topic eInvalidInput error in .NET Forum

eInvalidInput error

Anonymous — Thu, 29 Mar 2018 10:15:05 GMT

Line line = new Line(ptStart, ptEnd);
Vector3d vec = new Vector3d();
Point3d ptd1 = new Point3d();
vec = line.GetFirstDerivative(ptd1).GetNormal(); //eInvalidInput

ptStart coordinate is (655275600.438987,36192616.8036762,0)

ptEnd coordinate is (655295053.399808,36189883.0214896,0)

Why am I always getting "eInvalidInput" error. Any help would be greatly appreciated.

Re: eInvalidInput error

BKSpurgeon — Thu, 29 Mar 2018 11:23:09 GMT

 [CommandMethod("testDerivative")]
        public static void TestLine()
        {
            //// Your start points
            // Point3d p1 = new Point3d(655275600.438987, 36192616.8036762, 0);
            // Point3d p2 = new Point3d(655295053.399808, 36189883.0214896, 0);

            Point3d p1 = new Point3d(0, 0, 0);
            Point3d p2 = new Point3d(10, 0, 0);
            Point3d p3 = new Point3d(5, 0, 0);            

            using (Line ln = new Line(p1, p2))
            {
                Vector3d vec = new Vector3d();
                vec = ln.GetFirstDerivative(p3).GetNormal(); 
            }
        }

Try the above code - it works for me.
I think the error is coming because the p3 (i.e. the point supplied to the GetFirstDerivative method is not on the line itself. Solution is to: supply a point on the line.
Also I think you should dispose of your Line explicitly or use a `using` statement..

Re: eInvalidInput error

_gile — Thu, 29 Mar 2018 12:00:27 GMT

Hi,

It's because ptd1 (initialized to 0,0,0) does not lies on the line.

If you want to get the direction vector of the line, you can simply do:

vec = ptStart.GetVectorTo(ptEnd).GetNormal();

or:

vec = (ptEnd - ptStart).GetNormal()