The Fascinating Oloid: Parametrization and Surface Area

The Fascinating Oloid: Parametrization and Surface Area


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Some geometric shapes captivate not only for their beauty, but also for the unexpected mathematical properties they hide. The oloid, discovered by Paul Schatz in 1929, is one such shape.

At first glance, it may look like a sculptural curiosity. But the oloid is a ruled surface, generated by line segments between two congruent circles in perpendicular planes, where the center of each lies on the circumference of the other. In this article, we’ll explore:

  • How to define the oloid from two simple circles
  • How to parameterize its entire surface
  • Why all generating segments have the same length
  • And how to compute its surface area β€” which turns out to be exactly the same as a sphere of radius rr

Let’s dive in.

Oloid circles

From circles to surface

The oloid is generated from two circles:

ka:{x2+(yβˆ’r2)2=r2z=0 k_a: \begin{cases} x^2+ \left( y-\frac{r}{2} \right)^2 = r^2 \\ z=0 \end{cases} kb:{(yβˆ’r2)2+z2=r2x=0 k_b: \begin{cases} \left( y-\frac{r}{2} \right)^2 + z^2 = r^2 \\ x=0 \end{cases}

These circles lie in perpendicular planes, and the oloid surface is generated by connecting corresponding points from each one.

Generating segments

A=(r sin⁑(Ξ±),βˆ’r2βˆ’r cos⁑(Ξ±),0)A = \left(\begin{array}{ccc}r\,\sin\left(\alpha\right), & -\dfrac{r}{2}-r\,\cos\left(\alpha\right), & 0 \end{array}\right) Ξ²=Ο€βˆ’Ξ±/2\beta = \pi - \alpha/2 sin⁑(Ξ²)=sin(Ο€βˆ’Ξ±/2)=cos(Ξ±)\sin(\beta) = sin(\pi - \alpha/2) = cos(\alpha) ∣TMAβ†’βˆ£β€‰sin⁑(Ξ²)=rβ€…β€ŠβŸΉβ€…β€Šβˆ£TMAβ†’βˆ£=∣rcos⁑(Ξ±)∣|\overrightarrow{\rm TM_A}|\, \sin(\beta) = r \implies |\overrightarrow{\rm TM_A}| = \left| \dfrac{r}{\cos(\alpha)}\right | T=(0,βˆ’r2βˆ’rcos⁑(Ξ±),0)T = \left(\begin{array}{ccc} 0, & -\dfrac{r}{2}-\dfrac{r}{\cos\left(\alpha\right)}, & 0 \end{array}\right)

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∣TMBβ†’βˆ£2=∣TBβ†’βˆ£2+r2|\overrightarrow{\rm TM_B}|^2 = |\overrightarrow{\rm TB}|^2 + r^2 ∣TMBβ†’βˆ£2=(r2+rcos⁑(Ξ±)+r2)2=(r+rΒ cos(Ξ±)cos⁑(Ξ±))2|\overrightarrow{\rm TM_B}|^2 = \left( \dfrac{r}{2}+\dfrac{r}{\cos\left(\alpha\right)}+\dfrac{r}{2} \right)^2 = \left( \dfrac{r + r\ cos(\alpha)}{\cos\left(\alpha\right)} \right)^2 cos⁑(Ξ³)=βˆ’r∣TMBβ†’βˆ£=βˆ’cos⁑(Ξ±)1+cos(Ξ±)\cos(\gamma) = \dfrac{-r}{|\overrightarrow{\rm TM_B}|} = \dfrac{-\cos\left(\alpha\right)}{1 + cos(\alpha)} By=r2+rΒ cos⁑(Ξ³)=r2βˆ’rΒ cos⁑(Ξ±)1+cos(Ξ±)B_y = \dfrac{r}{2}+r\ \cos(\gamma) = \dfrac{r}{2} - \dfrac{r\ \cos\left(\alpha\right)}{1 + cos(\alpha)} Bz=rΒ sin⁑(Ξ³)B_z = r\ \sin(\gamma) sin⁑(Ξ³)2=1βˆ’cos⁑(Ξ³)2=1βˆ’(cos⁑(Ξ±)1+cos(Ξ±))2=(2Β cos⁑(Ξ±)+1(cos⁑(Ξ±)+1)2)\sin(\gamma)^2 = 1 - \cos(\gamma)^2 = 1 - \left( \dfrac{\cos\left(\alpha\right)}{1 + cos(\alpha)} \right)^2 = \left( \dfrac{2\ \cos(\alpha) + 1}{(\cos(\alpha) + 1)^2} \right) B=(0,r2βˆ’rΒ cos⁑(Ξ±)1+cos(Ξ±),Β±Β r 2 cos⁑(Ξ±)+1cos⁑(Ξ±)+1)B = \left(\begin{array}{ccc} 0, & \dfrac{r}{2} - \dfrac{r\ \cos\left(\alpha\right)}{1 + cos(\alpha)}, & \dfrac{ \pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right)

