Hi I'm trying to rotate the openGL camera using my phone's gyroscope. I am able to get the three angles (X,Y,Z), but I can't find a way to convert them to the direction vectors required by gluLookAt.
The camera is placed at (0,0,0) and should make the same rotations as the phone itself.
gluLookAt needs three vector parameters to calculate correct view matrix. The eye position (in your case (0,0,0)), the center position (where the camera is looking at) and the up vector (where is 'up' direction for the camera).
After choosing initial camera orientation, for many applications the center would be (0,0,1) and up direction (0,1,0). you need to rotate these vectors by desired angle. To do so, you can multiply them with a rotation matrix R = Rx * Rz * Ry where Rx,z,y are rotation matrices along given axes by given angle. If you are using any math package like glm, eigen or similar you can construct these easily. You can even implement them yourself (http://en.wikipedia.org/wiki/Rotation_matrix).
Note that in case eye position is not (0,0,0), you have to rotate the direction vector, not the center vector. Direction vector = center - eye.
So, a little pseudocode:
vec3 eye = .....
vec3 initCenter = .....
vec3 initDirection = initCenter - eye;
vec3 initUp = .....
mat3 rotation = rotate((1,0,0),xAngle) * rotate((0,1,0),yAngle) * rotate((0,0,1),zAngle);
gluLookAt(eye,eye + rotation*initDirection,rotation*initUp)
Implementation might look a little different, you might have to use vec4/mat4 as for example glm generates rotation matrices of dimension 4x4.
Related
I want to rotate the whole value of a 3d vector into one axis using quaternion rotations.
The reason behind is that I want to align the X and Y Axis of my smartphone with the X and Y Axis of my vehicle in order to detect lateral and longitudinal acceleration separated on these two axis. Therefore I want to detect the first straight acceleration of the car and rotate the whole acceleration value into the heading axis (X-Axis) of the phone assuming a straight forward motion.
How do I achieve this?
I got it myself while this Link helped me a lot.
In the end I just needed to define the destination axis as a Vector V(Magnitude_Source, 0, 0), calculate the angle between V and the source vector and rotate the source vector.
I have a square that rotates to a random angle and then travels in a straight line in the direction it is pointing. It does this by using a variable as its x axis and then calling
Variable++
Each frame.
unfortunatley i cannot work out how to return the exact position of the square because the square can be travelling at any angle and therefore doesn't rigidly follow the world coordinte grid. This means that the x variable is not the shapes x coordinate.
How do i return the shapes exact coordinates and how do i do it in such a way that i can have two squares drawn from the same class behaving differently.
So you've got a measure of distance from where the object started along its internal sideways axis and a measure of the angle between that axis and the horizontal?
If so then the formula you want is simple trigonometry. Assuming the object started at (x, y) and has travelled 'distance' units along an axis at an angle of 'angle' with the horizontal then the current position (x', y') is:
x' = x + distance * cos(angle)
y' = y + distance * sin(angle)
If you have the origin in the lower left of the screen and axes arranged graph paper style with x increasing to the right and y increasing as you go upward, that assumes that the angle is measured anticlockwise and that the object is heading along positive x when angle is zero.
If you'll permit a hand waving explanation, the formula works because one definition of sine and cosine is that they're the (x, y) coordinates of the point on the outside of a unit circle at the angle specified. It also matches with the very first thing most people learn about trigonometry, that sine is 'opposite over hypotenuse', and cosine is 'adjacent over hypotenuse'. In this case your hypotenuse has length 'distance' and and you want to get the 'opposite' and 'adjacent' lengths of a right angled triangle that coincides with the axes.
Assuming Android follows J2SE in this area, the one thing to watch out for is that Math.sin and Math.cos take an angle in radians, whereas OpenGL's rotatef takes an argument in degrees. Math.toDegrees and Math.toRadians can do the conversion for you.
When you made the shape you should have already specified its X & Y coordinates. Im not too sure what you mean when you say you cant find the coordinates?
Also make sure you do fame independent movement; currently you are adding one to your variable on every loop of your program. This means if it runs a 60 Frame Per Second(FPS) it will move 60 units, but if it runs at 30FPS it will move at half the speed
I want to rotate the cube when the user is touching it. To rotate it,is quite simple but actual problem is if the user is tilting it then always it should stop in particuler face(cube has 6 faces and i mean that only one face should be visible at one time). Please give your suggetion if anyone worked on that.
In the case of a cube this is simple: Face normals are the cartesian axes. So one looks straigt on a face if you constrain the rotations around the cartesian axes (X, Y, Z) and the rotation angles are multiple of pi/2 = 90°.
So in your code when the user stops interacting, set the rotation angles to the next multiple of 90°
fmod(round(angle/90) * 90, 360); // degrees
fmod(round(angle/(pi/2)) * pi/2, 2*pi); // radians
Either do it hard, or animate it.
If your object is not a cube, but arbitrary, you need to find the additional rotation for the face to get perpendicular to the view axis. This angle is determined by acos( scalar_product(normalize(face_normal), normalize(view_axis)) ) the axis of rotation is given by cross_product(normalize(face_normal), normalize(view_axis))
I'm trying to get billboarding to work, but having trouble with the last step.
