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Epicycloids are most commonly described by the parametric equations,

$x(t) = (R + a)\cos(t) – a \cos \left(\frac{R + a}{a} t \right),$

$y(t) = (R + a)\sin(t) – a \sin \left(\frac{R + a}{a} t \right).$

Where $R$ is the radius of the fixed circle and $a$ is the radius of the rolling circle.

With $R = ka$ we also have,

$x(t) = a[(k + 1)\cos(t) - \cos((k + 1)t)],$

$y(t) = a[(k + 1)\sin(t) - \sin((k + 1)t)].$

Several books discussing epicycloids mention that if the ratio of the radii of the circles $\left( \frac{R}{a} = k \right)$ is rational, then they are algebraic curves. However, I’ve only been able to find the Cartesian equations for the cardioid, nephroid and ranunculoid. With the cardioid being,

$(ax + x^2 + y^2)^2 = a^2(x^2 + y^2).$

The nephroid,

$(-4a^2 + x^2 + y^2)^3 = 108a^4y^2.$

And the ranunculoid,

$-52521875a^{12} – 1286250a^{10} (x^2 + y^2) – 32025a^8 (x^2 + y^2)^2 + 93312a^7 (x^5 – 10x^3y^2 + 5xy^4) – 812a^6 (x^2 + y^2)^3 – 21a^4 (x^2 + y^2)^4 – 42a^2(x^2 + y^2)^5 + (x^2 + y^2)^6 = 0.$

Clearly this doesn’t cover all epicycloids where $\frac{R}{a}$ is rational.

What is the proof that shows that epicycloids, where the ratio of the radii are rational, are algebraic curves?

J. M. ain't a mathematician
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Chelle
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    If $z=e^{it}$ then the equations are $x=P(z)$ and $y=Q(z)$ for some polynomials $P,Q$. One now needs to eliminate the variable $z$ to get a polynomial equation $S(x,y)=0$. A classical way is to compute the resultant of $P(z)-x$ and $Q(z)-y$ (understood as polynomials in $z$), which is our $S$. – user8268 Jun 26 '16 at 19:17

1 Answers1

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If the number is rational, using multiplication formulas for $\cos$ and $\sin$ you can express the original parametrization in terms of $\cos$ and $\sin$ of multiples of a single argument. Then you can use a rational parametrization of the circle to replace those $\cos$ and $\sin$ by rational functions of the parameters, and then you can eliminate the parameter.

The following Mathematica code does this:

R = 1;
a = 4;
u[t_] := (R + a) Cos[t] - a Cos[(R + a) t/a];
v[t_] := (R + a) Sin[t] - a Sin[(R + a) t/a];
Eliminate[
    TrigExpand[{x - u[t] == 0, y - v[t] == 0} 
       /. t -> a s] 
       /. {Cos[s] -> (1 - m^2)/(1 + m^2), Sin[s] -> 2 m/(1 + m^2)},
    {m}
    ]

With $R=1$ and $a=3$, we get the equation $$x^8+x^6 \left(4 y^2-84\right)+x^4 \left(6 y^4-252 y^2+1974\right)+x^2 \left(4 y^6-252 y^4+3948 y^2-10612\right)+13824 x=-y^8+84 y^6-1974 y^4+10612 y^2+5103$$

With $R=1$ and $a=4$, $$x^{10}+x^8 \left(5 y^2-180\right)+x^6 \left(10 y^4-720 y^2+10710\right)+x^4 \left(10 y^6-1080 y^4+32130 y^2-231060\right)+x^2 \left(5 y^8-720 y^6+32130 y^4-462120 y^2+1230705\right)-1600000 x=-y^{10}+180 y^8-10710 y^6+231060 y^4-1230705 y^2-58982$$

Mariano Suárez-Álvarez
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  • For larger values of $a$ and $R$ one need a better Groebner basis engine. Macaulay2 has no trouble computing the equation. I won't include the results because they are too large. – Mariano Suárez-Álvarez Jun 26 '16 at 21:24
  • Thank you for the insightful answer and the examples with the Mathematica code, it’s very much appreciated! – Chelle Jun 28 '16 at 19:59
  • I'd have used GroebnerBasis[] directly: With[{R = 1, a = 4}, GroebnerBasis[Append[Thread[{x, y} == {(R + a) Cos[t] - a Cos[(R + a) t/a], (R + a) Sin[t] - a Sin[(R + a) t/a]} // TrigExpand], Cos[t/a]^2 + Sin[t/a]^2 == 1], {x, y}, {Cos[t/a], Sin[t/a]}]]. – J. M. ain't a mathematician Dec 26 '16 at 19:21