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So I've been working on a game for three months -some kind of air hockey- and players play in fields with various shapes in a two-dimensional plane. Recently, I wanted to make one of these fields looking like an epicycloid, the one with four "petals". A "four petals" epicycloid.

When I create a field, I have to get its shape's parametric equation to draw it, and its shape's cartesian equation for everything about the game's physics.
By definition, an epicycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circle which rolls without slipping around a fixed circle. A "three petal" epicycloid as example The parametric representation of an epicycloid centered in $A(a, b)$, with the static circle of radius $R$ and the moving one of radius $r$ is :

$$x = a + m\left((R + r)\cos(t) - r\cos\left(\frac {R + r}{r}t\right)\right)$$ $$y = b + m\left((R + r)\sin(t) - r\sin\left(\frac {R + r}{r}t\right)\right)$$ where $m$ is a constant. According to the value of $\frac{R}{r}$, we have $\frac{R}{r}$ "petals".

Using this, I can draw the field, but I couldn't find a cartesian equation for a "four petal" epicycloid, so I can't work on the field's physics. Can anyone help me find a cartesian equation ?

Hoël Aviël OGOULA ANOTHO
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  • I do not understand the problem. You have drawn for 3 petals with $R/r= 3$, and for 4 petals use $R/r= 4$, and it is done exactly same way. You have already drawn it without the base circle at start of your question. You need as $y=f(x) $ without parametrization? – Narasimham Dec 16 '23 at 22:57
  • Yes, I need a cartesian equation, or $y = f(x)$, or at least a way to find them, so I'll be able to simulate the physics of the field. The drawings you see above were made using the parametric equation, but that's not even my problem. – Hoël Aviël OGOULA ANOTHO Dec 18 '23 at 06:56
  • (At first calculate without curve translation around origin). Square, add, simplify the two parametric equations for x,y to get polar coordinates, convert to Cartesian, add back translations. – Narasimham Dec 18 '23 at 22:13
  • Polar coordinate use is decidedly more convenient than Cartesian. – Narasimham Dec 19 '23 at 08:53
  • I'm going to try squaring as you said and follow your tip. Thanks. – Hoël Aviël OGOULA ANOTHO Dec 20 '23 at 16:57
  • So I tried your method and I ended up with : $$3125(X^2 - Y^2)^2 = (9 - 4(X^2 + Y^2))(16(X^2 + Y^2)^2 + 3(X^2 + Y^2) + 6)^2$$ where $$X = \frac{x - a}{m} $$ and $$ Y = \frac{y-b}{n}$$. It worked when I found the tangent. Thanks. – Hoël Aviël OGOULA ANOTHO Jan 06 '24 at 11:11
  • Good. This is the same equation as in my answer with $X=x/4, Y=y/4$ up to a factor of $2^{10}.$ – Jan-Magnus Økland Jan 09 '24 at 12:46
  • Oh ok. I was just scared of it because it was too long. – Hoël Aviël OGOULA ANOTHO Jan 13 '24 at 06:40

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In Macaulay2

R=4
r=1
S=QQ[i,c,s,c2,s2,x,y,MonomialOrder=>Eliminate 5]
I=ideal(x-((R+r)*c-r*c2),y-((R+r)*s-r*s2),c^2+s^2-1,c2^2+s2^2-1,i^2+1,(c+i*s)^(R+r)-(c2+i*s2)^r)
gens gb I

$$x^{10}+5x^8y^2+10x^6y^4+10x^4y^6+5x^2y^8+y^{10}\\-30x^8-120x^6y^2-180x^4y^4-120x^2y^6-30y^8\\-15x^6-45x^4y^2-45x^2y^4-15y^6\\+5840x^4-38320x^2y^2+5840y^4\\-11520x^2-11520y^2\\-331776=0$$

Jan-Magnus Økland
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