Signed Distance Function - 3D Plane - glsl

I truly love IQ's page and the information within regarding SDF's:
(https://www.iquilezles.org/www/articles/distfunctions/distfunctions.htm)
I have a few figured out but but his sdPlane is getting me..
float sdPlane( vec3 p, vec4 n )
{
// n must be normalized
return dot(p,n.xyz) + n.w;
}
He doesn't identify what the input 'n' is being used for and I am not the greatest in math.
Something like the Width, Height, Depth from point p, but then n.w doesn't make sense..
Any help will be awesome, Thanks!

n is a Plane Equation. .xyz is the normal vector of the plane and .w is the distance from origin.

Related

Project a 3D vertex to screen coordinates independently from OpenGL?

I have a vertex (x, y, z) and I want to calculate the screen location where this point would be rendered on my viewport. Something like Ray Picking, just more or less the other way around. I don't think I can use gluProject because at the time I need the projected point my matrices are restored to identities.
I would like to stay independent from OpenGL, so no extra render pass. This way I'm sure it would only be some math like the ray picking thing. I've implemented that one and it works well, so I want to project a vertex the same way.
Of course I have camera pos, up and lookAt vectors and fovy. Is there any source of information about this? Or does anyone know how to work this out?
If your know your matrices (or at least know how to construct them), you can compute screen location for a vertex by multiplying its position with the matrices and then performing viewport transformation:
vProjected = modelViewPojectionMatrix * v;
if (
// check that vertex shouldn't be clipped.
-vProjected.w <= vProjected.x && vProjected.x <= vProjected.w &&
-vProjected.w <= vProjected.y && vProjected.y <= vProjected.w &&
-vProjected.w <= vProjected.z && vProjected.z <= vProjected.w
) {
vProjected /= vProjected.w;
vScreen.x = VIEWPORT_W * vProjected.x / 2 + VIEWPORT_CENTER_X;
vScreen.y = VIEWPORT_H * vProjected.y / 2 + VIEWPORT_CENTER_Y;
}
Note that, as per OpenGL convention, (0, 0) is lower left corner, not upper left one.
Any math library with verctor and matrix operations can help you with that. For example, mathfu or glm.
UPD. How you can construct modelViewProjectionMatrix given camera position and orientation and projection params? We need two matrices (let's assume that model matrix is just an identity, i.e. vertex positions a given already in world coordinate system). First one would be the view matrix, which takes into account camera position and orientation. Here I'll be using mathfu since I'm more familiar with it, but almost every math library design with 3D graphics in mind has the same functions:
viewMatrix = mathfu::mat4::LookAt(
cameraLookAtPosition,
cameraPosition,
cameraUpVector
);
The second one would be projection matrix:
projectionMatrix = mathfu::mat4::Perspective(fovy, aspect, zNear, zFar);
Now modelViewProjectionMatrix is just a product of those two:
modelViewProjectionMatrix = projectionMatrix * viewMatrix;
Note that matrix multiplication is not commutative, in other words A * B != B * A. So order in which matrices are multiplied is important.

Getting the Tangent for a Object Space to Texture Space

A university assignment requires me to use the Vertex Coordinates I have to calculate the Normals and the Tangent from the Normal values so that I can create a Object Space to Texture Space Matrix.
I have the code needed to make the Matrix, and the binormal but I don't have the code for calculating the Tangent. I tried to look online, but the answers usually confuse me. Can you explain to me clearly how it works?
EDIT: I have corrected what I wrote previously as clearly I misunderstood the assignment. Thank you everyone for helping me see that.
A tangent in the mathematical sense is a property of a geometric object, not of the normalmap. In case of normalmapping, we are in addition searching for a very specific tangent (there are infinitely many in each point, basically every vector in the plane defined by the normal is a tangent).
But let's go one step back: We want a space where the u-direction of the texture is mapped on the tangent direction, the v-direction on the bitangent/binormal and the up-vector of the normalmap to the normal of the object. Thus the tangent for a triangle (v0, v1, v2) with uv-coordinates (uv1, uv2, uv3) can be calculated as:
dv1 = v1-v0
dv2 = v2-v0
duv1 = uv1-uv0
duv2 = uv2-uv0
r = 1.0f / (duv1.x * duv2.y - duv1.y * duv2.x);
tangent = (dv1 * duv2.y - dv2 * duv1.y) * r;
bitangent = (dv2 * duv1.x - dv1 * duv2.x) * r;
When having this done for all triangles, we have to smooth the tangents at shared vertices (quite similar to what happens with the normal). There are several algorithms for doing this, depending on what you need. One can, for example, weight the tangents by the surface area of the adjacent triangles or by the incident angle of them.
An implementation of this whole calculation can be found [here] along a more detailed explaination: (http://www.opengl-tutorial.org/intermediate-tutorials/tutorial-13-normal-mapping/)

