Ray Tracing 004The Ray Class

Summary

  • Math foundations: ray as a line in 3D space from an origin pointing into a direction.
  • Added a ray class to help compute coordinates of a point on the ray at a given time.
  • Updated to let the ray draw a gradient from blue to white along the y-coordinate.

A ray can be thought of as a function:

p(t)=A+tāˆ—B\begin{equation} p(t) = A + t * B \end{equation}

where p is a 3D position along a line, A is the ray origin, B is the ray direction, and t is the ray parameter which moves p(t) along the ray.

image/svg+xml

If t is positive you get the points in front of the origin A, also called a half-line or half-ray.

The ray class is a literal translation of this equation in 3D vectors:

// ray.h
#ifndef RAYH
#define RAYH
#include "vec3.h"

class ray {
public:
  vec3 A;
  vec3 B;

  ray() {}
  ray(const vec3& a, const vec3& b) { A = a; B = b; }

  vec3 origin() const { return A; }
  vec3 direction() const { return B; }
  vec3 point_at_parameter(float t) const { return A + t * B; }
};
#endif

The new loop in the main program moves the ray across the entire image and determines a color at each pixel. This example will only return a gradient based on the y-position.

image/svg+xml

The size of the image is set to 4 by 2, where the origin is set into the center of image (0,0,0), the lower left corner is at (-2,-1,-1), the upper right corner is at (2,1,-1). The z-coordinate is set at -1 into the screen to respect the convention of a right-handed coordinate system.

The screen is 1 unit apart from the origin (eye or camera). Positive values for z are behind the camera.

Click to view C++ source code

The color() function takes the current ray's direction and computes a unit-direction. This unit-direction is a vector of length 1.

The following math is more verbose than it needs to be, but because unit vectors play such an important role, I thought it would be a good idea to at least once describe in full detail how the unit vector is calculated. It can also serve as a refresher on notation and what exactly is behind the common vector symbols.

u^=uāˆ£uāˆ£\begin{equation} \hat{\boldsymbol{u}} = \frac{\boldsymbol{u}}{|\boldsymbol{u}|} \end{equation}
āˆ£uāˆ£=ux2+uy2+uz2\begin{equation} \boldsymbol{|u|} = \sqrt{u_x^2 + u_y^2 + u_z^2} \end{equation}

Example: For vector u\boldsymbol{u}:

u=(234)\begin{equation} \boldsymbol{u} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \end{equation}

the length of the vector āˆ£uāˆ£|\boldsymbol{u}| is approximately 5.39:

āˆ£uāˆ£=22+32+42Ā =4+9+16Ā =29ā‰ˆ5.39\begin{equation} \begin{aligned}|\boldsymbol{u}| & = \sqrt{2^2 + 3^2 + 4^2} \ & = \sqrt{4+9+16} \ & = \sqrt{29} \approx 5.39 \end{aligned} \end{equation}

The unit vector u^\hat{\boldsymbol{u}} is:

u^=uāˆ£uāˆ£=(229329429)ā‰ˆ(0.370.560.74)\begin{equation} \hat{\boldsymbol{u}} = \frac{\boldsymbol{u}}{|\boldsymbol{u}|} = \begin{pmatrix} \frac{2}{\sqrt{29}} \\ \frac{3}{\sqrt{29}} \\ \frac{4}{\sqrt{29}} \end{pmatrix} \approx \begin{pmatrix} 0.37 \\ 0.56 \\ 0.74 \end{pmatrix} \end{equation}

To double-check correctness, the length of the unit vector should be 1:

āˆ£(0.370.560.74)āˆ£=0.372+0.562+0.742=0.14+0.31+0.55=1\begin{equation} \begin{aligned}\left|\begin{pmatrix} 0.37 \\ 0.56 \\ 0.74 \end{pmatrix}\right| &= \sqrt{0.37^2 + 0.56^2 + 0.74^2} \\ &= \sqrt{0.14 + 0.31 + 0.55} \\ &= 1 \end{aligned} \end{equation}

Vector example, its components and the resulting unit vector

The y-component of this unit vector can take values from -1 to +1. To turn this into a range from 0 to 1, I add 1 to the value and divide it by 2.

The next line,

return (1.0f - t) * vec3(1.0f, 1.0f, 1.0f) + t * vec3(0.5f, 0.7f, 1.0f);

creates a linear interpolation (or "lerp") between white and blue. The first half of the equation turns 0 and leaves blue (0.5, 0.7, 1.0) when t is 1. The second half turns 0 and leaves white (1.0, 1.0, 1.0) when t is 0.

Each half will contribute to the final color for every t in between. Smaller t (y-coordinates towards the bottom of the image) result in brighter blue to white colors. Larger t (y-coordinates towards the top of the image) produce darker tones of blue.

Inside the loop I'm converting the pixel coordinates (from 0-199 in x-direction, from 0-99 in y-direction) to the world coordinate system (from -2 to +2 in x-direction, -1 to +1 in y-direction).

ray r(origin, lower_left_corner + u * horizontal + v * vertical);
vec3 col = color(r);

Javascript and Canvas

Here's the example in Javascript.

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Figure RT004: Simple canvas image rendered with Javascript

Click to view Javascript source code

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