I have a transform matrix 4x4 built in the following way:
glm::mat4 matrix;
glm::quat orientation = toQuaternion(rotation);
matrix*= glm::mat4_cast(orientation);
matrix= glm::translate(matrix, position);
matrix= glm::scale(matrix, scale);
orientation = toQuaternion(localRotation);
matrix*= glm::mat4_cast(orientation);
matrix= glm::scale(matrix, localScale);
where rotation, position, scale, localRotation, and localScale are all vector3's. As per the advice of many different questions on many different forums, it should be possible to fish local directions out of the resulting matrix like so:
right = glm::vec3(matrix[0][0], matrix[0][1], matrix[0][2]);
up = glm::vec3(matrix[1][0], matrix[1][1], matrix[1][2]);
forward = glm::vec3(matrix[2][0], matrix[2][1], matrix[2][2]);
where all these directions are normalized. I then use these directions to get a view matrix like so:
glm::lookAt(position, position + forward, up);
It all works great - EXCEPT: when I'm flying around a scene, the right and up vectors are totally erroneous. The forward direction is always exactly as it should be. When I'm at 0, 0, -1 looking at 0,0,0, all the directions are correct. But when I look in different directions (except for the inverse of looking at the origin from 0,0,-1), right and up are inconsistent. They aren't world vectors, nor are they local vectors. They seem to be almost random. Where am I going wrong? How can I get consistent local up and local right vectors from the 4x4 transform matrix?
matrix is a local-to-world transformation, whereas a view matrix is world-to-local, which means that in this case matrix is the inverse of the view matrix. The code you currently have only works for a view matrix. Simply exchange the rows and columns:
right = glm::vec3(matrix[0][0], matrix[1][0], matrix[2][0]);
up = glm::vec3(matrix[0][1], matrix[1][1], matrix[2][1]);
forward = glm::vec3(matrix[0][2], matrix[1][2], matrix[2][2]);
A possible reason that it works for (0, -1, 0) is because the rotation/scaling part of the view matrix looks like:
1 0 0
0 0 1
0 1 0
The corresponding part of matrix is the inverse of the above, which is of course identical. Try it with another direction.
Related
I am following the tutorials at LearnOpenGL.com and I am confused about the order of Matrices.
The Transformations chapter tells:
Matrix multiplication is not commutative, which means their order is important. When multiplying matrices the right-most matrix is first multiplied with the vector so you should read the multiplications from right to left. It is advised to first do scaling operations, then rotations and lastly translations when combining matrices otherwise they might (negatively) affect each other. For example, if you would first do a translation and then scale, the translation vector would also scale!
So If I am not wrong, the order is Translate * Rotate * Scale * vector_to_transform.
But immediately in the next Chapter, when calculating the LookAt matrix, the multiplication order is flipped. Here is the code snippet from the website:
// Custom implementation of the LookAt function
glm::mat4 calculate_lookAt_matrix(glm::vec3 position, glm::vec3 target, glm::vec3 worldUp)
{
// 1. Position = known
// 2. Calculate cameraDirection
glm::vec3 zaxis = glm::normalize(position - target);
// 3. Get positive right axis vector
glm::vec3 xaxis = glm::normalize(glm::cross(glm::normalize(worldUp), zaxis));
// 4. Calculate camera up vector
glm::vec3 yaxis = glm::cross(zaxis, xaxis);
// Create translation and rotation matrix
// In glm we access elements as mat[col][row] due to column-major layout
glm::mat4 translation = glm::mat4(1.0f); // Identity matrix by default
translation[3][0] = -position.x; // Third column, first row
translation[3][1] = -position.y;
translation[3][2] = -position.z;
glm::mat4 rotation = glm::mat4(1.0f);
rotation[0][0] = xaxis.x; // First column, first row
rotation[1][0] = xaxis.y;
rotation[2][0] = xaxis.z;
rotation[0][1] = yaxis.x; // First column, second row
rotation[1][1] = yaxis.y;
rotation[2][1] = yaxis.z;
rotation[0][2] = zaxis.x; // First column, third row
rotation[1][2] = zaxis.y;
rotation[2][2] = zaxis.z;
// Return lookAt matrix as combination of translation and rotation matrix
return rotation * translation; // Remember to read from right to left (first translation then rotation)
}
At the end of the code snippet, the matrix is calculated as rotation * translation, even though the matrix is going to be multiplied as,
gl_position = projection * lookAt * model * vec4(vertexPosition, 1.0);
as Column-major matrices must be pre-multiplied to the vector.
