Once more something relatively simple, but confused as to what they want.
the method to find distance on cartesian coordinate system is
distance=sqrt[(x2-x1)^2 + (y2-y1)^2]
but how do i apply it here?
//Requires: testColor to be a valid Color
//Effects: returns the "distance" between the current Pixel's color and
// the passed color
// uses the standard method to calculate "distance"
// uses the same formula as finding distance on a
// Cartesian coordinate system
double colorDistance(Color testColor) const;
The color class has defined colors:
int red,green,blue
Do i define something like 'oldGreen' 'oldRed' 'oldBlue' and get the distance that way? The passed color being red,green,blue ?
http://pastebin.com/v9K30dc7
I'm guessing they want you to use:
distance = sqrt[(r2-r1)^2 + (g2-g1)^2 + (b2-b1)^2]
break the color into it's red green and blue components, and use the same method, ie sqrt(sqr(delta red)+sqr(delta blue)+sqr(delta green))
note, this is not a really great method, as it doesn't allow for gamma or even the more complicated case of human perception. read http://en.wikipedia.org/wiki/Color_difference for more exotic methods.
Related
I'm just wondering if there was any way which one can perform mouse picking detection onto any object. Whether it would be generated object or imported object.
[Idea] -
The idea I have in mind is that, there would be iterations with every object in the scene. Checking if the mouse ray has intersected with an object. For checking the intersection, it would check the mouse picking ray with the triangles that make up the object.
[Pros] -
I believe the benefit of this approach is that, every object can be detected with mouse picking since they all inherit from the detection method.
[Cons] -
I believe this drawbacks are mainly the speed and the method being very expensive. So would need fine tuning of optimization.
[Situation] -
In the past I have read about mouse picking and I too have implemented some basic form of mouse picking. But all those were crappy work which I am not proud of. So again today, I have re-read some of the stuff from online. Nowadays I see alot of mouse picking using color ids and shaders. I'm not too keen for this method. I'm more into a mathematical side.
So here is my mouse picking ray thingamajig.
maths::Vector3 Camera::Raycast(s32 mouse_x, s32 mouse_y)
{
// Normalized Device Coordinates
maths::Vector2 window_size = Application::GetApplication().GetWindowSize();
float x = (2.0f * mouse_x) / window_size.x - 1.0f;
float y = 1.0f;
float z = 1.0f;
maths::Vector3 normalized_device_coordinates_ray = maths::Vector3(x, y, z);
// Homogeneous Clip Coordinates
maths::Vector4 homogeneous_clip_coordinates_ray = maths::Vector4(normalized_device_coordinates_ray.x, normalized_device_coordinates_ray.y, -1.0f, 1.0f);
// 4D Eye (Camera) Coordinates
maths::Vector4 camera_ray = maths::Matrix4x4::Invert(projection_matrix_) * homogeneous_clip_coordinates_ray;
camera_ray = maths::Vector4(camera_ray.x, camera_ray.y, -1.0f, 0.0f);
// 4D World Coordinates
maths::Vector3 world_coordinates_ray = maths::Matrix4x4::Invert(view_matrix_) * camera_ray;
world_coordinates_ray = world_coordinates_ray.Normalize();
return world_coordinates_ray;
}
I have this ray plane intersection function which calculates if a certain ray as intersected with a certain plane. DUH!
Here is the code for that.
bool Camera::RayPlaneIntersection(const maths::Vector3& ray_origin, const maths::Vector3& ray_direction, const maths::Vector3& plane_origin, const maths::Vector3& plane_normal, float& distance)
{
float denominator = plane_normal.Dot(ray_direction);
if (denominator >= 1e-6) // 1e-6 = 0.000001
{
maths::Vector3 vector_subtraction = plane_origin - ray_origin;
distance = vector_subtraction.Dot(plane_normal);
return (distance >= 0);
}
return false;
}
There are many more out there. E.g. Plane Sphere Intersection, Plane Disk Intersection. These things are like very specific. So it feel that is very hard to do mouse picking intersections on a global scale. I feel this way because, for this very RayPlaneIntersection function. What I expect to do with it is, retrieve the objects in the scene and retrieve all the normals for that object (which is a pain in the ass). So now to re-emphasize my question.
Is there already a method out there which I don't know, that does mouse picking in one way for all objects? Or am I just being stupid and not knowing what to do when I have everything?
Thank you. Thank you.
Yes, it is possible to do mouse-picking with OpenGL: you render all the geometry into a special buffer that stores a unique id of the object instead of its shaded color, then you just look at what value you got at the pixel below the mouse and know the object by its id that is written there. However, although it might be simpler, it is not a particularly efficient solution if your camera or geometry constantly moves.