The square root in the z coordinate of B creates the following restriction:

2Β cos⁑(Ξ±)+1β‰₯0β€…β€ŠβŸΉβ€…β€Šβˆ’2Ο€3≀α≀2Ο€32\ \cos(\alpha) + 1 \geq 0 \implies -\dfrac{2 \pi}{3} \leq \alpha \leq \dfrac{2 \pi}{3}

But we have to avoid zero denominators in the y coordinate of B, so the domain of Ξ±\alpha) becomes:

βˆ’2Ο€3<Ξ±<2Ο€3-\dfrac{2 \pi}{3} < \alpha < \dfrac{2 \pi}{3}

What follows is a complete derivation of the parametrization of the oloid, the computation of the length of each generating segment AB→\overrightarrow{AB}, and the beautiful integral that gives its total surface area.

Parametrization

We define points A∈kaA \in k_a and B∈kbB \in k_b based on a variable angle Ξ±\alpha, derive their coordinates, and connect them via a segment. Using vector operations, we describe each point on the surface with parameters α∈(βˆ’2Ο€/3,2Ο€/3)\alpha \in (-2\pi/3, 2\pi/3) and v∈[0,1]v \in [0,1].

A+v⋅AB→A + v \cdot \overrightarrow{AB}
Click to expand full parametrizationABβ†’=(βˆ’r sin⁑(Ξ±),r2+r cos⁑(Ξ±)βˆ’r (cos⁑(Ξ±)βˆ’1)2 (cos⁑(Ξ±)+1),Β±Β r 2 cos⁑(Ξ±)+1cos⁑(Ξ±)+1)\overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right), & \dfrac{r}{2}+r\,\cos\left(\alpha \right)-\dfrac{r\,\left(\cos\left(\alpha \right)-1\right)}{2\,\left(\cos\left(\alpha \right)+1\right)}, & \dfrac{\pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right)ABβ†’=(βˆ’r sin⁑(Ξ±),r (cos⁑(Ξ±)2+cos⁑(Ξ±)+1)cos⁑(Ξ±)+1,Β±Β r 2 cos⁑(Ξ±)+1cos⁑(Ξ±)+1)\overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right), & \dfrac{r\,\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)}{\cos\left(\alpha \right)+1}, & \dfrac{ \pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right)A+vΒ ABβ†’=(βˆ’r sin⁑(Ξ±) (vβˆ’1),r (2 vβˆ’3 cos⁑(Ξ±)βˆ’2 cos⁑(Ξ±)2+2 v cos⁑(Ξ±)+2 v cos⁑(Ξ±)2βˆ’1)2 (cos⁑(Ξ±)+1),Β±Β r v 2 cos⁑(Ξ±)+1cos⁑(Ξ±)+1)A + v\ \overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right)\,\left(v-1\right), & \dfrac{r\,\left(2\,v-3\,\cos\left(\alpha \right)-2\,{\cos\left(\alpha \right)}^2+2\,v\,\cos\left(\alpha \right)+2\,v\,{\cos\left(\alpha \right)}^2-1\right)}{2\,\left(\cos\left(\alpha \right)+1\right)}, & \dfrac{\pm\ r\,v\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right)0≀v≀1,Β βˆ’2Ο€3<Ξ±<2Ο€30 \leq v \leq 1,\ -\dfrac{2 \pi}{3} < \alpha < \dfrac{2 \pi}{3}

