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Flight  Dynamics  and  ControlLecture  1:Introduction

G.  DimitriadisUniversity  of  Liege

Reference  material• Lecture  Notes• Flight  Dynamics  Principles,  M.V.  Cook,  Arnold,  1997

• Fundamentals  of  Airplane  Flight  Mechanics,  David  G.  Hull,  Berlin,  Heidelberg  :  Springer-­Verlag Berlin  Heidelberg,  2007,  http://dx.doi.org/10.1007/978-­3-­540-­46573-­7

What  is  it  about?

Introduction• The  study  of  the  mechanics  and  dynamics  of  flight  is  the  means  by  which  :– We  can  design  an  airplane  to  accomplish  efficiently  a  specific  task

– We  can  make  the  task  of  the  pilot  easier  by  ensuring  good  handling  qualities

– We  can  avoid  unwanted  or  unexpected  phenomena  that  can  be  encountered  in  flight

Aircraft  description

Pilot Flight  ControlSystem Airplane Response Task

The  pilot  has  direct  control  only  of  the  Flight  Control  System.  However,  he  can  tailor  his  inputs  to  the  FCS  by  observing  the  airplane’s  response  while  always  keeping  an  eye  on  the  task  at  hand.

Control  Surfaces

• Aircraft  control  is  accomplished  through  control  surfaces  and  power– Ailerons– Elevators– Rudder– Throttle

• Control  deflections  were  first  developed  by  the  Wright  brothers  from  watching  birds

Wright  FlyerThe  Flyer  did  not  have  separate  control  surfaces.The  trailing  edges  of  the  windtips  could  be  bent  by  a  system  of  cables

Modern  control  surfaces

Elevator

Rudder

Aileron

Elevon  (elevator+aileron)

Rudderon  (rudder+aileron)

Other  devicesFlaps

Spoilers

•Combinations  of  control  surfaces  and  other  devices:  flaperons,  spoilerons,  decelerons  (aileron  and  airbrake)•Vectored  thrust

Airbreak

Mathematical  Model

InputAileronElevatorRudderThrottle

Aircraft  equations  of  motion

OutputDisplacementVelocityAcceleration

Flight  Condition

Atmospheric  Condition

Aircraft  degrees  of  freedomSix  degrees  of  freedom:

3  displacements

x:  horizontal  motion

y:  side  motion

z:  vertical  motion

3  rotations

Around  x:  roll

Around  y:  pitch

Around  z:  yaw

x

z

y

Uw

U:  resultant  linear  velocity,  cg:  centre of  gravityw:  resultant  angular  velocity

cg

Aircraft  frames  of  reference• There  are  many  possible  coordinate  systems:

– Inertial  (immobile  and  far  away)– Earth-­fixed  (rotates  with  the  earth’s  surface)– Vehicle  carried  vertical  frame  (fixed  on  aircraft  cg,  vertical  axis  parallel  to  gravity)

– Air-­trajectory  (fixed  on  aircraft  cg,  parallel  to  the  direction  of  motion  of  the  aircraft)

– Body-­fixed  (fixed  on  aircraft  cg,  parallel  to  a  geometric  datum  line  on  the  aircraft)

– Stability  axes  (fixed  on  aircraft  cg,  parallel  to  a  reference  flight  condition)

– Others

Airplane  geometry

cg

c

c

c /4

s = b /2

c /4

lT

lt

y

c(y)

xMAC

x y( )

Airplane  references  (1)

• Standard  mean  chord  (SMC)

• Mean  aerodynamic  chord  (MAC)

• Wing  area

• Aspect  Ratio

c = c 2 y( )−s

s

∫ dy / c y( )−s

s

∫ dy

c = c y( )−s

s

∫ dy / dy−s

s

∫

AR = b2 /S

S = bc

xMAC = c y( ) x y( )−s

s

∫ dy / c y( )−s

s

∫ dy

Airplane  references  (2)