Instead, doing an analytical ray-object intersection is the way to go. However, you don't need to check the intersection of every triangle of every object against the ray. That would be inefficient indeed. You should cull entire objects by their bounding boxes, or even portions of the whole scene. Game engines have their own spacial index data structure to speed-up ray-object intersections. They need it not only for mouse picking, but also for collision-detection, physics simulations, AI, and what-not.
Also note that the geometry used for the picking might be different from the one used for rendering. One example that comes to mind is that of semi-transparent objects.
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I am looking for a way to do a wormhole effect like this:
- https://www.youtube.com/watch?v=WeOBXmLeJWo&feature=youtu.be&t=43s
I have already found nice tunnels in the examples, but here it is a little bit more involved. Actually the space appears to be warped somehow and the movement is high velocity, so it is not just entering into a simple tunnel. Any idea how to do the space warping part of it?
I decided to add more info because this was too broad:
I have a galaxy and each star has a 3d coord, size, etc. in this galaxy. I can visit these stars with a space ship. There are very distant stars and it would take a lot of time to get to them, that's why I need warp (faster than light) speed. This does not necessarily requires a wormhole according to physics, but this app does not have to be overly realistic. I don't want to solve this with pure OpenGL so ofc. we can use shaders. I'd like to warp the space in the middle of the screen when accelerating to warp speeds. After that a tunnel effect can come, because I think it would consume a lot of resources to update every star by very high speeds, so I'd like to update only the close stars. This can't be a prerendered animation, because the destination is not always certain, so this has sometimes exploration purposes and sometimes traveling purposes. I don't think warping only the sky box is enough, but I am not sure about this.
There are 2 things going on there:
space curvature around the hole
You need to construct equation that describe the curvature of space around hole parametrized by hole parameters (mass,position,orientation) and time so you can animate it. Then from this curvature you can compute relative displacement of pixel/voxel around it. I would start with cylindrical cones with radius modulated by sin of the distance from hole +/- some animation parameters (need experimentation).
Something like this:
and for example start with (in wormhole local coordinates LCS):
r = R * sin(z*0.5*M_PI/wormhole_max_depth)
Then modulate it by additional therms. The wormhole_max_depth,R should be functions of time even linear or with some periodic therm so it is pulsating a bit.
The displacement can be done by simply computing distance of concerned point to cone surface and push it towards it the more the closer to the cone it is (inside cone voxels are considered below surface so apply max displacement strength)
particle/light/matter bursting out of the hole
I would go for this only when #1 is alredy done. It should be simple particle effect with some nice circular blended alpha texture animated on the surface of the cone from #1. I see it as few for loops with pseudo random displacement in position and speed ...
Techniques
This topic depends on how you want to do this. I see these possibilities:
Distort geometry during rendering (3D vector)
So you can apply the cone displacement directly on rendered stuff. This would be best applicable in GLSL but the geometry rendered must have small enough primitives to make this work on vertex level ...
Distort skybox/stars only (3D vector or 2D raster but objects stay unaffected)
So you apply the displacement on texture coordinates of skybox or directly on the star positions.
Distort whole rendered scene in second pass (2D raster)
This need to use 2 pass rendering and in the second pass just wrap the texture coordinates near hole.
As you got different local stars in each sector I would use star background generated from star catalogue (list of all your stars) And apply the distortion on them directly in 3D vector space (so no skybox.. option #2). And also because my engines already use such representation and rendering for the same reasons.
