I've implemented a fps camera based on the up, right and view vectors from this.
Right now I want to be able to interact with the world by placing cubes in a minecraft style.
My lookAt vector is the sum of the view vector and the camera position, so my first attempt was to draw a cube at lookAt, but this is causing a strange behaviour.
I compute every vector like in the web I mentioned (such that lookAt = camera_position + view_direction) but the cube drawn is always arround me. I've tried several things like actually placing it (rounding the lookAt) and it appears near the wanted position but not at the place i'm looking at.
Given these vectors, how can I draw that's centered at the position that my camera is looking but a little bit further (exactly like minecraft)?
but the cube drawn is always arround me.
Yeah and that's obvious. You place cubes on the sphere surface of radius view_direction with center at camera_position.
Given these vectors, how can I draw that's centered at the position
that my camera is looking but a little bit further (exactly like
minecraft)?
You need to place cubes at the intersection of the view vector with the scene geometry. In the simplest case, it can be just "ground" plane, so you need intersect view vector with "ground" plane. Then you need to round the intersection xyz coordinates to the nearest grid node xyz = round(xyz / cubexyz)*cubexyz where cubexyz - cube size.
Approximate code:
Vector3D intersectPoint(Vector3D rayVector, Vector3D rayPoint, Vector3D planeNormal, Vector3D planePoint) {
Vector3D diff = rayPoint - planePoint;
double prod1 = diff.dot(planeNormal);
double prod2 = rayVector.dot(planeNormal);
double prod3 = prod1 / prod2;
return rayPoint - rayVector * prod3;
}
.......
Vector3D cubePos = intersectPoint(view_direction, camera_position, Vector3D(0, 1, 0), Vector3D(0, 0, 0));
cubePos = round(cubePos / cubeSize) * cubeSize;
AddCube(cubePos);
It's hard to tell without having images to look at, but lookAt is most likely your normalized forward vector? If i understood you correctly, you'd want to do something like objectpos = camerapos + forward * 10f (where 10f is the distance you want to place the object in front of you in 3d space units) to make sure that it's placed a few units in front of your fps controller.
actually, if view_direction is your normalized forward vector and your lookAt is camera_pos + view_direction, then you'd end up with something very close to your camera position, which would explain why the cube spawns inside you. either way, my suggestion should still work :)
Related
I am trying to rotate a "cube" full of little cubes using keyboard which works but not so great.
I am struggling with setting the pivot point of rotation to the very center of the big "cube" / world. As you can see on this video, center of front (initial) face of the big cube is the pivot point for my rotation right now, which is a bit confusing when I rotate the world a little bit.
To explain it better, it looks like I am moving initial face of the cube when using keys to rotate the cube. So the pivot point might be okay from this point of view, but what is wrong in my code? I don't understand why it is moving by front face, not the entire cube by its very center?
In case of generating all little cubes, I call a function in 3 for loops (x, y, z) and the function returns cubeMat so I have all cubes generated as you can see on the video.
cubeMat = scale(cubeMat, {0.1f, 0.1f, 0.1f});
cubeMat = translate(cubeMat, {positioning...);
For rotation itself, a short example of rotation to left looks like this:
mat4 total_rotation; //global variable - never resets
mat4 rotation; //local variable
if(keysPressed[GLFW_KEY_LEFT]){
timer -= delta;
rotation = rotate(mat4{}, -delta, {0, 1, 0});
}
... //rest of key controls
total_rotation *= rotation;
And inside of those 3 for cycles is also this:
program.setUniform("ModelMatrix", total_rotation * cubeMat);
cube.render();
I have read that I should use transformation to set the pivot point to the middle but in this case, how can I set the pivot point inside of little cube which is in center of world? That cube is obviously x=2, y=2, z=2 since in for cycles, I generate cubes starting at x=0.
You are accumulating the rotation matrices by right-multiplication. This way, all rotations are performed in the local coordinate systems that result from all previous transformations. And this is why your right-rotation results in a turn after an up-rotation (because it is a right-rotation in the local coordinate system).
