Fixed-point-NR-Secant

Fixed-point-NR-Secant - v= [ gm 1 − e −( c / m ) t c m=...

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Unformatted text preview: v= [ gm 1 − e −( c / m ) t c m= 9= t= v= 68.1 9.8 10 40 ⇒ 40 = What is the value of c when v=40 m/sec and t=10 sec? [ 9.81 * 68.1 1 − e− c or f (c ) = [ 9.81 * 68.1 1− e c or c = g (c ) = Fixed point c 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 g( c) 20 15.8 15.04 14.85 14.8 14.79 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 ea(%) 15.8 15.04 14.85 14.8 14.79 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 14.78 et(%) 4.78 1.28 0.35 0.1 0.03 0.01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6.9 1.79 0.49 0.14 0.04 0.01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [ 9.81 * 68.1 1 40 31 32 33 ∆c= 14.78 14.78 14.78 14.78 14.78 14.78 0 0 0 0 0 0 1 c g( c) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 0 2.28 4.25 5.94 7.41 8.68 9.77 10.72 11.53 12.23 12.84 13.37 13.82 14.21 14.55 14.84 15.09 15.31 15.5 15.66 15.8 15.92 16.02 16.11 16.19 c 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 20 15 10 5 0 0 5 10 15 [ 81 * 68.1 1 − e −( c / 68.1)10 c [ 9.81 * 68.1 1 − e −( c / 68.1)10 − 40 = 0 c )= [ 9.81 * 68.1 1 − e −( c / 68.1)10 40 SOLUTION 14.78 0 15 20 25 [ gm v= 1 − e −( c / m ) t c m= 9= t= v= 68.1 9.8 10 40 f (c ) = f ' (c ) = − 9. 9. What is the value of c when v=40 m/sec and t=10 sec? ci +1 = ci − Newton Raphson c 1 2 3 4 5 6 7 8 ∆c= f( c) 10 13.95 14.75 14.78 14.78 14.78 14.78 14.78 f ' ( c) 11.37 1.68 0.05 0 0 0 0 0 ea(%) et(%) 32.34 39.47 5.63 5.78 0.18 0.18 0 0 0 0 0 0 0 0 0 -2.88 -2.08 -1.95 -1.95 -1.95 -1.95 -1.95 -1.95 1 c 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 f( c) 51.14 44.92 39.26 34.11 29.42 25.14 21.23 17.65 14.38 11.37 8.61 6.07 3.73 1.57 -0.42 -2.27 -3.98 -5.56 -7.03 25 20 15 10 5 0 -5 -10 -15 -20 0 5 10 15 f 0 -5 20 21 22 23 24 25 -8.4 -9.68 -10.86 -11.97 -13.01 -13.98 -10 -15 -20 0 5 10 15 [ [ 9.8 * 68 .1 1 − e −( c / 68 .1)10 − 40 = 0 c 9.8 * 68 .1 f ' (c ) = − 1 − e −( c / 68 .1)10 2 c 9.8 * 10 −( c / 68 .1)10 + e c f (c i ) ci +1 = ci − f ' (c i ) f (c ) = SOLUTION 14.78 15 20 25 15 20 25 [ gm v= 1 − e −( c / m ) t c m= 9= t= v= 68.1 9.8 10 40 f (c ) = f ' (c ) = − 9. 9. What is the value of c when v=40 m/sec and t=10 sec? ci +1 = ci − f Secant Need two points to start c f( c) ea(%) et(%) 1 5 29.42 66.17 2 8 17.65 60 45.87 3 12.5 4.87 56.25 15.43 4 14.22 1.13 13.73 3.82 5 14.73 0.1 3.62 0.33 6 14.78 0 0.33 0.01 7 14.78 0 0.01 0 8 14.78 0 0 0 ∆c= SOLUTION 14.78 1 c 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 f( c) 51.14 44.92 39.26 34.11 29.42 25.14 21.23 17.65 14.38 11.37 8.61 6.07 3.73 1.57 -0.42 -2.27 -3.98 -5.56 -7.03 25 20 15 10 5 0 -5 -10 -15 -20 0 5 10 15 0 -5 20 21 22 23 24 25 -8.4 -9.68 -10.86 -11.97 -13.01 -13.98 -10 -15 -20 0 5 10 15 [ [ 9.8 * 68 .1 1 − e −( c / 68 .1)10 − 40 = 0 c 9.8 * 68 .1 f ' (c ) = − 1 − e −( c / 68 .1)10 2 c 9.8 * 10 −( c / 68 .1)10 + e c ci − ci −1 ci +1 = ci − f (ci ) f (ci ) − f (ci −1 ) f (c ) = SOLUTION 15 20 25 15 20 25 ...
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This note was uploaded on 08/31/2011 for the course ENGR 3723 taught by Professor Staff during the Summer '11 term at Oklahoma State.

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