File:Mrs Miniver's Problem.svg
Summary
| Description |
English: Three instances of Mrs. Miniver's problem, of arranging two circles so that their area of overlap equals the area of their symmetric difference. In each case the inner yellow area equals the total area of the surrounding blue regions. The left case shows two circles of equal areas, the right case shows one circle with twice the area of the other, and the middle case is intermediate between these two. |
| Date | |
| Source | Own work |
| Author | David Eppstein |
| SVG development |
Licensing
David Eppstein, the copyright holder of this work, hereby publishes it under the following license:
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This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication. |
| The person who associated a work with this deed has dedicated the work to the public domain by waiving all of their rights to the work worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law. You can copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission.
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Source code
This media was created with Python (general-purpose programming language)Category:Images with Python source code
Here is a listing of the source used to create this file.
Here is a listing of the source used to create this file.
from math import pi,asin,sin,cos,acos
from PADS.SVG import *
import sys
# Circular segment formulas from https://en.wikipedia.org/wiki/Circular_segment
def areaFromRadius(r): return pi*r**2
def twoCircleArea(r1,r2): return areaFromRadius(r1)+areaFromRadius(r2)
def angleFromChord(r,c):
return 2*asin(c/(2*r))
def areaFromAngle(r,theta):
return r**2*(theta-sin(theta))/2
def areaFromChord(r,c):
theta = angleFromChord(r,c)
return areaFromAngle(r,theta)
def distanceToChord(r,c):
theta = angleFromChord(r,c)
return r*cos(theta/2)
def separationFromChord(r1,r2,c):
return distanceToChord(r1,c)+distanceToChord(r2,c)
# Bisection to solve Mrs. Miniver's problem within floating point accuracy
def miniverSeparation(r1,r2):
if r1 > r2: r1,r2 = r2,r1 # make sure small circle first
totalArea = twoCircleArea(r1,r2)
targetArea = totalArea/3
def miniverTest(x):
# x = cos(chord on small circle), -1: outside, 1: inside
chord = 2*r1*sin(acos(x))
area1 = areaFromChord(r1,chord)
if x > 0:
area1 = areaFromRadius(r1) - area1
area2 = areaFromChord(r2,chord)
return area1 + area2 < targetArea
lo,hi = -1.0,1.0
for i in range(100):
mid = (hi+lo)/2
if miniverTest(mid):
lo = mid
else:
hi = mid
chord = 2*r1*sin(acos(mid))
return distanceToChord(r2,chord) - r1*mid
# Make it into a nice picture
# Monochromatic and with a loose frame for now; we'll fix it up later
gridUnit = 150
targetArea = 100000.0
boundingBox = (6+4j)*gridUnit
output = SVG(boundingBox,sys.stdout)
output.group(fill="none",stroke="#000")
for i in (0,1,2):
ratio = 2.0**(i*0.25)
areaWhenSmall = twoCircleArea(1.0,ratio)
factor = (targetArea/areaWhenSmall)**0.5
r1,r2 = factor,factor*ratio
s = miniverSeparation(r1,r2)
o = (1+2j+2*i)*gridUnit
output.circle(o + (s+r2-r1)*0.5j, r1)
output.circle(o - (s+r1-r2)*0.5j, r2)
output.ungroup()
output.close()