2. 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
3. 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
4. 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.
5. 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 2
nd
Segment · 8
References · 63
S
Summary · 59
T
Take-off Distance · 8
W
Weight and Balance · 29
Wing area and Takeoff-thrust · 14
6. 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].
7. 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
8. 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 + ( CL
2
/ . 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
9. 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
10. 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]:
𝑚 𝑀𝑇𝑂 𝑆 𝑤 =
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’.
11. 6
𝐶𝐿,𝑚 = 𝐶 𝐷0. 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
𝑚 𝑀𝑇𝑂
𝑆 𝑊
and
𝑇 𝑇𝑂
𝑚 𝑀𝑇𝑂 .𝑔
can be now drawn in a layout
diagram.
Design diagram
Figure 2.2: Design Graph
12. 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 𝑚 𝑀𝑇𝑂 𝑆 𝑤 = 495𝑘𝑔/𝑚2
Thrust-weight ratio 𝑇𝑇𝑂 𝑚 𝑀𝑇𝑂 . 𝑔 = 0.285
Raymer [Tab.5.3 u. 5.4] are as typical values for jet airliners 𝑚 𝑀𝑇𝑂 𝑆 𝑤 = 586𝑘𝑔/𝑚2
and
𝑇𝑇𝑂 𝑚 𝑀𝑇𝑂 . 𝑔 = 0.25 in the data obtained from the diagram is plausible.
The actual values of the F100 lie in:
𝑚 𝑀𝑇𝑂 𝑆 𝑤 = 489.9𝑘𝑔/𝑚2 and 𝑇𝑇𝑂 𝑚 𝑀𝑇𝑂 . 𝑔 = 0.293
Maximum take-off mass 𝑚 𝑀𝑇𝑂
The maximum Take-Off mass consists of the units operating mass 𝑚 𝑂𝐸, Fuel mass 𝑚 𝐹 and
maximum payload 𝑚 𝑀𝑃𝐿, by rearranging we obtain [Eq. 5.47]
𝑚 𝑀𝑇𝑂 =
𝑚 𝑀𝑃𝐿
1 −
𝑚 𝐹
𝑚 𝑀𝑇𝑂
−
𝑚 𝑂𝐸
𝑚 𝑀𝑇𝑂
As we know the payload; the value of the original F100 used: 𝑚 𝑀𝑃𝐿 = 12228𝑘𝑔. The mass
fraction after statistics is estimated: According to Torenbeek (Figure 5.15), this is for short-haul
jet airliners 𝑚 𝑂𝐸 𝑚 𝑀𝑇𝑂 = 0.53. Loftin (Eq. 5.50) shows
𝑚 𝑂𝐸
𝑚 𝑀𝑇𝑂
= 0.23 + 1.04 .
𝑇𝑇𝑂
𝑚 𝑀𝑇𝑂 . 𝑔
= 0.23 + 1.04 . 0.285 = 0.53
From the equation of Marckwardt [Eq. 5:48] 𝑚 𝑂𝐸 𝑚 𝑀𝑇𝑂 = 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.
13. 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:
𝑚 𝐹
𝑚 𝑀𝑇𝑂
= (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 . 𝑣 𝐶𝑟
𝑆𝐹𝐶 𝑇. 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
𝐵 𝑡 = 𝑒−
2700 𝑠
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
14. 9
The fuel portion is the difference to 1: 𝑚 𝑂𝐸 𝑚 𝑀𝑇𝑂 = 1 − 0.801 = 0.199. Used in [Eq. 5:47],
we obtain the maximum take-off weight:
𝑚 𝑀𝑇𝑂 =
𝑚 𝑀𝑃𝐿
1 −
𝑚 𝐹
𝑚 𝑀𝑇𝑂
−
𝑚 𝑂𝐸
𝑚 𝑀𝑇𝑂
=
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:
𝑆 𝑊 =
𝑚 𝑀𝑇𝑂
𝑚 𝑀𝑇𝑂
𝑆 𝑊
=
46832𝑘𝑔
495𝑘𝑔/𝑚2
= 95𝑚2
The takeoff thrust is calculated by multiplying the thrust-to-weight ratio:
𝑇𝑇𝑂 = 𝑚 𝑀𝑇𝑂 . 𝑔.
