MIT2_017JF09_p33

MIT2_017JF09_p33 - 33 POSITIONING USING RANGING: 2D CASE...

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Unformatted text preview: 33 POSITIONING USING RANGING: 2D CASE 119 33 Positioning Using Ranging: 2D Case 1. Two sensors, denoted 1 and 2 and located at different locations in the x y plane, make a range measurements to a target t . Let the range from Sensor 1 to the target be r 1 and let the range from Sensor 2 to the target be r 2 ; we call the sensor locations [ x 1 , y 1 ] and [ x 2 , y 2 ], and the target location [ x t , y t ]. sensor 1 sensor 2 target y x What are the two equations describing the targets [ x t , y t ] position in the horizontal plane, based on the range measurements from the two devices? 2. Sketch these constraints for the fixed sensor locations x 1 = 0, y 1 = 0, x 2 = 1, and y 2 = 0, for two different target locations a) x t = 2 , y t = 0, and b) x t = 0 . 5 , y t = 0 . 5. This is a total of four circles to draw. 3. Consider your two constraints for a given target location; there are two unknowns [ x t , y t ], so we ought to be able to solve for them. First, solve for x t by subtracting the two constraint equations, such that the x 2 t and y t 2 terms go away; be sure you take advantage of the fact that x 1 = y 1 = y 2 = 0! This will let you derive a clean expression for x t . Then, put this value for x t into the constraint equation for Sensor 1, and solve for y t . There will be two solutions for y t , because this array cant distinguish which side the target is on....
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This note was uploaded on 11/29/2011 for the course CIVIL 1.00 taught by Professor Georgekocur during the Spring '05 term at MIT.

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MIT2_017JF09_p33 - 33 POSITIONING USING RANGING: 2D CASE...

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