Suppose $\mathbf{F}(u,v)$ is a parametric surface. We can define another surface $\mathbf{G}(u,v)$ by $$ \mathbf{G}(u,v) = \mathbf{F}(u,v) + r\mathbf{N}(u,v) $$ where $r \ge 0$ and $\mathbf{N}(u,v)$ is the unit normal of $\mathbf{F}(u,v)$ given by $$ \mathbf{N}(u,v) = \frac {\mathbf{F}_u(u,v) \times \mathbf{F}_v(u,v) } { \| \mathbf{F}_u(u,v) \times \mathbf{F}_v(u,v) \| } $$ The surface $\mathbf{G}(u,v)$ is called an offset surface in my field (CAGD), or a parallel surface in traditional differential geometry. I'm trying to relate the curvature properties of $\mathbf{G}$ to those of $\mathbf{F}$.
Specifically, I'm wondering about the principal directions at corresponding points of $\mathbf{F}$ and $\mathbf{G}$. My intuition says that these ought to be related in some simple way, maybe even parallel. Can anyone throw any light on this?
The application is in design/manufacturing, so we can restrict our attention to "nice" surfaces like roofs of cars or wings of airplanes. For example, we can assume that $r$ is small, so that no unpleasant cusps or self-intersections occur on $\mathbf{G}$.
It's not too bad to check that the normal of $\mathbf G$ at $(u,v)$ is just $\mathbf N(u,v)$. The principal directions do correspond (i.e., agree), and, moreover, if $k_i$ are the principal curvatures of the original surface, then $\tilde k_i = \dfrac{k_i}{1-rk_i}$ are the principal curvatures of the parallel surface. (So you can deduce the appropriate facts about mean and Gaussian curvature.)
EDIT: Assume that we have a parametrization where the $u$- and $v$-curves are lines of curvature in the original surface. (This is not necessary, but it sure makes things more evident. Such a parametrization can be found away from any umbilic points. See, for example, Theorem 3.3 on p. 119 of my differential geometry text.) So we'll have diagonal matrices for the first and second fundamental forms, hence a diagonal matrix for the shape operator.
In particular, \begin{align*} \mathbf I &= \begin{bmatrix} E & 0 \\ 0 & G \end{bmatrix} \quad\text{for some } E,G>0, \\ \mathbf{II} &= \begin{bmatrix} k_1E & 0 \\ 0 & k_2G\end{bmatrix}, \end{align*} where $k_1$ and $k_2$ are the principal curvatures. This follows because, by definition, the principal curvatures are the eigenvalues of the shape operator $S=\mathbf I^{-1}\mathbf{II}$. (This is the matrix representation relative to the basis $\{\mathbf F_u,\mathbf F_v\}$.) By construction, the principal directions, the eigenvectors, are the coordinate directions (which are orthogonal).
Recall that the shape operator is given by $S_p(X) = -d\mathbf N_p(X)$, and so we have $\mathbf N_u = -k_1\mathbf F_u$ and $\mathbf N_v = -k_2\mathbf F_v$. Thus, $\mathbf G_u = \mathbf F_u + r\mathbf N_u = (1-rk_1)\mathbf F_u$, $\mathbf G_{uu} = \mathbf F_{uu} - r\big(k_1\mathbf F_{uu} + (k_1)_u\mathbf F_u\big)$, $\mathbf G_{uv} = \mathbf F_{uv} - r\big(k_1\mathbf F_{uv} + (k_1)_v\mathbf F_u\big)$, and so on. Note, for example, that the $uu$-entry of the second fundamental form of $\mathbf G$ will be $$\mathbf G_{uu}\cdot \mathbf N = \mathbf F_{uu}\cdot\mathbf N - rk_1\mathbf F_{uu}\cdot\mathbf N -r(k_1)_u\mathbf F_u\cdot\mathbf N = (1-rk_1)\mathbf F_{uu}\cdot\mathbf N = (1-rk_1)(k_1E),$$ and the $uv$-entry will be $0$, since $\mathbf F_{uv}\cdot\mathbf N = 0$.
By the way, we've verified that $\mathbf G_u$ (resp., $\mathbf G_v$) is a scalar multiple of $\mathbf F_u$ (resp., $\mathbf F_v$), which tells us that the tangent planes are the same at corresponding points — and the same for the unit normal vectors $\mathbf N$.
Thus, we can compute the fundamental forms of the parallel surface to be \begin{align*} \tilde{\mathbf I} &= \begin{bmatrix} E(1-rk_1)^2 & 0 \\ 0 & G(1-rk_2)^2\end{bmatrix} \\ \widetilde{\mathbf{II}} &= \begin{bmatrix} k_1E(1-rk_1) & 0 \\0 & k_2G(1-rk_2)\end{bmatrix}. \end{align*} It follows that the matrix representation of the shape operator of the parallel surface is $$\tilde S_p = \tilde{\mathbf I}^{-1}\widetilde{\mathbf{II}} = \begin{bmatrix} \frac{k_1}{1-rk_1} & 0 \\ 0 & \frac{k_2}{1-rk_2} \end{bmatrix}.$$ The fact that the matrix is diagonal tells us that the eigenvectors (principal directions) are once again the basis vectors $\mathbf F_u,\mathbf F_v$, so the principal directions in the two surfaces match up perfectly. And we get the (slightly edited) formula for the principal curvatures I promised.