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There are surfaces which you can "slide" onto themselves so that you see no change.

In particular, the plane, the cylindre, the surfaces of revolution (circular cone, sphere, torus...), the helicoid. The sliding motion can be a combination of translation and rotation.

I want to know if there is some theory about these, and in particular if they have a name, if there are others than those I listed and if they can be characterized by some property of their principal curvatures. (I guess that at least one of them must be constant.)

Update:

If I am right, those surfaces are obtained by sweeping an arbitrary curve with a constant rigid motion, one of translation, rotation about an axis or "screw displacement". Then during surface motion, every point follows a straight line, a circular arc or an helix.


Using cylindrical coordinates $(r,\theta,z)$, where the $z$ axis is that of the helix, and using the parametric equation of the section curve of the surface by the plane $xz$, $(r(u),0,z(u))$, the equation of the swept surface is

$$(r(u),v,z(u)+\sigma v)$$ where $\sigma$ is the pitch.

In Cartesian coordinates,

$$(r(u)\cos v,r(u)\sin v,z(u)+\sigma v).$$

Then the first fundamental form is

$$\begin{align}E&=r'^2(u)+z'^2(u),\\F&=\sigma z'(u),\\G&=r^2(u)+\sigma^2.\end{align}$$

The normal vector,

$$\frac1{\sqrt{\sigma^2r'^2(u)+z^2(u)r^2(u)+r^2(u)r'^2(u)}}\\(\sigma r'(u)\sin v-z'(u)r(u)\cos v,-\sigma r'(u)\cos v-z'(u)r(u)\sin v,r(u)r'(u)).$$

And the second fundamental form,

$$\begin{align} L&=-z'(u)r(u)r''(u)-\sigma r'(u)r''(u)+r(u)r'(u)z'(u),\\ M&=-\sigma r'(u),\\N&=-z'(u)r^2(u),\end{align}$$ scaled by $\dfrac1{\sqrt{\sigma^2r'^2(u)+z^2(u)r^2(u)+r^2(u)r'^2(u)}}$.


I guess that a line of attack is to determine the position of the helix axis and the value of the step from the local differential forms, and to show that they are constant. Given the complexity of the expressions, this looks like a serious endeavor.

