Hi,

I am playing with he surface graph. I'm trying to display an arch just now with this code:

int maxj = 10;
for (int i = 0; i <= 180; i++)
      {
       for (int j = 0; j < maxj; j++)
       {
          //x
          surface.XValues.Add(j);
          //y
          surface.Values.Add(Math.Round(Math.Sin(i * Math.PI / 180) * 100, 2));
          //z
          surface.ZValues.Add(Math.Round(Math.Cos(i * Math.PI / 180) * 100 + 100, 2));
          surface.Colors.Add(returnColor(0, 20, 10));
       }
      }

My problem is that if the maxj is 10 then I cannot see the graph, if I decrease it to 3 then it is displayed.

It is not the number of points as if display the same or even more points without drawing the arc, using Sin/Cos then it works.

Any idea?

Thanks,
Daniel

Hello Daniel,

In general, triangulation algorithms implemented with double precision arithmetic are not stable with certain kinds of input, especially ordered sets of points like the ones that you are displaying. In other words, sometimes the surface series simply fails to triangulate the points that you specify. You can configure the chart to use geometric calculations with adaptive precision - this will make the triangulation somewhat slower, but more robust. To enable this feature, you have to set the UsePreciseGeometry property to true.

By the way, the occurrence of this problem is often a sign that the triangulated surface might not be appropriate. You can display the same surface with the Mesh Surface series - no issues with the precision and it renders much faster.

Best Regards,
Nevron Support Team

Thank you, I need to color each point the way I want to so I think I just going to stay with the triangulated surface.

I am just wondering can nevron display a torus? or at least the top half of it?

No worries.
It can display torus.