Calculating vertex normals fron Normal map - graphics

I want to generate the normals for vertices displaced by a height map. I have the corresponding normal map however examples I've seen show the the normal vector is equal to the RGB values of the pixel on the normal map. But this means all the normal vectors are positive, when some vectors should have negative values. How would you calculate these using the normal map?
Thanks!

You just shift the range to <-1.0,+1.0> so
if you got color channels in range <0.0,1.0>
// a)
nx=(2.0*r)-1.0
ny=(2.0*g)-1.0
nz=(2.0*b)-1.0
or:
// b)
nx=2.0*(r-0.5)
ny=2.0*(g-0.5)
nz=2.0*(b-0.5)
if you got 8bit per channel then the range is <0,255>
nx=(float(r)/127.5)-1.0
ny=(float(g)/127.5)-1.0
nz=(float(b)/127.5)-1.0
If you look at the normal map image you should see the bluish colors because neutral normal=(0,0,1) pointing up from a flat face is encoded as color=(r=0.5,g=0.5,b=1.0) like here:
also have a look here: Normal mapping gone horribly wrong where the normal is computed from such texture in GLSL by #1b method:
const vec4 v05=vec4(0.5,0.5,0.5,0.5);
texture2D(txr_normal,pixel_txr.st)-v05)*2.0;
Also the (r,g,b) can be mapped to (nz,ny,nx) instead of (nx,ny,nz) in that case just swap r,b (the normal map is then red-ish instead)

Related

What's the different between using modelViewmatrix directly and using normalMatrix instead? [duplicate]

I am working on some shaders, and I need to transform normals.
I read in few tutorials the way you transform normals is you multiply them with the transpose of the inverse of the modelview matrix. But I can't find explanation of why is that so, and what is the logic behind that?
It flows from the definition of a normal.
Suppose you have the normal, N, and a vector, V, a tangent vector at the same position on the object as the normal. Then by definition N·V = 0.
Tangent vectors run in the same direction as the surface of an object. So if your surface is planar then the tangent is the difference between two identifiable points on the object. So if V = Q - R where Q and R are points on the surface then if you transform the object by B:
V' = BQ - BR
= B(Q - R)
= BV
The same logic applies for non-planar surfaces by considering limits.
In this case suppose you intend to transform the model by the matrix B. So B will be applied to the geometry. Then to figure out what to do to the normals you need to solve for the matrix, A so that:
(AN)·(BV) = 0
Turning that into a row versus column thing to eliminate the explicit dot product:
[tranpose(AN)](BV) = 0
Pull the transpose outside, eliminate the brackets:
transpose(N)*transpose(A)*B*V = 0
So that's "the transpose of the normal" [product with] "the transpose of the known transformation matrix" [product with] "the transformation we're solving for" [product with] "the vector on the surface of the model" = 0
But we started by stating that transpose(N)*V = 0, since that's the same as saying that N·V = 0. So to satisfy our constraints we need the middle part of the expression — transpose(A)*B — to go away.
Hence we can conclude that:
transpose(A)*B = identity
=> transpose(A) = identity*inverse(B)
=> transpose(A) = inverse(B)
=> A = transpose(inverse(B))
My favorite proof is below where N is the normal and V is a tangent vector. Since they are perpendicular their dot product is zero. M is any 3x3 invertible transformation (M-1 * M = I). N' and V' are the vectors transformed by M.
To get some intuition, consider the shear transformation below.
Note that this does not apply to tangent vectors.
Take a look at this tutorial:
https://paroj.github.io/gltut/Illumination/Tut09%20Normal%20Transformation.html
You can imagine that when the surface of a sphere stretches (so the sphere is scaled along one axis or something similar) the normals of that surface will all 'bend' towards each other. It turns out you need to invert the scale applied to the normals to achieve this. This is the same as transforming with the Inverse Transpose Matrix. The link above shows how to derive the inverse transpose matrix from this.
Also note that when the scale is uniform, you can simply pass the original matrix as normal matrix. Imagine the same sphere being scaled uniformly along all axes, the surface will not stretch or bend, nor will the normals.
If the model matrix is made of translation, rotation and scale, you don't need to do inverse transpose to calculate normal matrix. Simply divide the normal by squared scale and multiply by model matrix and we are done. You can extend that to any matrix with perpendicular axes, just calculate squared scale for each axes of the matrix you are using instead.
I wrote the details in my blog: https://lxjk.github.io/2017/10/01/Stop-Using-Normal-Matrix.html
Don't understand why you just don't zero out the 4th element of the direction vector before multiplying with the model matrix. No inverse or transpose needed. Think of the direction vector as the difference between two points. Move the two points with the rest of the model - they are still in the same relative position to the model. Take the difference between the two points to get the new direction, and the 4th element, cancels out to zero. Lot cheaper.