• Centre  of  gravity  (cg)• Tailplane  area  (ST)• Tail  moment  arm  (lT)• Tail  volume  ratio:  A  measure  of  the  aerodynamic  effectiveness  of  the  tailplane

V T =ST lTSc

Airplane  references  (3)

• Fin  moment  arm  (lF)• Fin  volume  ratio

c /4

c /4cg

lF

lf

V F =SF lF

Sc

Aerodynamic  Reference  Centres

• Centre  of  pressure  (cp):  The  point  at  which  the  resultant  aerodynamic  force  F acts.  There  is  no  aerodynamic  moment  around  the  cp.

• Half-­chord:  The  point  at  which  the  aerodynamic  force  due  to  camber,  Fc,  acts

• Quarter-­chord  (or  aerodynamic  centre):  The  point  at  which  the  aerodynamic  force  due  to  angle  of  attack,  Fa,  acts.  The  aerodynamic  moment  around  the  quarter-­chord,  M0,  is  constant  with  angle  of  attack

Airfoil  with  centres

ac

Dc

cpD Da

Lc L La

Fc F Fa

L

DM0

c /4

c /2

hnc

c

Camber  line

V0

By  placing  all  of  the  lift  and  drag  on  the  aerodynamic  centre  we  move  the  lift  and  drag  due  to  camber  from  the  half-­chord  to  the  quarter  chord.  This  is  balanced  by  the  moment  M0

Full  description  of  aircraft  movement

• The  static  stability  analysis  presented  in  the  aircraft  design  lectures  is  good  for  the  preliminary  design  of  aircraft

• Aircraft  flight  is  a  dynamic  phenomenon:– Every  control  input  or  external  excitation  results  in  a  dynamic  response

– The  dynamic  response  may  be  oscillatory  and  have  a  single  or  several  frequency  components

– The  dynamic  response  may  be  damped  (stable)  or  undamped (unstable)

• The  modelling  of  this  dynamic  response  necessitates  the  derivation  of  the  full  equations  of  motion  of  the  aircraft

Nomenclature• Here  is  a  definition  of  the  degrees  of  freedom  of  an  aircraft  and  the  forces  and  moments  acting  on  it.

• All  degrees  of  freedom  are  relative  to  the  aircraft’s  centre  of  gravity  and  use  aircraft  geometrical  axes.

Symbols Definitionx, U, X translation,  velocity  and  force  applied  in  the  direction  parallel  to  

the  axis  of  the  fuselagey, V, Y translation,  velocity  and  force  applied  in  the  direction  

perpendicular  to  the  plane  of  symmetry  of  the  aircraftz, W, Z translation,  velocity  and  force  applied  in  the  direction  

perpendicular  to  both  x and  yp, L angular  velocity  and  moment  in  roll  directionq, M angular  velocity  and  moment  in  pitch  directionr, N angular  velocity  and  moment  in  yaw  direction

Body  and  axesAxis  system

Could  be  any  body  but  in  this  case  it  is  an  aircraft  of  mass  m.

For  the  moment  it  is  a  flexible  body

Any  point  p on  the  body  can  have  a  velocity  and  acceleration  with  respect  to  the  c.g.

Vector  notation• We  define  the  following  vector  notation

• Noting  that  u and  a are  velocities  and  accelerations  with  respect  to  the  center  of  gravity

x =xyz

!

"

###

$

%

&&&, w =

pqr

!

"

###

$

%

&&&, U =

UVW

!

"

###

$

%

&&&, F =

XYZ

!

"

###

$

%

&&&, M =

LMN

!

"

###

$

%

&&&

u =uvw

!

"

###

$

%

&&&, a =

axayaz

!

"

####

$

%

&&&&

Developing  the  equations  of  motion

• All  equations  of  motion  of  dynamic  systems  can  be  derived  using  Newton’s  Second  Law.

