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[2862] | 1 | """ |
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| 2 | Determining maximum of SMF slump surface elevation function. |
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| 3 | Jane Sexton, GA 2006 |
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| 4 | """ |
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| 5 | |
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| 6 | from math import exp, cosh |
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[2869] | 7 | import Numeric |
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[2862] | 8 | |
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[2878] | 9 | |
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[2862] | 10 | # Grilli and Watts 2005 example |
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| 11 | dx = 2.0 |
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| 12 | x0 = 10.0 |
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| 13 | y0 = 0.0 |
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| 14 | w = 2.0 |
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| 15 | kappa = 3 |
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| 16 | g = 9.8 |
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| 17 | d = 0.259 |
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| 18 | T = 0.052 |
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| 19 | theta = 15.0 |
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| 20 | lam0 = 5.0 |
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| 21 | |
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| 22 | def func(x,y,kappad): |
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[2869] | 23 | numerator = exp(-((x-x0)/lam0)**2.0) - kappad*exp(-((x-dx-x0)/lam0)**2.0) |
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| 24 | denominator = cosh(kappa*(y-y0)/(w+lam0))**2.0 |
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[2862] | 25 | return -numerator/denominator |
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| 26 | |
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| 27 | def dfuncdx(x,kappad): |
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[2869] | 28 | dfdx = (x-x0)*exp(-((x-x0)/lam0)**2.0)-kappad*(x-dx-x0)*exp(-((x-dx-x0)/lam0)**2.0) |
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[2862] | 29 | return dfdx |
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| 30 | |
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| 31 | step = 0.001 |
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| 32 | x = x0 |
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| 33 | deriv = 10 |
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| 34 | tol = 0.001 |
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| 35 | c = 0 |
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[2878] | 36 | direction = 'negative' |
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[2862] | 37 | while c < 100000 and deriv > 0: |
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[2878] | 38 | deriv = dfuncdx(x,0.83) |
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[2862] | 39 | if deriv < 0: xstar = x |
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| 40 | if direction == 'positive': x += step |
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| 41 | if direction == 'negative': x -= step |
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| 42 | c += 1 |
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| 43 | |
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[2869] | 44 | print 'location of maximum of surface elevation function: xstar = %f' % (xstar/lam0) |
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[2862] | 45 | const = 1.0 #a3D = ? #sydney = 86? and kappad=1 |
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| 46 | |
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[2869] | 47 | x = arange(-20,25,0.001) |
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| 48 | y = arange(-10,10,0.1) |
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| 49 | X,Y = meshgrid(x,y) |
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[2862] | 50 | |
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[2869] | 51 | test = 0.0 |
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| 52 | for xi in x: |
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[2878] | 53 | testi = func(xi,0.0,0.83) |
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[2869] | 54 | if direction == 'positive': |
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| 55 | if testi > test: |
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| 56 | test = testi |
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| 57 | xstar = xi |
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| 58 | else: |
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| 59 | if testi < test: |
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| 60 | test = testi |
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| 61 | xstar = xi |
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| 62 | |
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| 63 | print 'check: xstar = %f and eta(xstar) = %f' %(xstar/lam0, test) |
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