% -
2 2 P--~L ~ \ - - f f - ] (2.21)
o r
where
Under these conditions D + R
Wro so that Eq. (2.5) becomes
rsL
WTO
(CD -~- CDR -- I~roCL)qS -t- tXTo~WTo t ~ WTO
{ fl('~OTO) l d V } fl ~ro q S + / X r o + - -
ot go
~ro = CD + CoR -/zroCL (2.24)
which can be rearranged and integrated, as in Case 5, to yield
{ / I ( ~ "~CLmaxl}
( 2 . 2 5 )_ fi(Wro/S) In 1 - ~ro TSL IZro]
SG Pgo~ro WTO
provided that representative takeoff values of ot and fl are used.
For this case TSL/Wro must increase continuously with Wro/S, but in a more complex manner than in Case 5. That result is again obtained in the limit as all terms in ~ro approach zero and In (1 - e) approaches - e , whence
~( Wro / S) ~ro
s G PgO~TO (Ol/~)(ZsL/WTo)fLmax/k20
WTO ~ SGPgoCLmax
(2.22)(2.23) 2.2.6 Case 6: Takeoff Ground Roll
($G)
Given: dh/dt = 0 and values of p, D = qCDS, CLmax, VTO
= kToVsTALL,
and R = qCDRS + #ro(fiWro - qCLS).with the final result that
WTO 0l SGPgoCLmax
For this limiting case TSL/WTO is directly proportional to Wro/S and inversely proportional to s~.
30 AIRCRAFT ENGINE DESIGN a)
V=0
Rotation Ground Roll V~o
S G ..I ~1 SR
Transition gel
~ hTR
~1- Sre ~ SoL
STO
Fig. 2.7a Takeoff t e r m i n o l o g y (hrR < hobs).
hobs
Note: Total takeoff distance Two cases arise:
Case A (Fig. 2.7a): The total takeoff distance (sro) can be analyzed as ground roll (SG) plus three other distances: the first (SR) to rotate the aircraft to the takeoff lift condition (traditionally CL = 0 . 8 C L m a x ) while still on the ground; the second (SrR) to transit to the angle of climb direction; and the last (scD to clear an obstacle of given height. These distances may be estimated as followsl:
SR = tR Vro = tRkro~/{2fl /(pCL max)}(WTo/ S ) (2.26) where tR is a total aircraft rotation time based on experience (normally 3 s),
STR = Rc sin OCL -- Vr2° sin OCL g o ( n - 1)
VT20 sin OCL k2ro sin OCL 2fl ( Wro
SrR = go(O.8k2o _ 1) = g0~.8~ro --- 1) ~ \ - S - , ] (2.27) where Occ is the angle of climb, which can in turn be obtained from Eq. (2.2a) as
1 dh T - D
- sin Ocz - - -
V dt W
and, if hobs > hrR
hobs -- hTR
SCL -- (2.28)
tan OcL
where hobs is the required clearance height and hrR is given by the following expression, provided that hobs > hrR:
VZo(1--cosOcL) kZo(1--cOSOcL) 2fl ( t ~ o )
h r R = go(O.8k2o - 1) ---- g0(0---~8"-k2~o--1) P C L m a x - - (2.29)
CONSTRAINT ANALYSIS 31
b)
Rotation V = 0 Ground Roll V~o
SG • S R
STO
Transition
V " C l i m /
bs
~ - Sob ~ ~
Fig. 2.7b Takeoff terminology (hTR > hobs).
Case B (Fig. 2.7b): The distances SG and sn are the same as in case A, but the obstacle is cleared during transition so STO = Sa + SR + Sobs where Sobs is the distance from the end of rotation to the point where the height hobs is attained.
There follows
V2o sin Oobs
Sobs = Rc sinOobs go(O.8k2o - 1) (2.30) where
hobs Oobs = COS -1 1 - - Rc I
2.2.7 Case 7: Braking ROII(SB)
Given: ot < 0 (reverse thrust), d h / d t = 0 and values of p, VrD = krDVsTAIZ, D = qCDS, and R = qCDRS + Izs(flWro -- qCLS).
