I'm trying to make do water simulation. But I'm restricted to use 2D, So i started with just making the boundary of the sea by using sine wave, through Gl_Line_loop. but I'm just unable to fill it. I have tried changing it to the Gl_polygon mode but then i don't get the proper shape. here is the code:
here is the image of wave, i want to get filled
To tessellate the above, specify a top then a bottom vertex right along the line, then draw a triangle strip. i.e. for each (x, y) position along the sin wave, emit two vertices, the same x but y = 0 (the bottom). Then render a triangle strip.
Something like this:
glBegin(GL_TRIANGLE_STRIP);
for(x=-50;x<=50;x+=inc){
k = 2 * 3.14 / wavelength;
y = amplitude * sin(k * x);
glVertex3f(x, y-35, 0);
glVertex3f(x, y, 0);
}
glEnd();
Related
I'm trying to draw quadrics in C ++ with OpenGL.
Quadrics are the Sphere, and Hyperboloides of a leaf and two leaves. To draw them I use only the parametric equations of each. The sphere if I drew
Example: The parametric equations of the Sphere are:
To draw the sphere with these equations, what I do in OpenGL is simply this for a sphere that is draw on the origin x0, y0 and z0 equal to zero:
void Sphere(){
GLfloat x, y, z, alpha, beta; // Storage for coordinates and angles
GLfloat radius = 1.0f;
int gradation = 20;
glPolygonMode( GL_FRONT_AND_BACK, GL_LINE );
for (float alpha = 0.0; alpha < PI; alpha += PI/gradation)
{
glBegin(GL_TRIANGLE_STRIP);
for (beta = 0.0; beta < 2.01*PI; beta += PI/gradation)
{
x = radius*cos(beta)*sin(alpha);
y = radius*sin(beta)*sin(alpha);
z = radius*cos(alpha);
glVertex3f(x, y, z);
}
glEnd();
}
glutSwapBuffers();
}
I the result is this (a sphere):
Problem:
By applying the same logic to the draw hyperboloids, I do not get what I want. The parametric equations of the Hyperboloides are as follows:
When attempting to draw the hyperboloid of one sheet with the equations shown in the picture, this is the result:
If anyone knows anything about it, please if you can help me with this. Thank you all.
Note: I know there are other ways to draw a hyperboloid, but the way I draw them is through its parametric equations.
What is the code that produces this output, not the code that produces your sphere? Please provide a MCVE.
Here is my guess: the range of your loops produces only positive values of z, and you might try the range -PI to PI instead.
This question already has an answer here:
OpenGL stretched shapes - aspect ratio
(1 answer)
Closed 7 years ago.
I have been trying to generate an ellipse using OpenGL and I have a feeling I have got something very wrong. I am trying to use an ellipse generating code but for simplicity, I have set the length of the major and minor axes equal. This should give me a circle but somehow that is not what is rendered with OpenGL and I am not sure what is wrong.
So the code is as follows:
glPushAttrib(GL_CURRENT_BIT);
glColor3f(1.0f, 0.0f, 0.0f);
glLineWidth(2.0);
// Draw center
glBegin(GL_POINTS);
glVertex2d(0, 0);
glEnd();
glBegin(GL_LINE_LOOP);
// This should generate a circle
for (GLfloat i = 0; i < 360; i++)
{
float x = cos(i*M_PI/180.f) * 0.5; // keep the axes radius same
float y = sin(i*M_PI/180.f) * 0.5;
glVertex2f(x, y);
}
glEnd();
glPopAttrib();
This should generate a circle as far as I can think. However. I get something like the attached image, which is not a circle. I am not sure what I am doing wrong.
It is a circle in clip space. Note that the horizontal extent is half the screen's width and the vertical extent is half the screen's height. The viewport transformation that maps clip space (-1 to 1 on both axes) to screen space basically performs a scaling and translation, which causes the deformation of the circle.
To prevent this from happening, you need to set up an appropriate projection transform, e.g. with glOrtho.
I have some questions about the screen set up. Originally when I would draw a triangle the x vector 1 would be all the way to the right and -1 would be all the way to the left. Now I have adjusted it to account for the different aspect ratio of the window. My new question how do I make the numbers which are used to render a 2d tri go along with the pixel values. If my window is 480 pixels wide and 320 tall I want to have to enter this to span the screen with a tri
glBegin(GL_TRIANGLES);
glVertex2f(240, 320);
glVertex2f(480, 0);
glVertex2f(0, 0);
glEnd();
but instead it currently looks like this
glBegin(GL_TRIANGLES);
glVertex2f(0, 1);
glVertex2f(1, -1);
glVertex2f(-1, -1);
glEnd();
Any ideas?
