The plot code below depicts two point through which profiles of constant $K$ are drawn positive and negative.

    Show[{GCneg,GCpos},PlotRange->All,AspectRatio-> 0.42]

This seems acceptable, one would be lead to assume that such orthogonal crossings repeat for all intersections.

However it does not occur that way invariably, as can be seen in updated plots.

Your help to include a set of points on a few meridian profiles forming a mesh/net would be highly appreciated.

I have not seen plots Shown together in textbooks or articles of Differential Geometry. Cusp points may not be a problem as proper initial condition choice is made.


There are three classical groups of Gauss curvature surface meridians. The motivation of the post is to see if for each group the lines of positive and negative $K$ intersect at right angles.

In the central cases (sphere and tractrix) it happens only once for reasons not clear to me so far. For hyper/hypo cases the same should be now verified.

GC POS & NEG Meridians by hand


Meridians of a sphere and pseudo-sphere where integration starts at a common initial boundary point are set out orthogonally. Integration done by using NDSolve of Mathematica

$$ K= k_1. k_2 = + 1, K= k_1. k_2 = -1 \, \tag1$$

that integrates to ($\phi $ is slope)

$$ \sin^2 \phi = \lambda^2 - K r^2,\, \cos^2 \phi = \lambda^2 - K r^2 \tag2 $$

For clarity (cluttered graphics) I have omitted hypo group including only the hyper cases.

Visibly convincing violations of orthogonality of red/blue curves representing sphere and pseudo-sphere at intersections ( some violations marked by green vertical lines) do occur.

Quite a few intersections are visibly okay, but why should there be some non-orthogonal exceptions at all ?

This happens even between circles centered on x-axis and central pseudo-sphere meridians (tractrices)

For this reason a full perfect orthogonal net cannot exist and consequently an attempt to find such a net requested in my original question has been temporarily given up.

GC POS & NEG Meridians by Mma


Resuming this post after an interregnum..

From 1) and 2) we can write two slopes $\tan \phi $ as

for $K>0$

$$ \dfrac{dr}{dz}= \sqrt{\dfrac{\lambda^2-Kr^2}{1-\lambda^2-Kr^2} }\tag3 $$ $$ \dfrac{dr}{dz}=-\sqrt{\dfrac{\lambda^2-Kr^2}{1-\lambda^2-Kr^2} } \tag4 $$

and for $K<0$ reciprocal slopes of above of either sign with the rule for orthogonal trajectories

$$ \dfrac{dr}{dz} \rightarrow -\dfrac{dz}{dr} \tag 5 $$

$$ \dfrac{dr}{dz}= \sqrt{\dfrac{1-\lambda^2+Kr^2}{\lambda^2-Kr^2} }\tag6 $$ $$ \dfrac{dr}{dz}=-\sqrt{\dfrac{1-\lambda^2+Kr^2}{\lambda^2-Kr^2} } \tag7 $$

The plots of the above are carefully drawn.

It is recognized that the slope correspondence should be considered for each half of its profile. for valid orthogonality.

Correct pairing of half profiles is done as follows:

Equations 3 & 7 in first group and Equations 4 & 6 in second group. Cross matching is not valid. No interpretation of geometrical meaning can be possible by a cross-match.

Results of plotting and slope assignment confirm that the slopes should have opposite sign as per requirement 5) for orthogonal intersection to take place.

When K changes occur sign of slope of the integrands of "individual parts" indeed determine orthogonal trajectories in the manner suggested.

When proper matching is done from out of two $\pm $ signs along both meridians the question posed here is answered and now resolved.

Extraneous or spurious intersections are recognized with this criterion.

Valid red and blue meridian profiles intersect at:

$$(1- 2- 3- 5- 7)$$

and the same with z- displacement as constants of integration at:

$$(8 -9 -10- 11- 14)$$

So one of several valid curvilinear rectangles seen in a net of orthogonal trajectories is between $(1-2-9-8).$

The intersections at

$$(4 -6 - 12- 13)$$

are invalid/illegal in this context.

Constants chosen for graphical profiles

$$ K= \pm 1/a^2,\, a=1,\, \lambda = \sqrt 2. $$

SPH PS OP plots

Shall later post images for $ \lambda<\sqrt 2. $

  • $\begingroup$ I don't know what you mean. Of course such plots exist in the literature. Google for "curvature line parameterization". $\endgroup$ – Henrik Schumacher Apr 3 '19 at 8:22
  • $\begingroup$ You mean curvature as a function of arc distance, intrinsic diffrl equation $\kappa=f(s)?$ $\endgroup$ – Narasimham Apr 3 '19 at 9:09
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    $\begingroup$ My best guess of the OP's intent is that he wishes to show both circles and tractrices, which are indeed the respective meridian curves of the sphere and pseudosphere. That being said, a lot of knowledge is being assumed, and the question could stand to be clarified a bit more. $\endgroup$ – J. M.'s technical difficulties Apr 3 '19 at 9:28
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    $\begingroup$ @HenrikSchumacher "By mentioning" meridians" it is expected to be sufficiently clear." Now you know that expectation was wrong. $\endgroup$ – Henrik Schumacher Apr 3 '19 at 9:55
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    $\begingroup$ "it is expected to be sufficiently clear" - you are demonstrating exactly what I meant by "a lot of knowledge is being assumed". $\endgroup$ – J. M.'s technical difficulties Apr 3 '19 at 10:06

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