123 lines
5.2 KiB
Python
123 lines
5.2 KiB
Python
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# Code below available according to below conditions on https://github.com/MarDiehl/3Drotations
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####################################################################################################
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# Copyright (c) 2017-2019, Martin Diehl, Max-Planck-Institut für Eisenforschung GmbH
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# Copyright (c) 2013-2014, Marc De Graef/Carnegie Mellon University
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# All rights reserved.
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#
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# Redistribution and use in source and binary forms, with or without modification, are
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# permitted provided that the following conditions are met:
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#
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# - Redistributions of source code must retain the above copyright notice, this list
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# of conditions and the following disclaimer.
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# - Redistributions in binary form must reproduce the above copyright notice, this
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# list of conditions and the following disclaimer in the documentation and/or
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# other materials provided with the distribution.
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# - Neither the names of Marc De Graef, Carnegie Mellon University nor the names
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# of its contributors may be used to endorse or promote products derived from
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# this software without specific prior written permission.
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#
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# THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
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# AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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# IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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# ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE
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# LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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# DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR
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# SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER
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# CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY,
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# OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE
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# USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
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####################################################################################################
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import numpy as np
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sc = np.pi**(1./6.)/6.**(1./6.)
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beta = np.pi**(5./6.)/6.**(1./6.)/2.
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R1 = (3.*np.pi/4.)**(1./3.)
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def CubeToBall(cube):
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if np.abs(np.max(cube))>np.pi**(2./3.) * 0.5:
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raise ValueError
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# transform to the sphere grid via the curved square, and intercept the zero point
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if np.allclose(cube,0.0,rtol=0.0,atol=1.0e-300):
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ball = np.zeros(3)
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else:
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# get pyramide and scale by grid parameter ratio
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p = GetPyramidOrder(cube)
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XYZ = cube[p] * sc
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# intercept all the points along the z-axis
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if np.allclose(XYZ[0:2],0.0,rtol=0.0,atol=1.0e-300):
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ball = np.array([0.0, 0.0, np.sqrt(6.0/np.pi) * XYZ[2]])
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else:
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order = [1,0] if np.abs(XYZ[1]) <= np.abs(XYZ[0]) else [0,1]
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q = np.pi/12.0 * XYZ[order[0]]/XYZ[order[1]]
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c = np.cos(q)
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s = np.sin(q)
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q = R1*2.0**0.25/beta * XYZ[order[1]] / np.sqrt(np.sqrt(2.0)-c)
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T = np.array([ (np.sqrt(2.0)*c - 1.0), np.sqrt(2.0) * s]) * q
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# transform to sphere grid (inverse Lambert)
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# note that there is no need to worry about dividing by zero, since XYZ[2] can not become zero
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c = np.sum(T**2)
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s = c * np.pi/24.0 /XYZ[2]**2
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c = c * np.sqrt(np.pi/24.0)/XYZ[2]
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q = np.sqrt( 1.0 - s )
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ball = np.array([ T[order[1]] * q, T[order[0]] * q, np.sqrt(6.0/np.pi) * XYZ[2] - c ])
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# reverse the coordinates back to the regular order according to the original pyramid number
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ball = ball[p]
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return ball
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def BallToCube(ball):
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rs = np.linalg.norm(ball)
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if rs > R1:
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raise ValueError
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if np.allclose(ball,0.0,rtol=0.0,atol=1.0e-300):
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cube = np.zeros(3)
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else:
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p = GetPyramidOrder(ball)
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xyz3 = ball[p]
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# inverse M_3
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xyz2 = xyz3[0:2] * np.sqrt( 2.0*rs/(rs+np.abs(xyz3[2])) )
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# inverse M_2
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qxy = np.sum(xyz2**2)
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if np.isclose(qxy,0.0,rtol=0.0,atol=1.0e-300):
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Tinv = np.zeros(2)
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else:
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q2 = qxy + np.max(np.abs(xyz2))**2
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sq2 = np.sqrt(q2)
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q = (beta/np.sqrt(2.0)/R1) * np.sqrt(q2*qxy/(q2-np.max(np.abs(xyz2))*sq2))
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tt = np.clip((np.min(np.abs(xyz2))**2+np.max(np.abs(xyz2))*sq2)/np.sqrt(2.0)/qxy,-1.0,1.0)
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Tinv = np.array([1.0,np.arccos(tt)/np.pi*12.0]) if np.abs(xyz2[1]) <= np.abs(xyz2[0]) else \
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np.array([np.arccos(tt)/np.pi*12.0,1.0])
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Tinv = q * np.where(xyz2<0.0,-Tinv,Tinv)
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# inverse M_1
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cube = np.array([ Tinv[0], Tinv[1], (-1.0 if xyz3[2] < 0.0 else 1.0) * rs / np.sqrt(6.0/np.pi) ]) /sc
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# reverst the coordinates back to the regular order according to the original pyramid number
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cube = cube[p]
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return cube
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def GetPyramidOrder(xyz):
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if (abs(xyz[0])<= xyz[2]) and (abs(xyz[1])<= xyz[2]) or \
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(abs(xyz[0])<=-xyz[2]) and (abs(xyz[1])<=-xyz[2]):
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return [0,1,2]
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elif (abs(xyz[2])<= xyz[0]) and (abs(xyz[1])<= xyz[0]) or \
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(abs(xyz[2])<=-xyz[0]) and (abs(xyz[1])<=-xyz[0]):
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return [1,2,0]
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elif (abs(xyz[0])<= xyz[1]) and (abs(xyz[2])<= xyz[1]) or \
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(abs(xyz[0])<=-xyz[1]) and (abs(xyz[2])<=-xyz[1]):
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return [2,0,1]
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