You need to use functions glViewport and glOrtho with correct values. Basically glViewport sets the part of your window capable of rendering 3D-Graphics using OpenGL. glOrtho establishes coordinate system within that part of a window using OpenGL's coordinates.
So for your task you need to know exact width and height of your window. If you are saying they are 480 and 320 respectively then you need to call
glViewport(0, 0, 480, 320)
// or: glViewport ( 0,0,w,h)
somewhere, maybe in your SizeChanging-handler(if you are using WINAPI it is WM_SIZE message)
Next, when establishing OpenGL's scene you need to specify OpenGL's coordinates. For orthographic projection they will be the same as dimensions of a window so
glOrtho(-240, 240, -160, 160, -100, 100)
// or: glOrtho ( -w/2, w/2, -h/2, h/2, -100, 100 );
is siutable for your purppose. Not that here I'm using depth of 200 (z goes from -100 to 100).
Next on your rendering routine you may draw your triangle
Since the second piece of code is working for you, I assume your transformation matrices are all identity or you have a shader that bypasses them. Also your viewport is spanning the whole window.
In general if your viewport starts at (x0,y0) and has WxH size, the normalized coordinates (x,y) you feed to glVertex2f will be transformed to (vx,vy) as follows:
vx = x0 + (x * .5f + .5f) * W
vy = y0 + (y * .5f + .5f) * H
If you want to use pixel coordinates you can use the function
void vertex2(int x, int y)
{
float vx = (float(x) + .5f) / 480.f;
float vy = (float(y) + .5f) / 320.f;
glVertex3f(vx, vy, -1.f);
}
The -1 z value is the closest depth to the viewer. It's negative because the z is assumed to be reflected after the transformation (which is identity in your case).
The addition of .5f is because the rasterizer considers a pixel as a 1x1 quad and evaluates the coverage of your triangle in the middle of this quad.
I'm having little trouble whit trying to compare rotated 2D Quads coordinates to rotated x and y coordinates. I'm trying to determine if mouse was clicked inside the quad.
1) the rot's are this classes objects: (note : the operator << is overloaded for the use of the rotate coords func)
class Vector{
private:
std::vector <float> Vertices;
public:
Vector(float, float);
float GetVertice(unsigned int);
void SetVertice(unsigned int, float);
std::vector<float> operator <<(double);
};
Vector::Vector(float X,float Y){
Vertices.push_back(X);
Vertices.push_back(Y);
}
float Vector::GetVertice(unsigned int Index){
return Vertices.at(Index);
}
void Vector::SetVertice(unsigned int Index,float NewVertice){
Vertices.at(Index) = NewVertice;
}
//Return rotated coords:D
std::vector <float> Vector::operator <<(double Angle){
std::vector<float> Temp;
Temp.push_back(Vertices.at(0) * cos(Angle) - Vertices.at(1) * sin(Angle));
Temp.push_back(Vertices.at(0) * sin(Angle) + Vertices.at(1) * cos(Angle));
return Temp;
}
2) Comparasion and rotation of the coordinates THE NEW VERSION
Vector Rot1(x,y),Rot3(x,y);
double Angle;
std::vector <float> result1,result3;
Rot3.SetVertice(0,NewQuads.at(Index).GetXpos() + NewQuads.at(Index).GetWidth());
Rot3.SetVertice(1,NewQuads.at(Index).GetYpos() + NewQuads.at(Index).GetHeight());
Angle = NewQuads.at(Index).GetRotation();
result1 = Rot1 << Angle; // Rotate the mouse x and y
result3 = Rot3 << Angle; // Rotate the Quad x and y
//.at(0) = x and .at(1)=y
if(result1.at(0) >= result3.at(0) - NewQuads.at(Index).GetWidth() && result1.at(0) <= result3.at(0) ){
if(result1.at(1) >= result3.at(1) - NewQuads.at(Index).GetHeight() && result1.at(1) <= result3.at(1) ){
when i run this it works perfectly at 0 angle but when you rotate the quad, it fails.
and by failing I mean the activation area seem to just disappear.
am I doing the rotation of the coordinates correctly? or is it the comparison?
if it's the comparison how would you do it properly, I have tried changing the if's but whit out any luck...
edit
the drawing of the quad(Happens before the testing):
void Quad::Render()
{
if(!CheckIfOutOfScreen()){
glPushMatrix();
glLoadIdentity();
glTranslatef(Xpos ,Ypos ,0.f);
glRotatef(Rotation,0.f,0.f,1.f); // same rotation is used for the testing later...
glBegin(GL_QUADS);
glVertex2f(Zwidth,Zheight);
glVertex2f(Width,Zheight);
glVertex2f(Width,Height);
glVertex2f(Zwidth,Height);
glEnd();
if(State != NOT_ACTIVE)
RenderShapeTools();
glPopMatrix();
}
}
basicly I'm trying to test if mouse was clicked inside this quad:
Image
There is more than one way to achieve what you want, But from the image you posted I assume you want to draw to a surface the same size as your screen (or window) using only 2D graphics.
