Page 173 - Sensing, Intelligence, Motion : How Robots and Humans Move in an Unstructured World

148 ACCOUNTING FOR BODY DYNAMICS: THE JOGGER’S PROBLEM

VisBug assumes that the intermediate target point is either on the obstacle bound-
ary or on the M-line and is visible. However, the robot’s inertia may cause it to
move so that the intermediate target T i will become invisible, either because it
goes outside the sensing range r v (as after point P , Figure 4.1) or due to occlud-
ing obstacles (as in Figure 4.6). The danger of this is that the robot may lose
from its sight point T i —and the path convergence with it. One possible solution
is to keep the velocity low enough to avoid such overshoots—a high price in
efficiency to pay. The solution we choose is to keep the velocity high and, if the
intermediate target T i does go out of sight, modify the motion locally until T i is
found again (Section 4.2.6).

4.2.3 Velocity Constraints. Minimum Time Braking
By substituting p max for p and r v for d into (4.1), one obtains the maxi-
mum velocity, V max . Since the maximum distance for which sensing informa-
tion is available is r v , the sensing range boundary, an emergency stop should
be planned for that distance. We will show that moving with the maximum
speed—certainly a desired feature—actually guarantees a minimum-time arrival
at the sensing range boundary. The suggested braking procedure, developed fully
in Section 4.2.4, makes use of an optimization scheme that is sketched briefly in
Section 4.2.4.

Velocity Constraints. It is easy to see (follow an example in Figure 4.3) that
in order to guarantee a safe stopping path, under discrete control the maximum
velocity must be less than V max . This velocity, called permitted maximum velocity,
V p max , can be found from the following condition: If V = V p max at point C 2 (and
thus also at C 1 ), we can guarantee the stop at the sensing range boundary (point
B 1 , Figure 4.3). Recall that velocity V is generated at C 1 by the control force p.
Let |C 1 C 2 |= δx;then

δx = V p max · δt

= V p max − p max t
V B 1
= 0, then t = V p max /p max . For the segment |C 2 B 1 |= r v −
Since we require V B 1
δx,we have
p max t 2
r v − δx = V p max · t −
2

From these equations, the expression for the maximum permitted velocity V p max
can be obtained:

2
V p max = p 2 δt + 2p max r v − p max δt
max
As expected, V p max <V max and converges to V max with δt → 0.