I'm currently working on a game which renders a textured sphere (representing Earth) and cubes representing player models (which will be implemented later).
When a user clicks a point on the sphere, the cube is translated from the origin (0,0,0) (which is also the center of the sphere) to the point on the surface of the sphere.
The problem is that I want the cube to rotate so as to sit with it's base flat on the sphere's surface (as opposed to just translating the cube).
What the best way is to calculate the rotation matrices about each axis in order to achieve this effect?
This is the same calculation as you'd perform to make a "lookat" matrix.
In this form, you would use the normalised point on the sphere as one axis (often used as the 'Z' axis), and then make the other two as perpendicular vectors to that. Typically to do that you choose some arbitrary 'up' axis, which needs to not be parallel to your first axis, and then use two cross-products. First you cross 'Z' and 'Up' to make an 'X' axis, and then you cross the 'X' and 'Z' axes to make a 'Y' axis.
The X, Y, and Z axes (normalised) form a rotation matrix which will orient the cube to the surface normal of the sphere. Then just translate it to the surface point.
The basic idea in GL is this:
float x_axis[3];
float y_axis[3];
float z_axis[3]; // This is the point on sphere, normalised
x_axis = cross(z_axis, up);
normalise(x_axis);
y_axis = cross(z_axis, x_axis);
DrawSphere();
float mat[16] = {
x_axis[0],x_axis[1],x_axis[2],0,
y_axis[0],y_axis[1],y_axis[2],0,
z_axis[0],z_axis[1],z_axis[2],0,
(sphereRad + cubeSize) * z_axis[0], (sphereRad + cubeSize) * z_axis[1], (sphereRad + cubeSize) * z_axis[2], 1 };
glMultMatrixf(mat);
DrawCube();
Where z_axis[] is the normalised point on the sphere, x_axis[] is the normalised cross-product of that vector with the arbitrary 'up' vector, and y_axis[] is the normalised cross-product of the other two axes. sphereRad and cubeSize are the sizes of the sphere and cube - I'm assuming both shapes are centred on their local coordinate origin.
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Can someone tell me how to make triangle vertices collide with edges of the screen?
For math library I am using GLM and for window creation and keyboard/mouse input I am using GLFW.
I created perspective matrix and simple array of triangle vertices.
Then I multiplied all this in vertex shader like:
gl_Position = projection * view * model * vec4(pos, 1.0);
Projection matrix is defined as:
glm::mat4 projection = glm::perspective(
45.0f, (GLfloat)screenWidth / (GLfloat)screenHeight, 0.1f, 100.0f);
I have fully working camera and projection. I can move around my "world" and see triangle standing there. The problem I have is I want to make sure that triangle collide with edges of the screen.
What I did was disable camera and only enable keyboard movement. Then I initialized translation matrix as glm::translate(model, glm::vec3(xMove, yMove, -2.5f)); and scale matrix to scale by 0.4.
Now all of that is working fine. When I press RIGHT triangle moves to the right when I press UP triangle moves up etc... The problem is I have no idea how to make it stop moving then it hits edges.
This is what I have tried:
triangleRightVertex.x is glm::vec3 object.
0.4 is scaling value that I used in scaling matrix.
if(((xMove + triangleRightVertex.x) * 0.4f) >= 1.0f)
{
cout << "Right side collision detected!" << endl;
}
When I move triangle to the right it does detect collision when x of the third vertex(bottom right corner of triangle) collides with right side but it goes little bit beyond before it detects. But when I tried moving up it detected collision when half of the triangle was up.
I have no idea what to do here can someone explain me this please?
Each of the vertex coordinates of the triangle is transformed by the model matrix form model space to world space, by the view matrix from world space to view space and by the projection matrix from view space to clip space. gl_Position is the Homogeneous coordinate in clip space and further transformed by a Perspective divide from clip space to normalized device space. The normalized device space is a cube, with right, bottom, front of (-1, -1, -1) and a left, top, back of (1, 1, 1).
All the geometry which is in this (volume) cube is "visible" on the viewport.
In clip space the clipping of the scene is performed.
A point is in clip space if the x, y and z components are in the range defined by the inverted w component and the w component of the homogeneous coordinates of the point:
-w <= x, y, z <= w
What you want to do is to check if a vertex x coordinate of the triangle is clipped. SO you have to check if the x component of the clip space coordinate is in the view volume.