As you know in 3D graphics we talk about 3 coordinate references. The first is the coordinate reference of the object or model to be drawn, the second is the coordinate reference of the camera or view and the third is the coordinate reference of the screen.
In OpenGL the first two coordinate references are established through the MODELVIEW matrix and the third is achieved by the PROJECTION matrix and the viewport transformation.
In your case you want to rotate a quad and place it somewhere on the screen. Your quad has it's own model coordinates. Let's assume that for this specific 2D quad the origin is at the center of the quad and it has the dimensions of 5 by 5. Also let's assume that if we look to the center of the quad then the X axis points to the RIGHT, the Y axis points UP and the Z axis points towards the viewer.
The unrotated coordinates of the quad will be (from bottom left clockwise): (-2.5,-2.5,0), (-2.5,2.5,0), (2.5,2.5,0), (2.5,-2.5,0)
Now we want to have a camera and projection matrices and viewport so to simulate a 2D surface with known dimensions.
//Assume WinW contains the window width and WinH contains the windows height
glViewport(0,0,WinW,WinH);//Set the viewport to the whole window
glMatrixMode (GL_PROJECTION);
glLoadIdentity ();
glOrtho (0, WinW, WinH, 0, 0, 1);//Set the projection matrix to perform a 2D orthogonal projection
glMatrixMode (GL_MODELVIEW);
glLoadIdentity ();//Set the camera matrix to be the Identity matrix
You are now ready to draw your quad an this 2D surface with dimensions WinW, WinH. In this context if you just draw your quad using it's current vertices you will have the quad drawn with it's center at the bottom left of the window with each side measuring 5 pixels so you will actually see only quarter of a quad. If you want to rotate and move it you will do something like this:
//Prepare matrices as shown above
//Viewport coordinates range from bottom left (0,0) to top right (WinW,WinH)
float dX = CenterOfQuadInViewportCoordinatesX, dY = CenterOfQuadInViewportCoordinatesY;
float rotA = QuadRotationAngleAroundZAxisInDegrees;
float verticesX[4] = {-2.5,-2.5,2.5,2.5};
float verticesY[4] = {-2.5,2.5,2.5,-2.5};
//Remember that rotate is done first and translation second
glTranslatef(dX,dY,0);//Move the quad to the desired location in the viewport
glRotate(rotA, 0,0,1);//Rotate the quad around it's origin
glBegin(GL_QUADS);
glVertex2f(verticesX[0], veriticesY[0]);
glVertex2f(verticesX[1], veriticesY[1]);
glVertex2f(verticesX[2], veriticesY[2]);
glVertex2f(verticesX[3], veriticesY[3]);
glEnd();
Now you want to know whether the click of the mouse was within the rendered quad.
Whereas the viewport coordinates start from the bottom left the window coordinates start from the top left. So when you get the mouse coordinates you have to translate them to viewport coordinates in the following way:
float mouseViewportX = mouseX, mouseViewportY = WinH - mouseY - 1;
Once you have the mouse location in viewport coordinates you need to transform it to model coordinates in the following way (Please double check the calculations since I generally use my own matrix library for that and don't calculate it by hand):
//Translate the mouse location to model coordinates reference
mouseViewportX -= dX, mouseViewportY -= dY;
//Unrotate the mouse location
float invRotARad = -rotA*DEG_TO_RAD;
float sinRA = sin(invRotARad), cosRA = cos(invRotA);
float mouseInModelX = cosRA*mouseViewportX - sinRA*mouseViewportY;
float mouseInModelY = sinRA*mouseViewportX + cosRA*mouseViewportY;
And now you can finally check if the mouse falls within the quad - as you can see this is done in quad coordinates:
bool mouseInQuad = mouseInModelX > verticesX[0] && mouseInModelY < verticesX[1] &&
mouseInModelY > verticesY[0] && mouseInModelY < verticesY[1];
Hope I didn't make too many mistakes and this puts you on the right track. If you want to deal with more complex cases and 3D then you should have a look at gluUnproject (maybe you will want to implement your own) and for even more complex scenes you may need to use a stencil or depth buffers
I want to know how to draw a spiral.