Ray tracing texture implementation for spheres

I'm trying to implement textures for spheres in my ray tracer. I managed to get something working, but I am unsure about its correctness. Below is the code for getting the texture coordinates. For now, the texture is random and is generated at runtime.
virtual void GetTextureCoord(Vect hitPoint, int hres, int vres, int& x, int& y) {
float theta = acos(hitPoint.getVectY());
float phi = atan2(hitPoint.getVectX(), hitPoint.getVectZ());
if (phi < 0.0) {
phi += TWO_PI;
}
float u = phi * INV_TWO_PI;
float v = 1 - theta * INV_PI;
y = (int) ((hres - 1) * u);
x = (int) ((vres - 1) * v);
}
This is how the spheres look now:
I had to normalize the coordinates of the hit point to get the spheres to look like that. Otherwise they would look like:
Was normalising the hit point coordinates the right approach, or is something else broken in my code? Thank you!
Instead of normalising the hit point, I tried translating it to the world origin (as if the sphere center was there) and obtained the following result:
I'm using a 256x256 resolution texture by the way.
It's unclear what you mean by "normalizing" the hit point since there's nothing that normalizes it in the code you posted, but you mentioned that your hit point is in world space.
Also, you didn't say what texture mapping you're trying to implement, but I assume you want your U and V texture coordinates to represent latitude and longitude on the sphere's surface.
Your first problem is that converting Cartesian to spherical coordinates requires that the sphere is centered at the origin in the Cartesian space, which isn't true in world space. If the hit point is in world space, you have to subtract the sphere's world-space center point to get the effective hit point in local coordinates. (You figured this part out already and updated the question with a new image.)
Your second problem is that the way you're calculating theta requires that the the sphere have a radius of 1, which isn't true even after you move the sphere's center to the origin. Remember your trigonometry: the argument to acos is the ratio of a triangle's side to its hypotenuse, and is always in the range (-1, +1). In this case your Y-coordinate is the side, and the sphere's radius is the hypotenuse. So you have to divide by the sphere's radius when calling acos. It's also a good idea to clamp the value to the (-1, +1) range in case floating-point rounding error puts it slightly outside.
(In principle you'd also have to divide the X and Z coordinates by the radius, but you're only using those for an inverse tangent, and dividing them both by the radius won't change their quotient and thus won't change phi.)
Right now your sphere intersection and texture-coordinate functions are operating in world space, but you'll probably find it useful later to implement transformation matrices, which let you transform things from one coordinate space to another. Then you can change your sphere functions to operate in a local coordinate space where the center is the origin and the radius is 1, and give each object an associated transformation matrix that maps the local coordinate space to the world coordinate space. This will simplify your ray/sphere intersection code, and let you remove the origin subtraction and radius division from GetTextureCoord (since they're always (0, 0, 0) and 1 respectively).
To intersect a ray with an object, you'd use the object's transformation matrix to transform the ray into the object's local coordinate space, do the intersection (and compute texture coordinates) there, and then transform the result (e.g. hit point and surface normal) back to world space.