Please help me understand this.
LearnOpenGL unfortunately doesn't explain where the Camera transform comes from.
You can see the Camera transform as an inverse model transform.
3D math doesn't care if you move the camera towards the Objects or the Objects towards the Camera.
Also if your objects are already scaled to have the proper "World Space" size you don't need the camera to scale them. The scaling for the "intrinsic Camera Parameters" are dealt with in the projection matrix (scale for aspect ration and field of view). Which is done after the Camera transform.
So we move the object points towards the "camera" instead of the camera towards the points. As I said you would not leave the scaling in the Camera Matrix, since you only want to orient the objects in front of the Camera.
Placing the Camera as Model in the world space would be:
M = TR (leave out S for above reasons)
Then you inverse the Camera Transform ->
C = M^-1 | M = TR
= (TR)^-1
= R^-1 * T^-1 | inverse of matrix product -> flip order and invert matrices
Lets assume R(angle) is the Matrix that rotates by angle angle and T(t) is the Matrix that translates by vector t then:
= R(angle)^-1 * T(t)^-1
= R^T(angle) * T(-t)
Which is exactly what you return in your lookAt-method. The basis vectors of R are set up by the column vectors of you new coordinate frame, but then transposed (so you have them as row vectors). That's because the camera frame vectors are orthogonal and unit length, so the resulting matrix is "orthonomal" which means it's inverse is it's transpose. And the inverse Translation Matrix T(t) gets the translation vector T(-t) (inverse of eye-Position of the Camera).
Hope my explanation clearifies more than it confuses :-)
Okay! After reading through the entire chapter again, I missed a crucial detail. The View matrix usually does not scale the objects. It's just a matrix to rotate and translate the model matrix in such a way to simulate an eye into the world. This chapter of LearnOpenGL.com has a block explaining about combining matrices and it shows how to combine a translate and rotate matric and it's how the lookAt function is implemented.
What I have currently is causing my 3D object to become flat. But it is looking at my target.
Vector4 up;
newMatrix.SetIdentity();
up.set_x(0);
up.set_y(1);
up.set_z(0);
Vector4 zaxis = player_->transform().GetTranslation() - spider_->transform().GetTranslation();
zaxis.Normalise();
Vector4 xaxis = CrossProduct(up, zaxis);
xaxis.Normalise();
Vector4 yaxis = CrossProduct(zaxis, xaxis);
newMatrix.set_m(0, 0, xaxis.x()); newMatrix.set_m(0, 1, xaxis.y()); newMatrix.set_m(0, 2, xaxis.z());
newMatrix.set_m(1, 0, yaxis.x()); newMatrix.set_m(1, 1, yaxis.y()); newMatrix.set_m(1, 2, yaxis.z());
newMatrix.set_m(2, 0, zaxis.x()); newMatrix.set_m(2, 1, zaxis.y()); newMatrix.set_m(2, 2, zaxis.z());
Excuse the method for putting values into the matrix, I'm working with what my framework gives me.
Vector4 Game::CrossProduct(Vector4 v1, Vector4 v2)
{
Vector4 crossProduct;
crossProduct.set_x((v1.y() * v2.z()) - (v2.y() * v2.z()));
crossProduct.set_y((v1.z() * v2.x()) - (v1.z() * v2.x()));
crossProduct.set_z((v1.x() * v2.y()) - (v1.x() * v2.y()));
return crossProduct;
}
What am I doing wrong here?
Note that I have added the forth line before with the 1 in the corner before just in case, with no change.
you got problem when (0,1,0) is close to parallel to the direction you want to look at. Than the cross product will fail leading in one or two basis vectors to be zero which could lead to 2D appearance. But that would happen only if your objects are offset only in y axis between each other. To avoid that you can test dot product between the up and view direction and if the abs of result is bigger than 0.7 use (1,0,0) instead (as a up or right or whatever...).
