3
$\begingroup$

A Cylinder is such a common surface. But is there a parametrization for an isometrically $ R^2 $ bent cylinder whose major and minor dimensions are along x, y axes? I used an approximation to parametrize, that is why the generators are bent between boundaries that seem to have elliptic profiles.

Most probably there is none in a closed form, please indicate in differential form.

EDIT1:

I used an approximation for shape determination/parametrization between ellipse end profiles for a rough picture here to illustrate what I mean.

I shall attempt to explain as much as I can with two examples.The question is motivated by common experience of deformation of surfaces.

If two opposite points on the circular boundary of a thin flexible hemispherical surface are brought together, a spindle results, whose parametrization is known as described by isometry. In the mapping, a surface of revolution passes to another surface of revolution, one radius of curvature increasing and another decreasing retaining the product constant as demanded by Gauss Egregium theorem.

If a thin cone is pressed together on its sides it deforms to an oblique cone retaining Gauss curvature K value at 0.

If two opposite points on the circular boundary of a thin cylindrical surface of given height and diameter are brought together, a warped cylinder results roughly as shown in the above sketch, which can be also described by isometry. The top "ellipse" points are joined to bottom "ellipse" points, I believe they should be straight. As K should be zero, there will always be one straight line seen throughout the deformation or isometric mapping.At top, the opposite points are squeezed together,at bottom they are squeezed out apart.Perhaps at the center the circle remains a circle.

I placed "ellipse" in quotes because the top and bottom boundary ellipse-like curves may not be ellipses contained in a plane during deformation.

As for what was so far done, I took the formula $ C = a E(\epsilon) $ for a quarter perimeter of ellipse as constant for the new shape. Semi major/minor axes are variable with z. Antisymmetric shape z= f(x,y) about central plane could be valid, so along x- and y- straight generators remain; but after that I looked up to post it here.

Sincerely hoped it would be of interest as a Differential geometry topic. Best Regards.

EDIT2:

PX and QY are two of four equally spaced points on the rigid central circle. During the deformation/isometric mapping the red arms (V1,V2) of the H-shape rotate around hinge H3 joining PX with QY. Updated static picture:

Cyl Deformation

Also an imperfect(non-isometric) animation as an approximate structrural definition, lacks mathematical rigour : SqueezedCylinder

EDIT3:

Isometric mapping / deformation between cylinder/tetrahedron_Conjecture

After above post it occurred to me that a squeezed cylinder is also a squeezed tetrahedron, or, in other words a circular cylinder can pass into a tetrahedron by a series of continuous deformations/isometric mappings.( Or even into other polyhedra for that matter but for now only tetrahedron is considered).

I shall at first describe it, and later on add an animation true to the above conjecture.

It can be more convincing to make a model using thick paper or thin flexible plastic for sides of a tetrahedron model and to deform it in your own hands.

In a tetrahedron ABCD horizontal edges AB and CD are of equal length, have their adjoining faces not glued together. And the rest of six edges AC,AD,BC and BD are of equal length, neighboring faces glued together and capable of rotation as an edge fulcrum/hinge.

During squeeze-bending, A is pressed or brought towards B, and, C towards D while opening/splitting edge AB into a slit between edges AB and $\overline{AB}$ to make two edges, which are 2 sided polygon " Diangles" edges.

$$ AB \rightarrow (AB, \overline {AB}) ; \, CD \rightarrow (CD, \overline {CD} ) $$

My conjecture is that the faces of tetrahedron develop into oblique cones with vertices at A,B,C and D, the generators/rulings of oblique cone form at cone vertices A,B,C and D.

At a certain stage of deformation pairs (A,B) and (C,D) are opposite points of arcs that may now be recognized as semi-circles. Edges AB,AC,BC,BD re-appear now as helices of the circular cylinder, conserving zero geodesic curvature $ k_g$ during this isometric mapping or bending deformation.The cones intersect along tetrahedral edges in intermediate situations, but are tangential in the end cylinder configuration.

Every edge of a polyhedron has a cuspidal edge due to one curvature going to infinity, tangent slope has sudden jump.The cuspidal edges re-appear as oblique cone end rulings, finally as regular helical geodesics of cylinder.

When squeezing_bending continues even beyond attainment of circular/cylindrical configuration, we see that another imagined set of diagonals of tetrahedral edges form edges of the alternate oblique cone set, completing symmetrical deformations with respect to central cylindrical configuration.

So we can deform Tetrahedron $ \rightarrow $ Cylinder $ \rightarrow $ Tetrahedron $\rightarrow $ Cylinder cyclically.

Let me have your views. Here physical models synthesize mathematical parts into useful ways, no?

In the event of above conjecture being valid,(just as in the case of catenoid/helicoid), we are still left with the task of identifying a morphing/squeezing parameter between a cylinder and a 2-Edges-open tetrahedron that includes these two parametrizations.

EDIT4:

Development to make a physical model:

Tetra_Cyl_Development_To_Deform

EDIT 5:

Considering tetrahedron deformation extreme coordinates of A and C can be found out. In one case it is helix on cylinder and the other case it is a side of tetrahedron.

If L = AC, a= cylinder radius,we can find their constants in terms of $ L , a $

For Helix parametrization $ ( a \cos t , a \sin t, b t) , L = 2 \pi \sqrt{a^2+b^2} \rightarrow b^2 = (L/\pi)^2-a^2 ; $

Tetrahedron corner points are $ ( \pi a/2,0,h), ( \pi a/2,0,-h),\rightarrow h^2 = L^2/4 - (\pi a)^2/8; $

EDIT 6:

In isometry/bending deformation straight lines remain straight, or the geodesic curvature in tangent plane is null and remains null during squeezing deformation.

Accordingly the helix which is a geodesic on cylinder becomes a ruled generator of an oblique cone maintaining straightness in tangent plane at each instant of deformation.

$\endgroup$
  • 4
    $\begingroup$ I don't understand your question. $\endgroup$ – Ted Shifrin Dec 26 '14 at 19:17
  • 2
    $\begingroup$ You might consider discretizing the surface into horizontal triangulated bands, and then deform each band by treating edges as hinges, while maintaining triangle rigidity. $\endgroup$ – Joseph O'Rourke Dec 27 '14 at 17:26
  • $\begingroup$ @JosephO'Rourke: What do you think of this,if we go in the direction of oblique cones,by considering bending along generator hinges of each narrow triangle during transition between helix to tetrahedron your suggestion is in operation. Can you please consider modeling this in animation? $\endgroup$ – Narasimham Dec 28 '14 at 19:12
  • $\begingroup$ OK, no takers on bounty also. Shall add a video about this phenomenon. $\endgroup$ – Narasimham Jan 5 '15 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.