I am trying to use quaternion rotation using the quaternion from JOML: https://github.com/JOML-CI/JOML/blob/main/src/org/joml/Quaternionf.java.
I can get objects to rotate but they get stuck and are unable to complete a full rotation. The object is being updated every frame.
Edit:
Removed all euler related code and am simply trying to get the object to rotate on a single axis based on a certain angle.
Edit 2:
Am trying to use the conjugate and multiplying the quaternions together like I have seen in some videos. I'm not quite there though as the model spins itself off the screen for some reason.
Edit 3:
Normalizing the quaternion fixed the disappearing behaviour. The issue seems to be that there's no simple way to rotate a certain amount without either having a timer to lerp over which will not work in my case as I am trying to rotate an object an arbitrary amount with no set beginning and end.
Rotation function
public void rotate(float angle, float x, float y, float z) {
Quaternionf rot = new Quaternionf();
rot.rotateAxis((float) Math.toRadians(angle), x, y, z);
Quaternionf conjugate = rot.conjugate();
rotation = rot.mul(rotation).mul(conjugate);
}
Calling the rotation function
entity.transform.rotate( 1,0,1, 0);
It does not matter whether you transform your Euler angles into a matrix or a quaternion or whatever representation: As long as you use Euler angles to represent an orientation, Gimbal Lock is unavoidable.
So, in order to avoid Gimbal Lock from happening, you must discard using Euler Angles and stay with one representation for an orientation (like a 3x3 matrix or a quaternion) and apply delta changes to them, instead of trying to represent a full orientation as three Euler angles. So, whenever you - let's say - rotate the object a few degrees around a certain axis, you apply that delta change or orientation to the matrix/quaternion.
I believe I have figured it out.
You need to create a quaternion and rotate it to your delta values, do not manipulate quaternion values directly (e.g. use rotationX function instead).
To add quaternions together you multiply them.
Finally you need to use the equation:
delta quaternion * quaternion to rotate * inverse of delta quaternion
Code:
public void rotate(float x, float y, float z) {
//Use modulus to fix values to below 360 then convert values to radians
float newX = (float) Math.toRadians(x % 360);
float newY = (float) Math.toRadians(y % 360);
float newZ = (float) Math.toRadians(z % 360);
//Create a quaternion with the delta rotation values
Quaternionf rotationDelta = new Quaternionf();
rotationDelta.rotationXYZ(newX, newY, newZ);
//Calculate the inverse of the delta quaternion
Quaternionf conjugate = rotationDelta.conjugate();
//Multiply this transform by the rotation delta quaternion and its inverse
rotation.mul(rotationDelta).mul(conjugate);
}
Essentially, what is happening is there is some strange warping of the 3D cube being rendered by my raytracer, which continues to worsen as the camera moves up, even if the cube is in the same location on the screen.
The code is at http://pastebin.com/HucgjRtx
Here is a picture of the output:
http://postimg.org/image/5rnfrlkej/
EDIT: Problem resolved as being that I was just calculating the angles for vectors wrong. The best method I have found is creating a vector based on your FOV (Z) current pixel X, and current pixel Y, then normalizing that vector.
It looks like you're calculating rays to cast based on Euler angles instead of the usual projection.
Typically a "3D" camera is modeled such that the camera is at a point with rays projecting through a grid spaced some distance from it... which is, incidentally, exactly like looking at a monitor placed some distance from your face and projecting a ray through each pixel of the monitor.
The calculations are conceptually simple in fixed cases.. e.g.
double pixelSpacing = 0.005;
double screenDistance = 0.7;
for (int yIndex= -100; yIndex<= 100; yIndex++)
for (int xIndex= -100; xIndex<= 100; xIndex++) {
Vector3 ray = new Vector3(
xIndex * pixelSpacing,
yIndex * pixelSpacing,
screenDistance
);
ray = vec.normalize();
// And 'ray' is now a vector with our ray direction
}
You can use one of the usual techniques (e.g. 4x4 matrix multiplication) if you want to rotate this field of view.
The question change a bit, I figured out how to rotate around a single axis
I want to rotate a box around the Y axis using an angle.
