Class I weight estimation: Theory: An iteration method is used to estimate the airplane take-off weight. The iteration starts with a guessed value of take-off weight (defined by users). The guessed take-off weight is used to solve for the airplane empty weight with the two equations shown below: log10 WE 

log 10 WTO  A B

Eqn. (1)

A & B Coefficient: A= 0.6632 (Single Engine Military Trainer) B=0.8640 (Single Engine Military Trainer) This equation represents a linear relationship between the logarithm of the airplane empty weight and the logarithm of the airplane take-off weight for airplanes of same type. The line that represents the relationship is called the Regression line. The take-off weight regression coefficients, A and B, for different types of airplane are listed in (1) 1; they can also be determined using regression techniques:

WE  1  1  M ff  1  M Fres  M tfo WTO  WPL  WCrew  WPLexp  WFrefuel n

M ff   M ff i  i 1

1 WTO

 n 1    WPLexpi 1   i 1  

  1  M  ff i     j  i 1   WTO n

 n 1    WFrefuel 1   i 1  

Eqn. (2)

  M  ff j   j  i 1   n

where : Wi  WFusedi M ff i  Wi n

WPLexp  WPLexp i 1

i

n

WFrefuel  WFrefueli i 1

The airplane empty weights calculated from the two equations are compared. If the following condition is satisfied, the guessed take-off weight will be accepted as the take-off weight for this particular airplane. If the condition is not satisfied, the program would adjust the guessed takeoff weight and repeat the calculation until the condition is satisfied: W E ( Eqn.2)  W E ( Eqn.1)  0.05lbs

Eqn. (3)

Once the take-off weight is determined, the weight of the fuel used in the mission is estimated from:

WFused  (1  M ff )WTO 1

Eqn. (4)

Roskam J., Airplain Design Part I; 1999 Section 2.7.1, P. 69

1

The total fuel weight is given by:

WF  1  M Fres WFused

Eqn. (5)

The fuel weight at the beginning of each segment is computed from:

WFbegini 1  WFusedi  WFrefueli  2

Eqn. (6)

The airplane weight at the beginning of each segment is computed from:

Wbegini  Wbegini 1  WFused i  WFrefueli 1  WPLexpi 1

Eqn. (7)

2

Mission Profile 1: Warm up:

 5 min. idle power Taxi:  3 min Taxi to Runway Take off:  Alt = 0 ft. Climb 1:  h  1500 ft  R / C  2000 ft. min .  L  15.27 D lb / hr  C j  0.323 lb Resulting in: E  0.8min

Climb 2:  h  8500 ft

R / C  3000 ft.

L

C j  0.350

D

min .

 15.00 lb / hr lb

Resulting in: E  2.8min

Climb 3:  h  31000 ft

R / C  3500 ft.

L

C j  0.323

D

min .

 14.00

lb / hr lb

Resulting in: E  8.9min

Cruise:  h  35000 ft  v  475kts

3

 

R  1200nm lb / hr C j  0.408 lb

Climb 4:  h  1000 ft

R / C  2000 ft.

L

C j  0.323

D

min .

 7.90 lb / hr lb

Resulting in: E  0.5min

Cruise:  h  36500 ft  v  475kts  R  1200nm lb / hr  C j  0.408 lb Descent:  h  18250 ft  R  100nm Descent, Land and Taxi:  h Final  0 ft Climb 5:  h  18500 ft

R / C  4400 ft.

L

D

min .

 15.00

C j  0.323

lb / hr lb

Resulting in: E  4.2min

Climb 6:  h  18500 ft

R / C  4400 ft.

min .

4

L

D

 16.00

C j  0.323

 Resulting in:

lb / hr lb

E  4.2min

Cruise:  h  38000 ft  v  280kts  R  132nm lb / hr C j  0.323 lb  Descent:  h  4375 ft  R  100nm Descent:  h  1125 ft  R  100nm Loiter:  t  30 min .

C j  0.323

L

D

lb / hr lb

 18

Fuel Fractions: Mission segment

Fuel Fraction M ff

Warm up Taxi Take off Climb Climb Climb Cruise Climb

0.9900 0.9950 0.9997 0.9989 0.9966 0.9423 0.9997

5

Cruise Descent Descent ,Land/Taxi Climb Climb Cruise Descent Descent Loiter Table 1

0.9427 0.9900 0.9920 0.9985 0.9986 0.9905 0.9900 0.9900 0.9911

Mission segment Wbegin (lb)

ď &#x201E;WFUsed (lb)

WFbegin (lb)

Warm-up Taxi Take off Climb Climb Climb Cruise Climb Cruise Descent Land/Taxi Climb Climb Cruise Descent Descent Loiter Table 2

1310.0 648.4 34.1 142.1 438.1 7404.1 41.2 6935.5 1140.5 903.2 168.9 158.1 1057.8 1106.2 1095.1 968.4

24729.3 23419.3 22770.9 22736.7 22594.7 22156.5 14752.4 14711.2 7775.7 6635.2 5732.0 5563.1 5405.0 4347.2 3241.1 2146.0

130998.8 129688.8 129040.3 129006.2 128864.1 128426.0 121021.9 120980.6 114045.1 112904.7 112001.4 111832.6 111674.5 110616.7 109510.5 108415.4

M ff

0.8202

WFused

23551.7 lb

WF WFmax

24729.3 lb 24729.3 lb

W Fres

1177.6 lb

Wtfo

662.8 lb

WCrew

1800.0 lb

Wuseful

59829.3 lb

WE

70514.5 lb

6

130998.8 lb WTO Table 4

Sensitivity Analysis: The reason for conducting a sensitivity study is to find:  which parameters drive the design  which areas of technological change must be pursued  which parameters were selected pessimistically or optimistically The sensitivity of take-off weight to payload weight, or airplane growth factor due to payload, is found from:

WTO  BWTO  WPL C .WTO (1  B )  D  B(1  M Fres )WF

Eqn. (8)

The variable C is given from: C  1  (1  M Fres )(1  M ff )  M tfo

Eqn. (9)

The variable D is equal to: D  WCrew  W PL  W PLexp  W Frefuel

Eqn. (10)

The fuel weight correction for the expended payload and/or refueled fuel weight is given by:

n   WF    WPL expi  WFrefuel i i 1  

1   M n

j  i 1

ff

j

   

Eqn. (11)

The sensitivity of take-off weight to empty weight, or airplane growth factor due to empty weight is solved from: WTO WTO B  WE WE

Eqn. (12)

The sensitivity of take-off weight to expended payload weight at k'th segment, also known as the airplane growth factor due to expended payload weight, is calculated from:

7

WTO WPL exp

n   BWTO  1  M Fres 1   M ff j  j  k 1   CWTO 1  B   D  B 1  M Fres

K

    1     WF

Eqn. (13)

The sensitivity of take-off weight to refueled fuel weight at k'th segment is calculated from:

WTO WFRe fuel

n     BWTO  1  M Fres 1   M ff i   1   j  k 1    CWTO (1  B )  D  B 1  M Fres WF

K

Eqn. (14)

The sensitivity of take-off weight to other factors at k'th segment, such as range, endurance, specific fuel consumption, and lift-to-drag ratio, are derived from the following equation for a particular mission segment: Range Partial:

BW

2 TO

WTO  y k

n k 1     W W M    PL exp i Frefuel i  ff j   i 1  j  i 1    R 1  M Fres  M ff i   y WTO   k    CWTO 1  B   D  B 1  M Fres W F

Eqn. (15)

Endurance Partial:

2 TO

BW WTO  y k

n k 1      WPLexpi  WFrefueli  M ff j    i 1 j i 1    E 1  M Fres  M ff i    y WTO   k    CWTO 1  B   D  B 1  M Fres WF

Eqn. (16)

n the range case for jet airplanes, the Breguet partial is found from:

8

cj R  y V L D

Eqn. (17)

For jet airplanes, the Breguet partial in the endurance case is determined from:

cj E  y L D

Eqn. (18)

The Breguet partial for jet-driven airplanes in the specific fuel consumption case is found from: Range:

R R  y V L D

Eqn. (19)

Endurance:

E E  y 60 L D

 

Eqn. (20)

In the lift-to-drag case, the Breguet partial for jet airplanes is determined from: Range: Rc j R  2 y L V  D Endurance:

Ec j E  2 y 60 L D Mission WTO c j lb  hr  Segment

Eqn. (21)

 

Warm-up Taxi Take off

-------------

Eqn. (22)

WTO -------------

R

lb nm

WTO -------------

L

WTO

D

 lb  E  hr 

-------------

9

Climb Climb Climb Cruise Climb Cruise Descent Land/Taxi Climb Climb Cruise Descent Descent Loiter Table 5

WTO WTO WTO

162.3 1340.3 5187.2 72688.3 217.7 73275.2 --------2298.3 2154.6 14495.0 --------13665.3

WPL WCrew

WE Table 6

------------24.7 ----24.9 ----------------35.5 -------------

-3.5 -33.3 -119.7 -1679.7 -3.7 -1707.0 ---------49.5 -43.5 -292.6 ---------245.2

10487.5 12223.1 11350.0 ----8438.0 ------------10593.3 9931.3 ------------8827.8

3.71 3.71 1.93

10

Weight Estimation: Theory: The Class I weight estimation method allows a rapid estimation of airplane component weights. The method relies on the assumption that within each airplane category it is possible to express the weight of major airplane components (or groups) as a simple fraction of the airplane flight design gross weight. The Class I weight estimation method is also referred to as the Weight Fraction method. The airplane flight design gross weight is that weight at which the airplane can sustain its design ultimate load factor. For civil airplanes, the airplane flight design gross weight and the airplane take-off weight are often the same, although there are exceptions. The weight fractions for the various components of the airplane are defined by taking an average of the selected airplanes. FWcomponent 

F

WComponent

# of Airplane

Eqn. (25)

Using Available data on similar aircrafts: W gross  FWgross WTO

Eqn. (26)

The Class I weight of each component is computed from:  Westimate  FWcomponent W gross Componnent

Eqn. (27)

When the first estimated component Class I weights are summed, they yield a weight which is slightly different from the airplane empty weight. The difference is due to round-off errors in the weight fractions used. WE  WE  WE

Eqn. (28)

This difference is to be 'distributed' over all items in proportion to their component weight value listed in the First Estimates column by computing the Adjustment by:

Wcomponent  WE .

 Wcomponent WE

Eqn. (29)

The 'actual' Class I weight of the components is therefore computed from:  Wcomponent  Westimate  Wcomponent component

Eqn. (30)

11

Component

Weight Fraction FW

Wing

FWw

First Estimate Westimate Ww

Ww

Ww

Empennage

FWemp

 Wemp

Wemp

Wemp

Fuselage

FW f

W f

W f

Wf

Nacelles

FWn

Wn

Wn

Landing Gear

FWgear

Wn  W gear

W gear

W gear

Structure

FWstructure

Wstructure

Wstructure

Power Plant

FWPP

 Wstructure  WPP

WPP

WPP

Fixed Equipment

FW fix

 W fix

W fix

W fix

WE

WE

WE

Empty Weight Table 7 Regression table: Airplane FWstructure Name Boeing 737-200 0.270 Boeing 727-100 0.317 Boeing 787 0.275

FWPP

FW fix

FWE

FWw

FWemp

FW f

FWn

FWgear

0.071 0.129 0.521 0.092 0.024 0.105 0.012 0.038 0.078 0.133 0.552 0.111 0.026 0.111 0.024 0.045 0.075 0.146 0.496 0.126 0.015 0.094 0.018 0.039

Table 8 Weight fractions: 0.287 FW Structure

FWPP

0.075

FW fix

0.136

FWE

0.523

FWw

0.110

FWemp

0.022

FW f

0.103

FWn

0.018

FWgear

0.041

FWgross

1.000

Table 9

12

Class I component weight breakdown

FW Westimate (lb) W (lb) Weight (lb) Fuselage 0.103 13554.0 908.6 14462.6 Wing 0.110 14370.6 963.3 15333.9 Empennage 0.022 2851.4 191.1 3042.6 Landing Gear 0.041 5331.6 357.4 5689.1 Nacelle 0.018 2362.3 158.4 2520.7 Structure 0.287 38470.0 2578.8 41048.8 Power plant 0.075 9781.2 655.7 10436.9 Fixed Equipment 0.136 17833.3 1195.5 19028.8 Empty Weight 0.523 066084.5 4430.0 70514.5 Table 10 Detailed weight information including approximate center of gravity locations: Component

Component

Weight

X CG  ft  63.65 65.54 119.85 41.73 113.31 131.56 51.65

Fuselage Group 14462.6 Wing Group 15333.9 Empennage Group 3042.6 Landing gear group 5689.1 Nacelle Group 2520.7 Power plant group 10436.9 Fixed Equipment Group 19028.8 Table 11 Class I Empty weight CG Location information:

Wstructure

41048.8 lb

WE X CGstructure

70514.5 68.53

X CGE

73.31

YCGstructure

0.00

YCGE

0.00

Z CGStructure

0.71

YCG  ft  0 0 0 0 0 0 0

Z CG  ft  1.20 -3.50 16.42 -6.28 20.31 15.44 -3.00

1.89 Z CGE Table 12

13

Class I Total CG Location information: Component

Weight(lb)

Crew Trapped Fuel and Oil Mission Fuel Group 1 Mission Fuel Group 2 Passenger Group 1 Passenger Group 2 Passenger Group 3 Passenger Group 4 Baggage Cargo Military Load Group 1 Military Load Group 2 Table 13

1800.0 662.8 25022.8 0 27750.0 0 0 0 3000 0 0 0

X CG  ft  20.63 61.23 61.23 0 59.15 0 0 0 77.34 37.22 0 0

YCG  ft  Z CG  ft  0 5.25 0 3.14 0 3.14 0 0 0 4.79 0 0 0 0 0 0 0 0 0 -1.59 0 0 0 0

Class I total weight CG Location information:

Wcurrent

133447.0 lb

X cg

69.50 ft

Ycg

0.00 ft

3.13 ft Z cg Table 14

14

Inertia Estimation: Theory: The Class I methods for determining the moments of inertia are based on the comparison with similar aircraft. It is assumed that within a category of airplane a radius of gyration can be identified. Therefore, by averaging the radius of gyration of similar airplanes, the moments of inertia can be found by: About the X-body axis:

I xxB ď&#x20AC;˝

bw2W gross R x2 4g

Eqn. (31)

I yy B ď&#x20AC;˝

L2W gross R z2 4g

Eqn. (32)

15

I ZZ B 

e 2W gross R z2

Eqn. (33)

4g

Where the parameter e is defined as: e

bw  L 2

Eqn. (34)

The non-dimensional radius of gyration is found by finding the average non-dimensional radius of gyration for the selected / defined aircraft.