• Two  sets  of  equations  are  derived:– Sum  of  forces  acting  on  the  system  (internal  and  external)  are  equal  to  its  mass  times  its  acceleration

– Sum  of  moments  acting  on  the  system  (internal  and  external)  are  equal  to  its  moment  of  inertia  times  its  angular  acceleration

• Therefore,  the  object  of  the  derivation  is  to  estimate  the  accelerations  (linear  and  angular  of  the  aircraft)

• As  usual,  the  same  equations  of  motion  can  be  obtained  using  Lagrange’s  equation  (i.e.  conservation  of  energy)

Local  velocities  (1)

• The  local  velocity  vector  u is  given  simply  by

• Substituting  for  the  vector  definitions

• Where  x  denotes  the  vector  (cross)  product  and

u = !x+w× x

u =!x!y!z

!

"

###

$

%

&&&+

pqr

!

"

###

$

%

&&&×

xyz

!

"

###

$

%

&&&

pqr

!

"

###

$

%

&&&×

xyz

!

"

###

$

%

&&&=

i j kp q rx y z

Local  velocities  (2)

• The  equations  for  the  local  velocities  at  point  p(x,y,z) are

• Now  assume  that  the  body  is  rigid,  i.e.  no  parts  of  it  are  moving  with  respect  to  the  c.g

u = x − ry + qzv = y − pz + rxw = z − qx + py

Total  local  velocities

• This  gives therefore

• The  total  local  velocities  u´́=u+U at  p(x,y,z) are  given  by

u =w× x, or, u = −ry+ qzv = −pz+ rxw = −qx + py

x = y = z = 0

′ u = U + u = U − ry + qz′ v = V + v = V − pz + rx′ w = W + w = W − qx + py

(1)

Local  accelerations  (1)

• Similarly,  the  local  accelerations  at  point  p(x,y,z) are  given  by

• Substituting  for  the  vector  definitions

• where

a = !u+w×u

a =!u!v!w

!

"

###

$

%

&&&+

pqr

!

"

###

$

%

&&&×

uvw

!

"

###

$

%

&&&=

− !ry+ !qz− !pz+ !rx− !qx + !py

!

"

###

$

%

&&&+

pqr

!

"

###

$

%

&&&×

uvw

!

"

###

$

%

&&&

pqr

!

"

###

$

%

&&&×

uvw

!

"

###

$

%

&&&=

i j kp q r

−ry+ qz −pz+ rx −qx + py

(2)

Local  accelerations  (2)

• Carrying  out  all  the  algebra  leads  to

• Remembering  that  this  is  only  part  of  the  acceleration  of  point  p.  The  acceleration  of  the  centre of  gravity  must  be  added.

ax = −x q2 + r2( )+ y pq− !r( )+ z pr + !q( )

ay = x pq+ !r( )− y p2 + r2( )+ z qr − !p( )

az = x pr − !q( )+ y qr + !p( )− z p2 + q2( )

Total  local  acceleration

• The  total  local  acceleration  at  point  p(x,y,z) is  defined  as

• So  that,  finally!a = !U+w×U+ a

!ax = !U − rV + qW − x q2 + r2( )+ y pq− !r( )+ z pr + !q( )

!ay = !V − pW + rU + x pq+ !r( )− y p2 + r2( )+ z qr − !p( )

!az = !W − qU + pV + x pr − !q( )+ y qr + !p( )− z p2 + q2( )(4)

(3)

Example

• A  pilot  in  an  aerobatic  aircraft  performs  a  loop  in  20s  at  a  steady  velocity  of  100m/s.  His  seat  is  located  5m  ahead  of,  and  1m  above,  the  c.g.  What  total  normal  load  factor  does  he  experience  at  the  top  and  the  bottom  of  the  loop?

