I have a 3D scene with a movable camera. I have the 3D coordinates for that camera (x, y, z -> Y being the height) and the X and Y rotation (Up/Down, Left/Right).
I want get the coordinates (x1, z1) in the floor where I'm looking at.
Basically if the camera is at (0, 4096, 0) (4096 is the height) and my xRotation is 45º and my yRotation is 0, I will be looking at the point on the floor (4096, 0, 0)
I was trying to program it but i got stuck with the trigonometry. Help me with it.
The following code is what I have right now and not fully working:
float x1, z1, anguloX, anguloY;
anguloX = (90 - Xrotation) / 180 * Pi;
anguloY = (90 - Yrotation) / 180 * Pi;
x1 = Yposition * tan(anguloX) * cos(anguloY);
z1 = Yposition * tan(anguloY) * cos(anguloY);
x1 += Xposition;
z1 += Zposition;
Do not bother yourself with this kind of trigonometry, It is often more practical and understandable to use matrices and transformation to do this kind of stuff. I suggest to try this approach instead which is called Ray Casting:
Calculate the direction of camera sight, you must get the looking
direction of camera
Use ray casting to determine the intersection point of sight ray and
scene meshes(or just a plane that represents the floor).
Now about the first, here is the pseudo code :
XRotMat = CreateRotationMatrixAroundXAxis(verticalCameraAngel);
YRotMat = CreateRotationMatrixAroundYAxis(horizentalCameraAngel);
CameraSightDir = YRotMat*XRotMa*initialCameraDir;
and about the second step:
SightRay.Source = Camera.Position;
SightRay.Direction = CameraSightDir;
Intersection = IntersectRayWithPlane(SightRay , FloorPlane);
IntersectRayWithPlane is quite a simple procedure, you can read about it here.
Related
The following link suggests that we can convert dual fisheye coordinates to equirectangular coordinates using the following equations:
// 2D fisheye to 3D vector
phi = r * aperture / 2
theta = atan2(y, x)
// 3D vector to longitude/latitude
longitude = atan2(Py, Px)
latitude = atan2(Pz, (Px^2 + Py^2)^(0.5))
// 3D vector to 2D equirectangular
x = longitude / PI
y = 2 * latitude / PI
I applied to above equations to write my source code like this:
const float FOV = 220.0f * PI / 180.0f;
float r = sqrt(u*u + v*v);
float theta = atan2(v, u);
float phi = r * FOV * 0.5f;
float px = u;
float py = r * sin(phi);
float pz = v;
float longitude = atan2(py, px); // theta
float latitude = atan2(pz, sqrt(px*px + py*py)); // phi
x = longitude / PI;
y = 2.0f * latitude / PI;
Unfortunately my math is not good enough to understand this and not sure if I write the above code correctly, where I tried to guess the values for px, py and pz.
Assume my camera FOV is 220 degrees, and the camera resolution is 2880x1440, I would expect the point (358, 224) for rear camera in the overlapped area and the point (2563, 197) for front camera in the overlapped area would both map to a coordinate close to (2205, 1009). However the actual mapping points are (515.966370,1834.647949) and (1644.442017,1853.060669) respectively, which are both very far away from (2205,1009). Please kindly suggest how to fix the above code. Many thanks!
You are building the equirectangular image, so I would suggest you to use the inverse mapping.
Start with pixel locations in the target image you are painting. Convert the 2D location to longitude/latitude. Then convert that to a 3D point on the surface of the unit sphere. Then convert from the 3D point to a location in the 2D fisheye source image. In Paul Bourke page, you would start with the bottom equation, then the rightmost one, then the topmost one.
Use landmark points like 90° long 0° lat, to verify the results make sense at each step.
The final result should be a location in the source fisheye image in the [-1..+1] range. Remap to pixel or to UV as needed. Since the source is split in two eye images you will also need a mapping from target (equirect) longitudes to the correct source sub-image.
