mws_com_inp_txt_direct_examples

# mws_com_inp_txt_direct_examples - Chapter 05.02 Direct...

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05.02.1 Chapter 05.02 Direct Method of Interpolation – More Examples Computer Engineering Example 1 A robot arm with a rapid laser scanner is doing a quick quality check on holes drilled in a " 10 " 15 rectangular plate. The centers of the holes in the plate describe the path the arm needs to take, and the hole centers are located on a Cartesian coordinate system (with the origin at the bottom left corner of the plate) given by the specifications in Table 1. Table 1 The coordinates of the holes on the plate. x (in.) y (in.) 2.00 7.2 4.25 7.1 5.25 6.0 7.81 5.0 9.20 3.5 10.60 5.0 Figure 1 Location of holes on the rectangular plate.

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05.02.2 Chapter 05.02 If the laser is traversing from 00 . 2 x to 25 . 4 x in a linear path, what is the value of y at 00 . 4 x using the direct method of interpolation and a first order polynomial? Solution For first order polynomial interpolation (also called linear interpolation), we choose the value of y given by  x a a x y 1 0 Figure 2 Linear interpolation. Since we want to find the value of y at 00 . 4 x , using the two points 00 . 2 0 x and 25 . 4 1 x , then  2 . 7 , 00 . 2 0 0 x y x  1 . 7 , 25 . 4 1 1 x y x gives   2 . 7 00 . 2 00 . 2 1 0 a a y   1 . 7 25 . 4 25 . 4 1 0 a a y Writing the equations in matrix form, we have 1 . 7 2 . 7 25 . 4 1 00 . 2 1 1 0 a a Solving the above two equations gives 2889 . 7 0 a 044444 . 0 1 a Hence  x a a x y 1 0  25 . 4 00 . 2 , 044444 . 0 2889 . 7 x x x y  0 0 , y x   1 1 , y x   x f 1 x y
Direct Method of Interpolation – More Examples: Computer Engineering 05.02.3    00

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mws_com_inp_txt_direct_examples - Chapter 05.02 Direct...

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