Surface Textures in the Fragment Shader - Graphics Shaders - page 194

Following the same patern as before, we see that the normal vector, n , is

π π x

2 2 201

n =−∗∗ − a sin π t , ., ..

P P

d d

This is true along just the X axis. Because the ripples are propagating out
in a circular patern, we need to rotate the normal vector into where it really is.
Because we are just talking about a rotation, the transformation is the same as
if we were rotating a vertex. Noting that the unrotated N y is equal to 0., we get

′ =∗ −∗ =∗
′ =∗ +∗ =∗
′ =

Nx

Nx cos Θ Ny sin Θ Nx cos Θ

Ny Nx sin Θ Ny cos Θ Nx sin Θ

Nz Nz =1.

As we said at the beginning, so far we have been working in 2D, not 3D;
for the final version of the fragment shader in the ripples case above, you need
to think in polar coordinates, so you need to substitute R , the polar coordinate
radius, for x in the slope equation.

uniform float uTime;
uniform float uAmp0, uAmp1;
uniform float uPhaseShift;
uniform float uPd;
uniform float uLightX, uLightY, uLightZ;
uniform vec4 uColor;

in vec3 vMCposition;
in vec3 vECposition;

out vec4 fFragColor;

const float TWOPI = 2.*3.14159265;

void main( )
{
const vec3 C0 = vec3( -2.5, 0., 0. );
const vec3 C1 = vec3( 2.5, 0., 0. );

// first set of ripples:

float rad0 = length( vMCposition - C0 ); // ripple center 0
float H0 = -uAmp0 * cos( TWOPI*rad0/uPd - TWOPI*uTime );

float u = -uAmp0 * (TWOPI/uPd) * sin(TWOPI*rad0/uPd -
TWOPI*uTime);