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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?
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
First, see:
https://math.stackexchange.com/questions/105180/positioning-a-widget-involving-intersection-of-line-and-a-circle
I have an algorithm that solves for the height of an object given a circle and an offset.
It sort of works but the height is always off:
Here is the formula:
and here is a sketch of what it is supposed to do:
And here is sample output from the application:
In the formula, offset = 10 and widthRatio is 3. This is why it is (1 / 10) because (3 * 3) + 1 = 10.
The problem, as you can see is the height of the blue rectangle is not correct. I set the bottom left offsets to be the desired offset (in this case 10) so you can see the bottom left corner is correct. The top right corner is wrong because from the top right corner, I should only have to go 10 pixels until I touch the circle.
The code I use to set the size and location is:
void DataWidgetsHandler::resize( int w, int h )
{
int tabSz = getProportions()->getTableSize() * getProportions()->getScale();
int r = tabSz / 2;
agui::Point tabCenter = agui::Point(
w * getProportions()->getTableOffset().getX(),
h * getProportions()->getTableOffset().getY());
float widthRatio = 3.0f;
int offset = 10;
int height = solveHeight(offset,widthRatio,tabCenter.getX(),tabCenter.getY(),r);
int width = height * widthRatio;
int borderMargin = height;
m_frame->setLocation(offset,
h - height - offset);
m_frame->setSize(width,height);
m_borderLayout->setBorderMargins(0,0,borderMargin,borderMargin);
}
I can assert that the table radius and table center location are correct.
This is my implementation of the formula:
int DataWidgetsHandler::solveHeight( int offset, float widthRatio, float h, float k, float r ) const
{
float denom = (widthRatio * widthRatio) + 1.0f;
float rSq = denom * r * r;
float eq = widthRatio * offset - offset - offset + h - (widthRatio * k);
eq *= eq;
return (1.0f / denom) *
((widthRatio * h) + k - offset - (widthRatio * (offset + offset)) - sqrt(rSq - eq) );
}
It uses the quadratic formula to find what the height should be so that the distance between the top right of the rectangle, bottom left, amd top left are = offset.
Is there something wrong with the formula or implementation? The problem is the height is never long enough.
Thanks
Well, here's my solution, which looks to resemble your solveHeight function. There might be some arithmetic errors in the below, but the method is sound.
You can think in terms of matching the coordinates at the point of the circle across
from the rectangle (P).
Let o_x,o_y be the lower left corner offset distances, w and h be the
height of the rectangle, w_r be the width ratio, dx be the desired
distance between the top right hand corner of the rectangle and the
circle (moving horizontally), c_x and c_y the coordinates of the
circle's centre, theta the angle, and r the circle radius.
Labelling it is half the work! Simply write down the coordinates of the point P:
P_x = o_x + w + dx = c_x + r cos(theta)
P_y = o_y + h = c_y + r sin(theta)
and we know w = w_r * h.
To simplify the arithmetic, let's collect some of the constant terms, and let X = o_x + dx - c_x and Y = o_y - c_y. Then we have
X + w_r * h = r cos(theta)
Y + h = r sin(theta)
Squaring and summing gives a quadratic in h:
(w_r^2 + 1) * h^2 + 2 (X*w_r + Y) h + (X^2+Y^2-r^2) == 0
If you compare this with your effective quadratic, then as long as we made different mistakes :-), you might be able to figure out what's going on.
To be explicit: we can solve this using the quadratic formula, setting
a = (w_r^2 + 1)
b = 2 (X*w_r + Y)
c = (X^2+Y^2-r^2)
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?
I'm trying to make a card game where the cards fan out. Right now to display it Im using the Allegro API which has a function:
al_draw_rotated_bitmap(OBJECT_TO_ROTATE,CENTER_X,CENTER_Y,X
,Y,DEGREES_TO_ROTATE_IN_RADIANS);
so with this I can make my fan effect easily. The problem is then knowing which card is under the mouse. To do this I thought of doing a polygon collision test. I'm just not sure how to rotate the 4 points on the card to make up the polygon. I basically need to do the same operation as Allegro.
for example, the 4 points of the card are:
card.x
card.y
card.x + card.width
card.y + card.height
I would need a function like:
POINT rotate_point(float cx,float cy,float angle,POINT p)
{
}
Thanks
First subtract the pivot point (cx,cy), then rotate it (counter clock-wise), then add the point again.
