Missing vertices in sphere model - c++

I'm learning OpenGL and I'm working on creating my own sphere model. I was able to draw a complete sphere, although with some puzzling results. I'm wondering if someone can explain (and possibly correct) my code.
The rationale: build triangles using carthesian coordinates calculated from polar coordinates. The number of subdivisions tells me the steps in phi or theta radians to generate the sphere point. From a particular point P(phi, theta), I build the other edges of the sector for [phi, delta_phi], [theta, delta_tetha], with phi varying from [0, pi] (180 degrees) and tetha from [0, 2*pi] (360 degrees).
This is the code I came up with (I'm using QT objects, but it should be pretty straitghforward):
QVector3D polarToCarthesian(float rho, float phi, float theta)
{
float r = qSin(phi) * rho;
float y = qCos(phi) * rho;
float x = r * qSin(theta);
float z = r * qCos(theta);
return QVector3D{x, y, z};
}
void make_sector(QVector<QVector3D>& mesh, float phi, float theta, float rho, float deltaPhi, float deltaTheta)
{
QVector3D p1 = polarToCarthesian(rho, phi, theta);
QVector3D p2 = polarToCarthesian(rho, phi, theta + deltaTheta);
QVector3D p3 = polarToCarthesian(rho, phi + deltaPhi, theta);
QVector3D p4 = polarToCarthesian(rho, phi + deltaPhi, theta + deltaTheta);
// First Triangle
mesh.push_back(p1);
mesh.push_back(p1); // Normal
mesh.push_back(p3);
mesh.push_back(p3); // Normal
mesh.push_back(p2);
mesh.push_back(p2); // Normal
// Second Triangle
mesh.push_back(p2);
mesh.push_back(p2); // Normal
mesh.push_back(p3);
mesh.push_back(p3); // Normal
mesh.push_back(p4);
mesh.push_back(p4); // Normal
}
void build_sphere(QVector<QVector3D>& mesh, int ndiv)
{
const float PHI_MAX = static_cast<float>(M_PI);
const float THETA_MAX = static_cast<float>(M_PI) * 2;
const float delta_phi = PHI_MAX / ndiv;
const float delta_theta = THETA_MAX / ndiv;
for (int i = 0; i < ndiv; ++i) {
float phi = i * delta_phi;
for (int j = 0; j < ndiv; ++j) {
float theta = j * delta_theta;
make_sector(mesh, phi, theta, 1.0f, delta_phi, delta_theta);
}
}
}
// Then I can generate the sphere with
build_sphere(sphere_mesh, 10);
However, I cannot get a complete sphere unless I change the iteration for phi from ndiv iterations to 3 * ndiv iterations. I don't understand why! Phi should vary from 0 to PI to cover the whole Y axis while Theta from 0 to 2 * pi should cover the XZ plane.
Can somebody explain what's happening and why 3 * ndiv works?

phi should go from -π/2 to +π/2, not from 0 to π:
float phi = i * delta_phi - (M_PI / 2);
Also, you appear to have your r and y calculations the wrong way around. You want r to be maximum at the equator (when phi == 0).
I think your code may have worked (albeit producing twice as many polygons as it should have) if you had stuck at 2 * ndiv. As it is, going from 0 to π only puts polygons in the northern hemisphere, so you have to keep going beyond that to have any polygons in the southern hemisphere.
p.s. there's no 'h' in cartesian ;)

Related

Need help expanding particle system spread / divergence from 2 to 3 dimensions

I need help. I've been struggling with this for a week now and getting nowhere. I am building a 3D particle system mainly for learning and I am currently working on particle spread / divergence. In specific, introducing random direction to the particle direction so as to create something that looks more like a fountain as opposed to a solid stream.
I have been successful in getting this to work in one axis but no matter what I do, I cannot get it to work in 3 dimensions.
Here is what I am doing:
// Compute a random angle between -180 to +180 for velocity angle x, y and z. spreadAmount is a float from 0.0 to 1.0 to control degree of spread.
float velangrndx = spreadAmount * ((((double)(rand() % RAND_MAX) / (RAND_MAX)) - 0.5) * 360.0 * 3.14159265359 / 180.0);
float velangrndy = spreadAmount * ((((double)(rand() % RAND_MAX) / (RAND_MAX)) - 0.5) * 360.0 * 3.14159265359 / 180.0);
float velangrndz = spreadAmount * ((((double)(rand() % RAND_MAX) / (RAND_MAX)) - 0.5) * 360.0 * 3.14159265359 / 180.0);
// Compute Angles
float vsin_anglex_dir = -PF_SIN(velangrndx);
float vcos_anglex_dir = -PF_COS(velangrndx);
float vsin_angley_dir = -PF_SIN(velangrndy);
float vcos_angley_dir = -PF_COS(velangrndy);
float vsin_anglez_dir = -PF_SIN(velangrndz);
float vcos_anglez_dir = -PF_COS(velangrndz);
// Assign initial velocity to velocity x, y, z. vel is a float ranging from 0.0 - 0.1 specified by user. velx, vely, and velz are also floats.
velx = vel; vely = vel; velz = vel;
And finally, we get to the particle spread / divergence function below. If I use only the first X axis (comment out the Y and Z) it works as it should (see images), but if I use the Y and Z axis, it works totally incorrectly. px0, py0, and pz0 are temporary float variables so as to preserve the velocity variables.
// X Divergence
px0 = (velx * vsin_anglex_dir);
py0 = (velx * vcos_anglex_dir);
pz0 = velz;
velx = px0; vely = py0; velz = pz0;
// Y Divergence
py0 = (vely * vsin_angley_dir);
pz0 = (vely * vcos_angley_dir);
px0 = velx;
velx = px0; vely = py0; velz = pz0;
// Z Divergence
pz0 = (velz * vsin_anglez_dir);
px0 = (velz * vcos_anglez_dir);
py0 = vely;
velx = px0; vely = py0; velz = pz0;
The velx, vely, and velz are then used to calculate for particle screen position.
This is what the particle spread looks like at 25%, 75% and 100% for the X axis only (if I comment out the Y and Z code). This works as it should and in theory, if the rest of my code was working correctly, I should get this same result for the Y and Z axis. But I don't.
I could really use some help here. Any suggestions on what I am doing wrong and how to correctly expand the currently working spread function from 2 dimensions to 3?
Thanks,
-Richard
Likely it is because the values of velx, vely and velz are getting overwritten on subsequent calculations. See whether the below works the way you are expecting.
// X Divergence
float velxXD = (velx * vsin_anglex_dir);
float velyXD = (velx * vcos_anglex_dir);
float velzXD = velz;
// Y Divergence
float velxYD = velx;
float velyYD = (vely * vsin_angley_dir);
float velzYD = (vely * vcos_angley_dir);
// Z Divergence
float velxZD = (velz * vcos_anglez_dir);
float velyZD = vely;
float velzZD = (velz * vsin_anglez_dir);
velx=velxXD+velxYD+velxZD;
vely=velyXD+velyYD+velyZD;
velz=velzXD+velzYD+velzZD;

Perlin Noise algorithm does not seem to produce gradient noise

I am attempting to implement Perlin Noise in c++.
Firstly, the problem (I think) is that the output is not what I expect. Currently I simply use the generated Perlin Noise values in a greyscaled image, and this is the results I get:
However, from my understanding, it's supposed to look more along the lines of:
That is, the noise I am producing currently seems to be more along the lines of "standard" irregular noise.
