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F = m˙Vc.g. (2.6)
2.1.2 General moment equation for a rigid body
The moment dM, which is measured around the center of gravity, is equal to the
time-derivative of the angular momentum of the mass point relative to the c.g.:
dM = d
dt
(r × V)dm = (˙r × V)dm + (r × ˙V)dm (2.7)
where:
˙r = V − Vc.g. (2.8)
and:
(r × ˙V )dm = r × dF = dMc.g. (2.9)
In this equation dMc.g. denotes the moment of the force dF about the center of gravity.
The angular momentum of the mass point relative to the c.g. will be denoted by dh,
which is defined as: dh  (r × V)dm. Writing this out yields:
dMc.g. = d˙h − (V − Vc.g.) × Vdm = d˙h + Vc.g. × Vdm (2.10)
The contributions of all mass points are once again summed across the whole rigid
body, yielding:
ådMc.g. = åd˙h + Vc.g. ×åVdm (2.11)
The equation for the resulting moment Mc.g. about the c.g. then becomes:
Mc.g. = ˙h (2.12)
where h denotes the resulting angular momentum of the body about the center of
gravity.
2.1.3 Angular momentum around the center of gravity
Consider a rigid body with angular velocity W, with components p, q, and r about
the X, Y, and Z axes of the right-handed reference frame respectively:
W = i p + j q + k r (2.13)
2.1. General rigid-body equations of motion 13
symbol definition
Ixx å(y2 + z2) dm
Iyy å(x2 + z2) dm
Izz å(x2 + y2) dm
Jxy å xy dm
Jxz å xz dm
Jyz åyz dm
Table 2.1: Moments and products of inertia
where i, j, and k are unity vectors along the X, Y, and Z-axes. The total velocity
vector of a mass point of a rigid body that both translates and rotates becomes:
V = Vc.g. + W × r (2.14)
hence, the angular momentum of the rigid body about the c.g. can be written as:
h  ådh = år × (Vc.g. + W × r)dm = år × Vc.g.dm +år × (W × r)dm
(2.15)
The first term of the right hand side of equation (2.15) is equal to zero:
􀀀
årdm


× Vc.g. = 0 (2.16)
and for the second term we can write:
år × (W × r)dm = å

W(r · r) − r(W · r)
 
dm = å

Wk r k2 − r (W · r)
 
dm
(2.17)
Substitution of r = i x + j y + k z, (2.16), and (2.17) in equation (2.15) yields:
h = Wå(x2 + y2 + z2)dm −år (px + qy + rz)dm (2.18)
The components of h along the X, Y, and Z axes will be denoted as hx, hy, and hz
respectively, yielding:
hx = på(y2 + z2)dm − qå xy dm − r å xz dm
hy = −på xy dm + qå(x2 + z2)dm − r åyz dm
hz = −på xz dm − qåyz dm + r å(x2 + y2)dm
(2.19)
The summations appearing in these equations are defined as the inertial moments
and products about the X, Y, and Z axes respectively; see table 2.1.1 Using these
definitions, equations (2.19) can be written in vector notation as a product of the
inertia tensor I with the angular velocity vector W:
h = I · W (2.20)
where I is defined as:
I =
2
4
Ixx −Jxy −Jxz
−Jyx Iyy −Jyz
−Jzx −Jzy Izz
3
5 (2.21)
1The summations across the body actually have to be written as integrals, but that further refinement
has been omitted here.
14 Chapter 2. Rigid body equations of motion
If the airplane is symmetrical, Jxy and Jyz are both identically zero, which would further
simplify the inertia tensor. Although this assumption is often made in literature,
this report will consider the more general case where the aircraft does not necessarily
have to be symmetrical.
So far, in this analysis of angular motion we have neglected the angular momentum
of any spinning rotors, assuming that the airplane is a single rigid body. However,
these effects can actually be quite significant in practice. For example, a number of
WorldWar I aircraft had a single ‘rotary’ engine that had a fixed crankshaft and rotating
cylinders; the gyroscopic effects caused by the angular momentum of the engine
gave these aircraft tricky handling characteristics. The effects are also noticeable in
maneuvering flight of a small jet with a single turbofan engine on the longitudinal
axis [33].
Each rotor has an angular momentum relative to the body axes. This can be
computed from equation (2.20) by interpreting the moments and products of inertia
therein as those of the rotor with respect to the axes parallel to the airplane bodyaxes
　
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