Trilinear interpolation

Trilinear interpolation is a method of multivariate interpolation on a 3-dimensional regular grid. It approximates the value of a function at an intermediate point ${\displaystyle (x,y,z)}${\displaystyle (x,y,z)} within the local axial rectangular prism linearly, using function data on the lattice points. Trilinear interpolation is frequently used in numerical analysis, data analysis, and computer graphics.

Trilinear interpolation is the extension of linear interpolation, which operates in spaces with dimension ${\displaystyle D=1}${\displaystyle D=1}, and bilinear interpolation, which operates with dimension ${\displaystyle D=2}${\displaystyle D=2}, to dimension ${\displaystyle D=3}${\displaystyle D=3}. These interpolation schemes all use polynomials of order 1, giving an accuracy of order 2, and it requires ${\displaystyle 2^{D}=8}${\displaystyle 2^{D}=8} adjacent pre-defined values surrounding the interpolation point. There are several ways to arrive at trilinear interpolation, which is equivalent to 3-dimensional tensor B-spline interpolation of order 1, and the trilinear interpolation operator is also a tensor product of 3 linear interpolation operators.

For an arbitrary, unstructured mesh (as used in finite element analysis), other methods of interpolation must be used; if all the mesh elements are tetrahedra (3D simplices), then barycentric coordinates provide a straightforward procedure.

On a periodic and cubic lattice, we want the value at ${\displaystyle x}${\displaystyle x}, ${\displaystyle y}${\displaystyle y}, ${\displaystyle z}${\displaystyle z}. In the general case, each coordinate (for example,${\displaystyle x}${\displaystyle x}) is not exactly at a lattice point, but some distance between one lattice point ${\displaystyle x_{\text{0}}}${\displaystyle x_{\text{0}}} and the next, ${\displaystyle x_{\text{1}}}${\displaystyle x_{\text{1}}}. Let ${\displaystyle x_{\text{d}}}${\displaystyle x_{\text{d}}} that be that fractional distance away from the lower lattice point: ${\displaystyle {\frac {x-x_{0}}{x_{1}-x_{0}}}}${\displaystyle {\frac {x-x_{0}}{x_{1}-x_{0}}}}. Take a similar approach for the other coordinates:

${\displaystyle {\begin{aligned}x_{\text{d}}={\frac {x-x_{0}}{x_{1}-x_{0}}}\\y_{\text{d}}={\frac {y-y_{0}}{y_{1}-y_{0}}}\\z_{\text{d}}={\frac {z-z_{0}}{z_{1}-z_{0}}}\end{aligned}}}${\displaystyle {\begin{aligned}x_{\text{d}}={\frac {x-x_{0}}{x_{1}-x_{0}}}\\y_{\text{d}}={\frac {y-y_{0}}{y_{1}-y_{0}}}\\z_{\text{d}}={\frac {z-z_{0}}{z_{1}-z_{0}}}\end{aligned}}}

First one interpolates along ${\displaystyle x}${\displaystyle x} (imagine one is "pushing" the face of the cube defined by ${\displaystyle C_{0jk}}${\displaystyle C_{0jk}} to the opposing face, defined by ${\displaystyle C_{1jk}}${\displaystyle C_{1jk}}), giving:

${\displaystyle {\begin{aligned}c_{00}&=c_{000}(1-x_{\text{d}})+c_{100}x_{\text{d}}\\c_{01}&=c_{001}(1-x_{\text{d}})+c_{101}x_{\text{d}}\\c_{10}&=c_{010}(1-x_{\text{d}})+c_{110}x_{\text{d}}\\c_{11}&=c_{011}(1-x_{\text{d}})+c_{111}x_{\text{d}}\end{aligned}}}${\displaystyle {\begin{aligned}c_{00}&=c_{000}(1-x_{\text{d}})+c_{100}x_{\text{d}}\\c_{01}&=c_{001}(1-x_{\text{d}})+c_{101}x_{\text{d}}\\c_{10}&=c_{010}(1-x_{\text{d}})+c_{110}x_{\text{d}}\\c_{11}&=c_{011}(1-x_{\text{d}})+c_{111}x_{\text{d}}\end{aligned}}}

Where ${\displaystyle c_{000}}${\displaystyle c_{000}} means the function value of ${\displaystyle (x_{0},y_{0},z_{0}).}${\displaystyle (x_{0},y_{0},z_{0}).} Then one interpolates these values (along ${\displaystyle y}${\displaystyle y}, "pushing" from ${\displaystyle C_{i0k}}${\displaystyle C_{i0k}} to ${\displaystyle C_{i1k}}${\displaystyle C_{i1k}}), giving:

${\displaystyle {\begin{aligned}c_{0}&=c_{00}(1-y_{\text{d}})+c_{10}y_{\text{d}}\\c_{1}&=c_{01}(1-y_{\text{d}})+c_{11}y_{\text{d}}\end{aligned}}}${\displaystyle {\begin{aligned}c_{0}&=c_{00}(1-y_{\text{d}})+c_{10}y_{\text{d}}\\c_{1}&=c_{01}(1-y_{\text{d}})+c_{11}y_{\text{d}}\end{aligned}}}

Finally one interpolates these values along ${\displaystyle z}${\displaystyle z} (walking through a line):

${\displaystyle c=c_{0}(1-z_{\text{d}})+c_{1}z_{\text{d}}.}${\displaystyle c=c_{0}(1-z_{\text{d}})+c_{1}z_{\text{d}}.}

This gives a predicted value for the point, which can also be written as follows:${\displaystyle {\begin{aligned}c&=c_{000}(1-x_{d})(1-y_{d})(1-z_{d})\\&\quad +c_{100},円x_{d}(1-y_{d})(1-z_{d})\\&\quad +c_{010}(1-x_{d}),円y_{d}(1-z_{d})\\&\quad +c_{110},円x_{d},円y_{d}(1-z_{d})\\&\quad +c_{001}(1-x_{d})(1-y_{d}),円z_{d}\\&\quad +c_{101},円x_{d}(1-y_{d}),円z_{d}\\&\quad +c_{011}(1-x_{d}),円y_{d},円z_{d}\\&\quad +c_{111},円x_{d},円y_{d},円z_{d}\end{aligned}}}${\displaystyle {\begin{aligned}c&=c_{000}(1-x_{d})(1-y_{d})(1-z_{d})\\&\quad +c_{100},円x_{d}(1-y_{d})(1-z_{d})\\&\quad +c_{010}(1-x_{d}),円y_{d}(1-z_{d})\\&\quad +c_{110},円x_{d},円y_{d}(1-z_{d})\\&\quad +c_{001}(1-x_{d})(1-y_{d}),円z_{d}\\&\quad +c_{101},円x_{d}(1-y_{d}),円z_{d}\\&\quad +c_{011}(1-x_{d}),円y_{d},円z_{d}\\&\quad +c_{111},円x_{d},円y_{d},円z_{d}\end{aligned}}}

This makes it obvious that result of trilinear interpolation is independent of the order of the interpolation steps along the three axes: any other order, for instance along ${\displaystyle y}${\displaystyle y}, then along ${\displaystyle z}${\displaystyle z}, and finally along ${\displaystyle x}${\displaystyle x}, produces the same value.

The above operations can be visualized as follows: First we find the eight corners of a cube that surround our point of interest. These corners have the values ${\displaystyle c_{000}}${\displaystyle c_{000}}, ${\displaystyle c_{100}}${\displaystyle c_{100}}, ${\displaystyle c_{010}}${\displaystyle c_{010}}, ${\displaystyle c_{110}}${\displaystyle c_{110}}, ${\displaystyle c_{001}}${\displaystyle c_{001}}, ${\displaystyle c_{101}}${\displaystyle c_{101}}, ${\displaystyle c_{011}}${\displaystyle c_{011}}, ${\displaystyle c_{111}}${\displaystyle c_{111}}.

Next, we perform linear interpolation between ${\displaystyle c_{000}}${\displaystyle c_{000}} and ${\displaystyle c_{100}}${\displaystyle c_{100}} to find ${\displaystyle c_{00}}${\displaystyle c_{00}}, ${\displaystyle c_{001}}${\displaystyle c_{001}} and ${\displaystyle c_{101}}${\displaystyle c_{101}} to find ${\displaystyle c_{01}}${\displaystyle c_{01}}, ${\displaystyle c_{011}}${\displaystyle c_{011}} and ${\displaystyle c_{111}}${\displaystyle c_{111}} to find ${\displaystyle c_{11}}${\displaystyle c_{11}}, ${\displaystyle c_{010}}${\displaystyle c_{010}} and ${\displaystyle c_{110}}${\displaystyle c_{110}} to find ${\displaystyle c_{10}}${\displaystyle c_{10}}.

Now we do interpolation between ${\displaystyle c_{00}}${\displaystyle c_{00}} and ${\displaystyle c_{10}}${\displaystyle c_{10}} to find ${\displaystyle c_{0}}${\displaystyle c_{0}}, ${\displaystyle c_{01}}${\displaystyle c_{01}} and ${\displaystyle c_{11}}${\displaystyle c_{11}} to find ${\displaystyle c_{1}}${\displaystyle c_{1}}. Finally, we calculate the value ${\displaystyle c}${\displaystyle c} via linear interpolation of ${\displaystyle c_{0}}${\displaystyle c_{0}} and ${\displaystyle c_{1}}${\displaystyle c_{1}}

In practice, a trilinear interpolation is identical to two bilinear interpolation combined with a linear interpolation:

${\displaystyle c\approx l\left(b(c_{000},c_{010},c_{100},c_{110}),,円b(c_{001},c_{011},c_{101},c_{111})\right)}${\displaystyle c\approx l\left(b(c_{000},c_{010},c_{100},c_{110}),,円b(c_{001},c_{011},c_{101},c_{111})\right)}

An alternative way to write the solution to the interpolation problem is

${\displaystyle f(x,y,z)\approx a_{0}+a_{1}x+a_{2}y+a_{3}z+a_{4}xy+a_{5}xz+a_{6}yz+a_{7}xyz}${\displaystyle f(x,y,z)\approx a_{0}+a_{1}x+a_{2}y+a_{3}z+a_{4}xy+a_{5}xz+a_{6}yz+a_{7}xyz}

