Fokker100

Fokker100

Shashank Datthatreya

Table of Content Introduction ............................................................................................................................................ 1

Dimensions.......................................................................................................................................... 2

Landing Distance ................................................................................................................................. 2

Take-off Distance ................................................................................................................................. 3

Rate of climb in 2nd Segment................................................................................................................ 3

Rate of climb go-around ...................................................................................................................... 4

Cruise .................................................................................................................................................. 5

Design diagram .................................................................................................................................... 6

Maximum take-off mass 𝑚𝑀𝑇𝑂 .......................................................................................................... 7

Wing area and Takeoff-thrust .............................................................................................................. 9

Fuselage design ....................................................................................................................................... 9

Requirements ...................................................................................................................................... 9

Fuselage cross-section ......................................................................................................................... 9

Cabin Layout...................................................................................................................................... 11

Emergency exits ................................................................................................................................ 12

Cargo volume .................................................................................................................................... 13

Waterline .......................................................................................................................................... 14

Blades and high-lift devices ................................................................................................................... 15

Pre-arranged parameters .................................................................................................................. 15

Sweep ............................................................................................................................................... 15

Relative thickness profile ................................................................................................................... 15

Airfoil ................................................................................................................................................ 16

Escalation .......................................................................................................................................... 16

Fuel Capacity ..................................................................................................................................... 16

Twist ................................................................................................................................................. 17

V shape ............................................................................................................................................. 17

Setting angle...................................................................................................................................... 17

High-lift systems ................................................................................................................................ 17

Ailerons & Spoilers ............................................................................................................................ 19

Empennage design 1 ............................................................................................................................. 20

Tail-plane (CPR) ................................................................................................................................. 20

Rudder (SLW) .................................................................................................................................... 22

Elevator and Rudder .......................................................................................................................... 23

Weight and Balance............................................................................................................................... 24

Mass prediction Class-I( Raymer ) ...................................................................................................... 24

Mass prediction Class-II (Torenbeek) .................................................................................................. 25

CG calculation.................................................................................................................................... 26

Empennage design II ............................................................................................................................. 28

Tail-plane (CPR) ................................................................................................................................. 28

Rudder (SLW) .................................................................................................................................... 32

Chassis .................................................................................................................................................. 34

Number and arrangement of legs and wheels .................................................................................... 34

Positioning ........................................................................................................................................ 34

Tire selection ..................................................................................................................................... 37

LCN-value .......................................................................................................................................... 37

Determination of polar .......................................................................................................................... 38

Fuselage ............................................................................................................................................ 39

Wing.................................................................................................................................................. 40

Tail-plane .......................................................................................................................................... 41

Rudder .............................................................................................................................................. 42

Engine nacelles .................................................................................................................................. 43

Direct operating costs (Direct Operating Costs, DOC) ............................................................................. 44

Depreciation ...................................................................................................................................... 45

Interest.............................................................................................................................................. 47

Insurance........................................................................................................................................... 47

Fuel costs .......................................................................................................................................... 47

Maintenance Costs ............................................................................................................................ 49

Personnel costs ................................................................................................................................. 51

Fees ................................................................................................................................................... 51

Overall presentation .......................................................................................................................... 53

Summary ............................................................................................................................................... 54

References ............................................................................................................................................ 58

Abstract Based on the actual design of the existing jet airliner, here the Fokker-100 is taken into

consideration for the aircraft design. The project follows the aspect ratio or payload of the

original aircraft, including the JAR/FAR 25 requirements. The pre-dimensioning of the fuselage,

the wing, the tail and the landing gear, along with the determination of aircraft mass and center

of gravity are made. As a result, a 3-side view of the designed Fokker-100 is drawn with the

considered dimensions and key variables of the aircraft.

INDEX

B

Blades and high-lift devices · 20

C

Chassis · 39

Cruise · 10

D

Design diagram · 12

Determination of polar · 44

Dimensions · 7

Direct operating costs · 49

E

Empennage design 1 · 25

Empennage design II · 33

F

Fuselage design · 15

I

Introduction · 6

L

Landing Distance · 7

M

Maximum take-off mass · 12

R

Rate of climb go-around · 9

Rate of climb in 2nd

Segment · 8

References · 63

S

Summary · 59

T

Take-off Distance · 8

W

Weight and Balance · 29

Wing area and Takeoff-thrust · 14

1

Introduction The aim of this project is to design an existing aircraft; medium-range airliner “Fokker F100”. As

required, the actual performance of the F100 in the areas of payload, number of passengers,

range, Cruise Mach number, aspect ratio, and take-off and landing distance is used.

“Dragon” construction designed as a cantilever low-wing monoplane angled swept-back, the

machine is driven by two ZTL engines(turbofan) installed at the rear of the fuselage. Input

parameters are first dimensioned, the results on this basis help for a more detailed design that

is shown in the following pages and therefore will be modified. The results that are previously

concluded are checked against the current readings and, if necessary, adapted.

The sources and equations used in the text are added at the appropriate place by italics and

square brackets such as Author names, [Author names] , [Eq. 1.1].

2

Dimensions The preliminary sizing is taken care of, and that determines an aircraft design point in terms of

wing loading(mMTO/SW) and thrust-to-weight ratio ( TTO/(mMTO·g) ) or power-to-weight ratio (

PTO/(mMTO·g ) in case of propeller aircraft. The dimensioning is done essentially by the method

of Loftin.

Landing Distance From the specification of the safety landing distance JAR 25.125 and the Maximum Lift

coefficient results in a maximum value for the surface loading. The actual safe Landing Distance

of the original aircraft(Slift=1350m) serves as input value for the calculation.

The surface load at Maximum Landing mass

mML / Sw = kL CL,max,L sLFL

where = 1 and kL = 0.119(kg/m3)

The maximum achievable lift coefficient is according to various sources [Dubs, image5.4 ; Raymer, Figure 5.5] with the existing high-lift devices (double fowler flaps) and the wing sweep (φ25 = 17°) estimated:

CLmax = 2.7. This gives us

mML / Sw = 433.8(kg/m3)

Using this ratio of the maximum landing mass and total mass, determining the surface load

according to the equation.

mMTO / SW = (mML / SW ) / (mML/mMTO)

According to Rosakam [Figure 5.6] (for a civilian transport aircraft mML/mMTO = 0.65,..0.84,… 1.0)

and Loftin [Table 5.1] (for short-haul aircraft up to 3700 km range mML/mMTO = 0.91) we can

finalize on mML/mMTO = 0.87.

So, the Maximum Area Load is:

mMTO / SW = 499 kg/m2

3

Take-off Distance With the requirements for the safe Take-off Distance as well as the Maximum Lift coefficient

with Flaps in take-off position results in a minimum value for the Thrust-to-Weight ratio

depending on the wing loading.

According to Loftin, the following relationship can be assumed:

a = [TTO / (mMTO.g) ] / (mMTO / SW)

= kTO / (STOFL..CLmax.TO)

Here kTO = 2.34m3/kg, = 1825m (according to data of the original aircraft) and

CLmax.TO (0.8).CLmaxL = (0.8).2.7 2.2

The value for the ratio of Thrust-to-Weight ratio to Wing-loading is thus

a = 0.58281.10-3 (m2/kg)

Rate of climb in 2nd Segment The prescribed rate of climb in the 2nd Segment is followed by a minimum of Shear-Weight

ratio. According to JAR 25,121(b) a twin-engine aircraft will be in 2nd Segment after the

retraction of the landing gear, and despite a failed engine a climb can meet a gradient of 2.4%.

For this, a thrust-to-weight ratio (with both engines) of the least is required. N denotes the

number of engines (here N = 2), the sine of the angle of rise results from the required climb

gradient to sin 0.024. The Glide ratio L / D is estimated according to an approximation

method:

L/D = CL / [ CDp + ( CL2/ . A . e )]

The Oswald factor, e = 0.7 is in the extended flaps, wings surface of the Original aircraft A = 8.4

and the lift coefficient CL = CLmax,TO / 1.44 = 2.2 / 1.44 = 1.53 . The factor 1.44 is because the

climb in 2nd Segment with v2 = 1.2vS,TO is carried out,

Profile resistance can be estimated:

CDp = CD0 + CD flap + CD gear. CD0 0.02

4

The landing gear is retracted in 2nd Segment, i.e. CD0 = 0 . CD flap depends on the valve position

and thus of the lift coefficient. CL = 1.5, a flap angle of approximately 25 ° is necessary; it follows

CD flap = 0.02 and the Profile drag, CDp = 0.04 .Using this we obtain the glide ratio L / D = 9.2 and

therefore,

(TTO / mMTO.g ) = 2. [ (1/9.2) + 0.024] = 0.266

Rate of climb go-around The JAR 25,121 (d) requires climb gradient of 2.1% for two-engine airplanes if one engine goes

inoperative, this leads to a minimum value for the shear.

Weight ratio:

(TTO / mMTO.g ) = ( N / (N – 1) . (1/(L/D) + sin) . (mML / mMTO)

sin, this case is approximately equal to 0.021; mML / mMTO , which was mentioned in the

section "landing distance" as 0.87. The calculation works on the same pattern as the previous

one.

The lift coefficient is now at

( CLmaxL/ 1.69 ) = 2.7 / 1.69 = 1.6

(Because vMA = 1.3vS, L); it follows

CD flap = 0.025

For the procedure, FAR also is a landing gear that must be taken into account:

CD gear = 0.015

This is CDp = 0.065 and L / D = 7.86

Using these results, we conclude:

TTO / mMTO.g = 0.2578

5

Cruise The cruising analysis requires a given surface load to have a minimal necessary Thrust-Weight

ratio which is necessary to achieve the desired cruise Mach number. Wing loading and thrust-

to-weight ratio are first separated as a function of altitude calculated in the correlation.

Thrust-to-weight ratio [Eq. 5.27]:

(TTO / mMTO.g ) = 1 / [(TCR / T0).(L/D)max]

It is believed that the cruise is performed at maximum glide ratio and (L / D)cr = (L/D)max

The Maximum Glide ratio can be estimated by the method of Raymer:

(L/D)max = 7.5 . [A / (Swet / Sw) + 8]

Depending on the plane-form Swet / Sw that is approximated [image5,10], the value thus

obtained Swet / Sw = 6 and by the extension A = 8.4 we obtain (L / D)cr = 18.5 , TCR / T0 is

dependent on the flight altitude and the Bypass ratio (BPR) = 5, which is determined in the

graph.

Figure 2.1: Shear decrease with altitude

Now, the calculated values of Thrust-to-Weight ratio are used as a function of the height in the

table below.

Wing loading: Wing loading as a function of height is obtained from [Eq. 5:34]:

𝑚𝑀𝑇𝑂 𝑆𝑤 = 𝐶𝐿 . 𝑀2

𝑔. 2 . 𝑝(𝑕)

This is the adiabatic exponent = 1.4; the desired cruise Mach number M = 0.77; the

Buoyancy at ‘Maximum Glide’.

6

𝐶𝐿,𝑚 = 𝐶𝐷0. 𝜋. 𝐴. 𝑒 = 0.016 . 𝜋 . 8.4 . 0.85 = 0.61

and the pressure as a function of the amount of ISA-conditions p(h) as shown in Table 2.1

Table 2.1: T / W and m / S as a function of altitude

The figures for individual phases of flight limits 𝑚𝑀𝑇𝑂

𝑆𝑊 and

𝑇𝑇𝑂

𝑚𝑀𝑇𝑂 .𝑔 can be now drawn in a layout

diagram.

Design diagram

Figure 2.2: Design Graph

7

The Design point of the aircraft should now be positioned so that, in a possible low thrust-to-

weight ratio as high wing loading is allowed. It has the selection of the lowest possible thrust-

to-weight ratio priority. With these demands resulting from the design diagram of the Design

point with coordinates,

Wing loading 𝑚𝑀𝑇𝑂 𝑆𝑤 = 495𝑘𝑔/𝑚2

Thrust-weight ratio 𝑇𝑇𝑂 𝑚𝑀𝑇𝑂 .𝑔 = 0.285

Raymer [Tab.5.3 u. 5.4] are as typical values for jet airliners 𝑚𝑀𝑇𝑂 𝑆𝑤 = 586𝑘𝑔/𝑚2 and

𝑇𝑇𝑂 𝑚𝑀𝑇𝑂 . 𝑔 = 0.25 in the data obtained from the diagram is plausible.