Calculate the clip space position of the vertices on the CPU, as it does the vertex shader.
The glm library is very suitable for things like that:
glm::vec3 triangleVertex = ... ; // new model coordinate of the triangle
glm::vec4 h_pos = projection * view * model * vec4(triangleVertex, 1.0);
bool x_is_clipped = h_pos.x < -h_pos.w || h_pos.x > h_pos.w;
If you don't know how the orientation of the triangle is transformed by the model matrix and view matrix, then you have to do this for all the 3 vertex coordinates of the triangle-
I need a method to find a set of homogenous transformation matrices that describes the position and orientation in a sphere.
The idea is that I have an object in the center of this sphere which has a radius of dz. Since I know the 3d coordinate of the object I know all the 3d coordinates of the sphere. Is it possible to determine the RPY of any point on the sphere such that the point always points toward the object in the center?
illustration:
At the origo of this sphere we have an object. The radius of the sphere is dz.
The red dot is a point on the sphere, and the vector from this point toward the object/origo.
The position should be relatively easy to extract, as a sphere can be described by a function, but how do I determine the vector, or rotation matrix that points such that it points toward origo.
You could, using the center of the sphere as the origin, compute the unit vector of the line formed by the origin to the point on the edge of the sphere, and then multiply that unit vector by -1 to obtain the vector pointing toward the center of the sphere from the point on the edge of the sphere.
Example:
vec pointToCenter(Point edge, Point origin) {
vec norm = edge - origin;
vec unitVec = norm / vecLength(norm);
return unitVec * -1;
}
Once you have the vector you can convert it to euler angles for the RPY, an example is here
Of the top of my head I would suggest using quaterneons to define the rotation of any point at the origin, relative to the point you want on the surface of the sphere:
Pick the desired point on the sphere's surface, say the north pole for example
Translate that point to the origin (assuming the radius of the sphere is known), using 3D Pythagorus: x_comp^2 + y_comp^2 + z_comp^2 = hypotenuse^2
Create a rotation that points an axis at the original surface point. This will just be a scaled multiple of the x, y and z components making up the hypotenuse. I would just make it into unit components. Capture the resulting axis and rotation in a quaterneon (q, x, y, z), where x, y, z are the components of your axis and q is the rotation about that axis. Hard code q to one. You want to use quaterneons because it will make your resulting rotation matricies easier to work with
Translate the point back to the sphere's surface and negate the values of the components of your axis, to get (q, -x, -y, -z).
This will give you your point on the surface of the sphere, with an axis pointing back to the origin. With the north pole as an example, you would have a quaternion of (1, 0, -1, 0) at point (0, radius_length, 0) on the sphere's surface. See quatrotation.c in my below github repository for the resulting rotation matrix.
I don't have time to write code for this but I wrote a little tutorial with compilable code examples in a github repository a while back, which should get you started:
https://github.com/brownwa/opengl
Do the mat_rotation tutorial first, then do the quatereons one. It's doable in a weekend, a day if you're focused.
I want to convert 2D screen coordinates to 3D world coordinates. I have searched a lot but I did not get any satisfying result.
Note: I am not using OpenGL nor any other graphics library.
Data which I have:
Screen X
Screen Y
Screen Height
Screen Width
Aspect Ratio
If you have the Camera world Matrix and Projection Matrix this is pretty simple.
If you don't have the world Matrix you can compute it from it's position and rotation.
worldMatrix = Translate(x, y, z) * RotateZ(z_angle) * RotateY(y_angle) * RotateX(x_angle);
Where translate returns the the 4x4 translation matrices and Rotate returns the 4x4 rotation matrices around the given axis.
The projection matrix can be calculated from the aspect ratio, field of view angle, and near and far planes.
This blog has a good explanation of how to calculate the projection matrix.
You can unproject the screen coordinates by doing:
mat = worldMatrix * inverse(ProjectionMatrix)
dir = transpose(mat) * <x_screen, y_screen, 0.5, 1>
dir /= mat[3] + mat[7] + mat[11] + mat[15]
dir -= camera.position
Your ray will point from the camera in the direction dir.
This should work, but it's not a super concreate example on how to do this.