 R  n

Rx 

i 1

n

 R  n

x i

,

Ry 

Eqn. (35)

i 1

 R  n

y i

Rz 

,

n

Eqn. (36)

z i

i 1

n

Eqn. (37)

For each similar airplane, the non-dimensional radius of gyration was obtained from: For the X-axis:

R 

2R x i bw i

x i

For the Y-axis:

R 

y i

Eqn. (37)

For the Z-axis:

2R y i

R 

L i

z i

Eqn. (38)

The radius of gyration for each airplane was calculated from: About the X-body axis:

R x i

( I xxB ) i g (W gross ) i

R 

y i

Eqn. (40)

Airplane Name

( I yyB ) i g (W gross ) i Eqn. (41)

W gross (lb)

Boeing 737-200 Heavy 113000.0 Boeing 737-200 Light 62000.0 Table 15

bW ( ft )

L( ft )

93.00 100.0

100.0 100.0

2R z i e i

Eqn. (39)

Rz i

( I zz B ) i g (W gross ) i

Eqn. (42)

Ry Rz Rx 0.246 0.456 0.456 0.246 0.517 0.517

16

Rx

0.255

I xxB

120326.7 slug  ft 2 0.419

Ry

I yy B

1708348.5 slug  ft 2 0.487

Rz I ZZ B 1588021.8 slug  ft 2 Table 16 Class I Performance Sizing:, Stall Speed: Theory: FAR 23 certified single engine airplanes may not have a stall speed greater than 61 knots at the airplane take-off weight. In addition, FAR 23 certified multi-engine airplanes with a take-off weight less than 6000 lbs must also have a stall speed of no more than 61 knots, unless they meet certain climb gradient criteria (FAR 23.49). These stall speed requirements can be met flaps-up or flaps-down at the option of the designer. There are no requirements for maximum stall speed in the case of FAR 25 certified airplanes. Given the maximum allowable stall speed for the flight condition at which the stall is to be evaluated, the maximum allowable wing loading at that certain flight condition can be computed from: 1 W  2    Vs C Lmax S 2 S  S

Eqn. (1)

The corresponding maximum allowable wing loading at take-off to meet the stall speed requirement can then be found from: WTO  W  W  Eqn. (2)       S  TO WS  S  S The maximum take-off wing loading to meet stall speed requirement is always a vertical line on the matching (carpet) plot. To meet the stall speed requirement, the take-off wing loading must be less than the value represented by the stall line. The stall speed performance is evaluated using the following equation:

VS 

2Wcurrent  Tset sin  current   T  SW C Lmax

Eqn. (3)

Which iteration required to solve. Flight condition: loiter, add missing parameters W 20.0 lb 2 S TOmin ft W 150.0 lb 2 S TOmax ft

   

17

T W  T W 

TOmin

TOmax

hS TS

0.0 lb lb 0.4 lb lb 0 ft

VSclean

27  F 170.00 kts

VS

130.00 kts

WS

0.990

WTO

C Lmax S ( Clean )

1.500

C Lmax( S )

1.750

Table 17 W S TOS C ln

  W S 

TOS

95.56 lb

ft 2 118.66 lb 2 ft

Table 18 Sizing for take off distance Take-off distance of any airplane is determined by the following factors: 1. Take-off weight. 2. Take-off speed (also called lift-off speed). 3. Thrust-to-weight ratio at take-off. 4. Aerodynamic drag coefficient and ground friction coefficient. 5. Pilot technique. Take-off requirements are normally given in terms of take-off field length requirements. These requirements differ widely and depend on the type of airplane under consideration. For civil airplanes, the requirements of FAR 23 and FAR 25 must be adhered to. FAR 25 Requirements have been used in order to size the take of and landing distance. For jet driven airplanes, the thrust to weight ratio to meet take-off distance requirements is plotted using the following relationship: W    T   S TO Eqn. (5)     W TO 0.0267 STOC Lmax FTO The take-off field length is determined from following table:

S TO  1.66S TOG

S TOFL  1.66S TOG

S TO  1.66S TOG

S TOFL  S TO

S TOG is Undefined

S TOFL  S TO 18

hTO

0 ft

 TO

1.000

TTO C LmaxTO

27 F 1.770

S TO

6330 ft

Table 20 Sizing for Maximum Cruise speed: For jet driven airplanes, the thrust to weight ratio to meet maximum cruise speed requirements is plotted using the following relationship: 2

CD0 ,Clean  W  BDPclean T    Cr    q W   W TO  WTO  q FCr FCr    S TO

W     S TO

Eqn. (7)

The dynamic pressure is found from:

q

1 VCr2 , Max 2

The 'B' of the drag polar is calculated from:

1 ARweclean In order to zero lift drag coefficient Wing aspect ratio assumed to be equal to 9.95 BDPClean 

Assuming a parabolic drag polar for class I analysis:

C D  C D 0clean  C D 0 

1 C L2 ARw e

Eqn. (8)

The zero-lift drag coefficient can be expressed as:

f Eqn. (9) SW The wetted area and equivalent parasite area can be related by a logarithmic equation: C D 0Clean 

19

log10 f  a  b log10 S wet

Eqn. (10)

Similarly, the airplane wetted area can be estimated from take-off weight by the following empirically obtained equation:

log10 S wet  c  d log10 WTO

Eqn. (11)

Oswald's efficiency factor and the change in zero-lift drag due to flight condition are highly dependent on flight condition. Wing area has been selected based on plot 1 & 2, Method used to estimate these parameters also require 4 regression coefficient a,b,c and d which are obtained from series of graphs and tables;

WTO

130998.8 lb

SW

1387.86 ft 2 9.95

ARW a -2.5229 b 1.0000 c 0.0199 d 0.7531 e Clean 0.9500 Table 22 S wet f

7476.77 ft 2

C D 0Clean

22.43 ft 2 0.0162

C D 0Clean , M

0.0175

0.0337 B DPClean Table 23 The drag polar is C D  0.0162  0.0337C L2 Altitude  cr

35000 ft 0.700

VCrmax

664.75 kts

WCr

0.980

WTO

ARW

9.95

C D 0Clean , M

0.0175

0.9500 eClean Table 24

20

M CrMax

1.117

B DPClean 0.0337 Table 25 Sizing to Landing Distance Requirement: For civil airplanes, the requirements of FAR 23 and FAR 25 are in force. The wing loading to meet landing distance requirements is plotted using the following relationship: W W     0.5  h , L ( ISA) C Lmax L S L F1 TO WL  S L

Eqn. (12)

The factor correct for FAR 25 of certification is: F1  9.365 The Landing field length for FAR 25 and military is S S FL  L 0.6 hL TL WL WTO

0 ft

C Lma , L

1.770

Eqn. (13)

Eqn. (14)

27 F 0.790

7000 ft SL Table 26 The result is plotted on matching plot Sizing for Climb Requirements: FAR 25.101, FAR 25.111, FAR 25.119, and FAR 25.121 requirements must be met for airplane climb sizing in the FAR 25 category. Federal Aviation Regulation Part 25.101: General (a) Unless otherwise prescribed, airplanes must meet the applicable performance requirements of this subpart for ambient atmospheric conditions and still air.

21

(b) The performance, as affected by engine power or thrust, must be based on the following relative humidities; (1) For turbine engine powered airplanes, a relative humidity of-( i ) 80%, at and below standard temperatures; and ( ii ) 34%, at and above standard temperatures plus 50 degrees F. Between these two temperatures, the relative humidity must vary linearly. (2) For reciprocating engine powered airplanes, a relative humidity of 80% in a standard atmosphere. Engine power corrections for vapor pressure must be made in accordance with the following table. (c) The performance must correspond to the propulsive thrust available under the particular ambient atmospheric conditions, the particular flight condition, and the relative humidity specified in paragraph (b) of this section. The available propulsive thrust must correspond to engine power or thrust, not exceeding the approved power or thrust less-(1) Installation losses; and (2) The power or equivalent thrust absorbed by the accessories and services appropriate to the particular ambient atmospheric conditions and the particular flight condition. (d) Unless otherwise prescribed, the applicant must select the take-off, en route, approach, and landing configurations for the airplanes. (e) The airplane configurations may vary with weight, altitude, and temperature, to the extent they are compatible with the operating procedures required by paragraph (f) of this section. (f) Unless otherwise prescribed, in determining the accelerate-stop distance, takeoff flight paths, take-off distances, and landing distances, changes in the airplane's configuration, speed, power, and thrust, must be made in accordance with procedures established by the applicant for operation in service. (g) Procedures for the execution of balked landings and missed approaches associated with the conditions prescribed in FAR 25.119 and FAR 25.121(d) must be established. (h) The procedures established under paragraphs (f) and (g) of this section must-(1) Be able to be consistently executed in service by crews of average skill; (2) Use methods or devices that are safe and reliable; and

22

(3) Include allowance for any time delays, in the execution of the procedures, that may reasonably be expected in service. Federal Aviation Regulation Part 25.111: Take-off Path (a) The take-off path extends from a standing start to a point in the take-off at which the airplane is 1500 feet above the take-off surface, or at which the transition from the take-off to the en route configuration is completed and a speed is reached at which compliance with FAR 25.121(f) is shown, whichever point is higher. In addition-(1) The take-off path must be based on the procedures prescribed in FAR 25.101(f); (2) The airplane must be accelerated on the ground to maximum extended flap speed, at which point the critical engine must be made inoperative and remain inoperative for the rest of the take-off; and (3) After reaching maximum extended flap speed, the airplane must be accelerated to take-off safety speed. (b) During the acceleration to take-off safety speed, the nose gear may be raised off the ground at a speed not less than the rotation speed. However, landing gear retraction may not be begun until the airplane is airborne. (c) During the take-off path determination in accordance with paragraphs (a) and (b) of this section -(1) The slope of the airborne part of the take-off path must be positive at each point; (2) The airplane must reach take-off safety speed before it is 35 feet above the take-off surface and must continue at a speed as close as practical to, but not less than the take-off safety speed, until it is 400 feet above the take-off surface; (3) At each point along the take-off path, starting at the point at which the airplane reaches 400 feet above the take-off surface, the available gradient of climb may not be less than-(i) 1.2% for two-engine airplanes; (ii) 1.5% for three-engine airplanes; and (iii) 1.7% for four-engine airplanes; and (4) Except for gear retraction and propeller feathering, the airplane configuration may not be changed, and no change in power or thrust that requires action by the pilot may be made, until the airplane is 400 feet above the take-off surface.

23

(d) The take-off path must be determined by a continuous demonstrated take-off or by synthesis from segments. If the take-off path is determined by the segmental method-(1) The segments must be clearly defined and must be related to the distinct changes in the configuration, power or thrust, and speed; (2) The weight of the airplane, the configuration, and the power or thrust must be constant throughout each segment and must correspond to the most critical condition prevailing in the segment; (3) The flight path must be based on the airplane's performance without ground effect; and (4) The take-off path data must be checked by continuous demonstrated take-offs up to the point at which the airplane is out of ground effect and its speed is stabilized, to ensure that the path is conservative relative to the continuous path. The airplane is considered to be out of the ground effect when it reaches a height equal to its wing span. (e) For airplanes equipped with standby power rocket engines, the user is advised to use the methods of Appendix E of the FAR's. Federal Aviation Regulation Part 25.119 Landing Climb: All-Engines-Operating In the landing configuration, the steady gradient of climb may not be less than 3.2%, with-(a) The engines at the power or thrust that is available eight seconds after initiation of movement of the power or thrust controls from the minimum flight idle to the take-off position; and (b) A climb speed of not more than 1.3 V_s. Federal Aviation Regulation Part 25.121: Climb: One-Engine-Inoperative (a) Take-off; landing gear extended. In the critical take-off configuration existing along the flight path (between the points at which the airplane reaches lift-off speed and at which the landing gear is fully retracted) and in the configuration used in FAR 25.111 but without ground effect, the steady gradient of climb must be positive for two-engine airplanes, and not less than 0.3% for three-engine airplanes or 0.5% for four-engine airplanes, at lift-off speed and with--

24

(1) The critical engine inoperative and the remaining engines at the power or thrust available when retraction of the landing gear is begun in accordance with FAR 25.111 unless there is a more critical power operating condition existing later along the flight path but before the point at which the landing gear is fully retracted; and (2) The weight equal to the weight existing when retraction of the landing gear is begun, determined under FAR 25.111. (b) Take-off; landing gear retracted. In the take-off configuration existing at the point of the flight path at which the landing gear is fully retracted, and in the configuration used in FAR 25.111 but without ground effect, the steady gradient of climb may not be less than 2.4% for two-engine airplanes, 2.7% for three-engine airplanes, and 3.0% for four-engine airplanes, at take-off safety speed and with-(1) The critical engine inoperative, the remaining engines at the take-off power or thrust available at the time the landing gear is fully retracted, determined under FAR 25.111 , unless there is a more critical power operating condition existing later along the flight path but before the point where the airplane reaches a height of 400 feet above the take-off surface; and (2) The weight equal to the weight existing when the airplane's landing gear is fully retracted, determined under FAR 25.111. (c) Final take-off. In the en route configuration at the end of the take-off path determined in accordance with FAR 25.111, the steady gradient of climb may not be less than 1.2% for twoengine airplanes, 1.5% for three-engine airplanes, and 1.7 % for four-engine airplanes, at not less than 1.25 times the stall speed and with-(1) The critical engine inoperative and the remaining engines at the available maximum continuous power or thrust; and (2) The weight equal to the weight existing at the end of the take-off path, determined under FAR 25.111 (d) Approach. In the approach configuration corresponding to the normal all-engines-operating procedure in which the stall speed for this configuration does not exceed 110% of the stall speed for the related landing configuration, the steady gradient of climb may not be less than 2.1% for two-engine airplanes, 2.4% for three-engine airplanes, and 2.7% for four-engine airplanes, with-(1) The critical engine inoperative, the remaining engines at the available take-off power or thrust; (2) The maximum landing weight; and (3) A climb speed established in connection with normal landing procedures, but not exceeding 1.5 times the stall speed.

