Hypocycloid

The curve produced by a small Circle of Radius $b$ rolling around the inside of a large Circle of Radius $a>b$. A hypocycloid is a Hypotrochoid with $h=b$. To derive the equations of the hypocycloid, call the Angle by which a point on the small Circle rotates about its center $\vartheta$, and the Angle from the center of the large Circle to that of the small Circle $\phi$. Then

$$(a-b)\phi=b\vartheta,$$ (1)

so

$$\vartheta={a-b\over b}\phi.$$ (2)

Call $\rho\equiv a-2b$. If $x(0)=\rho$, then the first point is at minimum radius, and the Cartesian parametric equations of the hypocycloid are

$$x = (a-b)\cos\phi-b\cos\vartheta$$

$$= (a-b)\cos\phi-b\cos\left({{a-b\over b}\phi}\right)$$ (3)

$$y = (a-b)\sin\phi+b\sin\vartheta$$

$$= (a-b)\sin\phi+b\sin\left({{a-b\over b}\phi}\right).$$ (4)

If $x(0)=a$ instead so the first point is at maximum radius (on the Circle), then the equations of the hypocycloid are

$$x = (a-b)\cos\phi+b\cos\left({{a-b\over b}\phi}\right)$$ (5)

$$y = (a-b)\sin\phi-b\sin\left({{a-b\over b}\phi}\right).$$ (6)

An $n$-cusped non-self-intersecting hypocycloid has $a/b=n$. A 2-cusped hypocycloid is a Line Segment, as can be seen by setting $a=b$ in equations (3) and (4) and noting that the equations simplify to

$$x = a\sin\phi$$ (7)

$$y = 0.$$ (8)

A 3-cusped hypocycloid is called a Deltoid or Tricuspoid, and a 4-cusped hypocycloid is called an Astroid. If $a/b$ is rational, the curve closes on itself and has $b$ cusps. If $a/b$ is Irrational, the curve never closes and fills the entire interior of the Circle.

$n$-hypocycloids can also be constructed by beginning with the Diameter of a Circle, offsetting one end by a series of steps while at the same time offsetting the other end by steps $n$ times as large in the opposite direction and extending beyond the edge of the Circle. After traveling around the Circle once, an $n$-cusped hypocycloid is produced, as illustrated above (Madachy 1979).

Let $r$ be the radial distance from a fixed point. For Radius of Torsion $\rho$ and Arc Length $s$, a hypocycloid can given by the equation

$$s^2+\rho^2=16r^2$$ (9)

(Kreyszig 1991, pp. 63-64). A hypocycloid also satisfies

$$\sin^2\psi = {\rho^2\over a^2-\rho^2} {a^2-r^2\over r^2},$$ (10)

where

$$r{dr\over d\theta} = \tan\psi$$ (11)

and $\psi$ is the Angle between the Radius Vector and the Tangent to the curve.

The Arc Length of the hypocycloid can be computed as follows

$$x' = -(a-b)\sin\phi-(a-b)\sin\left({{a-b\over b}\phi}\right)$$

$$= (a-b)\left[{\sin\phi+\sin\left({{a-b\over b}\phi}\right)}\right]$$ (12)

$$y' = (a-b)\cos\phi-(a-b)\cos\left({{a-b\over a}\phi}\right)$$

$$= (a-b)\left[{\cos\phi-\cos\left({{a-b\over b}\phi}\right)}\right]$$ (13)

$x'^2+y'^2 = (a-b)^2\left[{\sin^2\phi+2\sin\phi\sin\left({{a-b\over b}\phi}\right)}\right.$
$+\left.{\sin^2\left({{a-b\over b}\phi}\right)+\cos^2\phi-2\cos\phi\cos\left({{a-b\over b}\phi}\right)+\cos^2\left({{a-b\over b}\phi}\right)}\right]$
$= (a-b)^2\left\{{2+2\left[{\sin\phi\sin\left({{a-b\over a}\phi}\right)-\cos\phi\cos\left({{a-b\over b}\phi}\right)}\right]}\right\}$
$= 2(a-b)^2\left[{1-\cos\left({\phi+{a-b\over b}\phi}\right)}\right]= 4(a-b)^2 {... ...ft({{a\over b}\phi}\right)}\right]= 4(a-b)^2\sin^2\left({a\phi\over 2b}\right),$	(14)

so

$$ds = \sqrt{x'^2+y'^2}\,d\phi = 2(a-b)\sin\left({a\phi\over 2b}\right)\,d\phi$$ (15)

for $\phi\leq(b/2a)\pi$. Integrating,

$$s(\phi) = \int_0^\phi ds=2(a-b)\left[{-{2b\over a}\cos\left({a\phi\over 2b}\right)}\right]_0^\phi$$

$$= {4b(a-b)\over a}\left[{-\cos\left({{a\over 2b}\phi}\right)+1}\right]$$

$$= {8b(a-b)\over a}\sin^2\left({{a\over 4b}\phi}\right).$$ (16)

