10.41 in Airplane Design Part VI and is a function of wing aspect ratio, wing taper ratio, and wing quarter chord sweep angle:

 Cl r   CL 

     f  ARw ,  w ,  C  C  0  4w   L M 0

The increment of the rolling-moment-coefficient-due-to-roll-rate due to the wing dihedral is given by:

Cl

ARw Sin C r  0.083



Eqn.(7)

ARw  4Cos C

The increment in the rolling-moment-coefficient-due-to-yaw-rate due to wing twist is determined from Figure 10.42 in Airplane Design Part VI and is a function of wing aspect ratio and wing taper ratio: Cl

t



r  f AR ,  w w



Eqn.(8)

The effect of symmetric flap deflection on the rolling-moment-coefficient-due-to-yaw-rate derivative is obtained from Figure 10.43 in Airplane Design Partt VI and is a function of wing aspect ratio, wing taper ratio, inboard and outboard flap locations in terms of wing half span, and wing span: C l

  f f

 f  AR ,  , , , b   w w i, f o, f w 

The symmetric flap deflection effect on the rolling-moment-coefficient-due-to-yaw-rate is equal to the outboard flap effect minus the inboard flap effect as given from: C l

  f f





Cl

   f  f

  C    lr    Out   f f

    In

Eqn.(9)

The product of the change in airplane angle-of-attack due to flap deflection and the flap deflection angle is found from:

  f  f

cl f cl w

Eqn.(10)

The vertical tail contribution is found from:

lr v









l cos   Z v sin  Z v cos   lv sin  C y 2 v v bw

Eqn.(11)

Segment: C lr , flap rad 1

1 n/a

2 0.0005

3 0.0002

4 0.0000

5 0.0000

6 0.0000

7 0.0000

1

n/a

0.0674

0.1406

0.1883

0.3455

0.1542

0.2020

n/a

0.0899

0.0791

0.0456

0.0397

0.1146

0.1221

n/a

0.1573

0.2197

0.2289

0.3852

0.2689

0.3241

Segment: C lr , flap rad 1

8 0.0000

9 0.0006

10 0.0004

11 n/a

12 0.0006

13 0.0000

14 0.0000

1

0.1533

0.1140

0.3101

n/a

0.1880

0.2290

0.1854

0.0833

0.0826

0.0584

n/a

0.0675

0.0616

0.1248

0.2366

0.1966

0.3685

n/a

0.2555

0.2905

0.3102

Segment: C lr , flap rad 1

15 0.0000

16 0.0004

17 0.0000

1

0.1545

0.3799

0.1405

0.0480

0.0235

0.0826

0.2025

0.4034

0.2231

C lrw C lrv C lr

 rad rad rad  rad rad rad  rad rad rad

1 1

           

 Yawing Moment Coefficient due to Yaw Rate Derivative: Theory: The airplane yawing-moment-coefficient-due-to-yaw-rate derivative, also known as the yaw damping derivative, is determined by: C

C

nr w

C

nr v

Eqn.(1)

The wing contribution to the yaw damping derivative is found from: Cn

Cn 2 r r C   C C nr D0 w 2 Lw C D0 w C L

Eqn.(2)

The lifting effect for the wing yaw damping derivative is found from Figure 10.44 in Airplane Design Part VI and is a function of the aspect ratio, taper ratio, quarter chord sweep angle and the static margin:

 r  f  AR ,  ,   w w C , S .M  2 4w   CL Cn

The static margin is given by:

S .M  x ac  x cg

Eqn.(3)

The drag effect for the wing yaw damping derivative is found from Figure 10.45 in Airplane Design Part VI and is a function of the aspect ratio, quarter chord sweep angle and the static margin:

CD0

 f  ARw , C , SM  4w  

The vertical tail contribution is calculated from:

Cn  rv

2 l Cos  Z v Sin 2 C y 2 v v bw

Eqn.(4)

Segment: C nr , w rad 1

1 n/a

2 3 4 5 6 7 -0.0016 -0.0030 -0.0011 -0.0006 -0.0027 -0.0031

1

n/a

-0.2787 -0.2972 -0.1978 -0.2204 -0.4025 -0.4476

1

n/a

-0.2803 -0.3002 -0.1990 -0.2210 -0.4052 -0.4506

C nr , v

C nr

  rad  rad 

Segment: C nr , w rad 1

8 9 10 11 12 13 14 -0.0026 -0.0029 -0.0082 n/a -0.0037 -0.0051 -0.0032

1

-0.3175 -0.3017 -0.4129 n/a -0.3033 -0.3054 -0.4634

1

-0.3202 -0.3045 -0.4211 n/a -0.3070 -0.3105 -0.4666

C nr , v

C nr

  rad  rad 

Segment: C nr , w rad 1

15 16 17 -0.0033 -0.0095 -0.0005

1

-0.2027 -0.3192 -0.3399

1

-0.2060 -0.3287 -0.3404

C nr , v

C nr

  rad  rad 

-Longitudinal Control derivatives:



Drag Coefficient due to Elevator Deflection Derivative

Theory: The airplane drag-coefficient-due-to-elevator-deflection derivative is estimated from: C

D

 e

Eqn.(1)

 e Di h

The change in airplane angle-of-attack due to elevator deflection at zero deflection is found by solving this series of equations with the elevator deflection equal to zero. The change in airplane angle-of-attack due to elevator deflection is found from: Cl

 

 K b e



 Theory

Cl Theory

k c l

h M  0

 C L  c

Eqn.(2)

The elevator span factor is obtained from:

K b  K b0  K bi

Eqn.(3)

The inboard elevator span factor and the outboard elevator span factor are obtained from Figure 8.52 in Airplane Design Part VI and is a function of the elevator inboard and outboard stations, and the horizontal tail taper ratio:



K b0  f  0 w ,  h



K bi  f  ie ,  h



The correction factor for flap lift is obtained from Figure 8.15 in Airplane Design Part VI and is a function of the elevator chord to horizontal tail chord ratio and the sectional lift curve slope to theoretical lift curve slope ratio:



 Theory



 C Cl e  f , C    C lTheory  

     

The theoretical horizontal tail sectional lift curve slope at Mach equal to zero is given by: C l , h

Theory

t   2  5.0525   c h

Eqn.(4)

The lift effectiveness parameter is found from Figure 8.14 in Airplane Design Part VI and is a function of the elevator chord to horizontal tail chord ratio and the horizontal tail thickness ratio at the center of the elevator:



Cl Theory  f  CCe , ct  

 t     c  ht 

, hr

The correction factor accounting for nonlinearities at high elevator deflection angles is found from Figure 8.13 in Airplane Design Part VI and is a function of the elevator chord to horizontal tail chord ratio and the elevator deflection angle:

 Ce

K  f 





, e 



The three dimensional flap effectiveness parameter is determined from Figure 8.53 in Airplane Design Part VI and is a function of horizontal tail aspect ratio, and elevator chord to horizontal tail chord ratio:

 CL  Ce   f  AR ,   Cl  h C  The airplane drag-coefficient-due-to-horizontal tail-incidence derivative is estimated from: C Di , h  2

C L0

ARw e

 Sh  C Lh   Sw 

 h , P.Off 

Eqn.(5)

The Oswald efficiency factor is estimated from Figure 3.2 in Methods for Estimating Stability and Control Derivatives of Conventional Subsonic Airplanes (E. Torenbeek , Sec 3.1 PP 3.1-3.17) and is a function of the wing aspect ratio and the wing taper ratio:

e  f  ARw ,  w  Segment: C Di , h

C D e

rad  rad  1 1

1 2 3 4 5 6 7 n/a 0.0331 0.0279 0.0201 0.0239 0.0059 0.0194 n/a 0.0125 0.0106 0.0076 0.0100 0.0248 0.0240

Segment: C Di , h

rad  rad 

0.0244 0.0138 0.0118 n/a

1

C D e Segment: C Di , h

rad  rad  1 1

C D e 

8 9 10 11 12 13 14 0.0093 0.0365 0.0312 0.0188 0.0132 0.0086 0.0233

1

0.0350 0.0228 0.0180

15 16 17 0.0201 0.0117 0.0228 0.0076 0.0311 0.0086

Lift Coefficient due to Elevator Deflection Derivative

Theory: The airplane lift-coefficient-due-to-elevator-deflection is calculated from: C

C

 0

K 

Eqn.(1) e

The airplane lift-coefficient-due-to-elevator-deflection derivative is determined from: C

L

 e

Eqn.(2)

 e Li

Where K  is the slope at the given elevator deflection. K   K    e

dK  d e

Eqn.(3)

The airplane lift-coefficient-due-to-elevator-deflection derivative at zero deflection is found from:

C Le ,O  f bal C Lih  eO The airplane lift-coefficient-due-to-stabilizer-incidence derivative is found from:

C Li ,h   h , P.OFF   h

Sh CL S w h

Eqn.(4)

The elevator balance factor is dependant on nose shape and is calculated from: f bal e  0.83  0.19857Balance e  Eqn.(5)

f bale  1

Segment: C L ,e

rad  1

3 4 5 6 7 0.3075 0.3077 0.3516 0.4866 0.6521

n/a -0.0463 0.0000 0.0027 0.0000 0.0021 -0.0361

C Le Segment: C L ,e

rad  1

8 9 10 0.3145 0.3053 0.3042

11 12 13 14 n/a 0.3053 0.3076 0.3092

0.0000 0.0135 -0.0330 n/a 0.0000 0.0000 -0.0288

C Le Segment: C L ,e

rad  1

15 0.3076

16 17 0.3026 0.3075

-0.0366 0.0000 0.0000

C Le 

1 2 n/a 0.3042

Pitching Moment Coefficient Due To Elevator Deflection Derivative

Theory: The airplane pitching moment-coefficient-due-to-elevator-deflection is calculated from:

C me  C m eO K  e The airplane pitching determined from: C

m

 e

Eqn.(1)

moment-coefficient-due-to-elevator-deflection

derivative

Eqn.(2)

 e mi

Where K  is the slope at the given elevator deflection. dK  K   K    e d e

Eqn.(3)

Where the correction for nonlinear pitching moment behavior of plain flaps is found from Figure 8.13 in Airplane Design Part VI and is a function of elevator deflection angle and the average elevator chord to horizontal tail chord ratio aft of hinge line.  c K   f   e , e  ch

  

Eqn.(4)

The airplane pitching moment-coefficient-due-to-elevator-deflection derivative at zero deflection is found from: C m eO  f bal C mi , h   e ,O

Eqn.(5)

The airplane pitching-moment-coefficient-due-to-stabilizer-incidence derivative, also known as the stabilizer control power, is given from:

C mih  C Lh  hVh Segment: C m e

Eqn.(6)

rad  1

n/a 0.1476

C me Segment: C m e

rad  1

-0.0086 0.0000

-0.0067 0.1166

-0.0438 0.1086

n/a 0.0000

0.0000

0.0949

15 16 17 -1.0122 -0.9788 -0.9968 0.1204

C me

0.0000

8 9 10 11 12 13 14 -1.0221 -0.9936 -1.0006 n/a -0.9910 -1.0113 -1.0177 0.0000

C me Segment: C m e

1 2 3 4 5 6 7 n/a -0.9702 -0.9914 -0.9922 -1.1401 -1.5691 -2.1097

0.0000

-Lateral directional Control Derivatives:



Sideforce Coefficient due to Aileron Deflection Derivative

The airplane sideforce-coefficient-due-to-aileron-deflection derivative is negligible for most conventional aileron arrangements: C Y

 0.0000 Rad 1 e



Rolling Moment Coefficient due to Aileron Deflection Derivative

Theory: C l a 

C l al  aL  C l ar  ar

a

Eqn.(1)

The aileron deflection angle of the airplane is given by:

a 

 aL   aR 2

Eqn.(2)

The rolling-moment-coefficient-due-to-left-aileron-deflection derivative is calculated from:

1 f bala C l   al 2

C l al 

Eqn.(3)

The rolling-moment-coefficient-due-to-right-aileron-deflection calculated from:

1 f bala C l   ar 2

C l ar 

derivative

Eqn.(4)

The aileron balance factor is dependent on nose shape and is calculated from: For a Round Nose:

f bala  0.83  0.30714Balancea 

Eqn.(5)

The aileron balance based on control surface area forward and aft of hinge line is found from: Balancea 

S1 S2

Eqn.(6)

The rolling effectiveness of two full-chord ailerons is determined from: C l 

k  C l  k

Eqn.(7)

The Prandtl-Glauert transformation factor is derived from:

  1  M 12

Eqn.(8)

The ratio of incompressible aileron sectional lift curve slope to 2p is solved from: k

cl

, M 0

2

Eqn.(9)

The aileron rolling moment effectiveness parameter is obtained from Figures 10.46 in Airplane Design Part VI and is a function of the inboard and outboard aileron

stations, wing aspect ratio, Prandtl-Glauert transformation factor, wing quarter chord sweep angle, wing taper ratio and the ratio of the incompressible aileron sectional lift curve slope to:

C l k

  C l    k

  C      l     O  k i

Eqn.(10)

Where:

 C l   k 

   f  Oa , ARw ,  ,   ,  w , k  O

 C l   k 

   f  ia , ARw ,  ,   ,  w , k  i





The wing quarter chord sweep angle corrected for Mach effect is given by:

 tan  C 4w    tan 1    

   

Eqn.(11)

The change in airplane angle-of-attack due to left aileron deflection is found from:

  al

c  l

Eqn.(12)

cl a

The change in airplane angle-of-attack due to right aileron deflection is found from:

  ar

c  l

Eqn.(13)

cl a

The average airfoil lift curve slope of that part of the wing covered by the aileron can be computed from:

cl , a 

cl , M  0





cl , M  0



Eqn.(14)

For this calculation, the wing sectional lift curve slope is assumed to be constant over the wing span. Therefore, the aileron sectional lift curve slope is equaled to the wing sectional lift curve slope. The lift effectiveness of the left aileron is given by:

cl



cl cl

c 

l Theory

k l

Eqn.(15)

Theory

The lift effectiveness of the Right aileron is given by:

cl



cl cl

c 

l Theory

k r

Eqn.(16)

Theory

The correction factor for aileron lift effectiveness is obtained from Figure 8.15 in Airplane Design Part VI and is a function of the aileron chord to wing chord ratio, and the sectional lift curve slope to the theoretical sectional lift curve slope: cl cl

Theory

  c cl  f  a , W ,M 0  c w cl Theory 

    

Eqn.(17)

The theoretical wing sectional lift curve slope is given by: cl

Theory

t   2  5.0525   c w

Eqn.(18)

The lift effectiveness parameter for the aileron is found from Figure 8.14 in Airplane Design Part VI and is a function of the aileron chord to wing chord ratio and the thickness ratio of the wing root and tip sections: c t    f  a ,     cw  c  w  The correction factor accounting for nonlinearities at high aileron deflection angles is found from Figure 8.13 in Airplane Design Part VI and is a function of the aileron to wing chord ratio and the aileron deflection angle:

c  l

theory

c  k l  f  a ,  al   cw  c  k r  f  a ,  ar   cw 

Segment: C l a , 0

rad  rad 

n/a 0.0819 0.1119 0.1125 0.1651 0.3130 0.9731

1

C l a Segment: C l a , 0

8 9 10 11 12 13 14 0.2141 0.1651 0.1615 n/a 0.1651 0.1748 0.1834

rad  rad  1

0.1086 0.0837 0.0819 n/a 0.0837 0.0887 0.0930

1

C l a Segment: C l a , 0

rad  rad  1

1

C l a 

1 2 3 4 5 6 7 n/a 0.1615 0.1742 0.1751 0.3256 0.6173 0.9731

1

15 16 17 0.1748 0.1572 0.1572 0.0887 0.0797 0.0797

Yawing Moment Coefficient due to Aileron Deflection Derivative

The airplane yawing-moment-coefficient-due-to-aileron-deflection derivative, also called the adverse aileron yaw, consists of two components:



C na  C na



induced

 

 C na

profile

K

Eqn.(1)

n a

The contribution of induced drag to the airplane yawing-moment-coefficient-due-to-ailerondeflection derivative is computed from: C na



   K C  C

 K ai C la

induce

l a

Lw , Clean

Eqn.(2)

The correlation constant for yawing moment due to aileron deflection is obtained from Figure 10.48 in Airplane Design Part VI and is a function of aileron inboard and outboard stations, wing aspect ratio and wing taper ratio.



K ain  f ia , AR w , w And





K a in  f O a , AR w , w



The method of computing the rolling-moment-coefficient-due-to-aileron-deflection derivative has been discussed in another topic. However, several substitutions are needed.

While computing the rolling-moment-coefficient-due-to-aileron-deflection derivatives for a control surface which spans from the aileron inboard station to the wing tip, replace

 O with 100 % a

The profile drag contribution to the yawing moment due to aileron deflection derivative is computed from:

C  na

profile

 C D par C D pal     a  al r 

  Oa   i a   4 

Eqn.(3)

The variation of aileron profile drag coefficient with the aileron deflection angle is found from: C DPax  a x



 

  0.01  





C DPax @  a x  0.01  C DPax @  a x  0.01 ax



 0.01

Eqn.(4)

The aileron profile drag coefficient is given by:

C D , Pax   C d P 

 cos  S af S a c C  4 S 4  xS a Sw

Eqn.(5)

The two-dimensional profile drag increment due to left (or right) aileron deflection can be obtained from Figure 4.44 in Airplane Design Part VI and is a function of aileron chord to wing chord ratio and aileron deflection angles.   C d P 

 C 0 4

c    f  a ,  a x x  cw

  

The wing area over the span of left (or right) aileron is calculated from: S af 





bw c i  c Ow  Oa   i a 4 w



Eqn.(6)

The wing chord length at the aileron inboard station is given by:

ciw  c rw 1   O ,a 1   w 

Eqn.(7)

The wing root chord is calculated from: c rw 

Sw 2  w  1 ARw

Eqn.(8)

Segment: C n a

rad 

1 2 3 4 5 6 7 n/a -0.0005 -0.0046 -0.0059 -0.0163 -0.0099 -0.0993

Segment: C n a

rad 

8 9 10 11 12 13 14 -0.0080 -0.0021 -0.0042 n/a -0.0044 -0.0060 -0.0035

Segment: C n a

rad 

15 16 17 n/a n/a n/a

1

-Rudder related stability derivatives:



Sideforce Coefficient due to Rudder Deflection Derivative

Theory: The airplane sideforce-coefficient-due-to-rudder-deflection derivative is estimated from: C

Y

 C v L



S v

Eqn.(9)

v   r w

The change in sideslip due to rudder deflection is given by:



  cl c l   Theory 

  Kb  r

cl

 

 k  Theory   cl ,v

 

   C L      C l  

Eqn.(10)

The correction factor accounting for nonlinearities at high rudder deflection angles is found from Figure 8.13 in Airplane Design Part VI and is a function of the rudder to vertical tail chord ratio and the rudder deflection angle:

c  k   f  r ,  r   cv 

The rudder span factor is obtained from Figure 8.51 with data from Figure 8.52 in Airplane Design Part VI and is a function of the rudder inboard and outboard stations and vertical tail taper ratio:

K b  K bO  K bi Where: K bO  f Or , v

  f 

Eqn.(11)

