Hilariously Fast Volume Computation with the Divergence Theorem

16 Feb 2018

(No, there won’t be jokes.)

The following presents a fast algorithm for volume computation of a simple, closed, triangulated 3D mesh. This assumption is a consequence of the divergence theorem. Further extensions may generalise to other meshes as well, although that is presently out of scope.

We begin with the definition of volume as the triple integral over a region of the constant one:

V=∭R1dVV = \iiint_R 1 \mathrm{d}V

Let 𝐅\mathbf{F} be a function in ℝ3\mathbb{R}^3 such that its divergence is equal to one. For the purposes of this paper, we choose:

𝐅(x,y,z)=<x,0,0>\mathbf{F}(x, y, z) = <x, 0, 0>

It can easily be verified that

div𝐅=βˆ‚Fβˆ‚x+βˆ‚Fβˆ‚y+βˆ‚Fβˆ‚z=1+0+0=1\mathrm{div} \mathbf{F} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} = 1 + 0 + 0 = 1


V=∭R1dV=∭Rdiv𝐅(x,y,z)dVV = \iiint_R 1 dV = \iiint_R \mathrm{div} \mathbf{F}(x, y, z) \mathrm{d}V

By the Divergence Theorem, this is equal to the surface integral:

V=∬S𝐅(x,y,z)d𝐒V = \iint_S \mathbf{F}(x, y, z) \mathrm{d}\mathbf{S}

This surface integral, defined over the surface S of the 3D mesh, is equal to the sum of its piecewise triangle parts. Let TiT_i denote the surface of the ii’th triangle in the mesh. Then,

V=βˆ‘i=0∬Ti𝐅(x,y,z)d𝐒V = \sum_{i = 0} \iint_{T_i} \mathbf{F}(x, y, z) \mathrm{d}\mathbf{S}

Let TinT_{in} represent the nn’th vertice of the ii’th triangle. Let Ξ”1\Delta_1 equal the vector difference between Ti1T_{i1} and Ti0T{i0}, and Ξ”2\Delta_2 likewise equal to Ti2βˆ’Ti0T_{i2} - T{i0}. Each individual triangle TiT_i may thus be parametrised as:

𝐫(u,v)=Ti0+uΞ”1+vΞ”2\mathbf{r}(u, v) = T_{i0} + u\Delta_1 + v\Delta_2

Then, simple differentiation yields:

𝐫u=Ξ”1\mathbf{r}_u = \Delta_1 𝐫v=Ξ”2\mathbf{r}_v = \Delta_2


𝐫u×𝐫v=Ξ”1Γ—Ξ”2\mathbf{r}_u \times \mathbf{r}_v = \Delta_1 \times \Delta_2

Thus, the surface integral can be rewritten in terms of this parametrisation, substituting in the definition of 𝐅\mathbf{F} as needed:

V=βˆ‘i=0∬Ti𝐅(x,y,z)(𝐫u×𝐫v)dAV = \sum_{i = 0} \iint_{T_i} \mathbf{F}(x, y, z) (\mathbf{r}_u \times \mathbf{r}_v) dA =βˆ‘i=0∬Ti𝐅(x,y,z)(Μ‡Ξ”i1Γ—Ξ”i2)dA= \sum_{i = 0} \iint_{T_i} \mathbf{F}(x, y, z) \dot (\Delta_{i1} \times \Delta_{i2}) dA =βˆ‘i=0∬Ti<x,0,0>(Μ‡Ξ”i1Γ—Ξ”i2)dA= \sum_{i = 0} \iint_{T_i} <x, 0, 0> \dot (\Delta_{i1} \times \Delta_{i2}) dA

This cross product is constant throughout the triangle and easy to calculate from the vertex data. Only the X component of the cross product should be calculated; the others are equal to zero due to the dot product with the zero components of 𝐅\mathbf{F}. VV can be thus be rewritten as:

V=βˆ‘i=0(Ξ”i1Γ—Ξ”i2)x∬TixdAV = \sum_{i = 0} (\Delta_{i1} \times \Delta_{i2})_x \iint_{T_i} x dA