But you want your rotations to be in the global coordinate system. Thus, simply revert the multiplication order:
total_rotation = rotation * total_rotation;
I'm currently in the process of finishing the implementation for a camera that functions in the same way as the camera in Maya. The part I'm stuck in the tumble functionality.
The problem is the following: the tumble feature works fine so long as the position of the camera is not parallel with the up vector (currently defined to be (0, 1, 0)). As soon as the camera becomes parallel with this vector (so it is looking straight up or down), the camera locks in place and will only rotate around the up vector instead of continuing to roll.
This question has already been asked here, unfortunately there is no actual solution to the problem. For reference, I also tried updating the up vector as I rotated the camera, but the resulting behaviour is not what I require (the view rolls as a result of the new orientation).
Here's the code for my camera:
using namespace glm;
// point is the position of the cursor in screen coordinates from GLFW
float deltaX = point.x - mImpl->lastPos.x;
float deltaY = point.y - mImpl->lastPos.y;
// Transform from screen coordinates into camera coordinates
Vector4 tumbleVector = Vector4(-deltaX, deltaY, 0, 0);
Matrix4 cameraMatrix = lookAt(mImpl->eye, mImpl->centre, mImpl->up);
Vector4 transformedTumble = inverse(cameraMatrix) * tumbleVector;
// Now compute the two vectors to determine the angle and axis of rotation.
Vector p1 = normalize(mImpl->eye - mImpl->centre);
Vector p2 = normalize((mImpl->eye + Vector(transformedTumble)) - mImpl->centre);
// Get the angle and axis
float theta = 0.1f * acos(dot(p1, p2));
Vector axis = cross(p1, p2);
// Rotate the eye.
mImpl->eye = Vector(rotate(Matrix4(1.0f), theta, axis) * Vector4(mImpl->eye, 0));
The vector library I'm using is GLM. Here's a quick reference on the custom types used here:
typedef glm::vec3 Vector;
typedef glm::vec4 Vector4;
typedef glm::mat4 Matrix4;
typedef glm::vec2 Point2;
mImpl is a PIMPL that contains the following members:
Vector eye, centre, up;
Point2 lastPoint;
Here is what I think. It has something to do with the gimbal lock, that occurs with euler angles (and thus spherical coordinates).
If you exceed your minimal(0, -zoom,0) or maxima(0, zoom,0) you have to toggle a boolean. This boolean will tell you if you must treat deltaY positive or not.
It could also just be caused by a singularity, therefore just limit your polar angle values between 89.99° and -89.99°.
Your problem could be solved like this.
So if your camera is exactly above (0, zoom,0) or beneath (0, -zoom,0) of your object, than the camera only rolls.
(I am also assuming your object is at (0,0,0) and the up-vector is set to (0,1,0).)
There might be some mathematical trick to resolve this, I would do it with linear algebra though.
You need to introduce a new right-vector. If you make a cross product, you will get the camera-vector. Camera-vector = up-vector x camera-vector. Imagine these vectors start at (0,0,0), then easily, to get your camera position just do this subtraction (0,0,0)-(camera-vector).
So if you get some deltaX, you rotate towards the right-vector(around the up-vector) and update it.
Any influence of deltaX should not change your up-vector.
If you get some deltaY you rotate towards the up-vector(around the right-vector) and update it. (This has no influence on the right-vector).
https://en.wikipedia.org/wiki/Rotation_matrix at Rotation matrix from axis and angle you can find a important formula.
You say u is your vector you want to rotate around and theta is the amount you want to pivot. The size of theta is proportional to deltaX/Y.
For example: We got an input from deltaX, so we rotate around the up-vector.
up-vector:= (0,1,0)
right-vector:= (0,0,-1)
cam-vector:= (0,1,0)
theta:=-1*30° // -1 due to the positive mathematical direction of rotation
R={[cos(-30°),0,-sin(-30°)],[0,1,0],[sin(-30°),0,cos(-30°)]}
new-cam-vector=R*cam-vector // normal matrix multiplication
One thing is left to be done: Update the right-vector.
right-vector=camera-vector x up-vector .
For picking objects, I've implemented a ray casting algorithm similar to what's described here. After converting the mouse click to a ray (with origin and direction) the next task is to intersect this ray with all triangles in the scene to determine hit points for each mesh.
