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Unformatted text preview: ddition) so, assume w.l.o.g. first link is longest
n 0 i Theorem 8.6.3: Reachability region for nlink arm is origincentered annulus with outer radius r = l r = l and inner radius 0 if longest link is at most half the total length of links, and r = l  l otherwise.
i =1 i M i M iM i L4 L3 L1 L2 ri = l1  (l2 + l3 + l4 ) Robot Arm Motion: Finding Configurations Find a single solution Can arm tip reach p? 2Link Case C1 l1 C2 l2 Intersect circle C1 of radius l1 (centered on origin J0) with circle C2 of radius l2 (centered on origin p). In general there are 2 solutions (depends on how
circles intersect) p = point to be reached
Source: O'Rourke Robot Arm Motion: Finding Configurations
Align L1 with L2 Case 1 : R C 0 "AntiAlign" L1 with L2 / O I Theorem: Every 3link problem can be solved by one of these 2link problems: (l1 + l2 , l3) [Fig 8.22(a)] (l1, l2 + l3) [Fig 8.22(b),(c), Fig
8.23]
j0 = 1st joint angle C is centered at p and has radius l3. C does not enclose J0 C encloses J0 Case 2 : R C = 0 / Align L2 with L3 Solution exists for every j0! j0= 0 and (l2 , l3 ) [Fig 8.22(d)]
Alternative to "AntiAligning" L1 with L2 (align L2 with L3) Boundary of annulus represents extreme 2link configurations for "single link" of length l1+l2 or l1l2
Source: O'Rourke Robot Arm Motion: Finding Configurations Recursive, linear algorithm for nlink reachability: annulus R represents n1 links of nlink arm with circle C of radius ln centered on p cases of Figure 8.22 apply Case 1 : R C 0 / Case 2 : R C [Fig 8.22(a),(b)] Choose one of (in general) 2 points of intersection [Fig 8.22(c),(d)] Choose any point on C (e.g. furthest from J0) Recursively find configuration for An1=(l1,...,ln1) Append last link Ln to this solution to connect to p
Given point p to reach, first determine if p is reachable (via Theorem 8.6.3); if so, find configuration recursively. Source: O'Rourke Robot Arm Motion: nLink Reachability
Two Kinks Theorem: If an nlink arm A can reach a point, it can re...
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 Spring '10
 DANIELS
 Algorithms

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