## glsl – How to write a shader that only uses a relief map without a normal map

Relief mapping simulates surface displacement, that is, the shader pretends that the pixel is actually further behind the surface than normal. Then compare the depth with the pixels next to it on the map, and so you can set a gradient, using Pythagoras: if we know the height difference and the difference in pixels to the left / right / up / down / etc. Then we can calculate the slope in each direction. Then we can use those slopes to calculate a new normal vector for that pixel. This implies several texture samples per pixel, in the sobel matrix you mentioned, and in fact it is how it was done "in the day".

For this reason, it is expensive and suboptimal, and in the case of blurred height maps, it can cause artifacts like the ones you saw.

Normal mapping, on the other hand, uses a normal map, precooked from a protuberance or ma displacement, in exactly the same method as the previous one. The difference is that by doing it beforehand, you save at least 4-8 texture samples per pixel, in addition to not having to calculate the slopes per pixel.
Not having to do this AWAY exceeds the cost of matrix transformations from tangent to visual space per pixel, which makes lighting much more efficient. This is how it is done in today's hardware, although tomorrow's graphics engine will probably use ray tracing instead.

## GET – Ted McGrath – Marketing Master Map | Proxies123.com

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## dg differential geometry: effect of the inverse exponential map on the curvature of a given curve

Suppose you have a curve $$alpha$$ in a variety you are at one point $$alpha (t)$$ of that curve. The curvature of $$alpha (t)$$ it is the same as the curvature of the curve $$exp <-1> {(t)} (α (s))$$ to $$s = t$$. The previous curve is what you see from the tangent space of $$alpha (t)$$.

Now imagine that you move to another point $$x$$and you want to determine how the curvature of $$alpha (t)$$ has changed when viewed from the tangent space of $$x$$.

That is, what is the curvature of $$exp <-1x (α (s))$$ to $$s = t$$? (when $$x$$ Y $$alpha (t)$$ they are close enough that we have a local difeomorphism, for example).

For now it is good for me to know the answer when the multiple is a space of constant sectional curvature. I guess in that case it will only depend on the distance between $$x$$ Y $$alpha (t)$$ and the relative position between the geodetic junction $$x$$ Y $$alpha (t)$$ with respect to the hyperplane (in the tangent space of $$alpha (t)$$) tangent to $$alpha$$ to $$t$$.

## glsl: I would like to learn to write a shader that only uses a relief map without a normal map

Relief mapping simulates surface displacement, that is, the shader pretends that the pixel is actually further behind the surface than normal. Then compare the depth with the pixels next to it on the map, and so you can set a gradient, using Pythagoras: if we know the height difference and the difference in pixels to the left / right / up / down / etc. Then we can calculate the slope in each direction. Then we can use those slopes to calculate a new normal vector for that pixel. This implies several texture samples per pixel, in the sobel matrix you mentioned, and in fact it is how it was done "in the day".

For this reason, it is expensive and suboptimal, and in the case of blurred height maps, it can cause artifacts like the ones you saw.

Normal mapping, on the other hand, uses a normal map, precooked from a protuberance or ma displacement, in exactly the same method as the previous one. The difference is that by doing it beforehand, you save at least 4-8 texture samples per pixel, in addition to not having to calculate the slopes per pixel.
Not having to do this AWAY exceeds the cost of matrix transformations from tangent to visual space per pixel, which makes lighting much more efficient. This is how it is done in today's hardware, although tomorrow's graphics engine will probably use ray tracing instead.

## Does it cost anything to add a marker on a google map?

I have an Android application that shows a map with a single marker at a specific longitude and latitude for all users
So does this cost anything?
I read the price table but there is something unclear about this point.

## homological algebra – free DGA given a map and cohomology groups

Why do free DGAs in a morphism often give the same (co) homology as other (co) homologies?

