Method of multivariate interpolation on a 3-dimensional regular grid
Trilinear interpolation as two bilinear interpolations followed by a linear 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
(
x
,
y
,
z
)
{\displaystyle (x,y,z)}
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
D
=
1
{\displaystyle D=1}
bilinear interpolation , which operates with dimension
D
=
2
{\displaystyle D=2}
D
=
3
{\displaystyle D=3}
2
D
=
8
{\displaystyle 2^{D}=8}
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.
Eight corner points on a cube surrounding the interpolation point C On a periodic and cubic lattice, let
x
d
{\displaystyle x_{\text{d}}}
y
d
{\displaystyle y_{\text{d}}}
z
d
{\displaystyle z_{\text{d}}}
x
{\displaystyle x}
y
{\displaystyle y}
z
{\displaystyle z}
x
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=
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−
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0
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1
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{\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}}
x
{\displaystyle x}
x
1
{\displaystyle x_{1}}
x
{\displaystyle x}
y
0
,
y
1
,
z
0
{\displaystyle y_{0},y_{1},z_{0}}
z
1
{\displaystyle z_{1}}
First one interpolates along
x
{\displaystyle x}
C
0
j
k
{\displaystyle C_{0jk}}
C
1
j
k
{\displaystyle C_{1jk}}
c
00
=
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000
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1
−
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d
)
+
c
100
x
d
c
01
=
c
001
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1
−
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d
)
+
c
101
x
d
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10
=
c
010
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1
−
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d
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+
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}}
(
x
0
,
y
0
,
z
0
)
.
{\displaystyle (x_{0},y_{0},z_{0}).}
y
{\displaystyle y}
C
i
0
k
{\displaystyle C_{i0k}}
C
i
1
k
{\displaystyle C_{i1k}}
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}
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}
y
{\displaystyle y}
z
{\displaystyle z}
Algorithm visualization [ edit ] A geometric visualisation of trilinear interpolation. The product of the value at the desired point and the entire volume is equal to the sum of the products of the value at each corner and the partial volume diagonally opposite the corner. 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}}
c
100
{\displaystyle c_{100}}
c
00
{\displaystyle c_{00}}
c
001
{\displaystyle c_{001}}
c
101
{\displaystyle c_{101}}
c
01
{\displaystyle c_{01}}
c
011
{\displaystyle c_{011}}
c
111
{\displaystyle c_{111}}
c
11
{\displaystyle c_{11}}
c
010
{\displaystyle c_{010}}
c
110
{\displaystyle c_{110}}
c
10
{\displaystyle c_{10}}
Now we do interpolation between
c
00
{\displaystyle c_{00}}
c
10
{\displaystyle c_{10}}
c
0
{\displaystyle c_{0}}
c
01
{\displaystyle c_{01}}
c
11
{\displaystyle c_{11}}
c
1
{\displaystyle c_{1}}
c
{\displaystyle c}
c
0
{\displaystyle c_{0}}
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 [ edit ] 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
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]
[
a
0
a
1
a
2
a
3
a
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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
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1
−
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011
x
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0
(
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100
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101
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0
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c
110
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+
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111
x
0
y
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0
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−
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1
)
(
y
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1
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(
z
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1
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,
a
1
=
c
000
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−
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001
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−
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010
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0
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1
+
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1
)
(
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1
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(
z
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+
−
c
100
y
1
z
1
+
c
101
y
1
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0
+
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110
y
0
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1
−
c
111
y
0
z
0
(
x
0
−
x
1
)
(
y
0
−
y
1
)
(
z
0
−
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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
−
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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
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0
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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
−
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101
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c
110
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1
+
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111
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0
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x
0
−
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1
)
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1
)
(
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1
)
,
a
5
=
−
c
000
y
1
+
c
001
y
1
+
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010
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−
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011
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0
+
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100
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−
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101
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1
−
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110
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0
+
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111
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0
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x
0
−
x
1
)
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y
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−
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1
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(
z
0
−
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1
)
,
a
6
=
−
c
000
x
1
+
c
001
x
1
+
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010
x
1
−
c
011
x
1
+
c
100
x
0
−
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101
x
0
−
c
110
x
0
+
c
111
x
0
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0
−
x
1
)
(
y
0
−
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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
−
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1
)
(
z
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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}}}
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] . 
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2017eb73468e519de26c930bdde4137584100afb)