• Images
• Videos
• Documents
• Books
• Articles

Related methods

Trilinear interpolation is the extension of linear interpolation, which operates in spaces with dimension D = 1 {\displaystyle D=1} , and bilinear interpolation, which operates with dimension D = 2 {\displaystyle D=2} , to dimension D = 3 {\displaystyle D=3} . These interpolation schemes all use polynomials of order 1, giving an accuracy of order 2, and it requires 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.

Formulation

On a periodic and cubic lattice, let x d {\displaystyle x_{\text{d}}} , y d {\displaystyle y_{\text{d}}} , and z d {\displaystyle z_{\text{d}}} be the differences between each of x {\displaystyle x} , y {\displaystyle y} , z {\displaystyle z} and the smaller coordinate related, that is:

x d = x − x 0 x 1 − x 0 y d = y − y 0 y 1 − y 0 z d = z − z 0 z 1 − z 0 {\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}}}

where x 0 {\displaystyle x_{0}} indicates the lattice point below x {\displaystyle x} , and x 1 {\displaystyle x_{1}} indicates the lattice point above x {\displaystyle x} and similarly for y 0 , y 1 , z 0 {\displaystyle y_{0},y_{1},z_{0}} and z 1 {\displaystyle z_{1}} .

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

c 00 = c 000 ( 1 − x d ) + c 100 x d c 01 = c 001 ( 1 − x d ) + c 101 x d c 10 = c 010 ( 1 − x d ) + c 110 x d c 11 = c 011 ( 1 − x d ) + c 111 x d {\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 c 000 {\displaystyle c_{000}} means the function value of ( x 0 , y 0 , z 0 ) . {\displaystyle (x_{0},y_{0},z_{0}).} Then one interpolates these values (along y {\displaystyle y} , "pushing" from C i 0 k {\displaystyle C_{i0k}} to C i 1 k {\displaystyle C_{i1k}} ), giving:

c 0 = c 00 ( 1 − y d ) + c 10 y d c 1 = c 01 ( 1 − y d ) + c 11 y d {\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 z {\displaystyle z} (walking through a line):

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

This gives us a predicted value for the point.

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

Algorithm visualization

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 c 000 {\displaystyle c_{000}} , c 100 {\displaystyle c_{100}} , c 010 {\displaystyle c_{010}} , c 110 {\displaystyle c_{110}} , c 001 {\displaystyle c_{001}} , c 101 {\displaystyle c_{101}} , c 011 {\displaystyle c_{011}} , c 111 {\displaystyle c_{111}} .

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

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

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

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

Alternative algorithm

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

f ( x , y , z ) ≈ a 0 + a 1 x + a 2 y + a 3 z + a 4 x y + a 5 x z + a 6 y z + a 7 x y z {\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

[ 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 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 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 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 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 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 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 x 1 y 1 z 1 x 1 y 1 x 1 z 1 y 1 z 1 x 1 y 1 z 1 ] [ a 0 a 1 a 2 a 3 a 4 a 5 a 6 a 7 ] = [ c 000 c 100 c 010 c 110 c 001 c 101 c 011 c 111 ] , {\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&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&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&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&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&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&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&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

a 0 = − 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 ) + 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 ) , a 1 = 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 ) + − 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 ) , a 2 = 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 ) + − 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 ) , a 3 = 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 ) + − 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 ) , a 4 = − 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 ) , a 5 = − 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 ) , a 6 = − 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 ) , a 7 = 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 ) . {\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}}}

See also

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

External links

• 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].