C.3 Lateral-Directional Derivatives
《Collection of lateral derivatives》
Yv 
1 Y
1 L
1 L
1 L
1 L
, Lv 
, Lr 
, Lp 
, L a 
m v
Ix  v
Ix  r
Ix  p
I x  a
1 N
1 N
1 N
1 N
1 N
, Nr 
, Np 
, N a 
, N r 
Iz  v
Iz  r
Iz  p
I z  a
I z  r
C.3.1 Nondimensionalize the Derivatives
1 Y
(1) Yv 
m v
Y Y   1
C
○Define Y  12  air S wU 02CY , then

 2  air S wU 0 Y .
v  v

--- In general,  CY /  v depends on v while  CY /   depends on v / U 0 .
qS wCLtrim
W
mg
1


○Also, 2  air S wU 0 
; hence,
U 0CLtrim
U 0CLtrim U 0CLtrim
 CY
g
Yv 
CY , CY 
.
--- Normally , Yv  0 .

Nv 
U 0CLtrim

78
(2) Lv 
○Define L  21  airU 02 S wbCL 
○Then, Lv 
g
U 0CLtrim K x 2 / b
L
1 L
Ix  v
C
C
 21  a i rU02 Swb L  21  a i rU0 Swb L
v
v

CL , CL 
 CL
2
--- Note that I x  mK x .
.

1 L
(3) Lr 
Ix  r
○Similar procedure will lead to: Lr 
g
2
 CL
CLtrim K x / b  r
.
○But we would like to non-dimensionalize the derivative.
 CL
g
C .
 Define CLr 
, then Lr 
2 Lr
  rb / 2U 0 
2U 0CLtrim  K x / b
(4) L p 
○Similarly, we have: L p 
1 L
Ix  p
g
2U 0CLtrim  K x / b
C , CL 
2 Lp
p
 CL
  pb / 2U 0 
.
79
(5) L a 
○It is then easily seen that: L a 
1 L
I x  a
g
bCLtrim  K x / b
a
a

 CL
 a
.
1 N
Iz  v
(6) N v 
○Define N  12  airU 02 S wbCN , then N v 
C
, CL
2 L
g
CN 
2
U 0CLtrim K z / b
 CN
--- CN 
. Also, K z  I z / m .


(7) N r 
○We will have: N r 
1 N
Iz  r
g
2U 0CLtrim  K z / b
(8)
Np 
1 N
Iz  p
C
& Np 
2 Nr
---Nondimensionalized derivatives: CN r 
 CN
g
2U 0CLtrim  K z / b
  rb / 2U 0 
& CN p 
2
CN p .
 CN
  pb / 2U 0 
.
80
(9) N a 
1 N
I z  a
○Similar procedure gives us: N a 
&
(10) N r 
g
CLtrim K z 2 / b
and N r 
CN , CN

g
2
1 N
I z  r
CLtrim K z / b
a

a
 CN
CN , CN 

r
r
 a
 CN
 r
.
《Summary of Lateral Directional Derivatives》
Yv 
g
g
g
,
CY , Lv 
C
N

CN  ,
L
v
2
2
U 0CLtrim
U 0CLtrim K x / b
U 0CLtrim K z / b
g
g
Lr 
C
L

C ,
,
p
2 Lr
2 Lp
2U 0CLtrim  K x / b
2U 0CLtrim  K x / b
g
g
Nr 
C
N

C ,
,
p
2 Nr
2 Np
2U 0CLtrim  K z / b
2U 0CLtrim  K z / b
g
g
,
L a 
C
L

CL
L a
r
2
2
r
CLtrim K x / b
CLtrim K x / b
g
g
,
N a 
C
N

CN
N a
r
2
2
r
CLtrim K z / b
CLtrim K z / b
81
C.3.2 Aerodynamics of the coupling derivatives
○We will give methods to estimate the following coupling derivatives
CY 
CLr 
 CL
  rb / 2U 0 
, CL p 
 CY
, CL 

 CL
  pb / 2U 0 
 CL

, CN  
, CN r 
 CN
 CN

 rb / 2U 0 
, CN p 
 CN
  pb / 2U 0 
--- CL , CL , CN  , and CN  are not coupling parameters.
a
r
a
r
--------------------------------------------------------- CY
(1) CY 
: Side slip damping

○Side force comes mostly from fuselage and vertical tail (VT).
A. Side force from the VT: Yvt  qSvt avt 
--- avt : lift curve slope of VT.
 CY is defined as CY  Y / qS w . Then, CYvt  ( S vt / S w )avt  , and
CY
S vt
vt
CY 

avt
--- Normally CY  0
vt
vt

Sw
--- AR of the VT is always low, so estimate avt carefully.
--- Side wash will be strong; need actual wind tunnel data.
B. From slender body theory, CY
body
 0 ; therefore, CY  CY .
vt
82
(2) CL 
 CL

: Rolling moment derivative
○ CL is mostly the result of the redistribution of lift due to the dihedral
angle . We will examine the first order effect on straight wing:
 Change in Local AOA due to side slip:

x
  c  / c   
c
The resulting rolling moment:
L  2
b/ 2
0
qaw c y dy .
For a rectangular wing with constant airfoil,
L   41 qb2caw ;
then, CL  L / qS wb   41 a w , hence
CL 
 41 aw .

