An example of a hyperbolic paraboloid.

Ruled surfaces have been studied extensively in geometry. Formally, a ruled surface is a surface which is a union of straight lines. The straight lines are called rulings (or the generators) of the surface. An example to think about is a cylinder: this is a union of straight lines, each of which intersects a circle. Just as with the circle and the cylinder, in a general ruled surface there is a curve in space which intersects each of the lines. Thus the surface can easily be parametrized as follows. Let g:(a,b)->R^{3} be a parametrization of the curve on the surface, let v_{t} be a vector at point g(t) which points in the direction of the line passing through g(t). Then the surface is parametrized by f:(a,b)x(c,d)->R^{3} defined by f(t,s) = g(t)+sv_{t}.

A helicoid. Image from https://minimalsurfaces.blog/

A nice example of such a surface is the *helicoid* where lines parallel to the xy-place pass through each point of a helix (for example parametrized by g(t)=(cos(t), sin(t),t)). One possible parametrization of a helicoid is f(t,s) = (s.cos(t), s.sin(t), t). Ruled surfaces are very familiar to us, if the rulings are all parallel to each other, then the ruled surface is a *generalized cylinder.* For example, the surface given by y=x^{2} in R^{3} is a generalized cylinder parametrized by f(t,s)=(t,t^{2},s).

The two rulings on a hyperbolic paraboloid surface.

Now there are surfaces which are *doubly-ruled surfaces,* meaning that each point of this surface belongs to two distinct lines. In other words, this surface has two distinct rulings. You can use projective geometry to show that this surface, the plane, and the hyperboloic paraboloid are the only surfaces that have this property. Robert Byrant has a nice proof of this fact using differential geometry in a 2012 Math Overflow post

Let us focus our attention on the hyperboloid one sheet given by x^{2}+y^{2}-z^{2}=1. To give explicit equations for the two rulings, I will follow the text *Elementary Differential Geometry* by Andrew Pressley. For every t, we can show the straight line L_{t} given by \[(x-z)\cos t=(1-y)\sin t, \ \ \ (x+z)\sin t = (1+y)\cos t\] lies on the surface. To see this, multiply the two equations together to get \[(x^2-z^2)\sin t\cos t =(1-y^2)\sin t\cos t.\] in other words x^{2}+y^{2}-z^{2}=1 unless cos(t)=0 or sin(t)=0. If cos(t)=0, then x=-z and y=1, and if sin(t)=0 then x=z and y=-1, and both of these lines lie on the surface. A short computation reveals that for each t, line L_{t} contains the point (sin(2t),-cos(2t), 0) and is parallel to the vector (cos(2t), sin(2t), 1). We thus get all lines in one ruling for t in [0,π).

One set of rulings on a hyperboloid of one sheet.

We now show that any point on the hyperboloid lies on one of these lines. Take a point (x,y,z) on the surface x^{2}+y^{2}-z^{2}=1. If x does not equal z, then let t be such that cot(t) = (1-y)/(x-z). If x does not equal -z, then let t be such that tan(t) = (1+y)(x+z). In both cases the point is on L_{t}. The only cases left are the points (0,1,0) and (0,-1,0). But these points lie on the lines L_{t} when t=π/2 and t=0 respectively. In addition, we can check that the lines in this ruling do not intersect. Suppose (x,y,z) lies on L_{t} and L_{s} and t does not equal s. Then \[(1-y)\tan t = (1-y)\tan s, \ \ \text{and} \ \ (1+y)\cot t = (1+y)\cot s.\] Assuming the tan and cot functions are not zero or undefined gives both y=1 and y=-1, a contradiction. The case where t=0 and s=π/2 give disjoint lines too: \[ L_0(t) = (t,-1,t) \ \ \text{and} \ \ L_{\pi/2}(s)=(-s,1,s).\]

What is the other ruling? For every t, we can show the straight line M_{s} given by \[(x-z)\cos s=(1+y)\sin s, \ \ \ (x+z)\sin s = (1-y)\cos s\] lies on the surface. The computations are almost identical, so we omit them. Each point of the surface lies on a line in M_{s} and the lines in this ruling do not intersect.

The second set of rulings on a hyperboloid of one sheet.

Another computation shows that if t+s is not a multiple of π, then L_{t} and M_{s} intersect in the point \[\left( \frac{\cos(t-s)}{\sin(t+s)}, \frac{\sin(t-s)}{\sin(t+s)}, \frac{\cos(t+s)}{\sin(t+s)} \right). \] For each t in [0,π), there is one s in [0,π) such that t+s is a multiple of π, and the lines L_{t} and M_{s} do not intersect. Intuitively, the two lines are on opposite sides of the hyperboloid of one sheet.

There are many other interesting observations to be made about these surfaces. For example, any set of three skew lines generates such a surface and the three skew lines lie in one of the rulings. Ian Agol posted a nice proof of this in the same 2012 Math Overflow post.