Also as We know nothing about your notations we can not confirm your setting the matrix is correct (it might be or may be not, it could be transposed etc.) for more info take a look at:
Understanding 4x4 homogenous transform matrices
But most likely your matrix holds also the perspective transform and by setting its cells you are overriding it completely leading to no perspective hence 2D output. In such case you should multiply original perspective matrix with your new matrix and use the result.
I am writing software to determine the viewable locations of a camera in 3D. I have currently implement parts to find the minimum and maximum length of view based on the camera and lenses intrinsic characteristics.
I now need to work out that if the camera is placed at X,Y,Z and is pointing in a direction (two angles, one around the horizontal and one around the vertical axis) what the boundaries the camera can see at are (knowing the viewing angle). The output I would like is 4 3D locations, making a rectangle that show the minimum position, top left, top right, bottom left and bottom right. The same is also required for the maximum positions.
Can anyone help with the geometry to find these points?
Some code I have:
QVector3D CameraPerspective::GetUnitVectorOfCameraAngle()
{
QVector3D inital(0, 1, 0);
QMatrix4x4 rotation_matrix;
// rotate around z axis
rotation_matrix.rotate(_angle_around_z, 0, 0, 1);
//rotate around y axis
rotation_matrix.rotate(_angle_around_x, 1, 0, 0);
inital = inital * rotation_matrix;
return inital;
}
Coordinate CameraPerspective::GetFurthestPointInFront()
{
QVector3D camera_angle_vector = GetUnitVectorOfCameraAngle();
camera_angle_vector.normalize();
QVector3D furthest_point_infront = camera_angle_vector * _camera_information._maximum_distance_mm;
return Coordinate(furthest_point_infront + _position_of_this);
}
Thanks
A complete answer with code will be probably way too long for SO, I hope that this will be enough. In the following we work in homogeneous coordinates.
I have currently implement parts to find the minimum and maximum length of view based on the camera and lenses intrinsic characteristics.
That isn't enough to fully define your camera. You also need a field of view angle and the width/height ratio.
With all these information (near plane + far plane + fov + ratio), you can build a 4x4 matrix known as perspective matrix. Google for it or check here for some references. This matrix maps the pyramidal region of the space which your camera "sees" (usually simply called frustrum) to the [-1,1]x[-1,1]x[-1,1] cube. Call it P.
Now you need a 4x4 camera matrix which transform points in world space to points in camera space. Since you know the camera position and the camera orientation this can be constructed easily (there is no room here to full explain how transformation matrices in homogeneous coordinates work, google for it). Call this matrix C.
Now consider the matrix A = P * C.
This matrix transforms points in world coordinates to points in the perspective space. Your camera will "see" those points if they are inside the [-1,1]x[-1,1]x[-1,1] cube. But you can invert this matrix in order to map points inside the cube to points in world space. So in order to obtain the 8 points you need in world space you can simply do:
y = A^(-1) * x
Where x =
[-1,-1,-1, 1] left - bottom - near
[-1,-1, 1, 1] left - bottom - far
etc.
In my opengl app, i want to orientate figures to look at the camera, to make this, i define for all the objects 2 vectors, front and up.
Im using gluLookAt to control the camera, so the vectors newFront and newUp i need are easily known.
The code i use to control the orientation for each figure is :
m4D orientate(v3D newFront, v3D newUp)
{
double angle = angle_between(front, newFront);
v3D cross = normalize(cross_product(front, newFront));
m4D matrix = rotate_from_axis(angle, cross);
up = normalize(up * matrix);
angle = angle_between(up, newUp);
cross = normalize(cross_product(up, newUp));
return(rotate_from_axis(angle, cross) * matrix);
}
This code works well when the matrix stack has only this matrix, but if i push a previous matrix rotation (rotating of course front and up vectors) it fails.
What's my fault?
Why always those complicated "I solve for an inverse rotation and multiply that onto the modelview" billboard/impostor solutions, when there's a much simpler method?
Let M be the modelview matrix from which a billboard matrix is to be determined. The matrix is a 4×4 real valued type. The upper left 3×3 defines rotation and scaling. For a billboard this part is to be identity.
So by replacing the upper left part of the current modelview matrix with identity, and keeping the rest as is i.e.
1 0 0 tx
0 1 0 ty
0 0 1 tz
wx wy wz ww
and using that matrix for further transformations you get exactly the desired effect. If there was a scaling applied, replace the upper left identity with a scaling matrix.