 , 

K bi Or v The correction factor for sectional rudder lift is obtained from Figure 8.15 in Airplane Design Part VI and is a function of the rudder chord to vertical tail chord ratio and the sectional lift curve slope to theoretical lift curve slope ratio:   cl  c a clW , M  0   f ,  cl  c w cl  Theory Theory   The theoretical vertical tail sectional lift curve slope at zero-Mach is given by:

C



lV M  0 Theory

t  2  5.0525   c v

Eqn.(12)

The lift effectiveness parameter is found from Figure 8.14 in Airplane Design Part VI and is a function of the rudder chord to vertical tail chord ratio and the thickness ratio of the vertical tail at the center of the rudder:  cr  t    ,    f  theory  cv  c  v  The “three dimensional rudder effectiveness parameter” is determined from Figure 8.53 in Airplane Design Part VI and is a function of the vertical tail aspect ratio, and rudder chord to vertical tail chord ratio:

c  l

  CL   Cl

Cr   f  ARv , eff C v 

Segment: Cy r

C y , Rudder

  

rad  1

1 2 3 4 5 6 7 n/a 0.1885 0.1906 0.1907 0.1999 0.2546 0.3335 n/a 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Segment: Cy r

8 9 10 11 12 13 14 0.1918 0.1893 0.1885 n/a 0.1893 0.1907 0.1913

rad  1

0.0000 0.0000 0.0000 n/a 0.0000 0.0000 0.0000

C y , Rudder Segment: Cy r

rad  1

0.0000 0.0000 0.0000

C y , Rudder 

15 16 17 0.1907 0.1874 0.1874

Rolling Moment Coefficient due to Rudder Deflection Derivative

Theory: The airplane rolling-moment-coefficient-due-to-rudder-deflection derivative is determined from:

Cl  r

z r cos   lr sin  Cy r bw

Eqn.(1)

Where:

l r  X acr  X cg And z r  Z acr  Z cg

Eqn.(2) Eqn.(3)

The center of pressure of the vertical tail area affected by rudder deflection is assumed to be at the leading edge of the rudder along the rudder mean geometric chord. Based on the above mentioned assumption, the Z-location of the center of pressure of the vertical tail area affected by rudder deflection measured from the reference line is calculated from:

Z acr  Z apexv   ir bv 





  ir bv 1  2 r  31   r 

Eqn.(4)

The rudder taper ratio is defined as:

r 

Eqn.(5)

Where the vertical tail chord length at rudder outboard station is given by:

cv O

 c rv 1   Or 1  v 

Eqn.(6)

And the vertical tail chord length at rudder inboard station is given by:

cv O

 crv 1  Or 1  v 

Eqn.(7)

The X-location of the center of pressure of the vertical tail area affected by rudder deflection measured from the reference line is calculated from:





X acr  X apexv  Z acr  Z apexv tan  LEv

Segment: Cl r

rad  1

Segment: Cl

rad  1

r

Eqn.(8)

1 2 3 4 5 6 7 n/a 0.0244 0.0203 0.0184 0.0199 0.0310 0.0355

8 9 10 11 12 13 14 0.0154 0.0211 0.0167 n/a 0.0166 0.0148 0.0199

15 16 17 0.0194 0.0038 0.0194 0.0000 0.0000 0.0000

C l , Rudder 

  

0.0000 0.0000 0.0000 n/a 0.0000 0.0000 0.0000

C l , Rudder Segment: Cl



n/a 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

C l , Rudder

r



2c v ir 1   r  2r  c r 1   31   r   cv

Yawing Moment Coefficient due to Rudder Deflection Derivative

The airplane yawing-moment-coefficient-due-to-rudder-deflection derivative, also called the rudder control power, is given by: C n  C y r

l r cos   z r sin  bw

Eqn.(1)

Where:

l r  X acr  X cg And z r  Z acr  Z cg

Eqn.(2) Eqn.(3)

The airplane sideforce-coefficient-due-to-rudder-deflection derivative is estimated from: Sv  Sw r The change in sideslip due to rudder deflection is given by: C y   v C L r



Eqn.(4)

  k     C L    c Eqn.(5)   l  Theory  r  c c  l Theory   l ,v    C l  The correction factor accounting for nonlinearities at high rudder deflection angles is found from Figure 8.13 in Airplane Design Part VI and is a function of the rudder to vertical tail chord ratio and the rudder deflection angle: c  Eqn.(6) k   f  r ,  r   cv 

  Kb 

cl

 

The rudder span factor is obtained from Figure 8.51 with data from Figure 8.52 in Airplane Design Part VI and is a function of the rudder inboard and outboard stations and vertical tail taper ratio:

K b  K bO  K bi Where: K bO  f Or , v

  f 

 , 

Eqn.(7) Eqn.(8)

Eqn.(9) K bi Or v The correction factor for sectional rudder lift is obtained from Figure 8.15 in Airplane Design Part VI and is a function of the rudder chord to vertical tail chord ratio and the sectional lift curve slope to theoretical lift curve slope ratio:   cl  c a clW , M  0   f ,  cl  c w cl  Theory Theory  

The theoretical vertical tail sectional lift curve slope at zero-Mach is given by:

C



lV M  0 Theory

t  2  5.0525   c v

Eqn.(10)

c  l

theory

c t    f  r ,     cv  c  v 

The “three dimensional rudder effectiveness parameter” is determined from Figure 8.53 in Airplane Design Part VI and is a function of the vertical tail aspect ratio, and rudder chord to vertical tail chord ratio:

  CL   Cl

Cr   f  ARv , eff C v 

  

The airplane sideforce-coefficient-due-to-rudder-deflection derivative is estimated from: C y   ,v wfh C y r  2  C y v eff 

  Sv   v C L v Sw r  

Eqn.(11)

The wing-fuselage-horizontal tail interference on the airplane sideforce-coefficient-due-tosideslip derivative of twin vertical tails is obtained from Figure 10.17 in Airplane Design Part VI and is a function of the vertical tail span, the fuselage depth at the quarter chord point of the vertical tail, the distance between the two vertical tails, and the fuselage length:

C y ,v C y v

wfh



 f bv , h f v , y vtwin , L f



eff

Segment: Cn r

rad  1

n/a 0.0000

C n , Rudder Segment: Cn r

rad  1

r

C n , Rudder

rad  1

0.0000

8 9 10 11 12 13 14 -0.0907 -0.0882 -0.0895 n/a -0.0890 -0.0909 -0.0904 0.0000

C n , Rudder Segment: Cn

1 2 3 4 5 6 7 n/a -0.0857 -0.0885 -0.0890 -0.0938 -0.1174 -0.1552

0.0000

n/a 0.0000

0.0000

15 16 17 -0.0900 -0.0892 -0.0884 0.0000

0.0000

-Hingemoment Derivatives Hingemoment derivatives are used for two purposes: 1. Computing stick, wheel and pedal cockpit control forces so they can be checked against airworthiness requirements. 2. Computing actuator force levels so that hydraulic or electro-mechanical actuators can be properly sized. The method applies only in the linear range of control surface deflections (<20 degrees at best) and in the linear range of angles of attack (roughly 12 degrees). This method is applicable only to plain flap type control surface. The control surface hingemoment along the control surface hinge line is expressed with the help of coefficients:

HM  C hC . S  C h l . S  l .S  C h C . S  c.s.  C h c . s .t .  c.s.t .

Eqn.(1)

The control surface area is calculated from: S C .S 



 

bc.s. c f iC . S  cbiC . S .  c f OC . S  cbO ,C . S 2



Eqn.(2)

The control surface span is given by:





bC .S .   OC . S .   iC . S ARl .S S l .S Eqn.(3) Since the area and span of the aileron are for ONE PANEL ONLY, the aileron span is computed from:



ba   Oa   ia



ARl .S S l .S

Eqn.(4)

The control surface mean geometric chord is calculated from:





1  c.s  c2.s 2 c f iC . S  cbic . s 3 1  c.s The control surface taper ratio is given by: cC .S 

C .S 

C fO c fi

c. s

 cbO

c.s

 cbi

c. s

Eqn.(5)

Eqn.(6)

c.s

The control-surface-hingemoment-coefficient-due-to-lifting-surface-angle-of-attack Derivative is given by:

C h

l .s

 cos c

4 l .s.

C



h uncorr



 C h



horn



 C h



PartSpan



Eqn.(7)

Depending on the application, the following substitutions should be made: For a wing,

l.s. = w,

c.s. = a (aileron) or (flap);

For a horizontal tail,

l.s. = h,

c.s. = e (elevator);

For a vertical tail,

l.s. = v,

c.s. = r (rudder);

C h   C h

l .S

The uncorrected control-surface-hingemoment-coefficient-due-to-lifting-surface-angle-ofattack derivative is determined by integrating the sectional coefficient over the span of the control surface using the methods outlined in NACA Technical Note 925 (Hildenbran.F.E , Least square procedure for the solution to lifting line integral ,NACA technical note 925, 1944) and NACA Technical Note 1175 (Swanson R.S & Crandall, Lifting surface theory and aspect ratio correction to lift and hinge moment parameters for full span elevator and horizontal tail, NACA Technical note 1175, 1947):

C 

h uncorr



 O

O c C .S .

