I'm very new to using raphael js library and I'm trying to figure it all out. I'm trying to create a chart based on percentages where 100% would be a full circle. The circle part I have figured out, but how would I go about changing it to show a half-circle for 50% or a quarter of a circle for 25%?
I recommend looking at the code behind this example on the Raphael home page. It should be easy enough to modify it to suit your needs.
This function in particular is what you're looking for
var rad = Math.PI / 180;
function sector(cx, cy, r, startAngle, endAngle, params) {
var x1 = cx + r * Math.cos(-startAngle * rad),
x2 = cx + r * Math.cos(-endAngle * rad),
y1 = cy + r * Math.sin(-startAngle * rad),
y2 = cy + r * Math.sin(-endAngle * rad);
return paper.path(["M", cx, cy, "L", x1, y1, "A", r, r, 0, +(endAngle - startAngle > 180), 0, x2, y2, "z"]).attr(params);
}
so, a 50% slice would be
var fifty = sector(100,100,50,0,180,{"fill":"red"});
var twentyfive = sector(100,100,50,180,270,{"fill":"red"});
Of course, this is working with degrees - you may want to wrap it so that you can use percentages.
You have to use path() and specify it using SVG's path syntax. Here is an example of creating a closed quarter-circle (upper left quadrant):
var arcPath = paper2.path("M200,200 v-150 a150,150 0 0,0 -150,150 z");
arcPath.attr("fill", "red");
See this link for more on SVG Paths.
Related
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?
The language I use doesn't have the built ability to create charts of any kind. The following code I found here. It's very old c++ code.
Center of the circle (origin)= (h,k)
Radius = r
...
void Circular_arc(constint h, constint k, constint r, constint start_angle, constint end_angle)
{
int color = getcolor();
float angle = (((start_angle <= end_angle) ? start_angle : end_angle)*(M_PI / 180));
float range = (((end_angle > start_angle) ? end_angle : start_angle)*(M_PI / 180));
float x = (r*cos(angle));
float y = (r*sin(angle));
do
{
putpixel((int)(h + x + 0.5), (int)(k - y + 0.5), color);
angle += 0.001;
x = (r*cos(angle));
y = (r*sin(angle));
} while (angle <= range);
}
...
I converted it to the language I'm using. I've put it in a loop which runs.
The issue with the code is that the slices always start from the same location and draw over what has been previously drawn.
The end result looks like this:
You can see from the differently colored numbers on the screen that there are 3 other arcs being covered up. First the black arc is drawn. Then the red. Then the green. The the yellow one. As the arcs are drawn, they cover up the previously drawn arc. I've inverted the length of the angles to show that there are, in fact 4 total arcs. Again, the black is drawn first, then the red, then the green, then the yellow.
Is there a way to start the next pie slice at the end of the current pie slice?
Because I really don't understand Turbo C++ I can't say for sure. But I think this is your solution:
void Circular_arc(constint h, constint k, constint r, constint start_angle, constint end_angle)
{
static int offset = 0;
start_angle = (start_angle + offset) % 360;
end_angle = (end_angle + offset) % 360;
offset = end_angle;
int color = getcolor();
float angle = (((start_angle <= end_angle) ? start_angle : end_angle)*(M_PI / 180));
float range = (((end_angle > start_angle) ? end_angle : start_angle)*(M_PI / 180));
float x = (r*cos(angle));
float y = (r*sin(angle));
do
{
putpixel((int)(h + x + 0.5), (int)(k - y + 0.5), color);
angle += 0.001;
x = (r*cos(angle));
y = (r*sin(angle));
} while (angle <= range);
}
Let me also say that this is a horrible solution in that it is in no way object oriented. But it seems to be the best you can do without starting over (which you should do.)
I've revamped this and created a method out of it. I did the following to fix my issue. Firstly, I got rid of the process that converted the beginning and ending angles from degree to radians by creating another function that does converts just the ending angle to radians. After Circular_arc is called, I set start_angle equal to the value of end_angle. Less code is written/used. Secondly, I removed the first calculation for finding x,y and moved the second one to the first thing in the loop. This wasn't a necessity, but I'm a big fan of writing as little code as possible. I'm lazy like that. The third biggest part and the whole reason for the OP in the first place was starting the current arc and the end of the previous arc. To do this, I created a variable and set it equal to the value of the start_angle. I then set parameters in the do-while loop to while (start_angle <= (end_angle+angle)). This started the current arc at the end of the previous arc.
