I have a 3D scene with an infinite horizontal plane (parallel to the xz coordinates) at a height H along the Y vertical axis.
I would like to know how to determine the intersection between the axis of my camera and this plane.
The camera is defined by a view-matrix and a projection-matrix.
There are two sub-problems here: 1) Extracting the position and view-direction from the camera matrix. 2) Calculating the intersection between the view-ray and the plane.
Extracting position and view-direction
The view matrix describes how points are transformed from world-space to view space. The view-space in OpenGL is usually defined such that the camera is in the origin and looks into the -z direction.
To get the position of the camera, we have to transform the origin [0,0,0] of the view-space back into world-space. Mathematically speaking, we have to calculate:
camera_pos_ws = inverse(view_matrix) * [0,0,0,1]
but when looking at the equation we'll see that we are only interrested in the 4th column of the inverse matrix which will contain 1
camera_pos_ws = [-view_matrix[12], -view_matrix[13], -view_matrix[14]]
The orientation of the camera can be found by a similar calculation. We know that the camera looks in -z direction in view-space thus the world space direction is given by
camera_dir_ws = inverse(view_matrix) * [0,0,-1,0];
Again, when looking at the equation, we'll see that this only takes the third row of the inverse matrix into account which is given by2
camera_dir_ws = [-view_matrix[2], -view_matrix[6], -view_matrix[10]]
Calculating the intersection
We now know the camera position P and the view direction D, thus we have to find the x,z value along the ray R(x,y,z) = P + l * D where y equals H. Since there is only one unknown, l, we can calculate that from
y = Py + l * Dy
H = Py + l * Dy
l = (H - Py) / Dy
The intersection point is then given by pasting l back into the ray equation.
Notes
1 The indices assume that the matrix is stored in a column-major linear array.
2 Note, that the inverse of a matrix of the form
M = [ R T ]
0 1
, where R is a orthogonal 3x3 matrix, is given by
inv(M) = [ transpose(R) -T ]
0 1
For a general line-plane intersection there are lot of answers and tutorials.
Your case is simple due to the plane is horizontal.
I suppose the camera is at C(cx, cy, cz) and it looks at T(tx, ty,tz).
Then the line camera-target can be defined by:
cx - x cy - y cz - z
------ = ------ = ------ /// These are two independant equations
tx - cx ty - cy tz - cz
For a horizontal plane, only a equation is needed: y = H.
Substitute this value in the line equations and you get
(cx-x)/(tx-cx) = (cy-H)/(ty-cy)
(cz-z)/(tz-cz) = (cy-H)/(ty-cy)
So
x = cx - (tx-cx)*(cy-H)/(ty-cy)
y = H
z = cz - (tz-cz)*(cy-H)/(ty-cy)
Of course if your camera looks in an also horizontal line then ty=cy and there is not solution.
Related
I have two questions. (I marked 1, 2 below)
In OpenGl, the clipping is done by sutherland-hodgman.
However, I wonder how to work sutherland-hodgman algorithm in homogeneous system (4D)
I made a situation.
In VCS, there is a line, R= (0, 3, -2, 1), S = (0, 0, 1, 1) (End points of the line)
And a frustum is right = 1, left = -1, near = 1, far = 3, top = 4, bottom = -4
Therefore, the projection matrix P is
1 0 0 0
0 1/4 0 0
0 0 -2 -3
0 0 -1 0
If we calculate the line with the P, then the each end points is like that
R' = (0, 3/4, 1, 2), S' = (0, 0, -5, -1)
I know that perspective division should not be done now, because if we do perspective division, the clipping result is not correct.
Here I am curious. What makes a correct clipping because we did not just do perspective division. What mathematical properties are here?
How to calculate the clipping result in above situation?
(The fact that two intersections occur in the w-y coordinate system makes me confused. I thought the result line is one, not divided two parts)
I'm not quite sure whether you understood the sutherland-hodgman algorithm correctly (or at least I didn't get your example). Thus I will prove here, that it doesn't make any difference whether clipping happens before or after the perspective divide. The proof is only shown for one plane (clipping has to be done against all 6 planes), since applying multiple such clipping operations after each other makes not difference here.
Let's assume we have two points (as you described) R' and S' in clip space. And we have a clipping plane P given in hessian normal form [n, p] (if we take the left plane this is [1,0,0,1]).
If we would be calculating in pure 3d space (R^3), then checking whether a line crosses this plane would be done by calculating the signed distance of both points to the plane and checking if the sign is different. The signed distance for a point X = [x/w,y/w,z/w] is given by
D = dot(n, X) + p
Let's write down the actual equation we have (including the perspective divide):
d = n_x * x/w + n_y * y/w + n_z * z/w + p
In order to find the exact intersection point, we would, again in R^3 space, calculate for both points (A = R'/R'w, B = S'/S'w) the distance to the plane (da, db) and perform a linear interpolation (I will only write the equations for the x-coordinate here since y and z are working similar):
x = A_x * (1 - da/(da - db)) + A_y * (da/(da-db))
x = R'x/R'w * (1 - da/(da - db)) + S'x/S'w * (da/(da-db))
And w = 1 (since we interpolate between two points both having w = 1)
Now we already know from the previous discussion, that clipping has to happen before the perspective divide, thus we have to adapt this equation. This means, that for each point, the clipping cube has a different scaling w. Lt's see what happens when we try to perform the the same operations in P^3 (before the perspective divide):
First, we "revert" the perspective divide to get to X=[x,y,z,w] for which the distance to the plane is given by
d = n_x * x/w + n_y * y/w + n_z * z/w + p
d = (n_x * x + n_y * y + n_z * z) / w + p
d * w = n_x * x + n_y * y + n_z * z + p * w
d * w = dot([n, p], [x,y,z,w])
d * w = dot(P, X)
Since we are only interested in the sign of the whole calculation, which we haven't changed by our operations, we can compare the D*ws and get the same inside-out result as in R^3.
