Unlike the curvature comb, the length of the handles is the non-inverted radius of curvature, which is the more natural choice.

These handles are very stable and can be converted just-in-time to classic bezier control points, without needing to round-trip back and forth between the two representations. Helpfully, this also avoids numerical drift.

To actually pull this off, we need to solve for the lengths $l_0$ and $l_1$ of the tangents, given the desired curvatures $k_0$ and $k_1$, the start/end points $A$ and $D$, and the unit-length tangents at the start/end.

Given a curve $\gamma\left(t\right)$, we can express the unit tangent vector $\mathbf{T}\left(t\right)$ as the normalized derivative:

This can be used to find the curvature vector $\mathbf{K}\left(t\right)$ via two vector cross products using the first and second derivative:

The cross products ensure that $\mathbf{K}\left(t\right)$ is perpendicular to $\mathbf{T}\left(t\right)$, i.e. they extract the normal vector component of the middle term. Now we can solve for $\mathbf{K}\left(0\right) = k_0$ and $\mathbf{K}\left(1\right) = k_1$.

After working through the math, we end up with a quadratic system of equations in $l_0$ and $l_1$:

Where:

This captures a few things:

• The sign of the curvature $k_i$ defines whether the curve turns clockwise or counterclockwise. When moving curve points around, the curve may be forced to flip, hence the $±$ is necessary, and both signs can change independently.

• The coupling between $l_0$ and $l_1$ is influenced only by $b$. If $b = 0$, then the two equations are independent and the problem is trivial to solve. This corresponds to the situation where the two tangents are parallel.

• The constant term $c_i$ is influenced only by the perpendicular distance $D - A$ (or $A - D$) to the tangent $\mathbf{T_i}$. This matches the earlier finding that only perpendicular distance affects curvature.

Hence, the problem is reduced to finding the intersection of two double-parabolas, one horizontal and one vertical: