测试结果:

AutoCad  .Net二次开发求两曲线最小距离-LMLPHP

主要思路:假设有两条曲线分别是c1和c2,把c1按照1的距离划分我这里用变量jd表示,得到一个曲线集合coll,然后遍历coll,得到coll中每一个曲线的两个端点,再用这两个端点分别求离曲线c2的最短距离,直接使用开发库的GetClosestPointTo方法就可以了,直到遍历完整个coll集合就能得到最短距离和其对应的点。

主要代码得到曲线集合coll:

 public List<Curve> GetCurves(Curve curve ,double jd)
{
List<Curve> lstCurves = new List<Curve>(); double totalLength = curve.GetDistanceAtParameter(curve.EndParam); if (totalLength < jd)
{
lstCurves.Add(curve);
return lstCurves;
}
double addLength = ; Point3dCollection pt3dCol = new Point3dCollection(); while (addLength < totalLength)
{
pt3dCol.Add(curve.GetPointAtDist(addLength));
addLength += jd; }
if (addLength != totalLength)
pt3dCol.Add(curve.GetPointAtDist(totalLength)); DBObjectCollection dbObjColl= curve.GetSplitCurves(pt3dCol); foreach (var item in dbObjColl)
{
lstCurves.Add((Curve)item);
} dbObjColl.Dispose(); return lstCurves;
}

主要代码得到最短距离和最近点:

public Line GetMinLine(Curve curve1,Curve curve2,double jd)
{
List<Curve> lstCurves = GetCurves(curve1, jd); double minVal = double.MaxValue;
Point3d ptMin1 = Point3d.Origin;
Point3d ptMin2 = Point3d.Origin;
foreach (var c in lstCurves)
{
Point3d pt1 = c.StartPoint;
Point3d pt2 = c.EndPoint; var pt11=curve2.GetClosestPointTo(pt1, false);
var pt22= curve2.GetClosestPointTo(pt2, false); var l1 = pt11.DistanceTo(pt1);
var l2 = pt22.DistanceTo(pt2); if (l1 < minVal)
{
minVal = l1;
ptMin1 = pt11;
ptMin2 = pt1;
}
if (l2 < minVal)
{
minVal = l2;
ptMin1 = pt22;
ptMin2 = pt2;
} }
ed.WriteMessage("\n最短距离:" + minVal + "\n"); return new Line(ptMin1,ptMin2);
}

关于GetClosestPointTo介绍如下:

AutoCad  .Net二次开发求两曲线最小距离-LMLPHP

05-11 20:35