Solution

100m/s

100m/s

cg

cg

5m

1m

2R

Movement  only  in  the  plane  of  symmetry:

V = p = p = r = 0Normal  acceleration:

′ a z = W − qU + xq − zq2

For  a  steady  manoeuvre:

W = q = 0

Pitch  rate:

q =2π20

= 0.314rad/s

Solution  (2)

• Substituting  into  equation  for  normal  acceleration  at  the  seat:

• Normal  load  factor  definition:

• Total  normal  load  factor  at  top  of  loop:

• Total  normal  load  factor  at  bottom  of  loop:

′ a z = −qU − zq2 = −0.314 ×100 − −1( ) × 0.3142 = −31.3m/s2

′ n =′ a z

g=31.39.81

= 3.19

n = ′ n −1= 2.19

n = ′ n +1= 4.19

Generalized  Force  Equations

• Assume  that  point  p(x,y,z) has  a  small  mass  dm.

• Applying  Newton’s  2nd law  to  the  entire  body  yields

• where  the  subscript  Vol denotes  that  the  integral  is  taken  over  the  entire  volume

!a dmVol∫ = F (5)

Force  equations  (2)

• Remember  from  equation  (3)  that

• Substituting  from  equations  (2)  and  (1)

• Putting  this  last  result  back  into  Newton’s  2nd Law,  equation  (5)

!a = !U+w×U+ a

!a = !U+w×U+ !w× x+w× w× x( )

!U+w×U+ !w× x+w× w× x( )( )dmVol∫ = F

(6)

Centre  of  gravity

• As  far  as  the  integral  over  the  volume  is  concerned,  w and  U are  constants

• The  generalized  force  equation  becomes

• The  definition  of  the  centre of  gravity  is

• The  force  equation  becomes

!U dmVol∫ +w×U dm

Vol∫ + !w× xdm

Vol∫ +w× w× xdm

Vol∫

#

$%

&

'(= F

xdmVol∫ = 0

m !U+w×U( ) = F (7)

Generalized  Moment  Equations

• The  angular  acceleration  of  point  p(x,y,z)around  the  centre of  gravity  is  given  by

• Again,  use  Newton’s  second  law,  this  time  in  moment  form,  to  obtain

• Substitute  from  equation  (6)

x× "a

x× "a dmVol∫ =M (8)

x× !U+w×U+ !w× x+w× w× x( )( )dmVol∫ =M

Center  of  gravity

• Using  the  definition  of  the  centre of  gravity,  the  moment  equation  becomes

• Now  remember  the  matrix  form  of  the  cross  product

xVol∫ × !w× x( )dm+ x

Vol∫ × w× w× x( )#$ %&dm =M

x×w =Xww× x =XTw

, where X =0 −z yz 0 −x−y x 0

#

$

%%%

&

'

(((

Moments  of  inertia

• The  first  term  in  the  moment  equation  becomes

• where

• is  the  system’s  inertia  matrix

xVol∫ × !w× x( )dm = XXT !w

Vol∫ dm = XXT

Vol∫ dm

#

$%

&

'( !w

Ic = XXT

Vol∫ dm =

y2 + z2 −xy −xz

−xy x2 + z2 −yz

−xz −yz x2 + y2

#

$

%%%%

&

'

((((

Vol∫ dm

Moments  of  inertia  (2)

• The  individual  moments  and  products  of  inertia  are  defined  as

• So  that  the  inertia  matrix  becomes

Ix = y2 + z2( )dmVol∫ , Iy = x2 + z2( )dm

Vol∫ , Iz = x2 + y2( )dm

Vol∫

Ixy = xydmVol∫ , Ixz = xzdm

Vol∫ , Iyz = yzdm

Vol∫

Ic =

Ix −Ixy −Ixz−Ixy Iy −Iyz−Ixz −Iyz Iz

"

#

$$$$

%

&

''''

(8)

Moment  equation

• Using  the  definition  of  the  inertia  matrix,  the  first  term  in  the  moment  equation  becomes  simply

• Similarly,  the  second  term  is

• The  full  moment  equation  becomes

xVol∫ × !w× x( )dm = Ic !w

xVol∫ × w× w× x( )#$ %&dm =w× Icw( )

Ic !w+w× Icw( ) =M (9)

Complete  equations  of  motion

• Assembling  equations  (7)  and  (9)  we  get  the  complete  equations  of  motion

• This  is  a  set  of  6  equations  of  motion  with  6  unknowns,  U,  V,  W,  p,  q,  r.