I'm working on a simple OpenGL world- and so far I've got a bunch of cubes randomly placed about and it's pretty fun to go zooming about. However I'm ready to move on. I would like to drop blocks in front of my camera, but I'm having trouble with the 3d angles. I'm used to 2d stuff where to find an end point we simply do something along the lines of:
endy = y + (sin(theta)*power);
endx = x + (cos(theta)*power);
However when I add the third dimension I'm not sure what to do! It seems to me that the power of the second dimensional plane would be determined by the z axis's cos(theta)*power, but I'm not positive. If that is correct, it seems to me I'd do something like this:
endz = z + (sin(xtheta)*power);
power2 = cos(xtheta) * power;
endx = x + (cos(ytheta) * power2);
endy = y + (sin(ytheta) * power2);
(where x theta is the up/down theta and y = left/right theta)
Am I even close to the right track here? How do I find an end point given a current point and an two angles?
Working with euler angles doesn't work so well in 3D environments, there are several issues and corner cases in which they simply don't work. And you actually don't even have to use them.
What you should do, is exploit the fact, that transformation matrixes are nothing else, then coordinate system bases written down in a comprehensible form. So you have your modelview matrix MV. This consists of a model space transformation, followed by a view transformation (column major matrices multiply right to left):
MV = V * M
So what we want to know is, in which way the "camera" lies within the world. That is given to you by the inverse view matrix V^-1. You can of course invert the view matrix using Gauss Jordan method, but most of the time your view matrix will consist of a 3×3 rotation matrix with a translation vector column P added.
R P
0 1
Recall that
(M * N)^-1 = N^-1 * M^-1
and also
(M * N)^T = M^T * N^T
so it seems there is some kind of relationship between transposition and inversion. Not all transposed matrices are their inverse, but there are some, where the transpose of a matrix is its inverse. Namely it are the so called orthonormal matrices. Rotations are orthonormal. So
R^-1 = R^T
neat! This allows us to find the inverse of the view matrix by the following (I suggest you try to proof it as an exersice):
V = / R P \
\ 0 1 /
V^-1 = / R^T -P \
\ 0 1 /
So how does this help us to place a new object in the scene at a distance from the camera? Well, V is the transformation from world space into camera space, so V^-1 transforms from camera to world space. So given a point in camera space you can transform it back to world space. Say you wanted to place something at the center of the view in distance d. In camera space that would be the point (0, 0, -d, 1). Multiply that with V^-1:
V^-1 * (0, 0, -d, 1) = (R^T)_z * d - P
Which is exactly what you want. In your OpenGL program you somewhere have your view matrix V, probably not properly named yet, but anyway it is there. Say you use old OpenGL-1 and GLU's gluLookAt:
void display(void)
{
/* setup viewport, clear, set projection, etc. */
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
gluLookAt(...);
/* the modelview matrix now holds the View transform */
At this point we can extract the modelview matrix
GLfloat view[16];
glGetFloatv(GL_MODELVIEW_MATRIX, view);
Now view is in column major order. If we were to use it directly we could directly address the columns. But remember that transpose is inverse of a rotation, so we actually want the 3rd row vector. So let's assume you keep view around, so that in your event handler (outside display) you can do the following:
GLfloat z_row[3];
z_row[0] = view[2];
z_row[1] = view[6];
z_row[2] = view[10];
And we want the position
GLfloat * const p_column = &view[12];
Now we can calculate the new objects position at distance d:
GLfloat new_object_pos[3] = {
z_row[0]*d - p_column[0],
z_row[1]*d - p_column[1],
z_row[2]*d - p_column[2],
};
There you are. As you can see, nowhere you had to work with angles or trigonometry, it's just straight linear algebra.
Well I was close, after some testing, I found the correct formula for my implementation, it looks like this:
endy = cam.get_pos().y - (sin(toRad(180-cam.get_rot().x))*power1);
power2 = cos(toRad(180-cam.get_rot().x))*power1;
endx = cam.get_pos().x - (sin(toRad(180-cam.get_rot().y))*power2);
endz = cam.get_pos().z - (cos(toRad(180-cam.get_rot().y))*power2);
This takes my camera's position and rotational angles and get's the corresponding points. Works like a charm =]