This is the Perlin Noise Algorithm I have implemented so far:
float perlinNoise2D(float x, float y)
{
// Find grid cell coordinates
int x0 = (x > 0.0f ? static_cast<int>(x) : (static_cast<int>(x) - 1));
int x1 = x0 + 1;
int y0 = (y > 0.0f ? static_cast<int>(y) : (static_cast<int>(y) - 1));
int y1 = y0 + 1;
float s = calculateInfluence(x0, y0, x, y);
float t = calculateInfluence(x1, y0, x, y);
float u = calculateInfluence(x0, y1, x, y);
float v = calculateInfluence(x1, y1, x, y);
// Local position in the grid cell
float localPosX = 3 * ((x - (float)x0) * (x - (float)x0)) - 2 * ((x - (float)x0) * (x - (float)x0) * (x - (float)x0));
float localPosY = 3 * ((y - (float)y0) * (y - (float)y0)) - 2 * ((y - (float)y0) * (y - (float)y0) * (y - (float)y0));
float a = s + localPosX * (t - s);
float b = u + localPosX * (v - u);
return lerp(a, b, localPosY);
}
The function calculateInfluence has the job of generating the random gradient vector and distance vector for one of the corner points of the current grid cell and returning the dot product of these. It is implemented as:
float calculateInfluence(int xGrid, int yGrid, float x, float y)
{
// Calculate gradient vector
float gradientXComponent = dist(rdEngine);
float gradientYComponent = dist(rdEngine);
// Normalize gradient vector
float magnitude = sqrt( pow(gradientXComponent, 2) + pow(gradientYComponent, 2) );
gradientXComponent = gradientXComponent / magnitude;
gradientYComponent = gradientYComponent / magnitude;
magnitude = sqrt(pow(gradientXComponent, 2) + pow(gradientYComponent, 2));
// Calculate distance vectors
float dx = x - (float)xGrid;
float dy = y - (float)yGrid;
// Compute dot product
return (dx * gradientXComponent + dy * gradientYComponent);
}
Here, dist is a random number generator from C++11:
std::mt19937 rdEngine(1);
std::normal_distribution<float> dist(0.0f, 1.0f);
And lerp is simply implemented as:
float lerp(float v0, float v1, float t)
{
return ( 1.0f - t ) * v0 + t * v1;
}
To implement the algorithm, I primarily made use of the following two resources:
Perlin Noise FAQ
Perlin Noise Pseudo Code
It's difficult for me to pinpoint exactly where I seem to be messing up. It could be that I am generating the gradient vectors incorrectly, as I'm not quite sure what type of distribution they should have. I have tried with a uniform distribution, however this seemed to generate repeating patterns in the texture!
Likewise, it could be that I am averaging the influence values incorrectly. It has been a bit difficult to discern exactly how it should be done from from the Perlin Noise FAQ article.
Does anyone have any hints as to what might be wrong with the code? :)
It seems like you are only generating a single octave of Perlin Noise. To get a result like the one shown, you need to generate multiple octaves and add them together. In a series of octaves, each octave should have a grid cell size double that of the last.
To generate multi-octave noise, use something similar to this:
float multiOctavePerlinNoise2D(float x, float y, int octaves)
{
float v = 0.0f;
float scale = 1.0f;
float weight = 1.0f;
float weightTotal = 0.0f;
for(int i = 0; i < octaves; i++)
{
v += perlinNoise2D(x * scale, y * scale) * weight;
weightTotal += weight;
// "ever-increasing frequencies and ever-decreasing amplitudes"
// (or conversely decreasing freqs and increasing amplitudes)
scale *= 0.5f;
weight *= 2.0f;
}
return v / weightTotal;
}
For extra randomness you could use a differently seeded random generator for each octave. Also, the weights given to each octave can be varied to adjust the aesthetic quality of the noise. If the weight variable is not adjusted each iteration, then the example above is "pink noise" (each doubling of frequency carries the same weight).
Also, you need to use a random number generator that returns the same value each time for a given xGrid, yGrid pair.

How to get vertices for a sphere? [duplicate]

Are there any tutorials out there that explain how I can draw a sphere in OpenGL without having to use gluSphere()?
Many of the 3D tutorials for OpenGL are just on cubes. I have searched but most of the solutions to drawing a sphere are to use gluSphere(). There is also a site that has the code to drawing a sphere at this site but it doesn't explain the math behind drawing the sphere. I have also other versions of how to draw the sphere in polygon instead of quads in that link. But again, I don't understand how the spheres are drawn with the code. I want to be able to visualize so that I could modify the sphere if I need to.
One way you can do it is to start with a platonic solid with triangular sides - an octahedron, for example. Then, take each triangle and recursively break it up into smaller triangles, like so:
Once you have a sufficient amount of points, you normalize their vectors so that they are all a constant distance from the center of the solid. This causes the sides to bulge out into a shape that resembles a sphere, with increasing smoothness as you increase the number of points.
Normalization here means moving a point so that its angle in relation to another point is the same, but the distance between them is different.
Here's a two dimensional example.
A and B are 6 units apart. But suppose we want to find a point on line AB that's 12 units away from A.
