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cleaned out the original anisotropy repo. Going to merge others prett…

…y soon
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1 parent b2e02d3 commit d60018ba6181f65617e26261de35bd0626ce8f90 @jfhbrook committed May 20, 2011
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-"""
-Trying to do something with the not-so-ellipsis
-"""
-
-from __future__ import division
-from sympy import *
-
-# conductivity matrix
-k_11 = Symbol('k__11')
-k_12 = Symbol('k__12')
-k_13 = Symbol('k__13') #
-k_22 = Symbol('k__22')
-k_23 = Symbol('k__23') #
-k_33 = Symbol('k__33')
-
-# Symmetrical 3x3 matrix K
-# K = Matrix([[k_11, k_12, k_13],
-# [k_12, k_22, k_23],
-# [k_13, k_23, k_33]])
-
-K = Matrix([[k_11, 0, 0],
- [0, k_11, 0],
- [0, 0, k_33]])
-
-
-
-# Let's parameterize.
-
-theta = Symbol('theta')
-phi = Symbol('phi')
-phi = 0
-
-#z is now in terms of theta and phi
-z_1 = sin(theta)*cos(phi)
-z_2 = sin(theta)*sin(phi)
-z_3 = cos(theta)
-
-# Some z vector
-#z_1 = Symbol('z_1')
-#z_2 = Symbol('z_2')
-#z_3 = Symbol('z_3')
-
-Z = Matrix(3,1,[z_1,z_2,z_3])
-
-# According to reasoning on paper--comes from dot and cross prod, respectively.
-cosz = z_3
-sinz = sqrt(z_1**2 + z_2**2)
-S = Matrix([[0, -z_3, z_2],
- [z_3, 0, -z_1],
- [z_2, z_1, 0 ]])
-
-#rotation matrix
-R = Z*Z.T + (eye(3) - Z*Z.T)*cosz + S*sinz
-
-#A-prime, the rotated matrix
-Kp = R*K*(R.T)
-
-# Now, supposing we have this rotated matrix,
-# let's assume what I did on paper is right
-
-req = cse(Kp[0,0]*Kp[1,1] - Kp[1,2]*Kp[2,1])
-
-#print("req as a function of theta and phi")
-printing.print_latex( req )
-
@@ -1,63 +0,0 @@
-"""
-Playing with the Heat Equation using sympy
-"""
-from __future__ import division
-from sympy import *
-
-# Mad props
-rho = Symbol('rho')
-cp = Symbol('Cp')
-
-# conductivity matrix
-k_11 = Symbol('k_11')
-k_12 = Symbol('k_12')
-k_13 = Symbol('k_13') #
-k_22 = Symbol('k_22')
-k_23 = Symbol('k_23') #
-k_33 = Symbol('k_33')
-
-# Symmetrical 3x3 matrix K
-K = Matrix([[k_11, k_12, k_13],
- [k_12, k_22, k_23],
- [k_13, k_23, k_33]])
-
-# Coordinates
-x = Symbol('x')
-y = Symbol('y')
-z = Symbol('z')
-
-x0 = Symbol('x0')
-y0 = Symbol('y0')
-z0 = Symbol('z0')
-
-
-#Scaling factors
-a = Symbol('a')
-b = Symbol('b')
-c = Symbol('c')
-
-A = Symbol('A')
-
-#thyme
-t = Symbol('t')
-
-#Proposed solution
-exppart = exp(-(1/(4*t))*( ((x-x0)/a)**2 + ((y-y0)/b)**2 + ((z-z0)/c)**2 ))
-T = A/(t**(3/2)) * exppart
-
-f = Function('f')
-
-delT = Matrix(3,1,[T.diff(x),T.diff(y),T.diff(z)])
-f = K * delT
-
-print('lhs:')
-lhs = T.diff(t)/cp/rho
-lhs = powsimp(collect(lhs,exppart)/exppart)
-pprint(lhs)
-
-print('rhs:')
-rhs = f[0].diff(x)+f[1].diff(y)+f[2].diff(z)
-rhs = powsimp(collect(rhs,exppart)/exppart)
-pprint(rhs)
-
-solve(lhs-rhs,[a,b,c,A])
@@ -1,38 +0,0 @@
-#!/usr/bin/env python
-
-from matplotlib import pyplot
-from numpy import array, linspace, ones
-
-#n=t_t/(t_1+t_2)
-def kxk2_planar(n,rk):
- return (rk-1)*n + 1
-
-def kzk2_planar(n,rk):
- return ((1./rk)*n + 1.)**(-1.)
