My application is a vector drawing application. It works with OpenGL. I will be modifying it to instead use the Cairo 2D graphics library. The issue is with zooming. With openGL camera and scale factor sort of work like this:
float scalediv = Current_Scene().camera.ScaleFactor / 2.0f;
float cameraX = GetCameraX();
float cameraY = GetCameraY();
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
float left = cameraX - ((float)controls.MainGlFrame.Dimensions.x) * scalediv;
float right = cameraX + ((float)controls.MainGlFrame.Dimensions.x) * scalediv;
float bottom = cameraY - ((float)controls.MainGlFrame.Dimensions.y) * scalediv;
float top = cameraY + ((float)controls.MainGlFrame.Dimensions.y) * scalediv;
glOrtho(left,
right,
bottom,
top,
-0.01f,0.01f);
// Set the model matrix as the current matrix
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
hdc = BeginPaint(controls.MainGlContext.mhWnd,&ps);
Mouse position is obtained like this:
POINT _mouse = controls.MainGlFrame.GetMousePos();
vector2f mouse = functions.ScreenToWorld(_mouse.x,_mouse.y,GetCameraX(),GetCameraY(),
Current_Scene().camera.ScaleFactor,
controls.MainGlFrame.Dimensions.x,
controls.MainGlFrame.Dimensions.y );
vector2f CGlEngineFunctions::ScreenToWorld(int x, int y, float camx, float camy, float scale, int width, int height)
{
// Move the given point to the origin, multiply by the zoom factor and
// add the model coordinates of the center point (camera position)
vector2f p;
p.x = (float)(x - width / 2.0f) * scale +
camx;
p.y = -(float)(y - height / 2.0f) * scale +
camy;
return p;
}
From there I draw the VBO's of triangles. This allows me to pan and zoom in. Given that Cairo only can draw based on coordinates, how can I make it so that a vertex is properly scaled and panned without using transformations. Basically GlOrtho sets the viewport usually but I dont think I could do this with Cairo.
Well GlOrtho is able to change the viewport matrix instead of modifying the verticies but how could I instead modify the verticies to get the same result?
Thanks
*Given vertex P, which was obtained from ScreenToWorld, how could I modify it so that it is scaled and panned accordng to the camera and scale factor? Because usually OpenGL would essentially do this
I think Cairo can do what you want ... see http://cairographics.org/matrix_transform/ . Does that solve your problem, and if not, why ?
Related
As I have it understood, a projection matrix scales a polygon depending on how far away or close it is from the camera. Though I might be completely wrong. My question is, how does the projection matrix "know" to show the sides of the following cube, as the camera moves, when the matrix is only supposed "scale" polygons?
Notice in the image, the cube is off to the right side of the screen, by moving the camera to the left. If the camera is moved in the opposite direction (to the right) in order to center the cube, the side of the cube will disappear as expected.
Here is my matrix code:
private void createProjectionMatrix(){
float aspectRatio = (float) Display.getWidth() / (float) Display.getHeight();
float y_scale = (float) ((1f / Math.tan(Math.toRadians(FOV/2f))) * aspectRatio);
float x_scale = y_scale / aspectRatio;
float frustum_length = FAR_PLANE - NEAR_PLANE;
projectionMatrix = new Matrix4f();
projectionMatrix.m00 = x_scale;
projectionMatrix.m11 = y_scale;
projectionMatrix.m22 = -((FAR_PLANE + NEAR_PLANE) / frustum_length);
projectionMatrix.m23 = -1;
projectionMatrix.m32 = -((2 * NEAR_PLANE * FAR_PLANE) / frustum_length);
projectionMatrix.m33 = 0;
}
The function of a projection matrix (in the context of graphics APIs, such as OpenGL) is to transform vertex positions from view-space into clip-space.
Clip space is generally a unit box (although in D3D it's a half-unit box). If a vertex position after being transformed into clip-space does not lie within that unit box, then it is clipped. This is essentially how the system "knows" the cube is visible on the screen.
I am working on some really simple ray-tracer.
For now I am trying to make the perspective camera works properly.
I use such loop to render the scene (with just two, hard-coded spheres - I cast ray for each pixel from its center, no AA applied):
Camera * camera = new PerspectiveCamera({ 0.0f, 0.0f, 0.0f }/*pos*/,
{ 0.0f, 0.0f, 1.0f }/*direction*/, { 0.0f, 1.0f, 0.0f }/*up*/,
buffer->getSize() /*projectionPlaneSize*/);
Sphere * sphere1 = new Sphere({ 300.0f, 50.0f, 1000.0f }, 100.0f); //center, radius
Sphere * sphere2 = new Sphere({ 100.0f, 50.0f, 1000.0f }, 50.0f);
for(int i = 0; i < buffer->getSize().getX(); i++) {
for(int j = 0; j < buffer->getSize().getY(); j++) {
//for each pixel of buffer (image)
double centerX = i + 0.5;
double centerY = j + 0.5;
Geometries::Ray ray = camera->generateRay(centerX, centerY);
Collision * collision = ray.testCollision(sphere1, sphere2);
if(collision){
//output red
}else{
//output blue
}
}
}
The Camera::generateRay(float x, float y) is:
Camera::generateRay(float x, float y) {
//position = camera position, direction = camera direction etc.