We can say that C is the normalized form of B with respect to A, with distance 12. We can obtain C with code like this:
#returns a point collinear to A and B, a given distance away from A.
function normalize(a, b, length):
#get the distance between a and b along the x and y axes
dx = b.x - a.x
dy = b.y - a.y
#right now, sqrt(dx^2 + dy^2) = distance(a,b).
#we want to modify them so that sqrt(dx^2 + dy^2) = the given length.
dx = dx * length / distance(a,b)
dy = dy * length / distance(a,b)
point c = new point
c.x = a.x + dx
c.y = a.y + dy
return c
If we do this normalization process on a lot of points, all with respect to the same point A and with the same distance R, then the normalized points will all lie on the arc of a circle with center A and radius R.
Here, the black points begin on a line and "bulge out" into an arc.
This process can be extended into three dimensions, in which case you get a sphere rather than a circle. Just add a dz component to the normalize function.
If you look at the sphere at Epcot, you can sort of see this technique at work. it's a dodecahedron with bulged-out faces to make it look rounder.
I'll further explain a popular way of generating a sphere using latitude and longitude (another
way, icospheres, was already explained in the most popular answer at the time of this writing.)
A sphere can be expressed by the following parametric equation:
F(u, v) = [ cos(u)*sin(v)*r, cos(v)*r, sin(u)*sin(v)*r ]
Where:
r is the radius;
u is the longitude, ranging from 0 to 2π; and
v is the latitude, ranging from 0 to π.
Generating the sphere then involves evaluating the parametric function at fixed intervals.
For example, to generate 16 lines of longitude, there will be 17 grid lines along the u axis, with a step of
π/8 (2π/16) (the 17th line wraps around).
The following pseudocode generates a triangle mesh by evaluating a parametric function
at regular intervals (this works for any parametric surface function, not just spheres).
In the pseudocode below, UResolution is the number of grid points along the U axis
(here, lines of longitude), and VResolution is the number of grid points along the V axis
(here, lines of latitude)
var startU=0
var startV=0
var endU=PI*2
var endV=PI
var stepU=(endU-startU)/UResolution // step size between U-points on the grid
var stepV=(endV-startV)/VResolution // step size between V-points on the grid
for(var i=0;i<UResolution;i++){ // U-points
for(var j=0;j<VResolution;j++){ // V-points
var u=i*stepU+startU
var v=j*stepV+startV
var un=(i+1==UResolution) ? endU : (i+1)*stepU+startU
var vn=(j+1==VResolution) ? endV : (j+1)*stepV+startV
// Find the four points of the grid
// square by evaluating the parametric
// surface function
var p0=F(u, v)
var p1=F(u, vn)
var p2=F(un, v)
var p3=F(un, vn)
// NOTE: For spheres, the normal is just the normalized
// version of each vertex point; this generally won't be the case for
// other parametric surfaces.
// Output the first triangle of this grid square
triangle(p0, p2, p1)
// Output the other triangle of this grid square
triangle(p3, p1, p2)
}
}
The code in the sample is quickly explained. You should look into the function void drawSphere(double r, int lats, int longs):
void drawSphere(double r, int lats, int longs) {
int i, j;
for(i = 0; i <= lats; i++) {
double lat0 = M_PI * (-0.5 + (double) (i - 1) / lats);
double z0 = sin(lat0);
double zr0 = cos(lat0);
double lat1 = M_PI * (-0.5 + (double) i / lats);
double z1 = sin(lat1);
double zr1 = cos(lat1);
glBegin(GL_QUAD_STRIP);
for(j = 0; j <= longs; j++) {
double lng = 2 * M_PI * (double) (j - 1) / longs;
double x = cos(lng);
double y = sin(lng);
glNormal3f(x * zr0, y * zr0, z0);
glVertex3f(r * x * zr0, r * y * zr0, r * z0);
glNormal3f(x * zr1, y * zr1, z1);
glVertex3f(r * x * zr1, r * y * zr1, r * z1);
}
glEnd();
}
}
The parameters lat defines how many horizontal lines you want to have in your sphere and lon how many vertical lines. r is the radius of your sphere.
Now there is a double iteration over lat/lon and the vertex coordinates are calculated, using simple trigonometry.
The calculated vertices are now sent to your GPU using glVertex...() as a GL_QUAD_STRIP, which means you are sending each two vertices that form a quad with the previously two sent.
