Lecture 6 : Aircraft orientation in 3 dimensions


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1 Lecture 6 : Aircraft orientation in 3 dimensions Or describing simultaneous roll, pitch and yaw
2 1.0 Flight Dynamics Model For flight dynamics & control, the reference frame is aligned with the aircraft and moves with it. (Why?) X b Y b Z b Question: Where s the origin located for this body axes?
3 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}
4 Figure 1.1 : Six Degrees of Freedom u p  roll rate Y b v X b q  pitch rate r  yaw rate Note the right hand rule for rotation rates Z b w
5 1.1 Defining the aircraft orientation  Euler angles The local horizon axes is aligned with the Earth fixed axes but translated to the aircraft s cg. i body axes j I K I J Earth fixed axes local horizon axes k J K
6 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? i j I K k J
7 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 Yaw Euler angle (Greek : psi) Pitch Euler angle (theta) Roll Euler angle (phi) NB : The order of the rotations is important!
8 Step 1) Rotate IJK by an angle about the K axis (yaw) This yields the intermediate axes i 1 j 1 k 1 J i 1 I j 1 K K = k 1
9 Step 1) Yaw about the K axis I i 1 i 1 j 1 k 1 = [ R ] I J K J = cos sin 0 sin cos I J K j 1
10 Step 2) Rotate i 1 j 1 k 1 by an angle about the j 1 axis (pitch) This yields the intermediate axes i 2 j 2 k 2 i 2 i 1 j 1 i 1 j 1 = j 2 k 2 k 1 k 1
11 Step 2) Pitch about the j 1 axis i 2 i 2 j 2 k 2 = [ R ] i 1 j 1 k 1 i 1 k 2 = cos 0 sin sin 0 cos i 1 j 1 k 1 k 1
12 Step 3) Rotate i 2 j 2 k 2 by an angle about the new i 2 axis (roll) This yields the body axes i j k i 2 i 2 = i j 2 j 2 j k 2 k k 2
13 Step 3) Roll about the i 2 axis i j k = [ R ] i 2 j 2 k 2 k j j 2 k 2 = cos sin 0 sin cos i 2 j 2 k 2
14 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 Question : What happens if we want IJK in terms of ijk? Hint : The rotation matrices have a special property [ R ] 1 = [ R ] T
15 Inverting the matrix product to get IJK in terms of ijk : I J K = [ R ] [ R ] [ R ] 1 i j k = [ R ] T [ R ] T [ R ] T i j k
16 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
17 Using the relationship derived : I J K = cos sin 0 sin cos cos 0 sin sin 0 cos cos sin 0 sin cos i j k Premultiply both sides by {0 0 1} K = {sin 0 cos } cos sin 0 sin cos = {sin cos sin cos cos } i j k i j k What happens if the Euler angles are small?
18 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? How about p = d / dt q = d / dt r = d / dt
19 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 + j 1 + i 2 Noting that K = sin i + cos sin j + cos cos k j 1 = j 2 = cos j  sin k i 2 = i
20 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 What happens if is 90 o? NB : These are the EOMs relating the rate of change of aircraft orientation to body rotational rates
21 Ex : Rotation rates during a sustained level turn 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. j k 2. Hence state the body axes rotation rates. Are they all zero?
22 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?
23 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)
24 5. Recall may be written in rotation rates as : = p i + q j + r 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 Interpret this What is the implication?
25 6. What happens if we substitute these sustained, level turn rotation rates p =  V/R sin q = V/Rcos sin r = V/Rcos cos in the equations for change in orientation? ' = p + (q sin + r cos ) tan ' = q cos  r sin ' = (q sin + r cos ) sec Physically what do you expect?
26 Followup exercise : Determine the rotation rates for a loop
27 3.0 Aerodynamic forces in the body axes For a general aircraft orientation, the angle of attack a and sideslip b are defined as follows: Y b b a X b Z b V direction of flight
28 3.1 : Writing body velocity components with aerodynamic angles Y b b a X b Resolving the velocity vector u = V T cos b cosa Z b v = V T sinb w = V T cosb sina V T Question : What about the aerodynamic forces?
29 3.2 Writing aerodynamic forces in the body axes Often aerodynamic forces are specified in terms of 3 mutually perpendicular forces L D D : drag, aero force opposite to V T L : lift, aero force perpendicular to V T S S : side force Y Z b a X NB : L, D & S defines an axes system i.e. the flight path axes V T
30 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) Y L sin(a) 0 cos(a) Z
31 Multiplying the rotation matrices yields the body axes components for the aerodynamic forces D =  (X cosa + Z sina) cosb  Y sinb S = (X cosa + Z sina) sinb  Y cosb L = X sina  Z cosa
32 or the inverse relation X = L sina + (S sinb  D cosb) cosa Y =  (S cosb + D sinb) Z =  L cosa + (S sinb  D cosb) sina Question : What do you expect if a and b are small?
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