44f09d6777
This includes `arcs`, `circle_inversion`, `circles`, `circles_apollonius`, `circles_incomplete`, `cuboids`, `edges`, `faces`, `fillets`, `general`, `geometry`, `intersections`, `linear_geometry`, `offsets`, `sort_edges`, `wires`. These are added to the `draftgeoutils` Doxygen group so that the functions contained in each module are listed appropriately in the automatically generated documentation.
226 lines
8.4 KiB
Python
226 lines
8.4 KiB
Python
# ***************************************************************************
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# * Copyright (c) 2009, 2010 Yorik van Havre <yorik@uncreated.net> *
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# * Copyright (c) 2009, 2010 Ken Cline <cline@frii.com> *
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# * *
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# * This file is part of the FreeCAD CAx development system. *
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# * *
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# * This program is free software; you can redistribute it and/or modify *
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# * it under the terms of the GNU Lesser General Public License (LGPL) *
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# * as published by the Free Software Foundation; either version 2 of *
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# * the License, or (at your option) any later version. *
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# * for detail see the LICENCE text file. *
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# * *
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# * FreeCAD is distributed in the hope that it will be useful, *
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# * but WITHOUT ANY WARRANTY; without even the implied warranty of *
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# * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the *
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# * GNU Library General Public License for more details. *
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# * *
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# * You should have received a copy of the GNU Library General Public *
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# * License along with FreeCAD; if not, write to the Free Software *
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# * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 *
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# * USA *
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# * *
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# ***************************************************************************
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"""Provides various functions for Apollonius and Soddy circle operations.
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The 'problem of Appollonius' consists of finding a circle that is
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simultaneously tangent to three circles on a plane.
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- http://en.wikipedia.org/wiki/Problem_of_Apollonius#Inversive_methods
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- http://mathworld.wolfram.com/ApolloniusCircle.html
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- http://mathworld.wolfram.com/ApolloniusProblem.html
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If each circle only has one intersection with each other, the problem is that
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of finding a 'Soddy circle', which has two solutions, and inner circle
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and an outer circle.
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- http://en.wikipedia.org/wiki/Problem_of_Apollonius#Mutually_tangent_given_circles:_Soddy.27s_circles_and_Descartes.27_theorem
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- http://mathworld.wolfram.com/SoddyCircles.html
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- http://mathworld.wolfram.com/InnerSoddyCenter.html
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- http://mathworld.wolfram.com/OuterSoddyCenter.html
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"""
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## @package circles_apollonius
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# \ingroup draftgeoutils
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# \brief Provides various functions for Apollonius and Soddy circles.
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import cmath
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import math
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import lazy_loader.lazy_loader as lz
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import FreeCAD as App
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from draftgeoutils.general import geomType, NORM
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from draftgeoutils.intersections import findIntersection
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# Delay import of module until first use because it is heavy
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Part = lz.LazyLoader("Part", globals(), "Part")
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## \addtogroup draftgeoutils
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# @{
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def outerSoddyCircle(circle1, circle2, circle3):
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"""Compute the outer soddy circle for three tightly packed circles.
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Original Java code Copyright (rc) 2008 Werner Randelshofer
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Converted to python by Martin Buerbaum 2009
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http://www.randelshofer.ch/treeviz/
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Either Creative Commons Attribution 3.0, the MIT license,
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or the GNU Lesser General License LGPL.
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"""
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if (geomType(circle1) != "Circle"
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or geomType(circle2) != "Circle"
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or geomType(circle3) != "Circle"):
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print("debug: outerSoddyCircle bad parameters! Must be circles.")
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return None
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A = circle1.Curve.Center
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B = circle2.Curve.Center
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C = circle3.Curve.Center
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ra = circle1.Curve.Radius
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rb = circle2.Curve.Radius
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rc = circle3.Curve.Radius
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# Solution using Descartes' theorem, as described here:
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# http://en.wikipedia.org/wiki/Descartes%27_theorem
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k1 = 1 / ra
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k2 = 1 / rb
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k3 = 1 / rc
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k4 = abs(k1 + k2 + k3 - 2 * math.sqrt(k1 * k2 + k2 * k3 + k3 * k1))
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q1 = (k1 + 0j) * (A.x + A.y * 1j)
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q2 = (k2 + 0j) * (B.x + B.y * 1j)
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q3 = (k3 + 0j) * (C.x + C.y * 1j)
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temp = ((q1 * q2) + (q2 * q3) + (q3 * q1))
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q4 = q1 + q2 + q3 - ((2 + 0j) * cmath.sqrt(temp))
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z = q4 / (k4 + 0j)
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# If the formula is not solvable, we return no circle.