All you have to understand now is how the trigonometry functions work, but I guess you can figure it out easily.
If you wanted to be sly like a fox you could half-inch the code from GLU. Check out the MesaGL source code (http://cgit.freedesktop.org/mesa/mesa/).
See the OpenGL red book: http://www.glprogramming.com/red/chapter02.html#name8
It solves the problem by polygon subdivision.
My example how to use 'triangle strip' to draw a "polar" sphere, it consists in drawing points in pairs:
const float PI = 3.141592f;
GLfloat x, y, z, alpha, beta; // Storage for coordinates and angles
GLfloat radius = 60.0f;
int gradation = 20;
for (alpha = 0.0; alpha < GL_PI; alpha += PI/gradation)
{
glBegin(GL_TRIANGLE_STRIP);
for (beta = 0.0; beta < 2.01*GL_PI; beta += PI/gradation)
{
x = radius*cos(beta)*sin(alpha);
y = radius*sin(beta)*sin(alpha);
z = radius*cos(alpha);
glVertex3f(x, y, z);
x = radius*cos(beta)*sin(alpha + PI/gradation);
y = radius*sin(beta)*sin(alpha + PI/gradation);
z = radius*cos(alpha + PI/gradation);
glVertex3f(x, y, z);
}
glEnd();
}
First point entered (glVertex3f) is as follows the parametric equation and the second one is shifted by a single step of alpha angle (from next parallel).
Although the accepted answer solves the question, there's a little misconception at the end. Dodecahedrons are (or could be) regular polyhedron where all faces have the same area. That seems to be the case of the Epcot (which, by the way, is not a dodecahedron at all). Since the solution proposed by #Kevin does not provide this characteristic I thought I could add an approach that does.
A good way to generate an N-faced polyhedron where all vertices lay in the same sphere and all its faces have similar area/surface is starting with an icosahedron and the iteratively sub-dividing and normalizing its triangular faces (as suggested in the accepted answer). Dodecahedrons, for instance, are actually truncated icosahedrons.
Regular icosahedrons have 20 faces (12 vertices) and can easily be constructed from 3 golden rectangles; it's just a matter of having this as a starting point instead of an octahedron. You may find an example here.
I know this is a bit off-topic but I believe it may help if someone gets here looking for this specific case.
Python adaptation of #Constantinius answer:
lats = 10
longs = 10
r = 10
for i in range(lats):
lat0 = pi * (-0.5 + i / lats)
z0 = sin(lat0)
zr0 = cos(lat0)
lat1 = pi * (-0.5 + (i+1) / lats)
z1 = sin(lat1)
zr1 = cos(lat1)
glBegin(GL_QUAD_STRIP)
for j in range(longs+1):
lng = 2 * pi * (j+1) / longs
x = cos(lng)
y = sin(lng)
glNormal(x * zr0, y * zr0, z0)
glVertex(r * x * zr0, r * y * zr0, r * z0)
glNormal(x * zr1, y * zr1, z1)
glVertex(r * x * zr1, r * y * zr1, r * z1)
glEnd()
void draw_sphere(float r)
{
float pi = 3.141592;
float di = 0.02;
float dj = 0.04;
float db = di * 2 * pi;
float da = dj * pi;
for (float i = 0; i < 1.0; i += di) //horizonal
for (float j = 0; j < 1.0; j += dj) //vertical
{
float b = i * 2 * pi; //0 to 2pi
float a = (j - 0.5) * pi; //-pi/2 to pi/2
//normal
glNormal3f(
cos(a + da / 2) * cos(b + db / 2),
cos(a + da / 2) * sin(b + db / 2),
sin(a + da / 2));
glBegin(GL_QUADS);
//P1
glTexCoord2f(i, j);
glVertex3f(
r * cos(a) * cos(b),
r * cos(a) * sin(b),
r * sin(a));
//P2
glTexCoord2f(i + di, j);//P2
glVertex3f(
r * cos(a) * cos(b + db),
r * cos(a) * sin(b + db),
r * sin(a));
//P3
glTexCoord2f(i + di, j + dj);
glVertex3f(
r * cos(a + da) * cos(b + db),
r * cos(a + da) * sin(b + db),
r * sin(a + da));
//P4
glTexCoord2f(i, j + dj);
glVertex3f(
r * cos(a + da) * cos(b),
r * cos(a + da) * sin(b),
r * sin(a + da));
glEnd();
}
}
One way is to make a quad that faces the camera and write a vertex and fragment shader that renders something that looks like a sphere. You could use equations for a circle/sphere that you can find on the internet.
