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8/12/2019 441 Lecture 9
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8/12/2019 441 Lecture 9
2/24
Aircraft DynamicsLecture 9
In this Lecture we will cover:
Newtons Laws
Mi=
d
dt
H
Fi =m
ddt
v
Rotating Frames of Reference
Equations of Motion in Body-Fixed Frame
Often Confusing
M. Peet Lecture 9: 2 / 24
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Review: Coordinate RotationsPositive Directions
If in doubt, use the right-hand rules.
Figure: Positive Directions
Figure: Positive Rotations
M. Peet Lecture 9: 3 / 24
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Review: Coordinate RotationsRoll-Pitch-Yaw
There are 3 basic rotations an aircraft can make: Roll = Rotation aboutx-axis Pitch = Rotation about y-axis Yaw = Rotation aboutz-axis Each rotation is a one-dimensional transformation.
Any two coordinate systems can be related by a sequence of 3 rotations.M. Peet Lecture 9: 4 / 24
8/12/2019 441 Lecture 9
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Review: Forces and MomentsForces
These forces and moments have standard labels. The Forces are:
X Axial Force Net Force in the positive x-directionY Side Force Net Force in the positive y-directionZ Normal Force Net Force in the positive z-direction
M. Peet Lecture 9: 5 / 24
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Review: Forces and MomentsMoments
The Moments are called, intuitively:
L Rolling Moment Net Moment in the positive p-directionM Pitching Moment Net Moment in the positive q-directionN Yawing Moment Net Moment in the positive r-direction
M. Peet Lecture 9: 6 / 24
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6DOF: Newtons LawsForces
Newtons Second Law tells us that for a particle F =ma. In vector form:
F =i
Fi =mddt
V
That is, if F = [Fx Fy Fz] and V = [u v w], then
Fx =mdu
dt Fx =m
dv
dt Fz =m
dw
dt
Definition 1.L= m V is referred to as Linear Momentum.
Newtons Second Law is only valid if F and Vare defined in an Inertialcoordinate system.
Definition 2.
A coordinate system is Inertialif it is not accelerating or rotating.
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Newtons LawsMoments
Using Calculus, this concept can be extended to rigid bodies by integration over
all particles.M=
i
Mi = d
dtH
Definition 3.
Where H= (rc vc)dm is the angular momentum.Angular momentum of a rigid body can be found as
H=I I
where I= [p, q, r]T is the angular rotation vector of the body about the
center of mass. p is rotation about the x-axis.
qis rotation about the y-axis.
r is rotation about the z-axis.
I is defined in an Inertial Frame.
The matrix I is the Moment of Inertia Matrix.M. Peet Lecture 9: 8 / 24
8/12/2019 441 Lecture 9
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Newtons LawsMoment of Inertia
The moment of inertia matrix is defined as
I=
Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz
Ixy =Iyx = xydm I xx = y2 +z2 dmIxz =Izx =
xzdm I yy =
x2 +z2
dm
Iyz =Izy =
yzdm I zz =
x2 +y2
dm
So HxHy
Hz
=
Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz
pIqI
rI
where pI, qI and rIare the rotation vectors as expressed in the inertial frame
corresponding to x-y-z.M. Peet Lecture 9: 9 / 24
8/12/2019 441 Lecture 9
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Moment of InertiaExamples:
Homogeneous Sphere
Isphere =2
5mr2
1 0 00 1 0
0 0 1
Ring
Iring =mr2
1
2 0 0
0 12
00 0 1
M. Peet Lecture 9: 10 / 24
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Moment of InertiaExamples:
Homogeneous Disk
Idisk =1
4mr2
1 + 13
hr2
0 00 1 + 1
3
hr2
00 0 1
2
F/A-18
I=
23 0 2.970 15.13 0
2.97 0 16.99
kslug f t2
M. Peet Lecture 9: 11 / 24
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Problem:The Body-Fixed Frame
The moment of inertia matrix, I, is fixed in the body-fixed frame. However,
Newtons law only applies for an inertial frame:M=
i
Mi = d
dtH
If the body-fixed frame is rotating with rotation vector , then for any vector,a,d
dta in the inertial frame is
da
dt
I
= da
dt
B
+ a
Specifically, for Newtons Second Law
F =mdV
dt
B
+m V
and
M=d H
dtB
+ H
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Equations of Motion
Thus we have
FxFyFz
= m u
vw
+m detx y z
p q ru v w
= m u+qw rv
v+ru pww+pv qu
and
LMN
= Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz
pqr
+ Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz
pqr
=
Ixxp Ixyq IxzrIxyp+Iyy q IyzrIxzp Iyz q+Izzr
+
pIxx qIxy rIxzpIxy+qIyy rIyzpIxz qIyz +rIzz
=
Ixxp Ixyq Ixzr+q(pIxz qIyz +rIzz ) r(pIxy+qIyy rIyz )Ixyp+Iyy q Iyzr p(pIxz qIyz +rIzz ) +r(pIxx qIxy rIxz)Ixzp Iyz q+Izzr+p(pIxy+qIyy rIyz ) q(pIxx qIxy rIxz)
Which is too much for any mortal. For aircraft, we have symmetry about the
x-z plane. Thus Iyz =Ixy = 0. Spacecraft?M. Peet Lecture 9: 13 / 24
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Equations of MotionReduced Equations
With Ixy =Iyz = 0, we have, in summary:
FxFy
Fz
= m
u+qw rvv+ru pw
w+pv qu
and
LMN
= Ixxp
Ixzr
qpIxz+ qrIzz
rqIyyIyyq+p2Ixz prIzz +rpIxx r2IxzIxzp+Izzr+pqIyy qpIxx+qrIxz
Right now,
Translational variables (u,v,w) depend on rotational variables (p,q,r). Rotational variables (p,q,r) do not depend on translational variables
(u,v,w). For aircraft, however, Moment forces (L,M,N) depend on rotational and
translational variables.M. Peet Lecture 9: 14 / 24
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EOMs in Rotating FrameExample: Snipers
Question: Consider a sniper firing a rifle due east at the equator. Ignoring
gravity and drag, what are the equations of motion of the bullet? Use theNorth-East-Up local coordinate system. Muzzle velocity: 1000m/s. Range:4km.