The actual values of the F100 lie in:

𝑚𝑀𝑇𝑂 𝑆𝑤 = 489.9𝑘𝑔/𝑚2 and 𝑇𝑇𝑂 𝑚𝑀𝑇𝑂 . 𝑔 = 0.293

Maximum take-off mass 𝑚𝑀𝑇𝑂 The maximum Take-Off mass consists of the units operating mass 𝑚𝑂𝐸 , Fuel mass 𝑚𝐹 and

maximum payload 𝑚𝑀𝑃𝐿 , by rearranging we obtain [Eq. 5.47]

𝑚𝑀𝑇𝑂 = 𝑚𝑀𝑃𝐿

1 −𝑚𝐹

𝑚𝑀𝑇𝑂−

𝑚𝑂𝐸

𝑚𝑀𝑇𝑂

As we know the payload; the value of the original F100 used: 𝑚𝑀𝑃𝐿 = 12228𝑘𝑔 . The mass

fraction after statistics is estimated: According to Torenbeek (Figure 5.15), this is for short-haul

jet airliners 𝑚𝑂𝐸 𝑚𝑀𝑇𝑂 = 0.53. Loftin (Eq. 5.50) shows

𝑚𝑂𝐸

𝑚𝑀𝑇𝑂= 0.23 + 1.04 .

𝑇𝑇𝑂

𝑚𝑀𝑇𝑂 . 𝑔= 0.23 + 1.04 . 0.285 = 0.53

From the equation of Marckwardt [Eq. 5:48] 𝑚𝑂𝐸 𝑚𝑀𝑇𝑂 = 0.56 is accepted.

According to the results of these three approximations is a mass fraction of 0.54 expected. To

determine the mass of fuel through the flight phases during the engine starts and running-hot

(1), rolling (2), Begin (3), Climb (4) cruise (5), Hold (6), descent (7) and landing (8). Out of the

number of the flight phases, the mass at its ‘Start’ phase is specified.

8

The product of the mass ratios of the end / beginning of each phase of flight provides what we

can call "mission fuel fraction" Mff, Looking at [Eq. 5.53] is calculated from the mass of fuel:

𝑚𝐹

𝑚𝑀𝑇𝑂= (1 − 𝑀𝑓𝑓 )

The individual "mission segment mass fractions" that are for Phases 1-4, 7 and 8 specified by

Rosakam (Figure 5.19) is assumed from experience; Climb and descent will be considered twice

to account for the coupled approach of alternate aerodromes:

Table 2.2: Mass ratios of all the phases of flight

The missing mass ratios refer to Breguet to be determined. The Breguet factor for the cruise of

a jet after [Eq. 5:54] is calculated:

𝐵𝑠 = 𝐿 𝐷 . 𝑣𝐶𝑟

𝑆𝐹𝐶𝑇 . 𝑔

The speed of the cruise glide ratio is 18.5, at an altitude of 10670m(Default for original aircraft)

in ISA conditions and the cruise Mach number MCr = 0.77, at vCr = 229m/s. As a specific fuel

consumption (1.75).(10-5) kg/(N.s) is adopted. It follows a Breguet factor of BS = 24608689m.

When considering the original aircraft predetermined range of 2984km, the mass ratio of the

phase "cruise"

𝑚6

𝑚5= 𝑒−

𝑆𝐶𝑅

𝐵𝑥 = 𝑒−2984000 𝑚

24608689 𝑚 = 0.886

Duration of 45 min is given to the flight by FAR 121.

𝑚7

𝑚6

= 𝑒−

𝑡

𝐵𝑡 = 𝑒−2700𝑠

107762 𝑠 = 0.975

Mff is now the product of all individual mass ratios:

Mff = 0.99 0.99 0.995 0.998 0.99 0.998 0.99 0.992 0.886 0.975 = 0.801

9

The fuel portion is the difference to 1: 𝑚𝑂𝐸 𝑚𝑀𝑇𝑂 = 1 − 0.801 = 0.199. Used in [Eq. 5:47],

we obtain the maximum take-off weight:

𝑚𝑀𝑇𝑂 = 𝑚𝑀𝑃𝐿

1 −𝑚𝐹

𝑚𝑀𝑇𝑂−

𝑚𝑂𝐸

𝑚𝑀𝑇𝑂

= 12228𝑘𝑔

1 − 0.199 − 0.54= 46832𝑘𝑔

Wing area and Takeoff-thrust Division of the take-off weight by the wing loading provides the Wing area:

𝑆𝑊 =𝑚𝑀𝑇𝑂

𝑚𝑀𝑇𝑂

𝑆𝑊

=46832𝑘𝑔

495𝑘𝑔/𝑚2= 95𝑚2

The takeoff thrust is calculated by multiplying the thrust-to-weight ratio:

𝑇𝑇𝑂 = 𝑚𝑀𝑇𝑂 . 𝑔. 𝑇𝑇𝑂

𝑚𝑀𝑇𝑂 . 𝑔 =

46832𝑘𝑔9.81𝑚

𝑠20.285 = 130935𝑁

The comparative data of the actual F100 are:

MMTO = 45810kg (Deviation 2.2%)

SW = 93,5m² (Deviation 1.1%)

TTO = 134400N (Deviation 2.6%)

Fuselage design

Requirements The F100 fuselage will be capable of 107 passengers in economy class seating + transport

luggage and supply. An additional cargo space or volume wire gauge is not specified.

Fuselage cross-section A cheaper manufacturing technology is chosen and whereas the pressurized cabin is concerned,

a circular cross section is taken, which is constant over the length. A slenderness ratio lf/df = 10

10

is to be taken. This optimum value 8 up is departed, as in the case of the F100-fuselage,it is a

stretched version of the previous model F28 Fellowship.

With the number of passengers, the slenderness ratio is given by Marckwardt [Figure 6.1] the

Number of seats per row: nsa = 5.

The rollover formula for average Slenderness ratios [Eq. 6.1]

𝑛𝑠𝑎 = 0.45 𝑛𝑝𝑎𝑥

supplies with n = 107 = the value of 4.65; rounded to nsa = 5. This is in accordance with JAR

25,817 a sufficient transition.

According to the cabin standards of Airbus Industries[Table 6.1](width of a three-seat bench

including bending in the Y-Class 60'', a two-seater of 40 '') and typical cabin dimensions by

Raymer [Figure 6.4] (seat width 17 '' - 22 '', aisle width 18 '' - 20 ''), both seats and aisle width

with 20 '' is assumed. These also comply with the required width JAR 25,815 (for more than 20

passengers minimum 15 '' at the bottom, 20 '' from 25 '' height above the floor). It follows with

an additional interval between 0.025M cabin wall and outdoor seating.

The required maximum fuselage interior width

5 20'' + 20'' + 2 0.025m = 3.10m

The body’s outer diameter is therefore according to [Eq. 6.2]:

DF, O = dF, I + 0.084m + 0.045 dF, I = 3.32m

Schmitt [Figure 6.3] gives:

Dext = 1.07 Dint = 3.31m

To ensure sufficient pitch of minimum 76 '' = 1,93m [Raymer] and to ensure adequate

headroom to the outer seats, the cabin floor must be below the center line. In the given

fuselage cross-section of the floor, it has a width of at this level 2,86m. According to Schmitt,

the floor thickness required is 0.035 DF = 0.1m

With this information now, the complete fuselage cross-section are drawn. (Fig.3.1)

11

Figure 3.1: fuselage cross-section

Cabin Layout 107 seats with 5 seats per row require all 21 rows of seats + 1 row with only one 2-seater

bench.

22 rows of seats need at a pitch of 32 '' = 0.813m with the length of the cabin that is 18,04m.

60'' 32 '' = 1.24m2

According to the cabin standards by Schmitt [Figure 6.5] the commuter aircraft in Y-class per 60

passengers, a toilet is provided. These two toilets would have a bottom surface of about

1.2m2[Marckwardt].

A kitchen floor area of

𝑆𝑔𝑎𝑙𝑙𝑒𝑦 = 𝐾𝑔𝑎𝑙𝑙𝑒𝑦 𝑛𝑝𝑎𝑥

1000+ 0.5𝑚2

With kgalley = 16 m2 for short-haul flights [Marckwardt, Table 6.2] the required kitchen floor

area of Sgalley = 2,21m2 , Wardrobes are not required in Y-class seating, but a small space

remains available for the purpose(see Fig.)

From these values and the bottom width of the cabin length 2,86m, we get the sum of the

length of the seat portion (22 0,813m = 17,89m), the transition input width range about 0.6m

and the sum of the areas listed above, divided by the bottom width (2.35m).

1.50m added for the two emergency exits inside the cabin so a length of 22.86m is noted.

According to [Eq. 6.8], the entire fuselage is then

12

𝑙𝐹 = 𝑙𝐶𝐴𝐵𝐼𝑁 + 1.6 𝑑𝐹 + 4𝑚 = 22.86𝑚 + 1.6 3.3𝑚 + 4𝑚 = 32.14𝑚 long.

The elements determined in this way should now save space housed in the cabin.(Fig. 3.2)

Figure 3.2: cabin design

Figure 3.3: Hull side view

Emergency exits In the cabin according to JAR 25,807 are three exits for aircraft 80-109 passengers on each side.

Emergency exits(min. 24''x48 '', single level) have the entry and supply door at the bow; two

type III outputs are next to each other on the wing attached (Figure 3.2, 3.3). Since the wing

position is not yet established, the location of emergency exits can still be moved. AC(Advisory

Circular) 25807-1 provides a method by which the required "Uniform Distribution "of the

emergency exits can be checked. In the first step, the passenger distribution controlled in terms

of the emergency exits. For this purpose, the aircraft "Zones" on both sides of the fuselage

outputs are divided. In this case, zone A extends from the center line of the front (type I). The

output pair to the midline between the two Type III exits, Zone B is at the rearmost row of

seats. In each zone, the number of seats is the sum of the "Ratings" that do not exceed the

limiting zone outputs.

Table 3.1: Zone capacity

13

In the second step, the distribution of the outputs with respect to the fuselage and to each

other is checked. First, the length of the passenger cabin is determined, this ranges from the

center line of the front output until the last row of seats and is therefore 20,1m long. Sum of

the "exit unit" values for both zones is determined, the sum of the values of the limiting zone

outputs. Each emergency exit type has a different "exit unit " value; in type-I it is 1.25, a double

type-III is 2.0 , thus, the "exit unit "value for zone A is 3.25, for Zone B is 2.0, a total of 5.25.

Next, we obtain the hull length factor (fuselage length factor, flf) by dividing the length of the

cabin by the total number of "exit units":

flf = 20,1m / 5.52 = 3,83m

At the front-end of Passenger cabin (at 4.80m from the fuselage nose), the "nominal positions"

of the emergency exits(their center lines) is determined. To this end, the flf with the "exit unit"

value, the corresponding zone is multiplied.

Emergency exit Nominal location

Output 1 (type I) Output 2 (double type III)

4.80m 4.80m + 3.83m + 3.25 = 16,25m

Table 3.2: Nominal emergency position

The difference to the actual position should not exceed 15% of the length of the cabin. The

actual position of the type III outputs is 17,60m, which means a deviation of

(17.6 to 16.25)m/20.1m = 0.07.

As a further requirement is requested, with the exception of a double-type III output, two

adjacent outputs should not be closer together than the flf dimension(3,83m in this case).

Cargo volume Assuming a baggage weight per passenger of 40lb = 18kg on short-haul flights [Nikolai] and an

average density of 170 kg luggage/m3

A total of at least 18𝑘𝑔

𝑝𝑎𝑥 107𝑝𝑎𝑥

1

170𝑘𝑔/𝑚3= 11.3𝑚3 cargo volume is required.