2 f C .S

  

C h  1  F1 K 1  c 2f

C .S

 d

Eqn.(8)

The control surface mean geometric chord aft of hinge is found from: 2 2 1   fC .S   fC .S c fC .S  c f Eqn.(1) 3 1   fC .S The taper ratio of the control surface aft of hinge is given by:

f

C.S



C f O ,C . S

Eqn.(2)

C fi ,C . S

The control surface chord aft of hinge at h station is given by:

C f C . S    c f i

 



    iC . S 1  1   fC .S  C .S  OC . S   iC . S 



  



Eqn.(3)

The induced angle of incidence correction to the hingemoment derivatives is obtained from NACA Technical Note 1175 (Swanson R.S & Crandall Lifting surface theoryaspect ratio correction to lift and hinge moment parameters for full span elevator and horizontal tail, NACA Technical note 1175, 1947) as a function of the lifting surface aspect ratio and the lifting surface sectional lift curve slope at the lifting surface mean geometric chord:



F1  f ARl .s , C l l . S



The lifting surface sectional lift curve slope at the lifting surface mean geometric chord is given by:

C l

l .s



C l

c 

Eqn.(4)

l theory

The ratio of the sectional lift curve slope to the theoretical sectional lift curve slope of the lifting surface at the lifting surface mean geometric chord is obtained from Figure 10.64a of Airplane Design Part VI as a function of local sectional trailing edge angle, local Reynolds number and the location of the flow transition. cl

c 

l Theory

 x    l . s Re,  lam    f   TE  c  l .s  

The trailing edge angle of the lifting surface section at the lifting surface mean geometric chord is given by: 1  2l .s.  l . S   TE  r l . S   TE  r l . S   TE t  TE Eqn.(5) l .S 31  l .s  The Reynolds number at the lifting surface mean geometric chord is computed from:



Re 



1.689U 1cl .s

Eqn.(6)



The theoretical lifting surface sectional lift curve slope at the lifting surface mean geometric chord is computed from:

c 





t  L.S Eqn.(7)  2  4.7  1  0.000375 TE  c  l .s The thickness ratio of the lifting surface airfoil at the lifting surface mean geometric chord can be computed from: l Theory

t     c  l .s.

 1  2l .s  t  t t         l .s   c  rl . s 31  l .s   c  rl . S  c  tl . s   1  2l .s  1  l .s  1     3 1   l .s  

Eqn.(8)

The ratio of induced angle of incidence at any section to induced angle of incidence of equivalent elliptic wing, from lifting line theory is obtained from NACA Technical Note 925 (Hildenbran.F.E , Least square procedure for the solution to lifting line integral ,NACA technical note 925, 1944) and NACA Technical Note 1175 (Swanson R.S & Crandall Lifting surface theoryaspect ratio correction to lift and hinge moment parameters for full span elevator and horizontal tail, NACA Technical note 1175, 1947) as a function of the span wise station at which the sectional hingemoment coefficient is to be evaluated, and the lifting surface taper ratio: K 1  f  , l .s , Shape  The change in control-surface-hingemoment-coefficient-due-to-lifting-surface-angle-ofattack-derivative due to lifting surface sweep and control surface partial span is given by:

C  h

PartSpan

 F2 F3 K  cl ,l . S

Eqn.(9)

The computation of the lift curve slope of the lifting surface section at the lifting surface mean geometric chord has been shown above.

The streamline curvature correction factor is obtained from NACA Technical Note 1175 (Swanson R.S & Crandall Lifting surface theory for aspect ratio correction to lift and hinge moment parameters for full span elevator and horizontal tail, NACA Technical note 1175, 1947) as a function of lifting surface aspect ratio and the section lift curve slope at the lifting surface mean geometric chord:



F2  f ARl .S , cl

l .s



The nose balance correction factor is obtained from NACA Technical Note 1175 (Swanson R.S & Crandall Lifting surface theoryaspect ratio correction to lift and hinge moment parameters for full span elevator and horizontal tail, NACA Technical note 1175, 1947) as a function of chord ratios:

 c f F3  f  c . s  cl .s.

cbc . s   ,   @ cfc.s c fc.s

  

The ratio between the control surface chord aft of hinge line and the lifting surface chord at the mean geometric chord station of the control surface aft of hinge line is computed from:

 c fc.s  c  l .s

c fC .S      @ c f ,C . S c 1     iC . S   OC . S . 1  2l .S  1    rl . s .  iC . S l .S  31  l .S   





Eqn.(10)

The ratio between the average control surface chord forward of hinge line and the average control surface chord aft of hinge line is given by: cb cf

c. s



c. s .

cbi c fi

 cbO

C .S

 c fO

C .S

Eqn.(11)

c.s

C .S

The effect of the control surface span is calculated from:

K 



K i 1   i ,C .S   K  ,O 1   OC . S



Eqn.(12)

O

  iC . S C .S . The inboard and outboard station control surface station factors are found from Figure 10.77b in Airplane Design Part VI (J. Roskam) and are functions of the inboard and outboard control surface stations, respectively: K  ,i  f  i ,C .S  and



K O  f  OC . S



The sectional hingemoment-coefficient-due-to-lifting-surface-angle-of-attack derivative at h station is computed from: C h   

c 

h bal

1  M 12

c h

c



h uncorr



 c h



TEShape



Eqn.(13)

The change in the control-surface-sectional-hingemoment-coefficient-due-to-lifting-surfaceangle-of-attack derivative at h station due to trailing edge shape can be computed from:

c 

h TEShape

 1  cl Theory  cl Theory 

 

 2 cl

 

  l .s  t    tan  TE      2  c  l .s  

Eqn.(14)

The theoretical sectional lift curve slope coefficient at h station is given by:

c 

l Theory

t  l .S   2  4.7  1  0.000375 TE  c  l .s

Eqn.(15)

The lifting surface trailing edge angle at h station is given by:



 l . S   TE  l . S    TE r  TE

l .s

t   TE

l .s



Eqn.(16)

The correction factor account for the control surface nose shape and the amount of balance at h station is found from Figure 10.65a in Airplane Design Part VI and is a function of the control surface balance ratio and control surface section nose shape:

c 

h bal

c h

 f Balance Ratio, Nose Shape 

That is, For round, low drag nose,

c 

h bal

c h

 1.32 

1.16 Balance Ratio 0.45

Eqn.(17)

The control surface balance ratio at h station is defined as: c Balance Ratio   b c  f

    C .S

  t     2c   f

Eqn.(18)

    i ,C . S   cbi ,C . S  cbi ,C . S  cbO ,C . S       cb  i ,C . S   O ,C . S    Eqn.(19) c      i ,C . S   f  C .S c   f i ,C . S  c f i , C . S  c f O , C . S      i ,C . S   O ,C . S The ratio of half of the control surface sectional thickness to control surface chord aft of the hinge line at h station is given by:

 t   2c f 

 t  c 1 f

     C .S 2









c fO ,C . S   t         t      c f     iC . S  c f  OC . S c f i ,C . S     i ,C . S   c f O , C . S   1  1     O   i  c f  C .S  i ,C . S   C .S

    i ,C . S       iC . S   OC . S   iC . S

   

Eqn.(20)

Elevator Hinge Moment:

Segment: Balancee C h

C h e

rad  rad 

C h e

3 2.33

4 2.33

5 2.33

6 2.33

7 2.33

256.0378 272.2689 273.9063 273.3695 271.0561 274.4819 272.1009

1

178.0539 201.0558 203.5493 202.8037 200.7602 206.4657 203.6400

rad  rad 

8 2.33

9 2.33

10 2.33

11 2.33

12 2.33

13 2.33

14 2.33

1

273.3620 272.8631 272.2121 256.0378 272.8631 273.5225 273.6448

1

203.2496 201.9460 200.9761 178.0539 201.9460 203.0163 203.3040

Segment: Balancee C h

2 2.33

1

Segment: Balancee C h

1 2.33

C h e

rad  rad 

15 2.33

16 2.33

17 2.33

1

273.5225 270.6716 270.6187

1

203.0163 198.7402 198.6658

Aileron Hinge Moment: Segment: Balancea C h

C h a

rad  rad 

C h a

2 0.11

3 0.11

4 0.11

5 0.11

6 0.11

7 0.11

1

-0.0625 -0.0744 -0.0791 -0.0791 -0.1031 -0.1234 -0.1329

1

-0.2073 -0.2265 -0.2370 -0.2372 -0.2988 -0.3459 -0.3721

Segment: Balancea C h

1 0.11

rad  rad 

8 0.11

9 0.11

10 0.11

11 0.11

12 0.11

13 0.11

14 0.11

1

-0.0886 -0.0758 -0.0744 -0.0625 -0.0758 -0.0791 -0.0816

1

-0.2608 -0.2295 -0.2264 -0.2073 -0.2295 -0.2372 -0.2432

Segment: Balancea C h

rad  rad 

C h a

15 0.11

16 0.11

17 0.11

1

-0.0791 -0.0723 -0.0723

1

-0.2372 -0.2221 -0.2221

Rudder Hinge Moment: Segment: Balancer

rad  rad  1

Ch C h

1

rad  rad  1

Ch

C h

1



5 0.02 0.1335

6 0.02 0.1651

7 0.02 0.1751

n/a

-0.3990 -0.4186 -0.4210 -0.5233 -0.6103 -0.6543

8 0.02 0.1161

9 0.02 0.0983

10 0.02 0.0990

11 12 0.02 0.02 n/a 0.1007

13 0.0.2 0.1032

14 0.02 0.1067

rad  rad  1

-0.4579 -0.4019 -0.3998 n/a

15 0.02 0.1032

16 0.02 0.0917

-0.4048 -0.4160 -0.4268

17 0.02 0.0916

1

4 0.02 0.1074

Segment: Balancer Ch

3 0.02 0.1060

Segment: Balancer

C h

1 2 0.02 0.02 n/a 0.0983

-0.4160 -0.3873 -0.3873

Trim satisfaction:

Theory: The elevator deflection angle for the trimmed lift condition is found by iterating the following equation:

e 

C Lh

Eqn.(1)

f bale C Li , h K 

Where the correction for nonlinear lift behavior of plain flaps is found from Figure 8.13 in Airplane Design Part VI and is a function of elevator deflection angle and the average elevator chord to horizontal tail chord ratio aft of hinge line.  c K   f   e , e  ch

  

The horizontal tail lift coefficient is computed from:

C Lh  

C

 C1C L , wf  C 3 C 4 

Eqn.(2)

C 2  C3C5

The wing-fuselage lift coefficient with flap effects is given by:  Sh   h  S w  C L1  C m  C 3 C 4   C 2  C3C5 Eqn.(3) C Lwf   Sh   C1  h S w   1 C 2  C3C5 The following coefficients are used in the equations shown above.

