The Dynamic Equations of Motion

an element of mass on an airplane The dynamic rigid body equations of motion are obtained from Newton second law, which state

a) Summation of all external forces acting on a body equal to the time rate of change of linear momentum of the body.

b) Summation of all external moments acting on the body equal to time rate

of change of moment of momentum (angular momentum).

Force Equations Newtons second law

)(mVdtdF =

Where

& 321

321

eFeFeFFeweveuV

zyx

KKKKKKKK

++=++=

The vector equation can be rewritten in scalar form i.e. three force equations

)(mudtdF x = )(mvdt

dF y = )(mwdtdFz =

The force components are composed contributions due to:

a) Aerodynamic forces. b) Propulsive force (Thrust). c) Gravitational forces.

Consider the airplane shown in figure If m : an element of mass of airplane v : The velocity of elemental mass relative to the absolute or inertial frame vc : Velocity of the center of the mass.

dtdr : Velocity of the element relative to the center of the mass

F : Resulting force acting on the elemental mass

Therefore

v = vc+ dtdr

= FF And according to Newton second low

dtdvmF =

+== mdrdvdtdFF c )( Assume constant mass of vehicle

+= mrdtddtdvmF c 22

The center of mass is that point which summation of mass*distance to that point equal zero

Then mr =0 Force equation yield

dtdv

mF c=

The derivatives of an arbitrary vector (A) referred to a rotating body frame having an angular velocity can be represented by

AdtdA

dtdA

BI

+= Where I: Inertial frame & B: Body fixed frame of the reference. Applying the identify

)( cB

c mdtdvmF +=K

where, F are the forces components except the ground force.

(i.e., Aerodynamics forces and Thrust along the body fixed axes)

+++=

wvurqpwvumFeee

eeeKKK

KKKK 321321

....

=FK ]).().().[( 321 eee qupvwpwruvrvqwum KKK +++++

Take the gravity force effect into consideration, so

Where:

sin=kK 1eK + cossin 2eK + coscos 3eK Where X, Y and Z are the forces components expect the ground force (i.e., Aerodynamic forces and thrust along the body fixed axes). The three scalar force equations are:

coscos)(sincos)(

sin)(

mgqupvwmFmgpwruvmFmgrvqwumF

Z

Y

X

+=+=++=

kmgeZeYeXFGGGGG +++= 321

Momentum Equations

= ).( VmdtdFKK

= ).( VrmdtdFrKKKK

)()( HdtdM =

The vector equation can be rewritten in scalar

xHdtdL = yHdt

dM = zHdtdN =

Where: L, M, and N : components of the moment along the x, y, and z axes respectively. Hx, Hy and Hz : components of angular momentum along x, y, z axes respectively. To develop the moment equations

M= Fr KK H= dtd

( VrKK ) m

The total angular momentum can be written as

H= H = + mrrvmr c )]([)..( KKK

The first term in RHS will be eliminated as ).( mr =0

)()()( rrrrrr KKKKKKKKK =

where

K = p. 1eG + q. 2eG + r. 3eG & rK = x . 1eG + y . 2eG + z . 3eG

Thus angular momentum can be written as:

H= [(x2 + y2 + z2) m].(p. 1eG

+q. 2eG

+r. 3eG

) [(PX+qy+rz) m].(x. 1eG

+y. 2eG

+z. 3eG

)

The scalar components of H are

Hx = p + mzy ).( 22 q mxy ).( -r mxz ).(

Hy = -p mxy ).( + q + mzx ).( 22 -r myz ).(

Hz = -p mxz ).( q myz ).( +r + mxy ).( 22 Note that:

mzyI x += )( 22 = mxyI xy mzxI y += )( 22 = mxzI xz myxI z += )( 22 = myzI yz Hence

XH = p Ix q xyI -r Ixz

yH = -p Ixy+ q Iy -r Iyz

zH = -p Ixz q Iyz +r Iz

HdtdHM

B

c +=

HHH

eeeeHeHeH

ZYX

ZYX rqp

KKKKKK 321

321)....( +++=

The scalar components of M is:

x z y

y z x

z y x

L q r

M p r

N p q

H H H

H H HH H H

= +

= += +

Since the moments and products of inertia doesnt change with time (good assumption for small aircraft) so:

x x xy xz

y y xy yz

y z xz yz

p qH I I rIq pH I I rI

p qH rI I I

= = =

Finally; (symmetry Ixy = Iyz = 0 )

)(..

yzxzxzx IIqrpqIrIpIL +=

)()( 22.

zxxzy IIprrpIqIM ++=)(

..

xyxzxzz IIpqqrIpIrIN ++=