𝑇𝑇𝑂
𝑚 𝑀𝑇𝑂 . 𝑔
=
46832𝑘𝑔9.81𝑚
𝑠2
0.285 = 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
15. 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]
𝑛 𝑠𝑎 = 0.45 𝑛 𝑝𝑎𝑥
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)
16. 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
𝑛 𝑝𝑎𝑥
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
17. 12
𝑙 𝐹 = 𝑙 𝐶𝐴𝐵𝐼𝑁 + 1.6 𝑑 𝐹 + 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
18. 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 18
𝑘𝑔
𝑝𝑎𝑥
107𝑝𝑎𝑥
1
170 𝑘𝑔/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
19. 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 °
20. 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
= 𝑡𝑎𝑛 𝐿𝐸
−
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
21. 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
𝑀 𝐷𝐷,𝑒𝑓𝑓 = 𝑀 𝐷𝐷 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:
22. 17
𝑉𝑇𝑎𝑛𝑘 = 0.54 𝑆 𝑊1.5 𝑡 𝑐 𝑟
1
𝐴
1 + + 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
23. 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
𝐶 𝐿 ,𝑚𝑎𝑥
𝐶 𝐿 ,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
𝑆 𝑊,𝑓
𝑆 𝑊
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 ΔC 𝐿,𝑚𝑎𝑥 𝑏𝑎𝑠𝑒
ΔC 𝐿,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 ΔC 𝐿,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
2 𝑆
𝑏 1 +
=
2.95𝑚2
28.2𝑚 1.23
= 5.48𝑚
24. 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
𝑆 𝑊,𝑓 𝑆 𝑊 = 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.
25. 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%
26. 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]
𝐶 𝐻 =
𝑆 𝐻 𝑙 𝐻
𝑆 𝑊 𝐶 𝑀𝐴𝐶
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
𝐶 𝑀𝐴𝐶 =
2
3
𝐶𝑟
1 + + 2
1 +
Here is the chord at the root
𝐶𝑟 =
2 𝑆 𝑊
𝑏(1 + )
=
2.95𝑚2
28.3𝑚 (1 + 0.23)
= 5.46𝑚
27. 22
Hence,
𝐶 𝑀𝐴𝐶 =
2
3
5.46𝑚
1 + 0.23 + 0.232
1 + 0.23
= 3.80𝑚
With these values can now be determined CPR area:
𝑆 𝐻 =
𝐶 𝐻 𝑆 𝑊 𝐶 𝑀𝐴𝐶
𝑙 𝐻
=
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
𝑀 𝐷𝐷,𝑒𝑓𝑓 = 𝑀 𝐷𝐷 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.
28. 23
• Surface
The SLW-surface, like the CPR area over the corresponding coefficient volume is done [Eq. 9.5]:
𝐶𝑣 =
𝑆 𝑣 𝑙 𝑣
𝑆 𝑤 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:
𝑆 𝑉 =
𝐶 𝑉 𝑆 𝑊 𝑏
𝑙 𝑉
=
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.
29. 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:
𝑆 𝑤𝑒𝑡 = 𝑆𝑧𝑦𝑙 + 𝑆 𝐾𝑜𝑛 = 2 𝑟𝑓,𝑒𝑥𝑡 𝑙 𝑧𝑦𝑙 + 𝑟𝑓,𝑒𝑥𝑡
2 + 𝑟𝑓,𝑒𝑥𝑡 𝑟𝑓,𝑒𝑥𝑡
2 + 𝑙 𝑘𝑜𝑛
2
= 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:
𝑚 𝐸 =
0.0724
𝑔
𝑇𝑇𝑂
1.1
𝑒−0.045 𝐵𝑃𝑅
=
0.0724
9.81 𝑚
𝑠2
65.5 103
𝑁1.1
𝑒−0.0455
= 1170𝑘𝑔
Both engines together thus have a mass of m2E = 2340kg
30. 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:
𝑆 𝑊,𝑛𝑒𝑢 =
𝑚 𝑀𝑇𝑂
𝑚 𝑠
=
44060𝑘𝑔
495𝑘𝑔/𝑚2
= 89𝑚2
𝑇𝑇𝑂,𝑛𝑒𝑢 = 𝑚 𝑀𝑇𝑂 𝑔
𝑇
𝑚 𝑔
= 44060𝑘𝑔9.81
𝑚
𝑠2
0.285 = 123185𝑁
31. 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).
32. 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.
𝑥 𝐿𝐸𝑀𝐴𝐶 = 𝑋 𝐹𝐺 − 𝑋 𝐶𝐺,𝐿𝐸𝑀𝐴𝐶 +
𝑀 𝑊𝐺
𝑀 𝐹𝐺
𝑋 𝑊𝐺,𝐿𝐸𝑀𝐴𝐶 − 𝑋 𝐶𝐺,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.