  • Would you include or exclude involute gears whose profile is made of circle involutes (https://en.wikipedia.org/wiki/Involute_gear) from the "sliding" category ? – Jean Marie Dec 23 '16 at 23:39
  • The surface of a screw should work. I think you can ask for rather crazy cross sections too : if you have a curve that isn't too wild contained in the plane $x=0$, and you look at all its translates under the action of $\mathbb{R}$ where $t$ acts by translating by $t$ along the $z$ axis, and rotates by $t$ degrees along that same axis, you will get a surface that you can slide along itself. – Olivier Bégassat Dec 23 '16 at 23:40
  • @OlivierBégassat: quite right, this is a generalization of the helicoid. –  Dec 23 '16 at 23:45
  • @JeanMarie: if an involute keeps overlapping itself by some combination of translation and rotation, then yes. But I don't think it is the case because the curvature isn't constant. –  Dec 23 '16 at 23:53
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    @YvesDaoust: I've added the riemannian-geometry tag so that your question will draw the attention of some relevant people. – levap Dec 24 '16 at 00:10
  • "I guess that at least one of [the principal curvatures] must be constant." This is not true of the helicoid, nor of an arbitrary surface of revolution. –  Dec 24 '16 at 00:15
  • @Rahul: I agree for the helicoid but have a doubt for the surface of revolution. –  Dec 24 '16 at 00:43
  • Consider the surface of revolution produced by rotating $y=10+\sin x$ around the $x$-axis: At the largest thickness the principal curvatures are 11 and 1; at the smallest thickness they are 9 and -1. The reason your chosen examples in the question (cone, sphere, torus) all happen to have one constant principal curvature is because their generating curves themselves have constant curvature! –  Dec 24 '16 at 01:36
  • @Rahul: quite right. Actually the curvature must be constant in the direction of the motion. Can we translate this statement in a local differential property ? –  Dec 24 '16 at 11:07
  • [The analogous for curves instead of surfaces is visited informally in Martin Gardner's "Aha! A two volume collection"](https://books.google.com.ar/books?id=3HzwNVMQOZkC&pg=RA1-PA39&lpg=RA1-PA39&dq=martin+gardner+sword+scabbard&source=bl&ots=VaUChrmtYb&sig=FFtS_06CfG3AwKjb66047VSZKnQ&hl=en&sa=X&ved=0ahUKEwivgZDm0bjRAhXGGJAKHWPFB8gQ6AEIGDAA#v=onepage&q=martin%20gardner%20sword%20scabbard&f=false) – Santropedro Jan 10 '17 at 22:39
  • @Santropedro: we are close. If we knew the individual helixes that make up the surface, finding the helix parameters (axis and step) from the Frenet-Serret elements seems quite doable. –  Jan 10 '17 at 23:03
  • A nice 4D curve is the one parametrized by $(A\cos(\alpha t),A\sin(\alpha t),B\cos(\beta t),B\sin(\beta t))$. It fits your criterion. It can be thought of as what comes after the line, circle, and helix. Just as the line is a limiting case of the circle, the helix is a limiting case of this. – Akiva Weinberger Jan 10 '17 at 23:51
  • Going by Andrew D. Hwang's answer, one can always choose a generating curve that is everywhere perpendicular to the helical orbits. Perhaps an arc length parametrization of such a curve, rather than the one lying on the $\theta=0$ half-plane, would yield simpler expressions for the differential forms? –  Jan 11 '17 at 23:33
  • @Rahul: do you think that the direction of the local helix can be derived from the osculating paraboloid ? –  Jan 12 '17 at 07:55
  • I don't know yet. I'm just responding to your last sentence by suggesting a possible way to reduce the complexity of the expressions of the differential forms. –  Jan 12 '17 at 08:25
  • @Rahul: I know. –  Jan 12 '17 at 08:30
  • Given the qualitative features of the available surface mesh this idea may or may not help: If the principal curvature functions can be calculated sufficiently accurately, it's possible to detect whether either family of level curves consists of helices (constant curvature and torsion). If so, the helix axis and pitch can be recovered. – Andrew D. Hwang Jan 12 '17 at 14:24
  • @AndrewD.Hwang: is there a relation between the principal curvatures and curvature/torsion of an embedded curve ? (and which level curves ?) –  Jan 12 '17 at 14:32
  • @Yves: I was unclear. :) The prospective strategy is to calculate the principal curvature functions (two continuous functions) on the surface accurately enough to find level curves of the principal curvature functions, and to calculate the curvature and torsion of the level curves of the principal curvatures. Constancy of the curvature and torsion detects helices. If a surface has helical symmetry, the principal curvatures are constant along helices, so a positive answer to the proposed test gives a prospective helix axis and pitch that can be tested further if necessary. – Andrew D. Hwang Jan 12 '17 at 14:46
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    @AndrewD.Hwang: that makes a lot of sense. If I understand, any local invariant property (such as the principal curvatures) will repeat along the local helix, and by following this helix, we can determine the motion parameters. Thanks for the hint, I will accept it as an answer. –  Jan 12 '17 at 14:56
  • @AndrewD.Hwang Can we now perhaps define a generalized ODE for asymptotics reducing to Sine-Gordon when $k=0?$ – Narasimham Jan 14 '17 at 14:51
  • @Yves Daoust. Bottle openings / cap threads , matching steel Bolt/Nut threads are some examples. – Narasimham Mar 01 '17 at 20:33

3 Answers3

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$\newcommand{\Reals}{\mathbf{R}}$Edit of January 12, 2017, to address clarifications from the comments.

If $S \subset \Reals^{3}$ is a smooth surface invariant under some one-parameter group $\Gamma$ of ambient (Euclidean) isometries, then "the geometry of $S$ is $\Gamma$-invariant". Specifically, scalar geometric quantities such as principal curvatures (or mean curvature, or Gaussian curvature) are constant along the orbits of $\Gamma$. Assuming at least one principal curvature is non-constant (i.e., that $S$ is not a portion of a sphere, plane, or cylinder), level curves of the principal curvature functions are $\Gamma$-invariant, i.e., lines, circles, or right circular helices.

To detect numerically whether or not a surface $S$ approximated by a discrete surface mesh is "$\Gamma$-invariant", one might calculate some scalar geometric quantity $K$, approximate some level curve of $K$, and then determine whether or not this level curve has constant curvature and torsion:

  • If yes, the curvature and torsion can be used to recover the (approximate) axis and pitch of the helical motion. If necessary, these data can be used to confirm the surface is "helical" with the stated axis and pitch.

  • If no, the surface $S$ is not helical.

The prospects of numerical success presumably depend on the fineness of the surface mesh and how closely the mesh approximates any helical structure the surface may have. It's entirely possible a more accurate and/or efficient strategy can be found with more analytic work.


(Original answer of December 24, 2016, with a few minor edits for attribution and notational consistency.)

I don't know of published literature, much less a standard name for these surfaces, but here are some observations:

Extrinsically, a self-sliding surface $S$ in Euclidean three-space is invariant under a one-parameter group $\Gamma$ of Euclidean isometries.