One nice thing is that the silhouette of a sphere looks the same from any angle. However, if the sphere is not in the center of a perspective view, then it would appear perhaps more like an ellipse. You could work out the equations for this and put them in the fragment shading. Then the light shading needs to changed as the player moves, if you do indeed have a player moving in 3D space around the sphere.
Can anyone comment on if they have tried this or if it would be too expensive to be practical?

Equirectangular to Cubic with point to point mapping?

I need to convert a panorama in equirectangular projection to 6 cubic faces and then to spherical projection and back, however I need to keep a track of how each point is mapped in each projection like
Equirectangular Point(x,y) <---> Cubic face Point (x, y) <---> Sphere Point(x, y, z)
How can I accomplish this in C++ and OpenCV?
These transformations are required because I need to find out the good matching key-points between two such images by comparing angles between keypoints when the two panoramas, projected on a sphere, are placed side by side.
Here is the panorama:
Solved, below is the function to convert 2d Panoramic to 3d spherical coordinates.
vector<int> getSphericalPoint3D(int x, int y, int cols, int rows)
{
//introduce a radius to
static const int radius = 128;
vector<int> point3D;
// the center
double c_x = (double)cols / 2;
double c_y = (double)rows / 2;
double X = (((double)x - c_x) * CV_PI) / c_x;
double Y = (((double)y - c_y) * CV_PI) / c_y;
int x3D = round(radius * cos(X) * cos(Y)) + radius;
int y3D = round(radius * cos(X) * sin(Y)) + radius;
int z3D = round(radius * sin(X)) + radius;
point3D = { x3D, y3D, z3D };
return point3D;
}
The idea is to normalize 2d pixels from -Pi to Pi on both X and Y axis
Use the following relationship to get spherical co-ordinates
x = R * cos(x)cos(y) + R (this R is added to avoid the negative values)
y = R * cos(x)sin(y) + R
z = R * sin(x) + R
A similar transformation is used for Cubic transformation

Drawing Sphere in OpenGL without using gluSphere()?

Are there any tutorials out there that explain how I can draw a sphere in OpenGL without having to use gluSphere()?
Many of the 3D tutorials for OpenGL are just on cubes. I have searched but most of the solutions to drawing a sphere are to use gluSphere(). There is also a site that has the code to drawing a sphere at this site but it doesn't explain the math behind drawing the sphere. I have also other versions of how to draw the sphere in polygon instead of quads in that link. But again, I don't understand how the spheres are drawn with the code. I want to be able to visualize so that I could modify the sphere if I need to.
One way you can do it is to start with a platonic solid with triangular sides - an octahedron, for example. Then, take each triangle and recursively break it up into smaller triangles, like so:
Once you have a sufficient amount of points, you normalize their vectors so that they are all a constant distance from the center of the solid. This causes the sides to bulge out into a shape that resembles a sphere, with increasing smoothness as you increase the number of points.
Normalization here means moving a point so that its angle in relation to another point is the same, but the distance between them is different.
Here's a two dimensional example.
A and B are 6 units apart. But suppose we want to find a point on line AB that's 12 units away from A.
We can say that C is the normalized form of B with respect to A, with distance 12. We can obtain C with code like this:
#returns a point collinear to A and B, a given distance away from A.
function normalize(a, b, length):
#get the distance between a and b along the x and y axes
dx = b.x - a.x
dy = b.y - a.y
#right now, sqrt(dx^2 + dy^2) = distance(a,b).
#we want to modify them so that sqrt(dx^2 + dy^2) = the given length.
dx = dx * length / distance(a,b)
dy = dy * length / distance(a,b)
point c = new point
c.x = a.x + dx
c.y = a.y + dy
return c
If we do this normalization process on a lot of points, all with respect to the same point A and with the same distance R, then the normalized points will all lie on the arc of a circle with center A and radius R.
Here, the black points begin on a line and "bulge out" into an arc.