 X cg  X acwf , P.OFF C m  C mO  C mO , n  C mO , py  C LO , n  C LO , py  wf  cw     X ac , wf , P.Off  X C  4 w  i   C Lwf , Power   cw    



The first coefficient is given from:

C1 

X cg  X ac , wf , P.OFF

cw The second coefficient is equivalent to:



   Cm  C mN , prop  C mO , power T , prop  

Eqn.(4)

Eqn.(5)

X cg  X ach

Sh cw Sw The third coefficient is found from: C2 

h

Eqn.(6)

For chosen configuration:

C3  0 The fourth coefficient is given by:

C 4  C LC @ C Lh  0 The fifth coefficient is estimated from:

C5  1 The change in airplane lift coefficient due to flap deflection is given by: C Lw  C Lwf  C Lwf ,Clean

Eqn.(7)

The wing-fuselage lift coefficient without flap effects but including power effects is computed from:

C L , wf  C L

wf , P . OFF

 C Lwf , Power

Eqn.(8)

The wing lift coefficient without flap effects or power effects is given by:



C Lwf , P.OFF  C L ,Wf    w0 ,Clean  i w



Eqn.(9)

The airplane angle of attack can be found from:

   0, wf 

C L , wf  C Lwf , Power C L , wf

Eqn.(10)

The change in wing lift coefficient due to flap deflection is given by: C Lf  C Lw  C Lw ,Clean

Eqn.(11)

The wing lift coefficient without flap effects but including power effects is given by:





C Lw  C Lw ,    w0 ,Clean  i w  C Lw , power

Eqn.(12)

The increment of wing lift coefficient due to power is determined from:

C Lw , Power 

C L , wf , power

Eqn.(13)

K wf The wing lift coefficient without flap effects but including power effects is given by: C Lwclean  C Lw , c ln, P .OFF  C Lw , Power

Eqn.(14)

The wing lift coefficient without flap effects and without power effects is found from: C Lw ,Clean , P .OFF  C Lw , ,Clean   i w 

Eqn.(15)

The horizontal tail downwash angle is calculated from:  d h    hg   d  P.Off

 h   h ,O    Segment: C Lh

e

trim

(deg) -1876.2

Segment: C Lh

e

trim

8 (N/A) 17.3023

(deg) -1876.2

Segment: C Lh

e

1 (N/A) 17.3023

15 (N/A) 0.1082

(deg) 8.62

Eqn.(16)

2 -0.1014

3 -0.0188

4 -0.0156

5 0.1082

6 0.0318

7 0.0560

11.38

9.54

6.04

8.62

12.11

4.30

9 -0.1014

10 -0.0188

11 -0.0156

12 0.1082

13 0.0318

14 0.0560

11.38

9.54

6.04

8.62

12.11

4.30

16 0.0318

17 0.0560

12.11

4.30

Trim diagrams: The trim diagram, sometimes called the "trim triangle", describes the relationships between the airplane lift coefficient and the airplane pitching moment. It is a graphical solution based on the following equations. The trim diagram is comprised of a lift coefficient vs. angle of attack graph and a lift coefficient vs. pitching moment coefficient graph. The "trim triangle" is defined as the triangular area bound between the forward and aft center of gravity lines and by the maximum airplane angle of attack line. The trim diagram is useful in determining: 1) Whether or not an airplane can be trimmed at any center of gravity location with reasonable surface deflections at different flight conditions. 2) Whether or not tail stall is a limiting factor in trim. 3) The control surface deflection and lift coefficient at different angles of attack and center of gravity locations. The equations used in the construction of the trim diagram are as follows: The lift coefficient equation is used to plot the lift coefficient -vs- alpha curves: C L  C L0  C L   C Li ih  C L K  e h

Eqn.(1)

The pitching moment coefficient equation: C m  C m0  C m   C mi ih  C m e K  el  C mT 1 h

Eqn.(2)

The correction factor for non-linear behavior of plain flaps is defined as:  c K   f   e , e  cw

  

The steady state thrust pitching moment coefficient for a jet airplane is found from:

C mT 1 

 TavailTrim d T

q1 S w c w The available trimmed thrust is found from: TavailTrim 

cos  T q1 S w C D  W sin   cos   T 

Eqn.(3)

Eqn.(4)

The airplane drag coefficient at the design point using the Class II Drag Polar can be found from the following equation: C D  C D 0  BC D1 C L  BC D 2 C L2  BC D 3 C L3  BC D 4 C L4  BC D 5 C L5

Eqn.(5)

The perpendicular distance from the thrust line to the airplane center of gravity is found from: dT  Z T  Z cg cos  T  X T  X cg sin  T Eqn.(6) The trimmed aircraft power setting is calculated from:

SHPsettrim 

TavailTrim U 1

5501  K Loss  prop

Eqn.(7)

The trimmed aircraft installed power is determined from:

SHPavailtrim 

TavailTrim U 1

550

Eqn.(8)

The trim point for the current flight condition is located using the following:

C m  0.0 CL 

W cos   Tavailtrim sin    T  q1 S w

Eqn.(9)

Where the thrust is defined above. The stability surface stall lines are constructed using the previous equations and: The angle of attack of the stabilizer is found from:



 h    ih    O  

d h   d 

Eqn.(10)

T.O Rotation performance: The horizontal tail area required for take-off rotation with the given elevator geometry is calculated from:

  zT cos   T   x mg sin    T   xT sin    T    M acwf , g  T     z cg cos   T   z cg  g sin    T   z mg  g sin    T        D g z D  z cg   W xcg  x mg   g z cg   g z mg        Lw f , g x mg  x acwf   g z cg   g z mg  I yymg      Sh   xmg   g z cg   g z mg  C Lh, g  hg qrotation









Eqn.(1)

The equation below has been developed to make easy use of the equation above by braking it down into individual terms which act on the airplane during take-off rotation.

Sh 

 M acwfg  AA  BB  CC  I yymg  mg

Eqn.(2)

The wing-fuselage pitching moment term: M acwf  c m0 q rotation S w c w

Eqn.(3)

The AA Term:





 Z cg  Z T cos   T   X gearaft  X T cos   T     AA  Tset   X gear  X T  M uz sin    T   aft  





Eqn.(4)

The BB Term:



BB  D g Z D  Z cg   W X cg  X gearaft  M uz



Eqn.(5)

Where:

D g  C D q rotation S w

Eqn.(6)

The CC Term:



CC  Lwf g x gearaft  x acwf  M uz



Eqn.(7)

Where:



Lwf g  C Lwf , g X gearaft  X acwf  M uz



Eqn.(8)

And C Lwf , g  C LO , wf  C L , w. f   C Lwf , g

Eqn.(9)

The Airplane Moment of Inertia in the Y axis about the main gear.:

I yymg  I yy B 



Wcurrent z cg  Z gearaft g

  X 2

gearaft

 X cg

 2

Eqn.(10)

The EE Term:



EE  C Lh , g q rotation h X ach  X gearaft  M uz



Eqn.(11)

Where: C Lh , g  C Lh  C Lh , g

Eqn.(12)

And C Lh , g  C L , h  h , g

Eqn.(13)

The ground friction term is found from:



M uz   G Z cg  Z gearaft



Eqn.(14)

Following gear geometry have been selected considering ground – strike, tip over angle and static and dynamic structural considerations: Nose Gear X (ft) n/a Y (ft) n/a Z (ft) n/a

Main Gear 1 n/a n/a n/a

Main gear 2 n/a n/a n/a

Lateral Tip-Over The lateral tip-over angle can be computed from:

h   tan 1    y 

Eqn.(1)

The first intermediate parameter, h, is given by:





h  Z cg  Z gear ,dft  X gear ,aft  X cg  tan i cos i

Eqn.(2)

Where the intermediate calculation parameter, i, can be found by:  Z gear forw  Z gearaft i  tan 1   X gear  X gear aft forw 

   

Eqn.(3)

For tricycle configuration The second intermediate parameter, y', is given by:





  X gear ,aft  X gear , forw  Ygear forw  Ycg  cos i  y    Ygearaft  Ygear forw  

      l  sin    

Eqn.(4)

The intermediate parameter, l, is given by:

l

X cg  X gear , forw

cos i

 h tan i

Eqn.(5)

The intermediate parameter, a, is given by:

 Ygear  Ygear , forw   aft  cos i   X gearaft  X gear forw  

  tan 1 

Eqn.(6)

 Take off 61.3 Landing 62.4 Required horizontal tail surface area to initiate take off rotation:

 

S hreq ft 2

26.62

The designed surface area is suitable for initiating the take off rotation.

