RESOLUTION OF INERTIAL FORCES AND MOMENTS
The right-hand side of Equation 3.8 contains the term V, which represents the time derivative of velocity with respect to the body axis system. The cross product terms (v x V) arise from the centripetal acceleration along any given axis due to angular velocities about the remaining two axes. In component form we have
V = ui + vj + wj
v = pi + qj + rj (3.10)
H = Hxj + Hyj + Hzj
Using the above expressions in Equation 3.8, we have
F = m [[Ui + vj + Wj] + (pi + qj + rj) x (ui + vj + wj)]]
F = m[U + qw — rv]i + m[v + ru — pw] j + m[W + pv — qu]k
Thus, the force components acting along the X, Y, and Z body axes are
Fx = m(u + qw — rv)
Fy = m(v + ru — pw) (3.11)
Fz = m(w + pv — qu)
Example 3.1
Assume two-dimensional accelerated flight of an aircraft. In this situation U + DU and V + Д V would have rotated by an incremental angle R Dt. Of course, the aircraft would also have rotated through the same angle. Here U and V are the total velocities of the aircraft in forward and lateral directions. Then the equation for the rectilinear longitudinal acceleration can be easily derived.
,• ,* ™(U + DU)cos(R Dt) — U — (V + Д V) sin(R Dt)
aX = lim(Dt! 0)—————————- d—————————-
Since the limit of sin(RtDt) = R, we finally obtain for small changes/perturbation
aX = U_ — VR
Example 3.2
Assume two-dimensional accelerated flight of an aircraft. Derive the equation for the corresponding rectilinear lateral acceleration.
We have for this case the following formula:
, (U + DU) sin(R At) + (V + DV) cos(R At) – V
ay = lim(At! 0)————————- A—————————–
Using the same reasoning as the previous case, we get
ay = V + UR
Such small perturbation terms appear in Equation 3.11 as can be clearly seen.
From Equation 3.9, we have
M = H + (v x H)
The total moment about a given axis is due to both direct angular acceleration about that axis and those arising from linear acceleration gradients due to combined rotations about all axes. The components of angular momentum about the three axes are given by
Hx = pix q1xy r1xz
Hy = – pixy + qly – rlyz (3.12)
Hz = pizx qizy A riz
Even though there could be variation in the mass and its distribution during a flight due to fuel consumption, use of external stores, etc., the assumption that the mass and mass distribution of the air vehicle are constant is reasonable and simplifies the expression for the rate of change of angular momentum. The time derivatives of the angular momentum components expressed in body axis are given by
Hx = pix – q ixy – rixz
Hy = – pixy + qiy – riyz (3.13)
Hz pizx qizy A riz
To get the total time rate of change of angular momentum, and thus the scalar components of the moments, we must add to these contributions arising because of steady rotations (v x H).
L = Hx – rHy + qHz
M = rHx + Hy – pHz (3.14)
N = —qHx + pHy + Hz
Substituting for angular momentum components and regrouping results, the moment equations can now be written (with respect to the body axis system) as
L — plx qIxy rIxz T qr(Iz Iy) F (r q )Iyz pqIxz F rplxy
M — – pixy + qIy – rIyz + rp(Ix – /z) + (p2 – r2)/xz – qrIxy + pqlyz (3.15)
N pIxz qIyz F rIz F pq(Iy Ix) T (q p )Ixy rpIyz T qrIxz
The total forces and moments acting on the aircraft are made up of contributions from the aerodynamic, gravitational, and aircraft thrust.