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if not z or not (1 / k4):
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return None
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X = -z.real
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Y = -z.imag
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print("Outer Soddy circle: " + str(X) + " " + str(Y) + "\n") # Debug
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# The Radius of the outer soddy circle can also be calculated
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# with the following formula:
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# radiusOuter = abs(r1*r2*r3 / (r1*r2 + r1*r3 + r2*r3 - 2 * math.sqrt(r1*r2*r3 * (r1+r2+r3))))
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circ = Part.Circle(App.Vector(X, Y, A.z), NORM, 1 / k4)
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return circ
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def innerSoddyCircle(circle1, circle2, circle3):
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"""Compute the inner soddy circle for three tightly packed circles.
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Original Java code Copyright (rc) 2008 Werner Randelshofer
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Converted to python by Martin Buerbaum 2009
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http://www.randelshofer.ch/treeviz/
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"""
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if (geomType(circle1) != "Circle"
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and geomType(circle2) != "Circle"
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and geomType(circle3) != "Circle"):
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print("debug: innerSoddyCircle bad parameters! Must be circles.")
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return None
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A = circle1.Curve.Center
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B = circle2.Curve.Center
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C = circle3.Curve.Center
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ra = circle1.Curve.Radius
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rb = circle2.Curve.Radius
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rc = circle3.Curve.Radius
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# Solution using Descartes' theorem, as described here:
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# http://en.wikipedia.org/wiki/Descartes%27_theorem
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k1 = 1 / ra
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k2 = 1 / rb
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k3 = 1 / rc
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k4 = abs(k1 + k2 + k3 + 2 * math.sqrt(k1 * k2 + k2 * k3 + k3 * k1))
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q1 = (k1 + 0j) * (A.x + A.y * 1j)
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q2 = (k2 + 0j) * (B.x + B.y * 1j)
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q3 = (k3 + 0j) * (C.x + C.y * 1j)
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temp = ((q1 * q2) + (q2 * q3) + (q3 * q1))
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q4 = q1 + q2 + q3 + ((2 + 0j) * cmath.sqrt(temp))
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z = q4 / (k4 + 0j)
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# If the formula is not solvable, we return no circle.
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if (not z or not (1 / k4)):
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return None
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X = z.real
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Y = z.imag
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print("Outer Soddy circle: " + str(X) + " " + str(Y) + "\n") # Debug
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# The Radius of the inner soddy circle can also be calculated
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# with the following formula:
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# radiusInner = abs(r1*r2*r3 / (r1*r2 + r1*r3 + r2*r3 + 2 * math.sqrt(r1*r2*r3 * (r1+r2+r3))))
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circ = Part.Circle(App.Vector(X, Y, A.z), NORM, 1 / k4)
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return circ
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def circleFrom3CircleTangents(circle1, circle2, circle3):
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"""Return the circle that is tangent to three other circles.
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This problem is called the 'Problem of Appollonius'.
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A special case is that of 'Soddy circles'.
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To Do
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-----
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Currently not all possible solutions are found, only the Soddy circles.
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* Calc all 6 homothetic centers.
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* Create 3 lines from the inner and 4 from the outer h. center.
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* Calc. the 4 inversion poles of these lines for each circle.
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* Calc. the radical center of the 3 circles.
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* Calc. the intersection points (max. 8) of 4 lines (through each
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inversion pole and the radical center) with the circle.
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* This gives us all the tangent points.
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"""
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if (geomType(circle1) != "Circle"
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and geomType(circle2) != "Circle"
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and geomType(circle3) == "Circle"):
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print("debug: circleFrom3CircleTangents bad input! Must be circles")
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return None
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int12 = findIntersection(circle1, circle2, True, True)
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int23 = findIntersection(circle2, circle3, True, True)
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int31 = findIntersection(circle3, circle1, True, True)
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if int12 and int23 and int31:
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if len(int12) == 1 and len(int23) == 1 and len(int31) == 1:
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# If only one intersection with each circle, Soddy circle
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# r1 = circle1.Curve.Radius
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# r2 = circle2.Curve.Radius
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# r3 = circle3.Curve.Radius
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outerSoddy = outerSoddyCircle(circle1, circle2, circle3)
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# print(str(outerSoddy)) # Debug
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innerSoddy = innerSoddyCircle(circle1, circle2, circle3)
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# print(str(innerSoddy)) # Debug
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circles = []
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if outerSoddy:
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circles.append(outerSoddy)
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if innerSoddy:
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circles.append(innerSoddy)
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return circles
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# Here the rest of the circles should be calculated
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# ...
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else:
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# Some circles are inside each other or an error has occurred.
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return None
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## @}
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