Stability derivatives table: 1 1.84

2 7.14

3 2.76

4 1.15

5 2.51

6 0.74

7 5.92

C Tx , 1

22.8377

0.5286

0.1222

0.0408

0.1714

0.0390

0.1430

C mT

0.4421

0.0073

0.0043

0.0007

0.0005

0.0013

0.0031

C Du

-0.0191

0.0033

0.0000

0.0587

0.0000

-0.1412

C Lu

0.0163

0.0202

0.1155

0.0272

0.0164

0.0140

C mu

0.0057

0.0056

0.0069

0.0372

0.0094

0.0056

0.0048

CTx ,U

-45.949

-1.1050

-0.2675

-0.0850

-0.3765

-0.0954

-0.3154

C mTU

-0.8860

-0.0151

-0.0093

-0.0014

-0.0010

-0.0031

-0.0068

36.5178

0.5660

0.1886

0.1233

0.2345

0.0704

0.4847

5.4200

5.4446

5.5120

6.1966

5.5198

5.6200

5.4447

1

-1.8645

-1.8954

-1.8688

-1.9578

-2.2150

-2.2402

-2.3107

1

0.0000

1

0.0000

1

2.2839

2.3090

2.3592

2.4254

1.854

2.5135

2.3799

1

-5.6624

-5.7372

-5.8491

-7.3922

-6.1652

-6.3958

-6.0949

1

0.0000

1

10.1408

10.2497

10.3070

11.7219

11.0742

11.3037

11.1528

 h (deg)

C D C L C m C mT ,

C D C L C m C Dq C Lq C mq C y C l C n C nT , 

C y C l C n C yP ClP C nP

rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad rad

1 1

                   

-15.7040 -15.8572 -15.9283

-17.8050 -17.0709 -17.3843 -17.2271

1

-0.7285

-0.7287

-0.7329

-0.7722

-0.7334

-0.7395

-0.7252

1

-3.1489

-0.0882

-0.0852

-0.0905

-0.0827

-0.0803

-0.0936

1

0.0932

0.0729

0.0628

0.0765

0.0739

0.0644

0.0780

1

0.0000

1

0.0190

0.0144

0.0213

0.0245

0.0206

0.0221

0.0186

1

0.0014

0.0008

0.0021

0.0025

0.0017

0.0023

0.0013

1

0.0053

0.0041

0.0058

0.0066

0.0057

0.0059

0.0052

-0.0482

-0.0271

0.0058

-0.0869

-0.0592

-0.0839

-0.0431

1

-0.6036

-0.5082

-0.49440

-0.5425

-0.4963

-0.4999

-0.5015

1

-0.0886

-0.1352

-0.0298

-0.0256

-0.0674

-0.0161

-0.998

1

C yr C lr C nr C Di , h

C Lih C mih C D e C L ,eo

C L e C m e C h , h

C h e C yi

C li

C ni

C y r C y

C l

C n r C n

C y a C l a C n a C h

C h a

0.3048

0.3130

0.3043

0.3285

0.3170

0.3145

0.3195

1

0.200

0.2840

0.0969

0.1075

0.1661

0.0736

0.2188

1

-0.1005

-0.1092

-0.0924

-0.0995

-0.1024

-0.0966

-0.1084

1

0.0478

0.0480

0.0409

0.0316

0.0257

0.0266

0.0480

1

1.1317

1.1358

1.1468

1.2535

1.1481

1.1644

1.1358

1

-2.8050

-2.8220

-2.8433

-3.1244

-2.9183

-2.9628

-2.9087

1

0.0169

0.0196

0.0166

0.0125

0.0104

0.0106

0.0196

1

0.3371

0.3893

0.3902

0.4177

0.3902

0.3910

0.3893

1

0.1601

0.3893

0.3902

0.4177

0.3902

0.3910

0.3893

1

-0.3968

-0.9673

-1.0410

-0.9919

-0.9950

-0.9970

1

-0.0811

-0.1138

-0.1197

-0.1600

-0.1189

-0.1297

-0.1137

1

-0.3646

-0.4251

-0.4356

-0.5385

-0.4361

-0.4521

-0.4251

1

-0.4432

-0.4435

-0.4471

-0.4814

-0.4475

-0.4529

-0.4416

1

-0.0278

-0.0188

-0.0391

-0.0445

-0.0334

-0.0426

-0.0245

1

0.1347

0.1367

0.1329

0.1434

0.1384

0.1373

0.1384

1

0.1278

0.1277

0.1283

0.1419

0.1284

0.1292

0.1277

1

0.1278

0.1277

0.1283

0.1419

0.1284

0.1292

0.1277

1

0.0120

0.0086

0.0162

0.0187

0.0140

0.0174

0.0111

1

0.0120

0.0086

0.0162

0.0187

0.0140

0.0174

0.0111

1

-0.0497

-0.0505

-0.0487

-0.0539

-0.0505

-0.0497

-0.0511

1

-0.0497

-0.505

-0.0487

-0.0539

-0.0505

-0.0497

-0.0511

1

0.0213

0.0207

0.0219

0.0418

0.0225

0.0243

0.0207

1

-0.2501

-0.2700

-0.2772

-0.3436

-0.2773

-0.2878

-0.2700

1

0.0000

1

0.1474

0.0819

0.1563

0.2445

0.1571

0.1674

0.1498

1

-0.0123

-0.0088

-0.0049

-0.0073

-0.0101

-0.0032

-0.0138

1

-0.0807

-0.0793

-0.0807

-0.1025

-0.0812

-0.0834

-0.2884

1

-0.2660

-0.2884

-0.2961

-0.3588

-0.2957

-0.3066

-0.2884

C h ,V C h

1

C l r

rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad  rad 

Longitudinal Dynamic Stability and flight qualities: -Longitudinal dimensional stability derivatives:



Along X Axis:

Theory: The longitudinal dimensional stability derivatives about the X-axis are calculated as follows:

Xu 





 q1 S w C Du  2C D1 W  g

Eqn. (1)

 U 

The forward acceleration imparted to the airplane due to thrust as a result of a unit change in speed:

X TU 



q1 S w CTx ,U  2CTx



Eqn. (2)

W   U g The forward acceleration imparted to the airplane as a result of a unit change in angle of attack: X 



 q1 S w C D  C L:



Eqn. (3)

W    g The forward acceleration imparted to the airplane as a result of a unit change in control surface deflection angle: X

C .S





 q S w C D ,C . S W  g

Eqn. (4)

  

Along Y Axis:

The longitudinal dimensional stability derivatives about the Y-axis are calculated as follows: The pitch angular acceleration imparted to the airplane as a result of a unit change in speed: MU 



q S w c w C mU  2C m1 I yyB U



Eqn. (1)

The pitch angular acceleration imparted to the airplane due to thrust as a result of a unit change in speed:

M TU 



qS w c w C mT ,U  2C mT



Eqn. (2)

I yy B U The pitch angular acceleration imparted to the airplane as a result of a unit change in angle of attack: M 

q S w c w C m I yy B

Eqn. (3)

The pitch angular acceleration imparted to the airplane due to thrust as a result of a unit change in angle of attack: M T 

q S w c w C mT I yyB

Eqn. (4)

The pitch angular acceleration imparted to the airplane as a result of a unit change in angle of attack rate:

M  

q S w c w2 C m

2 I yyB U 1

Eqn. (5)

The pitch angular acceleration imparted to the airplane as a result of a unit change in pitch rate: Mq 

qS w c w2 C mq 2 I yy B U

Eqn. (6)

The pitch angular acceleration imparted to the airplane as a result of a unit change in control surface deflection: qS w c w2 Mq  2 I yy B U

Eqn. (7)

The pitch angular acceleration imparted to the airplane as a result of a unit change in control surface deflection:

M C.S . 

q S w c w C mC . S

Eqn. (8)

I yyB

 Along Z-Aixs The longitudinal dimensional stability derivatives about the Z-axis are calculated as follows: Zu 



 q S w C LU  2C L



Eqn. (1)

W   U g The vertical acceleration imparted to the airplane as a result of a unit change in angle of attack: Z 



 qS w C L  C D



Eqn. (2)

W    g The vertical acceleration imparted to the airplane as a result of a unit change in angle of attack rate:

Z  

 q S w c w C Lq W 2 g

 U 

Eqn. (3)

The vertical acceleration imparted to the airplane as a result of a unit change of pitch rate:

Zq 

 q S w c w C Lq

Eqn. (4)

W  2 U g The vertical acceleration imparted to the airplane as a result of a unit change in control surface deflection angle:  q C L , C . S Z C .S  Eqn. (5) W    g -Longitudinal Transfer Functions 

Longitudinal Transfer Functions

The speed to control surface transfer function is defined as:

   NU  C.S S  D1

Eqn. (1)

U S

Where: N  A S3  B S2  C S  D U U U U U

Eqn. (2)

Where:

(

A =X U Z U δC .S 1 α

{( {

)

(

) )(

B = X U Z M +Z + M α U1 +Z q +Zδ X U δC .S 1 α q α C .S α C

δC .S

M q Zα + M α g sin Θ1

(

)}

}

M α + M T U1 +Z q + α

{ Mα g cos Θ1 X α M q }+ MδC .S {X α M q }+ M X (U +Z ) (U1 Zα )g cos Θ1} δC .S { α 1 q

Z δC .S

(

)

D =X M + M T g sin Θ Z M g cos Θ + M U δC .S α δC .S 1 δC .S α 1 α

(Zα g cos Θ1

X α g sin Θ1

)

The pitch attitude to control surface transfer function is defined as: Θ(s ) δC .S (s )

NΘ

Eqn. (3)

Where:

= A S 2 + B S +C Θ Θ Θ

Eqn. (4)

Where:

(

)

A =Z M +M U Z δC .S α δC .S 1 α Θ

{

)(

(

)}

{(

B =X Z M + U1 Z α M + M T Z X +XT +M δC .S U α δC .S α U Θ U U

[(

)

C Θ = X δC . S M α + M Tα Z u

[ (

+ M δC . S Zα X u + X Tu

)

(

)]

Zα M u + M Tu + ZδC . S

X α Zu

]

[ (M + M )(X α

Tα

)

(

)

X α ZU

+ X Tu + X α M u + M Tu

)]

}

The angle of attack to control surface transfer function is defined as:

   N  C .S S  D1  s

Eqn. (4)

Where: N



 A S3  B S2  C S  D



Eqn. (5)



Where: A =Z α δC .S

B =X Z +Z α δC .S U δC .S

{

)}

(

)(

{(

C =X U +Z M U + M T α δC .S 1 q U

(

)

(

+M U +Z XU + X T δC .S 1 q U

)

M q ZU

(

}

)

Dα = X δC . S M u + M Tu g sin Θ1 + ZδC . S MU + M TU g cos Θ1 + M δ ,C .S

(

)

{(

)