This process can be extended into three dimensions, in which case you get a sphere rather than a circle. Just add a dz component to the normalize function.
If you look at the sphere at Epcot, you can sort of see this technique at work. it's a dodecahedron with bulged-out faces to make it look rounder.
I'll further explain a popular way of generating a sphere using latitude and longitude (another
way, icospheres, was already explained in the most popular answer at the time of this writing.)
A sphere can be expressed by the following parametric equation:
F(u, v) = [ cos(u)*sin(v)*r, cos(v)*r, sin(u)*sin(v)*r ]
Where:
r is the radius;
u is the longitude, ranging from 0 to 2π; and
v is the latitude, ranging from 0 to π.
Generating the sphere then involves evaluating the parametric function at fixed intervals.
For example, to generate 16 lines of longitude, there will be 17 grid lines along the u axis, with a step of
π/8 (2π/16) (the 17th line wraps around).
The following pseudocode generates a triangle mesh by evaluating a parametric function
at regular intervals (this works for any parametric surface function, not just spheres).
In the pseudocode below, UResolution is the number of grid points along the U axis
(here, lines of longitude), and VResolution is the number of grid points along the V axis
(here, lines of latitude)
var startU=0
var startV=0
var endU=PI*2
var endV=PI
var stepU=(endU-startU)/UResolution // step size between U-points on the grid
var stepV=(endV-startV)/VResolution // step size between V-points on the grid
for(var i=0;i<UResolution;i++){ // U-points
for(var j=0;j<VResolution;j++){ // V-points
var u=i*stepU+startU
var v=j*stepV+startV
var un=(i+1==UResolution) ? endU : (i+1)*stepU+startU
var vn=(j+1==VResolution) ? endV : (j+1)*stepV+startV
// Find the four points of the grid
// square by evaluating the parametric
// surface function
var p0=F(u, v)
var p1=F(u, vn)
var p2=F(un, v)
var p3=F(un, vn)
// NOTE: For spheres, the normal is just the normalized
// version of each vertex point; this generally won't be the case for
// other parametric surfaces.
// Output the first triangle of this grid square
triangle(p0, p2, p1)
// Output the other triangle of this grid square
triangle(p3, p1, p2)
}
}
The code in the sample is quickly explained. You should look into the function void drawSphere(double r, int lats, int longs):
void drawSphere(double r, int lats, int longs) {
int i, j;
for(i = 0; i <= lats; i++) {
double lat0 = M_PI * (-0.5 + (double) (i - 1) / lats);
double z0 = sin(lat0);
double zr0 = cos(lat0);
double lat1 = M_PI * (-0.5 + (double) i / lats);
double z1 = sin(lat1);
double zr1 = cos(lat1);
glBegin(GL_QUAD_STRIP);
for(j = 0; j <= longs; j++) {
double lng = 2 * M_PI * (double) (j - 1) / longs;
double x = cos(lng);
double y = sin(lng);
glNormal3f(x * zr0, y * zr0, z0);
glVertex3f(r * x * zr0, r * y * zr0, r * z0);
glNormal3f(x * zr1, y * zr1, z1);
glVertex3f(r * x * zr1, r * y * zr1, r * z1);
}
glEnd();
}
}
The parameters lat defines how many horizontal lines you want to have in your sphere and lon how many vertical lines. r is the radius of your sphere.
Now there is a double iteration over lat/lon and the vertex coordinates are calculated, using simple trigonometry.
The calculated vertices are now sent to your GPU using glVertex...() as a GL_QUAD_STRIP, which means you are sending each two vertices that form a quad with the previously two sent.
All you have to understand now is how the trigonometry functions work, but I guess you can figure it out easily.
If you wanted to be sly like a fox you could half-inch the code from GLU. Check out the MesaGL source code (http://cgit.freedesktop.org/mesa/mesa/).
See the OpenGL red book: http://www.glprogramming.com/red/chapter02.html#name8
It solves the problem by polygon subdivision.
