Category Archives: Surfaces

Torus knots on hyperboloids (part 2)

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In the post Hyperboloidal Representation of Torus Knots we gave an explicit construction of T(p,q) torus knots and links where the knots are polygonal and the edges lie on two distinct hyperboloids of one sheet. As the knot is traversed, the edges alternate between lying on the first then the second hyperboloid. Each hyperboloid of one sheet is a doubly ruled surface (as described in the post Doubly ruled surfaces). For each of the hyperboloids the edges of the torus knot lie in one of the rulings.

White curved cylinder shaped surface with line segments on the surface against a brown background

The T(3,5) torus knot lying on two hyperboloids of one sheet.

I wanted to be able to create a 3D-printable model that showed both the T(p,q) knot and the two hyperboloids of one sheet. I used Cinema 4D to do this, and ended up making several attempts to get things just right.

The T(3,5) torus knot was constructed as described in the post Hyperboloidal Representation of Torus Knots. To sum up, the vertices were computed, then the knot was constructed in the standard way.  There are two relatively easy ways to construct the two hyperboloids of one sheet. In the computation for the vertices, we can find an explicit formula for the hyperboloids. We could parametrize a generating hyperbola curve for the hyperboloid, put it into the Formula tool, then use a Lathe tool to rotate the curve around the y-axis. (Remember that in Cinema 4D, the y-axis points up.)
Grey curved cylinder shaped surface on a black background.

The edges of the T(3,5) torus knot lie on these two hyperboloids of one sheet.

Rather than do this, I kept things simple. I added an Empty Spline, opened the Structure Manager, then added in the first three vertices of the T(3,5) torus knot. This created the first two edges of the knot. However, these had the wrong orientation – the vertices were entered for the standard xyz-orientation, not the orientation in Cinema 4D. I thus went into Model mode, and rotated the vertices 90o in the y-direction. I then added a Lathe and moved the Spline under the Lathe. (The reason why this works can be found in the post Doubly ruled surfaces – the formulas explicitly show that you can take one line segment in a ruling and rotate it around the z-axis to create the surface.) At this point the surface had many obvious edges. I went back into the Spline and edited the Object Properties: the Type is Linear, the Intermediate Points is Uniform, the Number is 50. (For some reason Bezier gives a bad looking surface.) I went back into the Lathe and edited the Object Properties. Here I increased the Subdivision to 50.

One of the errors I first made when constructing this surface was to take the coordinates of the vertices of the T(3,5) knot from the 3D printable model. This was a mistake, as the original vertices had been replaced by two nearby vertices for the Chamfer at the corners. This meant the two hyperboloids of one sheet where not correct.

A grey curved cylindrical surface with line segments on it with a black background

The T(3,5) knot on the two hyperboloids of one sheet.

At this point, we can put the object in a better form for 3D printing. One of the adjustments I made was to scale the hyperboloid surface up slightly. When 3D printed, the top and bottom edges of this surface end up each print as one single filament. This can be problematic to print. So increasing the size of the surface slightly gives a bit of wiggle room.  I used the Boole tool (A union B)  to join the hyperboloid surface to the knot. In order to even see the union of these two surfaces, I needed to turn off High Quality in the Object Properties of the Boole. I also created one object from the Boole by going to the Object menu of Objects and then “Current State to Object”. On the new Boole that was created, I selected  “Connect objects + Delete”. I’m not sure these last two steps are entirely necessary. I’ve noticed that the slicing programs for 3D printers are increasingly sophisticated and are able to handle objects that aren’t completely correctly triangulated much more easily than in the past.

Grey curved cylindrical surface with grooves the surface following straight line pattern. Background is black.

The hyperboloid surface without the T(3,5) torus knot.

Finally, I wanted to be able to 3D print the knot in one color and the surface in a different color. I went back into Cinema 4D and created several files. The first was the knot, the second was the surface without the knot. I created this using the Boole tool and the “A subtract B” option. I then printed the surface and the knot on an Ultimaker 5 printer. I had trouble printing this surface as the printer added in unnecessary supports. This is a work in progress! I’ll post an update with a correctly printed knot when we make it happen.

 

 

Doubly-ruled surfaces

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three blue curved surfaces against a black background

An example of a hyperboloid of one sheet.

Some of the most interesting quadratic surfaces are the doubly-ruled surfaces: the hyperboloid of one sheet (like x2+y2-z2=1)  and the hyperbolic paraboloid (like z=x2-y2). These surfaces have made an appearance in this blog previously when we discussed how to create good 3D printable models of these surfaces.

Two grey curved surfaces with equations on them against a black background.

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)->R3 be a parametrization of the curve on the surface, let vt 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)->R3 defined by f(t,s) = g(t)+svt.

A two sided twisted surface. One side is in blue, the other red.

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=x2 in R3 is a generalized cylinder parametrized by f(t,s)=(t,t2,s).

Image of a saddle shaped surface with lines on it.

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  x2+y2-z2=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 Lt 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 x2+y2-z2=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 Lt 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).

Orange cylindrical shaped surface

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 x2+y2-z2=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 Lt. The only cases left are the points (0,1,0) and (0,-1,0). But these points lie on the lines Lt 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 Lt and Ls 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 Ms 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 Ms and the lines in this ruling do not intersect.

Orange colored curvy cylindrical surface

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 Lt and Ms 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 Lt and Ms 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.