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计算3D坐标,从2D图像图中计算,不直接查阅观察/预测矩阵
原标题:Calculate 3D coordinates from 2D Image plane accounting for perspective without direct access to view/projection matrix

第一次就交换问题提出问题,希望这是正确的。

我似乎无法为我的情况制定足够近似近似的近似近似算法,因为从3D 数学来看,I m并不是最好的。

我有3个环境,在这种环境中,我可以接触包括我照相机在内的任何物体的位置和轮换,以及从任何两点起的追踪线,以便在一个点到一个碰撞点之间实现距离。 我还有我的摄像场。 然而,我没有任何形式的进入世界/概览/预测矩阵。

I also have a collection of 2d images that are basically a set of screenshots of the 3d environment from the camera, each collection is from the same point and angle and the average set is taken at about an average of a 60 degree angle down from the horizon.

我已经到了使用“登记点实体”这个可以放在代表2个图像角的3个世界的位置上,然后,如果在2个图像上找到一个点,那么,它就被视为与0-1范围相协调,然后在登记点的3个位置之间被推测。 这似乎运作良好,但只有形象是完美的顶点。 当摄像机被利用和引入另一种观点时,结果会变得更严重不准确,因为对这一观点没有赔偿。

我无需能够计算一个点的高度,即一个天空报废器的窗口,但至少在图像平面的基点上进行协调,或者如果我从某个特定图像空间的图像中扩展一条线,至少需要一点,即如果没有任何东西,该线会与地面交叉。

我发现的所有材料都说,只是用世界/概览/预测矩阵来预测这个点,而我发现,除了我没有接触这些矩阵外,这个点本身是直截了当的,只有我可以收集的数据,而其他算法则使用复杂的数学,我根本不掌握。

这样做的一个最终目标,是能够在3d环境中安装标识器,用户在图像中点击,而不能从用户的角度来看进行简单的预测。

感谢任何帮助。

<><>Edit>: 草原,虽然由于我的情况有限,我执行这一办法是一个微不足道的奇迹,但解决办法基本上被挫败,只是 rec弃了观点/预测矩阵。

最佳回答

在摄像机的位置、轮换和排泄物之间,你能否计算照相机的视角/预测矩阵(songho.ca/opengl/gl_projectionmatrix.html),从而使你能够不了解3D点?

问题回答

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