[Edit1] cone geometry
I haven't much time for this recently until today so I did not make much of a progress. I decided to start with cone geometry so here it is:
class wormhole
{
public:
reper rep; // coordinate system transform matrix
double R0,R1,H,t; // radiuses,depth
wormhole(){ R0=10.0; R1=100.0; H=50.0; t=0.0; };
wormhole(wormhole& a){ *this=a; };
~wormhole(){};
wormhole* operator = (const wormhole *a) { *this=*a; return this; };
/*wormhole* operator = (const wormhole &a) { ...copy... return this; };*/
void ah2xyz(double *xyz,double a,double h) // compute cone position from parameters a=<0,2pi>, h=<0,1>
{
double r,tt;
tt=t; if (t>0.5) tt=0.5; r=2.0*R0*tt; // inner radius R0
tt=t; if (t>1.0) tt=1.0; r+=(R1-r)*h*h*tt; // outer radius R1
xyz[0]=r*cos(a);
xyz[1]=r*sin(a);
xyz[2]=H*h*tt;
rep.l2g(xyz,xyz);
}
void draw_cone()
{
int e;
double a,h,da=pi2*0.04,p[3];
glColor3f(0.2,0.2,0.2);
for (h=0.0;h<=1.0;h+=0.1){ glBegin(GL_LINE_STRIP); for (e=1,a=0.0;e;a+=da) { if (a>=pi2) { e=0; a=0.0; } ah2xyz(p,a,h); glVertex3dv(p); } glEnd(); }
for (e=1,a=0.0;e;a+=da){ glBegin(GL_LINE_STRIP); for (h=0.0;h<=1.0;h+=0.1) { if (a>=pi2) { e=0; a=0.0; } ah2xyz(p,a,h); glVertex3dv(p); } glEnd(); }
}
} hole;
Where rep is my class for homogenous 4x4 transform matrix (remembering both direct and inverse matrices at the same time) function l2g just transforms from local coordinates to global. The cone parameters are:
R0 - inner cone radius when fully grown
R1 - outer cone radius when fully grown
H - the height/depth of the cone when fully grown
t - is animation parameter values <0.0,1.0> are the growth and values above 1.0 are reserved for wormhole fully grown animation
Here how it looks like:
What I would do is simply calculate a vector from the texture coordinate of the screen center to the texture coordinate of the pixel you're shading.
Then modify that vector in any way you want (time based for example) and apply it to the texture coordinate of the pixel you're shading and then use the resulting coordinate to sample your texture.
In pseudocode this would be something like this:
vec2 vector_to_screen_center = vec2(0.5) - texture_coordinate;
texture_coordinate += vector_to_screen_center * sin(time) * 0.1; // Time based modulation of the vector.
gl_FragColor = texture2D(screen_texture, texture_coordinate);
Your question does not have a GLSL tag. If you plan to do this without shaders, this is going to be hard and/or inefficient.
To give some background to this question, I'm creating a game that needs to know whether the 'Orbit' of an object is within tolerance to another Orbit. To show this, I plot a Torus-shape with a given radius (the tolerance) using the Target Orbit, and now I need to check if the ellipse is within that torus.
I'm getting lost in the equations on Math/Stack exchange so asking for a more specific solution. For clarification, here's an image of the game with the Torus and an Orbit (the red line). Quite simply, I want to check if that red orbit is within that Torus shape.
What I believe I need to do, is plot four points in World-Space on one of those orbits (easy enough to do). I then need to calculate the shortest distance between that point, and the other orbits' ellipse. This is the difficult part. There are several examples out there of finding the shortest distance of a point to an ellipse, but all are 2D and quite difficult to follow.
If that distance is then less than the tolerance for all four points, then in think that equates to the orbit being inside the target torus.
For simplicity, the origin of all of these orbits is always at the world Origin (0, 0, 0) - and my coordinate system is Z-Up. Each orbit has a series of parameters that defines it (Orbital Elements).
Here simple approach:
Sample each orbit to set of N points.
Let points from first orbit be A and from second orbit B.
const int N=36;
float A[N][3],B[N][3];
find 2 closest points
so d=|A[i]-B[i]| is minimal. If d is less or equal to your margin/treshold then orbits are too close to each other.
speed vs. accuracy
Unless you are using some advanced method for #2 then its computation will be O(N^2) which is a bit scary. The bigger the N the better accuracy of result but a lot more time to compute. There are ways how to remedy both. For example:
first sample with small N
when found the closest points sample both orbits again
but only near those points in question (with higher N).
you can recursively increase accuracy by looping #2 until you have desired precision
test d if ellipses are too close to each other
I think I may have a new solution.
Plot the four points on the current orbit (the ellipse).
Project those points onto the plane of the target orbit (the torus).
Using the Target Orbit inclination as the normal of a plane, calculate the angle between each (normalized) point and the argument of periapse
on the target orbit.
Use this angle as the mean anomaly, and compute the equivalent eccentric anomaly.
Use those eccentric anomalies to plot the four points on the target orbit - which should be the nearest points to the other orbit.
Check the distance between those points.
The difficulty here comes from computing the angle and converting it to the anomaly on the other orbit. This should be more accurate and faster than a recursive function though. Will update when I've tried this.
EDIT:
Yep, this works!
