thesis - Computer Graphics Group - Charles University - Univerzita ...

5.1. CONCEPTS 39
(this is the length of the circle arc swept by p(t) per second, hence the velocity magnitude). The direction of
˙p(t) would have to be perpendicular to both ω(t) and r(t) and due to the counter-clockwise convention would
have to equal
ω(t)×r(t)
. Therefore, combining the two conditions on the velocity magnitude and the direction,
ω(t)×r(t)
˙p(t) = (ω(t) · r(t) · sin(α)) · ω(t)×r(t)
ω(t)×r(t) = ω(t) × r(t). If v(t) = 0 and angular velocity was still constant, then
the trajectory of p(t) would no longer correspond to a circle, but the trajectory of r(t) would. Hence we could write
that ˙ r(t) = ω(t) × r(t). Finally, noting that the appropriate magnitude of r(t) was derived from the instantaneous
ω(t), we can conclude that this equation holds even if the angular velocity is not constant.
In the previous paragraph we have shown that ˙ r(t) = ω(t) × r(t) for vectors r(t) attached to
the rigid body’s center of mass. This time, we will assert that the same equation holds for vectors
r(t) attached to arbitrary points on the rigid body. If r(t) is a world space vector attached to a
world space point p(t) of a rigid body, then a(t) = p(t) and b(t) = p(t) + r(t) are the world space
positions of the origin and tip of r(t), r(t) = b(t) − a(t) and ˙ r(t) equals
˙r(t) = ∂
∂t b(t) − ∂
a(t) = ω(t) × r(t), (5.5)
∂t
as can be verified by evaluating equation (5.4) for points a(t) and b(t).
We will now use this equation to derive a formula for ∂
∂t R(t) = ˙ R(t). R(t) can be writ-
ten in a block fashion as R(t) = (rx(t), ry(t), rz(t)) and so ˙ R(t) = ( ˙ rx(t), ˙ ry(t), ˙ rz(t)). Since
rx(t), ry(t), rz(t) are the body space axes expressed in the world coordinates and the axes are attached
to the rigid body’s center of mass, we get that ˙ R(t) = (ω(t)×rx(t), ω(t)×ry(t), ω(t)×rz(t)).
If we define matrix a ∗ as
a ∗ ⎛
⎞
0 −az ay
= ⎝ az 0 −ax ⎠
−ay ax 0
then a × b can be expressed in the form of matrix multiplication as a × b = a ∗ · b, therefore
˙R(t) = (ω(t) ∗ · rx(t), ω(t) ∗ · ry(t), ω(t) ∗ · rz(t)), finally obtaining
5.1.4 Force and Torque
˙R(t) = ω(t) ∗ · R(t). (5.6)
When we were working with particle systems we tracked directions and magnitudes of forces
exerted on the particles ( F total
i (t) vectors), but we did not care about the points of application,
because they always coincided with the particle positions. The only effect a force exerted on
a particle could produce was a translation. However since rigid bodies have volumes and can
rotate, we need know not only what the direction and magnitude of the exerted force is, but also
the point of its application so that both body translation and rotation could be affected by the
force.
We will conceptually work with a rigid body that consists of N particles constrained to remain
at the same relative positions in the body space. Particle positions in the body space will be
fixed and characterized by vectors r b i , 1 ≤ i ≤ N, their world space positions will be given by
ri(t) = R(t) · r b i + x(t).
To simplify further derivations we will assume that the points of force application coincide
with the positions of particles2 the rigid body is made of. This will allow to characterize the force
distribution over the volume of the body by a set of world space vectors F total
i (t), where F total
i (t)
denotes the total force exerted at the rigid body’s particle i, at the world space position ri(t).
2 We can always think that there happens to be a particle at the given world space position so that forces could
act at arbitrary points in the world space.