1 | #!/usr/bin/env python |
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2 | # |
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3 | # Author: Mike McKerns (mmckerns @caltech and @uqfoundation) |
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4 | # Copyright (c) 2009-2016 California Institute of Technology. |
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5 | # License: 3-clause BSD. The full license text is available at: |
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6 | # - http://mmckerns.github.io/project/mystic/browser/mystic/LICENSE |
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7 | |
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8 | """Original matlab code: |
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9 | |
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10 | function A=marc_surr(x) |
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11 | h=x(1)*25.4*10^(-3); |
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12 | a=x(2)*pi/180; |
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13 | v=x(3); |
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14 | Ho=0.5794; |
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15 | s=1.4004; |
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16 | n=0.4482; |
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17 | K=10.3963; |
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18 | p=0.4757; |
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19 | u=1.0275; |
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20 | m=0.4682; |
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21 | Dp=1.778; |
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22 | |
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23 | v_bl=Ho*(h/(cos(a))^(n))^s; |
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24 | |
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25 | if v<v_bl |
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26 | A=0 |
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27 | else |
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28 | A=K*((h/Dp)^p)*((cos(a))^u)*(tanh((v/v_bl)-1))^m; |
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29 | end |
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30 | """ |
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31 | |
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32 | ### NOTES ### |
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33 | # h = thickness = [60,105] |
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34 | # a = obliquity = [0,30] |
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35 | # v = speed = [2.1,2.8] |
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36 | |
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37 | # explore the cuboid (h,a,v), with |
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38 | # subdivisions at h=100, a=20, v=2.2 |
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39 | # due to ballistic limit: v(h=100,a=20) = 2.22, |
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40 | # perforation in this region should be zero. |
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41 | # NOTE: 'failure' is A < = t |
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42 | # |
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43 | # Calculate for each of the 8 subcuboids: |
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44 | # * probability mass, i.e. the product of the normalized side-lengths, |
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45 | # since we're taking h, a and v to be uniformly distributed in their |
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46 | # intervals |
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47 | # * McDiarmid diameter of the perforation area A when restricted to that |
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48 | # cuboid or subcuboid |
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49 | # * the mean value of the perforation area A on each (sub)cuboid |
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50 | |
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51 | from math import pi, cos, tanh |
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52 | |
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53 | def ballistic_limit(h,a): |
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54 | """calculate ballistic limit |
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55 | |
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56 | Inputs: |
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57 | - h = thickness in (unknown) units |
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58 | - a = obliquity in (unknown) units |
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59 | |
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60 | Outputs: |
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61 | - v_bl = velocity (ballistic limit) in (unknown) units |
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62 | """ |
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63 | #h = x[0] * 25.4 * 1e-3 |
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64 | #a = x[1] * pi/180.0 |
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65 | Ho = 0.5794 |
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66 | s = 1.4004 |
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67 | n = 0.4482 |
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68 | return Ho * ( h / cos(a)**n )**s |
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69 | |
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70 | |
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71 | def marc_surr(x): |
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72 | """calculate perforation area using a tanh-based model surrogate |
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73 | |
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74 | Inputs: |
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75 | - x = [thickness, obliquity, speed] in (unknown) units |
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76 | |
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77 | Outputs: |
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78 | - A = performation area in (unknown) units |
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79 | """ |
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80 | # h = thickness = [60,105] |
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81 | # a = obliquity = [0,30] |
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82 | # v = speed = [2.1,2.8] |
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83 | h = x[0] * 25.4 * 1e-3 |
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84 | a = x[1] * pi/180.0 |
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85 | v = x[2] |
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86 | |
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87 | K = 10.3963 |
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88 | p = 0.4757 |
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89 | u = 1.0275 |
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90 | m = 0.4682 |
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91 | Dp = 1.778 |
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92 | |
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93 | # compare to ballistic limit |
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94 | v_bl = ballistic_limit(h,a) |
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95 | if v < v_bl: |
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96 | return 0 |
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97 | |
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98 | return K * (h/Dp)**p * (cos(a))**u * (tanh((v/v_bl)-1))**m |
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99 | |
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100 | # EOF |
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