Basically you just need to do the following steps:
calculate camera's worldMatrix
calculate camera's projection matrix
multiply worldMatrix with inverse projection matrix.
create a point <Screen_X_Value, Screen_Y_Value, SOME_POSITIVE_Z_VALUE, 1>
apply this "inverse" projection to your point.
then subtract the cameras position form this point.
The resulting vector is the direction from the camera. Any point along that ray are the 3D coordinates corresponding to your 2D screen coordinate.
I want to code a first person camera with its rotation stored in a quaternion. Unfortunately there is something wrong with the rotation.
The following function is responsible to rotate the camera. The parameters Mouse and Speed pass the mouse movement and rotation speed. Then the function fetches the rotation quaternion, rotates it and stores the result. By the way, I'm using Bullet Physics that is where the types and functions come from.
void Rotate(vec2 Mouse, float Speed)
{
btTransform transform = camera->getWorldTransform();
btQuaternion rotation = transform.getRotation();
Mouse = Mouse * Speed; // apply mouse sensitivity
btQuaternion change(Mouse.y, Mouse.x, 0); // create quaternion from angles
rotation = change * rotation; // rotate camera by that
transform.setRotation(rotation);
camera->setWorldTransform(transform);
}
To illustrate the resulting camera rotation when the mouse moves, I show you a hand drawing. On the left side the wrong rotation the camera actually performs is shown. On the right side the desired correct case is shown. The arrows indicate how the camera is rotate when moving the mouse up (in orange) and down (in blue).
As you can see, as long as the yaw is zero, the rotation is correct. But the more yaw it has, the smaller the circles in which the camera rotates become. In contrast, the circles should always run along the whole sphere like a longitude.
I am not very familiar with quaternions, so here I ask how to correctly rotate them.
I found out how to properly rotate a quaternion on my own. The key was to find vectors for the axis I want to rotate around. Those are used to create quaternions from axis and angle, when angle is the amount to rotate around the actual axis.
The following code shows what I ended up with. It also allows to roll the camera, which might be useful some time.
void Rotate(btVector3 Amount, float Sensitivity)
{
// fetch current rotation
btTransform transform = camera->getWorldTransform();
btQuaternion rotation = transform.getRotation();
// apply mouse sensitivity
Amount *= Sensitivity;
// create orientation vectors
btVector3 up(0, 1, 0);
btVector3 lookat = quatRotate(rotation, btVector3(0, 0, 1));
btVector3 forward = btVector3(lookat.getX(), 0, lookat.getZ()).normalize();
btVector3 side = btCross(up, forward);
// rotate camera with quaternions created from axis and angle
rotation = btQuaternion(up, Amount.getY()) * rotation;
rotation = btQuaternion(side, Amount.getX()) * rotation;
rotation = btQuaternion(forward, Amount.getZ()) * rotation;
// set new rotation
transform.setRotation(rotation);
camera->setWorldTransform(transform);
}
Since I rarely found information about quaternion rotation, I'll spend some time further explaining the code above.
Fetching and setting the rotation is specific to the physics engine and isn't related to this question so I won't elaborate on this. The next part, multiplying the amount by a mouse sensitivity should be really clear. Let's continue with the direction vectors.
The up vector depends on your own implementation. Most conveniently, the positive Y axis points up, therefore we end up with 0, 1, 0.
The lookat vector represents the direction the camera looks at. We simply rotate a unit vector pointing forward by the camera rotation quaternion. Again, the forward pointing vector depends on your conventions. If the Y axis is up, the positive Z axis might point forward, which is 0, 0, 1.
Do not mix that up with the next vector. It's named forward which references to the camera rotation. Therefore we just need to project the lookat vector to the ground. In this case, we simply take the lookat vector and ignore the up pointing component. For neatness we normalize that vector.
The side vector points leftwards from the camera orientation. Therefore it lies perpendicular to both the up and the forward vector and we can use the cross product to compute it.
Given those vectors, we can correctly rotate the camera quaternion around them. Which you start with, Z, Y or Z, depends on the Euler angle sequence which is, again, a convention varying from application to application. Since I want to rotations to be applied in Y X Z order, I do the following.
First, rotate the camera around the up axis by the amount for the Y rotation. This is yaw.
Then rotate around the side axis, which points leftwards, by the X amount. It's pitch.
And lastly, rotate around the forward vector by the Z amount to apply roll.
To apply those rotations, we need to multiply the quaternions create by axis and angle with the current camera rotation. Lastly we apply the resulted quaternion to the body in the physics simulation.
Matrices and pitch/yaw/roll both having their limitations, I do not use them anymore but use instead quaternions. I rotate the view vector and recalculate first the camera vectors, then the view matrix in regard to the rotated view vector.
void Camera::rotateViewVector(glm::quat quat) {
glm::quat rotatedViewQuat;
quat = glm::normalize(quat);
m_viewVector = glm::normalize(m_viewVector);
glm::quat viewQuat(0.0f,
m_viewVector.x,
m_viewVector.y,
m_viewVector.z);
viewQuat = glm::normalize(viewQuat);
rotatedViewQuat = (quat * viewQuat) * glm::conjugate(quat);
rotatedViewQuat = glm::normalize(rotatedViewQuat);
m_viewVector = glm::normalize(glm::vec3(rotatedViewQuat.x, rotatedViewQuat.y, rotatedViewQuat.z));
m_rightVector = glm::normalize(glm::cross(glm::vec3(0.0f, 1.0f, 0.0f), m_viewVector));
m_upVector = glm::normalize(glm::cross(m_viewVector, m_rightVector));
}
I am currently working on ray-tracing techniques and I think I've made a pretty good job; but, I haven't covered camera yet.
Until now, I used a plane fragment for view plane which is located between (-width/2, height/2, 200) and (width/2, -height/2, 200) [200 is just a fixed number of z, can be changed].
Addition to that, I use the camera mostly on e(0, 0, 1000), and I use a perspective projection.
I send rays from point e to pixels, and print it to image's corresponding pixel after calculating the pixel color.
Here is a image I created. Hopefully you can guess where eye and view plane are by looking at the image.
My question starts from here. It's time to move my camera around, but I don't know how to map 2D view plane coordinates to the canonical coordinates. Is there a transformation matrix for that?
The method I think requires to know the 3D coordinates of pixels on view plane. I am not sure it's the right method to use. So, what do you suggest?
There are a variety of ways to do it. Here's what I do:
Choose a point to represent the camera location (camera_position).
Choose a vector that indicates the direction the camera is looking (camera_direction). (If you know a point the camera is looking at, you can compute this direction vector by subtracting camera_position from that point.) You probably want to normalize (camera_direction), in which case it's also the normal vector of the image plane.
Choose another normalized vector that's (approximately) "up" from the camera's point of view (camera_up).
camera_right = Cross(camera_direction, camera_up)
camera_up = Cross(camera_right, camera_direction) (This corrects for any slop in the choice of "up".)
Visualize the "center" of the image plane at camera_position + camera_direction. The up and right vectors lie in the image plane.
You can choose a rectangular section of the image plane to correspond to your screen. The ratio of the width or height of this rectangular section to the length of camera_direction determines the field of view. To zoom in you can increase camera_direction or decrease the width and height. Do the opposite to zoom out.
So given a pixel position (i, j), you want the (x, y, z) of that pixel on the image plane. From that you can subtract camera_position to get a ray vector (which then needs to be normalized).
Ray ComputeCameraRay(int i, int j) {
const float width = 512.0; // pixels across
const float height = 512.0; // pixels high
double normalized_i = (i / width) - 0.5;
double normalized_j = (j / height) - 0.5;
Vector3 image_point = normalized_i * camera_right +
normalized_j * camera_up +
camera_position + camera_direction;
Vector3 ray_direction = image_point - camera_position;
return Ray(camera_position, ray_direction);
}
This is meant to be illustrative, so it is not optimized.
For rasterising renderers, you tend to need a transformation matrix because that's how you map directly from 3D coordinates to screen 2D coordinates.
For ray tracing, it's not necessary because you're typically starting from a known pixel coordinate in 2D space.
Given the eye position, a point in 3-space that's in the center of the screen, and vectors for "up" and "right", it's quite easy to calculate the 3D "ray" that goes from the eye position and through the specified pixel.
I've previously posted some sample code from my own ray tracer at https://stackoverflow.com/a/12892966/6782