Here is an example:

Example: Leave $$R$$ be a ring, let $$A$$ bean $$R$$-algebra, leave $$M$$ bean $$A$$-module and leave $$d: A rightarrow M$$ Be a derivation. There is a categorical construction $$Lambda (d)$$, the free anticomutative DGA given an element of $$text {Der} _R (A, M)$$, which is a cochain complex with multiplication and unity in the category of cochain complexes, defined by the configuration $$Lambda (d) ^ i = 0$$ for $$i <0$$, $$Lambda (d) ^ i = Lambda ^ i_A M$$ in degree $$n$$ for $$n geq 0$$, and where the differential is specified exclusively by setting
$$d (x wedge y) = d (x) y + (-1) ^ { text {deg} (x) text {deg} (y)} x d (y)$$
By free, I mean that it is part of an attachment between a category whose objects are derivations and a category whose objects are DGA.

In particular (and very interesting in particular), Yes $$X$$ it is a smooth collector $$C ^ { infty} (X)$$ that's $$mathbb {R}$$-algebra of $$C ^ { infty}$$-functions to $$mathbb {R}$$Y $$T ^ * (X)$$ is your cotangent space, then we have a derivation $$d: C ^ { infty} (X) rightarrow T ^ * (X)$$. Taking the left attachment I defined above, you get the free anticomutative DGA in this derivation, which many of us know and love as the De Rham complex $$Omega ^ * (X)$$.

The interesting thing is that the cohomology of this resolution is the same as the cohomology of others:
$$H_ {dRh} ^ n (X) cong H ^ n_ {sing} (X; mathbb {R})$$
$$H_ {dRh} ^ n (X) cong hat {H} ^ n (X, mathbb {R})$$
Where the last cohomology group is supposed to be Čech cohomology.

Here is another example:

Example: Leave $$R$$ be a ring and leave $$A$$ bean $$R$$-algebra. Leave $$M$$ bean $$A$$-module and leave $$d: M rightarrow A$$ bean $$A$$-Map of modules. We can form the free DGA (this time, a chain complex, not a chain complex, and this time resulting in homology, not cohomology), with $$M$$ in degree $$1$$ Y $$R$$ in degree $$0$$. We can form the Koszul complex, which is the free DGA given the map $$d: M rightarrow A$$. Is $$0$$ in negative degrees, and $$Lambda ^ i_A M$$ in degree $$i$$. The differential is uniquely specified by requiring that $$d (x wedge y) = d (x) y + (-1) ^ { text {deg} (x) text {deg} (y)} x d (y)$$.

Yes $$M$$ it's free with base $$x_1, …, x_n$$ Y $$x_i$$ map to a regular sequence $$a_i$$ in $$R$$ so the resolution of koszul is acyclic except in degree $$0$$and therefore it is almost isomorphic to $$R / (x_1, …, x_n)$$. Therefore, this resolution can be used to calculate $$text {Tor}$$ Y $$text {Ext}$$.

These can only be related distantly. In my opinion, it is mainly the first thing that is explained.

I am particularly interested in any answer that can explain the first example in all its generality, that is, to include some theorem about the cohomology of free DGA given a derivation.

## web application – split map | Content design, should the map be on the left or right side?

It is divided by the medium; left side for the map, and the content is loaded on the right side when the user clicks on a marker.

I have some opinions, and his initial reaction is "Why is the map on the left? Hmm, I guess it makes sense since it's from left to right."

If I had to point my finger, I would say that the initial expectation for the map on the right and the content on the left comes from the design of Google Maps.

So, should the map remain on the left side? Or move it to the right side?

## 3d: OpenGL relief map problems

I am using Core-Profile OpenGL version 3.3 and I am trying to implement relief mapping.
I am loading the model thanks to assimp and adding 2 textures to the material, one for diffuse color and another for normal mapping.
(I am using this class to load and display the model https://learnopengl.com/code_viewer_gh.php?code=includes/learnopengl/model.h)

Textures extraction:

``````// Model.hpp
// diffuse std::maps
std::vector diffuseMaps = loadMaterialTextures(material, aiTextureType_DIFFUSE, "texture_diffuse");
textures.insert(textures.end(), diffuseMaps.begin(), diffuseMaps.end());
// normal maps
std::vector normalMaps = loadMaterialTextures(material, aiTextureType_NORMALS, "texture_normal");
textures.insert(textures.end(), normalMaps.begin(), normalMaps.end());
``````

My problem is that the relief mapping is not shown on the object …
In the Shader vertex below, you can see how I calculate the TNB matrix.