The length of a single cusp is then

$$s\left({2\pi{b\over a}}\right)= {8b(a-b)\over a}\sin^2\left({\pi\over 2}\right)={8b(a-b)\over a}.$$ (17)

If $n\equiv a/b$ is rational, then the curve closes on itself without intersecting after $n$ cusps. For $n\equiv a/b$ and with $x(0)=a$, the equations of the hypocycloid become

$$x = {1\over n}[(n-1)\cos\phi-\cos[(n-1)\phi]a,$$ (18)

$$y = {1\over n}[(n-1)\sin\phi+\sin[(n-1)\phi]a,$$ (19)

and

$$s_n=n {8b(bn-b)\over nb} = 8b(n-1)={8a(n-1)\over n}.$$ (20)

Compute

$$xy'-yx' = \left[{(a-b)\cos \phi+b\cos\left({{a-b\over a} \phi}\right)}\right](b-a)\left[{\sin \phi+\sin\left({{a-b\over b} \phi}\right)}\right]$$

$$\phantom{=} \mathop{-}\left[{(a-b)\sin\phi-b\sin\left({{a-b\over b} \phi}\right)}\right](a-b)\left[{\cos \phi-\cos\left({{a-b\over b} \phi}\right)}\right]$$

$$= 2(a^2-3ab+2b^2)\sin^2\left({a\phi\over 2b}\right).$$ (21)

The Area of one cusp is then

$$A = {\textstyle{1\over 2}}\int_0^{2\pi b/a} (xy'-yx')\,d\phi$$

$$= (a^2-3ab+2b^2)\left[{at-b\sin\left({at\over b}\right)\over 2a}\right]^{2\pi b/a}_a$$

$$= (a^2-3ab+2b^2)\left[{a\left({2\pi{b\over a}}\right)\over 2a}\right]$$

$$= {b(a^2-3ab+2b^2)\over a}\pi.$$ (22)

If $n=a/b$ is rational, then after $n$ cusps,

$$A_n = n\pi {b(a^2-3ab+2b^2)\over a} = n\pi {{a\over n}\left({a^2-3a{a\over n}+2{a^2\over n^2}}\right)\over a}$$

$$= {n^2-3n+2\over n^2} \pi a^2 = {(n-1)(n-2)\over n^2} \pi a^2.$$ (23)

The equation of the hypocycloid can be put in a form which is useful in the solution of Calculus of Variations problems with radial symmetry. Consider the case $x(0)=\rho$, then

$$r^2 = x^2+y^2$$

$$= \left[{(a-b)^2\cos^2\phi-2(a-b)b\cos\phi\cos\left({{a-b\over b}\phi}\right)+b^2\cos^2\left({{a-b\over b}\phi}\right)}\right.$$

$$\phantom{=} \mathop{+} \left.{(a-b)^2\sin^2\phi+2(a-b)b\sin\phi\sin\left({{a-b\over b}\phi}\right)+b^2\sin^2\left({{a-b\over b}\phi}\right)}\right]$$

$$= \left\{{(a-b)^2+b^2-2(a-b)b}\left[{\cos\phi\cos\left({{a-b\over b}\phi}\right)-\sin\phi\sin\left({{a-b\over b}\phi}\right)}\right]\right\}$$

$$= (a-b)^2+b^2-2(a-b)b\cos\left({{a\over b}\phi}\right).$$ (24)

But $\rho=a-2b$, so $b=(a-\rho)/2$, which gives

$$(a-b)^2+b^2 = [a-{\textstyle{1\over 2}}(a-\rho)]^2+[{\textstyle{1\over 2}}(a-\rho)]^2$$

$$= [{\textstyle{1\over 2}}(a+\rho)]^2+[{\textstyle{1\over 2}}(a-\rho)]^2$$

$$= {\textstyle{1\over 4}}(a^2+2a\rho+\rho^2+a^2-2a\rho+\rho^2)$$

$$= {\textstyle{1\over 2}}(a^2+\rho^2)$$ (25)

$$2(a-b)b = 2[a-{\textstyle{1\over 2}}(a-\rho)]{\textstyle{1\over 2}}(a-\rho)$$

$$= {\textstyle{1\over 2}}(a+\rho)(a-\rho)={\textstyle{1\over 2}}(a^2-\rho^2).$$ (26)

