planar loop shapes

Question

I know if I stick two pins on a paper, and trace a taut loop around them, I get an ellipse. With one pin, I get a circle. Question is, are there names for shapes I get if I trace a taut loop around 3, 4, 5, ..., k pins, assuming the pins are not collinear, and the polygon formed by joining them is convex i.e. every pin stretches the loop at least one point as I trace around the pins? General formula for the locus? Any good, readable references to this? (I am not a mathematician, just a hobbyist.) Thank you.

Answer 1

Someone will tell me if there is an exotic case I'm not foreseeing, but I think usually it will be a smooth union of "ellipse arcs."

Assuming the string is large enough to circumscribe the entire collection of pins, it could still be that it is taught around the outermost pins, and that would make a polygon. So we'll also assume there is a little slack to allow more than just that.

Pins within the interior of the convex hull of the set of pins won't be able to affect the shape at all, so you may as well assume the pins lie on the perimeter of a convex polygon.

Since the loop isn't taught, I think at any given time there will be two pins closest to the stylus drawing the arc, along the loop which act like the foci of an ellipse. As the stylus moves, the loop may contact a new pin which acts like the focus of a different ellipse, and likewise a pin that used to be in contact may lose contact with the string as another pin further along takes up the role of a focus.

I don't think a formula for the locus around $n$-pins sounds very feasible, but you could pursue trying to describe it as the boundary of a union of ellipses. The thing is that I think the number of ellipses involved depends on how much slack there is in the string. If it is very slack, then two pins very close to each other on the edge might never act as a pair of foci, but if you shrink the string enough, then they will.

For a little while I considered what just the union of all possible ellipses would tell us, but then I considered just a triangle. If two points form the foci of an ellipse, and the stylus is currently on the opposite side of the ellipse from the third triangle point, the third triangle point will be taking up slack, preventing the full ellipse from being drawn. So the answer is not just something as simple as "the boundary of the union of all possible ellipses."

That leads me to this: at any given point of time, the string will form a polygon, one of whose points is the stylus. Then the arc currently being drawn is a piece of an ellipse determined by two points of contact with the polygon. The convex hull of the polygon will be bound within the shape being drawn. So the shape is the boundary of the union of these "polygons with ellipse bumps."

Answer 2

If you keep the loop taut, the moving part will form a variable triangle defined by two pins and the pen for a while, then change one pin at a time. During this process, the pen draws arcs of ellipse, forming a continuous curve with continuous tangent.

The endpoints of the arcs will be found by lengthening the sides (when the loop is about to leave a pin, it is straight) and forming convex polygons that have the required perimeter. When you have the two pins of contact and the length, you have the ellipse equation. Additional geometry is needed to find the delimiting angles.

For the case of this figure, the trajectory is made of six elliptic arcs.