I am trying to implement a raytracer that uses an arbitrary camera position and perspective projection. I have the camera position, the look at position, the angle of field of view, but I cannot figure out the direction I have to shoot the rays so that each ray corresponds to a pixel. If I could find a way to find the coordinates of the image plane, or the direction vectors the rays should have, it would be downhill from there. Any help is appreciated.
I would do the following: imagine that there is a rectangular grid just in front of your eye. The grid is defined by one point (the (0;0) point of the grid) and two (three dimensional) base vectors (x,y); with this you can calculate a ray as (origin + Xcoordinate * x + Ycoordinate * y) - eye. By adjusting the distance between your eye point, and origin; or by adjusting the length of the base vectors you could get the desired angle of view.
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I am using orthographic projection glOrtho for my scene. I implemented a virtual trackball to rotate an object beside that I also implemented a zoom in/out on the view matrix. Say I have a cube of size 100 unit and is located at the position of (0,-40000,0) far from the origin. If the center of rotation is located at the origin once the user rotate the cube and after zoom in or out, it could be position at some where (0,0,2500000) (this position is just an assumption and it is calculated after multiplied by the view matrix). Currently I define a very big range of near(-150000) and far(150000) plane, but some time the object still lie outside either the near or far plane and the object just turn invisible, if I define a larger near and far clipping plane say -1000000 and 1000000, it will produce an ungly z artifacts. So my question is how do I correctly calculate the near and far plane when user rotate the object in real time? Thanks in advance!
Update:
I have implemented a bounding sphere for the cube. I use the inverse of view matrix to calculate the camera position and calculate the distance of the camera position from the center of the bounding sphere (the center of the bounding sphere is transformed by the view matrix). But I couldn't get it to work. can you further explain what is the relationship between the camera position and the near plane?
A simple way is using the "bounding sphere". If you know the data bounding box, the maximum diagonal length is the diameter of the bounding sphere.
Let's say you calculate the distance 'dCC' from the camera position to the center of the sphere. Let 'r' the radius of that sphere. Then:
Near = dCC - r - smallMargin
Far = dCC + r + smallMargin
'smallMargin' is a value used just to avoid clipping points on the surface of the sphere due to numerical precision issues.
The center of the sphere should be the center of rotation. If not, the diameter should grow so as to cover all data.
I'm creating a 360° image player using Oculus rift SDK.
The scene is composed by a cube and the camera is posed in the center of it with just the possibility to rotate around yaw, pitch and roll.
I've drawn the object using openGL considering a 2D texture for each cube's face to create the 360° effect.
I would like to find the portion in the original texture that is actual shown on the Oculus viewport in a certain instant.
Up to now, my approach was try to find the an approximate pixel position of some significant point of the viewport (i.e. the central point and the corners) using the Euler Angles in order to identify some areas in the original textures.
Considering all the problems of using Euler Angles, do not seems the smartest way to do it.
Is there any better approach to accomplish it?
Edit
I did a small example that can be runned in the render loop:
//Keep the Orientation from Oculus (Point 1)
OVR::Matrix4f rotation = Matrix4f(hmdState.HeadPose.ThePose);
//Find the vector respect to a certain point in the viewport, in this case the center (Point 2)
FovPort fov_viewport = FovPort::CreateFromRadians(hmdDesc.CameraFrustumHFovInRadians, hmdDesc.CameraFrustumVFovInRadians);
Vector2f temp2f = fov_viewport.TanAngleToRendertargetNDC(Vector2f(0.0,0.0));// this values are the tangent in the center
Vector3f vector_view = Vector3f(temp2f.x, temp2f.y, -1.0);// just add the third component , where is oriented
vector_view.Normalize();
//Apply the rotation (Point 3)
Vector3f final_vect = rotation.Transform(vector_view);//seems the right operation.
//An example to check if we are looking at the front face (Partial point 4)
if (abs(final_vect.z) > abs(final_vect.x) && abs(final_vect.z) > abs(final_vect.y) && final_vect.z <0){
system("pause");
}
Is it right to consider the entire viewport or should be done for each single eye?
How can be indicated a different point of the viewport respect to the center? I don't really understood which values should be the input of TanAngleToRendertargetNDC().
You can get a full rotation matrix by passing the camera pose quaternion to the OVR::Matrix4 constructor.
You can take any 2D position in the eye viewport and convert it to its camera space 3D coordinate by using the fovPort tan angles. Normalize it and you get the direction vector in camera space for this pixel.
If you apply the rotation matrix gotten earlier to this direction vector you get the actual direction of that ray.
Now you have to convert from this direction to your texture UV. The component with the highest absolute value in the direction vector will give you the face of the cube it's looking at. The remaining components can be used to find the actual 2D location on the texture. This depends on how your cube faces are oriented, if they are x-flipped, etc.
If you are at the rendering part of the viewer, you will want to do this in a shader. If this is to find where the user is looking at in the original image or the extent of its field of view, then only a handful of rays would suffice as you wrote.
edit
Here is a bit of code to go from tan angles to camera space coordinates.
float u = (x / eyeWidth) * (leftTan + rightTan) - leftTan;
float v = (y / eyeHeight) * (upTan + downTan) - upTan;
float w = 1.0f;
x and y are pixel coordinates, eyeWidth and eyeHeight are eye buffer size, and *Tan variables are the fovPort values. I first express the pixel coordinate in [0..1] range, then scale that by the total tan angle for the direction, and then recenter.
I have a geometry with its vertices in cartesian coordinates. These cartesian coordinates are the ECEF(Earth centred earth fixed) coordinates. This geometry is actually present on an ellipsoidal model of the earth using wgs84 corrdinates.The cartesian coordinates were actually obtained by converting the set of latitudes and longitudes along which the geomtries lie but i no longer have access to them. What i have is an axis aligned bounding box with xmax, ymax, zmax and xmin,ymin,zmin obtained by parsing the cartesian coordinates (There is no obviously no cartesian point of the geometry at xmax,ymax,zmax or xmin,ymin,zmin. The bounding box is just a cuboid enclosing the geometry).
What i want to do is to calculate the camera distance in an overview mode such that this geometry's bounding box perfectly fits the camera frustum.
I am not very clear with the approach to take here. A method like using a local to world matrix comes to mind but its not very clear.
#Specktre I referred to your suggestions on shifting points in 3D and that led me to another improved solution, nevertheless not perfect.
Compute a matrix that can transfer from ECEF to ENU. Refer this - http://www.navipedia.net/index.php/Transformations_between_ECEF_and_ENU_coordinates
Rotate all eight corners of my original bounding box using this matrix.
Compute a new bounding box by finding the min and max of x,y,z of these rotated points
compute distance
cameraDistance1 = ((newbb.ymax - newbb.ymin)/2)/tan(fov/2)
cameraDistance2 = ((newbb.xmax - newbb.xmin)/2)/(tan(fov/2)xaspectRatio)
cameraDistance = max(cameraDistance1, cameraDistance2)
This time i had to use the aspect ratio along x as i had previously expected since in my application fov is along y. Although this works almost accurately, there is still a small bug i guess. I am not very sure if it a good idea to generate a new bounding box. May be it is more accurate to identify 2 points point1(xmax, ymin, zmax) and point(xmax, ymax, zmax) in the original bounding box, find their values after multiplying with matrix and then do (point2 - point1).length(). Similarly for y. Would that be more accurate?
transform matrix
first thing is to understand that transform matrix represents coordinate system. Look here Transform matrix anatomy for another example.
In standard OpenGL notation If you use direct matrix then you are converting from matrix local space (LCS) to world global space (GCS). If you use inverse matrix then you converting coordinates from GCS to LCS
camera matrix
camera matrix converts to camera space so you need the inverse matrix. You get camera matrix like this:
camera=inverse(camera_space_matrix)
now for info on how to construct your camera_space_matrix so it fits the bounding box look here:
Frustrum distance computation
so compute midpoint of the top rectangle of your box compute camera distance as max of distance computed from all vertexes of box so
camera position = midpoint + distance*midpoint_normal
orientation depends on your projection matrix. If you use gluPerspective then you are viewing -Z or +Z according selected glDepthFunc. So set Z axis of matrix to normal and Y,X vectors can be aligned to North/South and East/West so for example
Y=Z x (1,0,0)
X = Z x Y
now put position, and axis vectors X,Y,Z inside matrix, compute inverse matrix and that it is.
[Notes]
Do not forget that FOV can have different angles for X and Y axis (aspect ratio).
Normal is just midpoint - Earth center which is (0,0,0) so normal is also the midpoint. Just normalize it to size 1.0.
For all computations use cartesian world GCS (global coordinate system).
I have a sphere in my program and I intend to draw some rectangles over at a distance x from the centre of this sphere. The figure looks something below:
The rectangles are drawn at (x,y,z) points that I have already have in a vector of 3d points.
Let's say the distance x from centre is 10. Notice the orientation of these rectangles and these are tangential to an imaginary sphere of radius 10 (perpendicular to an imaginary line from the centre of sphere to the centre of rectangle)
Currently, I do something like the following:
For n points vector<vec3f> pointsInSpace where the rectnagles have to be plotted
for(int i=0;i<pointsInSpace.size();++i){
//draw rectnagle at (x,y,z)
}
which does not have this kind of tangential orientation that I am looking for.
It looked to me of applying roll,pitch,yaw rotations for each of these rectangles and using quaternions somehow to make them tangential as to what I am looking for.
However, it looked a bit complex to me and I wanted to ask about some better method to do this.
Also, the rectangle in future might change to some other shape, so a kind of generic solution would be appreciated.
I think you essentially want the same transformation as would be accomplished with a LookAt() function (you want the rectangle to 'look at' the sphere, along a vector from the rectangle's center, to the sphere's origin).
If your rectangle is formed of the points:
(-1, -1, 0)
(-1, 1, 0)
( 1, -1, 0)
( 1, 1, 0)
Then the rectangle's normal will be pointing along Z. This axis needs to be oriented towards the sphere.
So the normalised vector from your point to the center of the sphere is the Z-axis.
Then you need to define a distinct 'up' vector - (0,1,0) is typical, but you will need to choose a different one in cases where the Z-axis is pointing in the same direction.
The cross of the 'up' and 'z' axes gives the x axis, and then the cross of the 'x' and 'z' axes gives the 'y' axis.
These three axes (x,y,z) directly form a rotation matrix.
This resulting transformation matrix will orient the rectangle appropriately. Either use GL's fixed function pipeline (yuk), in which case you can just use gluLookAt(), or build and use the matrix above in whatever fashion is appropriate in your own code.
Personally I think the answer of JasonD is enough. But here is some info of the calculation involved.
Mathematically speaking this is a rather simple problem, What you have is a 2 known vectors. You know the position vector and the spheres normal vector. Since the square can be rotated arbitrarily along around the vector from center of your sphere you need to define one more vector, the up vector. Without defining up vector it becomes a impossible solution.
Once you define a up vector vector, the problem becomes simple. Assuming your square is on the XY-plane as JasonD suggest above. Then your matrix becomes:
up_dot_n_dot_n.X up_dot_n_dot_n.Y up_dot_n_dot_n.Z 0
n.X n.y n.z 0
up_dot_n.x up_dot_n.x up_dot_n.z 0
p.x p.y p.z 1
Where n is the normal unit vector of p - center of sphere (which is trivial if sphere is in the center of the coordinate system), up is a arbitrary unit vector vector. The p follows form definition and is the position.
The solution has a bit of a singularity at the up direction of the sphere. An alternate solution is to rotate first 360 around up, the 180 around rotated axis dot up. Produces same thing different approach no singularity problem.
Heyo,
I'm currently working on a project where I need to place the camera such that the full motion of a character would be viewable without moving the camera. I have the position where the character starts, as well as the maximum distance that the character will travel in all three directions (X,Y, & Z). I also have the field of view (which is 90 degrees).
Is there an equation that'll figure out where I need to place the camera so it won't have to move to see the full motion?
Note: this is using OpenGL.
Clarification: The camera should be "in front" of the character that's in the motion, not above.
It'll also be moving along a ground plane.
If you make a bounding sphere of the points, all you need to do is keep the camera at a distance greater than or equal to the radius of the bounding sphere / sin(FOV/2).
For example, if you have a bounding sphere with radius Radius, and a specified Field of View FOV, your camera just needs to be at a point "Dist" away, pointing towards the center of the bounding sphere.
The equation for calculating the distance is:
Dist = Radius / sin( FOV/2 );
This will work in 3D, for a camera at any orientation.
Simply having the maximum range of (X, Y, Z) is not on its own sufficient, because the viewing port is essentially pyramid shaped, with the apex of the pyramid being at the eye position.
For the sake of argument, let's assume that all movement is in the (X, Z) plane (i.e. the ground), and the eye is directly above the origin 10m along the Y axis.
Assuming a square viewport, with your 90˚ field of view you'd be able to see from ±10m along both the X and Z axis, but only for objects who are on the ground (Y = 0). As soon as they come off the ground your view is reduced. If it's 1m of the ground then your (X, Z) extent is only ±9m.
Clearly a real camera could be placed anyway in the scene, facing any direction. Even the "roll" angle of the camera could change how much is visible. There are actually infinitely many such camera points, so you will need to constrain your criteria somewhat.
Take the line segment from the startpoint to the endpoint. Construct a plane orthogonal to this line segment through the midpoint of the line segment. Then position the camera somewhere in this plane at an distance of more than the following from the intersection point of plane and line looking at the intersection point. The up vector of the camera must be in the plane and the horizontal field of view must be 90 degrees.
distance = sqrt(dx^2 + dy^2 + dz^2) / 2
This camera positions will all have the startpoint and the endpoint on the left or right border of the view port and verticaly centered.
Another solution might be to write a function that takes the startpoint, the endpoint, and the desired position of both points on the screen. Then just solve the projection equation for the camera transformation.
It depends, for example, if the object is gonna move in a plane, you can just place the camera outside a ball circumscribed its movement area (this depends on the fact that FOV is 90, which is a fortunate angle).
If the object is gonna move in 3D, it's much more difficult. It would help if you'd specify the region where the object moves (cube vs. ball...) and the direction you want to see it from.