// The Four Locations we will use for the checks
TArray<FVector> CurrentOrbit_CheckPositions;
TArray<FVector> TargetOrbit_ProjectedPositions;
CurrentOrbit_CheckPositions.SetNum(4);
TargetOrbit_ProjectedPositions.SetNum(4);
// We first work out the plane of the target orbit.
const FVector Target_LANVector = FVector::ForwardVector.RotateAngleAxis(TargetOrbit.LongitudeAscendingNode, FVector::UpVector); // Vector pointing to Longitude of Ascending Node
const FVector Target_INCVector = FVector::UpVector.RotateAngleAxis(TargetOrbit.Inclination, Target_LANVector); // Vector pointing up the inclination axis (orbit normal)
const FVector Target_AOPVector = Target_LANVector.RotateAngleAxis(TargetOrbit.ArgumentOfPeriapsis, Target_INCVector); // Vector pointing towards the periapse (closest approach)
// Geometric plane of the orbit, using the inclination vector as the normal.
const FPlane ProjectionPlane = FPlane(Target_INCVector, 0.f); // Plane of the orbit. We only need the 'normal', and the plane origin is the Earths core (periapse focal point)
// Plot four points on the current orbit, using an equally-divided eccentric anomaly.
const float ECCAngle = PI / 2.f;
for (int32 i = 0; i < 4; i++)
{
// Plot the point, then project it onto the plane
CurrentOrbit_CheckPositions[i] = PosFromEccAnomaly(i * ECCAngle, CurrentOrbit);
CurrentOrbit_CheckPositions[i] = FVector::PointPlaneProject(CurrentOrbit_CheckPositions[i], ProjectionPlane);
// TODO: Distance from the plane is the 'Depth'. If the Depth is > Acceptance Radius, we are outside the torus and can early-out here
// Normalize the point to find it's direction in world-space (origin in our case is always 0,0,0)
const FVector PositionDirectionWS = CurrentOrbit_CheckPositions[i].GetSafeNormal();
// Using the Inclination as the comparison plane - find the angle between the direction of this vector, and the Argument of Periapse vector of the Target orbit
// TODO: we can probably compute this angle once, using the Periapse vectors from each orbit, and just multiply it by the Index 'I'
float Angle = FMath::Acos(FVector::DotProduct(PositionDirectionWS, Target_AOPVector));
// Compute the 'Sign' of the Angle (-180.f - 180.f), using the Cross Product
const FVector Cross = FVector::CrossProduct(PositionDirectionWS, Target_AOPVector);
if (FVector::DotProduct(Cross, Target_INCVector) > 0)
{
Angle = -Angle;
}
// Using the angle directly will give us the position at th eccentric anomaly. We want to take advantage of the Mean Anomaly, and use it as the ecc anomaly
// We can use this to plot a point on the target orbit, as if it was the eccentric anomaly.
Angle = Angle - TargetOrbit.Eccentricity * FMathD::Sin(Angle);
TargetOrbit_ProjectedPositions[i] = PosFromEccAnomaly(Angle, TargetOrbit);}
I hope the comments describe how this works. Finally solved after several months of head-scratching. Thanks all!
I am tracking a ball using the rgb data from kinect. After this I look up the corresponding depth data. Both of this is working splendid. Now I want to have the actual x,y,z world coordinates (i.e skeleton Space) instead of the x_screen, y_screen and depth values. Unfortunately the methods given by the kinect sdk (http://msdn.microsoft.com/en-us/library/hh973078.aspx) don`t help me. Basically i need a function "NuiImageGetSkeletonCoordinatesFromColorPixel" but i does not exist. All the functions basically go in the opposite direction
I know this can probably be done with openNI but i can not use it for other reasons.
Is there a function that does this for me or do i have to do the conversion myself? If I have to do it myself, how would i do this? I sketched up a little diagram http://i.imgur.com/ROBJW8Q.png - do you think this would work?
Check the CameraIntrinsics.
typedef struct _CameraIntrinsics
{
float FocalLengthX;
float FocalLengthY;
float PrincipalPointX;
float PrincipalPointY;
float RadialDistortionSecondOrder;
float RadialDistortionFourthOrder;
float RadialDistortionSixthOrder;
} CameraIntrinsics;
You can get it from ICoordinateMapper::GetDepthCameraIntrinsics.
Then, for every pixel (u,v,d) in depth space, you can get the coordinate in world space by doing this:
x = (u - principalPointX) / focalLengthX * d;
y = (v - principalPointY) / focalLengthY * d;
z = d;
For color space pixel, you need to first find its associated depth space pixel, which you should use ICoordinateMapper::MapCameraPointTodepthSpace. Since not all color pixel has its associated depth pixel (1920x1080 vs 512x424), you can't have the full-HD color point cloud.
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.