**Mathematical Derivation of the Equations (16) and (17).**

Equation (16) can be explained using the plane vector method. Let # # # *X<sup>L</sup>* (*k*+1,*o*) <sup>−</sup> *<sup>X</sup><sup>L</sup>* (*k*,*a*) × *X<sup>L</sup>* (*k*+1,*o*) <sup>−</sup> *<sup>X</sup><sup>L</sup>* (*k*,*b*) # # # be the area of the parallelogram formed by three points *X<sup>L</sup>* (*k*+1,*o*) , *X<sup>L</sup>* (*k*,*a*) and *<sup>X</sup><sup>L</sup>* (*k*,*b*) . Let the spatial coordinates of the three points be *X<sup>L</sup>* (*k*+1,*o*) (*x*0, *y*0, *z*0), *X<sup>L</sup>* (*k*,*a*) (*x*1, *y*1, *z*1) and *X<sup>L</sup>* (*k*,*b*) (*x*2, *y*2, *z*2), from which, the three vectors are constructed as:

$$\begin{cases} \begin{aligned} \overline{X^{L}\_{\langle k+1,o\rangle}X^{L}\_{\langle k,o\rangle}} &= X^{L}\_{\langle k+1,o\rangle} - X^{L}\_{\langle k,o\rangle} = (\mathbf{x}\_{1} - \mathbf{x}\_{0}, y\_{1} - y\_{0}, z\_{1} - z\_{0})} \\ \end{aligned} \\\begin{aligned} \begin{aligned} \overline{X^{L}\_{\langle k+1,o\rangle}X^{L}\_{\langle k,b\rangle}} &= X^{L}\_{\langle k+1,o\rangle} - X^{L}\_{\langle k,b\rangle} = (\mathbf{x}\_{2} - \mathbf{x}\_{0}, y\_{2} - y\_{0}, z\_{2} - z\_{0})} \\ \end{aligned} \\\end{cases} \\\begin{aligned} \overline{X^{L}\_{\langle k,o\rangle}X^{L}\_{\langle k,b\rangle}} &= X^{L}\_{\langle k,o\rangle} - X^{L}\_{\langle k,b\rangle} = (\mathbf{x}\_{2} - \mathbf{x}\_{1}, y\_{2} - y\_{1}, z\_{2} - z\_{1}) \end{aligned} \end{cases} \end{cases} \end{cases} \tag{A13}$$

The molecules of Equation (11) can be obtained as follows:

$$\left| \left( X\_{(k+1,\rho)}^{L} - X\_{(k,\rho)}^{L} \right) \times \left( X\_{(k+1,\rho)}^{L} - X\_{(k,b)}^{L} \right) \right| = \begin{vmatrix} X\_{(k+1,\rho)}^{L} & X\_{(k,\rho)}^{L} & X\_{(k,b)}^{L} \\ \left( x\_{1} - x\_{0} \right) & \left( y\_{1} - y\_{0} \right) & \left( z\_{1} - z\_{0} \right) \\ \left( x\_{2} - x\_{0} \right) & \left( y\_{2} - y\_{0} \right) & \left( z\_{2} - z\_{0} \right) \end{vmatrix} \tag{A14}$$

The distance between the point *X<sup>L</sup>* (*k*+1,*o*) and the line *<sup>X</sup><sup>L</sup>* (*k*,*a*) *X<sup>L</sup>* (*k*,*b*) represented by Equation (11) is:

*dL ek* = # # # *X<sup>L</sup>* (*k*+1,*o*) <sup>−</sup>*X<sup>L</sup>* (*k*,*a*) × *X<sup>L</sup>* (*k*+1,*o*) <sup>−</sup>*X<sup>L</sup>* (*k*,*b*) # # # # # #*XL* (*k*,*a*) −*X<sup>L</sup>* (*k*,*b*) # # # = *sqrt*{[(*y*<sup>1</sup> − *y*0) ∗ (*z*<sup>2</sup> − *z*0) − (*y*<sup>2</sup> − *y*0) ∗ (*z*<sup>1</sup> − *z*0)] ∗[(*y*<sup>1</sup> − *y*0) ∗ (*z*<sup>2</sup> − *z*0) − (*y*<sup>2</sup> − *y*0) ∗ (*z*<sup>1</sup> − *z*0)] +[(*x*<sup>2</sup> − *x*0) ∗ (*z*<sup>1</sup> − *z*0) − (*x*<sup>1</sup> − *x*0) ∗ (*z*<sup>2</sup> − *z*0)] ∗[(*x*<sup>2</sup> − *x*0) ∗ (*z*<sup>1</sup> − *z*0) − (*x*<sup>1</sup> − *x*0) ∗ (*z*<sup>2</sup> − *z*0)] +[(*x*<sup>1</sup> − *x*0) ∗ (*y*<sup>2</sup> − *y*0) − (*x*<sup>2</sup> − *x*0) ∗ (*y*<sup>1</sup> − *y*0)] <sup>∗</sup>[(*x*<sup>1</sup> <sup>−</sup> *<sup>x</sup>*0) <sup>∗</sup> (*y*<sup>2</sup> <sup>−</sup> *<sup>y</sup>*0) <sup>−</sup> (*x*<sup>2</sup> <sup>−</sup> *<sup>x</sup>*0) <sup>∗</sup> (*y*<sup>1</sup> <sup>−</sup> *<sup>y</sup>*0)] ] /*sqrt*[(*x*<sup>2</sup> − *x*1) ∗ (*x*<sup>2</sup> − *x*1) + (*y*<sup>2</sup> − *y*1) ∗ (*y*<sup>2</sup> − *y*1) + (*z*<sup>2</sup> − *z*1) ∗ (*z*<sup>2</sup> − *z*1)] (A15)

Similarly, the molecule of Equation (17) can be expressed as the volume of a triangular pyramid composed of four points: *X<sup>L</sup>* (*k*+1,*o*) , *X<sup>L</sup>* (*k*,*c*) , *X<sup>L</sup>* (*k*,*d*) and *<sup>X</sup><sup>L</sup>* (*k*, *<sup>f</sup>*) in a geometric sense. It can be known that # # # *X<sup>L</sup>* (*k*,*c*) <sup>−</sup> *<sup>X</sup><sup>L</sup>* (*k*,*d*) × *X<sup>L</sup>* (*k*,*c*) <sup>−</sup> *<sup>X</sup><sup>L</sup>* (*k*, *f*) # # # is twice the area of the base. Let the spatial coordinates of the four points be *X<sup>L</sup>* (*k*+1,*ρ*) (*x*3, *y*3, *z*3), *X<sup>L</sup>* (*k*,*c*) (*x*4, *y*4, *z*4), *X<sup>L</sup>* (*k*,*d*) (*x*5, *y*5, *z*5) and *X<sup>L</sup>* (*k*, *f*) (*x*6, *y*6, *z*6), so the three vectors required for constructing are:

$$\begin{cases} \begin{aligned} \overline{X^{L}\_{\langle k+1,\rho\rangle}X^{L}\_{\langle k,d\rangle}} = X^{L}\_{\langle k+1,\rho\rangle} - X^{L}\_{\langle k,d\rangle} = (\mathbf{x}\_{5} - \mathbf{x}\_{3}, y\_{5} - y\_{3}, z\_{5} - z\_{3})} \\ \end{aligned} \\\begin{aligned} \begin{aligned} \overline{X^{L}\_{\langle k,\varepsilon\rangle}X^{L}\_{\langle k,d\rangle}} = X^{L}\_{\langle k,\varepsilon\rangle} - X^{L}\_{\langle k,d\rangle} = (\mathbf{x}\_{5} - \mathbf{x}\_{4}, y\_{5} - y\_{4}, z\_{5} - z\_{4})} \\ \end{aligned} \\\end{cases} \\\begin{aligned} \overline{X^{L}\_{\langle k,\varepsilon\rangle}X^{L}\_{\langle k,f\rangle}} = X^{L}\_{\langle k,\varepsilon\rangle} - X^{L}\_{\langle k,f\rangle} = (\mathbf{x}\_{6} - \mathbf{x}\_{4}, y\_{6} - y\_{4}, z\_{6} - z\_{4}) \end{aligned} \end{cases} \end{cases} \end{cases} \tag{A16}$$

