Surface Construction

The general problem of determining the boundary curves and the surface curves are special cases of the following interpolation problem: given two points ${\bf p}_0$ and ${\bf p}_1$, as well as associated normals ${\bf n}_0$ and ${\bf n}_1$, find a function of the form

$${\bf p}(t) = (1 - t) {\bf p}_0 + t {\bf p}_1 + g(t)\left[(1 - t){\bf n}_0 + t{\bf n}_1\right].$$

which interpolates surface points ${\bf p}_0$ and ${\bf p}_1$ with tangent normal to ${\bf n}_0$ and ${\bf n}_1$ at the endpoints. If ${\bf p}(t)$ is to interpolate the endpoints, then $g(0) = g(1) = 0$. We compute the required values for $g'(t)$ at the endpoints and apply Hermite interpolation.

Derivative interpolation requires that the tangents ${\bf p}'(0)$ and ${\bf p}'(1)$ lie in the plane normal to the ${\bf n}_0$ and ${\bf n}_1$, respectively, which amounts to requiring ${\bf p}'(0)\cdot{\bf n}_0 = 0$ and ${\bf p}'(1)\cdot{\bf n}_1 = 0$. We have

$${\bf p}'(t) = {\bf p}_1 - {\bf p}_0 + g(t)({\bf n}_1 - {\bf n}_0) + g'(t)\left[(1 - t){\bf n}_0 + t{\bf n}_1\right]$$

and consequently

$$g'(0) =\!\!\!\! \frac{\left[{\bf p}_0 - {\bf p}_1 + g(0)({\bf n}_0 - {\bf n}_1)\r... ...ot{\bf n}_0} = \left[{\bf p}_0 - {\bf p}_1 \right]\cdot{\bf n}_0\makebox[1ex]{}$$ (1)

$$g'(1) =\!\!\!\! \frac{\left[{\bf p}_0 - {\bf p}_1 + g(1)({\bf n}_0 - {\bf n}_1)\r... ...t{\bf n}_1} = \left[{\bf p}_0 - {\bf p}_1 \right]\cdot{\bf n}_1.\makebox[1ex]{}$$ (2)

The height function $g(t)$ is then constructed from the endpoint derivative values and the Hermite basis functions:

$$g(t) = g'(0) H_1^3(t) - g'(1) H_2^3(t)$$ (3)

where $H_i^3$ are two of the cubic Hermite basis functions:

$$H_1^3(t) = (1 - t)^2 t$$

$$H_2^3(t) = (1 - t) t^2.$$