• They  are  nonlinear  Ordinary  Differential  Equations.

m !U+w×U( ) = FIc !w+w× Icw( ) =M

(10)

Scalar  form• Substituting  for  the  definitions  of  Ic,  U,  w,  F and  M we  get  a  nicer  form

m !U − rV + qW( ) = Xm !V − pW + rU( ) =Ym !W − qU + pV( ) = ZIx !p− Iy − Iz( )qr + Ixy pr − !q( )− Ixz pq+ !r( )+ Iyz r2 − q2( ) = LIy !q+ Ix − Iz( ) pr + Iyz pq− !r( )+ Ixz p2 − r2( )− Ixy qr + !p( ) =M

Iz !r − Ix − Iy( ) pq− Iyz pr + !q( )+ Ixz qr − !p( )+ Ixy q2 + p2( ) = N

(11)

Symmetric  aircraft

• Consider  an  aircraft  that  is  symmetric  about  the  x-z plane.

• For  ever  point  p(x,y,z) with  mass  dm,  there  is  a  point  p(x, -y,z) with  mass  dm.

• It  follows  that  

• Similarly,Ixy = xydm

Vol∫ = 0

Iyz = yzdmVol∫ = 0

p(x,y,z)p(x,-y,z)

x

y

z

O

The  elementary  mass  moment  xydm around  the  CG  is  cancelled  by  the  elementary  mass  moment  x(-y)dm.

Asymmetric  Aircraft

Blohm  und  Voss  141

Ruttan  Bumerang

Blohm  und  Voss  237

Symmetric  aircraft  (2)

• For  symmetric  aircraft,  the  equations  of  motion  become

m !U − rV + qW( ) = Xm !V − pW + rU( ) =Ym !W − qU + pV( ) = ZIx !p− Iy − Iz( )qr − Ixz pq+ !r( ) = L

Iy !q+ Ix − Iz( ) pr + Ixz p2 − r2( ) =MIz !r − Ix − Iy( ) pq+ Ixz qr − !p( ) = N

(12)

Discussion  of  the  equations• If  we  can  solve  for  U,  V,  W,  p,  q,  r as  functions  of  time,  then  we  know  the  complete  time  history  of  the  motion  of  the  aircraft.

• Unfortunately,  terms  such  as  rU,  pV,  qW,  etc and  pq,  r2,  qr etc are  nonlinear.

• Furthermore,  we  have  only  defined  the  inertial  loads  up  to  now.

• We  have  not  said  anything  about  the  external  loads  acting  on  the  aircraft.

External  Forces  and  Moments

• There  are  five  sources  of  external  forces  and  moments:– Aerodynamic– Gravitational– Controls– Propulsion– Atmospheric  Disturbances

External  Forces  and  moments• The  full  equations  of  motion  in  the  presence  of  external  forces  and  moments  are

m U − rV + qW( ) = Xa + Xg + Xc + X p + Xd

m V − pW + rU( ) = Ya + Yg + Yc + Yp + Yd

m W − qU + pV( ) = Za + Zg + Zc + Zp + Zd

Ix p − Iy − Iz( )qr − Ixz pq + r ( ) = La + Lg + Lc + Lp + Ld

Iyq + Ix − Iz( ) pr + Ixz p2 − r2( ) = Ma + Mg + Mc + M p + Md

Izr − Ix − Iy( ) pq + Ixz qr − p ( ) = Na + Ng + Nc + N p + Nd