25

Based on performed trade studies and mission requirements following decisions have been made about main characteristics of the aircraft:

SW 1387.86 ft 2 126 ft L T 41000 lbf WF 25031.7 lb Table 27 Engine has been selected to be P&W 1000G. Based on performed calculations Aircrafts Thrust vs. Velocity behavior for Cruise condition determined: For an aircraft equipped with jet engines, the maximum cruise speed is found when:

Tavail  Treq

Eqn. (14)

Tavail   Cr AThrustVCr2 max  BThrustVCrmax  CThrust

Eqn. (15)

  CD 0 Clean S wVCr2 max   2WCr2 BDPClean   Treq   2  2 cos      S V   cos   T w Cr T    max  The lift coefficient at the maximum cruise speed is found from:

C L|VCrMax 

C D  C D 0Clean , M BDPClean

Eqn. (16)

Eqn. (17)

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

Treq . cos(  T ) 0.5VCr2 max SW

M CrMax 



VCrmax

RTalt

C L @ VCrmax C L

Eqn. (18)

Eqn. (19,20)

 0

Power Available and Power Required vs. Velocity

26

Section Wing Horizontal tail Vertical tail Table 28

Airfoil Modified RAE 2512 3% Camber RAE 2512 NACA 0009

Sizing the Flaps: Theory: The outboard station of the flap is solved using the following relation for the flapped wing area ratio:

Swf Sw



Of

 i f

1  w

 2  1    w

of

if



Eqn. (22)

The flapped wing area ratio is solved from:

Swf Sw

C Lwf C lmax cl

3    cl f  cos 4  c 1.0  0.08 cos 2  c  4 w  4 w 

Eqn. (23)

Note that the flapped wing area ratio is sized for the condition, take-off or landing, that requires the most flap area to meet the required lift coefficient.. The increment in wing maximum lift coefficient due to flap deflection at ‘x’ flight condition is found using the following equation:

C Lwf  K trim C Lmax  C Lmax,Clean  C Lw da

Eqn. (24)

The ratio of the increment in the maximum sectional lift coefficient due to flaps to the increment in the sectional lift coefficient due to flaps. Can be found from Figure 7.4 in Airplane Design Part II as a function of flap chord ratio and the flap type: clmax

cf   f  , Type  c l  cw  Assuming use of plain flaps for cost reduction:

cl f  cl f K  f

Eqn. (25)

Eqn. (26)

180 The correction factor which accounts for non-linearities at high flap deflections is found from Figure 7.6 of Airplane Design Part II (J. Roskam) as a function of the flap deflection angle and the flap chord ratio:

27

cf   K   f   f ,  cw   1.770 C Lmax

Eqn. (27)

TO

f

35 Deg.

TO

C Lmax

f

1.770 L

15 Deg.

L

C Lmax,Clean

1.425

Cf

20%

i

Cw 24 %

f

K trim

1.050

t    c  w,avg ARw

11% 9.95

SW

1387.86

 C 4W

28

W

0.25

Table 29

clmax

cl

0.9766

cl f

 K TO K L

0.6223

clfTO

0.6223 2.6263

clf

1.3500 L

C LWfTO

0.3933

C LWfL

0.2883

SW f

0.256

O

f

SW 45.6 %

Table 30 Horizontal tail area estimation: Theory: For a tail-aft airplane, the horizontal tail area at a given static margin is calculated using the quadratic equation:

28

 b  b 2  4ac 2a where The a term is defined as: Sh 

a

xCG C Lh  d h  1 1   S h C L wf  d  S w

Eqn. (28)

Eqn. (29)

where:

X CG xcg S h  cw S h

Eqn. (30)

The b term can be expanded to the following:  C L  d  1  xcg h  S .M  xcg 0 h  x ach  b 1  h    C d  S w  S h  L wf  

Eqn. (31)

The c term contains the following: c  S .M  xcg 0 h  x acwf

Eqn. (32)

Aerodynamic Center Location:

Wing Aerodynamic Center Location:

The Z-location of the wing aerodynamic center is calculated from: Z acW 

Z Cr

Eqn. (33)  y mgcw tan w 4 w The X-location of the wing aerodynamic center in terms of the wing mean geometric chord is found from:

x acw 

X acW  X apexw  x mgcw

cw The X-coordinate of the wing aerodynamic center is given by: X acw  X apexw  nacw

Eqn. (34)

Eqn. (35)

The X-location of the wing aerodynamic center relative to the wing apex is found from Figure 8.100 in Airplane Design Part VI (J. Roskam) and is a function of the taper ratio, aspect ratio, leading edge sweep angle, root chord length and free stream Mach number:

nacw  f  w , ARw ,  LEw , c rw , 

Eqn. (36)

29

The wing leading edge sweep angle is given by:  1  w   LEw  tan 1 tan  c   4w ARw 1   w   

Eqn. (37)

The wing root chord length is found from: c rw 

2S w bw  w  1

Eqn. (38)

The intermediate calculation parameter is given by:

  1  M 12

Eqn. (39)

The X-location of the wing mean geometric chord leading edge relative to the wing apex is given by:

x mgcw  y mgcw tan  LEw

Eqn. (40)

The Y-distance from the wing mean geometric chord to the fuselage centerline is given by: y mgcw 

cw 

bw 1  2 w  61   w 

2c rw 1   w  2w

31   w 

Eqn. (41)

Eqn. (42)

Horizontal tail Aerodynamic Center Location:

The Z-location of the horizontal tail aerodynamic center is calculated from:

Z ach 

Z cr

4

 y mgch . tan h

Eqn. (43)

h

The X-coordinate of the horizontal tail aerodynamic center is given by:

X ach  X apexh  nach

Eqn. (44)

The X-location of the horizontal tail aerodynamic center relative to the lifting surface apex is found from Figure 8.100 in Airplane Design Part VI (J. Roskam) and is a function of the taper ratio, aspect ratio, leading edge sweep angle, root chord length and free stream Mach number:

30

nach  f h , ARh ,  LEh , c rh , 

Eqn. (45)

The horizontal tail leading edge angle is given by:  1  h   LEw  tan 1 tan  c   4h ARh 1   h   

Eqn. (46)

The horizontal tail root chord length is found from: 2S h c rh  bh  h  1 The intermediate calculation parameter is given by:

Eqn. (47)

  1  M 12

Eqn. (48)

The X-location of the horizontal tail mean geometric chord leading edge relative to the horizontal tail apex is given by:

x mgch  y mgch tan  LEh

Eqn. (49)

The Y-distance from the horizontal tail mean geometric chord to the fuselage centerline is given by: b 1  2 h  y mgcw  h Eqn. (50) 61   h  The horizontal tail mean geometric chord length is given by: cw 

2c rh 1   h  2h 31   h 

Eqn. (51)

Vertical tail Aerodynamic Center Location:

The Z-coordinate of the vertical tail aerodynamic center is given by:

Z acv  Z apexv  z mgcv

Eqn. (52)

The X-coordinate of the vertical tail aerodynamic center is given by:

X acv  X apexv  nacv

Eqn. (53)

31

The X-location of the vertical tail aerodynamic center relative to the vertical tail apex is found from Figure 8.100 in Airplane Design Part VI and is a function of the taper ratio, aspect ratio, leading edge sweep angle, root chord length and free stream Mach number:

nach  f v , ARv ,  LEv , c rv , 

Eqn. (54)

Effective Aspect ratio should be use as Aspect Ratio,

For single vertical tail, the effective aspect ratio is given by:

 ARv f ARveff    ARv

  ARv hf     ARv 1  K v   1 h    AR vf    

Eqn. (55)

The ratio of vertical tail aspect ratio in the presence of fuselage to that of an isolated vertical tail is found from Figure 10.14 in Airplane Design Part VI. It is a function of Z-location of the vertical tail tip relative to the fuselage centerline, fuselage depth in region of vertical tail and vertical tail taper ratio.

ARv f ARv

 f z v , h f v , v

Eqn. (56)

The Z-location of the vertical tail tip relative to the fuselage centerline is given by:

z v  bv  Z apexv  Z fch

Eqn. (57)

The taper ratio of the vertical tail extended to the fuselage centerline is computed from:

v  1

Z

v

fch

 Z apexv 1  v 

Eqn. (58)

bv The aspect ratio of the vertical tail extended to the fuselage centerline is computed from:

ARv 

z v2 S v

Eqn. (59)

The area of the vertical tail extended to the fuselage centerline is calculated from:

32

C rv Z apexv  Z fch  Z fch  Z apexv 1  v   2   2 bv   The vertical tail root chord is given by: S v  S v 

c rv 

2S v bv 1  v 

Eqn. (60)

Eqn. (61)

The correction factor accounts for the relative size of vertical tail and horizontal tail is found from Figure 10.16 in Airplane Design Part VI. It is a function of horizontal tail area and the area of the vertical tail extended to the fuselage centerline: Eqn. (62) K vh  f S h , S v  The ratio of vertical tail aspect ratio in the presence of fuselage and horizontal tail to the vertical tail aspect ratio in the presence of fuselage alone is found from Figure 10.15 in Airplane Design Part VI. It is a function of the relative position of horizontal tail and vertical tail, the Z-location of the horizontal tail root chord relative to the fuselage centerline and the Z-location of the vertical tail tip relative to the fuselage centerline:

ARvhh ARv f

 x  f  z v , z h , cv 

  

Eqn. (63)

The Z-location of the horizontal tail root chord relative to the fuselage centerline is given by:

z h  Z fch  Z ach

Eqn. (64)

The horizontal distance from the leading edge of the vertical tail, at the spanwise station at which the vertical tail intersects with horizontal tail, to the horizontal tail aerodynamic center is Eqn. (65) x  X ach  X apexv  Z ach  Z apexv tan  LEv The vertical tail chord length at the spanwise station at which the vertical tail intersects with horizontal tail is calculated from:

Z ach  Z apexv 1  v    cv  c rv 1   bv    Airplane Aerodynamic center location:

X ac  X apexw  x mgcw  x ac c w

Eqn. (66)

Eqn. (67)

The X-location of the airplane aerodynamic center in terms of wing mean geometric chord is determined by:

x ac  x acPoff  x ac power

Eqn. (68)

The airplane aerodynamic center in terms of wing mean geometric chord is calculated from:

33

x acP .off 

x acwfnpy , p .off  x achcont  x acvee

Eqn. (69)

C L , P .OFF

The wing-fuselage-nacelle-pylon aerodynamic center in terms of wing mean geometric chord is then computed from:

x acwfnpy P .OFF  x acwfP .OFF C L wf  x acn C L n  x ac py C LPY

Eqn. (70)

The horizontal tail aerodynamic center location in terms of the wing mean geometric chord is found from:   d   Sh  Eqn. (71)   hP C L 1   h   S x ach cont h d    P .OFF  W  For lifting surfaces (except canard), the aerodynamic center location in terms of the wing mean geometric chord is solved from: x ach

x acl . s . 

X acl . s  X apexw  x mgcw

Eqn. (72)

cw

Lift Curve Slope:

Wing Lift Curve Slope:

The wing-fuselage lift curve slope including any flap effects is determined from:

C L wf  K wf C LW

Eqn. (1)

The wing-fuselage lift curve slope without flap effects is determined from:

C L wf clean  K wf C LW

Eqn. (2)

The wing-fuselage interference factor is calculated from: 2

 D f max w   Eqn. (3) K wf  1  0.025  0.25  bw  bw   The ratio of the extended wing chord to the wing chord with flaps retracted is given by: D f max w

c w c c f  1 w . cw c f cw

Eqn. (4)

34

The increment in wing chord due to flap deflection in terms of flap chord is found from Figure G-7 in Synthesis of Subsonic Airplane Design (E. Torenbeek) as a function of flap deflection angle and the flap type:

c w  f  f , Type  cf

Eqn. (5)

The ratio of flapped wing area to wing area is calculated from:

Swf



of

  if  2  1   w   o f   i f



Eqn. (6)

Sw 1  w Note: The wing lift curve slope at M1 = 0, including flap effects is found by substituting the wing lift curve slope at M1 = 0, without flap effects into the equations above. cl 

cl M 0 1 M

Eqn. (7)

2

Horizontal tail lift curve slope:

The horizontal tail lift curve slope is calculated from:

C Lh  C Lh 

 exp

.

S hexp Sh

K

h( B)

 K B(h) 

Eqn. (8)

The horizontal tail-body interference factor is found from DATCOM Figure 4.3.1.2-10. The exposed horizontal tail lift curve slope may be estimated from: 2ARhexp f gaph Eqn. (9) C Lh exp  1 2 2  AR 2  2  tan  c     hexp  2h  2 1  4 2 2        k     The horizontal tail gap correction factor is found from:   x  f gaph  f  ARh ,  gap  ,  gap   Eqn. (10) C  C e  e   The Prandtl-Glauert transformation factor is derived from:

  1  M 12

Eqn. (11)

35

The ratio of the sectional lift curve slope to 2pi is solved from: cl |M 0 Eqn. (12) k  h0 2 The horizontal tail semi-chord sweep angle is calculated from: c

2h

 1  h    tan 1  tan  C   4h ARhexp 1   h   

Eqn. (13)

The exposed horizontal tail aspect ratio is solved from:

ARhexp 

bh2exp

Eqn. (14)

S hexp

The exposed horizontal tail area is calculated from:

S hexp 

bhexp

2

c

rh

 c rh  h

Eqn. (15)

The exposed horizontal tail span is derived from: bhexp  bh  w f h

Eqn. (16)

The width of the fuselage in the region of the horizontal tail value is governed by the following statement: if

Z cr 4

 Z fch  h

1 hf 2 h Eqn. (17)

then : w fh  0

The exposed horizontal tail taper ratio is found from:

h  exp

ct h c rhexp

Eqn. (18)

The horizontal tail airfoil gap correction factor is derived from:  x   f gapho  f   gqp  ,  gqp   C  C e e  

Eqn. (19)

36

Vertical tail lift curve slope:

C yv  

2ARveff f gapv  AR 2  2  veff 2 2  k 

 tan  C  2V 2 1     2

      4    

1 2

Eqn. (20)

The vertical tail gap correction factor is found from:   x  Eqn. (21) f gap x  f  ARveff ,  gap  ,  gap   C C v   v   Note: In the case of the vertical tail lift curve slope, the aspect ratio must be substituted by the Vertical Tail Effective Aspect Ratio.

The Prandtl-Glauert transformation factor is derived from:

  1 M 2

Eqn. (22)

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

cl v | M  0

Eqn. (23)

2

The vertical tail semi-chord sweep angle is calculated from: c

2v

 1  v    tan 1 tan  C   4v ARv 1  v   

Eqn. (24)

The vertical tail airfoil gap correction factor is derived from:  x  f gapvo  f   gap  ,  gqp  C C V v  

  

Eqn. (25)

Airplane Lift Curve Slope:

The airplane lift curve slope is solved from: C L  C L | P .OFF  C L POWER

Eqn. (26)

37

The airplane lift curve slope without power effects is solved from: C L

P . OFF

 C L

Wf n

 C L

Eqn. (27)

h

The wing-fuselage-nacelle lift curve slope including any flap effects is determined from:

C L wfn  C L wf  C L n  C L P

Eqn. (28)

The airplane lift curve slope contribution due to nacelles is determined from: n

C Ln  Fnf  C Ln i 1

wL  d n   1  Sw  d 

Eqn. (29)

The nacelle lift curve slope is found from:  w C Ln  f    L Eqn. (30)

w

4S max

The airplane lift curve slope contribution due to horizontal tail is found from:

C L h  C Lh  hP .OFF

S h  d h   1  Sw  d  P.OFF

Eqn. (31)

The downwash gradient at the horizontal tail including power effects is found from:

d h  d h   d  Eqn. (32)   h    d  d  P.OFF  d  POWER The downwash gradient at the horizontal tail including flap effects is found from:

 d h   d   h     d  P.OFF  d  Clean

 bf    f ARw  bw   1    C L  C Lw Eqn. (33) Clean C LW  b f  w     ARw  bw   

Where the incremental downwash angle at the horizontal tail due to flaps is found from Figure 8.70 in Airplane Design Part VI.