The problem is I have two points in 3D space where y+ is up, x+ is to the right, and z+ is towards you. I want to orientate a cylinder between them that is the length of of the distance between both points, so that both its center ends touch the two points. I got the cylinder to translate to the location at the center of the two points, and I need help coming up with a rotation matrix to apply to the cylinder, so that it is orientated the correct way. My transformation matrix for the entire thing looks like this:
translate(center point) * rotateX(some X degrees) * rotateZ(some Z degrees)
The translation is applied last, that way I can get it to the correct orientation before I translate it.
Here is what I have so far for this:
mat4 getTransformation(vec3 point, vec3 parent)
{
float deltaX = point.x - parent.x;
float deltaY = point.y - parent.y;
float deltaZ = point.z - parent.z;
float yRotation = atan2f(deltaZ, deltaX) * (180.0 / M_PI);
float xRotation = atan2f(deltaZ, deltaY) * (180.0 / M_PI);
float zRotation = atan2f(deltaX, deltaY) * (-180.0 / M_PI);
if(point.y < parent.y)
{
zRotation = atan2f(deltaX, deltaY) * (180.0 / M_PI);
}
vec3 center = vec3((point.x + parent.x)/2.0, (point.y + parent.y)/2.0, (point.z + parent.z)/2.0);
mat4 translation = Translate(center);
return translation * RotateX(xRotation) * RotateZ(zRotation) * Scale(radius, 1, radius) * Scale(0.1, 0.1, 0.1);
}
I tried a solution given down below, but it did not seem to work at all
mat4 getTransformation(vec3 parent, vec3 point)
{
// moves base of cylinder to origin and gives it unit scaling
mat4 scaleFactor = Translate(0, 0.5, 0) * Scale(radius/2.0, 1/2.0, radius/2.0) * cylinderModel;
float length = sqrtf(pow((point.x - parent.x), 2) + pow((point.y - parent.y), 2) + pow((point.z - parent.z), 2));
vec3 direction = normalize(point - parent);
float pitch = acos(direction.y);
float yaw = atan2(direction.z, direction.x);
return Translate(parent) * Scale(length, length, length) * RotateX(pitch) * RotateY(yaw) * scaleFactor;
}
After running the above code I get this:
Every black point is a point with its parent being the point that spawned it (the one before it) I want the branches to fit into the points. Basically I am trying to implement the space colonization algorithm for random tree generation. I got most of it, but I want to map the branches to it so it looks good. I can use GL_LINES just to make a generic connection, but if I get this working it will look so much prettier. The algorithm is explained here.
Here is an image of what I am trying to do (pardon my paint skills)
Well, there's an arbitrary number of rotation matrices satisfying your constraints. But any will do. Instead of trying to figure out a specific rotation, we're just going to write down the matrix directly. Say your cylinder, when no transformation is applied, has its axis along the Z axis. So you have to transform the local space Z axis toward the direction between those two points. I.e. z_t = normalize(p_1 - p_2), where normalize(a) = a / length(a).
Now we just need to make this a full 3 dimensional coordinate base. We start with an arbitrary vector that's not parallel to z_t. Say, one of (1,0,0) or (0,1,0) or (0,0,1); use the scalar product ·(also called inner, or dot product) with z_t and use the vector for which the absolute value is the smallest, let's call this vector u.
In pseudocode:
# Start with (1,0,0)
mindotabs = abs( z_t · (1,0,0) )
minvec = (1,0,0)
for u_ in (0,1,0), (0,0,1):
dotabs = z_t · u_
if dotabs < mindotabs:
mindotabs = dotabs
minvec = u_
u = minvec_
Then you orthogonalize that vector yielding a local y transformation y_t = normalize(u - z_t · u).
Finally create the x transformation by taking the cross product x_t = z_t × y_t
To move the cylinder into place you combine that with a matching translation matrix.
Transformation matrices are effectively just the axes of the space you're "coming from" written down as if seen from the other space. So the resulting matrix, which is the rotation matrix you're looking for is simply the vectors x_t, y_t and z_t side by side as a matrix. OpenGL uses so called homogenuous matrices, so you have to pad it to a 4×4 form using a 0,0,0,1 bottommost row and rightmost column.