Homogeneous coordinates and perspective-correctness?

Does the technique that vulkan uses (and I assume other graphics libraries too) to interpolate vertex attributes in a perspective-correct manner require that the vertex shader must normalize the homogenous camera-space vertex position (ie: divide through by the w-coordinate such that the w-coordinate is 1.0) prior to multiplication by a typical projection matrix of the form...
g/s 0 0 0
0 g 0 n
0 0 f/(f-n) -nf/(f-n)
0 0 1 0
...in order for perspective-correctness to work properly?
Or, will perspective-correctness continue to work on any homogeneous vertex position in camera-space (with a w-coordinate other than 1.0)?
(I didn't completely follow the perspective-correctness math, so it is unclear which to me which is the case.)
Update:
In order to clarify terminology:
vec4 modelCoordinates = vec4(x_in, y_in, z_in, 1);
mat4 modelToWorld = ...;
vec4 worldCoordinates = modelToWorld * modelCoordinates;
mat4 worldToCamera = ...;
vec4 cameraCoordinates = worldToCamera * worldCoordinates;
mat4 cameraToProjection = ...;
vec4 clipCoordinates = cameraToProjection * cameraCoordinates;
output(clipCoordinates);
cameraToProjection is a matrix like the one shown in the question
The question is does cameraCoordinates.w have to be 1.0?
And consequently the last row of both the modelToWorld and worldToCamera matricies have to be 0 0 0 1?
You have this exactly backwards. Doing the perspective divide in the shader is what prevents perspective-correct interpolation. The rasterizer needs the perspective information provided by the W component to do its job. With a W of 1, the interpolation is done in window space, without any regard to perspective.
Provide a clip-space coordinate to the output of your vertex processing stage, and let the system do what it exists to do.
the vertex shader must normalize the homogenous camera-space vertex position (ie: divide through by the w-coordinate such that the w-coordinate is 1.0) prior to multiplication by a typical projection matrix of the form...
If your camera-space vertex position does not have a W of 1.0, then one of two things has happened:
You are deliberately operating in a post-projection world space or some similar construct. This is a perfectly valid thing to do, and the math for a camera space can be perfectly reasonable.
Your code is broken somewhere. That is, you intend for your world and camera space to be a normal, Euclidean, non-homogeneous space, but somehow the math didn't work out. Obviously, this is not a perfectly valid thing to do.
In both cases, dividing by W is the wrong thing to do. If your world space that you're placing a camera into is post-projection (such as in this example), dividing by W will break your perspective-correct interpolation, as outlined above. If your code is broken, dividing by W will merely mask the actual problem; better to fix your code than to hide the bug, as it may crop up elsewhere.
To see whether or not the camera coordinates need to be in normal form, let's represent the camera coordinates as multiples of w, so they are (wx,wy,wz,w).
Multiplying through by the given projection matrix, we get the clip coordinates (wxg/s, wyg, fwz/(f-n)-nfw/(f-n)), wz)
Calculating the x-y framebuffer coordinates as per the fixed Vulkan formula we get (P_x * xg/sz +O_x, P_y * Hgy/z + O_y). Notice this does not depend on w, so the position in the framebuffer of a polygons verticies doesn't require the camera coordinates be in normal form.
Likewise calculation of the barycentric coordinates of fragments within a polygon only depends on x,y in framebuffer coordinates, and so is also independant of w.
However perspective-correct perspective interpolation of fragment attributes does depend on W_clip of the verticies as this is used in the formula given in the Vulkan spec. As shown above W_clip is wz which does depend on w and scales with it, so we can conclude that camera coordinates must be in normal form (their w must be 1.0)

How do you pack a 3-floats (space vector) into 4 bytes (pixel)?