38

  bf   f ARw   bw   C LW  

      f h , b  h w   

Eqn. (34)

The downwash gradient at the horizontal tail without flap effects is found from:

 d h   4.44 K A K  K h cos  c    4W    d  Clean

1.19

C C

L wClean

L wClean

 

M1

Eqn. (35)

M 0

The correction factor for the aspect ratio is solved from:

KA 

1 1  ARw 1  ARw1.7

Eqn. (36)

The correction factor for the taper ratio is found from: K 

10  3 w 7

Eqn. (37)

The factor for the distance between the wing chord plane and the horizontal tail chord plane is defined from:

1 Kh 

hh bw

Eqn. (38)

1 3

 ln   2   bw  The Z-distance between the wing root chord plane and the horizontal tail aerodynamic center is calculated from: hh  Z ach  Z Cr

Eqn. (39) 4 w

The Z-coordinate of the horizontal tail aerodynamic center measured from reference line is calculated from:

Z ach  Z cr

4 h

 y mgch tan h

Eqn. (40)

39

The Y-distance between the horizontal tail mean geometric chord and the fuselage centerline is given by: y mgch 

ARh S h 1  2 h 

Eqn. (41)

61   h  The X-distance between the wing aerodynamic center and the horizontal tail aerodynamic center is given by:

l h  X ach  X acw

Eqn. (42)

The airplane lift curve slope without flap effects

C LClean  C L CLEAN  C L POWER

Eqn. (43)

The airplane lift curve slope without flap effects, without power effects is solved from:

C L CleanP .OFF  C L wF nClean  C L h  C L n

Eqn. (44)

Aerodynanamics, lift, airplane SW 1387.86 ft 2 9.95 ARW

w

0.25

w

28 Deg.

X apexW

49.20 ft

C L wf

X acwf

68.87 ft

X CG S h

0.001 ft 1

C Lh

X ach

112.53 ft

 h 

0.2949 PO

X CGh0

72 ft

S.M (desired) 12% Table 31

40

X CG

xcg

0.1717

x acwf

.4340

x ac

0.6201

Sh

375.27 ft 2 3.7352

C L

x ach Table 32

Vertical tail Area Sizing: Theory: The vertical tail area, given a desired static directional stability, is calculated from:  C n  C n wf Sv    C L v 

 S b w w   X ac  X cg v 

Eqn. (45)

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

C n wf  C n w  C n f

Eqn. (46)

The wing contribution to the directional stability is usually neglected:

C n w  0.0

Eqn. (47)

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

C n  57.3K N K Rl f

S BS L f

Eqn. (48)

SW 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 N  f X CG , L f , S Bs , h f 0.25 , h f max , w f max

Eqn. (49)

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

41

 

K Rl  f R N f

Eqn. (50)

The Reynold's number for the fuselage is calculated from:

RN f 

U1L f

Eqn. (51)

The vertical tail contribution to this derivative is solved from:

C n   C y  v v

X

acv

 X cg cos   Z acv  Z cg sin  bw

Eqn. (52)

The ventral fin contribution to this derivative is solved from:

X

C n  C y  vf vf

acvf

 X cg cos   Z acvf  Z cg sin  bw

Eqn. (53)

The airplane sideforce-coefficient-due-to-sideslip derivative is found from:

C y  C y w  C y f

Eqn. (54)

The wing contribution to this derivative is given by:

C y  0.00573 w w

Eqn. (55)

The fuselage contribution to the airplane sideforce-coefficient-due-to-sideslip derivative is given by: C y f  2 K i

SO SW

Eqn. (56)

The wing-fuselage interference factor is found from Figure 10.8 in Airplane Design Part VI (J. Roskam)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 i  f z w , h fw

Eqn. (57)

For single vertical tails:

  d   S v   v C y  k v C yv 1   v    d  v  S w

Eqn. (58)

42

The empirical factor for estimating airplane sideforce-coefficient-due-to-sideslip derivative is found from Figure 10.12 in Airplane Design Part VI (J.Roskam) 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

Eqn. (59)

The vertical tail span is given by:

bv 

ARv S v

Eqn. (60)

The intermediate calculation parameter is given by:

C n

    3.06   v  0.724    1  cos v   C 4  0.2489 rad 1

C yv 

X CG

70.29 ft

X acv

118.95 ft

C n

  d 1     d

f

w

 Z fcw  Z Cr  S  4 v    0.40   zf  S w   

w

 0.009 ARw Eqn. (61)

SV 184.24 ft 2 Table 33 Velocity- Load Diagram; Theory: The +1g Stall Speed is solved from: 2 VS 

W gross

SW C N max

Eqn. (1)

The maximum normal force coefficient is calculated from: C N max  C L2max,Clean  C D2 ,Clean

Eqn. (2)

The design equivalent maneuvering speed is estimated from:

43

V Aeas  VS nlim it

Eqn. (3)

The limit maneuvering load factor for normal or commuter category:

nlimit  2.1 

24,000 W gross  10,000

Eqn. (4)

If nlimit  3.8, nlimit  3.8 The stall speed for maximum negative normal force coefficient is found from: 2 VS   

W gross Sw

C N

max

Eqn. (5)  

The maximum negative normal force coefficient is calculated from:

C N max(  )  C L2max(  )  C D2 ,C L , Max (  )

0.5

Eqn. (6)

The design equivalent maneuvering speed for negative load factor regime is given by:

V A(  ) eas  VS (  )  nlimit (  )

Eqn. (7)

For normal, utility or commuter category

nlimit(  )  0.4nlimit Eqn. (8) FAR 23 requires the design equivalent cruise speed to be greater than or equal to the minimum design equivalent cruise speed: VCeas  VCeas

Eqn. (9)

min

The minimum design equivalent cruise speed is determined from one of the two equations shown below, whichever gives a smaller speed:

VCeasmin  0.9VH eas VCeas , Min  K C

W gross

Eqn. (10)

SW

The constant in the equation shown above

44

If If

W gross SW W gross SW

 20, K C  33  20, K C  34.1  0.055

Eqn. (11)

W gross SW

However, FAR 23 requires the design equivalent dive speed to be greater than or equal to the minimum design equivalent dive speed. The minimum design equivalent dive speed is determined from one of the two equations shown below, whichever gives a larger speed:

V Deas min  1.25VCeas

Eqn. (12)

V Deas min  K DVCeasmin

The constant in the equation shown above for normal and commuter type of aircraft is: If

W gross SW

 20, K D  1.40 W gross

W gross

Eqn. (13)

 20, K D  1.4125  0.0000625 SW SW The maximum gust intensity design speed (for commuter category only) should be determined by the designer of the airplane. However, FAR 23 requires the maximum gust intensity design speed to be greater than or equal to certain minimum value: If

V Beas  VBeas

Eqn. (14)

min

Minimum value for maximum gust intensity design speed is computed from one of the following equations, which is determined from one of the three equations shown below, whichever gives a smaller speed:  V At the intersection of V B gust line and C N max . curve.

a)

V Beas

b)

 K g U de B VCeas C L SW  V Beasmin  VS 1   498W gross  

c)

V Beas

min

min

 Vceas

Eqn. (15)

Eqn. (16)

Eqn. (17)

The slope of the gust line for maximum gust intensity design speed is given by: n V

 VB

K g U de B C L SW 498W gross

Eqn. (18)

45

The slope of the gust line for design cruise speed is solved from: n V

 VC

K g U de C C L SW 498W gross

Eqn. (19)

The slope of the gust line for design dive speed is solved from: n V

 VD

K g U de D C L SW 498W gross

Eqn. (20)

The gust alleviation factor is calculated from:

Kg 

0.88 g 5.3   g

Eqn. (21)

Where:

g 

2W gross S w c w gC L

Eqn. (22)

The derived gust speed (Under 20,000ft Altitude) for the gust line is solved from: U de B  66 ft s U de C  50 ft s U de D  25 ft s

46

Altitude T W gross

36500 ft 27 deg. F 130998.8 lb

SW CLmax

1387.86 ft 2 1.425

C D ,CLmax

0.0387

C Lmax,(  )

-0.600

C D ,C Lmax(  )

0.0280

cw

13.25 ft

C L VC eas

VH eas

520 keas

VDeas Table 34

530 keas

1

C N max

1.426

nlimit

2.50 g

nlimit (  )

-1.00 g

C N max (  )

-0.601

VS

139.73 keas

VS ( )

215.29 keas

V Aeas

220.93 keas

VCeas (min)

281.92 keas

V Deas (max)

381.38 keas

 n     V VB

0.0228 keas 1

 n     V VC

0.0040 keas 1

 n     V VD

0.0020 keas 1

Table V-n Diagram:

47

Class II Weight Estimation:  Wing Weight Estimation: GD method The weight of normal high lift devices as well as ailerons is also accounted for. The wing weight is found from: WWGD 

0.0428SW0.48 ARW M H 0.76

0.43

(WTO nult ) 0.84 0.14

Eqn. (1)

  100 t   cos  c / 2 1.54 w   c  rw   The maximum Mach number at sea-level is defined as: VH M H  eas a@ SL

Torenbeek method The wing weight is determined from:

WwTorenbeek

 bw 0.55   0.00171  Fcorr WMZF nult   cos  C 2W 

   

0.75

  6.3 cos  c   2w 1   bw   

   

0.5

  bw S w      t rwWTO cos  c 2 w  

0.30

Eqn. (2)

The maximum thickness of the wing root chord for straight tapered wings is found from:

 t   2S w  t rw      c  rw  bw 1   w   The wing weight correction factor is given by: Fcorr  FcorrS  Fcorre  Fcorrg  Fcorrf

Eqn. (3)

where:

FcorrS

= 2% if the airplane has spoilers and speed brakes; Fcorre = -5%if the airplane has 2 wing-mounted engines; Fcorrg = -5% if the landing gear is not mounted under the wing; Fcorrf = 2% if the wing has fowler flaps.

48

WTO

135850

Sw ARW

1387.86 ft 2 9.95

w

0.25

C

28.0 Deg. 4w

t    c w nult

11.94 % 3.75 g

WWGD

10547.7 lb

WTorenbeek

15653.5 lb

Ww Table 37

13100.6 lb

Vertical tail weight :

GD Method: The vertical tail weight is determined from: 0.217  Sr  0.337 1  v 0.363 cos  c / 4v WvGD  0.19 1   ARv Sv  

0.484

 Z  1  h  bv  

0.5

WTO nult 0.363 S v1.089 M H 0.601lv0.726  

Eqn. (5)

Torenbeek method: The vertical tail weight is estimated from:   S v0.2VDeas WvTorenb  K v S v 3.81  0.287  1000 cos  c / 4v   where K v = 1.0 for fuselage mounted horizontal tails or for airplanes without horizontal tail. The Class II vertical tail weight is determined by averaging the weights calculated by each estimation method.

ARv

2.00

v

0.41

C

35.0 deg. 4v

t    c  rv

9%

nlim it 2.50 g Table 38 49

nult

3.75 g

WvGD

1699.0 lb

WVTorenb

2315.2 lb

1744.3 lb WV Table 39

Horizontal tail weight:

GD method The horizontal tail weight is found from: 0.915

0.28 0.033   c w   0.584  bh  0.813      WhGD  0.0034 (WTO nult ) S h  t rh   l h    The horizontal tail root maximum thickness is found from:

 t   2Sh   t rh      c  rh  bh 1  h  

Eqn. (7)

Eqn. (8)

The X-distance between the horizontal tail and wing mean geometric chord quarter chord points is determined from: l h  X Apexh  X mgch 

ch c  X apexw  x mgcw  w 4 4

Eqn. (10)

The X-location of lifting surface mean geometric chord leading edge relative to the lifting surface apex is given by:

x mgcl . s .  y mgcl . s tan  LEl . s

Eqn. (11)

The Y-distance between the lifting surface apex and the lifting surface mean geometric chord is given by:  1  l .s    L.El . s  tan 1 tan  c   4l .s ARl .s 1  l .s.   

Eqn. (12)

The lifting surface mean geometric chord is given by:

50

4 1  l .s  l2.s 3 1  l .s 2 Torenbeek method cl .s. 

S l .s ARl .s.

Eqn. (13)

The horizontal weight is calculated from:   S h0.2VDeas WhTorenb  K h S h 3.81  0.287  1000 cos  c / 2h   K h = 1.0 K h = 1.1

Eqn. (9)

The horizontal tail half chord sweep is found from the horizontal tail planform geometry.

VDeas is in KEAS. Sh ARh

375.27 ft 2 1.88

h

0.70

X apexh

106.16 ft

10 % t    c h 2.50 g nlimit Table 40 3.75 g nult

lh

47.37 ft

WhGD

1085.0 lb

WhTorenb

2743.3 lb

1914.1 lb Wh Table 41

Fuselage:

Theory: GD method: W fGD  10.43K

1.42 inl

 qD    100  

0.283

 WTO     1000 

0.95

 Lf   hf  max

   

0.71

Eqn. (15)

where:

K inl = 1.25 for airplanes with inlets in or on the fuselage for a buried engine installation. 51

K inl = 1.0 for inlets located elsewhere. The design dive dynamic pressure is given by: 1 2 q D   @ SL 1.689VDeas 2

Torenbeek Method: The fuselage weight for a tail-aft airplane is found from: 0.5

W fTorenb  0.021K f  K press

c c     VD  X apex  rh  X apex  rw   h w 4 4   1.2  eas   K gear   S wet f w f max  h f max      

Eqn. (15)

The fuselage weight for a canard airplane is determined from: 0.5

c c     VD  X apex  rw  X apex  rc   w c 4 4   1.2  eas  W fTorenb  0.021K f  K press  K gear   S wet f w f max  h f max       The fuselage weight for a three surface airplane is determined from:

Eqn. (16)

0.5

W fTorenb  0.021K f  K press

c c     VD  X apex  rh  X apex  rc   eas h c 4 4   1.2    K gear   S wet f Eqn. (16) w f max  h f max      

The Class II fuselage weight is determined by averaging the weights calculated by each estimation method.

lf

125.00 ft

h f max

10.45 ft

N Pax

150

N crew

6

VDeas

530 keas

nlimit

2.8

W fGD

11797.9 lb

52

W fTorenb

19508.5 lb

Wf Table 42

15653.2 lb

Landing gear:

GD method The gear weight is found from:

WgGD

W   62.21 TO   1000 

0.84

Eqn. (1)

Torenbeek method: The gear weight is computed from: WgTorenbeek  WmgTorenb  WngTorenb  WtgTorenb

Eqn. (5)

The gear weight is given by:

W xgTorenbeek  K g r AxgTorenb  B xg

Torenbeek

0.75 1.5 WTO  C xgTorenbeek WTO  D xgTorenb WTO

Where, xg = mg for main gear, xg = ng for nose gear, xg = tg for tail gear.

