lecture 7 : flight equations of motiondynlab.mpe.nus.edu.sg/mpelsb/me4241/l7n.pdf · lecture 7 :...

G. Leng, Flight Dynamics, Stability & Control

Lecture 7 : Flight Equations of Motion

Or the differential equations for a 6 DOF model


1.0 Recap - 6 DOF Dynamics Model

• For flight dynamics & control, the reference frame is locatedat the cg, aligned with the aircraft and moves with it.

Yb

Xb

Zb


The 6 DOFs correspond to :

• translation along the X,Y,Z directions

– velocity vector in body axes V = {u,v,w}

• rotation about the X,Y, Z axes

– angular velocity vector in body axes = {p,q,r}

Now we need to relate forces and moments to these 6 DOFs


• Denote the forces acting on the aircraft by F = {X,Y,Z}

– These forces act at the origin (c.g) of the reference frame

• Similarly denote the moments by Mcg = {l, m, n}

– These are the roll, pitch and yaw moments taken wrt the c.g.

Question: So how do we relate the force to the 3 translationalDOF’s ?

Answer : F = d(mV)/dt = m {u’, v’, w’} ?


2.0 Coriolis theorem

Coriolis Theorem states that the absolute or actual time rate ofchange of a vector r = xi + yj + zk where the reference frame ijkrotates with absolute angular velocity = pi + q j + rk is :

dr/dt = ( dx/dt i + dy/dt j + dz/dt k ) + x r


2.1 Translational equations of motion

The translational motion is governed by :

(mV)/t + x (mV) = F

Assuming the mass m is constant over the time interval ofinterest, this works out as ...


u' + q w – r v = X/m

v' + r u – p w = Y/m

w' + p v – q u = Z/m

Translational equations of motion

Note the nonlinear terms arising from the correction x (mV)


Notes:

We can decompose

F = Faero + T + W

i.e.

aero forces Faero = {Xaero, Yaero, Zaero}

thrust T = {Tx, TY, TZ }

weight W = Mg {-sin cos sin cos cos }


3.0 Angular momentum and moments

The moments acting on the aircraft may be related to the 3rotational DOFs by

dH/dt = M

where M is the moment of the forces acting on the aircraft

H is the angular momentum of the aircraft

Question : Where’s the catch ?


The relation holds only if both M and H are taken wrt

(a)

or

(b)

The second case holds for our reference frame.

So what’s Hcg for an aircraft ?


3.1 Computing the angular momentum Hcg of an aircraft

Consider a small piece of the aircraft of mass dm locatedat r = {x,y,z} wrt the body axes.

Xb

Yb

Zb

r

dm


The linear momentum of this small mass relative to the cg is

= {p, q, r} x { x, y, z} dm

= {qz - ry, rx - pz, py - qx} dm

dLcg = ( x r) dm


The angular momentum wrt the cg of this small mass is

= r x ( x r) dm

= { x, y, z} x {qz - ry, rx - pz, py - qx} dm

dHcg = r x dLcg

Hcg = { x, y, z} x {qz - ry, rx - pz, py - qx} dm

Finally we sum over the whole aircraft


Where

(y2 + z2)dm - xy dm - zx dm

Icg = - xy dm (z2 + x2)dm - yz dm

- zx dm - yz (x2 + y2)dm

Simplifying the rhs, the aircraft angular momentum is Hcg = Icg

is the moment of inertia matrix.


The principal moments of inertia are:

Ixx= (y2 + z2)dm Iyy= (z2 + x2)dm Izz= (x2 + y2)dm

The crossed moments of inertia are:

Ixy= xy dm Iyz= yz dm Izx= zx dm

What’s the difference with planar motion of rigid bodies ?


Ex : Simplifications for symmetric aircraft

Question : Which is the plane of symmetry ?


X

Y

Consider Ixy = xy dm

For each section along the X axis,the contribution from the left sidecancels the contribution from theright.

Ixy = 0 Is that all ?


Similarly for Iyz= yz dm

Z

Y

For each section along the Z axis,the contribution from the left sidecancels the contribution from theright.

Iyz = 0


Hence the moment of inertiamatrix simplifies to:

Ixx 0 -Ixz

Icg = 0 Iyy 0

-Ixz 0 Izz


3.2 Rotational equations of motion

Now apply Coriolis theorem to dHcg/dt = Mcg

where Hcg = Icg the rotational equations of motion are:

( Icg )/t + x ( Icg) = Mcg

Which simplifies ( Ixy = Iyz = 0 ) to :


Ixx p’ + ( Izz - Iyy ) qr - Ixz ( r’ + pq ) = l

Iyy q’ + ( Ixx - Izz ) pr - Ixz ( r2 – p2 ) = m

Izz r’ + ( Iyy - Ixx ) pq - Ixz ( p’ - qr ) = n

Assuming the XZ plane is a plane of symmetry


Summary : The basic 6 DOF model for symmetric aircraft

u' + q w – r v = (Xaero+Tx)/m - g sin

v' + r u – p w = (Yaero+Ty)/m + g cos sin

w' + p v – q u = (Zaero+Tz)/m + g cos cos

Ixx p’ + ( Izz - Iyy ) qr - Ixz ( r’ + pq ) = lIyy q’ + ( Ixx - Izz ) pr - Ixz ( r2 – p2 ) = mIzz r’ + ( Iyy - Ixx) pq - Ixz ( p’ - qr ) = n

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

' = q cos - r sin

' = (q sin + r cos ) sec


Homework

Assume the thrust vector lies in theplane of symmetry, act at the c.g.with an orientation angle from theX axis as shown.

x

y

z

thrust

u = VT cos cos

v = VT sin

w = VT cos sin

Derive expressions for ’, ’ and VT’

Using the relations:


Answer : These are the translational EOM in flight path axes

lecture 7 : flight equations of motiondynlab.mpe.nus.edu.sg/mpelsb/me4241/l7n.pdf · lecture 7 :...

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