From the fuselage cross-section outlined the cargo hold cross-sectional area can roughly be

taken as SCC = 1.5m2 .The present volume is VCC = LF SCC kCC with lf 33m and Kcc 0.35 for

regional aircrafts. It follows Vcc 17m3, spread over the front and rear cargo room. The cargo

14

compartment volume is therefore more than sufficient; for larger amounts than 18kg. (For

Comparison: Cargo volume of the original is 9.72m3 + 7.36m3 = 17.08m3)

Waterline In the event of a ditch 25,807(e)(2) is in accordance with JAR demanded that the water line of

the floating aircraft is below the lower door edge, so that no water gets into the cabin. In a

ditch shortly after the start of the 1st phase of the plane approximately the maximum Take-off

mass mMTO = 46832kg. According to Archimedes' Principle the same mass of water displaces it

in the floating state.

We have water at a density of Water = 1 kg/dm3 of volume 46800dm³ = 46,8m³. The door

bottom edges are h = 1.30 m above the hull bottom. The circular section of the fuselage cross

section thus an area of A = ½ r2 ( - sin) with

= 2 𝑎𝑟𝑐𝑐𝑜𝑠 1 −𝑕

𝑟 = 2 𝑎𝑟𝑐𝑐𝑜𝑠 1 −

1.3𝑚

1.65𝑚 = 2.714rad

The area of the Fuselage cross-section, A = 3,13m². The hull will have to be a length of about

21m cylindrical shape. Excluding the non-cylindrical part of the nose and tail cone and the wing

box is obtained as already a volume of V = 21m 3.13m² = 65.7m³ that can safely be below the

waterline; the claim is therefore met with sufficient certainty.

3.7 Compilation of important body dimensions(bracketed figures = original values of F100)

Number of seats per row(YC) 5 (5)

Number of speeds 1 (1)

No. of people(Cabin Crew) 3 (3)

Trunk diameter outside 3.31m (3.30m)

Fuselage length 32.14m (32.50m)

Length of the cabin 22.86m (21.19m)

Length of the nose section 5.6m

Length of the tail section 11.55m

Rear angle 13 °

15

Blades and high-lift devices

Pre-arranged parameters From the requirements and sizing, these are already known:

• Aspect ratio A = 8.4

Wing area S = 95m2

• Lift coefficient in cruise CL, cruise = CL(L / D)max = 0.61

Sweep The wing sweep is used mainly to increase the critical Mach number, as it has the increased

resistance to shift to higher flight Mach numbers. Hence, the desired Cruise Mach number is

relevant to the choice of wing sweep.

For a flight Mach number of Mamax = 0.77 is the appropriate sweep of the wing leading edge

according to a historical trend line according to Raymer (fig. 4.19) LE ≈ 25°. This leads us to

referring to the conversion equation

𝑡𝑎𝑛25 = 𝑡𝑎𝑛𝐿𝐸 − 1 −

𝐴 1 + ≈ 21.5

For an initially assumed escalation of = 0.25 (also slightly larger or smaller values for the

escalation will cause negligible changes)

Raymer (Fig.4.20) gives a maximum value for the sweep function for the surface to prevent

sudden “Tail-Heavy speed”. For A = 8.4, therefore, : 25 14 °. In accordance with the

constructed tail but this value can be exceeded; However, this result means that the originally

planned Sweep of 21 ° is revised downwards. Thus, a sweep of the c/4-line of 25 = 19 ° is

chosen.

Relative thickness profile The aim is to choose the profile thickness as large as possible to build a lighter wing and to be

able to increase the tank volume. However, it is delimited at the top by the Mach number of

16

the rise in resistance: The characteristic impedance by supersonic flow with a value of 0.0015

on the upper wing surface, which does not exceed. As a result, the Mach number of the

resistance increase by about 0.02 on the cruise Mach number:

MDD = MCr + 0.02 = 0.77 + 0.02 = 0.79

Through the wing sweep the effective flow velocity is reduced by

𝑀𝐷𝐷,𝑒𝑓𝑓 = 𝑀𝐷𝐷 𝑐𝑜𝑠25 = 0.79 𝑐𝑜𝑠19 = 0.77 [Eq. 7.32]

According to [Eq. 7.33] we can now get, kM = 1.2 (new profile), cL = 0.61 and 25 =19 the

maximum relative profile thickness(t/c) max = 0.123 at the wing root is a section thickness

chosen as 12.3%; Here, the thickness is typically 20% -60% higher than the one at the wing tip

(Raymer). At an assumed 30% thus the thickness at the top is 9.5%.

Airfoil The airfoil is considered for cruise conditions at Maximum Glide ratio, lift coefficient CL=0.61.

This is from the original aircraft flying with a ‘Fokker’ developed transonic profile. However,

since the lack of data pulls us back, the first step is to select a profile from the NACA catalog

that meets the specifications. The question comes to the wing root profile as the NACA 632415

with an appropriate design lift; However, it is a little too thick, so it can be modified. (Profile

data taken from Abbott/Doenhoff.)

Escalation The rear camber ensures that a greater proportion of the lift to the wingtip is generated. To

approximate the desired elliptical lift distribution the wing being too sharp to get back, the

escalation of the wing must be reduced. According to Raymer(fig.4.23), for the sweep of the

25% line of 19°, an intensification of ≈ 0.23 is required. Torenbeek shows the optimal

escalation may be as follows: opt = 0.45 e-0.036 25 = 0.45 e-0.036 19 = 0.227; (Raymer)

The value is therefore confirmed.

Fuel Capacity Stated by Torenbeek with the above-specified parameters, the volume of the wing tanks are

estimated as:

17

𝑉𝑇𝑎𝑛𝑘 = 0.54 𝑆𝑊1.5 𝑡 𝑐 𝑟 1

𝐴1 + + 2

1 + 2

with = 𝑡 𝑐 𝑡

𝑡 𝑐 𝑟

Therefore: VTank = 17.4m3

According to the results of the dimensioning is the ratio of fuel and other requirements

MTOW mF/ MMTO= 0.199; the derived MTOW of 46832 kg, should have a 9320 kgs of fuel to be

carried, which at an average density of 0.76 kg / dm3 of about 12.3m³ equivalent. Even if the

equation gives relatively inaccurate results, the Tank volume will certainly be adequate.(The

original aircraft has a tank volume of 13,465 litre ≈ 13.5m³.

Twist The twist root εt = iw,tip – iw,root is initially set to -2 °, which means the setting angle takes tip

from the rear.

V shape For a flying subsonic low-wing monoplane with swept wings acc. to Raymer(Tab.7.7) first a V

angle of 3.5° is adopted.

Setting angle The setting angle should be chosen so that the cabin is horizontal in the cruise. The selected

NACA 632415-profile reaches the required lift coefficient for the straight flight of CL = 0.6 at an

angle of about 3.5 °. If now the wing in this angle is "default" to the hull, the hull during the

cruising flight will be in a horizontal position.

High-lift systems High-lift systems is required for landing at a maximum lift coefficient CL,max,L = 2.7. The result

will be hitting a safety factor of 10% for the case, from which the trim on the empennage with

an output is generated must be compensated for:

CL,max = 1.1 2.7 = 2.97

18

The sum of the additionally needed lift coefficients by high-lift devices 0.95CL,max,f + CL,max,S

must be at least as large as the difference between the required Lift coefficient for landing

CL,max = 2.97 and lift coefficient of the pure wing CL,max,clean[Eq. 8.10].

This is according to Eq. 8.3

𝐶𝐿,𝑚𝑎𝑥 ,𝑐𝑙𝑒𝑎𝑛 = 𝐶𝐿,𝑚𝑎𝑥

𝐶𝐿,𝑚𝑎𝑥 𝐶𝐿,𝑚𝑎𝑥 ,𝑐𝑙𝑒𝑎𝑛 + 𝐶𝐿,𝑚𝑎𝑥 = 0.83 1.6 + −0.28 = 1.05

𝐶𝐿 ,𝑚𝑎𝑥

𝐶𝐿 ,𝑚𝑎𝑥 = 0.83 follows from this Figure 8.10 with the leading-edge sharpness parameter

Δy = 22.0 (t / c) = 2.6 and the sweepback angle of the leading edge of φLE ≈ 24 °

When dimensioning it was assumed that the necessary lift coefficient by a wing with double-slit

fowler flaps and slats can be still achieved. ΔCL, max, f must at least be equal to

(2.97 to 1.05) / 0.95= 2.02. The increase in Lift coefficient of the wing by flaps at the trailing

edge is [Eq. 8.6]:

Δ𝐶𝐿,𝑚𝑎𝑥 ,𝑓 = Δ𝐶𝐿,𝑚𝑎𝑥 ,𝑓 𝑆𝑊,𝑓

𝑆𝑊

𝐾Λ

It is Δ𝐶𝐿,𝑚𝑎𝑥 ,𝑓 that gives the increase of the lift coefficient of the profile [Eq. 8.4]:

Δ𝐶𝐿,𝑚𝑎𝑥 ,𝑓 = 𝐾1 + 𝐾2 + 𝐾3 ΔC𝐿,𝑚𝑎𝑥 𝑏𝑎𝑠𝑒

ΔC𝐿,𝑚𝑎𝑥 𝑏𝑎𝑠𝑒

denotes the maximum increase in the Lift coefficient; it is according to Figure

8.12 at a section thickness of about 12% and optimal double-slotted flaps ΔC𝐿,𝑚𝑎𝑥 = 1.5

By a factor of K1 = 1, and the allowable flaps depth profile would be at a depth of 0.25 amount

(Figure 8.13). This must be considered when choosing the rear spar location. K2 is for the fowler

flaps with a flap deflection of 40 ° to 1 (Figure 8.14). K3 in line with K2 that is adopted for the

flap angle is also equal to 1. The increase of the profile Lift coefficient is taken on 𝐶𝐿,𝑚𝑎𝑥 ,𝑓 = 1.5

The factor KΛ increases according to Figure 8.20 in a wing sweep of 25 = 19°, a value of about

0.89 in. SW, f refers to the part of the Wing area, which is the valve directly exposed “on”. In the

event of an extension of the flaps about 65% of the span (roughly equivalent to the value of MD

80) they would have a length of all in all 0.65 8.495𝑚 = 18.4𝑚 with 9.2m on each wing-half.

The chord of the Root is stated by Raymer

2𝑆

𝑏 1 + =

2.95𝑚2

28.2𝑚 1.23 = 5.48𝑚

19

For the single-tapered wing with the given aggravation, the wing is at the outer end

of neither the flap (9.2m away from the body) nor 2.38 m deep. The trapezoidal part thus

formed has an Area of ½ (5.48m + 2.38m) 9,2m = 36,2m². SW, f is exactly twice the area. Then

the Ratio

𝑆𝑊,𝑓 𝑆𝑊 = 2 36.2𝑚2 95𝑚2 = 0.76

The total to be achieved by the Fowler flaps, additional buoyancy is therefore at

Δ𝐶𝐿,𝑚𝑎𝑥 ,𝑓 = 1.5 0.76 0.89 = 1.02

After this, there will be a further increase in the lift coefficient therefore it is necessary that the

original aircraft functions without slats. Rather, seems either the calculation method or the

input parameters or inaccuracies exhibit fault. For example, the lift coefficient of the pure wing

here assuming the use of a NACA profile is specified, while the actual F100 flies with the

tailored supercritical profile.

Ailerons & Spoilers As the flaps extend from the fuselage at the wing trailing edge to about 9m outward stands for

the aileron or about the area between 75% and 95% of the mid-spans available, which

corresponds to a length of 2.8m. The profile depth is in the range of 30% of the respective

chord, the 2.35m at the inner edge of the rudder and at the outer edge is 1.5m. Exact spoiler

geometries cannot be determined at this point. There are, however, probably four to five

spoilers extending over each 30-40% semi-span that can be installed.

20

Empennage design 1 The design is intended for the mounting of the engines at the rear fuselage as a T-tail.