The molecules of Equation (17) can be obtained as follows:

$$\begin{aligned} \left| \left( X\_{(\vec{k},\varepsilon)}^{L} - X\_{(\vec{k},d)}^{L} \right) \times \left( X\_{(\vec{k},\varepsilon)}^{L} - X\_{(\vec{k},f)}^{L} \right) \right| &= \begin{array}{c} \left| X\_{(\vec{k},\varepsilon)}^{L} - X\_{(\vec{k},d)}^{L} \right| \quad & \left| X\_{(\vec{k},f)}^{L} \right| \\\\ \left( \mathbf{x}\_{5} - \mathbf{x}\_{4} \right) \quad & \left( \mathbf{y}\_{5} - \mathbf{y}\_{4} \right) \quad \left( \mathbf{z}\_{5} - \mathbf{z}\_{4} \right) \\\\ \left( \mathbf{x}\_{6} - \mathbf{x}\_{4} \right) \quad & \left( \mathbf{y}\_{6} - \mathbf{y}\_{4} \right) \quad \left( \mathbf{z}\_{6} - \mathbf{z}\_{4} \right) \end{array} \right| \\ &= \operatorname{spt}(\mathbf{s}\_{6} \ast \mathbf{s}\_{4}, \mathbf{s}\_{6} \ast \mathbf{s}\_{6}, \mathbf{s}\_{5} \ast \mathbf{s}\_{6}) \end{aligned} \tag{A17}$$

where *sa*, *sb* and *sc* represent the component vectors of x, y and z axes, respectively:

$$\begin{cases} s\_{\mathfrak{g}} = (y\_5 - y\_4) \ast (z\_6 - z\_4) - (y\_6 - y\_4) \ast (z\_5 - z\_4) \\\ s\_{\mathfrak{b}} = (z\_5 - z\_4) \ast (x\_6 - x\_4) - (z\_6 - z\_4) \ast (x\_5 - x\_4) \\\ s\_{\mathfrak{c}} = (x\_5 - x\_4) \ast (y\_6 - y\_4) - (x\_6 - x\_4) \ast (y\_5 - y\_4) \end{cases} \tag{A18}$$

Therefore, the point-to-surface distance can be obtained as follows:

$$\begin{split}d\_{pk}^{L} &= \frac{\left| \left( \mathbf{X}\_{(k+1,\rho)}^{L} - \mathbf{X}\_{(k,d)}^{L} \right) \cdot \left( \left( \mathbf{X}\_{(k,c)}^{L} - \mathbf{X}\_{(k,d)}^{L} \right) \times \left( \mathbf{X}\_{(k,c)}^{L} - \mathbf{X}\_{(k,f)}^{L} \right) \right) \right|}{\left| \left( \mathbf{X}\_{(k,c)}^{L} - \mathbf{X}\_{(k,d)}^{L} \right) \times \left( \mathbf{X}\_{(k,c)}^{L} - \mathbf{X}\_{(k,f)}^{L} \right) \right|} \\ &= \frac{\left( \mathbf{x}\_{4} - \mathbf{x}\_{3} \right) \ast \mathbf{s}\_{a} + \left( \mathbf{y}\_{4} - \mathbf{y}\_{3} \right) \ast \mathbf{s}\_{b} + \left( \mathbf{z}\_{4} - \mathbf{z}\_{3} \right) \ast \mathbf{s}\_{c}}{\left| \operatorname{spt} \right( \mathbf{s}\_{a} \ast \mathbf{s}\_{s}, \mathbf{s}\_{b} \ast \mathbf{s}\_{c} \ast \mathbf{s}\_{c}) \right|} \end{split} \tag{A19}$$