33. 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.
𝑆 𝐻
𝑆 𝑊
= 𝑎 𝑥 𝐶𝐺−𝐴𝐶 + 𝑏
According to Eq. 11:19
𝑎 =
𝐶𝐿
𝐶𝐿η 𝐻
𝑙 𝐻
𝐶 𝑀𝐴𝐶
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
34. 29
𝑏 =
𝐶 𝑀,𝑊 + 𝐶 𝑀,𝐸
𝐶𝐿,𝐻 η 𝐻
𝑙 𝐻
𝐶 𝑀𝐴𝐶
with the parameters of coefficient moments by the engines:
𝐶 𝑀,𝐸 =
−𝑇 𝑧 𝐸
𝑞 𝑆 𝑊 𝐶 𝑀𝐴𝐶
=
−123185𝑁0.9𝑚
1 2 1.225
𝑘𝑔
𝑚3
4369.2
𝑚2
𝑠2
89𝑚23.8𝑚
= −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]:
𝐶 𝑀,𝐸 = 𝐶 𝑀,0,𝑓𝑙𝑎𝑝𝑝𝑒𝑑
𝐴 𝑐𝑜𝑠225
𝐴 + 2𝑐𝑜𝑠25
+
𝐶 𝑀,𝑜
εt
ε 𝑡
𝐶 𝑀,𝑜 𝑀
𝐶 𝑀,𝑜 𝑀−𝑜
= −0.349
8.4 𝑐𝑜𝑠2 19
8.4 + 𝑐𝑜𝑠2 19
+ −0.005 2 1 = −0.2447
Coefficient moment of the profile in the neutral point [Eq. 11.30]:
𝐶 𝑀,0,𝑓𝑙𝑎𝑝𝑝𝑒𝑑 = 𝐶 𝑀,𝑜 + 𝐶 𝑀 = 𝐶 𝑀,𝑜 + 𝐶𝐿,𝑓𝑙𝑎𝑝𝑝𝑒𝑑
𝑥 𝐴𝐶
𝐶 𝑀𝐴𝐶
−
𝐶 𝑀,𝑜
𝐶 𝑀𝐴𝐶
𝑐′
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
Δ𝐶 𝑀,𝑜
εt
35. 30
The Mach number effect
𝐶 𝑀,𝑜 𝑀
𝐶 𝑀,𝑜 𝑀−0
,according to [Image 11,15]: the speed is negligible during the landing approach
⇒ b = 0.2097
This is the linear equation:
𝑆 𝐻
𝑆 𝑊
= −0.9074 𝑥 𝐶𝐺−𝐴𝐶 + 0.2097
Design for stability requirement
In this case, the linear equation is in the form
𝑆 𝐻
𝑆 𝑊
= 𝑎 𝑥 𝐶𝐺−𝐴𝐶
This is according to [Eq. 11:24]
𝑎 =
𝐶𝐿,,𝑊
𝐶𝐿,,𝑊η 𝐻1 −
e
𝑙 𝐻
𝐶 𝑀𝐴𝐶
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 =
2 𝐴
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]:
36. 31
e
= 4.44 𝐾𝐴 𝐾 𝐾 𝐻 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]
𝑘 𝐴 =
10 − 3
7
=
10 − 30.4
7
= 1.257
Location factor CPR after [Eq. 11.37]
𝑘 𝐻 =
1 −
𝑍 𝐻
𝑏
2𝑙 𝐻
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:
𝑆 𝐻
𝑆 𝑊
= 1.0245 𝑥 𝐶𝐺−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.
37. 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:
𝑆 𝑣 =
𝑁𝐸 + 𝑁 𝐷
1
2
𝑣 𝑀𝐶
2 𝐹
𝐶 𝐿,
𝐶 𝐿,α 𝑡 𝑒𝑜𝑟𝑦
𝐶𝐿, 𝑡𝑒𝑜𝑟𝑦
𝐾 𝐼 𝐾Λ 𝑙 𝑣
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:
38. 33
𝑁𝐸 + 𝑁 𝐷 = 1.25𝑁𝐸 = 1.25 𝑇𝑇𝑂 𝑦 𝐸 = 1.25
123185𝑁
2
2.7𝑚 = 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:
𝑣 𝑀𝐶 = 1.2𝑣 𝑆,𝑇𝑂 𝑣 𝐿𝑂𝐹 =
2𝑔
𝑚 𝑇𝑂
𝑆 𝑊
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.
