0
$\begingroup$

I have a dataset obtained from astronomical simulation. This data was created in a curvilinear grid with polar coordinates.

The form of this dataset is a simple array (1d) of intensities, i can visualize this data in python whit the next code.

densINcgs = 8888035.76
Munit = 1.0   #solar mass
Runit = 1.0   #au
DensUnit = Munit / Runit**2.0 * densINcgs # grams / (cm^2)
# grid specification
nrad = 128 #128 #500
nsec = 384 # 256 #1500
Rmin = 0.4
Rmax = 1.8

r = np.linspace(Rmin, Rmax, nrad)
print("rx: ",np.linspace(179517444792000.0,897587223960000.2,384))

rr = []
for i in range(0,nrad):
    rr.append(Rmin*exp(i*log(Rmax/Rmin)/nrad))
print("rr",rr)
print("lin",np.linspace(0.,2.*np.pi, nsec))
theta, rad   = np.meshgrid(np.linspace(0., 2.*np.pi, nsec), rr)
xi = rad * np.cos(theta)
yi = rad * np.sin(theta)

#READ DATA
rho =   fromfile("dens10.dat",dtype='float32')        
Rho =          rho.reshape(nrad,nsec)


#FIRST IMAGE
figure(100)
imshow(log10(Rho*DensUnit),origin='lower',
cmap=cm.Oranges_r,aspect='auto')
xlabel('x [AU]', fontsize=16)
ylabel('y [AU]', fontsize=16)   
cb = plt.colorbar()


#SECON iMAGE
cb.set_label('log Density [$\\rm g$ $\\rm cm^{-2}$]')
zc = np.linspace(0, 255, 49152).reshape(128, 384)
zc=np.random.rand(49152)*10
zc= zc.reshape(128, 384)
figure(17)How visualize a data generate in curvilinear grid in opengl
pcolormesh(xi,yi,log10(Rho*DensUnit))


show()

First: Fisrt

Second: Second

Ther first image is the direct visualization of the data, and the second image is ta correct visualization of the data on a curvilinear grid. I generated this visualization with pcolormesh.

But i want generate the second visualization with openGL and glsl, using texture mapping. i tried this with the next code ( a part).

Load textures

I define a quad with this vertices.

float vertices[] = {
    -0.5f, -0.5f, -0.5f,  0.0f, 0.0f,
     0.5f, -0.5f, -0.5f,  1.0f, 0.0f,
     0.5f,  0.5f, -0.5f,  1.0f, 1.0f,
     0.5f,  0.5f, -0.5f,  1.0f, 1.0f,
    -0.5f,  0.5f, -0.5f,  0.0f, 1.0f,
    -0.5f, -0.5f, -0.5f,  0.0f, 0.0f

    }

i load the data set ( x=128 , y=384 )

float *pData = new   float[XDIM * YDIM];
FILE *archivo;
archivo = fopen("dens10.dat", "rb");
//  archivo=fopen("avance/media/data5.out","rb");
std::cout << "A3" << std::endl;
if (archivo != NULL){
 std::cout << "A3.1" << std::endl;
}
for (int i = 0; i < XDIM * YDIM ; i++)
{   //std::cout << "A3.5" << std::endl;
float v;
fread((void*)(&v), sizeof(v), 1, archivo);
pData[i]=v;

}

and i define te textures

glGenTextures(1, &texture1);
glBindTexture(GL_TEXTURE_2D, texture1);
// set the texture wrapping parameters
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_CLAMP_TO_BORDER);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_CLAMP_TO_BORDER);
// set texture filtering parameters
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER,        GL_LINEAR_ATTENUATION);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER,    GL_LINEAR_ATTENUATION);

vertex shader

#version 330 core
layout (location = 0) in vec3 aPos;
layout (location = 1) in vec2 aTexCoord;

out vec2 TexCoord;

uniform mat4 model;
uniform mat4 view;
uniform mat4 projection;

void main()
{   

vec4 cord = projection * view * model * vec4(aPos, 1.0f);

gl_Position = vec4(cord);

TexCoord=aPos.xy;
}

fragment shader

#version 330 core
out vec4 FragColor;

in vec2 TexCoord;

// texture samplers
uniform sampler2D texture1;
// uniform sampler2D texture2;

#define PI 3.14159265358979323844


void main()
{

float c=texture(texture1,TexCoord ).r;
// vec4 col=vec4(10000*c,30*c,20*c,1.0);
vec4 col=vec4(c*1000,300*c,100*c,1.0);

FragColor = col;
}

The result is the next.

Opengl-glsl

As can be seen this image is very similar to firt image obtained in python but my intention is get a image similary to the second.

This is the dataset [data set] (https://drive.google.com/file/d/14C02cVNkwWMbrG9DAjC05y8YAfCUUAOJ/view?usp=sharing)!

Thank you very much in advance.

$\endgroup$
  • $\begingroup$ Perhaps it would be helpful to explain your attempt a bit, rather than just posting the code. Also, you define a quad with a lot of vertices and with 5 entries per row. This is not easily readable. For a quad two triangles, with three vertices per triangle and three coordinates per vertex suffice. Furthermore, you can share vertices, such that you only have vertices left. With this in mind, I don't see where you're putting any logic to transform a one dimensional data set into the second image you are aiming for. $\endgroup$ – Tare Sep 23 '19 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.