Proving that all AB segments length is constant for all Ξ±\alpha

∣ABβ†’βˆ£2=r2Β sin⁑(Ξ±)2+r2 (cos⁑(Ξ±)2+cos⁑(Ξ±)+1)2(cos⁑(Ξ±)+1)2+r2 (2 cos⁑(Ξ±)+1)(cos⁑(Ξ±)+1)2|\overrightarrow{\rm AB}|^2 = r^2\ \sin(\alpha)^2 + \dfrac{r^2\,\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)^2}{(\cos\left(\alpha \right)+1)^2} + \dfrac{r^2\,(2\,\cos\left(\alpha \right)+1)}{(\cos\left(\alpha \right)+1)^2}
Click to expand full derivation∣ABβ†’βˆ£2=r2 (1βˆ’cos⁑(Ξ±)2+(cos⁑(Ξ±)2+cos⁑(Ξ±)+1)2(cos⁑(Ξ±)+1)2+2 cos⁑(Ξ±)+1(cos⁑(Ξ±)+1)2)|\overrightarrow{\rm AB}|^2 = r^2\,\left(1 -{\cos\left(\alpha \right)}^2+\frac{{\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)}^2}{{\left(\cos\left(\alpha \right)+1\right)}^2} + \frac{2\,\cos\left(\alpha \right)+1}{{\left(\cos\left(\alpha \right)+1\right)}^2}\right)t=cos⁑(Ξ±)t = \cos(\alpha)∣ABβ†’βˆ£2=r2 (1βˆ’t2+(t2+t+1)2(t+1)2+2 t+1(t+1)2)|\overrightarrow{\rm AB}|^2 = r^2\,\left(1-t^2+\frac{{\left(t^2+t+1\right)}^2}{{\left(t+1\right)}^2}+\frac{2\,t+1}{{\left(t+1\right)}^2}\right)∣ABβ†’βˆ£2=r2 ((1βˆ’t2)(t+1)2+(t2+t+1)2+(2t+1)(t+1)2)|\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{(1-t^2)(t+1)^2+(t^2+t+1)^2+(2t+1)}{(t+1)^2} \right)∣ABβ†’βˆ£2=r2 (3t2+6t+3(t+1)2)|\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{3t^2 + 6t + 3}{(t+1)^2} \right)∣ABβ†’βˆ£2=r2 (3Β (t+1)2(t+1)2)|\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{3\ (t+1)^2}{(t+1)^2} \right)∣ABβ†’βˆ£2=3r2|\overrightarrow{\rm AB}|^2 = 3r^2
∣ABβ†’βˆ£=3Β r|\overrightarrow{\rm AB}| = \sqrt3 \ r

Meaning all segments have equal length β€” a surprisingly symmetric property.

Surface area of the oloid

Because the oloid is a ruled surface, we apply a surface area formula adapted from J. B. Reynolds (source).