39. 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
40. 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:
41. 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:
42. 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.
43. 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
𝐴 𝑐 = 2
𝑡𝑖𝑟𝑒𝑙𝑜𝑎𝑑
𝑡𝑖𝑟𝑒𝑝𝑟𝑒𝑠𝑠𝑢𝑟 𝑒
=
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
𝐶 𝐷 = 𝐶 𝐷𝑂 +
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]:
𝐶 𝐷𝑂,𝑐 = 𝐶𝑓,𝑐 𝐹𝐹𝑐 𝑄𝑐
𝑆 𝑤𝑒𝑡,𝑐
𝑆𝑟𝑒𝑓
44. 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
𝑆 𝑤𝑒𝑡 ,𝑐
𝑆 𝑟𝑒𝑓
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 =
𝑣 𝑙𝑓
𝑣
=
𝑣 𝑙𝑓
15 10−6 𝑚2/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
log4.38108 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
45. 40
[Eq. 13:23]
The interference factors are related to the hull, ⇒ Qf = 1. The wetted surface of fuselage with
cylindrical midsection from Torenbeek:
𝑆 𝑤𝑒𝑡 ,𝑓 = 𝑑𝑓 𝑙𝑓 1 −
2
𝑓
273
1 +
2
𝑓
2
[Eq. 13.7]
= 3.3𝑚 32.14𝑚 1 −
2
9.7
273
1 +
2
9.72
= 291.7𝑚2
𝑆 𝑤𝑒𝑡 ,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
𝑣 𝐶𝑟 𝐶 𝑀𝐴𝐶
𝑣
= 1.328
236.1𝑚/𝑠
15 10−6 𝑚2/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
46. 41
𝑆 𝑤𝑒𝑡 ,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]
𝑆 𝑤𝑒𝑡 ,𝑓
𝑆𝑟𝑒𝑓
=
167.3
89
= 1.88
According to [Eq. 13:22] calculating the shape factor;
𝐹𝐹 𝑊 = 1 +
0.6
𝑥 𝑡
𝑡 𝑐 + 100 𝑡 𝑐 4
1.34 𝑀0.18
𝑐𝑜𝑠 𝑚
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 =
𝑣 𝑐 𝑀𝐴𝐶,𝐻
𝑣
=
236.1
𝑚
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 + 0.144 𝑀2 0.65
= 2.43 10−3
47. 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]:
𝑆 𝑤𝑒𝑡 ,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
𝑆 𝑤𝑒𝑡 ,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 =
𝑣 𝑐 𝑀𝐴𝐶,𝐻
𝑣
=
236.1
𝑚
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 + 0.144 𝑀2 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]
48. 43
𝑆 𝑤𝑒𝑡 ,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
𝑆 𝑤𝑒𝑡 ,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 =
𝑣 𝑙 𝑛
𝑣
=
236.1𝑚/𝑠 4.8𝑚
15 10−6 𝑚2/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 + 0.144 𝑀2 0.65
= 2.17 10−3
The form factor for engine nacelles calculated according to Raymer [Eq. 13.24]
𝐹𝐹𝑛 = 1 +
0.35
(𝑙 𝑛/𝑑 𝑛 )
= 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]):
𝑆 𝑤𝑒𝑡 ,𝑛 = 2 𝑙 𝑛 𝐷 𝑛 2 + 0.35
𝑙𝑙
𝑙 𝑛
+ 0.8
𝑙𝑙 𝐷𝑙
𝑙 𝑛 𝐷 𝑛
+ 1.15 1 −
𝑙𝑙
𝑙 𝑛
𝐷𝑒𝑓
𝐷 𝑛
49. 44
= 2 4.8𝑚 1.7𝑚 2 + 0.35 0.37 + 0.8
1.75 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;
𝐶 𝐷 = 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.
50. 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.
