世界空间到相机空间 [英] World space to camera space

问题描述

我对如何将世界空间坐标转换为相机坐标感到困惑.

I am confused on how to convert world space coordinates to camera coordinates.

我目前的理解是我需要计算相机空间向量在哪里

My current understanding is that I would need to calculate the camera space vector where

n = 眼点 - 观察

n = eyepoint - lookat

u = up(0,1,0) X n(归一化)

u = up(0,1,0) X n(normalized)

v = n X u

然后一旦我有 <U、V、N > 我会简单地将每个点乘以吗?

Then once I have < U, V, N > would I simply multiply each point by ?

推荐答案

让我们假设:

• 眼睛位置是 E=(e_x, e_y, e_z),
• 观看方向是D=(d_x, d_y, d_z)
• 向上向量是向上=(up_x, up_y, up_z)

• Eye position is E=(e_x, e_y, e_z),
• Viewing direction is D=(d_x, d_y, d_z)
• Up-Vector is UP=(up_x, up_y, up_z)

现在首先构造一个正交框架:

Now first construct an orthonormal frame:

• R = D X UP
• U = R X D
• 现在对 D、R、U 进行标准化，您就有了相机的正交框架 (D、R、U)

为了将全局坐标系转换为凸轮坐标系，您可以应用以下矩阵M_R:

In order to transform the global coord frame into the cam-coord frame you can apply the following matrix M_R:

• |R_x, R_y, R_z, 0 |
• |U_x, U_y, U_z, 0 |
• |-D_x, -D_y, -D_z, 0|
• |0.0, 0.0, 0.0, 1.0|

如果您的摄像头未定位在全球原点，您还必须应用翻译M_T:

If your cam is not positioned at global origin you also have to apply a translation M_T:

• |1, 0, 0, -e_x |
• |0, 1, 0, -e_y |
• |0, 0, 1, -e_z|
• |0, 0, 0, 1|

最后，从全局到凸轮坐标的完整转换矩阵是:

In the end your complete transformation matrix from global to cam-coords is:

• M = M_R * M_T

• M = M_R * M_T

• |R_x, R_y, R_z, (R dot -E) |
• |U_x, U_y, U_z, (U dot -E) |
• |-D_x, -D_y, -D_z, (D 点 E)|
• |0.0, 0.0, 0.0, 1.0|

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