When I decode a PTH file (BL1.pth) I found follwing values for first node:

numnodes: 477
finishline: 431

node 1:
centre X=19.59
centre Y=550.94
centre Z=10.09

dir X=0.02
dir Y=1.00
dir Z=0.00

limit left=-23.67
limit right=28.65

drive left=-1.45
drive right=18.80

The X and Y values I understand, but what about the e.g. drive left? Or the dir X? What does those values mean? Anyone an idea?

Jaap

From memory:

numnodes: 477 // Number of nodes for the whole track
finishline: 431 // Node number where the finish line is in the current track config

node 1:
centre X=19.59 // Position of Node 1 in the World coordinates
centre Y=550.94
centre Z=10.09

dir X=0.02 // Vector giving the orientation of the node in the world coordinates
dir Y=1.00
dir Z=0.00

limit left=-23.67 // Distance from the center point to the outer edge of the driving area (usually the wall), on the left side
limit right=28.65 // same distance on the right side

drive left=-1.45 // Distance from the center point to the outer edge of the actual paved track, on the left side
drive right=18.80 // same distance on the right side

I think the term 'node' is not clear to me then. I thought it was one point on an X-Y (Z) axis. And when drawing a line from node (point) 1 to 2 etc. you get a line in the shape of the circuit. Is that correct?

Yes, a node has a certain coordinate, but you should think of it as a checkpoint. So it is actually a line, with that point as midpoint (or a point near the middle).

Ok thanks. I think I got it (concerning the node). But what about the "drive left=-1.45"? Is that -1.45 meter from the X,Y coordinate?

Are you sure you are reading the values correct? You should get integers as a result, and you are getting floating point numbers. If those numbers are the coordinates converted to meters, then I think that -1.45 means 1.45 meter in the (negative) direction perpendicular to the (dirX, dirY, dirZ) vector.

Yes I think it is clear. It seems to be simpler that I thought. Now only figure out what about the direction and I can draw the circuit. :-)