I have also implemented the triangle intersection test algorithm based on the one described here. My question is, how should we account for the objects' transforms when performing the intersection? Obviously, I don't want to apply the transformation matrix to all vertices and then do the intersection test (too slow).
EDIT:
Here is the UnProject implementation I'm using (I'm using OpenTK by the way). I compared the results, they match what GluUnProject gives me:
private Vector3d UnProject(Vector3d screen)
{
int[] viewport = new int[4];
OpenTK.Graphics.OpenGL.GL.GetInteger(OpenTK.Graphics.OpenGL.GetPName.Viewport, viewport);
Vector4d pos = new Vector4d();
// Map x and y from window coordinates, map to range -1 to 1
pos.X = (screen.X - (float)viewport[0]) / (float)viewport[2] * 2.0f - 1.0f;
pos.Y = 1 - (screen.Y - (float)viewport[1]) / (float)viewport[3] * 2.0f;
pos.Z = screen.Z * 2.0f - 1.0f;
pos.W = 1.0f;
Vector4d pos2 = Vector4d.Transform(pos, Matrix4d.Invert(GetModelViewMatrix() * GetProjectionMatrix()));
Vector3d pos_out = new Vector3d(pos2.X, pos2.Y, pos2.Z);
return pos_out / pos2.W;
}
Then I'm using this function to create a ray (with origin and direction):
private Ray ScreenPointToRay(Point mouseLocation)
{
Vector3d near = UnProject(new Vector3d(mouseLocation.X, mouseLocation.Y, 0));
Vector3d far = UnProject(new Vector3d(mouseLocation.X, mouseLocation.Y, 1));
Vector3d origin = near;
Vector3d direction = (far - near).Normalized();
return new Ray(origin, direction);
}
You can apply the reverse transformation of each object to the ray instead.
I don't know if this is the best/most efficient approach, but I recently implemented something similar like this:
In world space, the origin of the ray is the camera position. In order to get the direction of the ray, I assumed the user had clicked on the near plane of the camera and thus applied the 'reverse transformation' - from screen space to world space - to the screen space position
( mouseClick.x, viewportHeight - mouseClick.y, 0 )
and then subtracted the origin of the ray, i.e. the camera position, from
the now transformed mouse click position.
In my case, there was no object-specific transformation, meaning I was done once I had my ray in world space. However, transforming origin & direction with the inverse model matrix would have been easy enough after that.
You mentioned that you tried to apply the reverse transformation, but that it didn't work - maybe there's a bug in there? I used a GLM - i.e. glm::unProject - for this.
I'm a student new to opengl. Currently, I'm doing a project that creates a scene.
Right now, my team is using gluLookAt() for my camera. What I want to accomplish is to try and rotate the LookAt vector around a certain point, namely where the camera is looking at.
This accomplishes a sort of "swaying in a circle". I need this because I am making a dart game for the scene, and my camera stay still, but I need it to move in a circle, but still allow the user's mouse to influence it. I also need it to create a drunken movement. That is why I am not considering rotating the Up or Eye vectors.
Currently, my look at code is like this.
int deltax = x - mouse.mX;
int deltay = y - mouse.mY;
cameradart.mYaw -= ((deltax/360.0) * 3.142) * 0.5;
cameradart.mPitch -= deltay * 0.02;
mouse.mX = x;
mouse.mY = y;
cameradart.lookAt.x = sin (cameradart.mYaw);
cameradart.lookAt.y = cameradart.mPitch ;
cameradart.lookAt.z = cos (cameradart.mYaw);
gluLookAt (cameradart.eye.x, cameradart.eye.y, cameradart.eye.z,
cameradart.eye.x + cameradart.lookAt.x, cameradart.eye.y + cameradart.lookAt.y,
cameradart.eye.z + cameradart.lookAt.z,
cameradart.up.x, cameradart.up.y, cameradart.up.z);
I know that it could be done easier using a different camera, but I really don't want to mess with my team's code by not using gluLookAt().