Tail-plane (CPR) The CPR is a trimmable horizontal tail-plane (trimmable horizontal stabilizer, THS) provided that

it gives a further priority area.

• Extension

The stretching should be approximately half of the wing.

AH = AW/2 = 8.4/2 ≈ 4

• Escalation

The escalation in comparison with the usual values of other jet-powered Commercial aircraft

(H = 0.27 …., 0.62, Rosakam) on H = 0.4 set.

• Sweep

The sweep of the CPR front edge should be about 5° above the wing, as the higher critical Mach

number CPR is then at high speeds(shocks occur later on the tail than on the wings) and

angles(covering the CPR later) remains in effect.

φH, LE = φW,LE + 5° = 23° + 5° = 28°

• Relative thickness

The Relative Thickness of the CPR should be about 10% lower than that of the outer wing. This

is a higher critical Mach number that is reached, resulting in a loss of efficiency by shocks

prevented.

(t/c)H ≈ (t/c)wingtip 0.9 = 9.6% 0.9 ≈ 8.5%

21

• V-shape and setting angle

V-shape and setting angle may both be set to 0°. A V-angle should keep the CPR staying out of

the exhaust plume; this case with a T-tail and Stern thrusters are critical. A (fixed) setting angle

is not necessary, since the CPR runs as THS, the setting angle is depends on CG-variable.

• Profile

The CPR is a symmetric profile of the NACA four-digit series provided. In an endeavor to relative

thickness of 8.5% fits in the NACA 0009th specification.

• Surface

The required CPR area is estimated by means of the ‘Tail Volume adjustment factor’.

[Eq. 9.4]

𝐶𝐻 =𝑆𝐻 𝑙𝐻

𝑆𝑊 𝐶𝑀𝐴𝐶

With SH - Area of the CPR

LH - Lever arm of the CPR

SW - Wing area

CMAC - Average aerodynamic. Depth profile of the wing

Stated by Raymer [Tab. 9.4] a typical value for the CPR is at coefficient volume of the jet-

powered commercial aircraft CH, start This value can be in the order at a THS 10% ... 15%,

and a T-tail because of the favorable flow by a further 5% can be reduced: CH = CH, start 0.85 =

0.85. The wing area is SW = 95m². Since the wing position is not yet well defined, the lever arm

can only be estimated by conventional values. Acc. to Raymer (Tab.9.5), for a plane with rear

engines, a lever arm of 45% ... 50% of the body length were adopted: lH = 0.5 32m = 16m. The

mean-aerodynamic. chord is

𝐶𝑀𝐴𝐶 =2

3𝐶𝑟

1 + + 2

1 +

Here is the chord at the root

𝐶𝑟 =2𝑆𝑊

𝑏(1 + )=

2.95𝑚2

28.3𝑚 (1 + 0.23)= 5.46𝑚

22

Hence,

𝐶𝑀𝐴𝐶 =2

35.46𝑚

1 + 0.23 + 0.232

1 + 0.23= 3.80𝑚

With these values can now be determined CPR area:

𝑆𝐻 =𝐶𝐻 𝑆𝑊 𝐶𝑀𝐴𝐶

𝑙𝐻=

0.85 95𝑚2 3.8𝑚

16𝑚= 19.2𝑚2

Rudder (SLW) • Extension

The extension of the SLW T-arrangement is usually less than the case of conventional

arrangement. This is in the range between 0.7 ... 1.2 ( Raymer ).

We select, AV = 1

• Escalation

A SLW T-arrangement is in contrast to the conventional design being not so pointed, since the

fin has to bear the weight of the CPR. Chosen here is a value of V = 0.9.

• Sweep, relative thickness and profile

The sweep of the SLW is for airspeeds where compressibility effects occur between 35° and 55°.

The Mach number of the resistance to increase the SLW is to be approximately 0.05 over the

wing, so at MDD, V = 0.84. With a sweep of φ25 = 45° would then be

𝑀𝐷𝐷,𝑒𝑓𝑓 = 𝑀𝐷𝐷 𝑐𝑜𝑠𝜑25 = 0.7

With CL = 0 (SLW a symmetric profile with straight flow generates no Buoyancy), according to

Eq. 7:33 (see panel design) a profile with approximately 12% relative thickness can be used

without the effectiveness at high speeds, which would be jeopardized. For the SLW therefore

the NACA 0012 profile can be used.

23

• Surface

The SLW-surface, like the CPR area over the corresponding coefficient volume is done [Eq. 9.5]:

𝐶𝑣 =𝑆𝑣 𝑙𝑣

𝑆𝑤 𝑏

The SLW coefficient volume is stated by Raymer 0.09; this is reduced by the ‘Endplates effect’,

the T-tail by 5%: CV = 0.09 0.95 0.085. The lever arm is in comparison to CPR relatively

shorter due to the sweep of the SLW:

lV = 0.45 32m = 14.4m

With the Span b = 28.2m, the necessary rudder area is obtained:

𝑆𝑉 =𝐶𝑉 𝑆𝑊 𝑏

𝑙𝑉=

0.085 95𝑚2 28.2𝑚

14.4𝑚= 15.8𝑚2

Elevator and Rudder For elevator and rudder no exact geometries are determined at this point, since there are no

exact requirements. Their size is by comparison with other aircraft indicative. The rudder covers

about 80% of the semi-span of SLW, its extension is up through the T-tail-plane limit. Its tread

depth is approximately 35% of the vertical stabilizer. The deflection angle of the Rudder is at =

25 ° limited. The elevator covers about 90% of the CPR, its tread depth is approximately 30% of

the CPR.

24

Weight and Balance

Mass prediction Class-I( Raymer ) For this mass prediction, the aircraft in the groups of wing, fuselage, empennage, tail-plane,

main and nose landing gear, engines and systems divided. In the first four of these groups is the

reference parameter "flow-around surface" with an empirical factor multiplied, and the thus

obtained mass is added. The surfaces are made known previous design steps. For chassis and

systems, a fixed proportion of the intended MTOW accepted. The mass of the non-installed

engine is known and is provided with a factor for attachments. The flow around airfoil surface is

obtained from the double reference wing area minus the share "in the" Hull:

Sexposed = 2 (SW – Croot df, ext) = 2 (95m2 – 5.5m 3.3m) = 154m2

The wetted surface of the hull is roughly one as a cylinder surface with cover surface plus the

conical tail is calculated:

𝑆𝑤𝑒𝑡 = 𝑆𝑧𝑦𝑙 + 𝑆𝐾𝑜𝑛 = 2𝑟𝑓 ,𝑒𝑥𝑡 𝑙𝑧𝑦𝑙 + 𝑟𝑓,𝑒𝑥𝑡2 + 𝑟𝑓 ,𝑒𝑥𝑡 𝑟𝑓,𝑒𝑥𝑡

2 + 𝑙𝑘𝑜𝑛2

= 21.65𝑚20𝑚 + (1.65𝑚)2

+ 1.65𝑚 (1.65𝑚)2 + (12𝑚)2 236𝑚2

The flow-around fin surface is twice the sum of CPR and SLW surfaces obtained:

Sexposed = 2 (19,2m² + 15,8m²) = 70m².

The mass of the engines, after Raymer be estimated [Eq. 10.16] as follows:

𝑚𝐸 =0.0724

𝑔𝑇𝑇𝑂

1.1𝑒−0.045𝐵𝑃𝑅 =0.0724

9.81 𝑚𝑠2

65.5 103𝑁1.1𝑒−0.0455 = 1170𝑘𝑔

Both engines together thus have a mass of m2E = 2340kg

25

With these reference parameters, mass prediction can now be done:

Factor Reference Name Value

Mass [Kg]

Wing Hull Tail Nose wheel Main landing gear Engines Systems

49 24 27 0,006 0.037 1.3 0.17

Sexp[m²] 154 Swet[m²] 236 Sexp[M²] 70 mMTO[Kg] 46800 mMTO[Kg] 46800 mE[Kg] 2340 mMTO[Kg] 46800

7546 5664 1890 280.8 1731.6 3042 7956

mOE 0 0 0 28110.4 Table 6.1: Mass forecast Class I

Taking these parameters into consideration we can obtain the Take-off weight:

mMTO = mOE + mPL + mfuel = (28110kg + 12228kg) / (100-19.9) 100 = 50360kg

Mass prediction Class-II (Torenbeek) For Class II-mass initially taken into account are the masses of the mass-groups of wing,

fuselage, horizontal stabilizer, rudder, landing gear, engine nacelle, installed engines and

systems with empirical formulas and the input value mMTO = 46832kg is estimated. The sum of

which, the dry operating mass of the maximum Take-Off mass can be calculated. The value thus

obtained is then as Start value used for re-calculation of the individual masses. This iteration is

repeated until the change in the maximum Start mass between two steps under 0.5%.

This (inner) iteration gives the following results:

Step mMTO Deviation

1 44871 kg 4.2%

2 44254 kg 1.4%

3 44060 kg 0.4% Table 6.2: Mass forecast Class II, inner iteration

The last value is 5.9% below the starting value 46832 kg. When the limit is exceeded by 5%

deviation, the wing area and takeoff thrust can be adjusted:

𝑆𝑊,𝑛𝑒𝑢 =𝑚𝑀𝑇𝑂

𝑚 𝑠 =

44060𝑘𝑔

495𝑘𝑔/𝑚2= 89𝑚2

𝑇𝑇𝑂 ,𝑛𝑒𝑢 = 𝑚𝑀𝑇𝑂 𝑔𝑇

𝑚𝑔= 44060𝑘𝑔9.81

𝑚

𝑠20.285 = 123185𝑁

26

This results in new wing and engine mass, which again the inner Iteration is done, this time with

the start value mMTO = 44060kg. Since the wing and engine mass have only a slight change, this

already provides the first pass at a deviation of 0.3% a Maximum Take-off mass of mMTO =

43908kg, which is divided as shown in Table 6.3.

Mass Distribution

mW[Kg] 3,944.95 mN[Kg] 867.56

mF[Kg] 5,071.83 mE,inst[Kg] 3,180.81 mH[Kg] 639.65 mSYS[Kg] 6,877.36 mV[Kg] 614.60 mOE[Kg] 22,942.56 mLG[Kg] 1,745.81

The comparison of the method is as follows:

Dimensioning Class I Class II - a Class II – b

mMTO[Kg] 46832 50360 44871 43908

Deviation of Dimensioning

± 0% + 7.5%

-4.2%

-6.2%

Table 6.4: Comparison of processes

The maximum Take-off weight of the original aircraft is 43090kg, calculated in the Class II

method, so value differs by almost 2% of the actual.

CG calculation For the centroid calculation the plane is divided into two main groups; Fuselage (tail, fuselage,

engines and nacelles, systems, nose landing gear) and wings(Wings, landing gear). Then the

mass for both groups and the respective determined focus and wing group then moved so that

the overall center of gravity comes to lie at about 25% MAC. When zero is selected at the

Fuselage nose, the focus of the wing group relate to the leading edge at the location of the

middle aerodynamic chord(LEMAC).

27

Mass Group Weight [kg] SP Location of the fuselage nose[m]

SP-mass position

Hull TW nacelle

5,071.80 867.60

15.11 25.07

76,634.90 21,750.73

Nose gear 261.90 3.47 908.79

Systems 6,877.40 12.00 82,528.80

Engines 3,180.80 24.11 76,689.09

CPR 639.70 32.32 20,675.10

SLW 614.60 31.21 19,181.67

17,513.80 Xx 298,369.08

Overall center of gravity [m], of the fuselage nose

17.04

Table 6.5: Balance of trunk group

Mass Group Weight [kg]

SP Location LEMAC [m]

SP-mass position

Wing 3,945.00

2.03

8,008.35

Main landing gear

1484.00 1.52

2,255.68

5,429.00 Xx 10,264.03

Overall center of gravity[m] measured from LEMAC

1.89

Table 6.6: Balance of wing group

Moments of the equilibrium based on LEMAC, the distance between Zero(aircraft nose) and

LEMAC be determined which is necessary to the overall center of gravity to adjust to the

desired position of 30% MAC [Eq. 10:24]. Thus defined the position of the wing.