The box has a size, and a Vector3f to signal the rotation.
To rotate the box correctly what I do is rotate the origin position then rotate the origin position plus the size, and use those two references to render the box.
However this rotation does not work correctly and causes rendering artifacts.
This is my code to rotate the positions:
Matrix4f matrix = new Matrix4f();
// Rotate the origin position
Vector3f pos = new Vector3f(new Vector3f(blockX, blockY, blockZ));
matrix.m03 = pos.x;
matrix.m13 = pos.y;
matrix.m23 = pos.z;
Vector3f rot = new Vector3f(new Vector3f(0, 1f, 0f));
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
Vector3f locationMin = new Vector3f(matrix.m03, matrix.m13, matrix.m23);
// Rotate the position with the size
// Top left back is the position of the block
Vector3f sizeRot = new Vector3f(new Vector3f(blockX + size, blockY + size, blockZ + size));
matrix = new Matrix4f();
matrix.m03 = sizeRot.x;
matrix.m13 = sizeRot.y;
matrix.m23 = sizeRot.z;
rot = new Vector3f(new Vector3f(0, 1f, 0f));
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
Vector3f locationMax = new Vector3f(matrix.m03, matrix.m13, matrix.m23);
// Then here I use the locationMax and the locationMin to render the cube
What could be wrong with this code? Is the logic I am using to rotate the box correct? as in rotate the origin position then rotate the origin position plus the size..
EDIT: I released that rotating after translating is stupid so instead I just rotated the locationMax which is not translated (it is only the size) then I translated and I still get the same result (Graphical Artifacts).
New Code:
float rx = blockX, ry = blockY, rz = blockZ;
Matrix4f matrix = new Matrix4f();
Vector3f rot = new Vector3f(0, 1f, 0f);
matrix = new Matrix4f();
matrix.m03 = size;
matrix.m13 = size;
matrix.m23 = size;
Matrix4f.rotate((float) Math.toRadians(45f), rot, matrix, matrix);
matrix.translate(new Vector3f(rx, ry, rz), matrix);
float mx = matrix.m03;
float my = matrix.m13;
float mz = matrix.m23;
// Here is use rx, ry, rz and mx, my, mz to render the box
============ * I figured it out (See below)* =============
EDIT:
This is what I ended up doing:
// Origin point
Vector4f a = new Vector4f(blockX, blockY, blockZ, 1);
// Rotate a matrix 45 degrees
Matrix4f mat = new Matrix4f();
mat.rotate((float) Math.toRandians(45f), new Vector3f(
0, 1f, 0), mat);
/* Transform the matrix to each point */
Vector4f c = new Vector4f(size.x, 0, size.z, 1);
Matrix4f.transform(mat, c, c);
Vector4f.add(c, a, c);
Vector4f b = new Vector4f(size.x, 0, 0, 1);
Matrix4f.transform(mat, b, b);
Vector4f.add(b, a, b);
Vector4f d = new Vector4f(0, 0, size.z, 1);
Matrix4f.transform(mat, d, d);
Vector4f.add(d, a, d);
// Here is use a, b, c, and d to render the box.
The problem with this is that I want to rotate around all axises and not only around the Y axis. This makes the code very long and unreadable and There are a lot of bugs when I try to rotate around all axises.
Update Question:
How do I take the above code and make it so I can rotate around all 3 axises. I want to do this so I can have a billboard that will always face the camera.
This is how I calculate the angle between the camera and the object:
Vector3f angle = new Vector3f();
// Calculate the distance between camera and object
Vector3f.sub(game.getCamera().getLocation(),
new Vector3f(blockX, blockY, blockZ), angle);
// Calculate the angle around the Y axis.
float vectorAngle = (float) ((float) Math.atan2(angle.z, angle.x) * -1 + (Math.PI / 2.0f));
Billboards are a very common application of computer graphics (as I'm sure you've noticed, since you're asking the question!)
Ultimately I think you are over complicating the problem, based on:
as in rotate the origin position then rotate the origin position plus the size..