There's a couple of solutions in my mind, I'll tell you the easiest to understand/implement as a new graphics student
Assuming at first you're looking at (0,0,1) -store that vector-:
Think of a point that's drawing a circle and you're looking at it,
-Do it first to turn right and left (2D on X & Z)
-Let HDiff be the horizontal difference between old mouse position and the new one
-Update the x = cos(HDiff)
-Update the z = sin(HDiff)
*I didn't try it but it should work :)
If you want to be able to manipulate the camera, using a camera matrix is a much more effective mechanism than working with gluLookat. GLM is a good library for matrix and vector math and includes a lookat mechanism you can use to initialize the matrix, or you can just initialize it with a series of operations. However, remember that lookat produces a view matrix, and the view matrix is the inverse of the camera matrix.
This piece of code has a demonstration of what I'm talking about. Specifically look at the player member variable and how it's manipulated
glm::mat4 player;
...
glm::vec3 playerPosition(0, eyeHeight, ipd * 4.0f);
player = glm::inverse(glm::lookAt(playerPosition, glm::vec3(0, eyeHeight, 0), GlUtils::Y_AXIS));
This approach lets you apply changes like rotation and translation directly to the player matrix
// Rotate on the Y axis
player = glm::rotate(player, angle, glm::vec3(0, 1, 0));
This is much more intuitive than manipulating the view matrix, since changes to the view matrix always have to be the inverse of what you'd do to the player matrix.
When you're ready to render you need to convert the player matrix to a view matrix by taking it's inverse. In my example it's done like this:
gl::Stacks::modelview().top() = riftOrientation * glm::inverse(player);
This is because I'm using an application based modelview matrix stack that gets applied to the Shader programs I'm running.
For an OpenGL 1.x program, you'd instead use LoadMatrix
glMatrixMode(GL_MODELVIEW);
glm::mat4 modelview = glm::inverse(player);
glLoadMatrixf(&modelview);
I am developing a small tool for 3D visualization of molecules.
For my project i choose to make a thing in the way of what Mr "Brad Larson" did with his Apple software "Molecules". A link where you can find a small presentation of the technique used : Brad Larsson software presentation
For doing my job i must compute sphere impostor and cylinder impostor.
For the moment I have succeed to do the "Sphere Impostor" with the help of another tutorial Lies and Impostors
for summarize the computing of the sphere impostor : first we send a "sphere position" and the "sphere radius" to the "vertex shader" which will create in the camera-space an square which always face the camera, after that we send our square to the fragment shader where we use a simple ray tracing to find which fragment of the square is included in the sphere, and finally we compute the normal and the position of the fragment to compute lighting. (another thing we also write the gl_fragdepth for giving a good depth to our impostor sphere !)
But now i am blocked in the computing of the cylinder impostor, i try to do a parallel between the sphere impostor and the cylinder impostor but i don't find anything, my problem is that for the sphere it was some easy because the sphere is always the same no matter how we see it, we will always see the same thing : "a circle" and another thing is that the sphere was perfectly defined by Math then we can find easily the position and the normal for computing lighting and create our impostor.
For the cylinder it's not the same thing, and i failed to find a hint to modeling a form which can be used as "cylinder impostor", because the cylinder shows many different forms depending on the angle we see it !
so my request is to ask you about a solution or an indication for my problem of "cylinder impostor".
In addition to pygabriels answer I want to share a standalone implementation using the mentioned shader code from Blaine Bell (PyMOL, Schrödinger, Inc.).
The approach, explained by pygabriel, also can be improved. The bounding box can be aligned in such a way, that it always faces to the viewer. Only two faces are visible at most. Hence, only 6 vertices (ie. two faces made up of 4 triangles) are needed.
See picture here, the box (its direction vector) always faces to the viewer:
Image: Aligned bounding box
For source code, download: cylinder impostor source code
The code does not cover round caps and orthographic projections. It uses geometry shader for vertex generation. You can use the shader code under the PyMOL license agreement.
I know this question is more than one-year old, but I'd still like to give my 2 cents.