𝑥𝐿𝐸𝑀𝐴𝐶 = 𝑋𝐹𝐺 − 𝑋𝐶𝐺 ,𝐿𝐸𝑀𝐴𝐶 +𝑀𝑊𝐺

𝑀𝐹𝐺 𝑋𝑊𝐺,𝐿𝐸𝑀𝐴𝐶 − 𝑋𝐶𝐺 ,𝐿𝐸𝑀𝐴𝐶

= 17.04𝑚 − 0.33.80𝑚 +5429𝑘𝑔

175138𝑘𝑔 1.89𝑚 − 0.33.80𝑚 = 16.13𝑚

By comparison, the distance xLEMAC lies with the original aircraft at about 16m.

28

Empennage design II

Tail-plane (CPR) The designs on controllability and stability data need a linear equation with the

Variables xCG-AC(AC distance to the center of gravity based on MAC). These can be added in a

common chart; may depend on the required priority area, the CPR area can be determined.

Interpretation by controllability

Dimension case of flight mode "Lift Off" at maximum damper position. The CPR face after

controllability requirement is given by a linear equation.

𝑆𝐻

𝑆𝑊= 𝑎 𝑥𝐶𝐺−𝐴𝐶 + 𝑏

According to Eq. 11:19

𝑎 =𝐶𝐿

𝐶𝐿η𝐻 𝑙𝐻

𝐶𝑀𝐴𝐶

with the parameters in the lift coefficient dimensioning air condition:

CL = 1.6 (Lift coefficient during go-around 1.3vs)

Lift coefficient of CPR:

CL, H ≈ -0.5 (conservative assumption; ‘-ve’ value as CPR output produced)

ηH = 0.9

Horizontal stabilizer level arm, from the centroid calculation:

lH = xAC,HLW – xLEMAC - xAC,LEMAC = 31.97m – 16.13m – 0.25 3.80m = 14.89m

Mean aerodynamic Chord:

cMAC = 3.8m

⇒ a = -0.9074

29

𝑏 = 𝐶𝑀,𝑊 + 𝐶𝑀 ,𝐸

𝐶𝐿,𝐻 η𝐻 𝑙𝐻

𝐶𝑀𝐴𝐶

with the parameters of coefficient moments by the engines:

𝐶𝑀,𝐸 =−𝑇𝑧𝐸

𝑞𝑆𝑊𝐶𝑀𝐴𝐶=

−123185𝑁0.9𝑚

1 2 1.225𝑘𝑔

𝑚34369.2

𝑚2

𝑠289𝑚23.8𝑚

= −0.1225

The ‘Go-around’ thrust is T = TTO = 123185N

z-distance from the centroid of the engines for E ≈ 0.9m ,other variables such

[Eq. 11:33]:

𝐶𝑀,𝐸 = 𝐶𝑀,0,𝑓𝑙𝑎𝑝𝑝𝑒𝑑 𝐴𝑐𝑜𝑠225

𝐴 + 2𝑐𝑜𝑠25

+ 𝐶𝑀 ,𝑜

εt ε𝑡

𝐶𝑀,𝑜 𝑀

𝐶𝑀,𝑜 𝑀−𝑜

= −0.3498.4 𝑐𝑜𝑠2 19

8.4 + 𝑐𝑜𝑠2 19 + −0.005 2 1 = −0.2447

Coefficient moment of the profile in the neutral point [Eq. 11.30]:

𝐶𝑀 ,0,𝑓𝑙𝑎𝑝𝑝𝑒𝑑 = 𝐶𝑀,𝑜 + 𝐶𝑀 = 𝐶𝑀,𝑜 + 𝐶𝐿,𝑓𝑙𝑎𝑝𝑝𝑒𝑑 𝑥𝐴𝐶

𝐶𝑀𝐴𝐶−𝐶𝑀,𝑜

𝐶𝑀𝐴𝐶𝑐′

𝑐

= −0.067 + 1.5 0.25 − 0.438 1 = −0.349

CM, 0 = - 0.067 (Abbott)

ΔcM for [Eq. 11.31] with xAC/CMAC = 0.25, xcp/CMAC = 0.44 and c '/ c = 1

ΔCL,Flapped =1.5 as calculated in Sect. 4.10

"High-lift devices"

Extension A = 8.4, sweep φ25 = 19 ° and Wing-restriction

εt = -2 ° to the Section. 4 "wing design"

Wing restriction is accordingly [image 11.14a] for φ25 = 19°,

A = 8.4 and = 0.23 Δ𝐶𝑀 ,𝑜

εt

30

The Mach number effect

𝐶𝑀,𝑜 𝑀

𝐶𝑀,𝑜 𝑀−0

,according to [Image 11,15]: the speed is negligible during the landing approach

⇒ b = 0.2097

This is the linear equation:

𝑆𝐻

𝑆𝑊= −0.9074 𝑥𝐶𝐺−𝐴𝐶 + 0.2097

Design for stability requirement

In this case, the linear equation is in the form 𝑆𝐻

𝑆𝑊= 𝑎 𝑥𝐶𝐺−𝐴𝐶

This is according to [Eq. 11:24]

𝑎 =𝐶𝐿,,𝑊

𝐶𝐿,,𝑊η𝐻 1 −e

𝑙𝐻

𝐶𝑀𝐴𝐶

with the parameters of the wing buoyancy gradient by [Eq. 7.27] the idealization of the

following is done

𝐶𝐿, = 2

1 − 𝑀2:

𝐶𝐿,,𝑊 = 2𝐴

2 + 𝐴2 1 + 𝑡𝑎𝑛225 − 𝑀2 + 4

𝐶𝐿,,𝑊 = 28.4

2 + 8.42 1 + 𝑡𝑎𝑛215.2 − 0.772 + 4= 6.47

1

𝑟𝑎𝑑

The buoyancy gradient of the horizontal stabilizer according to [Eq. 7.27]:

𝐶𝐿,,𝐻 = 24

2 + 42 1 + 𝑡𝑎𝑛217.6 − 0.772 + 4= 4.58

1

𝑟𝑎𝑑

Down-draft gradient on CPR after [Eq. 11:34]:

31

e

= 4.44 𝐾𝐴 𝐾 𝐾𝐻 𝑐𝑜𝑠25

1.19

𝐶𝐿, 𝑀

𝐶𝐿, 𝑀=0

= 0.609

(Conversion of wing sweep on the 50% -line with [Eq. 7.12])

Factor for the Stretching [Eq. 11.35]

𝐾𝐴 =1

𝐴−

1

𝑎 + 𝐴1.7=

1

4−

1

1 + 41.7= 0.163

Factor for the Worsening [Eq. 11.36]

𝑘𝐴 =10 − 3

7=

10 − 30.4

7= 1.257

Location factor CPR after [Eq. 11.37]

𝑘𝐻 =1 −

𝑍𝐻

𝑏

2𝑙𝐻

𝑏

3

= 1 −

5.6𝑚

28.2𝑚

2 13.05𝑚

28.2𝑚

3

= 0.822

with the difference in height between the wing root chord and medium aerodynamic Wing

depth of CPR for ZH ≈ 5.60m

ηH, LH, CMAC ,so

⇒ a = 1.0245

The linear equation is:

𝑆𝐻

𝑆𝑊

= 1.0245 𝑥𝐶𝐺−𝐴𝐶

Reunification

The two linear equations can now be entered into a common chart. Note that the rear center of

gravity of a "Safety margin" to the natural stability limit as above must comply calculated. This

level of static longitudinal stability of jet airliners by Rosakam is 0.05 MAC. This value is by

Raymer reduced even to 2% MAC, due to neglecting of several pitching moment influencing

engine effects at the beginning of the invoice. The allowable priority areas are now between

the lines from the controllability (blue) and the stability requirement, less stability measure

(green). Between these lines, the required focal area now, according to loading chart fitted so

that we get the smallest possible tail surface results.

32

Figure 7.1: Determination of the CPR area

In this case, a range of 0.2 SP-MAC has been specified. This results in a minimum Horizontal tail

surface SH = 0.22 SW = 0.22 = 89m² 19.58m². The CPR face of the original F100 is 21.72m²; so

that the deviation is below 10%. For comparison, the Tail forecast I gave SH = 19,2m², the

deviation from the old value is 2%. Carrying out a recalculation of the CPR mass is not

necessary. Also this can be read off the front-most and rear-most CG: The front is 0.015 SP MAC

before the AC, the rear 0.185 MAC behind it.

Rudder (SLW) The SLW is designed only after horizontal stabilizer. The dimensions are to case then the failure

of one engine (TW) at the start; which failed to TW symmetric generates a yaw moment NE,

that the resistance of the failed TW yet to ND is increased. It follows by [Eq. 11:43] a required

SLW area:

𝑆𝑣 =𝑁𝐸 + 𝑁𝐷

1

2 𝑣𝑀𝐶

2𝐹 𝐶𝐿 ,

𝐶𝐿 ,α 𝑡𝑕𝑒𝑜𝑟𝑦

𝐶𝐿, 𝑡𝑕𝑒𝑜𝑟𝑦 𝐾𝐼 𝐾Λ 𝑙𝑣

The yaw moment by the resistance of the failed TW is about 25% of the moment by the

opposite TW (Jet with rotating fan and high BPR). The Total yaw moment is thus:

33

𝑁𝐸 + 𝑁𝐷 = 1.25𝑁𝐸 = 1.25𝑇𝑇𝑂 𝑦𝐸 = 1.25123185𝑁

22.7𝑚 = 207.9𝑘𝑁𝑚

with the distance of the TW from the plane of symmetry yE = 2.7m

It is assumed to start at sea level, = 1.225kg / m³. The minimum flight speed for TW failure

vMC is about 20% higher than the stall speed in launch configuration:

𝑣𝑀𝐶 = 1.2𝑣𝑆,𝑇𝑂 𝑣𝐿𝑂𝐹 = 2𝑔

𝑚𝑇𝑂

𝑆𝑊

1

𝐶𝐿,𝐿𝑂𝐹=

29.81𝑚/𝑠2

1.225𝑘𝑔/𝑚3

4390𝑘𝑔

89𝑚2

1

2.2 60𝑚/𝑠

The rudder is required to δF = 25 ° = 0.4363rad

𝑐

𝐶𝐿,α ,theory

= 0.85

for the selected NACA 0012 profile. (CL, ) = 4.6 per rad for the relative rudder (cf/ C) = 0.3 and

relative profile thickness of the SLW (t/c) = 0.12. The correction factor for large flap value is

K '= 0.67 at a deflection angle F = 25 ° and (Cf / C) = 0.3. KΛ = 0.74 for 25 = 45 ° (Figure 8.20).

The actual SLW lever arm is lV = x0.25MAC, S - x0.25MAC, W = 13.51m.

After substituting these values results, SLW area:

SV = 8.25m²

This calculated surface is very low and well below the tail section I determined (15,8m²). The

cause is possibly the case of aircraft’s stern thrusters, much smaller yaw moment by a failed

engine is justified by the very small lever arm compared to TW under the wing; i.e. the

calculated case is probably not what the dimensions are. As in Chap. 5.2 "tail I" area

determined based on statistics, the number of aircraft engines will be discussed under the

wings, the value calculated here is likely to be great. In the absence of a better alternative to

the average of the two is

SV = (SVI + SVII) / 2 = 12.03m².

By comparison, the vertical tail of the original aircraft has an area of 12,3m².

34

Chassis

Number and arrangement of legs and wheels The landing gear assembly is elected as the previous model F28 Fellowship:

Nose landing gear (NG) with two wheels and main landing gear (MG) with two legs under the

wings, also with two wheels.