Point2D xy = fromImageToPlaneSpace({ x, y });
Vector3D imagePoint = right * xy.getX() + up * xy.getY() + position + direction;
Vector3D rayDirection = imagePoint - position;
rayDirection.normalizeIt();
return Geometries::Ray(position, rayDirection);
}
Point2D fromImageToPlaneSpace(Point2D uv) {
float width = projectionPlaneSize.getX();
float height = projectionPlaneSize.getY();
float x = ((2 * uv.getX() - width) / width) * tan(fovX);
float y = ((2 * uv.getY() - height) / height) * tan(fovY);
return Point2D(x, y);
}
The fovs:
double fovX = 3.14159265359 / 4.0;
double fovY = projectionPlaneSize.getY() / projectionPlaneSize.getX() * fovX;
I get good result for 1:1 width:height aspect (e.g. 400x400):
But I get errors for e.g. 800x400:
Which is even slightly worse for bigger aspect ratios (like 1200x400):
What did I do wrong or which step did I omit?
Can it be a problem with precision or rather something with fromImageToPlaneSpace(...)?
Caveat: I spent 5 years at a video company, but I'm a little rusty.
Note: after writing this, I realized that pixel aspect ratio may not be your problem as the screen aspect ratio also appears to be wrong, so you can skip down a bit.
But, in video we were concerned with two different video sources: standard definition with a screen aspect ratio of 4:3 and high definition with a screen aspect ratio of 16:9.
But, there's also another variable/parameter: pixel aspect ratio. In standard definition, pixels are square and in hidef pixels are rectangular (or vice-versa--I can't remember).
Assuming your current calculations are correct for screen ratio, you may have to account for the pixel aspect ratio being different, either from camera source or the display you're using.
Both screen aspect ratio and pixel aspect ratio can be stored a .mp4, .jpeg, etc.
I downloaded your 1200x400 jpeg. I used ImageMagick on it to change only the pixel aspect ratio:
convert orig.jpg -resize 125x100%\! new.jpg
This says change the pixel aspect ratio (increase the width by 125% and leave the height the same). The \! means pixel vs screen ratio. The 125 is because I remember the rectangular pixel as 8x10. Anyway, you need to increase the horizontal width by 10/8 which is 1.25 or 125%
Needless to say this gave me circles instead of ovals.
Actually, I was able to get the same effect with adjusting the screen aspect ratio.
So, somewhere in your calculations, you're introducing a distortion of that factor. Where are you applying the scaling? How are the function calls different?
Where do you set the screen size/ratio? I don't think that's shown (e.g. I don't see anything like 1200 or 400 anywhere).
If I had to hazard a guess, you must account for aspect ratio in fromImageToPlaneSpace. Either width/height needs to be prescaled or the x = and/or y = lines need scaling factors. AFAICT, what you've got will only work for square geometry at present. To test, using the 1200x400 case, multiply the x by 125% [a kludge] and I bet you get something.
From the images, it looks like you have incorrectly defined the mapping from pixel coordinates to world coordinates and are introducing some stretch in the Y axis.
Skimming your code it looks like you are defining the camera's view frustum from the dimensions of the frame buffer. Therefore if you have a non-1:1 aspect ratio frame buffer, you have a camera whose view frustum is not 1:1. You will want to separate the model of the camera's view frustum from the image space dimension of the final frame buffer.
In other words, the frame buffer is the portion of the plane projected by the camera that we are viewing. The camera defines how the 3D space of the world is projected onto the camera plane.
Any basic book on 3D graphics will discuss viewing and projection.
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'd like to convert a 3d position into 2d screen position. I had a look at a similar question: Projecting a 3D point to a 2D screen coordinate , but I dont understand it completely. I thought in order to calculate the 2d position I would need the projection matrix, but I dont see how it is used, apart from converting a point into the location coordinate space. Besides, is cam.FieldOfView equal to farZ in OpenGL?
Could someone please help me complete this function. Are the parameters sufficient to calculate the 2d position? Pos is already a vector relative to the camera position.
