Simply as variant location of the curve type of spiral, and as an entertainment. For example, on the cylinder of radius 1. On this curve may be an infinite number of points of self-intersection.

SPIRAL.mw     

If I needed an angle between the vectors, I'd programmed the following algorithm: put vectors at one point, then set aside for each of the vectors equidistant point, construct a circle, and define the angle by circumferential length.

    The crank rotates uniformly around the axis oX2. The red line is defined by two equations:
(x1+.5*sin(5*x3))^2+(x2-2)^2+(x3+.5*sin(2*x1))^2-9=0;
x1^6+x2^6+x3^6-12=0;
Today make kinematic analysis of this mechanism is only possible using the proposed method.
   
     Kinematic analysis.  
General view:



    and the plane X1oX3:

@tomleslie  Geodesic itself is not the shortest distance between two points on the surface. To do this, choose the direction. (It is about surfaces at all.)

Backup plan (if you like).

POINTS_ON_sphere.mw 

@necron  For other solutions you can try to organize a loop, when you see the graph

http://www.mapleprimes.com/posts/204684-Lever-Mechanisms-  

Planar slider-crank mechanism.  

(x+0.125sin(10y))^4+(y+0.125sin(10x))^4 - 0.4^4=0 

     It was proposed for members of the department  http://malplab.ru/ and their colleagues from other departments that they to make the calculation of the kinematics of almost classical crank mechanism. Example  performed  in analogy to the examples in the description of the method in the first message.
     The crank rotates uniformly variable length, its point is fixed at the origin, the movable point moves along the curve defined by the equation (x + sin (1.5y)) ^ 4 + y ^ 4 - 0.25 ^ 4 = 0, the conrod length is equal to 1, the second conrod point moves along the axis oX.
     The same proposal I want to make here in the forum for all those wishing to try their hand.

From a certain point in time I began to think about moving to Kazan.

@shadi1386 Basically, if alpha = const, then try  solve ([f1, f2], [x, y, beta]);

But this, I think, you will not be happy much. Maple 15

>restart:
f1 := (1/2)*x+2*alpha*beta*(x^2+3*y^2)^beta*x^2/(x^2+3*y^2)-(3/2)*y-(1/2)*alpha*(x^2+3*y^2)^beta = kappa*rho*c0^2;
f2 := -(1/2)*y+2*alpha*beta*(x^2+3*y^2)^beta*y^2/(x^2+3*y^2)-(1/2)*x-(1/2)*alpha*(x^2+3*y^2)^beta = 0;
solve([f1, f2], [x, y, beta]);

The approach to the solution is possible only numerically.

@Markiyan Hirnyk I did not apply to you in any way, and answered ThU message in its proposed form.

@ThU  I do not know what makes Markyan and I do not want to know.  To You I will answer. I downloaded your file equidistant_curve_MP_2.mw  and checked. Everything is working. I checked on Mapl 15 and on Maple 17. Looks like one in one in my file in the first message.
 I bad user of Maple  (and do not speak English), but I can describe to you algorithm in detail if  you want made it yourself, if you are interested, but does not work text.