Positioning When determining the mounting location of the legs as well as the tire selection done by the

method of Currey procedure. For this, the following values are first needed:

• LEMAC distance from the fuselage nose: 16.13m

• MAC distance from the center fuselage:

• H = [b / 2 (Cr - CMAC)] / (Cr - Ct) = 13.67m (5.48m - 3.80 m) / (5.48m – 1.26m) = 5.44m

• Front center of gravity location: 16.13m + 0.27 3.8m - 0.015 3.8m = 17.10m from the

fuselage nose

• Rear Total balance: 16.13m + 0.27 3.8m + 0.185 3.8m = 17.86m

• z-position of the overall center of gravity: 2.10 m of fuselage top

Location of the main landing gear on the longitudinal axis

Location of main landing gear on the longitudinal axis of the MAC-position and the approximate

location are in a plan view of the aircraft of wing box rear and a side view of the front and rear

center of gravity and also enters the MAC layer. At first by comparable aircraft (eg. BAe RJ85)

adopted track width of 5.0 m, the fitting for the main landing gear in the top view is set directly

behind the rear spar. This creates no difficulties at its attachment to the structure and the

suspension settle inward in a panel between the wing and fuselage be swiveled without

disturbing the wing box. Note that the position is at a vertical line entered in the side view

(Fig.7.1). The main landing gear position located at xMLG = 19.0 m, measured from the fuselage

nose.

In the longitudinal direction, is the rear-center of gravity, a tilt angle of at least 15°. For this

purpose, a line at an angle of 15 ° is from the rear to the center of gravity vertically drawn. The

intersection of this line with the previously determined chassis centerline is thus the bottom

35

line at suspension, where the total length of the main landing gear is fixed. The fuselage is

therefore at the spring-loaded chassis 1.4 m above the ground.

From the rear edge of the ground contact surface of the wheel is now to an angle of

approximately 15° are adhered to the rear.

The lateral positioning of the main landing gear may only be checked after the determination of

the wheelbase. However, the angle of nose gear on the aircraft longitudinal axis is to be known.

Figure 8.1: Positioning of the main landing gear

Nose gear

The nose gear must be attached to the hull as forward as possible for stress and stability

reasons, on the other hand, the load must not be too low to ensure steer-ability at the bottom.

Currey recommends a load of about 8% of the overall aircraft weight on the nose gear in rear

SP-location. For the calculation of the wheelbase F, the distances between main gear and

CG positions are needed:

M = xMLG - xCG,aft = 19.0m – 17.86m = 1.14m

(F-L) = xMLG - xCG, fwd = 19,0m-17,1m = 1.9m

If the load on the nose gear is at rear, the center of gravity should be 8% of the total weight, the

corresponding lever arms in inverse ratio must be:

36

(M / F) = 0.08

⇒ Wheel-base F = M / 0.08 = 1,14m / 0.08 = 14,25m. The wheelbase of the model aircraft is

14.01m. Installation of the nose gear is measured from the fuselage nose

x = 19.0m-14.25m = 4.75m

With the exact suspension positions, the center of gravity calculation must be checked:

The modified installation of the nose gear, the focus shifts to the trunk group only very slightly

by 0.02 m to the rear. The wing group has 1.89m of LEMAC to 2.16m behind LEMAC. Thus, the

wing moves slightly to the Rear: xLEMAC increases from 16.13m to 16.22m; the influence of the

respective changes in tail lever arms is negligible.

Lateral position of the main landing gear

Compliance with what is called the "Tip-over criterion" must be checked. For this purpose, the

compound bow line to the main landing gear, the solder on the projection of the front SP

(Critical case) is made on the ground. The starting point of the solder on the connecting line

itself is connected to the SP by a straight line. To tipping over of the aircraft to prevent the

angle must be between and precisely maximum of 55 °. (Fig.7.2)

Figure 8.2: lateral tip-over criterion (schematic)

With half the gauge (2.5 m) and the x-distance between the nose gear and main landing gear

(14.25m) the angle is calculate:

α = arc tan(2.5 / 14.25) = 10 °.

From the x-distance between the nose gear and the front focus follows (12.35m) the length of

the perpendicular h:

37

h = 12.35m sin10 ° = 2.14m

With the z-position of the center of gravity, the tilt angle is obtained:

Ψ = arc tan(2.5 / 2.14) = 49°

As per Currey, Tab.3.3 this is a quite typical value for a low-wing transport aircraft.(DC-9 and

727 have approximately the same value.)

In addition, an angle slope of at least 7.5° when placed with only one main beam is possibly

measured from the touchdown point of the outer wheel. (Fig.7.3)

Difficulties in meeting these criteria generally related to aircraft whose engines are mounted

under the wings.

Figure 8.3: Ground clearance of the wing tips (to scale)

Tire selection The calculated maximum take-off mass is mMTO = 43908kg, corresponding to 96800lb. Stated by

Rosakam Vol.II, Tab.9.2 for commercial aircraft of this size usually have 40''x14 '' - tires on the

main landing gear and 24''x7.7 '' - on the nose gear tires. From Rosakam Bd.IV, Tab.2.4 the

required tire pressure is then close to 140psi on all the tires.

LCN-value The "Load Classification Number" followed by Torenbeek , fig.10-1 from the tire pressure and

the "Equivalent Single Wheel Load"(ESWL). ESWL is the quotient of the load on a chassis frame

and a reduction factor.

38

The maximum load on a main chassis frame is available in rear center of gravity;

LMLG, max = mMTO[Wheelbase (xMLG - xCG,aft)]/ (2 wheelbase)

= 43908kg (14.25m – 1.14m) / (2 14.25m) = 20198kg ≡ 44536lb

(in the diagrams Anglo-Saxon units are used)

The reduction factor obtained from fig. 10.2 Assuming a tire contact area (On a chassis frame)

of

𝐴𝑐 = 2 𝑡𝑖𝑟𝑒𝑙𝑜𝑎𝑑

𝑡𝑖𝑟𝑒𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒=

44536𝑙𝑏

140 𝑙𝑏/𝑖𝑛2= 318𝑖𝑛2

"radius of relative stiffness" L = 45in ( Torenbeek ) and a distance between the two wheels on a

main chassis frame of ST = 24in gives the reduction factor the value of 1.25. ESWL is then

𝐸𝑆𝑊𝐿 =44536𝑙𝑏

1.25= 35629𝑙𝑏

Torenbeek (Fig. 10-1) returns to the tire pressure 140psi and ESWL = 35629lb and LCN

of 46.

Determination of polar Approximately true for profiles with small curvature of a polar form

𝐶𝐷 = 𝐶𝐷𝑂 +𝐶𝐿

2

𝐴 𝑒

The zero resistance is the sum of the resistances of the zero components wings, fuselage, tail,

engine nacelles, other components as well as leaks calculated.

For the latter two standard values are accepted. The remaining resistors calculated according to

[Eq. 13:15]:

𝐶𝐷𝑂,𝑐 = 𝐶𝑓,𝑐 𝐹𝐹𝑐 𝑄𝑐 𝑆𝑤𝑒𝑡 ,𝑐

𝑆𝑟𝑒𝑓

39

This is Cf, c the respective coefficient of frictional resistance, FFc a factor that takes the form of

resistance into account, Qc a factor for the interference based on the resistance of the hull and 𝑆𝑤𝑒𝑡 ,𝑐

𝑆𝑟𝑒𝑓 is the ratio of flow on the surface of the component and reference wing area.

Fuselage The flow is turbulent around the fuselage over the whole length. The Reynolds number is

𝑅𝑒 =𝑣𝑙𝑓

𝑣=

𝑣𝑙𝑓

15 10−6𝑚2/𝑠= 5.06 108

This would be called “Cut-off Reynolds number”

for M < 0.9: Recut-off = 38.21 (l/k)1.053 = 38.21 (32140mm / 0.00635mm)1.053 = 4.38 108 with

the surface, roughness k = 0.00635mm for smooth color (calculated according DATCOM). Thus,

it occurs that in the calculation of the frictional resistance, the cut-off Reynolds number is taken

in place of the actual one. It is considered that the frictional resistance in rougher surfaces is

greater than what would be calculated by the equation.

𝐶𝑓,𝑡𝑢𝑟𝑏 =0.455

log 𝑅𝑒 2.58 (1 + 0.144 𝑀2)0.65=

0.455

log 4.38108 2.58(1 + 0.1440.772)0.65

= 1.654 10−3

[Eq. 13:17]

The form factor for the hull from DATCOM

𝐹𝐹𝐹 = 1 +60

𝑙𝑓 𝑑𝑓 3 +

𝑙𝑓 𝑑𝑓

400= 1.09

40

[Eq. 13:23]

The interference factors are related to the hull, ⇒ Qf = 1. The wetted surface of fuselage with

cylindrical midsection from Torenbeek:

𝑆𝑤𝑒𝑡 ,𝑓 = 𝑑𝑓 𝑙𝑓 1 −2

𝑓

273

1 +2

𝑓2

[Eq. 13.7]

= 3.3𝑚 32.14𝑚 1 −2

9.7

273

1 +2

9.72 = 291.7𝑚2

𝑆𝑤𝑒𝑡 ,𝑓

𝑆𝑟𝑒𝑓

=291.7𝑚2

89𝑚2= 3.28

⇒ CDO,f = 1.654 10-3 1.09 1 3.28 = 5.91 10-3

Wing It is believed that the flow through the forward 10% of the wing is laminar, rear section is

turbulent. According to [Eq. 13.21] the frictional resistance is calculated

Cf = klam Cf,lam + (1 – klam) Cf,turbl

𝐶𝑓 ,𝑙𝑎𝑚 =1.328

𝑅𝑒= 1.328

𝑣𝐶𝑟 𝐶𝑀𝐴𝐶

𝑣

= 1.328

236 .1𝑚/𝑠

15 10−6𝑚2 /𝑠

= 0.1717 10−3

[Eq. 13:16]

𝐶𝑓 ,𝑡𝑢𝑟𝑏 =0.455

log 𝑅𝑒 2.58 (1 + 0.144 𝑀2)0.65=

0.455

log 4.628107 2.58 (1 + 0.1440.772)0.65

= 2.25 10−3

[Eq. 13:17]

⇒ Cf = 0.1 0.1717 10-3 + 0.9 2.25 10-3 = 2.045 10-3

The interference factor is existing wing-fuselage transition fairing at QW = 1.0

[Tab.13.5]. Wetted surface of the wing as per Torenbeek

41

𝑆𝑤𝑒𝑡 ,𝑊 = 2 𝑆𝑒𝑥𝑝 𝑡 𝑐 𝑟 1 +

1 + = 2 81.01 1 + 0.25 0.123

1 + 1.2950.23

1 + 0.23

= 167.3𝑚2

[Eq. 13.8]

𝑆𝑤𝑒𝑡 ,𝑓

𝑆𝑟𝑒𝑓

=167.3

89= 1.88

According to [Eq. 13:22] calculating the shape factor;

𝐹𝐹𝑊 = 1 +0.6

𝑥𝑡 𝑡 𝑐 + 100 𝑡 𝑐 4 1.34𝑀0.18 𝑐𝑜𝑠𝑚

0.28

= 1.356

with xt = 0.35 3.8m (Thickness from Abbott )

(t / c) = 0.12

M = 0.77

φm = 17.5 ° (angle of the line of maximum thickness section)

This means

cd0, w = 2.045 10-3 1.88 1.356 1 = 5.21 10-3

Tail-plane The frictional resistance at the tail-plane (and also to the vertical stabilizer) is a purely turbulent

flow calculation.

