如何使用校准矩阵估算基线(how to estimate baseline using calibration matrix)

如何使用校准矩阵估算基线(how to estimate baseline using calibration matrix)

如果给出两个摄像机的4乘4校准矩阵，是否有可能计算两个摄像机的基线值？ 基准值是以米或毫米计算的吗？

希望有人可以帮助我吗？ TT

is there any possibility to calculate the baseline value of two cameras if the 4 by 4 calibration matrix of both cameras are given? will the baseline value calculated in meter or milimeter?

Hope anyone can help me pls ? TT

最满意答案

在减去平移向量时，请注意不要混淆参考帧。 如果你的外在矩阵是[R | t] [R | t] ，取t1和t2之间的欧几里德距离不会给你基线。 这是因为t是世界原点在相机坐标上的位置，并且参考帧对于每个相机是不同的。 减去具有不同参考帧的向量是无效操作。

你想要的是世界坐标中两个摄像机中心之间的差异，即|c1 - c2| 其中c = -R't 。

外在矩阵中的这些参考框架问题可能是微妙和混乱的。 有关该主题的更长时间的讨论，请查看我去年写的这篇博客 。 它包括一个交互式演示，演示了t和c不同之处。

Be careful not to confuse your reference frames when subtracting translation vectors. If your extrinsic matrix is [R | t], taking the Euclidean distance between t1 and t2 will not give you the baseline. This is because t is the position of the world origin on camera coordinates, and the reference frame is different for each camera. Subtracting vectors with different reference frames is an invalid operation.

What you want is the difference between the two cameras' centers in world coordinates, i.e. |c1 - c2| where c = -R't.

These reference frame issues in the extrinsic matrix can be subtle and confusing. For a longer discussion on the topic, check out this blog I wrote last year. It includes an interactive demo that illustrates how t and c differ.