I'm not sure if the problem comes from the way I calculated the TNB matrix or the way I loaded the textures.

If you think it comes from the matrix, how could I calculate it correctly?

Thanks in advance.

Vertices shader:

``````#version 330 core
layout (location = 0) in vec3 aPos;
layout (location = 1) in vec3 aNormal;
layout (location = 2) in vec2 aTexCoords;
layout (location = 3) in vec3 aTangent;
layout (location = 4) in vec3 aBitangent;

out VS_OUT {
vec3 FragPos;
vec2 TexCoords;
vec3 TangentLightPos;
vec3 TangentViewPos;
vec3 TangentFragPos;
mat3 TBN;
} vs_out;

uniform mat4 Projection;
uniform mat4 View;
uniform mat4 Model;

uniform vec3 lightPos;
uniform vec3 viewPos;

void main()
{
vs_out.FragPos = vec3(Model * vec4(0.05 * aPos, 1.0));
vs_out.TexCoords = aTexCoords;

mat3 normalMatrix = transpose(inverse(mat3(Model)));
vec3 T = normalize(normalMatrix * aTangent);
vec3 N = normalize(normalMatrix * aNormal);
T = normalize(T - dot(T, N) * N);
vec3 B = cross(N, T);

vs_out.TBN = transpose(mat3(T, B, N));
vs_out.TangentLightPos = vs_out.TBN * lightPos;
vs_out.TangentViewPos  = vs_out.TBN * viewPos;
vs_out.TangentFragPos  = vs_out.TBN * vs_out.FragPos;

gl_Position = Projection * View * vec4(vs_out.FragPos, 1.0);
}
``````

Shader Fragment:

``````#version 330 core
out vec4 FragColor;

in VS_OUT {
vec3 FragPos;
vec2 TexCoords;
vec3 TangentLightPos;
vec3 TangentViewPos;
vec3 TangentFragPos;
mat3 TBN;
} fs_in;

//Material
uniform sampler2D texture_diffuse;
uniform sampler2D texture_normal;

uniform vec3 lightPos;
uniform vec3 viewPos;

void main()
{
// obtain normal from normal map in range (0,1)
vec3 normal = texture(texture_normal, fs_in.TexCoords).rgb;
// transform normal vector to range (-1,1)
normal = normalize(normal * 2.0 - 1.0);  // this normal is in tangent space
//

// get diffuse color
vec3 color = texture(texture_diffuse, fs_in.TexCoords).rgb;
// ambient
vec3 ambient = 0.3 * color;
// diffuse
vec3 lightDir = normalize(fs_in.TangentLightPos - fs_in.TangentFragPos);
float diff = max(dot(lightDir, normal), 0.0);
vec3 diffuse = diff * color;
// specular
vec3 viewDir = normalize(fs_in.TangentViewPos - fs_in.TangentFragPos);
vec3 reflectDir = reflect(-lightDir, normal);
vec3 halfwayDir = normalize(lightDir + viewDir);
float spec = pow(max(dot(normal, halfwayDir), 0.0), 32.0);

vec3 specular = vec3(0.2) * spec;
FragColor = vec4(ambient + diffuse + specular, 1.0);
}
``````

## site map: should i have a site map or not?

Articles published on our website can be viewed through information channels like this for the Content Creation category: https://digitalsplendid.net/category/content-creation/.

Therefore, my question is whether I should have a separate Site Map that includes articles in different categories in addition to links to all pages and publications. I think this makes the process manual, but I can still consider whether the consensus is to have a site map that helps improve the user experience and is good for SEO.

## How to make a map of my blog

I have a blog site if I check the Google search engine, I will see the address of my blog, but I can't see my post why.
Please, someone should help me.
Thank you.