Free Flight

Let us begin by summing Euler’s law Eq. (6.32) over all particles of a rigid body

and do the same for its alternate form Eq. (6.33)

(6.34)

where all internal moments cancel each other and only the external moments remain. The linear velocity was replaced by its time derivative of the displacement vector s,/. Introduce for the time being an arbitrary reference point R (see Fig. 6.11) of the rigid body В into Eq. (6.34):

*;/ = Sm + Sri

We obtain six terms:

^2 D’ (m. SjtD1 slR) Term (1) + D1 (т^тВ’sRI) Term (2)

+ ]T D’ (m, SmD’sK,) Term (3) + J2D’ (m‘ Term (4) (6.35)

І І

= ]T(W,)Term (5) + £($«/,)Term (6)

At this point we split the treatment into the two cases. First, we confine the reference point R to the c. m. В and develop the free-flight attitude equations. Afterward, we let R be any point of body В and assign it also as a point of the inertial frame I, thus addressing the dynamics of the top.

Let us modify the six terms of Eq. (6.35). The inner rotational derivative of the first term is transformed to frame B, and because В is a point of frame B, DBsiB = 0.

Term (1):

Y^D^nuSmD’sis) = ^Г/У[тЛл(/Лш + ГУл)] =

І І І

Referring back to Eq. (6.22), we conclude that the term in parentheses is the MOI IBB of the vehicle multiplied by its inertial angular velocity uinl, and therefore

Term (1) = D,{lBBu:BI)

Term (2):

D’^SbiD^b!) = SBID,(mBvIB)+mBD, SBIvIB = SBID,(mBvIB)

+ mBVBvB = SB1DI(mBv, B)

because the cross product is zero.

Term (3):

= 0

because В is the c. m. Term (4):

‘ED’imiSBi&Su,) = D1 БШ = Dl SBiD‘ y^ ni, slB

because mt is constant and В is the c. m. Term (5):

total external moment. Term (6):

£>B//,) = SB, f

І

because all internal forces cancel. The modified Terms (2) and (6) express Newton’s second law premultiplied by SBj and are therefore satisfied identically (SBi is
generally not zero). From the remaining Terms (1) and (5) we receive our final result:

D‘(lBBu>Bl) =mB (6.36)

where according to Eq. (6.23) = lBB is the angular momentum of body В

wrt the inertial frame and referred to the c. m. Euler’s law for rigid bodies states therefore that the time rate of change relative to the inertial frame of the angular momentum lBB of a rigid body referred to its c. m. is equal to the externally applied moment m B with the c. m. as reference point

D’lBB’ = mB (6.37)

Equation (6.36) does not include any reference to the linear velocity or acceler­ation of the vehicle. What a fortuitous characteristic! Euler’s law is applied as if the vehicle were not translating. This feature is referred to as the separation theo­rem. Just as linear and angular momenta can be calculated separately, then so can the translational equations of motion be formulated separately from the attitude equations. Newton’s second law, Eq. (5.9), and Euler’s law, Eq. (6.36), deliver the fundamental equations of aerospace vehicle dynamics

mB D’vg = /, D1 (Іввшв1) = mB

and with the compact nomenclature of linear and angular momenta

D1 pB = f, D‘lBI = mB (6.38)

The key point is the c. m. В. It serves as the focal point for the linear momentum pB, encompassing all mass of body В as if it were a particle. For the angular momentum lBB it is that reference point which separates the attitude motions from the translational degrees of freedom. As I will show, without the c. m. as reference point the equations of motion of aerospace vehicles are more complex.

As ahistorical tidbit, I want to mention that the equations of motion (6.38), which we like to call today the six-DoF equations, have been known for quite some time. In 1924, while aviation was still in its infancy, R. v. Mises published the “Bewe – gungsgleichungen eines Flugzeuges,” buried in his so-called Motor Rechnung? He presented the translational and attitude equations in one compact formalism, already transformed to body coordinates, and identified the key external forces and moments. There we even find the inception of small perturbations and linearized equations of motion for an airplane.

Example 6.7 Aero Data Reference Point

Frequently, the aerodynamic and propulsive data are not given relative to the c. m. but to an arbitrary reference point of the aircraft or missile. If you have been involved in wind-tunnel testing, you have dealt with the moment center of the sting balance, which is usually nowhere close to the yet unknown c. m. of the flight vehicle. Or, as the space shuttle bums fuel during its ascent, large shifts of c. m. occur. In each case we need to modify the right side of Eq. (6.37).

Figure 6.12 shows the aerodynamic force / and moment mBr acting on the fixed reference center Br. To calculate the moment mB, referred to the c. m. B, we

determine the torque SBref caused by changing the origin of the force vector /, and add the free moment vector mB, :

mB = mBr + SBref (6.39)

Substituting Eq. (6.39) into Eq. (6.37) yields Euler’s equation of motion referred to the c. m. B, but with the aerodynamics referenced to the arbitrary point Br

D! lg’ = mBr + SBrBf (6.40)

For an aircraft and missile the displacement vector s Br в most likely will change in time, as the c. m. shifts during flight. Similar adjustments are made if the propulsion moment center changes.

Example 6.8 Attitude Equations for Six-DoF Simulations

Missile simulations use Euler’s equation in a form that accommodates aerody­namic moment coefficients and the MOI tensor in body coordinates. We transfer the rotational time derivative of Eq. (6.36) to the body frame В

DB(lBBu>BI) + QBIIB u>Bl = mB

and pick body coordinates ]B

DB([lB]B[coBI]B) + [QS/]B[/#]%B/]B – [mB]B

Applying the chain rule to the first term and realizing that the MOI of a rigid body remains unchanged in time, [d/§/df]s = [0], we get the desired equations for programming:

This is the attitude equation most frequently found in six-DOF simulations.