Antonio Cañada Momblant
Toni Cañada

Toni Cañada

Oloid surface

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Antonio Cañada Momblant
·Sep 26, 2021·

5 min read

Oloid surface

An oloid is a three-dimensional curved geometric object that was discovered by Paul Schatz in 1929. It’s a ruled surface, the convex hull of a skeletal frame made by placing two linked congruent circles in perpendicular planes, so that the center of each circle lies on the edge of the other circle. The following 2 circles generate an oloid where \(r\) is the radius.

\begin{equation} k_a: \begin{cases} x^2+ \left( y-\frac{r}{2} \right)^2 = r^2 \newline z=0 \end{cases}\ \end{equation}

\begin{equation} k_b: \begin{cases} \left( y-\frac{r}{2} \right)^2 + z^2 = r^2 \newline x=0 \end{cases}\, \end{equation}

In this article we’ll parameterize this beautiful surface, and show that it’s surface is the same as the sphere (\(4 \ \pi \ r^2\)), apart of some other properties.

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As mention above, the oloid is a ruled surface, and it’s formed by the segments AB, where A belongs to \(k_a\) and B to \(k_b\), respectively, along both circles.

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\[ A = \left(\begin{array}{ccc}r\,\sin\left(\alpha\right), & -\dfrac{r}{2}-r\,\cos\left(\alpha\right), & 0 \end{array}\right) \]

\[ \beta = \pi - \alpha/2 \]

\[ \sin(\beta) = sin(\pi - \alpha/2) = cos(\alpha) \]

\[ |\overrightarrow{\rm TM_A}|\, \sin(\beta) = r \implies |\overrightarrow{\rm TM_A}| = \left| \dfrac{r}{\cos(\alpha)}\right | \]

\[ T = \left(\begin{array}{ccc} 0, & -\dfrac{r}{2}-\dfrac{r}{\cos\left(\alpha\right)}, & 0 \end{array}\right) \]

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\[ |\overrightarrow{\rm TM_B}|^2 = |\overrightarrow{\rm TB}|^2 + r^2 \] \[ |\overrightarrow{\rm TM_B}|^2 = \left( \dfrac{r}{2}+\dfrac{r}{\cos\left(\alpha\right)}+\dfrac{r}{2} \right)^2 = \left( \dfrac{r + r\ cos(\alpha)}{\cos\left(\alpha\right)} \right)^2 \] \[ \cos(\gamma) = \dfrac{-r}{|\overrightarrow{\rm TM_B}|} = \dfrac{-\cos\left(\alpha\right)}{1 + cos(\alpha)} \]

\[ B_y = \dfrac{r}{2}+r\ \cos(\gamma) = \dfrac{r}{2} - \dfrac{r\ \cos\left(\alpha\right)}{1 + cos(\alpha)} \]

\[ B_z = r\ \sin(\gamma) \]

\[ \sin(\gamma)^2 = 1 - \cos(\gamma)^2 = 1 - \left( \dfrac{\cos\left(\alpha\right)}{1 + cos(\alpha)} \right)^2 = \left( \dfrac{2\ \cos(\alpha) + 1}{(\cos(\alpha) + 1)^2} \right) \]

\[ B = \left(\begin{array}{ccc} 0, & \dfrac{r}{2} - \dfrac{r\ \cos\left(\alpha\right)}{1 + cos(\alpha)}, & \dfrac{ \pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right) \]

The square root in the z coordinate of B creates the following restriction: \[ 2\ \cos(\alpha) + 1 \geq 0 \implies -\dfrac{2 \pi}{3} \leq \alpha \leq \dfrac{2 \pi}{3} \]

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

\[ -\dfrac{2 \pi}{3} < \alpha < \dfrac{2 \pi}{3} \]


Oloid parameterization

The oloid is a ruled surface generated by the AB segments, by the following equation, where v is between 0 and 1.

\[ A + v\ \overrightarrow{\rm AB} \]

\[ \overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right), & \dfrac{r}{2}+r\,\cos\left(\alpha \right)-\dfrac{r\,\left(\cos\left(\alpha \right)-1\right)}{2\,\left(\cos\left(\alpha \right)+1\right)}, & \dfrac{\pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right) \]

\[ \overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right), & \dfrac{r\,\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)}{\cos\left(\alpha \right)+1}, & \dfrac{ \pm\ r\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right) \]

\[ A + v\ \overrightarrow{\rm AB} = \left(\begin{array}{ccc} -r\,\sin\left(\alpha \right)\,\left(v-1\right), & \dfrac{r\,\left(2\,v-3\,\cos\left(\alpha \right)-2\,{\cos\left(\alpha \right)}^2+2\,v\,\cos\left(\alpha \right)+2\,v\,{\cos\left(\alpha \right)}^2-1\right)}{2\,\left(\cos\left(\alpha \right)+1\right)}, & \dfrac{\pm\ r\,v\,\sqrt{2\,\cos\left(\alpha \right)+1}}{\cos\left(\alpha \right)+1} \end{array}\right) \]

\[ 0 \leq v \leq 1,\ -\dfrac{2 \pi}{3} < \alpha < \dfrac{2 \pi}{3} \]