The Reynolds number of the flow CPR

𝑅𝑒 =𝑣𝑐𝑀𝐴𝐶 ,𝐻

𝑣=

236.1𝑚

𝑠 2.36𝑚

15 10−6 𝑚2 𝑠 = 3.715 107

The actual Reynolds number is above the "cut-off Reynolds number" of 2,802 107, the

frictional resistance is calculated with the latter. According to [Eq. 13:17]

𝐶𝑓 ,𝐻 =0.455

𝑙𝑜𝑔 𝑅𝑒 2.58 1 + 0.144𝑀2 0.65= 2.43 10−3

42

With a relative profile thickness (t / c) = 0.09, a thickness reserve of 0.3 2.36m = 0,71m and a

sweep angle of 30% line of m = 22 is the form factor FFH = 1.79. The interference factor is a T-

tail, QH = 1.04. The wetted surface is obtained from the wing as

[Eq. 13.8]:

𝑆𝑤𝑒𝑡 ,𝐻 = 2 𝑆𝑒𝑥𝑝 1 + 0.25 (𝑡 𝑐) 𝑟

1 +

1 +

= 2 19.58𝑚2 1 + 0.25 0.10 1 + (0.1/0.08)0.4

1 + 0.4 = 40.2𝑚2

Relative to the reference wing area results in the

𝑆𝑤𝑒𝑡 ,𝐻

𝑆𝑟𝑒𝑓

=40.2𝑚2

89𝑚2= 0.452

⇒ cD0, H = 2.43 10-3 1.79 1.04 0.452 = 2.05 10-3

Rudder The Reynolds number of the flow SLW

𝑅𝑒 =𝑣𝑐𝑀𝐴𝐶 ,𝐻

𝑣=

236.1𝑚

𝑠 3.47𝑚

15 10−6 𝑚2 𝑠 = 5.46 107

Again, it exceeds the "cut-off Reynolds number" of 4.206 107; with this the frictional resistance

is thus determined:

𝐶𝑓,𝐻 =0.455

𝑙𝑜𝑔 𝑅𝑒 2.58 1 + 0.144𝑀2 0.65= 2.3 10−3

The form factor for [Eq. 13.22] is the maximum thickness position xt = 0.3 3.47m = 1.04m,

relative section thickness (t / c) = 0.12, and the sweep of the 30% line m = 44.7°. FFV = 2.15

The interference factor is also the rudder for [Tab.13.5] QV = 1.04

The wetted area is [Eq. 13.8]

43

𝑆𝑤𝑒𝑡 ,𝐻 = 2 𝑆𝑒𝑥𝑝 1 + 0.25 (𝑡 𝑐) 𝑟

1 +

1 +

= 2 12.03𝑚2 1 + 0.25 0.13 1 + (0.13/0.10)0.9

1 + 0.9 = 25𝑚2

Relative to the reference wing area results in

𝑆𝑤𝑒𝑡 ,𝐻

𝑆𝑟𝑒𝑓

=25𝑚2

89𝑚2= 0.28

⇒ cD0, V = 2.3 10-3 2.15 1.04 0.28 = 1.44 10-3

Engine nacelles The exact dimensions of the engine nacelles are not known, the required geometry data are

estimated relatively coarse for the calculation. The flow around the nacelle is turbulent is at

Reynolds number of;

𝑅𝑒 =𝑣𝑙𝑛

𝑣=

236.1𝑚/𝑠 4.8𝑚

15 10−6𝑚2/𝑠= 7.56 107

It is larger than the "cut-off Reynolds number": Recut-off = 5.92 107, so the drag coefficient is

determined:

𝐶𝑓 ,𝑛 =0.455

𝑙𝑜𝑔 𝑅𝑒 2.58 1 + 0.144𝑀2 0.65= 2.17 10−3

The form factor for engine nacelles calculated according to Raymer [Eq. 13.24]

𝐹𝐹𝑛 = 1 +0.35

(𝑙𝑛/𝑑𝑛 )= 1 +

0.35

(4.8𝑚/1.7𝑚)= 1.12

The interference factor for engines on the fuselage is [Tab. 13.5] Qn = 1.3, if the distance from

the fuselage is less than the engine diameter due to the shape of the nacelle on the original

F100. This is used to determine the wetted surface alone, the formula for the fan cowling and

the entire engine is applied [Eq. 13.10] (dimensions according to [Image 13:10]):

𝑆𝑤𝑒𝑡 ,𝑛 = 2 𝑙𝑛 𝐷𝑛 2 + 0.35𝑙𝑙

𝑙𝑛+ 0.8

𝑙𝑙 𝐷𝑕𝑙

𝑙𝑛 𝐷𝑛

+ 1.15 1 −𝑙𝑙

𝑙𝑛

𝐷𝑒𝑓

𝐷𝑛

44

= 2 4.8𝑚 1.7𝑚 2 + 0.35 0.37 + 0.81.75 1.2

4.8 1.7+ 1.15 1 − 0.37 0.5 = 44𝑚2

Because of the lack of precise dimensions or benchmarking unfortunately it is hardly possible to

examine this value for plausibility.

⇒ cD0, V = 2.17 10-3 1.12 1.3(44m2/89m2) = 1.56 10-3

9.6 Overall resistance

As for the two components of drag "other" and "leakage of the pressurized cabin" there are no

standard values or procedures, they are not taken into account at this point. If the overall

resistance of the aircraft is zero then the sum of the above individual resistances:

CDO = CD,F + CD,W + CD,H + CD,V + CD,N

= 5.91 10-3 + 5.21 10-3 + 2.05 10-3 + 1.44 10-3 + 1.56 10-3 = 16.17 10-3 161.7ct

(drag count)

With an assumed Oswald factor for the cruise of e = 0.85 and Aspect ratio A = 8.4, thus the

polar is;

𝐶𝐷 = 0.0162 +𝐶𝐿

2

22.43

Direct operating costs (Direct Operating

Costs, DOC) The calculation of the DOC is done according to the method of the Association of European

Airlines (AEA). This method takes into account the individual cost elements depreciation,

interest, insurance, fuel, maintenance, occupation and fees that are added to the total direct

operating costs of the Aircraft. The DOC can be related to various sizes, for instance, on the

distance flown, the flight time or the distance covered-kilometers. In the present case, the cost

of each aircraft per year (CA / C, a) is determined.

45

Depreciation The depreciation cost is in this case the impairment of the aircraft (difference of the purchase

price and the residual value) based in the period of use of the depreciation in years.

𝐶𝐷𝐸𝑃 =𝑃𝑡𝑜𝑡𝑎𝑙 − 𝑃𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙

𝑛𝐷𝐸𝑃

[Eq. 14.22]

The planned useful process is the AEA method as nDEP = 14 years for short-haul aircraft and the

residual value at 10% of the purchase price [Tab. 14.5]. The (total) purchase price consists of

shares for the delivery price Pdelivery and spare parts PS. The delivery rate can be estimated using

the maximum take-off mass, the dry operating mass or the number of seats. The prices are for

the year 1999, that have yet to be provided with an inflation surcharge. Average annual

inflation rate of 3.3% is assumed. The inflation factor then is

𝑘𝐼𝑁𝐹 = (1 + 0.033)(2001−1999) = 1.067

[Eq. 14:53]

• Estimation of m MTO:

According to [Eq. 14.24], for short and medium-haul aircrafts:

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦

𝑚𝑀𝑇𝑂𝑚𝑂𝐸 𝑘𝐼𝑁𝐹

with,

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦

𝑚𝑀𝑇𝑂 500$/𝑘𝑔

Therefore: Pdelivery = 500 $ / kg 43908kg 1.067 ≈ $23.4million

• Estimation of mOE:

From [Eq. 14.25]

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =𝑃𝑑𝑒𝑙𝑖𝑣 𝑒𝑟𝑦

𝑚𝑀𝑇𝑂

𝑚𝑂𝐸 𝑘𝐼𝑁𝐹

With

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦

𝑚𝑀𝑇𝑂 860$/𝑘𝑔

46

So, Pdelivery = 860$ / kg 22943kg 1.067 ≈ $21.1million

• Estimation of n :

From [Eq. 14.26]

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦

𝑛𝑝𝑎𝑥

𝑛𝑝𝑎𝑥 𝑘𝐼𝑁𝐹

With

𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦

𝑛𝑝𝑎𝑥= $265000

Therefore: Pdelivery = $265000 107 1.067 ≈ $30.1 million

Averaging over the three approaches provides

Pdelivery = 1/3 (23.4 + 21.1 + 30.1) million$ = $ 24.9 million

The price of spare parts is composed of an engine and cells share together, each given a

percentage of the delivery price, they are:

PS = (KS, AF PAF + KS, E nE PE) kINF [Eq. 14:27]

The cell price is the price excluding aircraft engines, according to [Tab.14.5] kS, AF = 0.1 and

kS,E = 0.3. Jenkinson gives the equation for the engine price estimate

Ps = [kS,AF (Pdel - nePE) + kS,EnEPE] KINF = [0.1($22.7mil. – 2 $3.9mil.) + 0.3 2 $3.9mil.]1.067

= $4.09mil.

Ptotal = Pdelivery + PS = $24.9mil. + $4.09mil. = $28.99mil.

Used in [Eq. 14.22] gives:

𝐶𝐷𝐸𝑃 =𝑃𝑡𝑜𝑡𝑎𝑙 1 −

𝑃𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙

𝑃𝑡𝑜𝑡𝑎𝑙

𝑛𝐷𝐸𝑃=

$28.99𝑚𝑖𝑙 . −0.9

14𝑦𝑒𝑎𝑟𝑠= $1.864𝑚𝑖𝑙 ./𝑦𝑒𝑎𝑟

47

Interest Assuming that the aircraft is on loan to be one hundred percent, the estimate of the annual

interest

CINT = pav Ptotal [Eq. 14:30]

This is the average interest rate with the given AEA data rate financing 8%, given that the

financing period of 14 years old, depreciation period of 14 years old and the relative residual

value of debt in the company after the end of the funding period will be 10%

pav = 0.0529 [Tab. 14.6]

CINT = 0.0529 $28.99mil./year = 1,534Mio $ / year

Insurance The insurance cost per year as a percentage of the delivery price is obtained.

CINS = KINS Pdelivery [Eq. 14:35]

The AEA method is based on kINS = 0.005

CINS = 0.005 = $ 24.9 million 0,125Mio $ / year.

Fuel costs Fuel costs per year resulting from the product of the sizes, number of flights, years, the mass of

the fuel consumed during a flight and the fuel price.

CF = nt,a mF PF [Eq. 14:36]

The fuel price is subject to significant fluctuations. As a basis for this statement a price of

$2.24/US gallon is considered (according to specification http://www.jetaviation.com, as of July

2001). Calculation should include 1 US gallon corresponding to 3785cc. At a density of Kerosene =

0.76g / cc has a US gallon a mass of 2.877kg. So the kg price for jet fuel will be $0.78.

The spent fuel mass per flight will be as described in Section 2.7 for [Eq. 5:53] calculated:

𝑚𝑓

𝑚𝑀𝑇𝑂= (1 − 𝑀𝑓𝑓 )

48

Mff consists of the mass ratios for the individual phases of flight Rosakam[image 5.19] together:

𝑀𝑓𝑓 =𝑚9

𝑚8𝑚8

𝑚7𝑚7

𝑚6𝑚6

𝑚5𝑚5

𝑚4𝑚4

𝑚3𝑚3

𝑚2𝑚2

𝑚1= 0.9920.99

𝑚7

𝑚6𝑚6

𝑚50.980.9950.990.99

The eventual holding pattern goes flying into the DOC invoice is not it, i.e. m7/m6 = 1.

The unknown mass ratio for the cruise is determined by Breguet.

The average flown flight segment scruise is by Obert [Tab.3.2] for short to medium-haul aircraft at

about 17-20% range. That is, scruise = 0.185 3167km = 586km. The mass ratio for the cruise is

[Eq. 5.56]

𝑚6

𝑚5= 𝑒

−𝑠𝑐𝑟

𝐵𝑠

The Breguet factor Bs has already been calculated in the dimensioning: Bs = 24608689m

This gives us,

𝑚6

𝑚5= 𝑒−

586000 𝑚

24608689 𝑚 = 0.976

By substituting we obtain Mff = 0.916 and

mF = (1 - 0.916) 43908kg = 3688kg

For the determination of the missing size nT, A (Number of flights per year) is the flight time per

average flight tf is required. The flight time is by itself [Image 14.5] composed of the duration of

the flight phases B and start climbing to 1500ft, C-Climb to cruise altitude, D-trip flight, E-

descent to 1500ft and F-landing and landing. The phases B, C, E and F are taken into

consideration and are simplified by a pure cruising gone forth 586km.

Under this assumption the time factor of Breguet rule is the flight time charge. The mass ratio

m1/m0 is obtained from the above calculated Mff without the phases of flight; engine start and

rolling.