I've successfully packed floats with values in [0,1] without losing too much precision using:
byte packedVal = floatVal * 255.0f ; // [0,1] -> [0,255]
Then when I want to unpack the packedVal back into a float, I simply do
float unpacked = packedVal / 255.0f ; // [0,255] -> [0,1]
That works fine, as long as the floats are between 0 and 1.
Now here's the real deal. I'm trying to turn a 3d space vector (with 3 float components) into 4 bytes. The reason I'm doing this is because I am using a texture to store these vectors, with 1 pixel per vector. It should be something like a "normal map", (but not exactly this, you'll see why after the jump)
So there, each pixel represents a 3d space vector. Where the value is very red, the normal vector's direction is mostly +x (to the right).
So of course, normals are normalized. So they don't require a magnitude (scaling) vector. But I'm trying to store a vector with arbitrary magnitude, 1 vector per pixel.
Because textures have 4 components (rgba), I am thinking of storing a scaling vector in the w component.
Any other suggestions for packing an arbitrary sized 3 space vector, (say with upper limit on magnitude of 200 or so on each of x,y,z), into a 4-byte pixel color value?
Storing the magnitude in the 4th component sounds very reasonable. As long as the magnitude is bounded to something reasonable and not completely arbitrary.
If you want a more flexible range of magnitudes you can pre-multiply the normalized direction vector by (0.5, 1.0] when you store it, and when you unpack it multiply it by pow(2, w).
Such method is used for storing high dynamic range images - RGBM encoding (M stands for magnitude). One of it's drawbacks is wrong results from interpolation so you can't use bilinear filtering for your texture.
You can look for other options from HDR encodings: here is a small list of few most popular

HLSL beginner needs some directions

Is there any example out there of a HLSL written .fx file that splats a tiled texture with different tiles?Like this: http://messy-mind.net/blog/wp-content/uploads/2007/10/transitions.jpg you can see theres a different tile type in each square and there's a little blurring between them to make a smoother transition,but right now I just need to find a way to draw the tiles on a texture.I have a 2D array of integers,each integer equals a corresponding tile type(0 = grass,1 = stone,2 = sand).I opened up a few HLSL examples and they were really confusing.Everything is running fine on the C++ side,but HLSL is proving to be difficult.
You can use a technique called 'texture splatting'. It mixes several textures (color maps) using another texture which contains alpha values for each color map. The texture with alpha values is an equivalent of your 2D array. You can create a 3-channel RGB texture and use each channel for a different color map (in your case: R - grass, G - stone, B - sand). Every pixel of this texture tells us how to mix the color maps (for example R=0 means 'no grass', G=1 means 'full stone', B=0.5 means 'sand, half intensity').
Let's say you have four RGB textures: tex1 - grass, tex2 - stone, tex3 - sand, alpha - mixing texture. In your .fx file, you create a simple vertex shader which just calculates the position and passes the texture coordinate on. The whole thing is done in pixel shader, which should look like this:
float tiling_factor = 10; // number of texture's repetitions, you can also
// specify a seperate factor for each texture
float4 PS_TexSplatting(float2 tex_coord : TEXCOORD0)
{
float3 color = float3(0, 0, 0);
float3 mix = tex2D(alpha_sampler, tex_coord).rgb;
color += tex2D(tex1_sampler, tex_coord * tiling_factor).rgb * mix.r;
color += tex2D(tex2_sampler, tex_coord * tiling_factor).rgb * mix.g;
color += tex2D(tex3_sampler, tex_coord * tiling_factor).rgb * mix.b;
return float4(color, 1);
}
If your application supports multi-pass rendering you should use it.
You should use a multi-pass shader approach where you render the base object with the tiled stone texture in the first pass and on top render the decal passes with different shaders and different detail textures with seperate transparent alpha maps.
(Transparent map could also be stored in your detail texture, but keeping it seperate allows different tile-levels and more flexibility in reusing it.)
Additionally you can use different texture coordinate channels for each decal pass one so that you do not need to hardcode your tile level.
So for minimum you need two shaders, whereas Shader 2 is used as often as decals you need.
Shader to render tiled base texture
Shader to render one tiled detail texture using a seperate transparency map.
If you have multiple decals z-fighting can occur and you should offset your polygons a little. (Very similar to basic simple fur rendering.)
Else you need a single shader which takes multiple textures and lays them on top of the base tiled texture, this solution is less flexible, but you can use one texture for the mix between the textures (equals your 2D-array).