For computer graphics, the most common transformations are Scaling, Translating, and Rotating, and you do these in an order to achieve a desired effect (traditionally you scale, then rotate about the origin, then translate the vertex's position).
Additionally, you will have three main matrices to render a model in 3d: World Matrix, View Matrix, and Projection Matrix. I believe you are having misunderstandings of transforming from Model Space to World Space.
Graphics TRS and Matrix info. If you are having conceptual problems, or this answer is insufficient, I highly recommend looking at this link. I have yet to find a better resource explaining the fundamentals of computer graphics.
So right at the moment, you have your three angles (in degrees, in a Vector3) corresponding to the angle difference in the X,Y, and Z coordinate spaces from your billboard and your camera. With this information, we generate the View matrix by first gathering all of our matrix transformations in one place.
I'm going to assume that you already have your Translation and Scaling matrices, and that they both work. This means that we only need to generate our Rotation matrix, and then transform that matrix with the scaling matrix, and then transforming that matrix by our translation matrix.
X Rotation Matrix
Y Rotation Matrix
Z Rotation Matrix
(Images taken from CodingLabs link above)
So you will generate these three matrices, using the X,Y, and Z angles you calculated earlier, and then transform them to consolidate them into a single matrix, transform that matrix by the scaling matrix, and then transform that matrix by the translation matrix. Now you have your awesome matrix that, when you multiply a a vertex by it, will transform that vertex into the desired size, rotation, and position.
So you transform every single vertex point by this generated matrix.
And then after that, you should be done! Using these techniques will hopefully simplify your code greatly, and set you on the right path :)
So now how about some code?
//I do not guarantee that this code compiles! I did not write it in an IDE nor did I compile it
float angleToRotX = 180f;
float angleToRotY = 90f;
float angleToRotZ = 0f;
// example vertex
Vector4f vertex = new Vector4f(0, 1, 0, 1);
// Rotate vertex's X coordinates by the desired degrees
Matrix4f rotationXMatrix = new Matrix4f();
rotationXMatrix.rotX(angleToRotX);
Matrix4f rotationYMatrix = new Matrix4f();
rotationYMatrix.rotY(angleToRotY);
Matrix4f rotationZMatrix = new Matrix4f();
rotationZMatrix.rotZ(angleToRotZ);
//now let's translate it by 1.5, 1, 1.5 in the X,Y,Z directions
Matrix4f translationMatrix = new Matrix4f();
translationMatrix.setTranslate(new Vector3f(1.5, 1, 1.5));
/*
Now we have our three rotational matrices. So we multiply them (transform them) to get a single matrix to transform all of the points in this model to the desired world coordinates
*/
Matrix4f rotationMatrix = new Matrix4f();
rotationMatrix.mul(rotationXMatrix);
rotationMatrix.mul(rotationYMatrix);
rotationMatrix.mul(rotationZMatrix);
Matrix4f worldMatrix = translationMatrix;
worldMatrix.mul(rotationMatrix);
//now worldMatrix, when applied to a vertex, will rotate it by X,Y,Z degrees about the origin of it's model space, and then translate it by the amount given in translationMatrix
worldMatrix.transform(vertex);
//now vertex should be (1.5, 0, 1.5, 1) with (x,y,z,1)
Now this code could really be simplified, and it is excessively verbose. Try it out! I don't have java downloaded on my machine, but I grabbed the methods from the java documentation Here
Here is an image of what is happening (again, taking from coding labs):
(Advanced Info: Quaternions. These are really cool way of orienting a model in 3d space, however I don't quite understand them to the degree I need to in order to explain it to someone else, and I also believe that your problem is more fundamental)
You could generate the matrix without much hassle. The OpenGL matrix looks like the following:
|lx,ux,vx,px| - lx,ly,lz = the left vector
|ly,uy,vy,py| - ux,uy,uz = the up vector
|lz,uz,vz,pz| - vx,vy,vz = the view vector
|0 ,0 ,0 ,1 | - px,py,pz = the translation
All you need to do, is set px,py,pz to the position of your box in the world,
your view vector to the normalized(camera position - box position), your up comes straight from your camera, and the left is calculated via normalized cross product. It's also good practice to reconstruct the up vector, after left one is derived (by another cross product). That's all there's to it.