Proving that all AB segments length is constant for all \( \alpha \)

\[ |\overrightarrow{\rm AB}|^2 = r^2\ \sin(\alpha)^2 + \dfrac{r^2\,\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)^2}{(\cos\left(\alpha \right)+1)^2} + \dfrac{r^2\,(2\,\cos\left(\alpha \right)+1)}{(\cos\left(\alpha \right)+1)^2} \] \[ |\overrightarrow{\rm AB}|^2 = r^2\,\left(1 -{\cos\left(\alpha \right)}^2+\frac{{\left({\cos\left(\alpha \right)}^2+\cos\left(\alpha \right)+1\right)}^2}{{\left(\cos\left(\alpha \right)+1\right)}^2} + \frac{2\,\cos\left(\alpha \right)+1}{{\left(\cos\left(\alpha \right)+1\right)}^2}\right) \] \[ t = \cos(\alpha) \] \[ |\overrightarrow{\rm AB}|^2 = r^2\,\left(1-t^2+\frac{{\left(t^2+t+1\right)}^2}{{\left(t+1\right)}^2}+\frac{2\,t+1}{{\left(t+1\right)}^2}\right) \] \[ |\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{(1-t^2)(t+1)^2+(t^2+t+1)^2+(2t+1)}{(t+1)^2} \right) \] \[ |\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{3t^2 + 6t + 3}{(t+1)^2} \right) \] \[ |\overrightarrow{\rm AB}|^2 = r^2\,\left( \dfrac{3\ (t+1)^2}{(t+1)^2} \right) \] \[ |\overrightarrow{\rm AB}|^2 = 3r^2 \]

\[ |\overrightarrow{\rm AB}| = \sqrt3 \ r \]


Computing the oloid’s surface

Due it’s a ruled surface, area can be computed by the following formula (see this publication by J. B. Reynolds):

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\[ \frac{d \ \overrightarrow{OB}}{d \ \alpha} = \left(\begin{array}{ccc} 0, & \dfrac{r\,\sin\left(\alpha \right)}{{\left(\cos\left(\alpha \right)+1\right)}^2}, & \dfrac{\pm\ r\,\sin\left(2\,\alpha \right)}{2\,{\left(\cos\left(\alpha \right)+1\right)}^2\,\sqrt{2\,\cos\left(\alpha \right)+1}} \end{array}\right) \]

\[ \frac{d \ \overrightarrow{OA}}{d \ \alpha} = \left(\begin{array}{ccc} r\,\cos\left(\alpha \right), & r\,\sin\left(\alpha \right), & 0 \end{array}\right) \]

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.

\[ (1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} = \left(\begin{array}{ccc} r\,v\,\cos\left(\alpha \right), & \dfrac{r\,\sin\left(\alpha \right)\,\left(v\,{\cos\left(\alpha \right)}^2+2\,v\,\cos\left(\alpha \right)+1\right)}{{\left(\cos\left(\alpha \right)+1\right)}^2}, & -\dfrac{r\,\sin\left(2\,\alpha \right)\,\left(v-1\right)}{2\,{\left(\cos\left(\alpha \right)+1\right)}^2\,\sqrt{2\,\cos\left(\alpha \right)+1}} \end{array}\right) \]


\[ \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} \right) = \] \[ = \left(\begin{array}{ccc} -\dfrac{r^2\,\sin\left(\alpha \right)\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\left(\cos\left(\alpha \right)+1\right)\,\sqrt{2\,\cos\left(\alpha \right)+1}}, & \dfrac{r^2\,\cos\left(\alpha \right)\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\left(\cos\left(\alpha \right)+1\right)\,\sqrt{2\,\cos\left(\alpha \right)+1}}, & -\dfrac{r^2\,\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}{\cos\left(\alpha \right)+1} \end{array}\right) \]


\[ \left| \overrightarrow{\rm AB} \times \left((1-v)\ \frac{d \ \overrightarrow{OB}}{\alpha} + v\ \frac{d \ \overrightarrow{OA}}{\alpha} \right) \right|^2 = \frac{2\,r^4\,{\left(3\,v\,\cos\left(\alpha \right)-\cos\left(\alpha \right)+1\right)}^2}{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1} \]


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\[ A =2 \ \sqrt{2} \ r^2 \int_{-2\pi/3}^{2\pi/3} \frac{\frac{1}{2} \cos{\alpha} + 1}{\sqrt{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1}} \,d\alpha\ \]

\[ A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \sqrt{2\ \cos(\alpha) + 1} \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{2\,{\cos\left(\alpha \right)}^2+3\,\cos\left(\alpha \right)+1}} \right|_{-2\pi/3}^{2\pi/3} \]

\[ A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \sqrt{2\ \cos(\alpha) + 1} \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{(2\ \cos(\alpha) + 1)(\cos(\alpha) + 1)}} \right|_{-2\pi/3}^{2\pi/3} \]

\[ A =\left. 2 \ \sqrt{2} \ r^2 \ \dfrac{\cos\left(\dfrac{\alpha}{2}\right) \left( \sin^{-1}\left( \dfrac{2 \sin\left( \dfrac{\alpha}{2}\right)}{\sqrt{3}}\right) + \tan^{-1} \left( \dfrac{\sin\left(\dfrac{\alpha}{2} \right)}{\sqrt{2\ \cos{\alpha} + 1} } \right)\right)}{\sqrt{\cos(\alpha) + 1}} \right|_{-2\pi/3}^{2\pi/3} \]

\[ A =2 \ \sqrt{2} \ r^2 \left( \dfrac{\sqrt{2} \pi}{2} + \dfrac{\sqrt{2} \pi}{2} \right) = 2 \ \sqrt{2} \ r^2 \left( \sqrt{2}\ \pi \right) = 4\ \pi \ r^2 \]

Which is the same area as the sphere.

You can find a parametrized oloid in this Geogebra link.

 
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