symmetric ranges of curve division on an oloid taken from wikipedia , I drew an Oloid by using the functions
          with
;
;
and
 with 
;
;
If I use any equal spaced range of t(x) on ;  I will end up with an unequal range t(y), which looks like this:

My goal is to have a symmetric, spacing on both of the curves like in this model:

I can only remap the values of t(x).
What function do I need to apply on the range t(x) to get a symmetric curve division also on  t(y) ?
Any expert out there?
Please help.
Best,
Phillip
 A: tl;dr
I do not have the ability to prove the non-existence of a solution to the OP's question, but I can offer justification for the existing equations. This should give some insight.
While they contain obvious singularities, the equations shown by the OP look to have originated loosely from Dirnbock and Stachel 1997 where there is no proof of (specifically) equations (2) and (4).
http://www.heldermann-verlag.de/jgg/jgg01_05/jgg0113.pdf
My proof of equations (2) and (4) follows. An inversion is required to get from point U to point T with a further inversion in going from T to V.
The spatial separation of points is not symmetric but the paths between them are reversible.
I provide this analysis and an alternative parametrization here:
https://github.com/logicmonkey/surfaces/blob/master/OloidPoints/OloidPoints.pdf
Inversion with Respect to a Circle
Point A
  on the unit circle centred at the origin O
  lies on the tangent line that crosses the x-axis at point P
  as shown below. The point P'
  is the inverse of P
  with respect to the circle. Similarly, P
  is the inverse of P'
 . If the x coordinate of P'
  is $\cos\left(t\right)$
 , then the x coordinate of P
  is $\frac{1}{cos(t)}$
 . This bidirectional relationship is exploited in the following sections.

Parametrization using Transcendental Functions
Points U
  and V
  lie on unit circles centred at $C_{U}=(0,-\frac{1}{2})$
  and $C_{V}=(0,\frac{1}{2})$
  respectively as shown below. U
  and V
  correspond to points A
  and B
  in Dirnbock and Stachel's paper of 1997.
$$U\left(t\right)=\left[\sin\left(t\right),-\frac{1}{2}-\cos\left(t\right),0\right]$$
The triangle T'UV
  is formed by the joining line UV
  and the intersection point T'
  of the tangent lines through U
  and V
 . Point T'
  is also the inverse of point $T_{U}=(0,-\frac{1}{2}-\cos\left(t\right))$
 . 
By inversion with respect to the circle, the distance $C_{U}T'$
  is$$|C_{U}T'|=\frac{1}{\cos\left(t\right)}$$


For simplicity, the circle centred at $C_{V}$
  is rotated from yz into the xy plane. Since the circle centres are distance 1 apart, length $T'C_{V}$
  is therefore
$$|T'C_{V}|=1+\frac{1}{\cos\left(t\right)}$$
 $T_{V}$
  is the inverse of T'
  with respect to the upper circle, making the displacement along the y axis
$$|C_{V}T_{V}|=\frac{1}{1+\frac{1}{\cos\left(t\right)}}=\frac{\cos\left(t\right)}{1+\cos\left(t\right)}$$
By Pythagoras, the squared distance of V
  to the y axis is $$|VT_{V}|^{2}=1-\frac{\cos^{2}\left(t\right)}{\left(1+\cos\left(t\right)\right)^{2}}=\frac{1+2\cos\left(t\right)}{\left(1+\cos\left(t\right)\right)^{2}}$$
Rotating back into the yz plane it follows that V has the parametrization
$$V\left(t\right)=\left[0,\frac{1}{2}-\frac{\cos\left(t\right)}{1+\cos\left(t\right)},\pm\frac{\sqrt{1+2\cos\left(t\right)}}{1+\cos\left(t\right)}\right]$$