That you can load then into OpenGL; if using fixed functio using glMultMatrix to apply the rotation, or if using shader to multiply onto the matrix you're eventually pass to glUniform.
Begin with a unit length cylinder which has one of its ends, which I call C1, at the origin (note that your image indicates that your cylinder has its center at the origin, but you can easily transform that to what I begin with). The other end, which I call C2, is then at (0,1,0).
I'd like to call your two points in world coordinates P1 and P2 and we want to locate C1 on P1 and C2 to P2.
Start with translating the cylinder by P1, which successfully locates C1 to P1.
Then scale the cylinder by distance(P1, P2), since it originally had length 1.
The remaining rotation can be computed using spherical coordinates. If you're not familiar with this type of coordinate system: it's like GPS coordinates: two angles; one around the pole axis (in your case the world's Y-axis) which we typically call yaw, the other one is a pitch angle (in your case the X axis in model space). These two angles can be computed by converting P2-P1 (i.e. the local offset of P2 with respect to P1) into spherical coordinates. First rotate the object with the pitch angle around X, then with yaw around Y.
Something like this will do it (pseudo-code):
Matrix getTransformation(Point P1, Point P2) {
float length = distance(P1, P2);
Point direction = normalize(P2 - P1);
float pitch = acos(direction.y);
float yaw = atan2(direction.z, direction.x);
return translate(P1) * scaleY(length) * rotateX(pitch) * rotateY(yaw);
}
Call the axis of the cylinder A. The second rotation (about X) can't change the angle between A and X, so we have to get that angle right with the first rotation (about Z).
Call the destination vector (the one between the two points) B. Take -acos(BX/BY), and that's the angle of the first rotation.
Take B again, ignore the X component, and look at its projection in the (Y, Z) plane. Take acos(BZ/BY), and that's the angle of the second rotation.
I am building a ray Tracer from scratch. My question is:
When I change camera coordinates the Sphere changes to ellipse. I don't understand why it's happening.
Here are some images to show the artifacts:
Sphere: 1 1 -1 1.0 (Center, radius)
Camera: 0 0 5 0 0 0 0 1 0 45.0 1.0 (eyepos, lookat, up, foy, aspect)
But when I changed camera coordinate, the sphere looks distorted as shown below:
Camera: -2 -2 2 0 0 0 0 1 0 45.0 1.0
I don't understand what is wrong. If someone can help that would be great!
I set my imagePlane as follows:
//Computing u,v,w axes coordinates of Camera as follows:
{
Vector a = Normalize(eye - lookat); //Camera_eye - Camera_lookAt
Vector b = up; //Camera Up Vector
m_w = a;
m_u = b.cross(m_w);
m_u.normalize();
m_v = m_w.cross(m_u);
}
After that I compute directions for each pixel from the Camera position (eye) as mentioned below:
//Then Computing direction as follows:
int half_w = m_width * 0.5;
int half_h = m_height * 0.5;
double half_fy = fovy() * 0.5;
double angle = tan( ( M_PI * half_fy) / (double)180.0 );
for(int k=0; k<pixels.size(); k++){
double j = pixels[k].x(); //width
double i = pixels[k].y(); //height
double XX = aspect() * angle * ( (j - half_w ) / (double)half_w );
double YY = angle * ( (half_h - i ) / (double)half_h );
Vector dir = (m_u * XX + m_v * YY) - m_w ;
directions.push_back(dir);
}
After that:
for each dir:
Ray ray(eye, dir);
int depth = 0;
t_color += Trace(g_primitive, ray, depth);
After playing a lot and with the help of the comments of all you guys I was able to create successfully my rayTracer properly. Sorry for answering late, but I would like to close this thread with few remarks.
So, the above mentioned code is perfectly correct. Based on my own assumptions (as mentioned in above comments) I have decided to set my Camera parameters like that.
The problem I mentioned above is a normal behaviour of the camera (as also mentioned above in the comments).