A=∫α1Ξ±2∫v1v2∣ABβ†’Γ—((1βˆ’v)Β dΒ OBβ†’dΞ±+vΒ dΒ OAβ†’dΞ±)βˆ£β€‰dΞ±Β dvA = \int_{\alpha_1}^{\alpha_2} \int_{v_1}^{v_2} \left| \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d \ \overrightarrow{OB}}{d\alpha} + v\ \frac{d \ \overrightarrow{OA}}{d\alpha} \right) \right| \,d\alpha\ dv A=βˆ«βˆ’2Ο€/32Ο€/3∫01∣ABβ†’Γ—((1βˆ’v)Β d OBβ†’dΞ±+vΒ d OAβ†’dΞ±)βˆ£β€‰dα dvA = \int_{-2\pi/3}^{2\pi/3} \int_{0}^{1} \left| \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d\,\overrightarrow{OB}}{d\alpha} + v\ \frac{d\,\overrightarrow{OA}}{d\alpha} \right) \right| \,d\alpha\, dv dΒ OBβ†’dΒ Ξ±=(0,r sin⁑(Ξ±)(cos⁑(Ξ±)+1)2,Β±Β r sin⁑(2 α)2 (cos⁑(Ξ±)+1)2 2 cos⁑(Ξ±)+1) \frac{d \ \overrightarrow{OB}}{d \ \alpha} = \left(\begin{array}{ccc} 0, & \dfrac{r\,\sin\left(\alpha \right)}{{\left(\cos\left(\alpha \right)+1\right)}^2}, & \dfrac{\pm\ r\,\sin\left(2\,\alpha \right)}{2\,{\left(\cos\left(\alpha \right)+1\right)}^2\,\sqrt{2\,\cos\left(\alpha \right)+1}} \end{array}\right) dΒ OAβ†’dΒ Ξ±=(r cos⁑(Ξ±),r sin⁑(Ξ±),0) \frac{d \ \overrightarrow{OA}}{d \ \alpha} = \left(\begin{array}{ccc} r\,\cos\left(\alpha \right), & r\,\sin\left(\alpha \right), & 0 \end{array}\right)
Click to expand full derivation

Since we’ll continue with the positive value of the 3rd coordinate of the derivative of OB, we are computing the oloid’s top surface. Then, in order to have the total surface, we’ll have to multiply by 2 the result of the integral.

