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我有兩個線段，在其起點/終點處用3D點表示。查找兩個3D線段之間的最短距離
線：
class Line
{
public string Name { get; set; }
public Point3D Start { get; set; } = new Point3D();
public Point3D End { get; set; } = new Point3D();
}
三維點是座標的X，Y和Z僅有3雙打
3DPoint：
class Point3D
{
public double X { get; set; }
public double Y { get; set; }
public double Z { get; set; }
}
問題：
我能找到兩條'線條'和''之間的距離嗎？距離'線'的端點。 [
1
我有什麼：
目前，我能順利拿到兩線之間，此代碼（Adapted From Here使用段與段段）的距離：
public double lineNearLine(Line l1, Line l2)
{
Vector3D uS = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
Vector3D uE = new Vector3D { X = l1.End.X, Y = l1.End.Y, Z = l1.End.Z };
Vector3D vS = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
Vector3D vE = new Vector3D { X = l2.End.X, Y = l2.End.Y, Z = l2.End.Z };
Vector3D w1 = new Vector3D { X = l1.Start.X, Y = l1.Start.Y, Z = l1.Start.Z };
Vector3D w2 = new Vector3D { X = l2.Start.X, Y = l2.Start.Y, Z = l2.Start.Z };
Vector3D u = uE - uS;
Vector3D v = vE - vS;
Vector3D w = w1 - w2;
double a = Vector3D.DotProduct(u, u);
double b = Vector3D.DotProduct(u, v);
double c = Vector3D.DotProduct(v, v);
double d = Vector3D.DotProduct(u, w);
double e = Vector3D.DotProduct(v, w);
double D = a * c - b * b;
double sc, sN, sD = D;
double tc, tN, tD = D;
if (D < 0.01)
{
sN = 0;
sD = 1;
tN = e;
tD = c;
}
else
{
sN = (b * e - c * d);
tN = (a * e - b * d);
if (sN < 0)
{
sN = 0;
tN = e;
tD = c;
}
else if (sN > sD)
{
sN = sD;
tN = e + b;
tD = c;
}
}
if (tN < 0)
{
tN = 0;
if (-d < 0)
{
sN = 0;
}
else if (-d > a)
{
sN = sD;
}
else
{
sN = -d;
sD = a;
}
}
else if (tN > tD)
{
tN = tD;
if ((-d + b) < 0)
{
sN = 0;
}
else if ((-d + b) > a)
{
sN = sD;
}
else
{
sN = (-d + b);
sD = a;
}
}
if (Math.Abs(sN) < 0.01)
{
sc = 0;
}
else
{
sc = sN/sD;
}
if (Math.Abs(tN) < 0.01)
{
tc = 0;
}
else
{
tc = tN/tD;
}
Vector3D dP = w + (sc * u) - (tc * v);
double distance1 = Math.Sqrt(Vector3D.DotProduct(dP, dP));
return distance1;
}
我需要什麼：
是否有任何方法來確定替代的端點nt矢量'dP'從上面的代碼？如果沒有，任何人都可以提出一個更好的方法來找到最短距離和距離的終點？
感謝您的閱讀，並提前感謝您的任何建議！
一個巨大的感謝你@Isaac麪包車Bakel的這一解決方案
這背後的理論是我的代碼完成：通過在連接它們的線表示兩條線之間最短的距離那最短的距離。
類：
- Sharp3D.Math：我使用的Vector3D此引用，但實際上任何3D向量類會工作。最重要的是，如果按元素進行減法運算，則甚至不需要使用矢量。
Point3D：我的個人Point3D類。隨意使用盡可能多或少用，如你所願。
class Point3D { public double X { get; set; } public double Y { get; set; } public double Z { get; set; } public Vector3D getVector() { return new Vector3D { X = this.X, Y = this.Y, Z = this.Z }; } }Line：My Personal Line class。隨意使用盡可能多或少用，如你所願。
class Line { public string Name { get; set; } public Point3D Start { get; set; } = new Point3D(); public Point3D End { get; set; } = new Point3D(); public double Length { get { return Math.Sqrt(Math.Pow((End.X - Start.X), 2) + Math.Pow((End.Y - Start.Y), 2)); } } }
功能：
ClampPointToLine：鉗位功能我寫的夾緊點到線。
public Point3D ClampPointToLine(Point3D pointToClamp, Line lineToClampTo) { Point3D clampedPoint = new Point3D(); double minX, minY, minZ, maxX, maxY, maxZ; if(lineToClampTo.Start.X <= lineToClampTo.End.X) { minX = lineToClampTo.Start.X; maxX = lineToClampTo.End.X; } else { minX = lineToClampTo.End.X; maxX = lineToClampTo.Start.X; } if (lineToClampTo.Start.Y <= lineToClampTo.End.Y) { minY = lineToClampTo.Start.Y; maxY = lineToClampTo.End.Y; } else { minY = lineToClampTo.End.Y; maxY = lineToClampTo.Start.Y; } if (lineToClampTo.Start.Z <= lineToClampTo.End.Z) { minZ = lineToClampTo.Start.Z; maxZ = lineToClampTo.End.Z; } else { minZ = lineToClampTo.End.Z; maxZ = lineToClampTo.Start.Z; } clampedPoint.X = (pointToClamp.X < minX) ? minX : (pointToClamp.X > maxX) ? maxX : pointToClamp.X; clampedPoint.Y = (pointToClamp.Y < minY) ? minY : (pointToClamp.Y > maxY) ? maxY : pointToClamp.Y; clampedPoint.Z = (pointToClamp.Z < minZ) ? minZ : (pointToClamp.Z > maxZ) ? maxZ : pointToClamp.Z; return clampedPoint; }distanceBetweenLines：返回表示兩行之間最短距離的行的函數。如果無法解決，則返回null。
public Line distBetweenLines(Line l1, Line l2) { Vector3D p1, p2, p3, p4, d1, d2; p1 = l1.Start.getVector(); p2 = l1.End.getVector(); p3 = l2.Start.getVector(); p4 = l2.End.getVector(); d1 = p2 - p1; d2 = p4 - p3; double eq1nCoeff = (d1.X * d2.X) + (d1.Y * d2.Y) + (d1.Z * d2.Z); double eq1mCoeff = (-(Math.Pow(d1.X, 2)) - (Math.Pow(d1.Y, 2)) - (Math.Pow(d1.Z, 2))); double eq1Const = ((d1.X * p3.X) - (d1.X * p1.X) + (d1.Y * p3.Y) - (d1.Y * p1.Y) + (d1.Z * p3.Z) - (d1.Z * p1.Z)); double eq2nCoeff = ((Math.Pow(d2.X, 2)) + (Math.Pow(d2.Y, 2)) + (Math.Pow(d2.Z, 2))); double eq2mCoeff = -(d1.X * d2.X) - (d1.Y * d2.Y) - (d1.Z * d2.Z); double eq2Const = ((d2.X * p3.X) - (d2.X * p1.X) + (d2.Y * p3.Y) - (d2.Y * p2.Y) + (d2.Z * p3.Z) - (d2.Z * p1.Z)); double[,] M = new double[,] { { eq1nCoeff, eq1mCoeff, -eq1Const }, { eq2nCoeff, eq2mCoeff, -eq2Const } }; int rowCount = M.GetUpperBound(0) + 1; // pivoting for (int col = 0; col + 1 < rowCount; col++) if (M[col, col] == 0) // check for zero coefficients { // find non-zero coefficient int swapRow = col + 1; for (; swapRow < rowCount; swapRow++) if (M[swapRow, col] != 0) break; if (M[swapRow, col] != 0) // found a non-zero coefficient? { // yes, then swap it with the above double[] tmp = new double[rowCount + 1]; for (int i = 0; i < rowCount + 1; i++) { tmp[i] = M[swapRow, i]; M[swapRow, i] = M[col, i]; M[col, i] = tmp[i]; } } else return null; // no, then the matrix has no unique solution } // elimination for (int sourceRow = 0; sourceRow + 1 < rowCount; sourceRow++) { for (int destRow = sourceRow + 1; destRow < rowCount; destRow++) { double df = M[sourceRow, sourceRow]; double sf = M[destRow, sourceRow]; for (int i = 0; i < rowCount + 1; i++) M[destRow, i] = M[destRow, i] * df - M[sourceRow, i] * sf; } } // back-insertion for (int row = rowCount - 1; row >= 0; row--) { double f = M[row, row]; if (f == 0) return null; for (int i = 0; i < rowCount + 1; i++) M[row, i] /= f; for (int destRow = 0; destRow < row; destRow++) { M[destRow, rowCount] -= M[destRow, row] * M[row, rowCount]; M[destRow, row] = 0; } } double n = M[0, 2]; double m = M[1, 2]; Point3D i1 = new Point3D { X = p1.X + (m * d1.X), Y = p1.Y + (m * d1.Y), Z = p1.Z + (m * d1.Z) }; Point3D i2 = new Point3D { X = p3.X + (n * d2.X), Y = p3.Y + (n * d2.Y), Z = p3.Z + (n * d2.Z) }; Point3D i1Clamped = ClampPointToLine(i1, l1); Point3D i2Clamped = ClampPointToLine(i2, l2); return new Line { Start = i1Clamped, End = i2Clamped }; }
實現：
Line shortestDistanceLine = distBetweenLines(l1, l2);
結果：
到目前爲止，這一直在我的測試準確。如果傳遞兩個相同的行，則返回null。我很欣賞任何反饋！
在這個圖像中，它看起來像所需的點是線的中心......就是在這個例子中？因爲如果它的隨機線路幾乎不會是這樣。那是兩條線之間可能的最短距離 – Neil
是的，它是用油漆作爲我試圖獲得的任意例子。這些線不是隨機產生的，但是它們的定位非常隨意。 – Kikootwo
只是爲了澄清，你是說它總是你正在尋找的線段的中點？ –