Vector2* convert(Vector3& pos, Matrix4& projectionMatrix, int screenWidth, int screenHeight)
{
float ratio = screenWidth / screenHeight;
...
screenX = screenWidth * ( 1.0f - screenX);
screenY = screenHeight * ( 1.0f - screenY);
return new Vector2(screenX, screenY);
}
Seems to me it would be something like that:
Vector2 Convert(Vector3 pos, const Matrix& viewMatrix, const Matrix& projectionMatrix, int screenWidth, int screenHeight)
{
pos = Vector3::Transform(pos, viewMatrix);
pos = Vector3::Transform(pos, projectionMatrix);
pos.X = screenWidth*(pos.X + 1.0)/2.0;
pos.Y = screenHeight * (1.0 - ((pos.Y + 1.0) / 2.0));
return Vector2(pos.X, pos.Y);
}
What are we doing here is just passing the Vector though the two transformation matrices: the view, then the projection. After the projection you get a vector with Y and X between -1 and 1. We do the appropriate transformation to obtain real pixel coordinates and return a new Vector2. Note that the Z component of 'pos' also store the depth of the point, in the screen space, at the end of the function.
You need the 'view' matrix because it defines where the camera is located and rotated. The projection only defines the way the 3D space is 'flattened' on the 2D space.
A field of view is not the farZ. A projection matrix has some parameters, among them:
the field of view, FOV, that is the horizontal angle of view, in radians;
the far plane, or farZ : this defines the maximum distance a point can be from the camera;
the near plane, nearZ: the minimum distance a point can be from the camera.
Besides the math problem, you may use directly the Vector2 instead of a heap allocation (returning a pointer). Vector2 is a light structure and pointers are very likely to cause headaches in this context (where are you going to delete it, and so on). Also note that I used 'const' references as we do not modify them, except the vector. For this one we want a local copy, this is why it is not a reference at all.
Previous code only work if you do not do any rotations (for eg. GL.Rotate(rotation_x, 1.0, 0.0, 0.0)).
But if you do here is the code:
private Vector2 Convert(Vector3 pos, Matrix4 viewMatrix, Matrix4 projectionMatrix, int screenWidth, int screenHeight)
{
pos = Vector3.Transform(pos, viewMatrix);
pos = Vector3.Transform(pos, projectionMatrix);
pos.X /= pos.Z;
pos.Y /= pos.Z;
pos.X = (pos.X + 1) * screenWidth / 2;
pos.Y = (pos.Y + 1) * screenHeight / 2;
return new Vector2(pos.X, pos.Y);
}
I think what you're looking for is a replacement for gluLookAt. Given a position and orientation it converts the scene geometry into screen coordinates for rendering. As the article says, it relies on a number of deprecated features of OpenGL, but it does provide a code sample you can implement using your vector / matrix library. More detailed information on the projection matrices is available from here.
Once you have the projection matrix you simply apply it to your vectors (post-multiply your scene's vectors by the projection matrix) and then just drop the Z component of the resulting vector ... that is, just use the X and Y components of the resultant vectors.
Here is my issue, I have a scale point, which is the unprojected mouse position. I also have a "camera which basically translates all objects by X and Y. What I want to do is achieve zooming into mouse position.
I'v tried this:
1. Find the mouse's x and y coordinates
2. Translate by (x,y,0) to put the origin at those coordinates
3. Scale by your desired vector (i,j,k)
4. Translate by (-x,-y,0) to put the origin back at the top left
But this doesn't factor in a translation for the camera.
How can I properly do this. Thanks
glTranslatef(controls.MainGlFrame.GetCameraX(),
controls.MainGlFrame.GetCameraY(),0);
glTranslatef(current.ScalePoint.x,current.ScalePoint.y,0);
glScalef(current.ScaleFactor,current.ScaleFactor,0);
glTranslatef(-current.ScalePoint.x,-current.ScalePoint.y,0);
Instead of using glTranslate to move all the objects, you should try glOrtho. It takes as parameters the wanted left coords, right coords, bottom coords, top coords, and min/max depth.
For example if you call glOrtho(-5, 5, -2, 2, ...); your screen will show all the points whose coords are inside a rectangle going from (-5,2) to (5,-2). The advantage is that you can easily adjust the zoom level.
If you don't multiply by any view/projection matrix (which I assume is the case), the default screen coords range from (-1,1) to (1,-1).
But in your project it can be very useful to control the camera. Call this before you draw any object instead of your glTranslate:
float left = cameraX - zoomLevel * 2;
float right = cameraX + zoomLevel * 2;
float top = cameraY + zoomLevel * 2;
float bottom = cameraY - zoomLevel * 2;
glOrtho(left, right, bottom, top, -1.f, 1.f);
Note that cameraX and cameraY now represent the center of the screen.
Now when you zoom on a point, you simply have to do something like this:
cameraX += (cameraX - screenX) * 0.5f;
cameraY += (cameraY - screenY) * 0.5f;
zoomLevel += 0.5f;