After following these directions from NeHe's tutorials (http://nehe.gamedev.net/data/articles/article.asp?article=19) I have my look, right, and up vectors, and I have already translated the modelview matrix to the centerpoint of the billboard by using glTranslatef().
float[] m = {right.x,right.y,right.z,0f,
up.x,up.y,up.z,0f,
look.x,look.y,look.z,0f,
pos.x,pos.y,pos.z,1f}; //pos is the centerpoint position
gl.glMultMatrixf(m, 0);
When I try to create a multiplication matrix out of these like so, the billboards are displayed all over the place in the wrong positions and orientations.
I guess my problem is that I don't know how to correctly create and multiply the matrix. I tried doing this instead, but then half the lines (the ones that need to rotate counter-clockwise) are rotated in the wrong direction:
//normal is the vector that the billboard faces before any manipulations.
float angle = look.getAngleDeg(normal); //returns angle between 0 and 180.
gl.glRotatef(angle, up.x,up.y,up.z);
Got it, using my second method. Calculating the angle between vectors (arccos of the dot product) can only give an angle between 0 and 180, so half the time you want to negate the angle so the rotation is in the opposite direction.
This is easy to check...since I already have the right vector, I can just check if the angle between the right vector and the normal is acute. If it's acute, then you want to negate the original angle.
I'm am working on a basic augmented reality application on Android. What I did so far is detect a square with opencv and then using cvFindExtrinsicCameraParams2() I calculated a rotation and translation vector. For this I used 4 object points, which are just the corners of a square around (0,0,0) and the 4 corners of the square in the image.
This yields me a pretty good rotation and translation matrix. I also calculated the rotation matrix with cvRodrigues2() since using this is easier than the rotation vector. As long as I use these to draw some points in the image everything works fine. My next step is however to pass these vectors and the matrix back to java and then use them with OpenGL to draw a square in an OpenGLView. The square should be exactly around the square in the image which is displayed behind the OpenGLView.
My problem is that I cannot find the correct way of using the rotation matrix and translation vector in OpenGL. I started of with exactly the same object points as used for the openCV functions. Then I applied the rotation matrix and translation vector in pretty much any possible way I could think of. Sadly none of these approaches produce a result which is anyway near what I hoped for. Can anyone tell me how to use them correctly?
So far the "closest" results I have gotten, was when randomly multiplying the whole matrix with -1. But most of the time the squares still look mirror inverted or rotated for 180 degrees. So I guess it was just a lucky hit, but not the right approach.
Okay after some more testing I finally managed to get it to work. While I don't understand it... it does 'work'. For anyone who will need to do this in the future here is my solution.
float rv[3]; // the rotation vector
float rotMat[9]; // rotation matrix
float tv[3]; // translation vector.
rv[1]=-1.0f * rv[1]; rv[2]=-1.0f * rv[2];
//Convert the rotation vector into a matrix here.
//Complete matrix ready to use for OpenGL
float RTMat[] = {rotMat[0], rotMat[3], rotMat[6], 0.0f,
rotMat[1], rotMat[4], rotMat[7], 0.0f,
rotMat[2], rotMat[5], rotMat[8], 0.0f,
tv[0], -tv[1], -tv[2], 1.0f};
As genpfault said in his comment everything needs to be transposed since OpenGL since OpenGL needs a column-major order. (Thanks for the comment, I saw that page earlier already.) Furthermore the y and z rotation angle as well as the y and z translation need to be multiplied by -1. This is what I find a bit weird. Why only those and not the x values too?
This works as it should I guess. But corners the don't match exactly. I guess this is caused by some wrong openGLView configurations. So even though I am still not a 100% happy with my solution I guess it is the answer to my question.
Pandoro's method really works! In case someone wondering "how to convert the rotation vector into a rotation matrix" here's how I did it. By the way, I've used these in OpenGL 2, not ES.
// use the rotation vector generated from OpenCV's cvFindExtrinsicCameraParams2()
float rv[] = {rotation->data.fl[0], rotation->data.fl[1], rotation->data.fl[2] };
// use the translation vector generated from OpenCV's cvFindExtrinsicCameraParams2()
float tv[] = {translation->data.fl[0], translation->data.fl[1], translation->data.fl[2]} ;
float rm[9];
// rotation matrix
CvMat* rotMat = cvCreateMat (3, 3, CV_32FC1);
// rotation vectors can be converted to a 3-by-3 rotation matrix
// by calling cvRodrigues2() - Source: O'Reilly Learning OpenCV
cvRodrigues2(rotation, rotMat, NULL);
for(int i=0; i<9; i++){
rm[i] = rotMat->data.fl[i];
}
rv[1]=-1.0f * rv[1]; rv[2]=-1.0f * rv[2];
//Convert the rotation vector into a matrix here.
//Complete matrix ready to use for OpenGL
float RTMat[] = {rm[0], rm[3], rm[6], 0.0f,
rm[1], rm[4], rm[7], 0.0f,
rm[2], rm[5], rm[8], 0.0f,
tv[0], -tv[1], -tv[2], 1.0f};
Good luck!