Or describing simultaneous roll, pitch and yawdynlab.mpe.nus.edu.sg/mpelsb/me4241/L6n.pdf · Or...

G. Leng, Flight Dynamics, Stability & Control

Lecture 6 : Aircraft orientation in 3 dimensions

Or describing simultaneous roll, pitch and yaw


1.0 Flight Dynamics Model

• For flight dynamics & control, the reference frame is

aligned with the aircraft and moves with it. (Why?)

Yb

Xb

Zb Question: Where’s the origin located

for this body axes ?


• The aircraft is modelled as a rigid body with __ degrees of

freedom

• The ___ DOFs correspond to

-

-

• Denote the translational velocity of the aircraft by V = {u,v,w}

• Denote the angular velocity of the aircraft by = {p,q,r}


Figure 1.1 : Six Degrees of Freedom

u v

w

p - roll rate

q - pitch rate

r - yaw rate

Xb

Yb

ZbNote the right hand

rule for rotation rates


1.1 Defining the aircraft orientation - Euler angles

K

J

I

k

j

i

Earth fixed axes

local horizon axes

body axes

I

JK

The local horizon axes is aligned with the Earth fixed axes but

translated to the aircraft’s cg.


Question : How do we express the basis vectors IJK of the

local horizon system in terms of the basis vectors ijk of the

body axes system or vice versa ?

KJ

I

k

j

i


Yaw Euler angle (Greek : psi)

Pitch Euler angle (theta)

Roll Euler angle (phi)

The local horizon axes system IJK can be rotated to coincide with

the body axes i j k by using three rotation angles or Euler angles

NB : The order of the rotations is important !


Step 1) Rotate IJK by an angle about the K axis (yaw)

This yields the intermediate axes i1 j1 k1

K

JI j1

i1

K = k1


Step 1) Yaw about the K axis

= [ R ]

I

J

i1

j1

i1

j1

k1

I

J

K

cos sin 0

-sin cos 0

0 0 1

=I

J

K


Step 2) Rotate i1 j1 k1 by an angle about the j1 axis (pitch)

This yields the intermediate axes i2 j2 k2

j1i1

k1

i2

k2

j1 = j2i1

k1


Step 2) Pitch about the j1 axis

= [ R ]

k1

i1

i2

k2

i2

j2

k2

i1

j1

k1

=i1

j1

k1

cos 0 -sin

0 1 0

sin 0 cos


j2

i2

k2

Step 3) Rotate i2 j2 k2 by an angle about the new i2 axis (roll)

This yields the body axes i j k

j

k

j2

i2 = i

k2


Step 3) Roll about the i2 axis

= [ R ]

k2

j2

k j

i

j

k

i2

j2

k2

=i2

j2

k2

1 0 0

0 cos sin

0 -sin cos


So finally, the basis vectors of the local horizon IJK and the

body axes ijk are related as follows :

i

j

k

= [ R ] [ R ] [ R ]I

J

K

Hint : The rotation matrices have a special property

[ R ]-1 = [ R ]T

Question : What happens if we want IJK in terms of ijk ?


Inverting the matrix product to get IJK in terms of ijk :

I

J

K

= [ R ]T [ R ]T [ R ]Ti

j

k

= [ R ] [ R ] [ R ]i

j

k

-1


Exercise : Express the weight force component in terms of the

body axes basis vectors i j k

Hint : The weight component points downwards ie W K

hence we need only express K in terms of i j k


Using the relationship derived :

cos -sin 0

sin cos 0

0 0 1

cos 0 sin

0 1 0

-sin 0 cos

1 0 0

0 cos -sin

0 sin cos

I

J

K

i

j

k

=

Premultiply both sides by {0 0 1}

K = {-sin 0 cos } 1 0 0

0 cos -sin

0 sin cos

i

j

k

= {-sin cos sin cos cos } i

j

k

What

happens if

the Euler

angles are

small ?


How about

p = d / dt

q = d / dt

r = d / dt

2.0 Rotation rates and change in aircraft orientation

How are the body rotational rates p,q,r related to the rate of

change of the Euler angles ?


The angular velocity vector is = p i + q j + r k written

using the body axes basis vectors

describes the rate of change in orientation which can also

be written as :

= ’ K + ’ j1 + ’ i2

Noting that

K = -sin i + cos sin j + cos cos k

j1 = j2 = cos j - sin k

i2 = i


p

q

r=

’

’

’

-sin 0 1

cos sin cos 0

cos cos -sin 0

Inverting yields :

' = p + (q sin + r cos ) tan

' = q cos - r sin

' = (q sin + r cos ) sec

NB : These are the EOMs relating the rate of change of

aircraft orientation to body rotational rates

What

happens if

is 90 o ?


Ex : Rotation rates during a sustained level turn

kj

1. What is the angular velocity

vector for an aircraft executing a

level sustained turn of radius R

at speed V ? The bank angle is

.

2. Hence state the body axes

rotation rates.

Are they all zero ?


1. The turn rate = V/R

2. Level turn the rotation axis is vertical

3. Hence the angular velocity vector is

= V/R {0, 0, 1} What is ambiguous ?

= V/R K not the body axes k !

So how do we write this in body axes ?


4. Recall the relationship derived via Euler angles

K = -sin i + cos sin j + cos cos k

Hence in body axes

= V/R(-sin i + cos sin j + cos cos k)


Hence the body axes rotational rates for a sustained level turn are :

p = - V/R sin

q = V/Rcos sin

r = V/Rcos cos

What is the implication ?

5. Recall may be written in rotation rates as :

= p i + q j + r k

Interpret this


' = p + (q sin + r cos ) tan

' = q cos - r sin

' = (q sin + r cos ) sec

q = V/Rcos sin

r = V/Rcos cos

6. What happens if we substitute these sustained, level turn

rotation rates

in the equations for change in orientation ?

p = - V/R sin

Physically what do you expect?


Follow-up exercise : Determine the rotation rates for a loop


3.0 Aerodynamic forces in the body axes

a

bXb

Yb

Zb

V direction of flight

For a general aircraft orientation, the angle of attack a and

sideslip b are defined as follows:


3.1 : Writing body velocity components with aerodynamic angles

a

bXb

Yb

Zb

VT

Resolving the velocity vector

u = VT cos b cosa

v = VT sinb

w = VT cosb sina

Question : What about the aerodynamic forces ?


3.2 Writing aerodynamic forces in the body axes

Often aerodynamic forces are specified in terms of 3 mutually

perpendicular forces

D : drag, aero force opposite to VT

L : lift, aero force perpendicular to VT

S : side forcea

b

X

YZ

VT

L

D

S

NB : L, D & S defines an axes system

i.e. the “flight path axes”


The transformation can be written in terms of two rotations

1) a rotation about the Y body axes by -a

2) a rotation about the resulting Z axis by b

and then inverting the components

D cos(b) sin(b) 0 cos(-a) 0 -sin(-a) X

S = - -sin(b) cos(b) 0 0 1 0 Y

L 0 0 1 sin(-a) 0 cos(-a) Z


D = - (X cosa + Z sina) cosb - Y sinb

S = (X cosa + Z sina) sinb - Y cosb

L = X sina - Z cosa

Multiplying the rotation matrices yields the body axes components

for the aerodynamic forces


X = L sina + (S sinb - D cosb) cosa

Y = - (S cosb + D sinb)

Z = - L cosa + (S sinb - D cosb) sina

or the inverse relation

Question : What do you expect if a and b are small ?

Or describing simultaneous roll, pitch and yawdynlab.mpe.nus.edu.sg/mpelsb/me4241/L6n.pdf · Or...

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