Question

I am wondering what algorithm (or formula) Inkscape uses to calculate the control points if the nodes on a path are made “smooth”.

That is, if I have a path with five nodes whose d attribute is

M 115.85065,503.57451
  49.653441,399.52543 
  604.56143,683.48319 
  339.41126,615.97628 
  264.65997,729.11336

And I change the nodes to smooth, the d attribute is changed to

M 115.85065,503.57451 
C                     115.85065,503.57451 24.747417,422.50451
  49.653441,399.52543 192.62243,267.61777 640.56491,558.55577
  604.56143,683.48319 580.13686,768.23328 421.64047,584.07809
  339.41126,615.97628 297.27039,632.32348 264.65997,729.11336
  264.65997,729.11336

Obviously, Inkscape calculates the control point coordinates (second last and last coordinate pair on lines on or after C). I am interested in the algorithm Inkscape uses for it.

Answer 1

I have found the corresponding piece of code in Inkscape’s source tree under
src/ui/tool/node.cpp, method Node::_updateAutoHandles:

void Node::_updateAutoHandles()
{

    // Recompute the position of automatic handles.
    // For endnodes, retract both handles. (It's only possible to create an end auto node
    // through the XML editor.)
    if (isEndNode()) {
        _front.retract();
        _back.retract();
        return;
    }

    // Auto nodes automaticaly adjust their handles to give an appearance of smoothness,
    // no matter what their surroundings are.
    Geom::Point vec_next = _next()->position() - position();
    Geom::Point vec_prev = _prev()->position() - position();
    double len_next = vec_next.length(), len_prev = vec_prev.length();
    if (len_next > 0 && len_prev > 0) {
        // "dir" is an unit vector perpendicular to the bisector of the angle created
        // by the previous node, this auto node and the next node.
        Geom::Point dir = Geom::unit_vector((len_prev / len_next) * vec_next - vec_prev);
        // Handle lengths are equal to 1/3 of the distance from the adjacent node.
        _back.setRelativePos(-dir * (len_prev / 3));
        _front.setRelativePos(dir * (len_next / 3));
    } else {
        // If any of the adjacent nodes coincides, retract both handles.
        _front.retract();
        _back.retract();
    }
}