Given a curve, $\gamma: \mathbb{R} \to \mathbb{R}^2$ define the flow in the normal direction by $\gamma(t) + \epsilon \, \mathbf{n}(t)$. This is different from the evolute which moves at speed proportional to the radius of curvature.
Example: The unit square with edge thickness, moves inwards, but the corners move diagonally.
If the polygon is piecewise linear, how to the cusps move under this unit speed flow?
For my application, $\gamma(t)$ is a polygon in the Euclidean plane, not necessarily convex. The points are $ \{ (a_1, b_1), \dots, (a_n, b_n)\} \in \mathbb{R}^2$. I need to account for the thickness of the boundary. The result is a polygon which is almost the same: $ \{ (a_1 \pm \epsilon , b_1\pm \epsilon), \dots, (a_n\pm \epsilon, b_n\pm \epsilon)\} \in \mathbb{R}^2$. How do I compute the envelope in this case?
This is similar to Ravi Vakil's The Mathematics of Doodling. In fact, I have even drawn the "unit-speed evolute" in this case...
Pez, if I understand correctly, you want to know how each polygon vertex $v = (a_i,b_i)$ moves as you move each edge along the edge normal by $\epsilon$?
First, compute the normals $N_1, N_2$ of the two edges adjacent to $v$. Then the inflated vertex position $v'$ is given by
$$v' = v + \frac{N_1+N_2}{\|N_1+N_2\|} \epsilon \sec \frac{\psi}{2},$$ where $\psi$ is the angle between the normals (see the diagram at https://www.dropbox.com/s/glpwplrtim8fkl7/kites.pdf). Applying the cosine half-angle formula I get
$$v' = v + \frac{(N_1+N_2)\epsilon\sqrt{2}}{\|N_1+N_2\|\sqrt{1+N_1\cdot N_2}} = v + \frac{(N_1+N_2)\epsilon}{1+N_1\cdot N_2}.$$
As a sanity check, notice that $(v'-v)\cdot N_1 = \epsilon$ and likewise for $N_2$.