441 Lecture 9

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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

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Review: Coordinate RotationsPositive Directions

If in doubt, use the right-hand rules.

Figure: Positive Directions

Figure: Positive Rotations


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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

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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


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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


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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

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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

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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


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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


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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


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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

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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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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

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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


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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


Summary

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Summary


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


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.


441 Lecture 9

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