Are higher-order Bézier curves envelopes?

Question

I only realized from this question and the answers to it that quadratic Bézier curves are the envelopes of the lines used to compute them iteratively. That is, if a quadratic Bézier curve for points $P_0$, $P_1$, $P_2$ is constructed by linearly interpolating between $P_0$ and $P_1$ and between $P_1$ and $P_2$ and then again linearly interpolating along the line between the two resulting points, then the Bézier curve is the envelope of those lines.

I'm wondering whether this generalizes to higher Bézier curves. An $n$-th order Bézier curve can be analogously constructed by $n$ iterations of linear interpolation. Is it the envelope of any of the lines used in that construction? Or of the higher-order curves that are generated in the intermediate stages of the construction? Or both? If so, how can we see this most easily?

An answer for cubic Bézier curves would already be interesting, even if it doesn't generalize to higher orders.

(I tried searching for “Bézier” and “envelope” but didn't find anything useful, partly because the top hits are all about an Inkscape extension called Bezier Envelope.)

Answer 1

As you said, a quadratic Bezier curve is the envelope of lines whose end-points are moving along two straight lines.

In just the same way, a cubic Bezier curve is the envelope of lines whose end-points are moving along two quadratic curves. You can see this by looking at the last linear interpolation in the de Casteljau algorithm.

And so on. In general, a Bezier curve of degree $m$ is the envelope of lines whose end-points are moving along two curves of degree $m-1$.

Here are the details for the cubic case. Suppose the cubic curve has control points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$.

Then its equation is $$ \mathbf{P}(t) = (1-t)^3\mathbf{A} + 3t(1-t)^2\mathbf{B} + 3t^2(1-t)\mathbf{C} + t^3\mathbf{D} $$ In the final step of the de Casteljau algorithm, this is expressed as $$ \mathbf{P}(t) = (1-t)\mathbf{H}(t) + t\mathbf{K}(t) $$ where \begin{align} \mathbf{H}(t) &= (1-t)^2\mathbf{A} + 2t(1-t)\mathbf{B} + t^2\mathbf{C} \\ \mathbf{K}(t) &= (1-t)^2\mathbf{B} + 2t(1-t)\mathbf{C} + t^2\mathbf{D} \end{align} Clearly $\mathbf{H}$ and $\mathbf{K}$ are quadratic Bezier curves, which are shown in green and red respectively in the picture above.

Differentiating the original curve equation, and re-arranging, we get $$ \mathbf{P}'(t) = 2\bigl[(1-t)^2(\mathbf{B} - \mathbf{A}) + 2t(1-t)(\mathbf{C} - \mathbf{B}) + t^2(\mathbf{D} - \mathbf{C}) \bigr] $$ Then, a little algebra confirms that, in fact $$ \mathbf{P}'(t) = 2\bigl[\mathbf{K}(t) - \mathbf{H}(t) \bigr] $$ This shows that the line joining $\mathbf{H}(t)$ and $\mathbf{K}(t)$, which is constructed in the final step of the de Casteljau algorithm, is tangent to the cubic curve $\mathbf{P}(t)$.

There is further discussion on this page.

Answer 2

This answer merely elaborates on the computation shown by @bubba for cubic Bezier curves. It is an alternative derivation for the hodograph as defined in the link provided in the comments.

For a sequence of control points $a_0, a_1, \ldots$ and indices $0 \leq k \leq m$ let's write $$p(k,m; t)$$ for the Bezier curve of degree $m-k$ with control points $a_k, a_{k+1}, \ldots, a_m$. These can be defined recursively for $0\leq k < m$ by $$\begin{eqnarray} p(k, k; t) &=& a_k\\ \tag1 p(k, m; t) &=& (1-t)\, p(k, m-1;t) + t\, p(k+1, m; t) \end{eqnarray} $$

for $t\in [0,1]$. By induction on $m-k$ it follows that $$\tag2 p'(k, m; t) = (m-k) \left(p(k+1, m; t) - p(k, m-1; t)\right)$$ (where this is interpreted as simply $0$ if $k=m$). Indeed $p'(k, k; t) = 0$ and for $m > k$ by the recursive definition $(1)$ above: $$\begin{eqnarray} p'(k, m; t) &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (1-t)\, p'(k,m-1; t) + t \,p'(k+1, m;t)\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (1-t)(m-k-1)\,(p(k+1,m-1;t)-p(k, m-2;t)) + \\ && t (m-k-1)\, (p(k+2, m;t) - p(k+1, m-1; t))\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (m-k-1)\,((1-t)\,p(k+1,m-1;t) + t \, p(k+2, m;t))-\\ && (m-k-1)\,((1-t)\, p(k, m-2; t) + t\, p(k+1, m-1;t))\\ &=& p(k+1, m; t) - p(k,m-1;t) + \\ && (m-k-1)\,(p(k+1, m; t)- p(k, m-1; t))\\ &=& (m-k)\,(p(k+1, m; t) - p(k,m-1;t)) \end{eqnarray} $$

Now $(1)$ and $(2)$ show that the segment $\left[p(k, m-1;t), p(k+1,m;t) \right]$ is indeed tangent to $p(k, m; s)$ at $s=t$.