The following is my attempt at converting my code to C++. If there are errors then please let me know and I'll do my best fix them. I hope this helps someone in the future!
...
void Circular_arc(constint h, constint k, constint r, constint start_angle, constint end_angle)
{
float angle
angle = start_angle
do
{
x = (r*cos(start_angle));
y = (r*sin(start_angle));
putpixel((int)(h + x + 0.5), (int)(k - y + 0.5), getcolor());
angle += 0.001;
} while (start_angle <= (end_angle+angle));
}
...
Here's what the pie chart looks like so far...
I'm trying to implement anti-aliased pixel rendering. My basic idea is to render 4 pixels instead of 1, and give each "real" pixel a weight based on its distance to the "virtual" pixel:
void put_aa_pixel(double x, double y)
{
int x0 = int(x);
int x1 = x0 + 1;
int y0 = int(y);
int y1 = y0 + 1;
double weight_x1 = x - x0;
double weight_x0 = 1 - weight_x1;
double weight_y1 = y - y0;
double weight_y0 = 1 - weight_x1;
put_pixel(x0, y0, int((weight_x0 * weight_y0) * 255));
put_pixel(x1, y0, int((weight_x1 * weight_y0) * 255));
put_pixel(x0, y1, int((weight_x0 * weight_y1) * 255));
put_pixel(x1, y1, int((weight_x1 * weight_y1) * 255));
}
Multiplying the x and y weights gives me the overlapping area of the virtual pixel inside each real pixel. I naively assumed this would give me a perfect anti-aliasing effect, but the moving pixels inside my test program just display an aweful flicker. It looks much worse then simple pixels without any anti-aliasing.
However, when I switch from multiplication to addition, it looks much better:
put_pixel(x0, y0, int((weight_x0 + weight_y0) * 127.5));
put_pixel(x1, y0, int((weight_x1 + weight_y0) * 127.5));
put_pixel(x0, y1, int((weight_x0 + weight_y1) * 127.5));
put_pixel(x1, y1, int((weight_x1 + weight_y1) * 127.5));
Adding the weights doesn't seem to have any geometric significance. So why does this work better? What's wrong with the first version? And is there an even better approach?
As requested :)
Intuitively: your x and y weights express distance along axis from virtual to real pixel. So, the actual distance is sqrt(w_x^2 + w_y^2). Explains why sum works better - it's way closer to this form than multiplication.
There was a bug lurking in my code for half a year:
double weight_x1 = x - x0;
double weight_x0 = 1 - weight_x1;
double weight_y1 = y - y0;
double weight_y0 = 1 - weight_x1; // BUG
Can you see the bug? Yes, it's a classic copy and paste error:
double weight_y0 = 1 - weight_y1; // FIXED
After fixing the bug, the original multiplication approach looks very nice.
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?
This question already has answers here:
OpenGL Line Width
(4 answers)
Closed 2 years ago.
What is the best way to draw a variable width line without using glLineWidth?
Just draw a rectangle?
Various parallel lines?
None of the above?
You can draw two triangles:
// Draws a line between (x1,y1) - (x2,y2) with a start thickness of t1 and
// end thickness t2.
void DrawLine(float x1, float y1, float x2, float y2, float t1, float t2)
{
float angle = atan2(y2 - y1, x2 - x1);
float t2sina1 = t1 / 2 * sin(angle);
float t2cosa1 = t1 / 2 * cos(angle);
float t2sina2 = t2 / 2 * sin(angle);
float t2cosa2 = t2 / 2 * cos(angle);
glBegin(GL_TRIANGLES);
glVertex2f(x1 + t2sina1, y1 - t2cosa1);
glVertex2f(x2 + t2sina2, y2 - t2cosa2);
glVertex2f(x2 - t2sina2, y2 + t2cosa2);
glVertex2f(x2 - t2sina2, y2 + t2cosa2);
glVertex2f(x1 - t2sina1, y1 + t2cosa1);
glVertex2f(x1 + t2sina1, y1 - t2cosa1);
glEnd();
}
Ok, how about this: (Ozgar)
A
/ \
/ \
. p1 \
/ \
/ D
B - .p2
- - - C
So AB is width1 and CD is width2.