I was able to produce cylinder impostors with another technique, I took inspiration from pymol's code. Here's the basic strategy:
1) You want to draw a bounding box (a cuboid) for the cylinder. To do that you need 6 faces, that translates in 18 triangles that translates in 36 triangle vertices. Assuming that you don't have access to geometry shaders, you pass to a vertex shader 36 times the starting point of the cylinder, 36 times the direction of the cylinder, and for each of those vertex you pass the corresponding point of the bounding box. For example a vertex associated with point (0, 0, 0) means that it will be transformed in the lower-left-back corner of the bounding box, (1,1,1) means the diagonally opposite point etc..
2) In the vertex shader, you can construct the points of the cylinder, by displacing each vertex (you passed 36 equal vertices) according to the corresponding points you passed in.
At the end of this step you should have a bounding box for the cylinder.
3) Here you have to reconstruct the points on the visible surface of the bounding box. From the point you obtain, you have to perform a ray-cylinder intersection.
4) From the intersection point you can reconstruct the depth and the normal. You also have to discard intersection points that are found outside of the bounding box (this can happen when you view the cylinder along its axis, the intersection point will go infinitely far).
By the way it's a very hard task, if somebody is interested here's the source code:
https://github.com/chemlab/chemlab/blob/master/chemlab/graphics/renderers/shaders/cylinderimp.frag
https://github.com/chemlab/chemlab/blob/master/chemlab/graphics/renderers/shaders/cylinderimp.vert
A cylinder impostor can actually be done just the same way as a sphere, like Nicol Bolas did it in his tutorial. You can make a square facing the camera and colour it that it will look like a cylinder, just the same way as Nicol did it for spheres. And it's not that hard.
The way it is done is ray-tracing of course. Notice that a cylinder facing upwards in camera space is kinda easy to implement. For example intersection with the side can be projected to the xz plain, it's a 2D problem of a line intersecting with a circle. Getting the top and bottom isn't harder either, the z coordinate of the intersection is given, so you actually know the intersection point of the ray and the circle's plain, all you have to do is to check if its inside the circle. And basically, that's it, you get two points, and return the closer one (the normals are pretty trivial too).
And when it comes to an arbitrary axis, it turns out to be almost the same problem. When you solve equations at the fixed axis cylinder, you are solving them for a parameter that describes how long do you have to go from a given point in a given direction to reach the cylinder. From the "definition" of it, you should notice that this parameter doesn't change if you rotate the world. So you can rotate the arbitrary axis to become the y axis, solve the problem in a space where equations are easier, get the parameter for the line equation in that space, but return the result in camera space.
You can download the shaderfiles from here. Just an image of it in action:
The code where the magic happens (It's only long 'cos it's full of comments, but the code itself is max 50 lines):
void CylinderImpostor(out vec3 cameraPos, out vec3 cameraNormal)
{
// First get the camera space direction of the ray.
vec3 cameraPlanePos = vec3(mapping * max(cylRadius, cylHeight), 0.0) + cameraCylCenter;
vec3 cameraRayDirection = normalize(cameraPlanePos);
// Now transform data into Cylinder space wherethe cyl's symetry axis is up.
vec3 cylCenter = cameraToCylinder * cameraCylCenter;
vec3 rayDirection = normalize(cameraToCylinder * cameraPlanePos);
// We will have to return the one from the intersection of the ray and circles,
// and the ray and the side, that is closer to the camera. For that, we need to
// store the results of the computations.
vec3 circlePos, sidePos;
vec3 circleNormal, sideNormal;
bool circleIntersection = false, sideIntersection = false;
// First check if the ray intersects with the top or bottom circle
// Note that if the ray is parallel with the circles then we
// definitely won't get any intersection (but we would divide with 0).
if(rayDirection.y != 0.0){
// What we know here is that the distance of the point's y coord
// and the cylCenter is cylHeight, and the distance from the
// y axis is less than cylRadius. So we have to find a point
// which is on the line, and match these conditions.
// The equation for the y axis distances:
// rayDirection.y * t - cylCenter.y = +- cylHeight
// So t = (+-cylHeight + cylCenter.y) / rayDirection.y
// About selecting the one we need:
// - Both has to be positive, or no intersection is visible.
