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Unformatted text preview: x t1 find the new pose x t Sampling from the Velocity Motion Model 2/13/2012 4 c c y x = − θ y x x t 1 ? = ′ ′ ′ = y x x t t ∆ ω v v r = cos sin v y y v x x c c + = − = Eqs 5.7, 5.8 Sampling from the Velocity Motion Model 2/13/2012 5 ∆ ∆ + − ∆ + + − + = ∆ + ∆ + − ∆ + + = ′ ′ ′ t t t y x t t y t x y x v v v v v c v c ω θ ) cos( cos ) sin( sin ) cos( ) sin( Eqs 5.9 *we already derived this for the differential drive!...
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This note was uploaded on 02/13/2012 for the course CSE 4421 taught by Professor Burton during the Winter '11 term at York University.
 Winter '11
 BURTON

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