3D 空间中的激光投影仪校准

我正在研究一种在现实世界中校准激光投影仪的解决方案。该项目有几个目标。

1. Take in a minimum of four points measured in the real world in 3d space that represent the projection surface. 
2. Take in coordinates from the laser that are equivalent to the points received in part 1
3. Determine if the calibration file matches the real world captured points and show a deviation between the coordinate spaces
4. Using the data from the previous steps, take coordinates in 3d real world space and translate them to laser coordinates.

Example:

长方形平板桌子的每个角都有一个目标。目标的一个角通过垫片在空中升高 50 毫米（约 2 英寸）。只是因为目标升高了 50mm，我不希望我的投影表面倾斜。我想考虑高度并仍然将最终产品投影在 Z = 0 平面上。这些目标在现实空间中的坐标如下所示（值均以毫米为单位）：

激光坐标范围从-32000到32000，他们使用光电传感器捕获现实世界中的目标，并在捕获时返回激光坐标系中的值。捕获的坐标如下所示。 （值范围从 -32000 到 32000，它们的单位我不知道）。

Goal:

确定捕获的激光坐标是否正确代表所提供的现实世界坐标（如果不是，则确定与这些坐标的偏差），然后提供变换矩阵或其他一些方法，将 3D 现实世界空间中的点准确地变换为 2D 激光空间。

我尝试过的：

我尝试过实施几种不同的透视校正解决方案。一份来自 OpenCvSharp3，一份来自代码项目线程。这两种方法同样有效，但都有问题。

1. I cannot determine a deviation from the real world. The points are perfectly mapped to the perspective representation so I cannot determine if they are misaligned from the real world expectation
2. I cannot represent source points in a 3d space. Targets may be placed arbitrarily in the Z-direction, and with perspective mapping I cannot seem to account for that Z Direction.

如果有人以前尝试/解决/遇到过这个问题并且能够提供任何见解，我们将不胜感激。

先感谢您！

这里有一个示例代码，显示了所需的正确计算。想象一下真实的表格 X,Y 是 X 与图片水平，Y 向上。为了方便和更容易遵循样本，原点为 0,0,0。然后第二张图像将正确的 X,Y 想象为左上角，其中 X 轴向下，Y 向右（-90 度旋转）。不用担心稍后可以轻松调整。下面的快速示例中的注释解释了每一行的作用及其原因。

// compute from the points the table height and width
var realTableDimX = 902d;
var realTableDimY = 597d;

// the real table matrix is 0,0,0 on bottom corner
// so matrix is identity (our source)
var realTableMatrix = new Matrix3D();

// the laser is rotated 90 degree and position at the top left based compared to the realTablematrix
var laserMatrix = new Matrix3D();

// rotate and translate the laser matrix into position
// 90 degree doesnt account for the little skew (see comment outside code for explaination on how to fix that)
laserMatrix.Rotate(new Quaternion(new Vector3D(0, 0, 1), -90d)); 
laserMatrix.Translate(new Vector3D(0, realTableDimY, 0d));

// compute the laser dimensions (i used your points you found out)
var laserDimX = 20392d - (-16300d);
var laserDimY = 12746 - (-11409d);

// calculate the ratio to factor in to displace the point to the real value onto the laser matrix
var laserXRatio = laserDimX / realTableDimX;
var laserYRatio = laserDimY / realTableDimY;

// since matrix are 1 to 1 when you compute if you have an equal ratio of 40 in X and 40 in Y then the point at 10,10,0 on
// the real table is correcponding to the point 400,400,0 on the laser matrix. But since it's rotated 90 degree you
// will need to apply the Y ration to the X point after the transform and the X ratio to the Y point to get the actual point.

// sample point on table. Logic can be flipped with laser matrix to do it the other way
var sampleRealTablePoint = new Point3D(450, 300, 0); // roughly centered

// transform the point from one UCS to another is simply multiplying by it's current UCS (matrix it's in)
// to transform it to world UCS then multiply by the inverse of other UCS we want it in (the other matrix)
var sampleRealTablePointWorld = realTableMatrix.Transform(sampleRealTablePoint);

// convert that point into the laser matrix but first create and inverted matrix of the laser matrix
// we copy a matrix to not modify the current one when inverting it
var laserInvertedMatrix = new Matrix3D() * laserMatrix;
laserInvertedMatrix.Invert();

// get the sample point in the world of the laser matrix
var sampleRealTablePointToLaserMatrix = laserInvertedMatrix.Transform(sampleRealTablePointWorld);

// not adjust the X and Y like said earlier
var finalAdjustedPoint = new Point3D(sampleRealTablePointToLaserMatrix.X * laserXRatio, sampleRealTablePointToLaserMatrix.Y * laserYRatio, 0d);

 // this is if you want the point in the world of the laser matrix and not it's offset from the 0,0,0
 // the vector here is the top left corner of your laser matrix (x, y, z)
 var laserWorldFinalPoint = finalAdjustedPoint + new Vector3D(-11409d, -16155d, 0d);

所以这里你有它。真实桌子上的样本中心点转换为{12203,5947,0}使用这个代码几乎是激光矩阵的死点。

现在来说说倾斜部分。这也很简单。这一行：

laserMatrix.Rotate(new Quaternion(new Vector3D(0, 0, 1), -90d));

这就是你需要改变的一切。你想要的是始终认为左上角是原点，所以你想要做的就是想象一条向右延伸的直线（在你的激光图像上），并且你想弄清楚这个完美的 X 轴之间的角度是多少（ vector(1,0,0)) 和左下点的向量，在本例中略大于 90 度（可能在 -91 到 -95 度之间）。您需要计算出 Z 轴的角度才能实现这一目标。您使用的公式是否返回 +270 度而不是 -90 度并不重要，因为它是一个矩阵，它会给出相同的结果。

这里是一种基于给定旋转轴计算两个向量之间角度的方法

public double AngleToInDegree(Vector3D v, Vector3D vector, Vector3D normal)
{
    var dotNormal = Vector3D.DotProduct(normal, Vector3D.CrossProduct(v, vector));

    var dotVector = Vector3D.DotProduct(v, vector);

    var angle = Math.Atan2(dotNormal, dotVector);

    return angle * 180d / Math.PI;
}

可用的示例是：

var angle = AngleToInDegree(new Vector3D(1,0,0), new Vector3D(0,1,0), new Vector3D(0,0,1));

这会检查完美 X 向量和完美 Y 向量相对于完美 Z 向量之间的角度，这将为您提供 90 度。