mws_ind_inp_txt_ndd_examples

mws_ind_inp_txt_ndd_examples - Chapter 05.03 Newtons...

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05.03.1 Chapter 05.03 Newton’s Divided Difference 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 Newton’s divided difference method of interpolation and a first order polynomial. 1 2 3 5 6 7 x y 4
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05.03.2 Chapter 05.03 Solution For linear interpolation, the value of y is given by ) ( ) ( 0 1 0 x x b b x y Since we want to find the value of y at 10 . 1 x , using the two points 28 . 1 x and 66 . 0 x , then , 28 . 1 0 x 88 . 0 ) ( 0 x y , 66 . 0 1 x 14 . 1 ) ( 1 x y gives ) ( 0 0 x y b 88 . 0 0 1 0 1 1 ) ( ) ( x x x y x y b 28 . 1 66 . 0 88 . 0 14 . 1 41935 . 0 Hence ) ( ) ( 0 1 0 x x b b x y ), 28 . 1 ( 41935 . 0 88 . 0 x 28 . 1 66 . 0 x At 10 . 1 x ) 28 . 1 10 . 1 ( 41935 . 0 88 . 0 ) 00 . 4 ( y in. 95548 . 0 If we expand ), 28 . 1 ( 41935 . 0 88 . 0 ) ( x x y 28 . 1 66 . 0 x we get , 41935 . 0 4168 . 1 ) ( x x y 28 . 1 66 . 0 x This is the same expression that was obtained with the direct method. Example 2 The geometry of a cam is given in Figure 2. A curve needs to be fit through the seven points given in Table 2 to fabricate the cam.
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Newton’s Divided Difference Interpolation-More Examples: Industrial Engineering 05.03.3 Figure 2 Schematic of cam profile. Table 2 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 quadratic profile from 20 . 2 x to 28 . 1 x to 66 . 0 x , what is the value of y at 10 . 1 x using Newton’s divided difference method of interpolation and a second order polynomial. Find the absolute relative approximate error for the second order polynomial approximation. Solution For quadratic interpolation, the value of y is chosen as ) )( ( ) ( ) ( 1 0 2 0 1 0 x x x x b x x b b x y Since we want to find the value of y at , 10 . 1 x using the three points 20 . 2 0 x , 28 . 1 1 x and 66 . 0 2
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mws_ind_inp_txt_ndd_examples - Chapter 05.03 Newtons...

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