My example how to use 'triangle strip' to draw a "polar" sphere, it consists in drawing points in pairs:
const float PI = 3.141592f;
GLfloat x, y, z, alpha, beta; // Storage for coordinates and angles
GLfloat radius = 60.0f;
int gradation = 20;
for (alpha = 0.0; alpha < GL_PI; alpha += PI/gradation)
{
glBegin(GL_TRIANGLE_STRIP);
for (beta = 0.0; beta < 2.01*GL_PI; beta += PI/gradation)
{
x = radius*cos(beta)*sin(alpha);
y = radius*sin(beta)*sin(alpha);
z = radius*cos(alpha);
glVertex3f(x, y, z);
x = radius*cos(beta)*sin(alpha + PI/gradation);
y = radius*sin(beta)*sin(alpha + PI/gradation);
z = radius*cos(alpha + PI/gradation);
glVertex3f(x, y, z);
}
glEnd();
}
First point entered (glVertex3f) is as follows the parametric equation and the second one is shifted by a single step of alpha angle (from next parallel).
Although the accepted answer solves the question, there's a little misconception at the end. Dodecahedrons are (or could be) regular polyhedron where all faces have the same area. That seems to be the case of the Epcot (which, by the way, is not a dodecahedron at all). Since the solution proposed by #Kevin does not provide this characteristic I thought I could add an approach that does.
A good way to generate an N-faced polyhedron where all vertices lay in the same sphere and all its faces have similar area/surface is starting with an icosahedron and the iteratively sub-dividing and normalizing its triangular faces (as suggested in the accepted answer). Dodecahedrons, for instance, are actually truncated icosahedrons.
Regular icosahedrons have 20 faces (12 vertices) and can easily be constructed from 3 golden rectangles; it's just a matter of having this as a starting point instead of an octahedron. You may find an example here.
I know this is a bit off-topic but I believe it may help if someone gets here looking for this specific case.
Python adaptation of #Constantinius answer:
lats = 10
longs = 10
r = 10
for i in range(lats):
lat0 = pi * (-0.5 + i / lats)
z0 = sin(lat0)
zr0 = cos(lat0)
lat1 = pi * (-0.5 + (i+1) / lats)
z1 = sin(lat1)
zr1 = cos(lat1)
glBegin(GL_QUAD_STRIP)
for j in range(longs+1):
lng = 2 * pi * (j+1) / longs
x = cos(lng)
y = sin(lng)
glNormal(x * zr0, y * zr0, z0)
glVertex(r * x * zr0, r * y * zr0, r * z0)
glNormal(x * zr1, y * zr1, z1)
glVertex(r * x * zr1, r * y * zr1, r * z1)
glEnd()
void draw_sphere(float r)
{
float pi = 3.141592;
float di = 0.02;
float dj = 0.04;
float db = di * 2 * pi;
float da = dj * pi;
for (float i = 0; i < 1.0; i += di) //horizonal
for (float j = 0; j < 1.0; j += dj) //vertical
{
float b = i * 2 * pi; //0 to 2pi
float a = (j - 0.5) * pi; //-pi/2 to pi/2
//normal
glNormal3f(
cos(a + da / 2) * cos(b + db / 2),
cos(a + da / 2) * sin(b + db / 2),
sin(a + da / 2));
glBegin(GL_QUADS);
//P1
glTexCoord2f(i, j);
glVertex3f(
r * cos(a) * cos(b),
r * cos(a) * sin(b),
r * sin(a));
//P2
glTexCoord2f(i + di, j);//P2
glVertex3f(
r * cos(a) * cos(b + db),
r * cos(a) * sin(b + db),
r * sin(a));
//P3
glTexCoord2f(i + di, j + dj);
glVertex3f(
r * cos(a + da) * cos(b + db),
r * cos(a + da) * sin(b + db),
r * sin(a + da));
//P4
glTexCoord2f(i, j + dj);
glVertex3f(
r * cos(a + da) * cos(b),
r * cos(a + da) * sin(b),
r * sin(a + da));
glEnd();
}
}
One way is to make a quad that faces the camera and write a vertex and fragment shader that renders something that looks like a sphere. You could use equations for a circle/sphere that you can find on the internet.
One nice thing is that the silhouette of a sphere looks the same from any angle. However, if the sphere is not in the center of a perspective view, then it would appear perhaps more like an ellipse. You could work out the equations for this and put them in the fragment shading. Then the light shading needs to changed as the player moves, if you do indeed have a player moving in 3D space around the sphere.
Can anyone comment on if they have tried this or if it would be too expensive to be practical?