I have got good results now but there are few things to check while coding a rayTracer:
Always make sure to take care of Radians to Degrees (or vice versa) conversion while computing FOV and ASPECT RATIO. I did it as follows:
double angle = tan((M_PI * 0.5 * fovy) / 180.0);
double y = angle;
double x = aspect * angle;
2) While computing Triangle intersections, make sure to implement cross product properly.
3) While using intersections of different objects make sure to find the intersection which is at a minimum distance from the camera.
Here's the result I got:
Above is a very simple model (courtesy UCBerkeley), which I rayTraced.
This is the correct behavior. Get a camera with a wide angle lens, put the sphere near the edge of the field of view and take a picture. Then in a photo app draw a circle on top of the photo of the sphere and you will see that it's not a circular projection.
This effect will be magnified by the fact that you set aspect to 1.0 but your image is not square.
A few things to fix:
A direction vector is (to - from). You have (from - to), so a is pointing backward. You'll want to add m_w at the end, rather than subtract it. Also, this fix will rotate your m_u,m_v by 180 degrees, which will make you about to change (j - half_w) to (half_w - j).
Also, putting all the pixels and all the directions in lists is not as efficient as just looping over x,y values.
I have a class tetronimo (a tetris block) that has four QRect types (named first, second, third, fourth respectively). I draw each tetronimo using a build_tetronimo_L type functions.
These build the tetronimo in a certain direction, but as in tetris you're supposed to be able to rotate the tetronimo's, I'm trying to rotate a tetronimo by rotating each individual square of the tetronimo.
I have found the following formula to apply to each (x, y) coordinate of a particular square.
newx = cos(angle) * oldx - sin(angle) * oldy
newy = sin(angle) * oldx + cos(angle) * oldy
Now, the QRect type of Qt, does only seem to have a setCoords function that takes the (x, y) coordinates of top-left and bottom-right points of the respective square.
I have here an example (which doesn't seem to produce the correct result) of rotating the first two squares in my tetronimo.
Can anyone tell me how I'm supposed to rotate these squares correctly, using runtime rotation calculation?
void tetromino::rotate(double angle) // angle in degrees
{
std::map<std::string, rect_coords> coords = get_coordinates();
// FIRST SQUARE
rect_coords first_coords = coords["first"];
//top left x and y
int newx_first_tl = (cos(to_radians(angle)) * first_coords.top_left_x) - (sin(to_radians(angle)) * first_coords.top_left_y);
int newy_first_tl = (sin(to_radians(angle)) * first_coords.top_left_x) + (cos(to_radians(angle)) * first_coords.top_left_y);
//bottom right x and y
int newx_first_bl = (cos(to_radians(angle)) * first_coords.bottom_right_x) - (sin(to_radians(angle)) * first_coords.bottom_right_y);
int newy_first_bl = (cos(to_radians(angle)) * first_coords.bottom_right_x) + (sin(to_radians(angle)) * first_coords.bottom_right_y);
//CHANGE COORDINATES
first->setCoords( newx_first_tl, newy_first_tl, newx_first_tl + tetro_size,newy_first_tl - tetro_size);
//SECOND SQUARE
rect_coords second_coords = coords["second"];
int newx_second_tl = (cos(to_radians(angle)) * second_coords.top_left_x) - (sin(to_radians(angle)) * second_coords.top_left_y);
int newy_second_tl = (sin(to_radians(angle)) * second_coords.top_left_x) + (cos(to_radians(angle)) * second_coords.top_left_y);
//CHANGE COORDINATES
second->setCoords(newx_second_tl, newy_second_tl, newx_second_tl - tetro_size, newy_second_tl + tetro_size);
first and second are QRect types. rect_coords is just a struct with four ints in it, that store the coordinates of the squares.
The first square and second square calculations are different, as I was playing around trying to figure it out.
I hope someone can help me figure this out?