+ X TU g sin Θ1

Z u g cos Θ1

X U + X T gSinΘ1 ZU gCosΘ1 Z M + M T gCosΘ + M 1 δC .S δC .S U U U

}

The characteristic equation is defined as:

D1  A1S 4  B1S 2  D1S  E1

Eqn. (6)

Where: A U Z  1 1 







B1  U 1  Z  X U  X T  M q  Z  M U 1  Z q U



C 1  XU  X T U

M q U1 Z Z  M U1 Zq  

 M q Z  ZU X   M g sin 1  M  M T 





D1  g sin 1 M  M T  M XU  X T  U







g cos  ZU M  MU  M T U

U1 Zq 



U1 Z MU  MTU 

 X U1 Zq  ZU X M q  XU  X TU  The longitudinal stability is determined by the characteristic equation. This equation is set to zero, and the roots are used to determine the longitudinal stability characteristics of the airplane. For an airplane with 2 complex pairs of roots, they are cast in the form:

s1, 2   1, 21, 2  j1, 2 1  1, 2

Eqn. (7)

The complex pair of roots with the highest frequency is referred to as the short period mode, and the lower frequency pair as the phugoid mode. For airplanes with one pair of complex roots and 2 real roots, a mode called the 'third' oscillation results from placing the complex pair in the same form as the short period and phugoid. The real roots are cast into time constants, which are defined as follows:

TC long  

1 s

Eqn. (8)

Dynamic Derivatives: Flight segment: W  lb  S  ft 2     (deg) X u s 1

1 25.50

2 25.00

3 24.64

4 24.60

5 21.83

5.0 -0.3937

8.9 -0.0540

-0.1 -0.0159

-0.6 -0.0264

-2.01 -0.0126

X TU s 1

-0.0065

-0.0093

-0.0074

-0.0008

-0.0085

 ft  X  2  s  Z u s 1

16.6506

16.7074

16.8008

14.7789

16.3435

-3.7175

-0.4713

-0.2603

-0.1351

-0.2541

 ft  Z  2  s   ft  Z     s  ft  Zq    s  1   M u   ft.s 

-5.5110

-146.8825

-450.1534

-863.5066

-0.2541

-0.1001

-0.8260

-1.3907

-1.2787

-1.1213

-0.4445

-3.6668

-6.0756

-5.0537

-5.1199

1.0162

-0.0009

0.0013

-0.0001

M  s 2

-1.2077

-11.3754

-34.6027

-62.2732

-28.5941

M T s 2

0.0000

M  s 2

-0.3989

-0.4681

-0.7852

-0.7306

-0.5770

M q s 2

-1.1062

-1.2937

-2.1383

-1.7598

-1.5978

Along

17.0

135.9

254.6

592.0

254.3

Blong

37.6

392.7

1194.4

2349.3

908.4

C long

99.1

1714.1

9544.8

38057.3

7681.6

Dlong

397.5

65.1

234.8

1071.2

203.4

Elong

369.5

162.5

275.2

303.0

231.5

RH long

 rad    s 

 n,S .P 

 SP n

P , long

 rad     s 

 P ,long

-1723525 18175322 2270637081 93423471587 1217726259 ---------

3.5320

6.1129

8.0105

5.4859

---------

0.407

0.382

0.246

0.323

2.7704

0.3096

0.1701

0.0893

0.1739

-0.238

0.026

0.062

0.55

0.066

TC long (1) (s)

0.433

--------

TC long ( 2 ) (s)

0.815

--------

TC long (3) (s)

--------

TC long ( 4 ) (s)

--------

 ft  X e  2  s   ft  Ze  2  s  M  e s 2

-0.0068

-0.5152

-1.3457

-1.7408

-0.6637

-0.0645

-10.2450

-31.7204

-57.9508

-24.8836

-0.2570

-5.8054

-17.9097

-33.1122

-12.8051

Flight segment: W  lb  S  ft 2     (deg) X u s 1

6 21.64

7 21.04

-1.3 -0.0185

5.8 0.0322

X TU s 1

-0.0105

-0.0067

 ft  X  2  s  Z u s 1

16.6139

16.7813

-0.1841

-0.4731

 ft  Z  2  s   ft  Z     s  ft  Zq    s  1   M u   ft.s 

-1259.1861

-167.8103

-2.7764

-0.9972

-12.4857

-4.6729

-0.0010

-0.1674

-- M  s 2 --

-104.1449

-117.2852

M T s 2

0.0000

M  s 2

-1.4698

-4.2056

M q s 2

-3.9951

-11.8871

Along

374.1

136.0

Blong

3292.0

2329.5

C long

42500.4

17217.9

Dlong

1317.6

-1029.7

Elong

574.3

875.9

RH long

177478360076 -46075057844 10.6457

11.2989

0.412

0.761

0.1164

0.2246

 P ,long

0.129

-0.147

TC long (1) (s)

--------

TC long ( 2 ) (s)

--------

TC long (3) (s)

--------

TC long ( 4 ) (s)

--------

 ft  X e  2  s   ft  Ze  2  s  M  e s 2

-2.3740

-0.6036

-87.3768

-11.9988

-46.2578

-50.6063

 rad    s 

 n,S .P 

 SP n

P , long

 rad     s 

Longitudinal flight qualities: Short Period and Long Period Frequency and Damping: MIL-F-8785C requires the equivalent short period undamped natural frequency of the short period mode to be within the following limits for the three Flight Phase Categories. Common design practice is to adopt the military requirements because the FAR/VLA requirements do not set specific limits on the undamped natural frequency. The short period undamped natural frequency and the steady state normal acceleration per unit of angle of attack are used to determine the flight phase level. Result has been plotted against the requirement to show the level of satisfactory. For Flight Phase Category C Requirements, the MIL-F-8785C Airplane Class must be known: Four airplane classes are defined in MIL-F-8785C. The classes are: Class I: Small, light airplanes Class II: Medium weight, low-to-medium maneuverability airplanes Class II-C Carrier Based Class II-L Land Based Class III: Large, heavy, low-to-medium maneuverability airplanes Class IV: High maneuverability airplanes It is considered good design practice to use the following military requirements for civilian airplanes. The FAR/VLA requirements only require the short period oscillation to be heavily damped. The equivalent short period damping ratio must be within the limits presented in the following table. Flight Phase

Category A and Category C

 SP

0.35 0.25 0.15

1.30 2.00 --

0.30 0.20 0.15

2.00 2.00 --

min

Level 1 Level 2 Level 3

Category B

max

min

max

100

Flying Quality Levels are defined in MIL-F-8785C, "Military Specification - Flying Qualities of Piloted Airplanes." Although the FAR/VLA requirements do not set specific flying quality levels, common design practice is to adopt the military definitions. Airplanes must be designed to satisfy the Level 1 flying quality requirements with all systems in their normal operating state. Level 1: Flying qualities are clearly adequate for the mission Flight Phase. Level 2: Flying qualities are adequate to accomplish the mission Flight Phase, but some increase in pilot workload or degradation in mission effectiveness, or both, exists. Level 3: Flying qualities such that the airplane can be controlled safely, but the pilot workload is excessive or mission effectiveness is inadequate, or both. Category A Flight Phases can be terminated safely, and Category B and Category C Flight Phases can be completed. The required levels of flying qualities are tied into the probability with which certain system failures can occur. For example, it is desired to have: At least Level 1 for airplane normal (no failure) state, At least Level 2 after failures that occur less than once per 100 flights, At least Level 3 after failures that occur less than once per 10,000 flights. Flying quality levels below Level 3 are not allowed except in special circumstances. Each airplane mission can be broken down into a number of sequential flight phases. The flight phase categories are defined in MIL-F-8785C. There are also suggested civilian equivalents to the military definitions. Longitudinal Flight qualities:

 g    rad  T2 P sec .

sec .

2 4.565 -----

3 4 5 6 7 14.013 27.134 11.048 39.149 11.3784 -----

-----

85.001 65.791 50.030 60.247 10.624

----83.529

Level P Level SP

II I

I I

I II

I I

II I

Level n , SP

III

101

Lateral Directional Dynamic stability and flying qualities: -Lateral-Directional Dimensional Stability Derivatives:



About X-axis:

The lateral-directional dimensional derivatives along the X-axis are calculated as follows: The roll angular acceleration imparted to the airplane as a result of a unit change in sideslip angle:

L 

q1 S w bw C l I xxS

Eqn. (1)

The roll angular acceleration imparted to the airplane as a result of a unit change in roll rate: LP 

q S w bw2 C lP 2 I xxS U

Eqn. (2)

The roll angular acceleration imparted to the airplane as a result of a unit change in yaw rate: Lr 

q S w bw2 C lr 2 I xxSU

Eqn. (3)

The roll angular acceleration imparted to the airplane as a result of a unit change in aileron angle:

L a 

q S w bw2 C l , a I xxS

Eqn. (4)

The roll angular acceleration imparted to the airplane as a result of a unit change in rudder angle:

L r 

q S w bw2 C l , r

Eqn. (5)

I xxS The roll angular acceleration is imparted to the airplane as a result of a unit change in spoileron angle: L spn 

q S w bw2 C l , spn I xxS

Eqn. (6)

102



About Y-Axis:

The lateral acceleration imparted to the airplane as a result of a unit change in sideslip angle:

Y 

q S w C y

W  g

Eqn. (7)

  

The lateral acceleration imparted to the airplane as a result of a unit change in roll rate:

YP 

q S wC yP

Yr 

q S w C yr

Eqn. (8)

W  2 U g The lateral acceleration imparted to the airplane as a result of a unit change in yaw rate:

W 2 g

Eqn. (9)

 U 

The lateral acceleration imparted to the airplane as a result of a unit change in aileron angle:

Y a 

q S w C y , a

Eqn. (10)