Then,
// find line between p1 and p2
Vector p1p2 = p2 - p1 ;
// find a perpendicular
Vector perp = p1p2.perpendicular().normalize()
// Walk from p1 to A
Vector A = p1 + perp*(width1/2)
Vector B = p1 - perp*(width1/2)
Vector C = p2 - perp*(width2/2)
Vector D = p2 - perp*(width2/2)
// wind triangles
Triangle( A, B, D )
Triangle( B, D, C )
Note there's potentially a CW/CCW winding problem with this algorithm -- if perp is computed as (-y, x) in the above diagram then it will be CCW winding, if (y, -x) then it will be a CW winding.
I've had to do the same thing earlier today.
For creating a line that spans (x1,y1) -> (x2,y2) of a given width, a very easy method is to transform a simple unit-sized square spanning (0., -0.5) -> (1., 0.5) using:
glTranslatef(...) to move it to your desired (x1,y1) location;
glScalef(...) to scale it to the right length and desired width: use length = sqrt( (x2-x1)^2 + (y2-y1)^2 ) or any other low-complexity approximation;
glRotatef(...) to angle it to the right orientation: use angle = atan2(y2-y1, x2-x1).
The unit-square is very simply created from a two-triangle strip GL_TRIANGLE_STRIP, that turns into your solid line after the above transformations.
The burden here is placed primarily on OpenGL (and your graphics hardware) rather than your application code. The procedure above is turned very easily into a generic function by surrounding glPushMatrix() and glPopMatrix() calls.
For those coming looking for a good solution to this, this code is written using LWJGL, but can easily be adapted to any implementation of OpenGL.
import java.awt.Color;
import org.lwjgl.opengl.GL11;
import org.lwjgl.util.vector.Vector2f;
public static void DrawThickLine(int startScreenX, int startScreenY, int endScreenX, int endScreenY, Color color, float alpha, float width) {
Vector2f start = new Vector2f(startScreenX, startScreenY);
Vector2f end = new Vector2f(endScreenX, endScreenY);
float dx = startScreenX - endScreenX;
float dy = startScreenY - endScreenY;
Vector2f rightSide = new Vector2f(dy, -dx);
if (rightSide.length() > 0) {
rightSide.normalise();
rightSide.scale(width / 2);
}
Vector2f leftSide = new Vector2f(-dy, dx);
if (leftSide.length() > 0) {
leftSide.normalise();
leftSide.scale(width / 2);
}
Vector2f one = new Vector2f();
Vector2f.add(leftSide, start, one);
Vector2f two = new Vector2f();
Vector2f.add(rightSide, start, two);
Vector2f three = new Vector2f();
Vector2f.add(rightSide, end, three);
Vector2f four = new Vector2f();
Vector2f.add(leftSide, end, four);
GL11.glBegin(GL11.GL_QUADS);
GL11.glColor4f(color.getRed(), color.getGreen(), color.getBlue(), alpha);
GL11.glVertex3f(one.x, one.y, 0);
GL11.glVertex3f(two.x, two.y, 0);
GL11.glVertex3f(three.x, three.y, 0);
GL11.glVertex3f(four.x, four.y, 0);
GL11.glColor4f(1, 1, 1, 1);
GL11.glEnd();
}
Assume your original points are (x1,y1) -> (x2,y2). Use the following points (x1-width/2, y1), (x1+width/2,y1), (x2-width/2, y2), (x2+width/2,y2) to construct a rectangle and then use quads/tris to draw it. This the simple naive way. Note that for large line widths you'll get weird endpoint behavior. What you really want to do then is some smart parallel line calculations (which shouldn't be that bad) using vector math. For some reason dot/cross product and vector projection come to mind.
A rectangle (i.e. GL_QUAD or two GL_TRIANGLES) sounds like your best bet by the sounds of it, not sure I can think of any other way.
Another way to do this, if you are writing a software rasterizer by chance, is to use barycentric coordinates in your pixel coloration stage and color pixels when one of the barycentric coordinates is near 0. The more of an allowance you make, the thicker the lines will be.