Untested:
POINT rotate_point(float cx,float cy,float angle,POINT p)
{
float s = sin(angle);
float c = cos(angle);
// translate point back to origin:
p.x -= cx;
p.y -= cy;
// rotate point
float xnew = p.x * c - p.y * s;
float ynew = p.x * s + p.y * c;
// translate point back:
p.x = xnew + cx;
p.y = ynew + cy;
return p;
}
If you rotate point (px, py) around point (ox, oy) by angle theta you'll get:
p'x = cos(theta) * (px-ox) - sin(theta) * (py-oy) + ox
p'y = sin(theta) * (px-ox) + cos(theta) * (py-oy) + oy
this is an easy way to rotate a point in 2D.
The coordinate system on the screen is left-handed, i.e. the x coordinate increases from left to right and the y coordinate increases from top to bottom. The origin, O(0, 0) is at the upper left corner of the screen.
A clockwise rotation around the origin of a point with coordinates (x, y) is given by the following equations:
where (x', y') are the coordinates of the point after rotation and angle theta, the angle of rotation (needs to be in radians, i.e. multiplied by: PI / 180).
To perform rotation around a point different from the origin O(0,0), let's say point A(a, b) (pivot point). Firstly we translate the point to be rotated, i.e. (x, y) back to the origin, by subtracting the coordinates of the pivot point, (x - a, y - b).
Then we perform the rotation and get the new coordinates (x', y') and finally we translate the point back, by adding the coordinates of the pivot point to the new coordinates (x' + a, y' + b).
Following the above description:
a 2D clockwise theta degrees rotation of point (x, y) around point (a, b) is:
Using your function prototype: (x, y) -> (p.x, p.y); (a, b) -> (cx, cy); theta -> angle:
POINT rotate_point(float cx, float cy, float angle, POINT p){
return POINT(cos(angle) * (p.x - cx) - sin(angle) * (p.y - cy) + cx,
sin(angle) * (p.x - cx) + cos(angle) * (p.y - cy) + cy);
}
float s = sin(angle); // angle is in radians
float c = cos(angle); // angle is in radians
For clockwise rotation :
float xnew = p.x * c + p.y * s;
float ynew = -p.x * s + p.y * c;
For counter clockwise rotation :
float xnew = p.x * c - p.y * s;
float ynew = p.x * s + p.y * c;
This is the answer by Nils Pipenbrinck, but implemented in c# fiddle.
https://dotnetfiddle.net/btmjlG
using System;
public class Program
{
public static void Main()
{
var angle = 180 * Math.PI/180;
Console.WriteLine(rotate_point(0,0,angle,new Point{X=10, Y=10}).Print());
}
static Point rotate_point(double cx, double cy, double angle, Point p)
{
double s = Math.Sin(angle);
double c = Math.Cos(angle);
// translate point back to origin:
p.X -= cx;
p.Y -= cy;
// rotate point
double Xnew = p.X * c - p.Y * s;
double Ynew = p.X * s + p.Y * c;
// translate point back:
p.X = Xnew + cx;
p.Y = Ynew + cy;
return p;
}
class Point
{
public double X;
public double Y;
public string Print(){
return $"{X},{Y}";
}
}
}
Ps: Apparently I can’t comment, so I’m obligated to post it as an answer ...
I struggled while working MS OCR Read API which returns back angle of rotation in range (-180, 180]. So I have to do an extra step of converting negative angles to positive. I hope someone struggling with point rotation with negative or positive angles can use the following.
def rotate(origin, point, angle):
"""
Rotate a point counter-clockwise by a given angle around a given origin.
"""
# Convert negative angles to positive
angle = normalise_angle(angle)
# Convert to radians
angle = math.radians(angle)
# Convert to radians
ox, oy = origin
px, py = point
# Move point 'p' to origin (0,0)
_px = px - ox
_py = py - oy
# Rotate the point 'p'
qx = (math.cos(angle) * _px) - (math.sin(angle) * _py)
qy = (math.sin(angle) * _px) + (math.cos(angle) * _py)
# Move point 'p' back to origin (ox, oy)
qx = ox + qx
qy = oy + qy
return [qx, qy]
def normalise_angle(angle):
""" If angle is negative then convert it to positive. """
if (angle != 0) & (abs(angle) == (angle * -1)):
angle = 360 + angle
return angle