测试结果：
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 = 0; 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：
原文：https://www.cnblogs.com/HelloQLQ/p/12112982.html
