mws_ind_inp_txt_direct_examples

mws_ind_inp_txt_direct_examples - Chapter 05.02 Direct...

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05.02.1 Chapter 05.02 Direct Method of Interpolation – More Examples Industrial Engineering Example 1 The geometry of a cam is given in Figure 1. A curve needs to be fit through the seven points given in Table 1 to fabricate the cam. Figure 1 Schematic of cam profile. Table 1 Geometry of the cam. Point x   in. y   in. 1 2.20 0.00 2 1.28 0.88 3 0.66 1.14 4 0.00 1.20 5 –0.60 1.04 6 –1.04 0.60 7 –1.20 0.00 If the cam follows a straight line profile from 28 . 1 x to 66 . 0 x , what is the value of y at 10 . 1 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 1 2 3 5 6 7 x y 4
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05.02.2 Chapter 05.02  x a a x y 1 0 Figure 2 Linear interpolation. Since we want to find the value of y at 10 . 1 x , and we are using a first order polynomial, using the two points 28 . 1 0 x and 66 . 0 1 x , then  88 . 0 , 28 . 1 0 0 x y x  14 . 1 , 66 . 0 1 1 x y x gives   88 . 0 28 . 1 28 . 1 1 0 a a y   14 . 1 66 . 0 66 . 0 1 0 a a y Writing the equations in matrix form, we have 14 . 1 88 . 0 66 . 0 1 28 . 1 1 1 0 a a Solving the above two equations gives, 4168 . 1 0 a 41935 . 0 1 a Hence  x a a x y 1 0 28 . 1 66 . 0 , 41935 . 0 4168 . 1 x x   10 . 1 41935 . 0 4168 . 1 10 . 1 y in. 95548 . 0  0 0 , y x   1 1 , y x   x f 1 x y
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Direct Method of Interpolation – More Examples: Industrial Engineering 05.02.3 Example 2 The geometry of a cam is given in Figure 3. A curve needs to be fit through the seven points given in Table 2 to fabricate the cam. Figure 3
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This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida - Tampa.

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mws_ind_inp_txt_direct_examples - Chapter 05.02 Direct...

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