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Spacecraft and Aircraft Dynamics

Illinois Institute of Technology^ Matthew M. Peet

Lecture 9: 6DOF Equations of Motion

Aircraft Dynamics

Lecture 9

In this Lecture we will cover: Newton’s Laws

• ∑^ M~i = (^) dtd H~
• ∑^ F~i = m (^) dtd ~v Rotating Frames of Reference
• Equations of Motion in Body-Fixed Frame
• Often Confusing

Review: Coordinate Rotations

Roll-Pitch-Yaw

There are 3 basic rotations an aircraft can make:

• (^) Roll = Rotation about x-axis
• (^) Pitch = Rotation about y-axis
• (^) Yaw = Rotation about z-axis
• (^) Each rotation is a one-dimensional transformation. Any two coordinate systems can be related by a sequence of 3 rotations.

Review: Forces and Moments

Forces

These forces and moments have standard labels. The Forces are: X Axial Force Net Force in the positive x-direction Y Side Force Net Force in the positive y-direction Z Normal Force Net Force in the positive z-direction

6DOF: Newton’s Laws

Forces Newton’s Second Law tells us that for a particle F = ma. In vector form: F^ ~ = ∑ i

F^ ~i = m d dt

V~

That is, if F~ = [Fx Fy Fz ] and V~ = [u v w], then Fx = m du dt Fx = m dv dt Fz = m dw dt

Definition 1.

L^ ~ = mV~ is referred to as Linear Momentum. Newton’s Second Law is only valid if F~ and V~ are defined in an Inertial coordinate system.

Definition 2.

A coordinate system is Inertial if it is not accelerating or rotating.

Newton’s Laws

Moments Using Calculus, this concept can be extended to rigid bodies by integration over all particles. M^ ~ = ∑ i

M^ ~i = d dt

H~

Definition 3.

Where H~ = ∫^ (~rc × ~vc)dm is the angular momentum. Angular momentum of a rigid body can be found as H^ ~ = I~ωI where ~ωI = [p, q, r]T^ is the angular rotation vector of the body about the center of mass.

• p is rotation about the x-axis.
• (^) q is rotation about the y-axis.
• r is rotation about the z-axis.
• (^) ωI is defined in an Inertial Frame. The matrix I is the Moment of Inertia Matrix.

Moment of Inertia

Examples:

Homogeneous Sphere

Isphere =^25 mr^2

Ring

Iring = mr^2

Moment of Inertia

Examples:

Homogeneous Disk

Idisk =^14 mr^2

1 + (^13) r^ h 2 0 0 0 1 + (^13) r^ h 2 0 (^0 0 )

F/A-

I =

 (^) kslug − f t^2

Equations of Motion

Thus we have  

Fx Fy Fz

 (^) = m

u˙ v ˙ w ˙

 (^) + m det

xˆ ˆy zˆ p q r u v w

 (^) = m

u˙ + qw − rv v ˙ + ru − pw w ˙ + pv − qu

and  

L

M

N

Ixx −Ixy −Ixz −Iyx Iyy −Iyz −Izx −Izy Izz

p˙ q ˙ r ˙

 (^) + ~ω ×

Ixx −Ixy −Ixz −Iyx Iyy −Iyz −Izx −Izy Izz

p q r

Ixx p˙ − Ixy q˙ − Ixz r˙ −Ixy p˙ + Iyy q˙ − Iyz r˙ −Ixz p˙ − Iyz q˙ + Izz r˙

 (^) + ~ω ×

pIxx − qIxy − rIxz −pIxy + qIyy − rIyz −pIxz − qIyz + rIzz

Ixx p˙ − Ixy q˙ − Ixz r˙ + q(−pIxz − qIyz + rIzz ) − r(−pIxy + qIyy − rIyz ) −Ixy p˙ + Iyy q˙ − Iyz r˙ − p(−pIxz − qIyz + rIzz ) + r(pIxx − qIxy − rIxz ) −Ixz p˙ − Iyz q˙ + Izz r˙ + 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?

Equations of Motion

Reduced Equations With Ixy = Iyz = 0, we have, in summary:  

Fx Fy Fz

 (^) = m

u˙ + qw − rv v ˙ + ru − pw w ˙ + pv − qu

and (^)  

L

M

N

Ixx p˙ − Ixz r˙ − qpIxz + qrIzz − rqIyy Iyy q˙ + p^2 Ixz − prIzz + rpIxx − r^2 Ixz −Ixz p˙ + Izz r˙ + 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). I (^) For aircraft, however, Moment forces (L,M,N) depend on rotational and translational variables.

EOMs in Rotating Frame

Example: 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) = − 12 v(0)pt^2 Conclusion: For a target at range E(ti) = 4km, we have ti = 4s and hence the error at target is: N (ti) = 0 U (ti) = − 12 ∗ 2000 ∗. 0000727 ∗ 16 = − 1. 1635 m Of course, if we were firing west, the error would be +1. 1635 m.

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 through a sequence of three rotations. Recall these transformations are:

Roll Rotation (φ) : R 1 (φ)

=

0 cos φ − sin φ 0 sin φ cos φ

Pitch Rotation (θ): R 2 (θ)

=

cos θ 0 sin θ 0 1 0 − sin θ 0 cos θ

Yaw Rotation (ψ): R 3 (ψ)

=

cos ψ − sin ψ 0 sin ψ cos ψ 0 0 0 1

Euler Angles

To move a vector Body-Fixed Frame ⇒ Inertial Frame we need to Invert the Rotations. Rotation matrices are easily inverted, however Ri(θ)−^1 = Ri(−θ) Thus V~I = (R 1 (φ)R 2 (θ)R 3 (ψ))−^1 V~BF , where (R 1 (φ)R 2 (θ)R 3 (ψ))−^1 = R 3 (ψ)−^1 R 2 (θ)−^1 R 1 (φ)−^1 = R 3 (−ψ)R 2 (−θ)R 1 (−φ)

=

cos ψ − sin ψ 0 sin ψ cos ψ 0 0 0 1

cos θ 0 sin θ 0 1 0 − sin θ 0 cos θ

0 cos φ − sin φ 0 sin φ cos φ

cos θ cos ψ sin φ sin θ cos ψ − cos ψ sin ψ cos φ sin θ cos ψ + sin φ sin ψ cos θ sin ψ sin φ sin θ sin ψ + cos φ cos ψ cos φ sin θ sin ψ − sin φ cos ψ − sin θ sin φ cos θ cos φ cos θ

These transformations now describe a Roll-Pitch-Yaw.

Euler Angles

Velocity vector

Thus to find the inertial velocity vector, we must rotate FROM the body-fixed coordinates to the inertial frame:  

dx dt dy dt dz dt

 (^) = R 3 (−ψ)R 2 (−θ)R 1 (−φ)

u v w