Now let

$$2\Omega t\equiv {a\over b}\phi,$$ (27)

so

$$\phi={a-\rho\over a} \Omega t$$ (28)

$${\phi\over a-\rho} = {\Omega t\over a},$$ (29)

then

$$r^2 = {\textstyle{1\over 2}}(a^2+\rho^2)-{\textstyle{1\over 2}}(a^2-\rho^2)\cos\left({{a\over b}\phi}\right)$$

$$= {\textstyle{1\over 2}}(a^2+\rho^2)-{\textstyle{1\over 2}}(a^2-\rho^2)\cos(2\Omega t).$$ (30)

The Polar Angle is

$$\tan\theta\equiv{y\over x} = {(a-b)\sin\phi+b\sin\left({{a-b... ...ght)\over (a-b)\cos\phi-b\cos\left({{a-b\over a}\phi}\right)}.$$ (31)

But

$$b = {\textstyle{1\over 2}}(a-\rho)$$ (32)

$$a-b = {\textstyle{1\over 2}}(a+\rho)$$ (33)

$${a-b\over b} = {a+\rho\over a-\rho},$$ (34)

so

$$\tan\theta = {{\textstyle{1\over 2}}(a+\rho)\sin\phi+{\textstyle{1\over 2}}(a-... ...\phi-{\textstyle{1\over 2}}(a-\rho)\cos\left({{a+\rho\over a-\rho}\phi}\right)}$$

$$= {(a+\rho)\sin\left({{a-\rho\over a}\Omega t}\right)+(a-\rho)\sin\... ...rho\over a}\Omega t}\right)-(a-\rho)\cos\left({{a+\rho\over a}\Omega t}\right)}$$

$$= {a\left[{\sin\left({{a-\rho\over a}\Omega t}\right)+\sin\left({{a... ...ho\over a} \Omega t}\right)+\cos\left({{a+\rho\over a}\Omega t}\right)}\right]}$$

$$= {2a\sin(\Omega t)\cos\left({{\rho\over a}\Omega t}\right)-2\rho\c... ...a}\Omega t}\right)+2\rho\cos(\Omega t)\cos\left({{\rho\over a}\Omega t}\right)}$$

$$= {a\tan(\Omega t)-\rho\tan\left({{\rho\over a}\Omega t}\right)\over a\tan(\Omega t)\tan\left({{\rho\over a}\Omega t}\right)+\rho}.$$ (35)

Computing

$$\tan\left({\theta+{\rho\over a}\Omega t}\right) = {\left[{a\tan(\Omega t)-\rho\tan\left({{\rho\over a}\Omega t}\rig... ...{{\rho\over a}\Omega t}\right)}\right]\tan\left({{\rho\over a}\Omega t}\right)}$$

$$= {a\tan(\Omega t)\left[{1+\tan^2\left({{\rho\over a}\Omega t}\right)}\right]\over\rho\left[{1+\tan^2\left({{\rho\over a}\Omega t}\right)}\right]}$$

$$= {a\over\rho}\tan(\Omega t),$$ (36)

then gives

$$\theta=\tan^{-1}\left[{{a\over\rho}\tan(\Omega t)}\right]-{\rho\over a}\Omega t.$$ (37)

Finally, plugging back in gives

$$\theta = \tan^{-1}\left[{{a\over\rho}\tan\left({{a\over a-\rho}\phi}\right)}\right]-{\rho\over a}{a\over a-\rho} \phi$$

$$= \tan^{-1}\left[{{a\over\rho}\tan\left({{a\over a-\rho}\phi}\right)}\right]-{\rho\over a-\rho}\phi.$$ (38)

This form is useful in the solution of the Sphere with Tunnel problem, which is the generalization of the Brachistochrone Problem, to find the shape of a tunnel drilled through a Sphere (with gravity varying according to Gauss's law for gravitation ) such that the travel time between two points on the surface of the Sphere under the force of gravity is minimized.

See also Cycloid, Epicycloid

References

Bogomolny, A. ``Cycloids.'' http://www.cut-the-knot.com/pythagoras/cycloids.html.

Kreyszig, E. Differential Geometry. New York: Dover, 1991.

Lawrence, J. D. A Catalog of Special Plane Curves. New York: Dover, pp. 171-173, 1972.

Lee, X. ``Epicycloid and Hypocycloid.'' http://www.best.com/~xah/SpecialPlaneCurves_dir/EpiHypocycloid_dir/epiHypocycloid.html.

MacTutor History of Mathematics Archive. ``Hypocycloid.'' http://www-groups.dcs.st-and.ac.uk/~history/Curves/Hypocycloid.html.

Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, pp. 225-231, 1979.

Wagon, S. Mathematica in Action. New York: W. H. Freeman, pp. 50-52, 1991.

Yates, R. C. ``Epi- and Hypo-Cycloids.'' A Handbook on Curves and Their Properties. Ann Arbor, MI: J. W. Edwards, pp. 81-85, 1952.