I wrote this code:
void RenderScene(void)
{
glClear(GL_COLOR_BUFFER_BIT);
GLfloat x,y,z = -50,angle;
glBegin(GL_POINTS);
for(angle = 0; angle < 360; angle += 1)
{
x = 50 * cos(angle);
y = 50 * sin(angle);
glVertex3f(x,y,z);
z+=1;
}
glEnd();
glutSwapBuffers();
}
If I don't include the z terms I get a perfect circle but when I include z, then I get 3 dots that's it. What might have happened?
I set the viewport using glviewport(0,0,w,h)
To include z should i do anything to set viewport in z direction?
You see points because you are drawing points with glBegin(GL_POINTS).
Try replacing it by glBegin(GL_LINE_STRIP).
NOTE: when you saw the circle you also drew only points, but drawn close enough to appear as a connected circle.
Also, you may have not setup the depth buffer to accept values in the range z = [-50, 310] that you use. These arguments should be provided as zNear and zFar clipping planes in your gluPerspective, glOrtho() or glFrustum() call.
NOTE: this would explain why with z value you only see a few points: the other points are clipped because they are outside the z-buffer range.
UPDATE AFTER YOU HAVE SHOWN YOUR CODE:
glOrtho(-100*aspectratio,100*aspectratio,-100,100,1,-1); would only allow z-values in the [-1, 1] range, which is why only the three points with z = -1, z = 0 and z = 1 will be drawn (thus 3 points).
Finally, you're probably viewing the spiral from the top, looking directly in the direction of the rotation axis. If you are not using a perspective projection (but an isometric one), the spiral will still show up as a circle. You might want to change your view with gluLookAt().
EXAMPLE OF SETTING UP PERSPECTIVE
The following code is taken from the excellent OpenGL tutorials by NeHe:
glViewport(0, 0, width, height);
glMatrixMode(GL_PROJECTION); // Select The Projection Matrix
glLoadIdentity(); // Reset The Projection Matrix
// Calculate The Aspect Ratio Of The Window
gluPerspective(45.0f,(GLfloat)width/(GLfloat)height,0.1f,100.0f);
glMatrixMode(GL_MODELVIEW); // Select The Modelview Matrix
glLoadIdentity(); // Reset The Modelview Matrix
Then, in your draw loop would look something like this:
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT); // Clear The Screen And The Depth Buffer
glLoadIdentity();
glTranslatef(-1.5f,0.0f,-6.0f); // Move Left 1.5 Units And Into The Screen 6.0
glBegin(GL_TRIANGLES); // Drawing Using Triangles
glVertex3f( 0.0f, 1.0f, 0.0f); // Top
glVertex3f(-1.0f,-1.0f, 0.0f); // Bottom Left
glVertex3f( 1.0f,-1.0f, 0.0f); // Bottom Right
glEnd();
Of course, you should alter this example code your needs.
catchmeifyoutry provides a perfectly capable method, but will not draw a spatially accurate 3D spiral, as any render call using a GL_LINE primitive type will rasterize to fixed pixel width. This means that as you change your perspective / view, the lines will not change width. In order to accomplish this, use a geometry shader in combination with GL_LINE_STRIP_ADJACENCY to create 3D geometry that can be rasterized like any other 3D geometry. (This does require that you use the post fixed-function pipeline however)
I recommended you to try catchmeifyoutry's method first as it will be much simpler. If you are not satisfied, try the method I described. You can use the following post as guidance:
http://prideout.net/blog/?tag=opengl-tron
Here is my Spiral function in C. The points are saved into a list which can be easily drawn by OpenGL (e.g. connect adjacent points in list with GL_LINES).
cx,cy ... spiral centre x and y coordinates
r ... max spiral radius
num_segments ... number of segments the spiral will have
SOME_LIST* UniformSpiralPoints(float cx, float cy, float r, int num_segments)
{
SOME_LIST *sl = newSomeList();
int i;
for(i = 0; i < num_segments; i++)
{
float theta = 2.0f * 3.1415926f * i / num_segments; //the current angle
float x = (r/num_segments)*i * cosf(theta); //the x component
float y = (r/num_segments)*i * sinf(theta); //the y component
//add (x + cx, y + cy) to list sl
}
return sl;
}
An example image with r = 1, num_segments = 1024:
P.S. There is difference in using cos(double) and cosf(float).
You use a float variable for a double function cos.