Physical
wing chord
c 
   cc 
For general wing shapes with taper ratio ,
the sweep angle  and at high Mach number:
1 1 2 
t a n
.
CL  
a w 
C
Ltrim 
6 1  

1 M 2 cos 2 
U rel
c
c 
U rel
Effective
wing chord
 
83
Or, when partial span flaps ( bF ) are down, add the increment
tan  F
1 b 1  2 F
CL ,flap   F
CLF
6 b 1   1 M 2 cos2 
F
F
--- The subscript F indicating the corresponding parameters for the portion of the
span where the flaps are down.
○The fuselage may also contributes:
zw
h*
zw h*  w*
CL ,b o d y 1.2
AR
b
b
--- z w < 0 for high wing and z w > 0 for low wing ( w* : width of the fuselage.)
○Vertical tail also may contribute to Cl  , left as an exercise!
(3) CN  
 CN

: Yawing moment derivative
○ CN  comes mostly from VT though fuselage & propeller may reduce it.
○For the contribution from VT:
a) Yawing moment from VT due to sideslip: N vt  Yvt lvt  qS vt avt  lvt
--- lvt : Distance between VT a.c. and CG
b) We normally define CN  N / qS wb ; hence, CN
 vt
l
 vt
b
S vt
avt .
Sw
84
○For the effects from the fuselage:
a) Yawing moment from the fuselage due to sideslip: N body  2qv* k*
--- k*: Parameter that depends on the fineness
1
ratio (
length
). Typical
diameter
k * : 0.8 - 0.9.
b) Then, CN  , body  2 v*k* / S wb .

0 fineness ratio
--- v* : volume of the fuselage.
○Contribution from the wing and from the power plant are both small.
For the effects from the wing, see NACA Technical Reports TR-635 and TR-1098.
As for the effects from the power plant, see NACA TR-819.
 CL
(4) CLr 
: Yaw to roll coupling
  rb / 2U 0 
○ CLr is mostly due to different velocity along wing as a result of yawing.
Rolling moment due to the wing area cdy :
x
dL   21  airU 2Cl  cdy  y
y
U  U 0  ry ; hence, U  U 0  2U 0 r y  r y
By neglecting the high order term, we thus have
2
L   q
b/ 2
b/ 2
2
2 2
cCl ydy   airU0 r 
b/ 2
b/ 2
cCl y 2 dy
r
Yaw to roll
coupling
c(y)
dy
U0  ry
U0
ry
85
3
1  U C
For a rectangular wing with constant airfoil: L  12
air 0 Ltrim cb r .
Then,
CL 
hence,
CLr
L
rb 1

 3 CLtrim ,
qSwb 2U 0
 CL

 13 CLtrim .
  rb / 2U 0 
--- S w  cb
To include the effect of taper ratio , sweep angle  and at high Mach No:
CLr 
 
1 1 3
12 1 
(5) CL p 
 2  M 2 cos2  

 CLtrim .
2
2
 1 M cos  
 CL
  pb / 2U 0 
: Roll damping
○ C L p is mostly due to wing though the vertical tail may also contributes.
○We shall estimate the wing contribution:
 Local A.O.A change due to rolling:   py /U0
 Differential roll moment element due to  :
dL  qawc d y y .
p b/2 2
 For the entire wing: L  2qaw
cy dy

0
U0
x
Roll
dampling
dy
c(y)
y
p
   py /U0
py
Urel
U0
86
--- For a rectangular wing with constant airfoil: L  
1
qaw pcb3 /U 0
12
pb
1 a ; hence, C
1a .
Then, CL  L  L  



w
L
p
6 w
qSwb qb2c
2U 0 6
 
1 1 3
--- And for a tapered wing: CL  
aw
p
12 1 
Notes: (a) Sweep should have no effect on this,
(b) Mach no. effect should show up in aw .
(6) CN r 
 CN
  rb / 2U 0 
: Yaw damping
CG
r
lvt
U0 Urel
vt
○Yaw damping is mostly due to vertical tail
lvt r
Effective VT AOA due to yawing:  vt  lvt r / U 0 .
Side force from VT: Fvt  qS vt avt  lvt r / U 0
2
Yawing moment due to Fvt : N vt   Fvt lvt  qSvt avt lvt r / U0
Fvt
N vt
lvt2 S vt
lvt2 S vt
rb
 Then, CN vt 
 avt
r  2avt 2 
qbS w
bU 0 S w
b S w 2U 0
 CN
lvt
l S
vt
C

 Hence, N r
--- CN   vt vt avt
 2 CN  .
vt
b
vt
  rb / 2U 
bS w
0
87
(7) CN p 
 CN
  pb / 2U 0 
: Roll to yaw coupling
○Contributions of CN p come from two sources: wing and the vertical tail.
○Wing contribution is mostly due to rotation of the lift vector.
Normally, U rel  U 0 so that
x
L
dLL  dLR  qCl cdy .
dy
Urel
dLL
However, dLL and dLR are
y
py
py U0
each rotated by an angle
p
py
py / U 0 , and the following
yawing moment results: dN  2  y   py / U0   qCl cdy
Overall yawing moment due to rolling: N  2q p / U0 
b/ 2
0
--- For rectangular wings with constant airfoil: N  
Then, CN 
1
qcCLtrim
12
c(y)
dLR
R
Urel
U0
py
cCl y 2 dy
pb3 / U 0
pb
N  1C
1C

C


;
hence,
.
L
N
6 Ltrim
trim
p
6
qbS w
2U 0
--- And for a wing with taper ratio : CN p  
 C
1 1 3
12 1 
Ltrim .
○Contribution from the vertical tail can be derived the same way.
88