Advanced moiré a pattern reduction in HLSL / GLSL procedural textures shader - antialiasing

I am working on a procedural texture, it looks fine, except very far away, the small texture pixels disintegrate into noise and moiré patterns.
I have set out to find a solution to average and quantise the scale of the pattern far away and close up, so that close by it is in full detail, and far away it is rounded off so that one pixel of a distant mountain only represents one colour found there, and not 10 or 20 colours at that point.
It is easy to do it by rounding the World_Position that the volumetric texture is based on using an if statement i.e.:
if( camera-pixel_distance > 1200 meters ) {wpos = round(wpos/3)*3;}//---round far away pixels
return texturefucntion(wpos);
the result of rounding far away textures is that they will look like this, except very far away:
the trouble with this is i have to make about 5 if conditions for the various distances, and i have to estimate a random good rounding value
I tried to make a function that cuts the distance of the pixel into distance steps, and applies a LOD devider to the pixel_worldposition value to make it progressively rounder at distance but i got nonsense results, actually the HLSL was totally flipping out. here is the attempt:
float cmra= floor(_WorldSpaceCameraPos/500)*500; //round camera distance by steps of 500m
float dst= (1-distance(cmra,pos)/4500)*1000 ; //maximum faraway view is 4500 meters
pos= floor(pos/dst)*dst;//close pixels are rounded by 1000, far ones rounded by 20,30 etc
it returned nonsense patterns that i could not understand.
Are there good documented algorithms for smoothing and rounding distance texture artifacts? can i use the scren pixel resolution, combined with the distance of the pixel, to round each pixel to one color that stays a stable color?
Are you familiar with the GLSL (and I would assume HLSL) functions dFdx() and dFdy() or fwidth()? They were made specifically to solve this problem. From the GLSL Spec:
genType dFdy (genType p)
Returns the derivative in y using local differencing for the input argument p.
These two functions are commonly used to estimate the filter width used to anti-alias procedural textures.
and
genType fwidth (genType p)
Returns the sum of the absolute derivative in x and y using local differencing for the input argument p, i.e.: abs (dFdx (p)) + abs (dFdy (p));
OK i found some great code and a tutorial for the solution, it's a simple code that can be tweaked by distance and many parameters.
from this tutorial:
http://www.yaldex.com/open-gl/ch17lev1sec4.html#ch17fig04
half4 frag (v2f i) : COLOR
{
float Frequency = 0.020;
float3 pos = mul (_Object2World, i.uv).xyz;
float V = pos.z;
float sawtooth = frac(V * Frequency);
float triangle = (abs(2.0 * sawtooth - 1.0));
//return triangle;
float dp = length(float2(ddx(V), ddy(V)));
float edge = dp * Frequency * 8.0;
float square = smoothstep(0.5 - edge, 0.5 + edge, triangle);
// gl_FragColor = vec4(vec3(square), 1.0);
if (pos.x>0.){return float4(float3(square), 1.0);}
if (pos.x<0.){return float4(float3(triangle), 1.0);}
}

What is the easiest way to align the Z axis with a vector?

Given a point such as (0, 0, 0) and a vector like (x, y, z). What is the easiest way to align the negative Z-axis centered at (0, 0, 0) to point in the direction of this vector? Examples using OpenGL would be welcome, but not neccessary.
There's lots of different ways to rotate a coordinate-frame to point in a given direction; they'll all leave the z-axis pointed in the direction you want, but with variations in how the x- and y-axes are oriented.
The following gets you the shortest rotation, which may or may not be what you want.
vec3 target_dir = normalise( vector );
float rot_angle = acos( dot_product(target_dir,z_axis) );
if( fabs(rot_angle) > a_very_small_number )
{
vec3 rot_axis = normalise( cross_product(target_dir,z_axis) );
glRotatef( rot_angle, rot_axis.x, rot_axis.y, rot_axis.z );
}
You probably want to have a look at Diana Gruber's article
To answer my own question, the best answer I've come up with is this:
Divide the vector up into "components". The x component is the displacement along the x axis. If we turn to trigonometry, we have that cos(alpha) = x / vector_magnitude. If we compute the RHS then we can derive alpha, which is the amount by which we'd have to rotate around the y axis.
Then the coordinate system can be aligned to the vector by a series of calls to glRotatef()
There are lots of resources out there about rotating your coordinates (or rotating objects, which amounts to the same thing). I learnt a lot from this site, both about how to program in multiple dimensions and especially how to manipulate vectors
The page here has a section "Transformations for moving a vector to the z-axis" that
seems to be what you want, or perhaps the inverse of it.