HLSL tex2d sampler seemingly using inconsistent rounding; why?

I have code that needs to render regions of my object differently depending on their location. I am trying to use a colour map to define these regions.
The problem is when I sample from my colour map, I get collisions. Ie, two regions with different colours in the colourmap get the same value returned from the sampler.
I've tried various formats of my colour map. I set the colours for each region to be "5" apart in each case;
Indexed colour
RGB, RGBA: region 1 will have RGB 5%,5%,5%. region 2 will have RGB 10%,10%,10% and so on.
HSV Greyscale: region 1 will have HSV 0,0,5%. region 2 will have HSV 0,0,10% and so on.
(Values selected in The Gimp)
The tex2D sampler returns a value [0..1].
[ I then intend to derive an int array index from region. Code to do with that is unrelated, so has been removed from the question ]
float region = tex2D(gColourmapSampler,In.UV).x;
Sampling the "5%" colour gave a "region" of 0.05098 in hlsl.
From this I assume the 5% represents 5/100*255, or 12.75, which is rounded to 13 when stored in the texture. (Reasoning: 0.05098 * 255 ~= 13)
By this logic, the 50% should be stored as 127.5.
Sampled, I get 0.50196 which implies it was stored as 128.
the 70% should be stored as 178.5.
Sampled, I get 0.698039, which implies it was stored as 178.
What rounding is going on here?
(127.5 becomes 128, 178.5 becomes 178 ?!)
Edit: OK,
http://en.wikipedia.org/wiki/Bankers_rounding#Round_half_to_even
Apparently this is "banker's rounding". I have no idea why this is being used, but it solves my problem. Apparently, it's a Gimp issue.
I am using Shader Model 2 and FX Composer. This is my sampler declaration;
//Colour map
texture gColourmapTexture <
string ResourceName = "Globe_Colourmap_Regions_Greyscale.png";
string ResourceType = "2D";
>;
sampler2D gColourmapSampler : register(s1) = sampler_state {
Texture = <gColourmapTexture>;
#if DIRECT3D_VERSION >= 0xa00
Filter = MIN_MAG_MIP_LINEAR;
#else /* DIRECT3D_VERSION < 0xa00 */
MinFilter = Linear;
MipFilter = Linear;
MagFilter = Linear;
#endif /* DIRECT3D_VERSION */
AddressU = Clamp;
AddressV = Clamp;
};
I never used HLSL, but I did use GLSL a while back (and I must admit it's terribly far in my head).
One issue I had with textures is that 0 is not the first pixel. 1 is not the second one. 0 is the edge of the texture and 1 is the right edge of the first pixel. The values get interpolated automatically and that can cause serious trouble if what you need is precision like when applying a lookup table rather than applying a normal texture. You need to aim for the middle of the pixel, so asking for [0.5,0.5], [1.5,0.5] rather than [0,0], [1, 0] and so on.
At least, that's the way it was in GLSL.
Beware: region in levels[region] is rounded down. When you see 5 % in your image editor, the actual value in the texture 8b representation is 5/100*255 = 12.75, which may be either 12 or 13. If it is 12, the rounding down will hit you. If you want rounding to nearest, you need to change this to levels[region+0.5].
Another similar thing (already written by Louis-Philippe) which might hit you is texture coordinates rounding rules. You always need to hit a spot in the texel so that you are not in between of two texels, otherwise the result is ill-defined (you may get any of two randomly) and some of your source texels may disapper while other duplicate. Those rules are different for bilinar and point sampling, you may need to add half of texel size when sampling to compensate for this.
GIMP uses banker's rounding. Apparently.
This threw out my code to derive region indicies.

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