http://www.astro.virginia.edu/~eww6n/math/Hypocycloid.html

Hypocycloid

The curve produced by fixed point P on the Circumference of a small Circle of Radiusb rolling around the inside of a large Circle of Radiusa>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

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

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

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

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

 

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

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

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

 
 

An n-cusped non-self-intersecting hypocycloid has a/b=n. A 2-cusped hypocycloid is a Line Segment (Steinhaus 1983, p. 145), as can be seen by setting a=b in equations (3) and (4) and noting that the equations simplify to

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

y = 0. (8)

A 3-cusped hypocycloid is called a Deltoid or Tricuspoid, and a 4-cusped hypocycloid is called anAstroid. 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 Lengths, 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

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

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

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

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

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

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

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

and

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

Compute
 
 

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

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

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

The Area of one cusp is then

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

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

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

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

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

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

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

r² = x²+y²

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

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

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

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

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

$$[a-{\textstyle{1\over 2}}(a-\rho)]^2+[{\textstyle{1\over 2}}(a-\rho)]^2$$ (a-b)²+b² =

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

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

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

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

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

$${\textstyle{1\over 2}}(a^2+\rho^2)-{\textstyle{1\over 2}}(a^2-\rho^2)\cos\left({{a\over b}\phi}\right)$$ r² =

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

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

$${\textstyle{1\over 2}}(a+\rho)$$ a-b = (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 theBrachistochrone 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 Astroid, Cycloid, Deltoid, 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.

Steinhaus, H. Mathematical Snapshots, 3rd American ed. New York: Oxford University Press, 1983.

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.
 

1999-06-06