build_twist_taper Subroutine

pure subroutine build_twist_taper(this, L, nc, twist_deg, taper)

Apply a z-progressive twist and linear taper to a NURBS hexahedron.

Let the control points be indexed by $k=1,\dots,n_c^{(z)}$ along $z$. Define the normalized axial coordinate $$t(k) = \begin{cases} \dfrac{k-1}{n_c^{(z)}-1}, & n_c^{(z)} > 1,\\[4pt] 0, & \text{otherwise}, \end{cases}$$ which varies from 0 at the bottom face to 1 at the top face.

The total twist angle (in radians) at level $t$ is $$\theta(t) = \Bigl(\dfrac{\pi}{180}\Bigr)\,\texttt{twist\_deg}\; t ,$$ i.e. a linear ramp from $0$ to $\alpha_{\max} = (\pi/180)\,\texttt{twist\_deg}$.

The in-plane (x–y) taper scale is chosen linear in $t$: $$s_{xy}(t) = (1-\texttt{taper})(1-t) + \texttt{taper},$$ so $s_{xy}(0)=1$ at the bottom and $s_{xy}(1)=\texttt{taper}$ at the top (shrinking if $\texttt{taper}<1$).

For each control point $\mathbf{X}_c=(x,y,z)$, first shift to the in-plane centroid $\mathbf{c}_{xy}=(c_x,c_y)$ of the box, apply the scale $s_{xy}(t)$ and rotation by $\theta(t)$ about $\mathbf{c}_{xy}$, and keep $z$ unchanged: $$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix}c_x\\c_y\end{bmatrix} + s_{xy}(t)\, \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} \!\left(\begin{bmatrix}x\\y\end{bmatrix}-\begin{bmatrix}c_x\\c_y\end{bmatrix}\right), \qquad z' = z .$$

Knots are preserved; only the control points are updated.

Arguments