𝐶 𝐷𝐸𝑃 =
𝑃𝑡𝑜𝑡𝑎𝑙 − 𝑃𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝑛 𝐷𝐸𝑃
[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 mMTO:
According to [Eq. 14.24], for short and medium-haul aircrafts:
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑚 𝑀𝑇𝑂
𝑚 𝑂𝐸 𝑘𝐼𝑁𝐹
with,
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑚 𝑀𝑇𝑂
500$/𝑘𝑔
Therefore: Pdelivery = 500 $ / kg 43908kg 1.067 ≈ $23.4million
• Estimation of mOE:
From [Eq. 14.25]
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =
𝑃𝑑𝑒𝑙𝑖𝑣 𝑒𝑟𝑦
𝑚 𝑀𝑇𝑂
𝑚 𝑂𝐸 𝑘𝐼𝑁𝐹
With
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑚 𝑀𝑇𝑂
860$/𝑘𝑔
51. 46
So, Pdelivery = 860$ / kg 22943kg 1.067 ≈ $21.1million
• Estimation of n :
From [Eq. 14.26]
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦 =
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑛 𝑝𝑎𝑥
𝑛 𝑝𝑎𝑥 𝑘𝐼𝑁𝐹
With
𝑃𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦
𝑛 𝑝𝑎𝑥
= $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:
𝐶 𝐷𝐸𝑃 =
𝑃𝑡𝑜𝑡𝑎𝑙 1 −
𝑃 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝑃 𝑡𝑜𝑡𝑎𝑙
𝑛 𝐷𝐸𝑃
=
$28.99𝑚𝑖𝑙. −0.9
14𝑦𝑒𝑎𝑟𝑠
= $1.864𝑚𝑖𝑙./𝑦𝑒𝑎𝑟
52. 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
𝑚 𝑀𝑇𝑂
= (1 − 𝑀𝑓𝑓 )
53. 48
Mff consists of the mass ratios for the individual phases of flight Rosakam[image 5.19] together:
𝑀𝑓𝑓 =
𝑚9
𝑚8
𝑚8
𝑚7
𝑚7
𝑚6
𝑚6
𝑚5
𝑚5
𝑚4
𝑚4
𝑚3
𝑚3
𝑚2
𝑚2
𝑚1
= 0.9920.99
𝑚7
𝑚6
𝑚6
𝑚5
0.980.9950.990.99
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
𝐵 𝑠
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 𝑙𝑛
𝑚1
𝑚0
= −
𝐿 𝐷
𝑆𝐹𝐶 𝑇g
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.
54. 49
𝑈 𝑎,𝑓 = 𝑡𝑓
𝑘 𝑢1
𝑡𝑓 + 𝑘 𝑈2
= 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 =
𝑈 𝑎,𝑓
𝑡𝑓
=
2727.3
2
= 1364
Insertion into [Eq. 14:36] yields
𝐶 𝑝 = 𝑛𝑡,𝑎 𝑚 𝑓𝑃𝐹 = 1364
1
𝑦𝑒𝑎𝑟
3688kg $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
𝐶 𝑀 = (𝑡 𝑀,𝑓 𝐿 𝑀 + 𝐶 𝑀,𝑀,𝑓)𝑡𝑓𝑛𝑡,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):
𝐶 𝑀 = ((𝑡 𝑀,𝐴𝐹,𝑓 + 𝑡 𝑀,𝐸,𝑓) 𝐿 𝑀 + 𝐶 𝑀,𝑀,𝐴𝐹,𝑓 + 𝐶 𝑀,𝑀,𝐸,𝑓 )𝑡𝑓𝑛𝑡,𝑎
[Eq. 14:43]
55. 50
The unknown quantities are Maintenance hours per flight hour [Eq. 14.44]:
𝑡 𝑀,𝐴𝐹,𝑓 =
1
𝑡𝑓
910−5
1
𝑘𝑔
𝑚 𝐴𝐹 + 6.7 −
350000𝑘𝑔
𝑚 𝐴𝐹 + 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]:
𝐶 𝑀,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 𝑛 𝐶 + 𝑘4 = 0.032 16 + 0.57 = 1.082 [Eq. 14.52]
Maintenance engine hours per flight hour [Eq. 14:46]:
𝑡 𝑀,𝐸,𝑓 = 𝑛 𝐸 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]:
𝐶 𝑀.𝑀,𝐸,𝑓 = 𝑛 𝐸
$2.56
h
𝑘1 𝑘2 + 𝑘3 1 + 1.02 10−4
1
𝑁
𝑇𝑇𝑂,𝐸
0.8
1 +
1.3
𝑡𝑓
𝑘𝐼𝑁𝐹
56. 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:
𝐶 𝑀 = 5.2 + 1.69 73.6 $/h + 73.5 $/h + 110 $/h 2h 1364
1
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:
58. 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
59. 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"
61. 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
63. 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