where the coefficients are found by solving the linear system

${\displaystyle {\begin{aligned}{\begin{bmatrix}1&x_{0}&y_{0}&z_{0}&x_{0}y_{0}&x_{0}z_{0}&y_{0}z_{0}&x_{0}y_{0}z_{0}\1円&x_{1}&y_{0}&z_{0}&x_{1}y_{0}&x_{1}z_{0}&y_{0}z_{0}&x_{1}y_{0}z_{0}\1円&x_{0}&y_{1}&z_{0}&x_{0}y_{1}&x_{0}z_{0}&y_{1}z_{0}&x_{0}y_{1}z_{0}\1円&x_{1}&y_{1}&z_{0}&x_{1}y_{1}&x_{1}z_{0}&y_{1}z_{0}&x_{1}y_{1}z_{0}\1円&x_{0}&y_{0}&z_{1}&x_{0}y_{0}&x_{0}z_{1}&y_{0}z_{1}&x_{0}y_{0}z_{1}\1円&x_{1}&y_{0}&z_{1}&x_{1}y_{0}&x_{1}z_{1}&y_{0}z_{1}&x_{1}y_{0}z_{1}\1円&x_{0}&y_{1}&z_{1}&x_{0}y_{1}&x_{0}z_{1}&y_{1}z_{1}&x_{0}y_{1}z_{1}\1円&x_{1}&y_{1}&z_{1}&x_{1}y_{1}&x_{1}z_{1}&y_{1}z_{1}&x_{1}y_{1}z_{1}\end{bmatrix}}{\begin{bmatrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\\a_{4}\\a_{5}\\a_{6}\\a_{7}\end{bmatrix}}={\begin{bmatrix}c_{000}\\c_{100}\\c_{010}\\c_{110}\\c_{001}\\c_{101}\\c_{011}\\c_{111}\end{bmatrix}},\end{aligned}}}${\displaystyle {\begin{aligned}{\begin{bmatrix}1&x_{0}&y_{0}&z_{0}&x_{0}y_{0}&x_{0}z_{0}&y_{0}z_{0}&x_{0}y_{0}z_{0}\1円&x_{1}&y_{0}&z_{0}&x_{1}y_{0}&x_{1}z_{0}&y_{0}z_{0}&x_{1}y_{0}z_{0}\1円&x_{0}&y_{1}&z_{0}&x_{0}y_{1}&x_{0}z_{0}&y_{1}z_{0}&x_{0}y_{1}z_{0}\1円&x_{1}&y_{1}&z_{0}&x_{1}y_{1}&x_{1}z_{0}&y_{1}z_{0}&x_{1}y_{1}z_{0}\1円&x_{0}&y_{0}&z_{1}&x_{0}y_{0}&x_{0}z_{1}&y_{0}z_{1}&x_{0}y_{0}z_{1}\1円&x_{1}&y_{0}&z_{1}&x_{1}y_{0}&x_{1}z_{1}&y_{0}z_{1}&x_{1}y_{0}z_{1}\1円&x_{0}&y_{1}&z_{1}&x_{0}y_{1}&x_{0}z_{1}&y_{1}z_{1}&x_{0}y_{1}z_{1}\1円&x_{1}&y_{1}&z_{1}&x_{1}y_{1}&x_{1}z_{1}&y_{1}z_{1}&x_{1}y_{1}z_{1}\end{bmatrix}}{\begin{bmatrix}a_{0}\\a_{1}\\a_{2}\\a_{3}\\a_{4}\\a_{5}\\a_{6}\\a_{7}\end{bmatrix}}={\begin{bmatrix}c_{000}\\c_{100}\\c_{010}\\c_{110}\\c_{001}\\c_{101}\\c_{011}\\c_{111}\end{bmatrix}},\end{aligned}}}