Under these conditions
D + R (Co -~- CDR -- lZBCL)q if" tZBflWTo
t~ WTO t ~ Wro
so that Eq. (2.5) becomes
TsL
- ~ ~-~ ~oro ~ --gi-
(2.31)WTO go
where
~L = CD -'[- CDR -- ].I"BCL (2.32)
which can be rearranged and integrated, as in Case 6, to yield
_ f l ( W T o / S ) l n { I + ~ L / [ ( I Z B + ( - - O r ) T s L ~ C L m a x ] I
sa Pgo~r 7 - W r o ]
k2--~-D ]
] (2.33)32 AIRCRAFT ENGINE DESIGN
provided that representative landing values of a and fi are used and where kTD is a constant greater than one (generally specified by appropriate flying regulations).
For this case reverse thrust can be used to great advantage. If the second term in the bracket of Eq. (2.33) is made much less than one by making ( - a ) very large, then
S B --~ ¢~(Wro/S) ~L
PgO~L [(--~)It~](TsLIWTo)(CLmax/k2D)
or
WTO (--Ol) SBPgoCLmax (2.34)
Note: Total landing distance
The total landing distance (sL) can be analyzed as braking roll (sB) plus two other distances (see Fig. 2.8): the first (SA) to clear an obstacle of given height and the second (SFR) a free roll traversed before the brakes are fully applied. These distances may be estimated as follows: 1
S A =
pgo(C~ + coR) - - \ k~ob, + k~D +
C L max 2h obs
( c . + co.)(ko~s + ~ )
(2.35) where hobs is the height of the obstacle and the velocity at the obstacle is
gobs = kobs VSTALL (2.36)
and
SFR = tFR VTD : tFRkTD~// { 2 fl / (pC L max) }( W T o / S ) (2.37)
gobs
-~-*~ ~ A p p r o a c h
h °b~
Free Roll
VrD k
Braking V = 0
I.
sA J Iv[ SFR I. I "~l S BSL Yl
Fig. 2.8 Landing terminology.
CONSTRAINT ANALYSIS 33 where tFR is a total system reaction time based on experience (normally 3 s) that allows for the deployment of a parachute or thrust reverser.
2.2.8 Case 8: Service Ceiling (Ps = dh/dt)
Given: d V / d t = O, n = 1 (L = W), and the values of h (i.e., or), d h / d t > O, and CL.
Under these conditions Eq. (2.11) becomes
TSL -- fl K1 - - + K2 + fl/q(WTo/S)
Wro ot
where
or
and
or
so that
qCL S = flWTo
,dh}
+ V ~- (2.38)
CL = fl-- ( ~ 0 (2.39)
q - - 2 - - ( L
v = (2.40)
Since
then and cr.
Under these conditions Eq. (2.11) becomes
_{
TSL fl K1 + K 2 +
W~o ,~ \ ~ ) C L max W CL -- k2 ° -- q S --
TSL fl [ CLmax + +
CDO "~ CDR }
( f l W r o / q S ) + sin 0 (2.42)
CDO "~ CDR I
CLm~x/k2 ° + sinO J (2.43) 2.2.9 Case 9: Takeoff Cfimb Angle
Given: O, n = 1 (L = W), d V / d t = 0, CDR , CLmax, kro, and the values o f h
____{ CDO"~-CDR l dh}
TSL fi K1CL + K2 + + (2.41)
Wro ~ CL V d[
34 AIRCRAFT ENGINE DESIGN and
V = V T ° = ~ O'PSL2t~k2OcL
max ( ~ O ) (2.44) is employed to find Mro for a given Wro/S and thus the applicable values of et, K1, K2, and Coo. Because they vary slowly with WTo/S, the constraint bound- ary is a line of almost constant TSL/Wro.2.2.10 Case 10: Carrier Takeoff
Given: n = 1 (L = W), Vro, dV/dt, CL m a x , kTo, 1~, and the values of h and a.
Solving Eq. (2.44) for wing loading gives
WTO PsL C L max V•O
T max = 2~k2ro (2.45)
where the takeoff velocity (Vro) is the sum of the catapult end speed (Ve.d) and the wind-over-deck (Vwod) or
Vro = Ve,d + Vwoa (2.46)
A typical value of kro is 1.1 and of Ve,d is 120 kn (nautical miles per hour).
Wind-over-deck can be 20 to 40 kn, but design specifications may require launch with zero wind-over-deck or even a negative value to ensure launch at anchor. This constraint boundary is simply a vertical line on a plot of thrust loading vs wing loading with the minimum thrust loading given, as already seen in Eq. (2.43), by
r l TSL fl K 1 - + K2 + + --
(2.47)L ~ Jmin =
~ k2o CLmax/k2o go
--~where or, K1, KE, and Coo are evaluated at static conditions. A typical value of the required minimum horizontal acceleration at the end of the catapult (dV/dt) is 0.3 go.