-
-def plot_planar():
- pyplot.figure()
- rts = linspace(0,1,60)
- rks = [100., 50.,10.,4.,2.,1.]
- rks = rks+[1./x for x in reversed(rks)]
- #annotate_text=linspace(-0.05,1.05,len(rks))
- for i, rk in enumerate(rks):
- kxk2 = kxk2_planar(rts,rk)
- pyplot.plot(rts,kxk2,'k')
- pyplot.axis([0,1,0,2])
- pyplot.title('Planar Geometry')
- pyplot.xlabel(r'$n$')
- pyplot.ylabel(r'$ k_x $ / $ k_2 $')
-
- pyplot.figure()
- for i, rk in enumerate(rks):
- kzk2 = kzk2_planar(rts,rk)
- pyplot.plot(rts,kzk2,'r')
- pyplot.axis([0,1,0,2])
- pyplot.title('Planar Geometry')
- pyplot.xlabel(r'$n$')
- pyplot.ylabel(r'$ k_z $ / $ k_2 $')
- pyplot.show()
-
-if __name__ == "__main__":
- plot_planar()
@@ -1,19 +0,0 @@
-#!/usr/bin/env python
-
-from itertools import cycle
-from numpy import linspace
-
-angles=linspace(0,90,15)
-hosts=cycle(['hookbill',
- 'magpie',
- 'mallard',
- 'puffin',
- 'ruddy',
- 'wood'])
-
-domain='arsc.edu'
-user='holbrook'
-
-for angle, host in zip(angles,hosts):
- print 'ssh '+'holbrook@'+host+'.arsc.edu comsol matlab -ml -nospash -ml -nodesktop -ml -r "cd scratch/anisotropy/mesh; mesh_generate(\'../data/mesh/\','+str(angle)+'); exit"\n'
-
View
@@ -1,44 +0,0 @@
-% Trying out different ways of getting k
-% Make sure data's loaded.
-
- opts = optimset('lsqcurvefit');
- opts.MaxFunEvals = 10e5;
- opts.MaxIter = 10e5;
- opts.TolFun = 10e-11;
- opts.TolX = 10e-11;
- opts.Display = 'off';
-
- [kxy,kz]=meshgrid(ks);
- for i=1:length(ks),
- for j=1:length(ks),
- fprintf('k results for kxy=%1.2f and kz=%1.2f',kxy(i,j),kz(i,j));
- % Some fits to try extracting k out of the heating curve
- t=solutions{i,j}{1}(1,:);
- T=solutions{i,j}{1}(2,:);
-
- % 1) Curve fit to expint() solution
- % Impression: Tweakable to get decent results
- Q = 0.12; %W/m
-
- figure;
- plot(log(t(t>0)),T(t>0), 'ko');
- hold on;
-
- %imo this results in a horrible curve fit, though it might be tweakable.
- [a,r2] = lsqcurvefit(@ (x,a) x(1).*real(expint(x(2)./a)),[0.05 -0.0001],t(t>0),T(t>0),[],[],opts);
-
- fprintf('\nResult of expint fit: %1.6f\n',Q/4/pi/a(1));
- fprintf('R^2: %3.3f', r2);
- plot(log(t(t>0)), a(1).*real(expint(a(2)./t(t>0))),'r-*');
-
- % 2) Naive linear fit to latter "half" of heating curve
- % Seems relatively close
-
- % 10 secs chosen from experience, not procedurally.
- % Extra param coerces polyfit to do a better job
- [b,S] = polyfit(log(t(t>=10)),T(t>=10),1);
-
- fprintf('\nResult of linear fit: %1.6f\n', Q/4/pi/b(1));
- plot(log(t(t>=10)),b(1)*log(t(t>=10))+b(2), 'b-*');
- end
- end
View
@@ -1,4 +0,0 @@
-function flat=flatten(x)
- % skooshes a given array into 1xn
- flat=reshape(x',1,size(x,1)*size(x,2));
-end
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