如何在光线追踪器中使用移动照相机？ [英] How to move a camera using in a ray-tracer?

问题描述

我目前正在研究射线追踪技术，我认为我做得很好。但是，我还没有遮盖相机。

I am currently working on ray-tracing techniques and I think I've made a pretty good job; but, I haven't covered camera yet.

直到现在，我仍使用一个平面片段作为视图平面，该片段位于（-width / 2，height / 2，200）和（宽度/ 2，-高度/ 2，200） [200只是z的固定数，可以更改]。

Until now, I used a plane fragment for view plane which is located between (-width/2, height/2, 200) and (width/2, -height/2, 200) [200 is just a fixed number of z, can be changed].

此外，我主要在 e（0，0，1000）上使用相机，透视投影。

Addition to that, I use the camera mostly on e(0, 0, 1000), and I use a perspective projection.

我将光线从点 e 发送到像素，并在计算出像素颜色后将其打印到图像的相应像素。

I send rays from point e to pixels, and print it to image's corresponding pixel after calculating the pixel color.

这是我的图片创建。希望您可以通过查看图像来猜测眼睛和视线的位置。

Here is a image I created. Hopefully you can guess where eye and view plane are by looking at the image.

我的问题从这里开始。是时候移动相机了，但是我不知道如何将2D视图平面坐标映射到规范坐标。是否有一个转换矩阵？

My question starts from here. It's time to move my camera around, but I don't know how to map 2D view plane coordinates to the canonical coordinates. Is there a transformation matrix for that?

我认为该方法需要知道视图平面上像素的3D坐标。我不确定这是正确的使用方法。那么，您有什么建议呢？

The method I think requires to know the 3D coordinates of pixels on view plane. I am not sure it's the right method to use. So, what do you suggest?

推荐答案

有多种方法可以做到这一点。这是我的工作：

There are a variety of ways to do it. Here's what I do:

1. 选择一个点来表示相机位置（ camera_position ）。


2. 选择一个指示摄像机注视方向的向量（ camera_direction ）。 （如果您知道相机正在注视的点，则可以通过从该点减去 camera_position 来计算该方向矢量。）您可能希望将其标准化（ camera_direction ），在这种情况下，它也是图像平面的法线向量。


3. 选择另一个归一化向量，该向量（大约）从相机的角度向上的视图（ camera_up ）。


4. camera_right =交叉（camera_direction，camera_up）


5. camera_up =十字（camera_right，camera_direction）（这会纠正 up选项中的任何斜率。）

1. Choose a point to represent the camera location (camera_position).
2. Choose a vector that indicates the direction the camera is looking (camera_direction). (If you know a point the camera is looking at, you can compute this direction vector by subtracting camera_position from that point.) You probably want to normalize (camera_direction), in which case it's also the normal vector of the image plane.
3. Choose another normalized vector that's (approximately) "up" from the camera's point of view (camera_up).
4. camera_right = Cross(camera_direction, camera_up)
5. camera_up = Cross(camera_right, camera_direction) (This corrects for any slop in the choice of "up".)

在 camera_position + camera_direction 可视化图像平面的中心。向上和向右的矢量位于图像平面中。

Visualize the "center" of the image plane at camera_position + camera_direction. The up and right vectors lie in the image plane.

您可以选择图像平面的矩形截面以与屏幕相对应。此矩形部分的宽度或高度与camera_direction的长度之比决定了视野。要放大，可以增加camera_direction或减小宽度和高度。

You can choose a rectangular section of the image plane to correspond to your screen. The ratio of the width or height of this rectangular section to the length of camera_direction determines the field of view. To zoom in you can increase camera_direction or decrease the width and height. Do the opposite to zoom out.

因此，给定像素位置（i，j），您需要<图像平面上该像素的code>（x，y，z）。从中可以减去 camera_position 来获得射线矢量（然后需要对其进行归一化）。

So given a pixel position (i, j), you want the (x, y, z) of that pixel on the image plane. From that you can subtract camera_position to get a ray vector (which then needs to be normalized).

Ray ComputeCameraRay(int i, int j) {
  const float width = 512.0;  // pixels across
  const float height = 512.0;  // pixels high
  double normalized_i = (i / width) - 0.5;
  double normalized_j = (j / height) - 0.5;
  Vector3 image_point = normalized_i * camera_right +
                        normalized_j * camera_up +
                        camera_position + camera_direction;
  Vector3 ray_direction = image_point - camera_position;
  return Ray(camera_position, ray_direction);
}

这只是为了说明，因此未进行优化。

This is meant to be illustrative, so it is not optimized.

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