  • If $\Gamma$ consists of translations, $S$ is a (generalized) cylinder.

  • If $\Gamma$ consists of rotations, $S$ is a surface of rotation.

Otherwise, the orbits of $\Gamma$ are helices. In the (unpublished) past, a student (J. M. Antonio) and I called these "pasta surfaces".

Intrinsically, a self-sliding surface admits local Clairaut coordinates $(u, v)$ in which the metric has the form $$ g = E(u)\, du^{2} + G(u)\, dv^{2}, $$ with the group action given by translation in $v$, and with $u$ a parameter for a "profile" (a curve transverse to the group orbits). (The group action foliates $S$ by curves. The field of orthogonals is one-dimensional, hence integrable: Through each point $p$, there exists a profile through $p$ that is everywhere perpendicular to the orbits. Let $u$ be a parameter for this curve, and let $v$ be induced by the group action.)

Thanks to the one-parameter family of isometries (which manifests analytically as $v$-independence of the metric components), the geometry of $S$ is "constant in $v$". Particularly, under an equivariant embedding (i.e., an embedding $i:S \hookrightarrow \mathbf{R}^{3}$ for which a one-parameter group of Euclidean isometries preserves the image and acts by intrinsic coordinate translation), the principal curvatures (and consequently the Gaussian and mean curvatures) are constant along $v$-coordinate curves.

It's also not difficult to show that one-parameter families of pasta surfaces of constant Gaussian curvature exist, analogous to the one-parameter families of surfaces of rotation having constant Gaussian curvature. (For studying Gaussian curvature, it's particularly convenient to arrange that $EG \equiv 1$, i.e., that $E(u) = 1/G(u)$. The Gaussian curvature in these "momentum" coordinates is linear in the metric components: $K = -\frac{1}{2}G''(u)$.) The only complete, embedded examples are spheres, cylinders, and planes, but examples based on the pseudosphere give isometric embeddings of arbitrarily large hyperbolic disks.

(I've never looked at constant mean curvature, but don't expect obstructions to existence.)


Edit of December 26, 2016: Up to a rigid motion of Euclidean three-space, every self-sliding surface (other than a generalized cylinder) arises as follows.

Fix a parametric curve $\gamma(u) = (r(u), 0, z(u))$ whose image lies in the half-plane $x > 0$, $y = 0$ and whose velocity is non-vanishing; a real number $k$ (the "pitch" or "slope" of a helix); and an arbitrary smooth function $\psi$ (the "gauge"); and put $$ X(u, v) = \left[\begin{array}{@{}c@{}} r(u) \cos(v + \psi(u)) \\ r(u) \sin(v + \psi(u)) \\ z(u) + k(v + \psi(u)) \\ \end{array}\right]. $$ The image of $X$ is the result of sliding the image of $\gamma$ (the green curve below) under the action $$ (x, y, z) \mapsto (x\cos t - y\sin t, x\sin t + y\cos t, kt) $$ of the additive group of real numbers. This action represents rotation about the $z$-axis if $k = 0$ and helical motion about the $z$-axis if $k \neq 0$. (Formally, translation along the $z$-axis arises by letting $k \to \infty$, but I assume you're mostly interested in helical motions.)

A self-sliding surface and a generator

(The gauge function $\psi$ does not change the surface geometrically, but "deforms" the $u$-coordinate curves "internally" by sliding them along the $v$-coordinates curves (circles or helices). If $\psi \equiv 0$, the $u$-coordinates curves ($v$ constant) are plane curves, the images of $\gamma$ under the group action. For some purposes it's more convenient to choose $\psi$ so that the $u$- and $v$-coordinate curves are pointwise orthogonal, as in the diagram below.)

