Hi!
How to write this math expression EAC projection formula in the material editor ?
What am I doing wrong?
Example Image EAC projection from youtube.
In fact, you should get an equirectangular projection.
Hey, We need to think about whether to use a cubic map ?
i find amazing theme http://paulbourke.net/miscellaneous/cubemaps Written by Paul Bourke
Converting to and from 6 cubic environment maps and a spherical projection
Algorithm
The conversion process involves two main stages. The goal is to determine the best estimate of the colour at each pixel in the final spherical image given the 6 cubic texture images. The first stage is to calculate the polar coordinates corresponding to each pixel in the spherical image. The second stage is to use the polar coordinates to form a vector and find which face and which pixel on that face the vector (ray) strikes. In reality this process is repeated a number of times at slightly different positions in each pixel in the spherical image and an average is used in order to avoid aliasing effects.
If the coordinates of the spherical image are (i,j) and the image has width “w” and height “h” then the normalised coordinates (x,y) each ranging from -1 to 1 are given by:
x = 2 i / w - 1
y = 2 j / h - 1
or y = 1 - 2 j / h depending on the position of pixel 0
The polar coordinates theta and phi are derived from the normalised coordinates (x,y) below. theta ranges from 0 to 2 pi and phi ranges from -pi/2 (south pole) to pi/2 (north pole). Note there are two vertical relationships in common use, linear and spherical. In the former phi is linearly related to y, in the later there is a sine relationship.
theta = x pi
phi = y pi / 2
or phi = asin(y) for spherical vertical distortion
The polar coordinates (theta,phi) are turned into a unit vector (view ray from the camera) as below. This assumes a right hand coordinate system, x to the right, y upwards, and z out of the page. The front view of the cubic map is looking from the origin along the positive z axis.
x = cos(phi) cos(theta)
y = sin(phi)
z = cos(phi) sin(theta)
The intersection of this ray is now found with the faces of the cube. Once the intersection point is found the coordinate on the square face specifies the corresponding pixel and therefore colour associated with the ray.