 Eqn. (6)

The landing gear weight wing location correction factor is determined from:     0.5 z f   Z f c w  Z Cr 4 w   K g r  1  0.08  zf     The above equation yields:

Eqn. (7)

K g r  1.0 For low wing airplanes. K g r  1.08 For high wing airplanes. The constants for the landing gear weight are obtained from Table 5.1 of Airplane Design Part V (J. Roskam). The Class II landing gear weight is determined by averaging the weights calculated by each estimation method.

h fw

16.47 ft

Z fcw

0.00 ft

53

Z Cr

-3.84 ft 4 w

AngTorenbeek

20.0

BngTorenbeek

0.10

C ngTorenbeek

0.000

DngTorenbeek 2.0 Table 43 WngGD

583.3 lb

WngTorenb

820.5 lb

Wng

820.5 lb

WmgGD

3152.3 lb

WmgTorenbeek

4434.6 lb

4434.6 lb Wmg Table 44 Total Structure:

Ww

13100.6 lb

Wh

1914.1 lb

Wv

1744.3 lb

Wf

15653.2 lb

4495.4 lb W gear Total Structure 39457.1 lb Table 45

Power Plant Weight:

Engine:

Engine with accessories 7622.12/2 lb

Fuel System:

GD method: For a fuel system with self-sealing bladder cells, the fuel system weight is found from:

54

0.818

 WF  W fsGD  41.6 max w   Wsup p  Wrfs  W fds Eqn. (3) 100 F  For a fuel system with non-self-sealing bladder cells, the fuel system weight is found from: 0.758

 WF  W fsGD  23.1 max w   Wsup p  Wrfs  W fds 100 F  where the weight of the bladder support structure is found from: 0.854  WFmax w  Wsup p  7.91  100 F  Torenbeek method For a fuel system with non-self-sealing bladder cells, the fuel system weight is found from: 0.727

 WF  W fsTorenb  1.6 maxW   Wrfs  W fds  100 F  For a fuel system with self-sealing bladder cells, the fuel system weight is found from: 0.727

 WF  W fsTorenb  3.2 maxW   Wrfs  W fds 100  F   For airplanes equipped with integral fuel tanks (wet wing), the fuel system weight is found by: 0.333 WFmax  0.5  w  Wrfs  W fds W fsTorenb  80N eng  N sft  1  15N sft     F   WFmaxW

28244.7 lb

F

6.82

N sft

4

K fds

7.38

FIntegral

1.0

W fsTorenb

921.1 lb

lb (JP-5) gallon

921.1 lb W fs Table 47

Propulsion System:

GD Method:

WPGD  WEngCtrl  Wess  WOilSys  WInjectSys

Eqn. (6)

55

Engine Controls Weight The engine controls weight for jet engines is given by:

WEngCtrl

N    88.46 K eng ins L f  bw  eng  100  

0.294

 1  K eng ins K EngCtrl L f N eng 

0.792

where:

K EngCtrl  0.686 for non-afterburning engines K EngCtrl  1.080 for afterburning engines The engine installation factor is defined as:

K Engins  1.0 if all the engines are mounted on the wing; K Eng ins  0.0 if all the engines are mounted on the fuselage/wing-root. Engine Starting System Weight The engine starting system weight is found from:

Wess

W   K ess  eng   1000 

n ess

Where the typical values for the factors are: For jet engines using electrical starting systems:

K ess  38.93 ness  0.918

Torenbeek method: For airplanes with turbojet or turbofan engines, the weight for accessory drives, powerplant controls, starting and ignition systems is found from:

WPTorenbeek  36 N eng m f ,TO

Eqn. (7)

Oil System and Oil Cooler

56

The oil system and oil cooler weight is estimated from:

WOilSys  0.00 or jet engines (weight included in the engine weight estimate) The Class II propulsion system weight is determined by averaging the weights calculated by each estimation method.

Weng

7622.1 lb

K OilSys

0.00

m f ,TO

3.3 lb/sec.

WEng ,Ctrl

54.4 lb

Wess

83.3 lb

WWaterInj

0 lb

WOil , Sys.

0 lb

WPGD

137.7 lb

WPTorenb

237.6 lb

187.7 lb WP Table 48 The nacelle weight GD method: The nacelle weight is found from:

WnGD  FBPRGD N n  A  lmc P2 0.5

0.731

Nacelle Weight GD Method: The nacelle weight is found from:

WnGD  FBPRGD N n  A  lmc P2 0.5

0.731

The bypass ratio factor is determined from the chart shown below:

57

Torenbeek method: The nacelle weight is calculated from:

Wn ,Torenb  FBPR ,TorenbTTO The bypass ratio factor is determined from the chart shown below:

lmc

11.0 17.46 ft 2 5.59 ft

P2 WnGD

40.00 Psi 1425.3 lb

WnTorenbeek

2860.0 lb

Wn

2142.6 lb

BPR A

Total Power Plant Weight:

58

Weng

7622.1 lb

W fs

921.1 lb

WP Wn

187.7 lb 107.9

W PP 8730.8 Table 49

Fixed Equipment Weight:

Flight Control Systems:

GD method The following equation applies to business jets as well as to commercial transport airplanes. The flight control system weight is found from:

W q  W fcsGD  56.01 TO D   100000 

0.576

 WcgCtrl

The design dive dynamic pressure is computed from:

qD 

1  1.689VDeas 2

2

Torenbeek method The flight control system weight is given from:

0.667 W fcsTorenb  K LD K LE 0.44  0.2 K fcsPower WTO  WcgCtrl

where: K LD = 1.15 for airplanes with lift dumpers (spoilers or speed brakes); K LE = 1.00 for airplanes without leading edge devices K fcs Power = 1.0 for airplanes with powered flight controls.

The center of gravity control system weight is calculated from: 0.442

WF max w    WcgCtrl  K cgCtrl  0.01  F   The Class II flight control system weight is determined by averaging the weights calculated by each estimation method.

59

WFmax

28244.7 lb

K fcsPower

1.0

K cgCtrl

0.00

W fcsGD

3398.1 lb

W fcs ,Torenb

2278.1 lb

W fc

2585.6 lb

0.00 lb WcgCtrl Table 50

Instrumentation, Avionics and Electronics:

GD method (modified) The weight of instruments is found from: 0.032WTO  0.006WTO  0.15WTO   WiaeGD  N Caption  N CopPilot  N FltEngr 15   0.012WTO    N eng  5  1000  1000  1000   Torenbeek method For jet transports, the instrumentation, avionics and electronics weight is determined from: 0.25 WiaeTorenbeek  0.575WE0.556 Rmax

Eqn. (4)

70514.5 lb WE Rmax 2600 nm Table 51

Wiae

1839.4 lb

Air-Condition Anti Icing and De-Icing system weight:

GD method For pressurized airplanes, the air-conditioning, pressurization, anti- and de-icing system weight is found from:

WapiGD

N  N pax    469V pax crew  10000  

0.419

Eqn. (5)

60

Torenbeek method: The air-conditioning, pressurization, anti- and de-icing system weight for pressurized airplanes is determined from: .28 Wapi ,Torenbeek  6.75 L1cabin

VPax N Crew

21264.00 ft 3 6

LCabin

75.00 ft

WapiTorenbeek

1695.8 lb

WapiGD

5338.0 lb

Eqn. (6)

3516.9 lb Wapi Table 52  Auxiliary Power Unit Weight: From the detailed weight statements in Appendix A of Airplane Design Part V, it is possible to derive weight fractions for auxiliary power units as a function of the take-off weight:

Wapu  0.07WTO Wapu

1703.0 lb Furnishings weight:

Theory: GD Method: The furnishings weight is found from: (1  Pc )   W furCessna  40N Captain  N CoPilot  N FltEngr   15N crew  32 N pax  K lavatory ( N pax )1.33  K buffet ( N pax )1.12  109  N pax  100   Eqn. (1)

0.505

where:

K lavatory =0.31

and

K buffet =1.02

The design ultimate cabin pressure is determined from: Pc  1.5P@ hPcabin

61

W   0.771 TO   1000 

Torenbeek method

W furTorenbeek  0.211 WE  WCrew  WPL  WPLexp  Wtfo

0.91

Eqn. (2)

The Class II furnishings weight is determined by averaging the weights calculated by each estimation method.

N Crew

6

N Pax

150

W furGD

6169.1 lb

W furTorenbeek

7943.5 lb

7056.3 lb W furnish Table 53

Oxygen System Weight:

GD method The oxygen system weight is found from:

WoxGD  7N crew  N pax 

0.702

Torenbeek method The following equation can be applied to transport airplanes and business type airplanes. WoxTorenbeek  Wox fixed  K ox N crew  N pax 

The typical values for the factors are: For short flights above 25,000 ft Wox fixed  30

K ox  1.2 Wox ,GD

242.5 lb

Wox ,Torenbeek

217.2 lb

Woxygen

229.8 lb

Total Fixed Equipment Weight:

62

W fcs

2838.1 lb

Wiae

1839.4 lb

Wapi

3516.9 lb

WFurnishing

7056.3 lb

Wapu

1703.0 lb

Wother

570.7 lb

W fix Table 54

26088.6 lb

Take-off Weight: Airplane empty weight, defined from:

WE  WStructure  WPP  W fix

Eqn. (2)

The Class II take-off weight is then calculated from:

WTO 

WE  WPL  WCrew  WPLexp  WFrefuel

M ff 1  M Fres  M Fres  M tfo

W fix

26088.6 lb

WStructure W PP W PL WCrew W PL exp

39457.1 lb 8730.8 lb 333750.0 lb 1800.0 lb 0 lb

W Frefule

0 lb

M ff

0.8202

M FRe s

5.0 %

M tfo

0.5 %

W FUsed

24491.0 lb

WF Wtfo

25715.5 lb 681.1 lb

Eqn. (3)

74276.6 lb WE 136223.2 lb WTO Table 55

63

Component CG Location: Theory:  Wing: C  c rw  0.4 c rw  ct w

b  y  0.4 w   2  X cg wing  X apexw  y tan  LEw  0.4C

Horizontal tail: C  c rh  0.4 c rh  cth

Eqn. (1)

b  y  0.38 h   2 X cg h  X apexh  y tan  LEh  0.42C

Eqn. (2)

Z cg h  Z cr  y tan h

Vertical tail: C  c rv   c rv  ct v

 bv  y      2  X cg v  X apex v  y tan  LE v  0.42C Z cg v  Z c r

v

Eqn. (3)

 y tan v 4

v

 Fuselage: The X-location of the fuselage CG are given as follows, As a fraction of the fuselage length: Single Engine Tractors: 0.32 - 0.35 X cg f  X nose f 

Fcg f l f 100

Eqn. (1)

Z cg f  Z f

64

Class II Empty Weight CG Location Table: Component

Weight (lb)

1-Wing 2-Horizontal tail 3-Vertical tail 4-Fuselage 5-Nacelles 6-Nose Landing Gear 7-Main Landing Gear 8-Engines 9-Fuel System 10-Air Induction System 11-Propulsion System 12-Flight Control System 13-Hydraulic and Pneumatic System 14-Instrumentation, Avionics, Electronics 15-Electrical System 16-Air Conditioning/Press./Anti Icing 17-Oxygen System 18-Auxilary Power Unit 19-Furnishings 20-Cargo Handling Equipment 21-Operational Items 22-Other Items Table 57

13561.4 1606.9 1848.1 15445.7 2549.6 778.5 4235.7 8655.0 960.1 0.0 197.4 2620.4 335.5 1715.2 3011.2 3190.6 207.7 1342.0 6479.3 1236.1 3774.2 559.2

X CG  ft  68.36 115.62 121.41 60.63 100.71 26.00 66.67 103.59 90.00 103.00 103.00 90.00 105.00 8.81 95.00 70.00 55.70 120.00 80.00 50.00 50.00 100.00

X cg Structure

70.78 ft

X cg PP

102.25 ft

X cg fix

73.27 ft

X cg E

75.76 ft

Ycg structure

0.00 ft

YcgPP

0.00 ft

Ycgfix

0.00 ft

Ycg E

0.00 ft

Z cg Structure

1.66 ft

Z cg PP

15.32 ft

Z cg fix

1.79 ft

Z cg E

3.51 ft

YCG  ft  0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Z CG  ft  -3.12 3.85 15.86 0.00 17.10 5.90 5.90 17.10 -0.80 15.44 15.44 4.17 -1.81 0.80 0.82 1.23 1.23 4.23 4.79 -2.00 -2.00 -2.00

Table 58

65

Total CG: Component

Weight (lb)

Crew Trapped Fuel and Oil Mission Fuel Group 1 Mission Fuel Group 2 Passenger Group 1 Passenger Group 2 Passenger Group 3 Passenger Group 4 Baggage Cargo Military Load Group 1 Military Load Group 2 Table 60

1350.0 638.6 23165.1 0.0 27750.0 0.00 0.00 0.00 0.0 4500.0 0.00 0.00

X CG ft 20.63 67.68 67.68 0.00 59.15 0.00 0.00 0.00 77.34 50.00 0.00 0.00

YCG ft 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Z CG ft 5.25 -2.27 -2.27 0.00 4.79 0.00 0.00 0.00 0.00 -1.59 0.00 0.00

X CG

69.35 ft

YCG

0 ft

Z CG Table 61

2.58 ft

xcg E

74.84

I xxB

133653.1 Slug  ft 2

I yy B

1645878.5 Slug  ft 2

I zz B

1512225.3 Slug  ft 2

I xz B Table 62

235951.2 Slug  ft 2

66

Detailed Mission Profile: Segment 1, Warm up Taxi: Taxi on a Normal constructed runway, 27 deg. F offset from ISA, No wind Sea Level. 0.0 deg. f V  VTO Alt. T Weight X CG Z CG Flight quality Category Max. Thrust Available Table 1

25 kts 0 ft 27 deg F 132605.0 lb 68.70 ft 3.13 ft C, Take off 42000 lbf

Segment 2, Take off: Take off from 7000 ft runway, no wind, 27 deg. F offset from ISA, Sea Level. 35.0 f

V Alt. T Weight X CG Z CG Flight quality Category Table 2

175 kts 35 ft 27 deg F 131402.6 lb 70.29 ft 3.10 ft C, Take off

Segment 3, Climb to 1500 ft: Climb, 27 deg. F offset from ISA

f

10.0

V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available Performance, max cruise speed CJ

250.00 kts 768 ft 27 deg F 120206.8 lb 69.80 ft 3.31 ft B, Climb lbf 0.466

lb lb, hr

Table 3

67

Segment 4, Climb to 10000 ft: Climb, 27 deg. F offset from ISA

f V

0

250 kts Alt. 5750 ft 27 deg F T Weight 121253.8 lb 69.78 ft X CG 3.29 ft Z CG Flight quality Category B, Climb Max Thrust Available 291.6 lbf lb CJ 0.643 lb, hr Table 4

Segment 6, Climb to Cruise Altitude: Climb, 27 deg. F offset from ISA 0 f

V Alt. T Weight X CG Z CG Flight quality Category Max. Thrust Available CJ

480 kts 22500 ft 27 deg F 120462.6 lb 69.77 ft 3.30 ft B, Climb 879 lbf lb 0.517 lb, hr

Segment 5, Accelerate to Climb Speed: Climb, 27 deg. F offset from ISA

f V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

0 400 kts 35000 ft 27 deg F 129987.8 lb 69.46 ft 3.17 ft B, Cruise 583.2 lbf lb 0.466 lb, hr

Table 5

Segment 7, Cruise: Climb, 27 deg. F offset from ISA 0.0 f

V Alt. T Weight X CG Z CG Flight quality Category Maximum Thrust Available CJ

Table 6

475 kts 35000 ft 27 deg F 127403.7 lb 69.72 ft 3.24 ft B, Cruise 1200 lbf lb 0.39 hp, hr

Table 7

68

Segment 8, Descend to 1500 ft: Descend, 27 deg. F offset from ISA

f V

0

340 kts Alt. 18250 ft 27 deg F T Weight 114899.6 lb 68.92 ft X CG 3.45 ft Z CG Flight quality Category B, Descent Max Thrust Available 291.6 lbf lb CJ 0.643 lb, hr Table 8

Segment 10, Land: Land, 27 deg. F offset from ISA

f V

20.0

175 kts Alt. 500 ft 27 deg F T Weight 118114.5 lb 68.97 ft X CG 3.47 ft Z CG Flight quality Category C, Landing Max Thrust Available 291.6 lbf lb CJ 0.643 lb, hr Table 10