Answer: The earth is rotating about its axis at angular velocity 2 radday
, or
.0000727rads
. The rotation is positive about the local North-axis. Thus
=
pq
r
=
.00007270
0
Since the bullet is in free-flight, there are no forces. Thus the Equations ofmotion are
00
0
= m
u+qw rvv+ru pw
w+pv qu
= m
uv pw
w+pv
M. Peet Lecture 9: 15 / 24
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EOMs in Rotating FrameExample: Snipers
Simplified EOMs: Using q=r = 0, we simplify to
u= 0 v=pw w= pv.
Solution: For initial condition u(0) = 0, v(0) =V and w(0) = 0 has solution
u(t) = 0 v(t) =v(0) cos(pt) w(t) = v(0) sin(pt)
Since p is very small compared to flight time, we can approximate
u(t) = 0 v(t) =v(0) w(t) = v(0)pt
Which yields displacement
N(t) = 0 E(t) =v(0)t U(t) = 1
2v(0)pt2
Conclusion: For a target at range E(ti) = 4km, we haveti = 4sand hence theerror at target is:
N(ti) = 0 U(ti) = 1
2 2000 .0000727 16 = 1.1635m
Of course, if we were firing west, the error would be +1.1635m.M. Peet Lecture 9: 16 / 24
E l A l
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Euler Angles
Issue: Equations of motion are expressed in the Body-Fixed frame.
Question: How do determine rotation and velocity in the inertial frame. For
intercept, obstacle avoidance, etc.Approach: From Lecture 4, any two coordinate systems can be related througha sequence of three rotations. Recall these transformations are:
Roll Rotation () :
R1()
=
1 0 00 cos sin 0 sin cos
Pitch Rotation ():
R2()
=
cos 0 sin
0 1 0 sin 0 cos
Yaw Rotation ():
R3()
=
cos sin 0sin cos 0
0 0 1
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E l A l
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Euler Angles
Definition 4.
The term Euler Anglesrefers to the angles of rotation (, , ) needed to gofrom one coordinate system to another using the specific sequence of rotationsYaw-Pitch-Roll:
VBF =R1()R2()R3()VI.
NOTE BENE: Euler angles are often defined differently (e.g. 3-1-3). We usethe book notation.The composite rotation matrix can be written
R1()R2()R3() =
1 0 00 cos sin 0 sin cos
cos 0 sin 0 1 0
sin 0 cos
cos sin 0sin cos 0
0 0 1
This moves a vector
Inertial Frame Body-Fixed Frame
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Euler Angles
To move a vector
Body-Fixed Frame Inertial Frame
we need toInvert the Rotations. Rotation matrices are easily inverted, however
Ri()1 =Ri()
Thus VI= (R1()R2()R3())1 VBF, where
(R1()R2()R3())1
=R3()1R2()
1R1()1
=R3()R2()R1()
=
cos sin 0sin cos 0
0 0 1
cos 0 sin 0 1 0
sin 0 cos
1 0 00 cos sin
0 sin cos
=
cos cos sin sin cos cos sin cos sin cos + sin sin cos sin sin sin sin + cos cos cos sin sin sin cos
sin sin cos cos cos
These transformations now describe a Roll-Pitch-Yaw.M. Peet Lecture 9: 19 / 24
E l A l
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Euler AnglesVelocity vector
Thus to find the inertial velocity vector, we must rotate FROM the body-fixedcoordinates to the inertial frame:
dxdtdydtdzdt
=R3()R2()R1()
uv
w
M. Peet Lecture 9: 20 / 24
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Euler Angles
The rate of rotation of the Euler Angles can be found by rotating the rotationvector into the inertial frame
pq
r
=
1 0 sin 0 cos cos sin
0 sin cos cos
This transformation can also be reversed as
=
1 sin tan cos tan 0 cos sin
0 sin sec cos sec
pq
r
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Summary
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Summary
M. Peet Lecture 9: 22 / 24
Conclusion
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Conclusion
In this lecture we have covered
Equations of Motion
How to differentiate Vectors in Rotating Frames
Derivation of the Nonlinear 6DOF Equations of Motion
Euler Angles
Definition of Euler Angles
Using Rotation Matrices to transform vectors
Derivatives of the Euler angles Relationship to p-q-r in Body-Fixed Frame
M. Peet Lecture 9: 23 / 24
Next Lecture
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Next Lecture
In the next lecture we will cover
Linearized Equations of Motion
How to linearize the nonlinear 6DOF EOM
How to linearize the force and moment contributions
Force and Moment Contributions
The gravity and thrust contributions
The full linearized equations of motion including forces and moments
How to decouple into Longitudinal and Lateral Dynamics Reminder on how to create a state-space representation.
M. Peet Lecture 9: 24 / 24