We now focus on the surface integral ∬TixdA\iint_{T_i} x dA. Expanding with the parametrisation yields:

∬TixdA=∫01∫0uxdvdu=∫01∫0u(Ti0x+uΞ”i1x+vΞ”i2x)dvdu\iint_{T_i} x dA = \int_{0}^{1} \int_{0}^{u} x dv du = \int_{0}^{1} \int_{0}^{u} (T_{i0x} + u \Delta_{i1x} + v \Delta_{i2x}) dv du

This integral can be directly evaluated, treating vertex data as constants:

∫01∫01βˆ’u(Ti0x+uΞ”i1x+vΞ”i2x)dvdu\int_{0}^{1} \int_{0}^{1-u} (T_{i0x} + u \Delta_{i1x} + v \Delta_{i2x}) dv du =Ti0x∫01∫01βˆ’udvdu+Ξ”i1x∫01∫01βˆ’uudvdu+Ξ”i2x)∫01∫01βˆ’uvdvdu= T_{i0x} \int_{0}^{1} \int_{0}^{1-u} dv du + \Delta_{i1x} \int_{0}^{1} \int_{0}^{1-u} u dv du + \Delta_{i2x}) \int_{0}^{1} \int_{0}^{1-u} v dv du =Ti0x(12)+Ξ”i1x(16)+Ξ”i2x(16)= T_{i0x} (\frac{1}{2}) + \Delta_{i1x} (\frac{1}{6}) + \Delta_{i2x} (\frac{1}{6}) =Ti0x(12)+(Ti1xβˆ’Ti0x)(16)+(Ti2xβˆ’Ti0x)(16)= T_{i0x} (\frac{1}{2}) + (T_{i1x} - T_{i0x})(\frac{1}{6}) + (T_{i2x} - T_{i0x})(\frac{1}{6}) =Ti0x(16)+(Ti1x)(16)+(Ti2x)(16)= T_{i0x} (\frac{1}{6}) + (T_{i1x})(\frac{1}{6}) + (T_{i2x})(\frac{1}{6}) =16(Ti0x+Ti1x+Ti2x)= \frac{1}{6}(T_{i0x} + T_{i1x} + T_{i2x})

Substituting into the original sum and pulling out a constant factor of 16\frac{1}{6} to avoid the inner loop division, this yields the following compact formula for the volume:

V=16βˆ‘i=0(Ξ”i1Γ—Ξ”i2)x(Ti0x+Ti1x+Ti2x)V = \frac{1}{6} \sum_{i = 0} (\Delta_{i1} \times \Delta_{i2})_x (T_{i0x} + T_{i1x} + T_{i2x})

A reference implementation is available under the AGPLv3.

Performance analysis

The final algorithm contains no numerical integration nor differentiation. In contrast to common naive algorithms for volume, which are equivalent to rendering the mesh and then sampling the render, an expensive operation, there is only a single loop in this algorithm, over the triangles. Thus, this algorithm for volume computation is O(n) to the number of the triangles. Furthermore, the per-triangle calculation is similarly efficient: given the natural expansion of the cross product, the inner part contains seven additions and three multiplications. On the outside of the loop is only a single multiplication. Thus, for a mesh of nn triangles, the algorithm requires 8nβˆ’18n - 1 additions and 3n+13n + 1 multiplications, or 11n11n floating point operations. This is very fast.

For a ballpark number, if volume needs to be calculated every frame in a high-performance 60 frames per second application, without the aid of a GPU, only using the CPU capabilities of a $35 Raspberry Pi, around 30 million triangles could be measured every frame.


The vector calculus exam is soon, and I need to study. Plus, who doesn’t love 3D graphics?!

I would be (pleasantly) surprised if the algorithm is novel. Further research after posting reveals the paper Efficient Feature Extraction for 2D/3D Objects in Mesh Representation by Cha Zheng and Tsuhan Chen, which appears to describe the same algorithm, although the derivation is different. It was fun while it lasted!

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