My solution aims to save you some time coding, rather than explain everything in detail. Hope that is useful to someone.
In my 3d application I wish to have an object (a tree, for example), and my camera to look at this object. Then, I want the camera to rotate about the object, in a circle, while looking at the tree the whole time. Imagine walking around a tree, while constantly changing your angle so that you are still looking at it. I know this requires both rotation of my camera, and translation of my camera, but the math is far beyond the level I have been taught in schooling thusfar. Can anyone point me in the right direction?
Here is one way with very simple math. First, you need a constant for the distance the camera is from the center of the tree (the radius of the circle path it travels on). Also, you need some variable to track it's angle around the circle.
static final float CAM_PATH_RADIUS = 5f;
static final float CAM_HEIGHT = 2f;
float camPathAngle = 0;
Now you can change the camPathAngle to anything you want from 0 to 360 degrees. 0 degrees corresponds with the location on the circle that is in the same direction as the world's X-axis from the tree's center.
On each frame, after you've update camPathAngle, you can do this to update the camera position.
void updateTreeCamera(){
Vector3 camPosition = camera.getPosition();
camPosition.set(CAM_PATH_RADIUS, CAM_HEIGHT, 0); //Move camera to default location on circle centered at origin
camPosition.rotate(Vector3.Y, camPathAngle); //Rotate the position to the angle you want. Rotating this vector about the Y axis is like walking along the circle in a counter-clockwise direction.
camPosition.add(treeCenterPosition); //translate the circle from origin to tree center
camera.up.set(Vector3.Y); //Make sure camera is still upright, in case a previous calculation caused it to roll or pitch
camera.lookAt(treeCenterPosition);
camera.update(); //Register the changes to the camera position and direction
}
I did it like that for the sake of commenting it. It's actually shorter than the above if you chain commands:
void updateTreeCamera(){
camera.getPosition().set(CAM_PATH_RADIUS, CAM_HEIGHT, 0)
.rotate(Vector3.Y, camPathAngle).add(treeCenterPosition);
camera.up.set(Vector3.Y);
camera.lookAt(treeCenterPosition);
camera.update();
}
I need to rotate a triangle so that it lies on a plane given by a normal n and a constant d.
I have the normal n1 of the plane that the two triangles lie in. Now i need to rotate the right red triangle so that it results in the orange one.
The points of the triangles and the normals are stored as 3-dimensional vectors.
Until now, I did the following:
Get the normalized rotation quaternion (rotQuat) between n1 and n2.
Multiply every point of the triangle by the quaternion. Therefore I convert the point to a quaternion(point.x, point.y, point.z, 0) and do the multiplcation as follows: resultQuat = rotQuat * point * conjugate(rotQuat). I then apply x, y and z of the result to the point.
This is how i get the rotation between two vectors:
public static Quaternion getRotationBetweenTwoVector3f(Vector3f vec1, Vector3f vec2)
{
Vector3f cross = Vector3f.cross(vec1, vec2);
float w = (float) (java.lang.Math.sqrt(java.lang.Math.pow(vec1.getLength(), 2) * java.lang.Math.pow(vec2.getLength(), 2)) + Vector3f.dot(vec1, vec2));
Quaternion returnQuat = new Quaternion(cross.x, cross.y, cross.z, w);
returnQuat.normalize();
return returnQuat;
}
The problem is that the triangle has the correct orientation after the rotation, but the triangle also moves it's position. I need a rotation that rotates the triangle so that it's still connected to the two points of the left red triangle (like the orange one).
How is this possible?
Your problem is that rotation matrix/quaternions rotate points around an axis that passes through the origin. To rotate around different point than the origin, you need to translate the triangle points to the origin (just Substract the rotation point value from the triangle points), then multiply by the quaternion and then translate back.
So the algorithm becomes:
translatedPoints[i] = triPoints[i] - rotationPoint;
translatedPoints rotate using quaternion
translate translatedPoints back by adding the rotation point value.