射线追踪相机 [英] Ray-Tracing Camera

问题描述

我目前正在研究射线追踪技术，我认为我已经做得非常好;但是，我还没有涵盖相机。到目前为止，我使用了一个位于（ - width / 2，height / 2，200）和 2，-height / 2,200） [200只是z的固定数，可以改变]。除此之外，我使用相机主要在 e（0,0,1000）并使用透视投影。我将点 e 的光线发送到像素，并在计算像素颜色后将其打印到图像的相应像素。

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. 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]. Addition to that, I use the camera mostly on e(0,0,1000) and use perspective projection. I send rays from point e to pixels, and print it to image's corresponding pixel after calculating the pixel color.

这是我创建的图片。希望你可以通过看图像来猜测眼睛和视图平面。

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。现在是时候移动我的相机，但我不知道如何映射2D视图平面坐标到规范坐标。有一个转换矩阵吗？该方法我认为需要知道视图平面上的像素的3D坐标。我不知道这是正确的方法使用。那么你建议什么呢？

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? 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?

PS：我希望我能表达自己，因为试图解释这个没有绘图是我真正的挑战。

P.S.: I hope I could express myself, because trying to explain this without drawing is a real challenge for me.

推荐答案

有多种方法可以做到。这里是我做的：

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

1. 选择一个点来代表相机位置（ camera_position ）。


2. 选择一个表示相机方向的向量（ camera_direction ）。 （如果你知道相机正在看的一个点，你可以通过从那个点减去 camera_position 来计算这个方向向量。）你可能想标准化（

3. 选择另一个标准化的向量，从相机的点（大约）上选择另一个标准化的向量， c $> camera_right = Cross（camera_direction，camera_up）


4. camera_up = Cross（camera_right，camera_direction）（这将纠正up选择中的任何偏差。 li>
   

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 处显示图像平面的向上和向右的矢量位于图像平面。

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

您可以选择一个矩形截面的图像平面对应于您的屏幕。此矩形部分的宽度或高度与camera_direction的长度的比率确定视野。要放大，您可以增加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，假设像素位置（i，j） code>（x，y，z）。从中你可以减去 camera_position 得到一个光线矢量（然后需要被归一化）。

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);
}

这是说明性的，因此未进行优化。

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

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