Segment 9 , Approach: Descend, 27 deg. F offset from ISA

f V

35.0

Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

200 kts 750 ft 27 deg F 113038.1 lb 69.41 ft 3.51 ft C, Approach 583.2 lbf lb 0.466 lb, hr

Table 9

Segment 11, Taxi in: Taxi in, 27 deg. F offset from ISA

f V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

0 25 kts 0 ft 27 deg F 118114.5 lb 68.97 ft 3.47 ft B 583.2 lbf lb 0.466 lb, hr

Table 11

69

Segment 12, Missed Approach: Climb, 27 deg. F offset from ISA 35.0 f

V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

200 kts 750 ft 27 deg F 112150.1 lb 69.53 ft 3.54 ft B, Climb 291.6 lbf lb 0.643 lb, hr

Segment 13, Economy Climb: Climb, 27 deg. F offset from ISA

f V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

Table 12

0 250 kts 4375 ft 27 deg F 111984.1 lb 68.95 ft 3.52 ft B, Climb 583.2 lbf lb 0.466 lb, hr

Table 13

Segment 14, LRC: Cruise, 27 deg. F offset from ISA 0 f

V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

280 kts 8000 ft 27 deg F 111937.3 lb 68.89 ft 3.52 ft B, Cruise 291.6 lbf lb 0.643 lb, hr

Segment 15, Economy Descent: Descend, 27 deg. F offset from ISA

f V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

Table 14

0 250 kts 4375 ft 27 deg F 111636.8 lb 68.96 ft 3.73 ft B, Descent 583.2 lbf lb 0.466 lb, hr

Table 15

70

Segment 16, Economy Descent: Descend, 27 deg. F offset from ISA 20.0 f

Segment 17, Accelerate to Climb Speed: Climb, 27 deg. F offset from ISA

f V

135 kts 1125 ft 27 deg F 108733.3 lb 69.71 ft 3.79 ft C, Approach 291.6 lbf lb 0.643 lb, hr

V Alt. T Weight X CG Z CG Flight quality Category Max Thrust Available CJ

0

135 kts Alt. 1500 ft 27 deg F T Weight 118892.2 lb 65.37 ft X CG 3.62 ft Z CG Flight quality Category B, Loiter Max Thrust Available 583.2 lbf lb CJ 0.466 lb, hr

Table 16

Table 17

Class II Drag Estimation:

Wing, Horizontal and Vertical tail :

Theory: The subsonic lifting surface zero-lift drag coefficient may be computed from:

C

D0 W

R

R

N fus

  C f 4  W   Lan   

 

t t R R  1  L   100 Wf Ls  c c 

 f RN

fus

,M

LS ( Lifting Surface Corection Factor )

 

L   1.6  t

c Max

at X

t

  

 f  C

4W

C

f

Tur

b SWet

W

SW

C

f

Turb

S Wet

  W    

 t  ,  t  , M      c Wr  r Wr 

,

 0.3 C

  t           r  C  t  2 Cl 2 2 W CD   2C L  t   4  tW  LW AR e W W W t

  t  0.5  C  C 

71

e  1.1

CL

W

ARW

 C L  W R   1 R   ARW   

R  Leading edge suction parameter of wing

 

R  f  ARW , W ,  C

l ,  ler 

4W

C

 , M Source : Airplane Design PartVI, Fig4 - 7  W 1 

  Induced drag factor due to inear twist

 

  f  ARW , W ,  C

4W

l ,  ler 

C

 : Zero lift drag

 , M  Source : Airplane Design Part VI, Fig 4 - 9  W 1 

Airfoil sections lift coefficient: C

l Max

R  er R  et

 f R e , t , Airfoil Type c

VS C r  VS C t 

 WTO  V  S  0.5 C L MaxClean  Calculatin g C

CL

Max W

L

0.5

   

: Max w

Clean

 fCouple C L Max

 lh   C   lh  f Couple  1.10  3.0  C  f Couple  f 

CL CL

Max W Unsweeped

maxW Cos C

Sweeped 4W

72

CL C

LMax W Unsweeped

Cos C C

K

LMax W

MaxW Sweeped Cos C

4W

 C  l Max  Cl Max  r t  4 2

K : Taper ratio factor

K  0.117  0.997  W CL MaxW Clean fCouple

C L MaxClean  0.05

CL MaxClean

C

L Max

 1.05 C L

X

Max X

C L

Max Clean

3 2   K   1.0  0.08Cos  C Cos 4 C  4W   4W K

C l

Max

C l

The transonic zero-lift drag coefficient is found from:

C D0w  C D0w @ M 0.6  C Dwwave The wing thickness ratio at the spanwise station of the wing mean geometric chord is given by:  1  2w  t  t t          w   c  rw 3(1  w )  c  rw  c  rw  t    1  2w  c  rw 1 (1  w ) 3(1  w ) The transonic wing drag coefficient due to lift is found by: CD C DL  2L C L2W w CL

K sand

0.00167x10-3 ft

 RLE     c w

1.500 %

73

Lw

1.2

 X lam   c  w 

60 %

cl

cl

rw , M  0

6.1020 rad 1 tw , M  0

f gapw

0.99

g

-3.0 deg

w

C Dgap

0.0002 a

C Dgap flap

0.0002

Class II Drag for Wing:

Flight Segment 1 : N/A Aerodynamically 0.037 M1

Flight Segment 2 : 0.177

cl rw

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

clW ,Clean

S wetw

S wetw

C D0 w

2302.33 ft 2 0.0056

C D0 w

2302.33 ft 2 0.0037

C DLW

N/A L  0

C DLW

0.0113

rw

74

Flight Segment 3:

Flight Segment 4: 0.375

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

clW ,Clean

S wetw

S wetw

C D0 w

2302.22 ft 2 0.0034

C D0 w

2302.33 ft 2 0.0035

C DLW

0.0136

C DLW

0.0134

M1 cl

0.369

rw

Flight Segment 5: M1 cl

0.674

rw

Flight Segment 6: 0.556

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

clW ,Clean

S wetw

S wetw

C D0 w

2302.33 ft 2 0.0042

C D0 w

2302.33 ft 2 0.0038

C DLW

0.0364

C DLW

0.0092

rw

rw

75

Flight Segment 7: M1 cl

0.801

Flight Segment 8: 0.532

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

clW ,Clean

S wetw

S wetw

C D0 w

2302.33 ft 2 0.0050

C D0 w

2302.33 ft 2 0.0037

C DLW

0.0149

C DLW

0.0090

rw

Flight Segment 9: M1 cl

0.207

rw

Flight Segment 10: 0.199

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

clW ,Clean

S wetw

S wetw

C D0 w

2302.33 ft 2 0.0036

C D0 w

2302.33 ft 2 0.0036

C DLW

0.0203

C DLW

0.0252

rw

rw

76

Flight Segment 11: M1 cl

0.037

Flight Segment 12: 0.295

M1 cl

cl tw

cl tw

clW

clW

clW ,Clean

cl

S wetw

2302.33 ft 2 0.0056

S wetw

rw

CD0

w

C DLW

W , Clean

C D0 w

2302.33 ft 2 0.0035

C DLW

0.0338

Flight Segment 13: M1 cl

0.374

rw

Flight Segment 14: 0.801

M1 cl

cl tw

cl tw

clW

clW

cl

cl

S wetw

C D0 w

2302.33 ft 2 0.0035

C D0 w

2302.33 ft 2 0.0050

C DLW

0.0219

C DLW

0.0136

rw

W , Clean

S wetw

W , Clean

Flight Segment 15: M1 cl

0.374

rw

Flight Segment 16: 0.200

M1 cl

cl tw

cl tw

clW

clW

cl

cl

S wetw

C D0 w

2302.33 ft 2 0.0035

C D0 w

2302.33 ft 2 0.0036

C DLW

0.0093

C DLW

0.0888

rw

W , Clean

S wetw

rw

W , Clean

77

Flight Segment 17: M1 cl

0.370

cl tw

clW

cl

rw

W , Clean

S wetw C D0 w

2302.33 ft 2 0.0035

C DLW

0.0088

Class II Drag for Horizontal tail:

K sand

 RLE     c w Lh

0.00167  103 0.742 % 2.0

 X lam   c  w 

80 %

cl rw , M  0

cl

5.9014 rad 1 tw , M  0

f gaph

0.97

g

0.0 deg

w

C Dgap flap

0.0002

78

Flight Segment 1: (N/A Aerodynamically)

Flight Segment 2: 0.177

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft 2 0.0010

C D0 h

610.32 ft2 0.0006

C DLh

N/A

C DLh

0.0060

M1 cl

0.037

rh

Flight Segment 3: M1 cl

0.369

rh

Flight Segment 4: 0.375

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0005

C D0 h

610.32 ft2 0.0006

C DLh

0.0016

C DLh

0.0002

rh

Flight Segment 5:

rh

Flight Segment 6: 0.556

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0007

C D0 h

610.32 ft2 0.0006

C DLh

0.0010

C DLh

0.0000

M1 cl

0.674

rh

rh

79

Flight Segment 7:

Flight Segment 8: 0.532

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0007

C D0 h

610.32 ft2 0.0006

C DLh

0.0002

C DLh

0.0000

M1 cl

0.801

rh

Flight Segment 9:

rh

Flight Segment 10: M1 cl

0.119

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0006

C D0 h

610.32 ft2 0.0006

C DLh

0.0024

C DLh

0.0011

M1 cl

0.207

cl th

rh

1

Flight Segment 11:

rh

Flight Segment 12: 0.295

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0010

C D0 h

610.32 ft2 0.0006

C DLh

N/A

C DLh

0.0005

M1 cl

0.037

rh

rh

80

Flight Segment 13:

Flight Segment 14: 0.801

M1 cl

cl th

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0005

C D0 h

610.32 ft2 0.0007

C DLh

0.0018

C DLh

0.0010

M1 cl

0.374

rh

Flight Segment 15:

rh

Flight Segment 16: M1 cl

0.200

cl th

clh

clh

cLh ,Clean

cLh ,Clean

S weth

S weth

C D0 h

610.32 ft2 0.0005

C D0 h

610.32 ft2 0.0006

C DLh

0.0003

C DLh

0.0000

M1 cl

0.374

cl th

rh

1

rh

Flight Segment 17: M1 cl

0.370

cl th

clh

cLh ,Clean

S weth C D0 h

610.32 ft2 0.0005

C DLh

0.0001

rh

81

Class II Drag for Vertical tail:

K sand

 RLE     c w Lh

0.00167  103 1.100 % 1.2

 X lam   c  w 

50 %

cl

cl

rw , M  0

5.6339 rad 1 tw , M  0

f gapv

0.95

C DgapRudder

0.0002

ARV ,eff

2.00

Flight Segment 1:

Flight Segment 2: 0.177

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

1227.00 ft 2

S wetV C D0V

1227.00 ft 2 0.0019

C DYV

0.0000

M1 cl

0.037

rV

C D0V C DYv

rV

82

Flight Segment 3: M1 cl

rV

Flight Segment 4:

cl tV

clV

c yv 

S wetV C D0V

1227.00 ft 2 0.0017

C DYV

0.0000

Flight Segment 5:

M1

0.375

cl rV

cl tV

clV

c yv 

S wetV

1227.00 ft 2

C D0V

0.0014

C DYV

0.0000

Flight Segment 6: 0.556

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

S wetV

C D0V

1227.00 ft 2 0.0025

C D0V

1227.00 ft 2 0.0023

C DYV

0.0000

C DYV

0.0000

M1 cl

0.674

rV

Flight Segment 7:

rV

Flight Segment 8: 0.532

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

S wetV

C D0V

1227.00 ft 2 0.0012

C D0V

1227.00 ft 2 0.0022

C DYV

0.0000

C DYV

0.0000

M1 cl

0.801

rV

rV

83

Flight Segment 9:

Flight Segment 10: 0.199

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

S wetV

C D0V

1227.00 ft 2 0.0023

C D0V

1227.00 ft 2 0.0017

C DYV

0.0000

C DYV

0.0000

M1 cl

0.207

rV

Flight Segment 11:

rV

Flight Segment 12: 0.295

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

1227.00 ft 2

S wetV C D0V

1227.00 ft 2 0.0018

C DYV

C DYV

0.0000

Flight Segment 13:

Flight Segment 14:

M1 cl

0.037

rV

C D0V

rV

0.801

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

S wetV

C D0V

1227.00 ft 2 0.0022

C D0V

1227.00 ft 2 0.0018

C DYV

0.0000

C DYV

0.0000

M1 cl

0.374

rV

rV

84

Flight Segment 15:

Flight Segment 16: 0.200

M1 cl

cl tV

cl tV

clV

clV

c yv 

c yv 

S wetV

S wetV

C D0V

1227.00 ft 2 0.0022

C D0V

1227.00 ft 2 0.0023

C DYV

0.0000

C DYV

0.0000

M1 cl

0.374

rV

rV

Flight Segment 17: M1 cl

0.370

cl tV

clV

c yv 

S wetV C D0V

1227.00 ft 2 0.0021

C DYV

0.0000

rV

Fuselage Drag:

Theory: The subsonic fuselage zero-lift drag coefficient is found from: C

D0 fus

C

D0 fus  base

C

Db fus

   Lf 60 C  R  1  0.0025 3 D0 fus  base wf df   Lf        df 

   X   

Eqn. (1)

The intermediate calculation parameter X can be expanded into the following equation:

85

 C   f fus Lam X   

S C f C f S fusTurb Wet fus FusTurb  Wet Fus Lam SW

    

Eqn. (2)

The wing-fuselage interference factor is found from Figure 4.1 in Airplane Design Part VI as a function of the Reynold's number based on the fuselage length and the steady state flight Mach number:

R  f RN ,M Wf fus

The fuselage base drag coefficient is determined from:

C

 db 0.029 d f 

Db fus

 S  Sf 

C D0

   

3

fusbase

Sf

   

1

Eqn. (3)

2 SW

The fuselage base diameter is found from: Sb

d  2. 0 b

d

f

f

Eqn. (4)

 2.0

Sf

Eqn. (5)

The fuselage drag coefficient due to lift is found from: C

DL



 2 2

Sb

f

SW

fus

C L C L 1 0

 C  3 dc

S Plf

f

Eqn. (6)

SW

Eqn. (7)

CL

The ratio of the drag coefficient of the finite cylinder to the drag coefficient of the infinite cylinder is obtained from Figure 4.19 in Airplane Design Part VI (J. Roskam) as a function of the fuselage length and the fuselage maximum diameter at the wing-fuselage intersection:

  f L f ,d f

 86

The steady state cross-flow drag coefficient is found from Figure 4.20 in Airplane Design Part VI (J.Roskam) as a function of the steady state flight Mach number and the airplane angle of attack: C

dc

 f M1,

The transonic fuselage zero-lift drag coefficient is found from:

C D0 f  Rw f C D f

 C D P f  C D b f  C D wave f

S f max

Sw The wing fuselage interference factor is based on the fuselage length and steady state flight Mach number: Rw f  f R N , M 1 The fuselage skin friction drag coefficient is: S wet CD f f  C f Sw The fuselage base pressure drag coefficient at M=0.6 is:      L f  S wet 60 . C D p f @ M 0.6  C f @ M 0.6   0.0025  3 D S   f w   Lf    D    f  The fuselage base drag coefficient is found using the relation: C D b f  f C D b f @ M 0.6 , d b , D f , M 1

f

f

f

f

f

f

max w

max w

max w

The fuselage wave drag coefficient is found through the relation: C D wave f  f L f , D f max w , M 1