(Yes, I can do this much simpler, but I'm trying to learn from this)
It seems more like a math question than a programming question. Just plug in values like 90 degrees for the angle to figure this out. For 90 degrees, a point (x,y) is mapped to (-y, x). You probably don't want to rotate around the origin but around a certain pivot point c.x, c.y. For that you need to translate first, then rotate, then translate back:
(x,y) := (x-c.x, y-c.y) // translate into coo system w/ origin at c
(x,y) := (-y, x) // rotate
(x,y) := (x+c.x, y+c.y) // translate into original coo system
Before rotating you have to translate so that the piece is centered in the origin:
Translate your block centering it to 0, 0
Rotate the block
Translate again the center of the block to x, y
If you rotate without translating you will rotate always around 0, 0 but since the block is not centered it will be rotated around the center. To center your block is quite simple:
For each point, compute the median of X and Y, let's call it m
Subtract m.X and m.Y to the coordinates of all points
Rotate
Add again m.X and m.Y to points.
Of course you can use linear algebra and vector * matrix multiplication but maybe it is too much :)
Translation
Let's say we have a segment with coordinates A(3,5) B(10,15).
If you want to rotate it around its center, we first translate it to our origin. Let's compute mx and my:
mx = (10 - 3) / 2
my = (15 - 5) / 2
Now we compute points A1 and B1 translating the segment so it is centered to the origin:
A1(A.X - mx, A.Y - my)
B1(B.X - mx, B.Y - my)
Now we can perform our rotation of A1 and B1 (you know how).
Then we have to translate again to the original position:
A = (rotatedA1.X + mx, rotatedA1.y + my)
B = (rotatedB1.X + mx, rotatedB1.y + my)
If instead of having two points you have n points you have of course do everything for n points.
You could use Qt Graphics View which does all the geometric calculations for you.
Or are you just wanting to learn basic linear geometrical transformations? Then reading a math textbook would probably be more appropriate than coding.
I'm primarily a Flash AS3 dev, but I'm jumping into openframeworks and having trouble using 3D (these examples are in AS)
In 2D you can simulate an object orbiting a point by using Math.Sin() and Math.cos(), like so
function update(event:Event):void
{
dot.x = xCenter + Math.cos(angle*Math.PI/180) * range;
dot.y = yCenter + Math.sin(angle*Math.PI/180) * range;
angle+=speed;
}
I am wondering how I would translate this into a 3D orbit, if I wanted to also orbit in the third dimension.
function update(event:Event):void
{
...
dot.z = zCenter + Math.sin(angle*Math.PI/180) * range;
// is this valid?
}
An help is greatly appreciated.
If you are orbiting around the z-axis, you are leaving your z-coordinate fixed and changing your x- and y-coordinates. So your first code sample is what you are looking for.
To rotate around the x-axis (or y-axes), just replace x (or y) with z. Use Cos on whichever axis you want to be 0-degrees; the choice is arbitrary.
If what you actually want is to orbit an object around a point in 3d-space, you'll need two angles to describe the orbit: its elevation angle and its inclination angle. See here and here.
For reference, those equations are (where θ and φ are your angles)
x = x0 + r sin(θ) cos(φ)
y = y0 + r sin(θ) sin(φ)
z = z0 + r cos(θ)
If you are orbiting around Z axis, then you just do your first code, and leave Z coordinate as is.
I would pick two unit perpendicular vectors v, w that define the plane in which to orbit, then loop over the angle and pick the proper ratio of these vectors v and w to build your vector p = av + bw.
More details are coming.
EDIT:
This might be of help
http://en.wikipedia.org/wiki/Orbit_equation
EDIT: I think it is actually
center + sin(angle) * v * radius1 + cos(angle) * w * radius2
Here v and w are your unit vectors for the circle.
In 2D they were (1,0) and (0,1).
In 3D you will need to compute them - depends on orientation of the plane.
If you set radius1 = radius 2, you will get a circle. Otherwise, you should get an ellipse.
If you just want the orbit to happen at an angled plane and don't mind it being elliptic you can just do something like z = 0.2*x + 0.2*y, or any combination you fancy, after you have determined the x and y coordinates.