Appendix: Proof of Smoothing Properties

To prove that the blending does create the proper surface function, let $h_b$ be the blended height function defined in equation (11), and suppose $h$ is an arbitrary height function with the desired boundary derivatives (consistent with the surface normal given in equation (10).) We show the derivatives of $h_b$ match those of $h$. We will restrict attention to the edge with $\gamma = 0$, and the other edges will follow from symmetry. For simplicity, we write $h_i$ for ${\cal P}_ih$, and leave off the explicit dependence on the barycentric coordinates in all expressions.

On the edge $\gamma = 0$ we have $\alpha + \beta = 1$ and as $f(\beta ) = f(1 - \alpha ) = 1 - f(\alpha )$ we have $f(\alpha ) + f(\beta ) = 1$ and $\sigma(\alpha , \beta , 0) = 1$. Thus $b_0 = f(\alpha )$, $b_1 = f(\beta )$ and $b_2 = f(\gamma ) = 0$. Consequently, we have

$$h_b = b_0 h_0 + b_1 h_1$$

$$= f(\alpha ) h_0 + f(\beta ) h_1$$

$$= \left[f(\alpha ) + f(\beta )\right] h$$

$$= h$$

the last equality coming from the fact both $h_0(\alpha ,\beta ,0)$ and $h_1(\alpha ,\beta ,0)$ interpolate the boundary height function on the edge.

We prove the derivative interpolation by showing all the directional derivatives of the blended $h_b$ surface match those of the boundary surface $h$ itself. In barycentric coordinates, a directional derivative in the direction ${\bf d}= (d_\alpha , d_\beta , d_\gamma )$ (where the components sum to zero) is

$$h_{\bf d}= d_\alpha \frac{\partial h}{\partial \alpha } + d_\beta... ...c{\partial h}{\partial \beta } + d_\gamma \frac{\partial h}{\partial \gamma }$$

It therefore suffices to show the partials of the blended height function $h_b$ are equal to those of the boundary height partials. (The partials of the boundary function $h$ have not been explicitly formulated, only implicitly in terms of the edge surface normal, but their explicit values are not needed to show that they match those of the blended function $h_b$.)

The partials of the blended surface evaluate to

$$\frac{\partial h_b}{\partial \alpha } =\!\!\! b_0 \frac{\partial h_0}{\partial \alpha } + \frac{\partial b_0}{\... ...rac{\partial h_2}{\partial \alpha } + \frac{\partial b_1}{\partial \alpha } h_2$$

$$\frac{\partial h_b}{\partial \beta } =\!\!\! b_0 \frac{\partial h_0}{\partial \beta } + \frac{\partial b_0}{\p... ...\frac{\partial h_2}{\partial \beta } + \frac{\partial b_1}{\partial \beta } h_2$$

$$\frac{\partial h_b}{\partial \gamma } =\!\!\! b_0 \frac{\partial h_0}{\partial \gamma } + \frac{\partial b_0}{\... ...rac{\partial h_2}{\partial \gamma } + \frac{\partial b_1}{\partial \gamma } h_2$$

The partials of the blending functions on the edge are

$$\begin{array}{lll} \displaystyle \frac{\partial b_0}{\partia... ...tyle \frac{\partial b_2}{\partial \gamma } \!=\! 0. \end{array}$$

Substitution, combined with the identities $f(\alpha ) + f(\beta ) = 1$, and the observation that the relevant partials of $h_0$ and $h_1$ are equal to those of $h$ on the edge (by construction) it follows that

$$\frac{\partial h_b}{\partial \alpha } = \frac{\partial h}{\partia... ...{} \frac{\partial h_b}{\partial \gamma } = \frac{\partial h}{\partial \gamma }$$

and thus the blending produces the proper surface.