A hyperbolic surface with helical symmetry

Andrew D. Hwang
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  • This answer is a little technical for me, and I could find no explanation of the Clairaut coordinates. Anyway I seem to understand that the surface can be obtained by sliding the normal plane to a directrix helix (with limit cases straight line and circle) that contains a generatrix curve. The generatrix should rotate simultaneously with the osculating plane. –  Dec 26 '16 at 11:20
  • I am not so much after surfaces of constant curvature. I am more after characterizing a "pasta" surface in terms of local differential properties. –  Dec 26 '16 at 11:21
  • My apology for being overly-technical. :) Clairaut coordinates $(u, v)$ are by definition those in which the metric has the form $du^{2} + G(u)\, dv^{2}$. I'll add details about ("equivariant") parametrization in a bit, with the hope that better addresses what you're after. – Andrew D. Hwang Dec 26 '16 at 12:05
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    I found quite interesting material in books by Stephen P. Radzevich (such as Kinematic Geometry of Surface Machining). He speaks of "surfaces sliding over themselves". Unfortunately, I cannot access his papers which seem to answer my question. –  Dec 26 '16 at 12:10
  • How can we check that a surface is self-sliding ? –  Dec 26 '16 at 13:34
  • Ah, I see. Practically, that may depend on how a surface comes to you (parametrically; as a level set; etc.), and the answer may be "it's clear by inspection or else it's messy". Loosely, you have to recognize that "the geometry (e.g., the family of tangent planes; the principal curvature functions; the Gaussian curvature function) is invariant under a helical motion". Are you looking for something along the lines of [exterior differential systems](https://en.wikipedia.org/wiki/Differential_ideal)? – Andrew D. Hwang Dec 26 '16 at 13:55
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    I will get discrete surface meshes. –  Dec 26 '16 at 14:07
  • For discrete surface meshes, it appears nothing in my answer is helpful: If the mesh "doesn't respect" the self-sliding property, then the mesh isn't self-sliding (even "discretely"), and therefore doesn't contain enough information to deduce analytically that the mesh came from approximating a self-sliding surface. (The mesh _does_ contain enough information to guess the likelihood of the underlying surface being self-sliding, but that's an issue of approximation and computational geometry.) Sorry not to be of more help.... – Andrew D. Hwang Dec 26 '16 at 14:34
  • This is why I am asking about local differential properties, which can very well be evaluated numerically. By the way, these surfaces can be characterized as being the result of an helicoidal sweep of a curve with a constant pitch. If the pitch and the axis can be defined and evaluated locally, I guess that we are done. –  Dec 26 '16 at 15:10
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    Dini's surface would probably be another interesting example of a constant GC surface that can move helically. – J. M. ain't a mathematician Dec 27 '16 at 20:46
  • $ ds^2=(1/G(u))du^2+G(u)dv^2 ; ds^2=(1/G(u))du^2+G(u)dv^2$ was a metric different from the above given metric for same $K<0$ sliding surface given here? Am a bit confused. constGCsurf with respect to earlier http://math.stackexchange.com/questions/2074863/surfaces-with-constant-gaussian-curvature/2074889#2074889 – Narasimham Jan 11 '17 at 20:46
  • @Narasimhan: It's not hard to show one form can be converted to the other by a change of coordinates in $u$ (though of course not for the same "$G$"). I'll edit this when I have time. :) – Andrew D. Hwang Jan 11 '17 at 21:42
  • The graphic appears to be same... awaiting this pasta' s $G$. – Narasimham Jan 12 '17 at 13:40
  • @Narasimham: I'd rather wait for the bounty to expire before editing, since editing bumps the thread/draws attention, and given the clarification in the comments, it's unlikely this answer will address Yves' question (how to detect helical symmetry numerically from a surface mesh). – Andrew D. Hwang Jan 12 '17 at 14:17
  • That would be fine. – Narasimham Jan 12 '17 at 14:23
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    The "level curve" approach seems a viable option. Anyway, it is not purely local, in the sense that after computing the curvature (which is a second order feature), you need to look in a neighborhood for equal curvatures. So this looks like a third order process. I wonder if this degree if avoidable or not. Maybe we can answer by looking at simpler cases: how do we check locally that a surface is prismatic or of revolution ? –  Jan 13 '17 at 05:50
  • About the last, I think Andrew Hwang says it is influenced through $k$ factor. It looks we are veering to the same thing. – Narasimham Jan 13 '17 at 06:21
  • @AndrewD.Hwang The green curve $\gamma$ slider could be drawn for the second surface as well. And also later on, Manipulation/simulation by changing k factor would be both fun as well as instructive I think from the lead you have given. – Narasimham Jan 13 '17 at 06:34
  • [Const_GC_related][2] [2]: http://math.stackexchange.com/questions/2074863/surfaces-with-constant-gaussian-curvature/2074889#2074889 – Narasimham Jan 14 '17 at 15:07
  • @AndrewD.Hwang An entire rigid profile with helical motion around an axis is a novel feature of surfaces swept out this way. – Narasimham Jan 16 '17 at 08:16
  • @AndrewD.Hwang Also do the $u-$ parameter lines lie in a plane? – Narasimham Jan 19 '17 at 19:56
  • @Narasimham: The $u$-parameter curves can be chosen to be planar (by taking $\psi \equiv 0$), or (for example) they can be chosen to be orthogonal to the $v$-parameter curves, but if the pitch $k$ is non-zero, then _both_ of these conditions cannot be satisfied. – Andrew D. Hwang Jan 19 '17 at 20:20
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Any arbitrary meridian of a surface of revolution contained in the plane of the axis can be twisted with a screw motion, to obtain these sliding surfaces.

