Suppose I have a surface in 3D space described by the implicit equation $F(x,y,z)=0$, and I have three points $P_1$, $P_2$, $P_3$ lying on this surface. I’m interested in bounding the maximum deviation between the surface and the triangle defined by the three points.
Motivation: in engineering and manufacturing computations, you often want to approximate a shape by a mesh of triangles. I’d like to understand how the approximation error depends on the size of the triangles.
My surfaces are fairly flat: think of the roof of a car, and a triangle about 5 mm across. So you can ignore problems arising from unpleasant surface wiggles inside the triangle.
I’m guessing that there’s some bound on the deviation that depends on the size of the triangle and some curvature measure of the surface. Some results for parametric surfaces are given here, and I’m seeking something similar.
Maybe I need to be more precise about the term “deviation”. At each point $P$ lying in the triangle, I can construct a line $L$ through $P$ that’s normal to the plane of the triangle. Let $d(P)$ be the distance from $P$ to the first intersection of $L$ and the surface. I’m interested in getting a bound for $$ \max \{ d(P) : P \text{ in the triangle}\} $$ Or, you can make up some other physically plausible definition of “deviation”, if that helps.
I thought there would be some standard result in approximation theory, but I can’t find one.
A problem with getting a bound in terms of the curvature of the surface is that curvature measures how fast tangent vectors turn as one moves with unit speed inside the surface, but moving with unit speed in the planar triangle translates (or "lifts") to moving faster than unit speed in the surface itself, by a factor which is the surface's slope. Maybe one can go in two stages: 1) get a bound in terms of moving in the planar triangle and 2) get a bound on the difference between motion speeds (I don't know how to do this fully).
What I mean by 1 is the following:
Assume (this is the "no weird wiggle" hypothesis) that the surface is graphical over the triangle. That is, after a change of coordinates putting the planar triangle is in the x-y plane, and the surface is given by $(x,y, h(x,y))$ for some sufficiently smooth $h$. Then what we want is a bound on $h$. We will find a bound in terms of the second partials of $h$, which are not directly "curvature" (this is the difference between moving in the x-y plane and in the graph that I stated in the beginning; basically when (x,y) is not an extremum the tangent plane to the graph is tilted, and x-y orthonormal vectors don't lift to orthonormal tangent vectors, so Hessian and second fundamental form don't agree). Ignoring this issue, we argue as follows.
Let $p=(x,y)$ be any point in the (transformed) triangle, and $l=p+s\hat{u}$ for unit vector $\hat{u}$ be the line through that point $p$ in the direction $\hat{u}$. Then the restriction $h_l$ of $h$ on the line has $h_l(s)''= \hat{u}^T H \hat{u}$ where $H$ is the Hessian of $h$ at $p$. As such $h_l(s)''$ is bounded by the largest eigenvalues of $H$, and by the Gershgorin circle theorem, $|h_l(s)''|$ is bounded by, say, the sum of absolute values of the three second partials of $h$. Let the maximum of this sum over the whole triangle be $M$. Then for any $p$ we have
$$ h(p) \leq \frac{1}{2} R^2 M$$
where $R$ is the circumradius. Indeed suppose p is a point where $h$ attains its maximum. Connect $p$ to the closest vertex $A$ by a segment (whose length $d$ is at most $R$). Let the line giving this segment be $p+s\hat{u}$ as above. Then $h_l'(0)=0$, and over the segment the value has to drop from $h(p)$ to $0$. But by repeated integration we get first from $h''(s)\geq -M$ to $h_l'(s)= h_l'(s)-h_l'(0) \geq -Ms$ and then to $h_l(s)-h_l(0) \geq -\frac{1}{2} M s^2$. So for $s=d$ we have $h_l(d)-h_l(0)= h(A)-h(p)=-h(p) \geq -\frac{1}{2} d^2 M \geq -\frac{1}{2} R^2 M$, as wanted. (This is basically the argument in the proof of Theorem 4 in your reference, in a different language.)
As for 2, the length of a lift of a unit tangent vector is at most $\sqrt{1+h_x^2+h_y^2}$, so I expect that the largest absolute value of the eigenvalue of the hessian is bounded by the largest principal curvature of the surface at the corresponding point times this stretching factor $\sqrt{1+h_x^2+h_y^2}$. If this holds, we could replace $M$ with the product of maximal principal curvature over the surface times the maximum of the stretching factor (but alas, that one is still in terms of $h_x$ and $h_y$; perhaps it, too, can be bounded in terms of the principal curvatures).