2.2.11 Case 11: Carrier Landing
Given: n = 1 ( L = W), VTD, CL max, kTD, fl, and the values o f h and ~r.
Rewriting Eq. (2.45) for the touchdown condition gives
WTO PsL C L
maxV~'D
- - g - max = 2flk2ro (2.48)
where the touchdown velocity (Vro) is the sum of the engagement speed (Ve.g, the speed of the aircraft relative to the carrier) and the wind-over-deck (Vwod), or
Vro = Veng + Vwod
A typical value o f k r o is 1.15 and of Ve.g is 140 kn (nautical miles per hour). As in Case 10, this constraint boundary is simply a vertical line on a plot of thrust
CONSTRAINT ANALYSIS 35 loading vs wing loading. The minimum thrust loading is given by Eq. (2.49):
o O,min = /K'
CDO ÷ CDR I
+ K 2 + CLmax/k2 D + sin 0 J (2.49) where - 0 is the glide-slope angle.
2.2.12 Case 12: Carrier Approach (Wave-off)
Given: O, n = 1 (L = W), VTD, d V / d t , el, fi, CDR, CLmax, kTD, and the values of h and a.
Because carrier pilots do not flair and slow down for landing but fly right into the carrier deck in order to make certain the tail hook catches the landing cable, the approach speed is the same as the touchdown speed (VrD). Rewriting Eq. (2.48) for the approach condition gives
[WTo] (rpsLCLmaxV2D (2.50)
- U max =
As in Cases 10 and 11, this constraint boundary is simply a vertical line on a plot of thrust loading vs wing loading.
Under these conditions Eq. (2.11) becomes
1 --~dV}
CL max CDO ÷ fOR __
L
IrSL1WTO..Jmin
=~--0/K1---~--TD +g2+
CLmax/k2D + sin 0 + go (2.51) where - 0 is the glide-slope angle and or, K1, K2, and CDO are evaluated at static conditions. Typical wave-off requirements are an acceleration of 1/8 go while on a glide-slope of 4 deg.2.3 Preliminary Estimates for Constraint Analysis
Preliminary estimates of the aerodynamic characteristics of the airframe and of the installed engine thrust lapse are required before the constraint analysis can be done. The following material is provided to help obtain these preliminary estimates.
2.3.1 Aerodynamics
The maximum coefficient of lift (CL max) enters into the constraint analysis dur- ing the takeoff and landing phases. Typical ranges for CL max divided by the cosine of the sweep angle at the quarter chord (Ac/4) are presented in Table 2.1, which is taken from Ref. 4. This table provides typical CL max for cargo- and passenger-type aircraft. For fighter-type aircraft a clean wing will have CL max ~ 1.0 --+ 1.2 and a wing with a leading edge slat will have CLmax ~ 1.2 -+ 1.6. The takeoff maximum lift coefficient is typically 80% of the landing value.
36 AIRCRAFT ENGINE DESIGN Table 2.1 CLmax for high lift devices 4
High lift device Typical flap angle, deg CL m a x / C O S ( A c / 4 )
Trailing edge Leading edge Takeoff Landing Takeoff Landing
Plain 20 60 1.4 ~ 1.6 1.7 ~ 2.0
Single slot 20 40 1.5 ~ 1.7 1.8 ~ 2.2
Fowler 15 40 2.0 ~ 2.2 2.5 ~ 2.9
Double sltd. 20 50 1.7 ~ 2.0 2.3 ~ 2.7
Double sltd. Slat 20 50 2.3 ~ 2.6 2.8 ---> 3.2
Triple sltd. Slat 20 40 2.4 ~ 2.7 3.2 ~ 3.5
The lift-drag polar for most large cargo and passenger aircraft can be estimated (Ref. 5) by using Fig. 2.9 and Eq. (2.9) with
0.001 _< K " < 0.03, 0.1 < C L m i n < 0.3, K ' -- 1 h A R e
where the wing aspect ratio (AR) is between 7 and 10 and the wing planform efficiency factor (e) is between 0.75 and 0.85. Note that
: ld'll("2 K 2 - 2 K " C L min
K1 = K' + K", Coo C D m i n + -~ " - ' L m i n ' :
The lift-drag polar for high-performance fighter-type aircraft can be estimated (Ref. 5) using Eq. (2.9), K : = 0, and Figs. 2.10 and 2.11.