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Sensing Intelligence Motion - How Robots & Humans Move - Vladimir J. Lumelsky Part 5

Tham khảo tài liệu 'sensing intelligence motion - how robots & humans move - vladimir j. lumelsky part 5', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 96 MOTION PLANNING FOR A MOBILE ROBOT T S Figure 3.9 Illustration for Theorem 3.3.4. defined hit point. Now move from Q along the already generated path segment in the direction opposite to the accepted local direction until the closest hit point on the path is encountered say that point is Hj. We are interested only in those cases where Q is involved in at least one local cycle that is when MA passes point Q more than once. For this event to occur MA has to pass point Hj at least as many times. In other words if MA does not pass Hj more than once it cannot pass Q more than once. According to the Bug2 procedure the first time MA reaches point Hj it approaches it along the M-line straight line Start Target or more precisely along the straight line segment Lj-1 T . MA then turns left and starts walking around the obstacle. To form a local cycle on this path segment MA has to return to point Hj again. Since a point can become a hit point only once see the proof for Lemma 3.3.4 the next time MA returns to point Hj it must approach it from the right see Figure 3.9 along the obstacle boundary. Therefore after having defined HJ in order to reach it again this time from the right MA must somehow cross the M-line and enter its right semiplane. This can take place in one of only two ways outside or inside the interval S T . Consider both cases. 1. The crossing occurs outside the interval S T . This case can correspond only to an in-position configuration see Definition 3.3.2 . Theorem 3.3.4 therefore does not apply. 2. The crossing occurs inside the interval S T . We want to prove now that such a crossing of the path with the interval S T cannot produce local cycles. Notice that the crossing cannot occur anywhere within the interval S Hj because otherwise at least a part of the straight-line segment Lj-1 Hj would be included inside the obstacle. This is impossible BASIC ALGORITHMS 97 because MA is known to have walked along the whole segment Lj-1 Hj . If the crossing occurs