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So far we have shape of pseudosphere defined below/upto its cuspidal boundary/edge. However,

FullPseudosphereModelBeltrami has detail around (what appears to me) material added outside cusp radius that appears with small amplitude frills.

Daina Taimina's Crochet Art gives detail of its hyperbolic geometry (adding more circumferential thread material over the radial) e.g., at 15 min 19 sec of video as large amplitude frills.

EDIT 1:

As per Hilbert's theorem (if stated in other words) when a smooth patch constant $K<0$ is sufficiently extended it should encounter a cuspidal boundary.

We see no cuspidal boundaries somewhere among the crocheted frills. Is $K$ constant? Can a parameterization be suggested for (what looks to me like a neat inner $K<0$ area of a toroid) surface she is demonstrating at 11 min 05 secs?

Nash embedding theorem states that every Riemannian manifold can be isometrically embedded into Euclidean space of sufficiently large dimension. Is embedding into $\mathbb E^3$ at all possible with such frills?

Is there yet no parametrization or differential equation describing full frilled/convoluted shape of a pseudosphere?

Or is this shape nearer to the Catalan's minimal surface?

Narasimham
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  • I love hyperbolic geometry, but was wondering about your previous post on Dido's problem variation, is that sorted out? – An aedonist Jul 28 '17 at 19:59
  • The frills are not part of the pseudosphere it is more that it frills because the real life surface gets to big to be nice and smooth – Willemien Jul 28 '17 at 22:59
  • The frills appear smooth beyond cusp but may be with another different parameterization. – Narasimham Jul 29 '17 at 05:44
  • Related https://math.stackexchange.com/q/954376/88985 it is more that the physical surface is to large and not bend far enough it to have no frills maybe easiest way to test : make an soft material saddle surface and try to flatten it frills or cusps are depend on the material used (hyperbolic geometry is geometry where every point is a saddle point) – Willemien Jul 29 '17 at 05:58
  • The present one continues from my earlier question quoted. Assuming Beltrami used a paper mache like material that lost stiffness due to passage of time and developed edge frills due to self-weight etc., what can be said of those prominent wavy crocheted frills? – Narasimham Jul 29 '17 at 21:27
  • Not sure if it is loss of stiffness that developed the frills it might as well be that Beltrami did not know what to do after the cusp and did not object showing that, the big wavy crocheted frills are just the same only in a more extreme manner, was thinking about a way to explore this frilling yourself: – Willemien Jul 30 '17 at 11:54
  • Take a peace of paper and draw a binary tree, all nodes the same size, don't stop when it gets crowded because the tree reached the sides of the paper you need to add at least 3 or 4 layers in depth ( binary tree 1 2 4 8 16 32 64 128 256 512 1024 nodes in a layer – Willemien Jul 30 '17 at 11:59
  • The exponential/binary additions to propagating/expanding crochet was explained by Prof. Daina in the TED talk. Maybe this would help further to propose/define / model its shape equation. – Narasimham Jul 30 '17 at 21:32

1 Answers1

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I will try to explain why the tractricoid has a cusp (and why there sometimes are frills beyond the cusp )

The tractricoid is the "real" name of what many people call the pseudosphere

The tractricoid is the best known but not the only pseudo-spherical surface, In the earlier post I think I even show that it was not the first known one)

The tractricoid is the revolution surface of a tractrix (good to know but not used in the proof)

I thought it was easy but while trying to write a good and simple explanation I realised it was more complex than I expected. but i think a good simple explanation is a good thing to have

In this proof I maybe sometimes state the obvious this is to make the proof better to understand. I hope it is not to annoying.

the whole proof might as well done a bit easier by just looking at a tractrix instead of an tractricoid, (replacing radius of h-circle with distance of the x-axis)

I did choose not to do so, to keep the idea of a tractricoid intact and to remove the attention of coordinates (I would like a proof that did not mention coordinates at all, I did get quite close, I think :)

Lets go into the maths:

Suppose we have a tractricoid where the axis of rotation/ revolution / symmetry is the x axis.

Define an h-circle as the circle that is the intersection of the on the plane perpendicular to the x axis and the tractricoid.

suppose $S_1 (x_1, y_1 , 0) $ is a point on the tractricoid.

The radius of the h-circle through $ S_1 $ is $ y_1 $ and the circumference of the h-circle through $ S_1 = 2 \pi y_1 $

Some fixed distance $ds$ from $S_1$ we have the point $ S_2 = (x_2, y_2 , 0) $

With distance here I mean the length of the arc that is the intersection between the tractricoid and the $ y >0, z=0 $ half plane, measuring the distance this way is very importand and we will come back to this later.

The circumference of the h-circle through $S_2 = 2 \pi y_1 $ times some (small) factor $ a > 1 $

A distance $ds$ from $S_2$ we have the point $S_3 = (x_3, y_3 , 0) $

The circumference of the h-circle through $S_3 = 2 \pi y_3 = 2 \pi y_2 a = 2 \pi y_1 a^2$ and $ y_3 = a y_2 = a^2 y_1 $

This can be generalised to $ y_n = a y_{n-1} $

(Now stating some obvious facts)

When $ i < j < k $ then $S_j$ is between $S_i$ and $S_k$

if $ i < j$ then the circomference of the h-circle through $S_i$ is smaller than the circumference of the h-circle through $S_j$

(End of obvious facts)

The difference between $y_n $ and $y_{n-1}$ is $ (a-1) y_{n-1}$

At some point $S_c$ , the difference between $y_c $ and $y_{c-1} $ is $ (a-1) y_{c-1}$ is $ ds$ (or to be more exact the difference is between $ \frac{ds}{a}$ (?) and $ ds$ )

The circumference of the h-circle through $S_c$ is $ 2 \pi y_c $

A step further at $S_{c+1}$ the circumference of the h-circle is $ 2 \pi y_c $ times $a$. and the radius at $S_{c+1}$ is $ a y_{c}$

Now an impossible situatie occurs that The difference between the radius at $S_{c+1}$ and $S_{c}$ is $(a-1) y_c $

And $(a-1 ) y_c > (a-1 ) y_{c-1} = ds $ so the necessary radius of the circle at $S_{c+1}$ is bigger than the radius at $S_{c} + ds $ and therefore this circle cannot be constructed.

(the maximum circle that can be made after $ S_c $ is one with radius $y_c + ds$)

This is why there is a cusp

But then what "beyond the cusp"? the circle is the shortest curve to surround a given area, and a way out of this is to change curve and opt for a longer curve than a circle to surround a given area, and this is your circle with frills, or cusps ,

Or as the tractricoid does: stop at the cusp.

Hope this helps

Willemien
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  • You are discussing situation on the meridian of tractricoid (spelling) aka pseudosphere as.. at the cusp two principal radii $R_1$ goes to zero and $R_2$ goes to $\infty.$ – Narasimham Jul 31 '17 at 17:09
  • Think I corrected it now every where :) I wanted to keep the proof simple (and still correct) hope I did that ( still not sure about where I put a question mark :) – Willemien Jul 31 '17 at 19:11