(1βˆ’v)Β dΒ OBβ†’Ξ±+vΒ dΒ OAβ†’Ξ±=(r v cos⁑(Ξ±),r sin⁑(Ξ±) (v cos⁑(Ξ±)2+2 v cos⁑(Ξ±)+1)(cos⁑(Ξ±)+1)2,βˆ’r sin⁑(2 α) (vβˆ’1)2 (cos⁑(Ξ±)+1)2 2 cos⁑(Ξ±)+1) (1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} = \left(\begin{array}{ccc} r\,v\,\cos\left(\alpha \right), & \dfrac{r\,\sin\left(\alpha \right)\,\left(v\,{\cos\left(\alpha \right)}^2+2\,v\,\cos\left(\alpha \right)+1\right)}{{\left(\cos\left(\alpha \right)+1\right)}^2}, & -\dfrac{r\,\sin\left(2\,\alpha \right)\,\left(v-1\right)}{2\,{\left(\cos\left(\alpha \right)+1\right)}^2\,\sqrt{2\,\cos\left(\alpha \right)+1}} \end{array}\right)
ABβ†’Γ—((1βˆ’v)Β dΒ OBβ†’Ξ±+vΒ dΒ OAβ†’Ξ±)= \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} \right) = =(βˆ’r2 sin⁑(Ξ±) (3 v cos⁑(Ξ±)βˆ’cos⁑(Ξ±)+1)(cos⁑(Ξ±)+1) 2 cos⁑(Ξ±)+1,r2 cos⁑(Ξ±) (3 v cos⁑(Ξ±)βˆ’cos⁑(Ξ±)+1)(cos⁑(Ξ±)+1) 2 cos⁑(Ξ±)+1,βˆ’r2 (3 v cos⁑(Ξ±)βˆ’cos⁑(Ξ±)+1)cos⁑(Ξ±)+1) = \left(\begin{array}{ccc} -\dfrac{r^2\,\sin\left(\alpha \right)\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\left(\cos\left(\alpha \right)+1\right)\,\sqrt{2\,\cos\left(\alpha \right)+1}}, & \dfrac{r^2\,\cos\left(\alpha \right)\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\left(\cos\left(\alpha \right)+1\right)\,\sqrt{2\,\cos\left(\alpha \right)+1}}, & -\dfrac{r^2\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\cos\left(\alpha \right)+1} \end{array}\right)∣ABβ†’Γ—((1βˆ’v)Β dΒ OBβ†’Ξ±+vΒ dΒ OAβ†’Ξ±)∣2=2 r4 (3 v cos⁑(Ξ±)βˆ’cos⁑(Ξ±)+1)22 cos⁑(Ξ±)2+3 cos⁑(Ξ±)+1\left| \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} \right) \right|^2 = \frac{2\,r^4\,{\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}^2}{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1}
A=2Β 2Β r2βˆ«βˆ’2Ο€/32Ο€/3∫013 v cos⁑(Ξ±)βˆ’cos⁑(Ξ±)+12 cos⁑(Ξ±)2+3 cos⁑(Ξ±)+1 dΞ±Β dvA =2 \ \sqrt{2} \ r^2 \int_{-2\pi/3}^{2\pi/3} \int_{0}^{1} \frac{{3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1}}{\sqrt{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1}} \,d\alpha\ dv
A=22 r2βˆ«βˆ’2Ο€/32Ο€/312cos⁑(Ξ±)+12cos⁑2(Ξ±)+3cos⁑(Ξ±)+1 dΞ±A = 2 \sqrt{2} \, r^2 \int_{-2\pi/3}^{2\pi/3} \frac{\tfrac{1}{2} \cos(\alpha) + 1}{\sqrt{2 \cos^2(\alpha) + 3 \cos(\alpha) + 1}} \, d\alpha
A=2Β 2Β r2Β cos⁑(Ξ±2)2Β cos⁑(Ξ±)+1(sinβ‘βˆ’1(2sin⁑(Ξ±2)3)+tanβ‘βˆ’1(sin⁑(Ξ±2)2Β cos⁑α+1))2 cos⁑(Ξ±)2+3 cos⁑(Ξ±)+1βˆ£βˆ’2Ο€/32Ο€/3 A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \sqrt{2\ \cos(\alpha) + 1} \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1}} \right|_{-2\pi/3}^{2\pi/3}A=2Β 2Β r2Β cos⁑(Ξ±2)2Β cos⁑(Ξ±)+1(sinβ‘βˆ’1(2sin⁑(Ξ±2)3)+tanβ‘βˆ’1(sin⁑(Ξ±2)2Β cos⁑α+1))(2Β cos⁑(Ξ±)+1)(cos⁑(Ξ±)+1)βˆ£βˆ’2Ο€/32Ο€/3 A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \sqrt{2\ \cos(\alpha) + 1} \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{(2\ \cos(\alpha) + 1)(\cos(\alpha) + 1)}} \right|_{-2\pi/3}^{2\pi/3}A=2Β 2Β r2Β cos⁑(Ξ±2)(sinβ‘βˆ’1(2sin⁑(Ξ±2)3)+tanβ‘βˆ’1(sin⁑(Ξ±2)2Β cos⁑α+1))cos⁑(Ξ±)+1βˆ£βˆ’2Ο€/32Ο€/3 A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{\cos(\alpha) + 1}} \right|_{-2\pi/3}^{2\pi/3}
A=2Β 2Β r2(2Ο€2+2Ο€2)=2Β 2Β r2(2Β Ο€)=4 π r2 A =2 \ \sqrt{2} \ r^2 \left( \dfrac{\sqrt{2} \pi}{2} + \dfrac{\sqrt{2} \pi}{2} \right) = 2 \ \sqrt{2} \ r^2 \left( \sqrt{2}\ \pi \right) = 4\ \pi \ r^2

Through careful differentiation and vector cross products, we derive the integral that gives the total surface. After evaluating the integral (included in full in the article), we obtain:

A=4Ο€r2A = 4\pi r^2

Exactly the same surface area as a sphere of radius rr!

Explore interactively

You can find a parametrized oloid in this GeoGebra link.