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enter image description hereHow the concentric super ellipses as shown in the figure can be divided into parts containing equal area such that the total area of the superellipse A = A1 + A2 + ... An where n = 60 in the shown figure given its semi-major axis a and semi-minor axis b. The origin of the sectors lies at the center of the superellipse. Any help is appreciated.

pretty
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  • ah sorry you are looking for superellipse! then of course the trick used earlier doesn't work here – user May 27 '18 at 22:08
  • @gimusi The explanation for the ellipse is precise and clear but I am searching an apt solution for super ellipse. – pretty May 27 '18 at 22:18

2 Answers2

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Solve the problem for concentric circles in the unit circle centered at the origin. Then stretch the $x$-axis by $a$ and the $y$-axis by $b$ to turn the circles into ellipses. All the areas will be scaled by the same stretch factor.

This is essentially the same question you asked here: Dividing an ellipse into equal area

Ethan Bolker
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  • @ Ethan: sorry for not mentioning as superellipse and have edited the question appropriately. – pretty May 26 '18 at 17:13
  • @pretty Thanks for accepting the answer, but is it an answer? Uniform stretching won't make a superellipse, variable stretching might, but wouldn't do the areas uniformly. – Ethan Bolker May 26 '18 at 22:47
  • @EthanThe answer holds for ellipse and it helped me but not for superellipse. I am still waiting if someone could answer. – pretty May 27 '18 at 06:58
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Following the same idea presented here Dividing an ellipse into equal area, it suffices consider the affinity by $X=ax$ and $Y=by$ such that

$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$

and divide the unit circle in 60 equals parts by the lines

  • $y=0$
  • $y=\pm\frac{\sqrt 3}{3}x$
  • $y=\pm{\sqrt 3}x$
  • $x=0$

and by the circles

  • $x^2+y^2=\frac{\sqrt 5}5$
  • $x^2+y^2=\frac{\sqrt {10}}5$
  • $x^2+y^2=\frac{\sqrt {15}}5$
  • $x^2+y^2=\frac{2\sqrt {5}}5$

enter image description here

thus the equations for the lines and ellipses dividing the main ellipse into $60$ sectors of equal area $A=\frac{\pi ab}{60}$ are

  • $Y=0$
  • $Y=\pm\frac{\sqrt 3}{3}\frac b a X$
  • $y=\pm{\sqrt 3}\frac b a X$
  • $X=0$

for the lines and

  • $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt 5}5$
  • $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {10}}5$
  • $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {15}}5$
  • $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{2\sqrt {5}}5$
user
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