𝑡𝑓 = −𝐵𝑡 𝑙𝑛𝑚1

𝑚0

= −𝐿 𝐷

𝑆𝐹𝐶𝑇 g 𝑙𝑛

𝑀𝑓𝑓

0.99 0.99= 7289𝑠 121𝑚𝑖𝑛

[Eq. D.14]

According to [Eq. 14.72] the aircraft can use Ua,f (flight time per year) that is determined.

49

𝑈𝑎 ,𝑓 = 𝑡𝑓 𝑘𝑢1

𝑡𝑓 + 𝑘𝑈2= 2𝑕

3750𝑕

2𝑕 + 0.750𝑕= 2727.3𝑕

The parameter kU1 and ku2 were from [Tab. 14:11] chosen for the process of AEA.

This will be calculated for a year

𝑛𝑡 ,𝑎 =𝑈𝑎 ,𝑓

𝑡𝑓=

2727.3𝑕

2𝑕= 1364

Insertion into [Eq. 14:36] yields

𝐶𝑝 = 𝑛𝑡 ,𝑎 𝑚𝑓 𝑃𝐹 = 13641

𝑦𝑒𝑎𝑟3688kg $0.78 / kg = $3.924mil./year

Maintenance Costs The maintenance cost CM consist of the shares personnel costs CM, L and material costs CM, M.

They are initially determined based on a one hour flight and then multiplied by the total flight

time per year. [Eq. 14:41] yields the maintenance costs, thus

𝐶𝑀 = (𝑡𝑀,𝑓 𝐿𝑀 + 𝐶𝑀,𝑀,𝑓)𝑡𝑓 𝑛𝑡,𝑎

with tM, f - Maintenance time per flight hour,

LM - Hourly rate, LM = 69 $ / hkINF = $69/h 1.067 = $73.6/h

CM, M, f - Material cost per flight hour

tf - Flying time per flight, tf = 121min

nT, A - Number of flights per year, nT, A= 1364 1 / year

Because the maintenance costs for engine and airframe in the composition will differ greatly

from wage and material proportion of the total costs for the Cell(airframe, AF) and for the

engines (engines, E):

𝐶𝑀 = ((𝑡𝑀,𝐴𝐹,𝑓 + 𝑡𝑀,𝐸,𝑓) 𝐿𝑀 + 𝐶𝑀 ,𝑀,𝐴𝐹 ,𝑓 + 𝐶𝑀,𝑀,𝐸,𝑓 )𝑡𝑓 𝑛𝑡,𝑎

[Eq. 14:43]

50

The unknown quantities are Maintenance hours per flight hour [Eq. 14.44]:

𝑡𝑀,𝐴𝐹 ,𝑓 =1

𝑡𝑓 910−5

1

𝑘𝑔𝑚𝐴𝐹 + 6.7 −

350000𝑘𝑔

𝑚𝐴𝐹 + 75000𝑘𝑔 (0.8h + 0.68𝑡𝑓)

=1

2𝑕 910−5

1

𝑘𝑔 22943 − 3181 kg + 6.7 −

350000𝑘𝑔

22943 − 3181 𝑘𝑔 + 75000𝑘𝑔 (0.8h

+ 0.682h)

= 5.2(𝑀𝑀𝐻 / 𝐹𝐻)

Material cost cell [Eq. 14:45]:

𝐶𝑀 ,𝑀,𝐴𝐹 ,𝑓 =1

𝑡𝑓 4.2 10−6 + 2.210−6

1

𝑕𝑃𝐴𝐹

=1

2𝑕 4.210−6 + 2.210−6

1

𝑕2h $24.9mil. −2$3.9mil. = $73.5/𝑕

To calculate the corresponding quantities for the engines we must use the first four factors

k1….k4 are determined (BPR: bypass ratio, OAPR: total pressure ratio, nC: Number of compressor

stages). The source of engine parameters served the Aviation Homepage.

𝑘1 = 1.27 − 0.2 𝐵𝑃𝑅0,2 = 1.27 − 0.2 3.040,2 = 1.020 [Eq. 14.49]

𝑘2 = 0.4 𝑂𝐴𝑃𝑅

20

1,3+ 0.4 = 0.4

15.8

20

1,3+ 0.4 = 0.694 [Eq. 14.50]

𝑘4 = 0.57 ns = 2 is the number of TW-waves [Eq. 14.51]

𝑘3 = 0.032 𝑛𝐶 + 𝑘4 = 0.032 16 + 0.57 = 1.082 [Eq. 14.52]

Maintenance engine hours per flight hour [Eq. 14:46]:

𝑡𝑀,𝐸,𝑓 = 𝑛𝐸 0.21 𝑘1 𝑘3 1 + 1.02 10−4 1

𝑁 𝑇𝑇𝑂 ,𝐸

0,4

1 +1.3𝑕

𝑡𝑓

= 2 0.21 1.02 1.082 1 + 1.02 10−4 1

𝑁 (123185N/2)

0,4

1 +1.3𝑕

2𝑕

= 1.69 𝑀𝑀𝐻 𝐹𝐻

Material costs of engine [Eq. 14:47]:

𝐶𝑀.𝑀,𝐸,𝑓 = 𝑛𝐸 $2.56

h 𝑘1 𝑘2 + 𝑘3 1 + 1.02 10−4

1

𝑁 𝑇𝑇𝑂 ,𝐸

0.8

1 +1.3𝑕

𝑡𝑓 𝑘𝐼𝑁𝐹

51

= 2 $2.56

h 1.02 0.694 + 1.082 1 + 1.02 10−4

1

𝑁

123185𝑁

2

0.8

1 +1.3𝑕

2𝑕 (1

+ 0.033)(2001 −1989) = $100/𝑕

Substituting in [Eq. 14:43] is now obtained overall maintenance costs:

𝐶𝑀 = 5.2 + 1.69 73.6 $/h + 73.5 $/h + 110 $/h 2h 13641

year= 1.884mil. $/year

Personnel costs Staff costs consist of payment for the Cockpit Crew (CO) and those for the Cabin Crew (CA) that

are paid at an average hourly rate LCO. LCA for the block time tb

CC = (nCO LCO + nCA LCA ) tb nt,a [Eq. 14:58]

According to the AEA block time (flight time plus ground times for roles, waiting, etc.) for short-

haul flights is 15 min on the time of flight. [Tab.14.4]

tb = Tf + 0.25 h = 2.25h

In the cockpit, the pilot and co-pilot are required, nCO = 2; going by the AEA method, a member

of the crew in the cabin per 35 passengers is assumed, nCA = 4. Hourly rates are then assumed in

the AEA method for short-haul aircraft:

LCO = $246.5 /h,

LCA = $ 81.0 / h [Tab. 14.9]

For the staff costs assuming the following values:

CC = (2 $246.5 / h + 4 $81.0 / h) 2.25h 1364 (1 / year) = $2.507mil. / year

Fees The fees include landing fees, air traffic control and clearance fees. These elements are first

calculated for 1 flight and then multiplied by the number of flights per year to the annual costs

that occur. The fees are based on the rates calculated from 1989 (year of AEA publication

method). It is assumed that there would be a higher inflation than in the other cost elements:

pINF = 6.5%. The inflation factor is thus:

52

𝑘𝐼𝑁𝐹 = (1 + 𝑝𝐼𝑁𝐹)(2001−1989) = 1.06512 = 2.13

The individual fees elements are:

Landing fees [Eq. 14.60]:

𝐶𝐹𝐸𝐸,𝐿𝐷 = 𝑘𝐿𝐷 𝑚𝑀𝑇𝑂 𝑘𝐼𝑁𝐹 𝑛𝑡 ,𝑎 = 0.0078$

𝑘𝑔 43908kg 2.13 1364

1

year

= 995021 $/𝑦𝑒𝑎𝑟

Fees air traffic control [Eq. 14.61]:

𝐶𝐹𝐸𝐸,𝑁𝐴𝑉 = 𝑘𝑁𝐴𝑉 R 𝑚𝑀𝑇𝑂 𝑘𝐼𝑁𝐹 𝑛𝑡 ,𝑎

= 0.00414$

𝑛𝑚 𝑘𝑔316.4nm 43908𝑘𝑔2.13 1364

1

year= 797448 $/year

Handling charges [Eq. 14.62]:

𝐶𝐹𝐸𝐸,𝐺𝑁𝐷 = 𝑘𝐺𝑁𝐷 𝑚𝑃𝐿 𝑘𝐼𝑁𝐹 𝑛𝑡 ,𝑎 = 0.1$

𝑘𝑔12228kg 2.13 1364

1

year

= 3.553𝑚𝑖𝑙 . $/𝑦𝑒𝑎𝑟

The total fees per year is therefore amounted to

𝐶𝐹𝐸𝐸 = 𝐶𝐹𝐸𝐸,𝐿𝐷 + 𝐶𝐹𝐸𝐸,𝑁𝐴𝑉 + 𝐶𝐹𝐸𝐸,𝐺𝑁𝐷 = 0.995 + 0.797 + 3.553 𝑚𝑖𝑙 .$

𝑦𝑒𝑎𝑟

= 5.345𝑚𝑖𝑙 .$

𝑦𝑒𝑎𝑟

53

Overall presentation The total annual direct operating cost is CDO = 17,183Mio $ / year and the division of all costs

into individual portions is illustrated in Fig.9.1.

Figure 10.1: Composition of the DOC

54

Summary At this point, the compilation of the most important geometric follows the Aircraft parameters,

determined as described in the previous contents. A Comparative data on the original Fokker

F100 is done.

The calculated parameters are a three-page view of the aircraft (Fig.10.1).

Design Original

General

Length 35.40m 35.53m

Overall height 8.10m 8.50m

Fuselage

Length 32.14m 32.50m

Maximum Diameter 3.31m 3.30m

Emergency exits, width 0.51m 0.51m

Height 0.91m 0.91m

Rear angle 13.0 ° -

Cabin length 22.86m 21.19m

Cabin height 2.01m 2.01m

Cabin width (floor) 2.86m 2.88M

Cargo volume ≈ 17m³ 17.08m³

Wing

Span 27.40m 28.08m

Area 89.0m² 93.5m²

Sweep (25% line) 19.0 ° 17.5 °

Avg. aerodynamic Chord 3.80m 3.80m

Worsening 0.230 0.235

Avg. relative Thickness 10.9% 10.3%

V-angle 3.5 ° 2.5 °

Tail-plane

Area 19.58m² 21.72m²

Span 8.85m 10.00m

Stretching 4.0 4.6

Worsening 0.40 0.39

Sweep (25% line) 23 ° 26 °

Fin

Area 12m² 12.3m²

Height 3.46m 3.30m

Stretching 1.00 0.89

Worsening 0.90 0.74

Sweep (25% line) 45 ° 41 °

Landing gear

Gauge 5.00m 5.04m

Wheelbase 14.25m 14.01m

Tire size, main-FW 40 "x14" 40 "x14"

Bug-FW 24 "X7.7" 24 "X7.7"

55

2-D drawings:

Figure: Three side view of the designed F100

56

3-D design in CATIA:

3-view drawings used as stickers for 3-D projection

Structural model design in accordance with the dimensions and 3-view reference

57

Shaping and Finishing done

Full model

58

References

Printed Sources

[1] Theory of Wing Sections, by IRA H. ABBOTT

[2] Aircraft Landing Gear Design: Principles and Practices, by N.S.CURREY

[3] USAF Stability and Control Datcom, by D.E.HOAK

[4] U.S. Department for Transportation, FEDERAL AVIATION ADMINISTRATION

[5] Joint Aviation Requirements, JAR-2, Large Aeroplanes, by JOINT AVIATION AUTHORITIES

[6] Subsonic Aircraft: Evolution and the Matching of Size to Performance, by L.K.LOFTIN

[7] Aircraft Design: A Conceptual Approach, by D.P.RAYMER

[8] Airplane Design, by J.ROSKAM

[9] Synthesis of Subsonic Airplane Design, by E.TORENBEEK

Internet sources

JetAviation http://www.jetaviation.com