W    g The lateral acceleration imparted to the airplane as a result of a unit change in rudder angle: Y r 

q S w C y , r

Eqn. (11)

W    g The lateral acceleration imparted to the airplane as a result of a unit change in spoileron angle: Y spn 

q S w C y apn

W  g

  

Eqn. (12)

103



About Z-Axis:

The lateral-directional dimensional derivatives about the Z-axis are calculated as follows: The yaw angular acceleration imparted to the airplane as a result of a unit change in sideslip angle: N 

q S w bw C n  I zz , S

Eqn. (1)

The yaw angular acceleration imparted to the airplane as a result of a unit change in sideslip angle due to thrust:

NT  

q S w bw C nT I zz , S

Eqn. (2)

The yaw angular acceleration imparted to the airplane as a result of a unit change in roll rate:

NT  

q S w bw2 C nTP

2 I zz ,S U

Eqn. (3)

The yaw angular acceleration imparted to the airplane as a result of a unit change in yaw rate:

Nr 

q S w bw2 C nr

2 I zz , S U

Eqn. (4)

The yaw angular acceleration imparted to the airplane as a result of a unit change in aileron angle:

Na 

q S w bw2 C n , a

Eqn. (5)

I zz , S The yaw angular acceleration imparted to the airplane as a result of a unit change in rudder angle: q S w bw C n , r Eqn. (6) Nr  I zz , S The yaw angular acceleration imparted to the airplane as a result of a unit change in spoileron angle:

N r 

q S wbwC n , spn I zz , S

Eqn. (7)

104

Lateral-Directional Transfer Functions In the following equations, the ratio of aircraft moments of inertias(s) is defined as: A  1

B  1

I xz

I xx

I xz

I zz

Eqn. (8) & Eqn. (9)

The sideslip-to-control surface transfer function is defined as:

   N

 s

 C.S

Eqn. (10)

Where: N





3 2  s A s  B s  C  s  D



Eqn. (11)

Where:

1 A1B1  B  Y N  L P  A1N P  B1L r   Y L A  Y L B  N  N U1LC .S B1  NC.S   C .S  r P r C .S C .S 1 C .S 1 C .S A Y



C .S

Y

C .S

L P N r  N P Lr  YP NC .S Lr  LC .S N r  g cos1LC .S  NC .S L P 



D  g cos N  L  L N 1  C .S r C .S r



The bank-angle-to-control surface transfer function is defined as:

   N  C.S s  D2  s

Where:



2 N  s A s  B s  C 

Eqn. (12)



Eqn. (13)

Where:

105



A  U L A  N 1  C .S C .S 1





 







B  U N L  L N  Y L L  N  A1  N T A1   N Y 1    C .S  C .S r C .S r C .S C .S 

U1 Yr N  LC .S  N T LC .S  L  N C .S  C

 Y





NC .S L r  LC .S N r  Y , C .S L   N  A1  N T A1 



 U 1  Yr

N  LC .S  N T LC .S  L  N C .S 

The heading-to-control surface transfer function is defined as:

   N  C.S s  D2  s

Eqn. (14)

Where:

 A s3  B s 2  C s  D

B  U L 1 C.S 1

N  N L  U L 1 C.S P C.S P





N  L

 C.S











C.S

 N T L



C.S

L  N P  N  LP 







 L N 

C.S





 gCos N  L  N T L  L N  1 C.S C.S  C.S



The lateral-directional stability is determined by the characteristic equation. This equation is set to zero, and the roots are used to determine the lateral-directional stability characteristics of the airplane.

106

For an airplane with complex pairs of roots, they are cast in the form: S     J 1 1,2 1,2 1,2 1,2 1,2

Eqn. (15)

For an airplane with real roots, these are cast in the form of time constants, defined as follows: TC  

Eqn. (16)

107

-Lateral – Directional Flying qualities:



Dutch roll performance

Theory: The maximum bank angle to maximum sideslip ratio during Dutch roll is found from:

 

 magnitude of D

 (s) solved for the Dutch roll roots.  (s)

The maximum bank angle to maximum sideslip transfer function during Dutch roll is:  ( s 2 A1  sLr ) s 2  sN r

L

 (s)   ( s)



N

s  L s   s s  s B  N s  2



A1  sLr

 sN r



Eqn. (1)

The ratio of aircraft moments of inertias are defined as: A1  B1 

I xzS I xxS I xzS I zz S

Eqn. (2)

Eqn. (3)

The Dutch roll roots are of the form: s   D  nD  j nD 1   D2

Eqn. (4)

The mode shape corresponding to the dutch roll can then be computed by converting the above equation into the form:

  s      re j   s   D

Eqn. (5)

Therefore, the magnitude of the maximum bank angle to maximum sideslip ratio during Dutch roll is found from:

 

r

Eqn. (6)

108

The magnitude of the transfer function is calculated in the same manner as for the dutch roll roots. The difference is that the spiral root is typically a real value. The spiral root is found from:

1 Eqn. (7) TS The lateral-directional characteristic equation has four roots, generally consisting of a pair of complex roots and two real roots which represent the following modes: s

-Dutch Roll Mode: The minimum Dutch roll frequency and damping characteristics are specified below: Minimum Dutch Roll Undamped Natural Frequency And Damping Ratio Requirements Min D  nD

Min nD

Min D *

(rad/sec)

0.4 0.19 0.19 0.08 0.08 0.08 0.02 0

--0.35 0.35 0.15 0.15 0.10 0.05

Level Flight Phase 1

2 3

Category Airplane Class A - combat & ground attack IV A - other I and IV II and III B All C I, II-C, and IV II-L and III All All All All

1.0 1.0 0.4** 0.4** 1.0 0.4** 0.4** 0.4**

* The governing requirement is that which yields the largest damping ratio value. Note: For Class III, the maximum damping ratio value required is 0.7. ** Class III airplanes may be excepted from these requirements, Subject to specific approval. Civilian Requirements

 D  0 .0

FAR 25

When the ratio of bank angle to sideslip angle magnitudes in the dutch roll mode exceeds the following value:  20  2 Eqn. (8)



n

The minimum damping requirements shall be increased by:

109

    D  D  Factor   n2D  20   

 

Eqn. (9)

Where the Factor is found from: Level 1 2 3

Factor 0.014 0.009 0.005

The civilian requirements in the minimum Dutch roll frequency and damping characteristics table are less specific than the military requirements. Good design practice is to use the military requirements, whenever there is doubt about which Dutch roll requirements should be used.

 Roll Performance Theory: The airplane roll mode time constant is a measure of rapidity of roll response. A small roll time constant signifies a rapid build up of roll rate following a lateral control input by the pilot. The roll mode time constant shall be no greater than the appropriate value shown below: TRmax, Re quired

Class I and IV II and III All I, II-C*, IV II-L*, III

Flight Phase Category A A B C C

Level 1

Level 2

Level 3

1.0 1.4 1.4 1.0 1.4

1.4 3.0 3.0 1.4 3.0

10 10 10 10 10

Following a full deflection of the lateral cockpit controls, airplanes must exhibit a minimum bank angle response within a certain specified time. The elapsed time is counted from the time of cockpit control force application.

For preliminary design the roll control performance is approximated from:

110

 L a  a  L span  span   L    L spn  spn  L t t    e P  1 Eqn. (10) 2 LP LP    

 t   





The roll performance requirements for the different class airplanes could be found below. FAR 23.157 - Rate of Roll Requirements specifies requirements for controllability and maneuverability for take-off and approach conditions which are listed below. It must be possible to roll the airplane from a steady 30 degree banked turn through an angle of 60 degrees, so as to reverse the direction of the turn within: Take-off Condition: 1. For an airplane of 6000 pounds or less maximum weight, 5 seconds from initiation of roll; and 2. For an airplane of over 6000 pounds maximum weight,

W  500 second 1300 Approach Condition 1. For an airplane of 6000 pounds or less maximum weight, 4 seconds from initiation of roll; and, 2. For an airplane of over 6000 pounds maximum weight,

W  2800 second 2200 For FAR 25 it is suggested to use the Class II or Class III military airplane requirements. MIL-F-8785C - Flying Qualities of Piloted Airplanes specifies requirements for dynamic lateral-directional roll mode stability which is listed below: Roll Performance For Class I Airplanes To Achieve The Following Bank Angle Change (Seconds) Category A Category B Category C Level 60 degrees 60 degrees 30 degrees 1 1.3 1.7 1.3 2 1.7 2.5 1.8 3 2.6 3.4 2.6

111



Spiral Mode

The time to double the amplitude in the spiral mode may be calculated from: T2, S 

ln 2 1 TS

Eqn. (1)

And:

TS  0 TS  0

Spiral is not damped. Spiral is damped.

There are no specific requirements for spiral stability in any airplane. However, the military requirements place limits on the allowable divergence of the spiral mode.

Following a disturbance in a bank of up to 20 degrees, the time for the bank angle to double will be greater than the values shown in the table below. This requirement shall be met with the airplane trimmed for wings-level, zero-yaw rate flight with the cockpit controls free:

T2 S Flight Phase Category A B C

Level 1 12 sec 20 sec 12 sec

Level 2 8 sec 8 sec 8 sec

Level 3 4 sec 4 sec 4 sec

Roll mode check:

LevelTR23

2 Met

3 N/R

4 N/R

LevelTR Levelt

1 1

5 6 7 N/R N/R N/R 1 1

1 1

 actual (deg) 59.9 158.8 527.4 120.9 94.0

1 1 79.0

N/R: No Requirement Exist

112

Spiral & Dutch Mode check: 2 0.8672

3 0.9042

4 1.4278

5 0.9938

6 0.7705

7 0.8373

T2 S (s)

14.630

------

64.175

-------

17.488

T 1 (s)

------

107.672

731.525

------

-------

Level S

Stable

Level D Level D , 23

1 Met

Level nD

Level nD  D





113