yielding the result

${\displaystyle {\begin{aligned}a_{0}={}&{\frac {-c_{000}x_{1}y_{1}z_{1}+c_{001}x_{1}y_{1}z_{0}+c_{010}x_{1}y_{0}z_{1}-c_{011}x_{1}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {c_{100}x_{0}y_{1}z_{1}-c_{101}x_{0}y_{1}z_{0}-c_{110}x_{0}y_{0}z_{1}+c_{111}x_{0}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{1}={}&{\frac {c_{000}y_{1}z_{1}-c_{001}y_{1}z_{0}-c_{010}y_{0}z_{1}+c_{011}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}y_{1}z_{1}+c_{101}y_{1}z_{0}+c_{110}y_{0}z_{1}-c_{111}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{2}={}&{\frac {c_{000}x_{1}z_{1}-c_{001}x_{1}z_{0}-c_{010}x_{1}z_{1}+c_{011}x_{1}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}x_{0}z_{1}+c_{101}x_{0}z_{0}+c_{110}x_{0}z_{1}-c_{111}x_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{3}={}&{\frac {c_{000}x_{1}y_{1}-c_{001}x_{1}y_{1}-c_{010}x_{1}y_{0}+c_{011}x_{1}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}x_{0}y_{1}+c_{101}x_{0}y_{1}+c_{110}x_{0}y_{0}-c_{111}x_{0}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{4}={}&{\frac {-c_{000}z_{1}+c_{001}z_{0}+c_{010}z_{1}-c_{011}z_{0}+c_{100}z_{1}-c_{101}z_{0}-c_{110}z_{1}+c_{111}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{5}=&{\frac {-c_{000}y_{1}+c_{001}y_{1}+c_{010}y_{0}-c_{011}y_{0}+c_{100}y_{1}-c_{101}y_{1}-c_{110}y_{0}+c_{111}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{6}={}&{\frac {-c_{000}x_{1}+c_{001}x_{1}+c_{010}x_{1}-c_{011}x_{1}+c_{100}x_{0}-c_{101}x_{0}-c_{110}x_{0}+c_{111}x_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{7}={}&{\frac {c_{000}-c_{001}-c_{010}+c_{011}-c_{100}+c_{101}+c_{110}-c_{111}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}.\end{aligned}}}${\displaystyle {\begin{aligned}a_{0}={}&{\frac {-c_{000}x_{1}y_{1}z_{1}+c_{001}x_{1}y_{1}z_{0}+c_{010}x_{1}y_{0}z_{1}-c_{011}x_{1}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {c_{100}x_{0}y_{1}z_{1}-c_{101}x_{0}y_{1}z_{0}-c_{110}x_{0}y_{0}z_{1}+c_{111}x_{0}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{1}={}&{\frac {c_{000}y_{1}z_{1}-c_{001}y_{1}z_{0}-c_{010}y_{0}z_{1}+c_{011}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}y_{1}z_{1}+c_{101}y_{1}z_{0}+c_{110}y_{0}z_{1}-c_{111}y_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{2}={}&{\frac {c_{000}x_{1}z_{1}-c_{001}x_{1}z_{0}-c_{010}x_{1}z_{1}+c_{011}x_{1}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}x_{0}z_{1}+c_{101}x_{0}z_{0}+c_{110}x_{0}z_{1}-c_{111}x_{0}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{3}={}&{\frac {c_{000}x_{1}y_{1}-c_{001}x_{1}y_{1}-c_{010}x_{1}y_{0}+c_{011}x_{1}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}+{}\\&{\frac {-c_{100}x_{0}y_{1}+c_{101}x_{0}y_{1}+c_{110}x_{0}y_{0}-c_{111}x_{0}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{4}={}&{\frac {-c_{000}z_{1}+c_{001}z_{0}+c_{010}z_{1}-c_{011}z_{0}+c_{100}z_{1}-c_{101}z_{0}-c_{110}z_{1}+c_{111}z_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{5}=&{\frac {-c_{000}y_{1}+c_{001}y_{1}+c_{010}y_{0}-c_{011}y_{0}+c_{100}y_{1}-c_{101}y_{1}-c_{110}y_{0}+c_{111}y_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{6}={}&{\frac {-c_{000}x_{1}+c_{001}x_{1}+c_{010}x_{1}-c_{011}x_{1}+c_{100}x_{0}-c_{101}x_{0}-c_{110}x_{0}+c_{111}x_{0}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}},\\[4pt]a_{7}={}&{\frac {c_{000}-c_{001}-c_{010}+c_{011}-c_{100}+c_{101}+c_{110}-c_{111}}{(x_{0}-x_{1})(y_{0}-y_{1})(z_{0}-z_{1})}}.\end{aligned}}}

• Linear interpolation
• Bilinear interpolation
• Tricubic interpolation
• Radial interpolation
• Tetrahedral interpolation
• Spherical linear interpolation

• pseudo-code from NASA, describes an iterative inverse trilinear interpolation (given the vertices and the value of C find Xd, Yd and Zd).
• Paul Bourke, Interpolation methods, 1999. Contains a very clever and simple method to find trilinear interpolation that is based on binary logic and can be extended to any dimension (Tetralinear, Pentalinear, ...).
• Kenwright, Free-Form Tetrahedron Deformation. International Symposium on Visual Computing. Springer International Publishing, 2015 [1].

• Multivariate interpolation
• Euclidean solid geometry

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