// - If both are positive, we need the smaller one.
float topT = (+cylHeight + cylCenter.y) / rayDirection.y;
float bottomT = (-cylHeight + cylCenter.y) / rayDirection.y;
if(topT > 0.0 && bottomT > 0.0){
float t = min(topT,bottomT);
// Now check for the x and z axis:
// If the intersection is inside the circle (so the distance on the xz plain of the point,
// and the center of circle is less than the radius), then its a point of the cylinder.
// But we can't yet return because we might get a point from the the cylinder side
// intersection that is closer to the camera.
vec3 intersection = rayDirection * t;
if( length(intersection.xz - cylCenter.xz) <= cylRadius ) {
// The value we will (optianally) return is in camera space.
circlePos = cameraRayDirection * t;
// This one is ugly, but i didn't have better idea.
circleNormal = length(circlePos - cameraCylCenter) <
length((circlePos - cameraCylCenter) + cylAxis) ? cylAxis : -cylAxis;
circleIntersection = true;
}
}
}
// Find the intersection of the ray and the cylinder's side
// The distance of the point and the y axis is sqrt(x^2 + z^2), which has to be equal to cylradius
// (rayDirection.x*t - cylCenter.x)^2 + (rayDirection.z*t - cylCenter.z)^2 = cylRadius^2
// So its a quadratic for t (A*t^2 + B*t + C = 0) where:
// A = rayDirection.x^2 + rayDirection.z^2 - if this is 0, we won't get any intersection
// B = -2*rayDirection.x*cylCenter.x - 2*rayDirection.z*cylCenter.z
// C = cylCenter.x^2 + cylCenter.z^2 - cylRadius^2
// It will give two results, we need the smaller one
float A = rayDirection.x*rayDirection.x + rayDirection.z*rayDirection.z;
if(A != 0.0) {
float B = -2*(rayDirection.x*cylCenter.x + rayDirection.z*cylCenter.z);
float C = cylCenter.x*cylCenter.x + cylCenter.z*cylCenter.z - cylRadius*cylRadius;
float det = (B * B) - (4 * A * C);
if(det >= 0.0){
float sqrtDet = sqrt(det);
float posT = (-B + sqrtDet)/(2*A);
float negT = (-B - sqrtDet)/(2*A);
float IntersectionT = min(posT, negT);
vec3 Intersect = rayDirection * IntersectionT;
if(abs(Intersect.y - cylCenter.y) < cylHeight){
// Again it's in camera space
sidePos = cameraRayDirection * IntersectionT;
sideNormal = normalize(sidePos - cameraCylCenter);
sideIntersection = true;
}
}
}
// Now get the results together:
if(sideIntersection && circleIntersection){
bool circle = length(circlePos) < length(sidePos);
cameraPos = circle ? circlePos : sidePos;
cameraNormal = circle ? circleNormal : sideNormal;
} else if(sideIntersection){
cameraPos = sidePos;
cameraNormal = sideNormal;
} else if(circleIntersection){
cameraPos = circlePos;
cameraNormal = circleNormal;
} else
discard;
}
From what I can understand of the paper, I would interpret it as follows.
An impostor cylinder, viewed from any angle has the following characteristics.
From the top, it is a circle. So considering you'll never need to view a cylinder top down, you don't need to render anything.
From the side, it is a rectangle. The pixel shader only needs to compute illumination as normal.
From any other angle, it is a rectangle (the same one computed in step 2) that curves. Its curvature can be modeled inside the pixel shader as the curvature of the top ellipse. This curvature can be considered as simply an offset of each "column" in texture space, depending on viewing angle. The minor axis of this ellipse can be computed by multiplying the major axis (thickness of the cylinder) with a factor of the current viewing angle (angle / 90), assuming that 0 means you're viewing the cylinder side-on.
Viewing angles. I have only taken the 0-90 case into account in the math below, but the other cases are trivially different.
Given the viewing angle (phi) and the diameter of the cylinder (a) here's how the shader needs to warp the Y-Axis in texture space Y = b' sin(phi). And b' = a * (phi / 90). The cases phi = 0 and phi = 90 should never be rendered.
Of course, I haven't taken the length of this cylinder into account - which would depend on your particular projection and is not an image-space problem.