It has a proportionality constant between $\theta$ rotation and $z$ axial progression. We need to simply add one $ h \cdot \theta $ term to the z-coordinate.

Constant Gauss curvature K surfaces are only interesting as, before and after twist, new $K$ comes out a different constant. But it is not otherwise relevant in general imho for this question of OP.

My best guess is that

1) They are Helical Canal surfaces ( not standard nomenclature, we may prepend "general" to it, or const. negative curvature helicoids or Hwang Helicoids ) created by motion of a rigid canal cross section along a helix in which each point describes a helix with constant radial parameter $u$ and variable parameter $\theta$ or arc.

$$ (f(u) \cos v, f(u) \sin v, \, g(u) + h \cdot v) $$

The bi-normal parametrization however does not have much of an advantage over the meridional section definition; in any case the two parametrizations are equivalent.

2) Along parametric lines screw motion all curvatures $ k_n, k_g, K$ are identical.

3) Tangent vector has a constant magnitude (speed) and constant inclination to a fixed direction, here z-axis. Hence $\dfrac{dr}{dz} = \cos \phi \cos \psi = \cos \alpha $ = constant inclination of swept parameter line tangent, to symmetry axis just as it is for a single sheet hyperboloid... Cannot prove this now, although it is necessary and sufficient for this class of surfaces.

4) Parametric lines cut orthogonally,$F=0$.

5) Rotationally symmetric cases are trivial.

A number of meridians are twisted in this manner and their warping effects shown below.

Twisted/Warped/Screw Motion Surfaces

Mathematica program lists meridians chosen and the effect of $ h\cdot \theta $ term.

Clear[X, Z, u, v]; R = 2; r = 1; (* Arbitrary Meridian Screw Motion *)
X[u_] = R Sin[u]; 
Z[u_] = -R (Cos[u] + Log[Tan[u/2]]); (* Dini with (graphics error) *)
X[u_] = r (u - Sin[u]); Z[u_] = r (1 - Cos[u]);(* Cycloid Axes Interchanged *)
X[u_] = r (1 - Cos[u]); Z[u_] = r (u - Sin[u]); (* Cycloid  *)
X[u_] = (r Cos[u] + R); Z[u_] = r Sin[u]; (* Toroid  *)
X[u_] = (-r u + 2 R); Z[u_] = 2 r - R u;(* Cone *)
X[u_] = Exp[-u/2]; Z[u_] = u;(* Exponential curve *)
X[u_] = Sin[ 2 u] + 3 ; Z[u_] = u; (* Sine curve *)
X[u_] = Sin[ 2 u] + 2 ; Z[u_] = Cos[3 u]; (* Lissajou curve *)
ParametricPlot[{Z[u], X[u]}, {u, -Pi, Pi  }, 
 PlotLabel -> "PRE-TWIST MERIDIAN"]
h = 0; ParametricPlot3D[{X[u] Cos[v], Z[u] + 0 h v, 
  X[u] Sin[v]}, {u, -Pi, Pi}, {v, 0, 3 Pi}, 
 PlotLabel -> " MERIDIAN SWEPT AROUND Z-AXIS"]
h = 1.5; ParametricPlot3D[{X[u] Cos[v], Z[u] + h v, 
  X[u] Sin[v]}, {u, -Pi, Pi}, {v, 0, 3 Pi}, Boxed -> False, 
 Axes -> None, PlotLabel -> "POST-TWIST ( h v) REVOLVED MERIDIAN"]

Mechanical fabrication may be also interesting. In a metal cutting machine-shop when meridian shape is supplied in the form of a template mounted on the carriage of a lathe, the operator sets $h$ value from feed/speed ratio forming the needed helix angle to produce helically slidable matching bolt and nut machined surfaces imparting a screw motion to cutting tool. The template block edge holding canal boundary profile has a constant angle to the axis of rotation.

Narasimham
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  • Thanks. If I see correctly, canal surfaces are defined by the sweeping of a sphere (constant radius or not ?). My main concern about these surfaces is how they can be proven self-sliding by local analysis. My current line of thoughts is that if we can compute the screw-motion parameters locally from the differntial forms, we are done. –  Jan 11 '17 at 11:09
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of quite an aside comment relevance (no answer) .. such dry bean pod examples are plenty in Nature. Similar to our geometry here when toroidal arching up is disregarded. Posted it elsewhere before.

Spiral canal pod

Narasimham
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