The transonic fuselage drag coefficient due to lift is found from: Sb 3 S plf f C D L f  2 2 f  cd c  Sw Sw

 K sand  SWet K instal

Class II Drag for Fuslage:

0.00167  103 -2.60deg 1387.86 ft 2 0.00

87

Segment 1:

M1 C D0 f

0.037

C DL f

Segment 2: 0.177

N/A

M1 C D0 f

N/A

C DL f

Segment 4:

M1 C D0 f

0.375

C DL f

0.369

0.0046

M1 C D0 f

0.001

C DL f

0.0000

Segment 5:

0.0068 0.0002

C DL f

0.0000

0.0012 0.0000

C DL f

0.801

C DL f

Segment 8:

0.0013 0.0000

C DL f

0.0000

0.0068 0.0000

C DL f

0.199

C DL f

Segment 11:

N/A N/A

C DL f

0.0000

0.0011 0.0000

C DL f

0.374

C DL f

Segment 12: 0.295

0.037

M1 C D0 f

0.0054

M1 C D0 f

M1 C D0 f

Segment 13:

Segment 9: 0.207

0.532

M1 C D0 f

0.0062

M1 C D0 f

M1 C D0 f

Segment 10:

Segment 6: 0.556

0.674

M1 C D0 f

0.0058

M1 C D0 f

M1 C D0 f

Segment 7:

Segment 3:

Segment 14:

0.0012

Segment 15: 0.374

0.0069

M1 C D0 f

0.0000

C DL f

0.0000

0.801

0.0012

M1 C D0 f

0.0001

C DL f

0.0012

88

Segment 16:

Segment 17:

M1 C D0 f

0.200

C DL f

0.370

0.0014

M1 C D0 f

0.0015

C DL f

0.0000

0.0012

Trailing edge flap Drag:

The drag coefficient due to flap deflection is estimated from: C

D flap

 C

 C

D prof

flap

Di

 C flap

D int

Eqn. (1) flap

The flap profile drag increment is found from: C

D prof

 C flap

dP C

Cos 4W

SW f C

Eqn. (2)

4W SW

The induced flap drag increment is calculated from: C

Di flap

 K 2 C

L

2Cos f

C

Eqn. (3) 4W

The induced drag factor follows from Figures 4.52 and 4.53 in Airplane Design Part VI and is a function of the wing aspect ratio and the inboard and outboard flap stations:   K  f  AR ,  ,   W i o   f f  

The interference drag increment due to flap deflection is estimated from: C

K

Dint flap

int

K

int

C

D prof

flap

Eqn. (4)

 0.00

clrW , M 0

cltW , M 0

 0r

 0t

-2.8 deg.

f gapW

0.99

W

W

89

f

25.0 deg

Cf

20 %

Cw

Segment 10:

Segment 16:

0.199 -2.8 deg

0

rW

-2.8 deg

0

tW

 0,Clean

-1.5 deg

 0,Clean

-1.5 deg

W

-4.0 deg

W

-3.6 deg

cl rw

cl rw

cl tw

cl tw

clW

clW

C LW ,Clean

C LW ,Clean

C LW

C LW

C LW 0 , f

0.2115

C LW 0 , f

0.1751

C LW , f

0.2232

C LW , f

0.1959

Sf

0.256

Sf

0.256

M1

0 0

rW

tW

0

SW

M1

0

0.200 -2.8 deg -2.8 deg

SW

C D0 f

0.0143

C D0 f

0.0073

CDf

0.0192

CDf

0.0110

Landing gear Drag:

Theory: The retractable landing gear drag coefficient is computed from:

CDretract

2    1.5      1   CL1  CL0 w ,f  S S      wf w   i CDbasic 1.0  0.04 Locationi   i L c strut gi            

Eqn.

(1)

90

The basic undercarriage drag coefficient is found from:

C Dbasic ,i ď&#x20AC;˝

1.5S FrontTirei ď&#x20AC;Ť 0.75S RearTire i

Eqn. (2)

SW

Gear

S Front ft 2 Nose Gear 4.097 Main Gear 8.860 Main Gear 8.860 Segment 1:

S Rear ft 2 4.097 8.860 8.860

Lstrut g

4.3900 5.8900 5.8900

Segment 2:

C L1

N/A

C L1

0.4204

C Dretract

N/A

CDretract

0.0354

Segment 10:

C L1

1.2807

CDretract

0.0354

Segment 16:

C L1

2.0528

CDretract

0.0354

Aileron Sizing: Theory: Steps taken in order to size appropriate aileron for this aircraft are as follows: 1-Type of the aircraft has been determined: This Aircraft could be classified as FAR-23 Type I aircraft (Weight is less than 12000 lb) 2- Critical flight segment should have been determined: Based on trade studies, Due to flap deflection and usual weight consideration Take off segment have been selected. (Phase A) 3- Rolling Time Constant TR should be determined based on type and critical flight phase:

91

TR = 1 Sec . Eqn. (1) 4-Calculating Rolling moment due to roll rate derivative C lP :

 

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

Cl = Cl P

PW

+ Cl

Eqn. (2) PV

The wing contribution to roll damping is given by:

C lP = w

βC l P K

C L =0

( ) (C ) ( ) (C )

C Lα w K . β C Lα W

lP Γ

CL

C L =0

l P Γ = 0

(

+ ΔC l P

)

drag

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 Prandtl-Glauert transformation factor, the sectional lift curve slope of the lifting surface, the lifting surface quarter chord sweep angle, and the lifting surface taper ratio:

βC l P K

= f ARW , β , Λ C C L =0

4W

, λW

The ratio of the incompressible sectional lift curve slope of the lifting surface to 2p is defined as: C  C   l   l  X M X  M 0    K 2 2

Eqn. (4)

  1 M 2

Eqn. (5)

 The Prandtl-Glauert transformation factor is given by:

The dihedral effect parameter is found from:

C  C 

lP 

l P  0

4Z Z  1  W Sin  12 W bW  bW

2

  Sin 2 

Eqn. (6)

92

The drag contribution is determined from:

C C



C 

lP C DL

l P drag w



C

2 LW

C 

lP C DL

l P drag h

C

2 Lh

C L2 h  0.125C D 0h Eqn. (7)

C L2 h  0.125C D 0h

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:

C 

lP C DL C LW 2

 f  ARW ,  C  4W  

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

lP V

 ZV   2   bW 

2 C Y

Eqn. (8) V

For Segment 1: M1 C L ,Clean

0.037 0.6678

C l rw

C l tw

C l

w

C l

rh

C l

th

C l

h

X acV

118.79 ft

Z acV

15.69 ft

ARveff

2.97

93

C l

rv

C l

tv

C l

V

C yv , 

x

c

d

v

d

C y v

C lPw

ClP ,h

C lP ,V

ClP

5- Assuming the most critical condition in take off and assuming speed to be 80 kts. , Roll Angular Acceleration Imparted to the Airplane as a Result of a Unit Change in Roll Rate due to wing, horizontal tail and vertical tail has been calculated from: LP

l .S

C LP qSb 2 2 I xx , M V

Eqn. (9) l .S

LP  LPW  LPv  LPh LPW

-38.72 sec 1

LPh

-0.648 sec 1

LPV

-0.134 sec 1

LP

-39.50 sec 1

Eqn. (10)

6-Calculating TR and comparison with requirement:

1 Eqn. (11) LP TR  0.0253 sec .  1.4 sec . Eqn. (12) 7- Determining required time and rolling angle during the qualification bank maneuver: TR 

Based on FAR-23:

94

Aircraft type I I I

Flight Phase A B C

Level I 60º In 1.3 Sec. 60º In 1.7 Sec. 30º In 1.3 Sec.

Level II 60º In 1.7 Sec. 60º In 2.5 Sec. 30º In 1.8 Sec.

Level III 60º In 2.6 Sec. 60º In 3.4 Sec. 30º In 2.6 Sec.

Aiming for the level I flight qualities:

t = 1.3Sec .

Eqn. (13)

φ = 30  = 0.5236 rad .

8- Calculating the minimum required rolling moment to satisfy the flight quality requirements: Since: La 



 L

a

   t   1  2  LP t  L    L  . e  1 P  P    

 a t

LP

L a  a 2 P

L

e

LP .t

1

Eqn. (14)

Eqn. (15)

9- Choosing Aileron Parameters based on wing design and statistic:

 a  20   0.349rad . a a

i

a

O

20º=0.349 rad 0.43 set  0.60

10- Solving Eqn. 15 For L a : L a  30.98 rad 1

using graphs 11- Calculating C l′ δ Clδ 12- Determining graphically Clδ Theory

13- Determining C l δ 14 – Calculating C l α

Theory

graphically

a

95

15- Calculating required Aileron Deflection: δa =

Cl δ Cl α

a

 a  0.48 rad 16- Calculating C L Graphically 17- Choosing the gearing ratio (Based on statistics): Ga  0.85

18- Calculating roll control power due to aileron deflection derivative C la : q1

20.45 lb

cl rw

cl tw

a

17.0 deg

l

ft 2

a

18.5 deg

f bala

0.85

clal

0.0215

clar

0.0107

cla

cl ar

cl a 0

cl a

19- Designed aileron rolling moment has been calculated:

La 

 .V 2 SbCl 2 I xxM

a

Eqn. (16)

96

27.16<58.231

La La  L a Desired,

Due to wing design, Outboard station of aileron could be changed to reach the desire level of rolling moment, increasing the produced rolling moment by the aileron surface. A MATLAB code has been developed to iterate the method and solve for outboard station ratio for the threshold of 0.001 rad 1 Result:

 O  0.89 Theory: C l a 

C l al  aL  C l ar  ar

a

Eqn. (17)

The aileron deflection angle of the airplane is given by:    aR  a  aL Eqn. (18) 2 The rolling-moment-coefficient-due-to-left-aileron-deflection derivative is calculated from:

C l al 

1 f bala C l   al 2

Eqn. (19)

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

C l ar 

1 f bala C l   ar 2

Eqn. (20)

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. (21)

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

S1 S2

Eqn. (22)

The rolling effectiveness of two full-chord ailerons is determined from:

97

C l 

k  C l  k

Eqn. (23)

The Prandtl-Glauert transformation factor is derived from:

  1  M 12

Eqn. (24)

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

k

a

, M 0

Eqn. (25)

2 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: Cl  Cl   Cl     Eqn. (26)     k  k k i  O 

Where:

 C l   k   C l   k 

   f  Oa , ARw ,  ,   ,  w , k  O    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. (27)

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

  al

c  l

cl a

l

Eqn. (28)

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

98

 

c  l

r

Eqn. (29)

cl a

ar

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. (30)

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

l

cl cl

c 

l Theory

k l

Eqn. (31)

Theory

The lift effectiveness of the Right aileron is given by:

cl

r

cl cl

c 

l Theory

k r

Eqn. (32)

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. (33)

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

Theory

t   2  5.0525   c w

Eqn. (34)

99

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 

c  l

theory

Eqn. (35)

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 k l  f  a ,  al    cw  Trim Drag : Theory:

c k r  f  a ,  ar  cw

  

The total trim drag coefficient is determined from:

C Dtrim  C Dtrim prof  C De  C Dcv  C Drv  C Dr  C Del

Eqn. (1)

The trim drag coefficient due to the profile drag is found by treating the control surface as a plain flap: C Dtrim

Pr of

 C DP

 cos  e

C

4

h

Se Sh Sh Sw

Eqn. (2)

The profile drag coefficient due to the lifting surface control surface at  C  0.0  may be 4

found from Figure 4.44 in Airplane Design Part VI by J. Roskam.

C 

DP C . S

 C  f   Control , Control  C surface 

   

The horizontal tail flapped area is defined in Figure 4.72 in Airplane Design Part VI:

S e f  f  ie , Oc ,  h

The vertical tail flapped area is defined in Figure 4.72 in Airplane Design Part VI:

S r f  f  iv , Or , v r

The airplane drag-coefficient-due-to-control surface-deflection is found from:

100

C DC . S .  C DP

Segment 1: N/A 0.0000 CD e

cos C c.s .

4 l .s

S c.s. f S l .s . S l .s. S w

Eqn. (3)

Segment 2:

Segment 3:

C De

0.0029

C De

0.0027

C Del

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0000

C Da

0.0046

C Da

0.0000

C Dr

0.0000

C Dr

0.0000

C Dr

0.0000

C DTrim

0.0000

C DTrim

0.0075

C DTrim

0.0027

Segment 4:

Segment 5:

Segment 6:

C De

0.0016

C De

0.0000

C De

0.0021

C Del

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0090

C Da

0.0000

C Da

0.0000

C Dr

0.0014

C Dr

0.0006

C Dr

0.0000

C DTrim

0.0016

C DTrim

0.0000

C DTrim

0.0021

Segment 7:

Segment 8

Segment 9:

C De

0.0016

C De

0.0000

C De

0.0021

C Del

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0090

C Da

0.0000

C Da

0.0000

C Dr

0.0014

C Dr

0.0006

C Dr

0.0000

C DTrim

0.0016

C DTrim

0.0000

C DTrim

0.0021

Segment 10:

Segment 11:

Segment 12:

C De

0.0016

C De

0.0000

C De

0.0021

C Del

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0090

C Da

0.0000

C Da

0.0000

C Dr

0.0014

C Dr

0.0006

C Dr

0.0000

C DTrim

0.0016

C DTrim

0.0000

C DTrim

0.0021

101

Segment 13:

Segment 14:

Segment 15:

C De

0.0016

C De

0.0000

C De

0.0021

C Del

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0090

C Da

0.0000

C Da

0.0000

C Dr

0.0014

C Dr

0.0006

C Dr

0.0000

C DTrim

0.0016

C DTrim

0.0000

C DTrim

0.0021

Segment 16:

Segment 17:

C De

0.0016

C De

0.0000

C Del

0.0000

C Del

0.0000

C Da

0.0090

C Da

0.0000

C Dr

0.0014

C Dr

0.0006

C DTrim

0.0016

C DTrim

0.0000

Total Drag Flight Segment 1 (N/A) 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

L

D ----4.85 16.44 24.60 9.33 16.34 16.18 19.00 26.34 14.19

CD

CL

----0.0867 0.0357 0.0225 0.0697 0.0288 0.0312 0.0202 0.0618 0.0902

---0.4204 0.5864 0.5536 0.6502 0.4708 0.5056 0.3841 0.0391 1.2807

8.77 13.78 14.39 18.56 7.15 22.80

0.0732 0.0424 0.0318 0.0216 0.2870 0.0179

0.6415 0.5836 0.4578 0.4012 2.0528 0.4074

102

Airplane Lift Coefficient at Zero Angle of Attack:

The airplane zero-angle-of-attack lift coefficient including flap effects is obtained from: Theory:

C L0 Wf h C C C L0 L0 L0 Clean Wf Clean h

C

C C

C

L0

L0 L0

L0

C

L0

Wf

K

Wf Clean

  Wf Clean

CW

C

0 AClean

L

C

L0

W

Wf



C L 0A Clean Wf   i W 0A 0W Clean

0A Clean

   0 l  W

0W

      0 l W   t        0   l W  M  0.3

  0W     t

  0W    Source : Airplane designVI Fig 8.41    f  , AR ,    t   W W C 4 W       0l t  t    W M  f  M,  C ,  ,   4 W  c r  c t      0l  W  M  0.3



0A

0W

0 

C

0W

L0

0h

i

W

W

CL

W

CL  C

h 

 h 

L

 h

0A

h

W

Sh SW

 

180

C L

0W

 0 A  0 A 

hf

103



hf

 f h h , b , AR , b f , C L f

Source : Airplane design VI Fig8.70

Airplane Lift Coefficient at Zero Angle of Attack:

Segment 1:

0

0

Segment 2:

0

clean

0

clean

0

C L0

C L0

hf

Segment 3: -4.9 º

0

-7.2º

0

-0.0255

C L0

hf

-4.9

clean

-5.9º -0.0112 hf

C L0 ,f

C L0 ,f

0.2373

C L0 ,f

0.1088

C L0 h

C L0 h

-0.0103

C L0 h

0.0036

C L0Clean , P .O

C L0Clean , P .O

0.4664

C L0Clean , P .O

0.4776

C L0 P .OFF

C L0 P .OFF

0.7036

C L0 P .OFF

0.5863

C L0Clean

C L0Clean

0.4664

C L0Clean

0.4776

C L0

C L0

0.7036

C L0

0.5863

Segment 4:

0

clean

0 C L0

Segment 5: -4.0º

0

-4.0º

0

0.0000

C L0

hf

clean

Segment 6: -4.1 º

0

-4.1 º

0

0.0000

C L0

hf

-4.9º

clean

-4.9º 0.0000 hf

C L0 ,f

0.0000

C L0 ,f

0.0000

C L0 ,f

0.0000

C L0 h

0.0150

C L0 h

0.0126

C L0 h

0.0139

C L0Clean , P .O

0.4212

C L0Clean , P .O

0.4875

C L0Clean , P .O

0.5095

C L0 P .OFF

0.4212

C L0 P .OFF

0.4875

C L0 P .OFF

0.5095

C L0Clean

0.4212

C L0Clean

0.4875

C L0Clean

0.5095

C L0

0.4212

C L0

0.4875

C L0

0.5095

104

Segment 7:

0

clean

0 C L0

Segment 8: -4.0 º

0

-4.0 º

0

0.0000

C L0

hf

clean

Segment 9: -4.9º

0

-4.9º

0

0.0000

C L0

hf

-4.9º

clean

-7.8º -0.0322 hf

C L0 ,f

0.0000

C L0 ,f

0.0000

C L0 ,f

-0.3015

C L0 h

0.0157

C L0 h

0.0141

C L0 h

-0.0172

C L0Clean , P .O

0.4634

C L0Clean , P .O

0.5042

C L0Clean , P .O

0.4696

C L0 P .OFF

0.4634

C L0 P .OFF

0.5042

C L0 P .OFF

0.7710

C L0Clean

0.4634

C L0Clean

0.5042

C L0Clean

0.4696

C L0

0.4634

C L0

0.5042

C L0

0.7710

Segment 10:

0

clean

0 C L0

Segment 11:

-4.9º

0

-6.8º

0

-0.0211

C L0

hf

clean

Segment 12: -3.9º

0

-3.9º

0

0.0000

C L0

hf

-4.9º

clean

-7.5º -0.0291 hf

C L0 ,f

-0.1965

C L0 ,f

0.0000

C L0 ,f

0.2701

C L0 h

-0.0059

C L0 h

0.0156

C L0 h

-0.0140

C L0Clean , P .O

0.4665

C L0Clean , P .O

0.4007

C L0Clean , P .O

0.4696

C L0 P .OFF

0.6630

C L0 P .OFF

0.4007

C L0 P .OFF

0.7398

C L0Clean

0.4665

C L0Clean

0.4007

C L0Clean

0.4696

C L0

0.6630

C L0

0.4007

C L0

0.7398

Segment 13:

0

clean

0 C L0

Segment 14:

-4.9º

0

-4.9º

0

0.0000

C L0

hf

clean

Segment 15: -4.0º

0

-4.0º

0

0.0000

C L0

hf

-4.0º

clean

-4.0º 0.0000 hf

C L0 ,f

0.0000

C L0 ,f

0.0000

C L0 ,f

0.0000

C L0 h

0.0149

C L0 h

0.0158

C L0 h

0.0150

C L0Clean , P .O

0.4781

C L0Clean , P .O

0.4718

C L0Clean , P .O

0.4210

C L0 P .OFF

0.4781

C L0 P .OFF

0.4718

C L0 P .OFF

0.4210

C L0Clean

0.4781

C L0Clean

0.4718

C L0Clean

0.4210

C L0

0.4781

C L0

0.4718

C L0

0.4210

105

Segment 16:

0

clean

0 C L0

Segment 17:

-4.9 º

0

-6.8º

0

-0.0209

C L0

hf

-4.9º

clean

-4.9º 0.0000 hf

C L0 ,f

0.1975

C L0 ,f

0.0000

C L0 h

-0.0055

C L0 h

0.0150

C L0Clean , P .O

0.4624

C L0Clean , P .O

0.4779

C L0 P .OFF

0.6599

C L0 P .OFF

0.4779

C L0Clean

0.4624

C L0Clean

0.4779

C L0

0.6599

C L0

0.4779

Zero lift pitching moment coefficient:

Theory: The wing-fuselage zero-lift pitching moment coefficient is given by: C

m0

C

m0 Wf

C

Eqn. (1)

m0 h

The fuselage zero-lift pitching moment coefficient is solved from:

C

m0

 f

K 2  K 1  i n W

36.5SW C W

 i 1  

fi

  2

0

W

 i W  i Cl fi

C X  m 0 M   i  C m

Eqn. (2)

0 M 0

The fuselage slenderness effect is obtained from Figure 8.111 in Airplane Design Part VI and is a function of the fuselage fineness ratio:

lf   Source : Airplain designVI Fig 8.111 df   

K 2  K 1   f 

The fuselage segment geometry is illustrated in Figure 8.112 in Airplane Design Part VI.

The effect of the Mach number on the fuselage zero-lift pitching moment coefficient is found from Figure 8.99 in Airplane Design Part VI and is a function of the steady state flight Mach number:

106

 C m 0  Cm

  M

 

 f M

0 M 0

The wing zero-lift pitching moment coefficient contribution to the zero-angle-of-attack pitching moment coefficient is determined from: 2

C

m0

W

  ARW  Cos C   C m0 4W   r    ARW  2Cos C 4W 

C m

0t

2

  C m 0    t  

  t  

Eqn. (3)

The linear twist effect on the wing zero-lift pitching moment coefficient is found from Figure 8.98 in Airplane Design Part VI and is a function of the wing taper ratio, wing quarter chord sweep angle, and the wing aspect ratio:

C m

0  f  , W C , ARW 4 t

The horizontal tail contribution to airplane zero-lift pitching moment coefficient is determined from: C

m0

 C h

L

 h

h

Sh SW

.

X AC  X CG h  0h CW

Eqn. (4)

The horizontal tail zero-lift pitching moment coefficient contribution to the zero-angle-of-attack pitching moment coefficient is determined from:

c mh0

2     AR h  cos  c 4   h     AR h  2 cos  c 4 h 

 c m0rh  c m0th   2 

  C m 0       th

   C m 0 M  t   h C   m 0 M 0 

Eqn. (5)

The linear twist effect on the horizontal tail zero-lift pitching moment coefficient is found from Figure 8.98 in Airplane Design Part VI and is a function of the horizontal tail taper ratio, horizontal tail quarter chord sweep angle, and the horizontal tail aspect ratio:

Zero lift pitching moment coefficient:

107

Segment 1:

Segment 2: 0.037

Segment 3: 0.117

0.124

M1 cl , rw

cl ,tw

cl ,tw

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C LW ,clean

C Lw0f

0.2795

C Lw0f

0.2790

C Lw0f

0.0000

C mW ref

0.8243

C mW ref

0.8224

C mW ref

0.0000

C m W

-0.1308

C m W

-0.1668

C m W

0.0000

M1 cl , rw

M1 cl , rw

cl ,tw

cl ,W

TE

Segment 4:

TE

Segment 5: 0.235

TE

Segment 6: 0.599

0.233

M1 cl , rw

cl ,tw

cl ,tw

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C LW ,clean

C Lw0f

0.1839

C Lw0f

0.0000

C Lw0f

0.0000

C mW ref

0.5411

C mW ref

0.0000

C mW ref

0.0000

C m W

-0.1098

C m W

0.0000

C m W

0.0000

M1 cl , rw

M1 cl , rw

cl ,tw

cl ,W

TE

Segment 7:

TE

Segment 8: 0.235

TE

Segment 9: 0.599

0.233

M1 cl , rw

cl ,tw

cl ,tw

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C LW ,clean

C Lw0f

0.1839

C Lw0f

0.0000

C Lw0f

0.0000

C mW ref

0.5411

C mW ref

0.0000

C mW ref

0.0000

C m W

-0.1098

C m W

0.0000

C m W

0.0000

M1 cl , rw

M1 cl , rw

cl ,tw

cl ,W

TE

TE

TE

108

Segment 10: 0.235

Segment 11: 0.599

Segment 12: 0.233

M1 cl , rw

cl ,tw

cl ,tw

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C LW ,clean

C Lw0f

0.1839

C Lw0f

0.0000

C Lw0f

0.0000

C mW ref

0.5411

C mW ref

0.0000

C mW ref

0.0000

C m W

-0.1098

C m W

0.0000

C m W

0.0000

M1 cl , rw

M1 cl , rw

cl ,tw

cl ,W

TE

Segment 13: 0.235

TE

Segment 14: 0.599

TE

Segment 15: 0.233

M1 cl , rw

cl ,tw

cl ,tw

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C LW ,clean

C Lw0f

0.1839

C Lw0f

0.0000

C Lw0f

0.0000

C mW ref

0.5411

C mW ref

0.0000

C mW ref

0.0000

C m W

-0.1098

C m W

0.0000

C m W

0.0000

M1 cl , rw

M1 cl , rw

cl ,tw

cl ,W

TE

Segment 16:

TE

Segment 17:

 MC1 m W TE cl , rw

M1 cl , rw

cl ,tw

clC,twm W

cl ,W

cl ,W

C LW ,clean

C LW ,clean

C Lw0f

0.1839

C Lw0f

0.0000

C mW ref

0.5411

C mW ref

0.0000

-0.1098 0.235

TE

109

Total Moment Coefficient:

Theory: The airplane zero-angle-of-attack pitching moment coefficient without power effects is estimated from: C m 0  C m 0wf  C m 0h

Eqn. (1)

The wing-fuselage contribution is determined from:  X cg  X ac wf C m 0 wf  C L 0wf ,Clean  cw 

  Cm  C m w ,TE 0 wf , Clean  

Eqn. (2)

The horizontal tail contribution to the zero-angle-of-attack pitching moment coefficient is calculated from:  X cg  X ac h C m 0 ,h  C L 0 ,h  cw 

   C m 0 h 

Eqn. (3)

The wing-fuselage zero-lift pitching moment coefficient including flap effects is found from:  X cg  X ac wf C m 0 ,Wf  C m 0 ,WF  C L 0 ,WF   cw  

   

Eqn. (4)

Total Moment Coefficient:

Segment 1:

Segment 2:

Segment 3:

C m0 , wf ,Clean

-0.0469

C m0 , wf ,Clean

-0.0483

C m0 , wf

C m0 , wf

-0.1075

C m0 , wf

-0.0846

C m0 , wf

C m0 , wf

-0.0088

C m0 , wf

-0.0281

C m0 , h

C m0 , h

0.0289

C m0 , h

-0.0167

C m0

C m0

0.0425

C m0

-0.0213

C m0 , wf ,Clean

Segment 4:

-0.0466

Segment 5:

Segment 6:

110

C m0 , wf ,Clean

-0.0483

C m0 , wf ,Clean

-0.0544

C m0 , wf ,Clean

-0.0513

C m0 , wf

-0.0396

C m0 , wf

-0.0454

C m0 , wf

-0.0513

C m0 , wf

-0.0268

C m0 , wf

-0.0619

C m0 , wf

-0.0035

C m0 , h

-0.0529

C m0 , h

-0.0467

C m0 , h

-0.0506

C m0

0.0576

C m0

-0.0457

C m0

-0.0288

Segment 7:

Segment 8:

Segment 9:

C m0 , wf ,Clean

-0.0597

C m0 , wf ,Clean

-0.0508

C m0 , wf ,Clean

-0.0474

C m0 , wf

-0.0597

C m0 , wf

-0.0442

C m0 , wf

-0.1708

C m0 , wf

-0.0005

C m0 , wf

-0.0300

C m0 , wf

-0.1472

C m0 , h

-0.0570

C m0 , h

-0.0520

C m0 , h

0.0503

C m0

-0.0501

C m0

-0.0563

C m0

-0.0704

Segment 10:

Segment 11:

Segment 12:

C m0 , wf ,Clean

-0.0474

C m0 , wf

C m0 , wf

-0.1476

-0.0810

C m0 , wf

C m0 , wf

-0.0966

C m0 , h

0.0152

C m0 , h

C m0 , h

0.0418

C m0

-0.0421

C m0

C m0

-0.0314

C m0 , wf ,Clean

-0.0469

C m0 , wf ,Clean

C m0 , wf

-0.0995

C m0 , wf

Segment 13:

-0.0466

-0.0570

Segment 14:

Segment 15:

C m0 , wf ,Clean

-0.0483

C m0 , wf ,Clean

-0.0597

C m0 , wf ,Clean

-0.0483

C m0 , wf

-0.0483

C m0 , wf

-0.0597

C m0 , wf

-0.0441

C m0 , wf

-0.0043

C m0 , wf

-0.0213

C m0 , wf

-0.1009

C m0 , h

-0.0536

C m0 , h

-0.0591

C m0 , h

-0.0551

C m0

-0.0344

C m0

-0.0565

C m0

-0.0087

Segment 16:

Segment 17:

111

C m0 , wf ,Clean

-0.0466

C m0 , wf ,Clean

-0.0483

C m0 , wf

-0.1212

C m0 , wf

-0.0483

C m0 , wf

-0.1056

C m0 , wf

-0.1020

C m0 , h

0.0141

C m0 , h

-0.0581

C m0

-0.0680

C m0

-0.1338

Airplane Pitching Moment Curve Slope:

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

C m , P .OFF  C LWF xcg  x acwf  C m h

Eqn. (5)

Where:

C ml . s  C L ,l . s x acl . s  xcg

Segment 1:

C m

N/A

Segment 4:

C m

Segment 7:

C m

Segment 10:

C m

Segment 13:

Eqn. (6)

Segment 2:

C m

Segment 5:

C m

Segment 8:

C m

Segment 11:

C m

Segment 14:

Segment 3:

C m

Segment 6:

C m

Segment 9:

C m

Segment 12:

C m

Segment 15:

112

C m

Segment 16:

C m

C m

C m

Segment 17:

C m