《Estimation of Longitudinal Derivatives -- An example》
Aircraft of interest: Grumann AO-1AF Mohawk
Aircraft parameters
Sw  330 ft 2 , St  85 ft 2
ARw  5.35, ARt  2.65
b  42 ft , c  98 inch
iw  15
.  , it  1
W  12000lb ,
m  372.67 slug
I y  20200slug - ft 2
Equilibrium conditions:
U 0  100 knots  167 ft / sec
Altitude: Sea level
 air  0.002377 slug / ft 3
q  3315
. lb / ft 2
52
《Computation of the trim lift coefficient》
.  , is
 Since St  14 Sw and the effective AOA of the wing,  w  iw   w 15
much larger than the effective AOA of the tail,  t  it   t 1 , we will
assume that the aircraft weight are supported by the wing. Therefore,
.
CLtrim  W  1097
.
q Sw
《Computation of the longitudinal derivatives》
○For the Mach number effect:
C
C
C
 At this low Mach number, we can set  ML   MD   MM  0 .
 As a result,
(1) Lu  U


   0 .
C     
C
L
2  32.2  0.3856
CLw  M


2  M  U0
167
2g
M CM
cCL ( K y c )2 U  M
2g
M  CD
Du  U C
D
2 M 
0 L
(2) M u 
 Also,
2g
0 CL
g
2 g CD
.
U 0 CL trim
--- The value of CDtrim will be estimated later.
53
(3) L 
g
a
CL w
 We will estimate aw from the right figure.
 The corresponding thrust coefficient at the
equilibrium: Tc '  0.4 (the left figure below).
 Then, the right figure shows that aw  5.7 / rad ;
.2  5.7  167.31.
hence, L  132
.097
1 T
(4) Tu  m
U
 Firstly, T  qS wTc ' ;
T
T '
hence,  U  qSw  Uc .
T '
 The left figure shows that  Uc  0.00562 .
qS  T '
 As a result, Tu  mw  Uc  0165
.
(6) And by assuming that e  0.75 
 
2 g CD
 01408
.
.
0 CL trim
g
D  2  e A aw  2912
. .
R
(5) It is also true that CDtrim  Tc ' ; hence, Du  U
54
(7) M 
g
cCL ( K y c )
2
C M


 First of all, K y 2  I y / m  54.2 ; therefore, g cCL ( K y c )2  4.424 .
C
 Also, C M   M
C
C
C
  CM  L  aw  CM
L
L
 From the right figure, and using the
curve for complete airplane, we can
C
estimate that  CM  0125
.
L
. .
 As a result, M  315
(8) M 
Xt
U 0 CL ( K y c ) 2 c
g
C M it
 We had define: C Mit  tVhat .
. / rad .
 The right fig. shows that at  393
 We will assume that  t  0.9, then
C M it  2.455; hence, M  1432
.
.
55
(9) M  M d
d
 We will estimate d from the following formula deduced from simple
d
horseshoe theory:
--- Estimated X t  22 ft

d
8aw   b
8 Xt
8 Xt 2


 1  0.6183,   1   b

d  3 AR  8 X t   b  
 Then, M  0.8855 .
 
(10) M 
g
cCL ( K y c )2
C M
 The right figure shows that
CM  161
. .
. .
 Therefore, M  712
《A set of longitudinal derivatives of Mohawk》
Lu  0.3856 , L  167.31, Du  01408
.
,
D  2912
. , Tu  0165
. ,
. , M  1432
.
Mu  0 , M  315
,
M  0.8855 , M  712
. .
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Lateral derivatives for Grumann “AO-1AF” MOHAWK
《Aircraft Data》
Sw  330ft 2 , AR  5.35, b  42 ft, c  98in ,
  0, aw  5.7 / rad.,   0.5,   6.5 ,
St  85ft 2 , Se  19.6 ft 2 , Ce / Ct  0.25, Svt  68.8 ft 2 , lvt  22 ft
m  372.7 slugs, I x  17,000slug  ft 2 , I z  36,400slug  ft 2
《Selected flight condition》
U 0  167ft/sec, q  33.15lb/ft2 , CLtrim  1.097, Tc'  0.4
《Estimation of lateral derivatives》
g
(1) Yv  U C
CY
L
0 trim
From right figure: CY  0.875(Tail on, the solid line) 
Hence, Yv  
32.2
0.875 
1671.097
0.154
--- With tail off, CY  0.363 (The dash line)

 Fuselage contributes about 40% of the total side force.
57
(2) Lv 
gb
U 0C Ltrim K x
2
CL
Upper figure shows that: CL  0.088
As a result, Lv  0.088
(3) N v 
32.242
1671.09745.6
gb
U 0C Ltrim K z
2

 0.0142.
CN 
From lower figure: CN   0.275 
Then,
0.27532.242
Nv 
1671.09797.7
Solid line -- Tail on
Dash line- Tail off
 0.021 .
gb2
(4) N r 
CN r
2
2U 0C Ltrim K z
First, CN r  CN r  2(lvt / b)  CN 
vt
Also from the lower figure: C N 
Then, CN 
As a result,
vt
tailon
vt
and CN 
vt
 CN 
 0.275 and CN 
 0.319 ; hence, C N r  2  0.319
32.2422
N r  0.334
21671.09797.7
22
42
tailon
tailoff
 CN 
tailoff
 0.044
 0.334 .
 0.531.
58
(5) Lr 
gb2
2U 0CLtrim K x
We have estimated CLr from theory:
C Lr
2
 
1 1 3
CLr 
12 1 
 2  M 2 C
 0.308

2  Ltrim
 1 M 
--- M ( Mach no.)  0.15, (swep angle )  0 and  ( taper ratio)  0.5 .
Then,
32.2422
Lr  0.308
21671.09745.6
 1.047 .
(6) L p 
gb2
2U 0CLtrim K x
2
CL p
 
a
We can estimate CLr from theory: CLr   w 13  0.792
12 1 
Then, Lp  2.692.
(7) N p 
gb 2
2U 0C Ltrim K z
2
CN p
 
Lets estimate the wing contribution: CN p   1 3 CLtrim  0.152
12 1 
1
By neglecting the VT:
32.2422
N p  0.152
21671.09797.7
 0.241
59
(8) L a , (9) L r , (10) N a , (11) N r
From the figures below, there were estimated that:
CL  0.269 , CL  0.0075, CN  0.005 and CN  0.275
a
r
a
r
As a result, we have the control derivatives:
.2 42  7.27 , L  0.0075 32.2 42  0.203
L a  0.269 1.32
r
097 45.6
1.097 45.6
.2 42  0.063,
N a  0.005 1.32
09797.7
.2 42  3.471
N r  0.275 1.32
09797.7
Obviously, aileron is mostly for roll control while rudder is for yaw.
《Lateral derivatives of MOHAWK》
Yv  0.154 , Lp  2.692, Lr  1.047 , Lv  0.0142,
N p  0.241 , N r  0.531 , N v  0.021 ,
L a  7.27 , L r  0.203, N a  0.063, N r  3.471
60