The following lines describe the details of the algorithms used by the software. In order to simplify the presentation, the following assumptions are made:

    1 step displacement value is identical for X and Y axis

    Same motors and leading screws on X and Y axis.

Algorithm description:
    Read A (Xa,Ya) starting point coordinates.

    read the coordinates of the following (destination) point B (Xb,Yb).

    Computation of the distance and the time TPS required to travel from starting point to destination (speed is known).

    Slope calculation  m = (Yb-Ya) / (Xb-Xa) = ref. slope

    First Stage:

      If m > 2 drive  1  Y step

      If  m >= 0.5 and  m <= 2 drive 1 X step + 1 Y step

      If m < 0.5 drive 1 X step


    When the line is stored, one imagine the result of the dispacement as a new starting point, then recalculate a new slope value from this new point, up to the destination point B.

      If ref slope >=1
        If m > ref. slope then drive 1 Y step

        If m <= ref. slope then drive 1 X+ 1 Y steps

      If ref. slope <1
        If m >= ref. slope then drive 1 X + 1 Y steps

        If m < ref. slope then drive 1 X step

    A each stage, we can count the number of X alone or Y alone steps (N1) and the number of simultaneous X+Y steps. To define the the time t to spend at each stage, we get the following relation  TPS = N1*t + (N2*t*1.414) ... from which t = TPS/(N1+N2*1.414).

    Once destination B is reached, B become the new starting point (B-> A)  and the next point is read (next destination)...and so on.

Gilles Muller has found a new algorithm, more powerfull: the trajectory is similar but remains centered on the theoretical curve. 
    Preliminary operations are identical
      If ref.slope = 1
        drive  (Xb-Xa)  X + Y steps
      Si ref; slope < 1
        m = ((Xb-Xa) mod (Yb-Ya))

        Do (Yb-Ya) times
        {
            (1/(pente ref) div 2) X steps
            1  X+Y step
            1/(ref. slope) -1 - (1/(ref. slope) div 2) X steps
            si (m >0)
            {
            drive 1 X step;
            m= m-1
            }
        }

      If ref. slope > 1
        m = ((Xb-Xa) mod (Yb-Ya))

        Do (Xb-Xa) times
        {
            (ref. slope div 2) Y steps
            1 X+Y step
            ref. slope -1 - (ref. slope div 2) Y steps
            si (m >0)
            {
            drive 1 X step;
            m= m-1
            }
        }

With these 2 methods you get these types of trajectories: 
traject.gif (3351 octets)
Starting point A(0,0) 

Destination B(27,3)

ref. slope = 3/27 = 0.111

->  X displacement (with respect to this new point B coordinates are now (26,3))

m = 3/26 = 0.115 ->  X+Y displacement ( B(25,2))

m = 2/25 = 0.08 ->  X displacement ( B(24,2))

et ainsi de suite...

m = 0 (That's not a good example! there is no remainder....but it works also!)

Do 3 times

9 / 2 = 4.5 -> 4  X steps

1  X+Y step

9 - 1 - 4 = 4  X steps

...
Remind that for a 48 steps/turn motor and a 6mm lead screw (1mm/trun), 1 step correspond to an absolute displacement value close to 2/100 mm (0.0208mm). To be compared with wire kerf value (and error)....