For the two points R' and S', the calculated distances in P^3 are dr = da * R'w and ds = db * S'w. When we now use the same interpolation equation as above but for R' and S' we get for x:
x' = R'x * (1 - (da * R'w)/(da * R'w - db * S'w)) + S'x * (da * R'w)/(da * R'w - db * S'w)
On the first view this looks rather different from the result we got in R^3, but since we are still in P^3 (thus x'), we still have to do the perspective divide on the result (this is allowed here, since the interpolated point will always be at the border of the view-frustum and thus dividing by w will not introduce any problems). The interpolated w component is given as:
w' = R'w * (1 - (da * R'w)/(da * R'w - db * S'w)) + S'w * (da * R'w)/(da * R'w - db * S'w)
And when calculating x/w we get
x = x' / w';
x = R'x/R'w * (1 - da/(da - db)) + S'x/S'w * (da/(da-db))
which is exactly the same result as when calculating everything in R^3.
Conclusion: The interpolation gives the same result, no matter if we perform the perspective divide first and interpolation afterwards or interpolating first and dividing then. But with the second variant we avoid the problem with points flipping from behind the viewer to the front since we are only dividing points that are guaranteed to be inside (or on the border) of the viewing frustum.
You speak of polygon clipping in a homogeneous system (4D) but from your question I assume that you actually mean homogeneous coordinates, which makes a lot more sense. (There are many possible homogenous systems.)
Ok, so you want to use "4D" coordinates, which are really "3D coordinates and a w term". The w term represents (projection transformations) the projective term that partially relates the screen-space coordinate to the original world space position. Assuming that you are NOT interested in projective space clipping, this term is not relevant.
I'm assuming this because the clipping box you describe is axis-aligned on planes in 3D. Even if it was rotated or scaled in 3D space, each of the planes would still be a 3D plane, the 4th coordinate always being '1'.
So how to clip:
clip line segment L against each of the planes of the clipping box, i.e. 6 clipping planes in total (you describe the normals of each clipping plane aptly), and see if any intersection point v is shared by the line and the tested plane P so that
v lies on the line segment (i.e. a t between 0 and 1)
v lies within the bounds of the plane P (i.e. the coordinate should not lie beyond any of the adjacent planes. Since you are using axis-aligned clipping planes, this is easy to check.)
Any of these intersections between a (3D + w) line and one of the 3D planes occurs in 3D, and intersection points have to be a 3D coordinates. You can extend each of these coordinates with a 4th w coordinate into a "4D" coordinate so that you can further transform them using 4x4 matrices for view and projection processing.
Firstly, if you would like an explanation of the GLM lookAt algorithm, please look at the answer provided on this question: https://stackoverflow.com/a/19740748/1525061
mat4x4 lookAt(vec3 const & eye, vec3 const & center, vec3 const & up)
{
vec3 f = normalize(center - eye);
vec3 u = normalize(up);
vec3 s = normalize(cross(f, u));
u = cross(s, f);
mat4x4 Result(1);
Result[0][0] = s.x;
Result[1][0] = s.y;
Result[2][0] = s.z;
Result[0][1] = u.x;
Result[1][1] = u.y;
Result[2][1] = u.z;
Result[0][2] =-f.x;
Result[1][2] =-f.y;
Result[2][2] =-f.z;
Result[3][0] =-dot(s, eye);
Result[3][1] =-dot(u, eye);
Result[3][2] = dot(f, eye);
return Result;
}
Now I'm going to tell you why I seem to be having a conceptual issue with this algorithm. There are two parts to this view matrix, the translation and the rotation. The translation does the correct inverse transformation, bringing the camera position to the origin, instead of the origin position to the camera. Similarly, you expect the rotation that the camera defines to be inversed before being put into this view matrix as well. I can't see that happening here, that's my issue.
Consider the forward vector, this is where your camera looks at. Consequently, this forward vector needs to be mapped to the -Z axis, which is the forward direction used by openGL. The way this view matrix is suppose to work is by creating an orthonormal basis in the columns of the view matrix, so when you multiply a vertex on the right hand side of this matrix, you are essentially just converting it's coordinates to that of different axes.
When I play the rotation that occurs as a result of this transformation in my mind, I see a rotation that is not the inverse rotation of the camera, like what's suppose to happen, rather I see the non-inverse. That is, instead of finding the camera forward being mapped to the -Z axis, I find the -Z axis being mapped to the camera forward.
If you don't understand what I mean, consider a 2D example of the same type of thing that is happening here. Let's say the forward vector is (sqr(2)/2 , sqr(2)/2), or sin/cos of 45 degrees, and let's also say a side vector for this 2D camera is sin/cos of -45 degrees. We want to map this forward vector to (0,1), the positive Y axis. The positive Y axis can be thought of as the analogy to the -Z axis in openGL space. Let's consider a vertex in the same direction as our forward vector, namely (1,1). By using the logic of GLM.lookAt, we should be able to map (1,1) to the Y axis by using a 2x2 matrix that consists of the forward vector in the first column and the side vector in the second column. This is an equivalent calculation of that calculation http://www.wolframalpha.com/input/?i=%28sqr%282%29%2F2+%2C+sqr%282%29%2F2%29++1+%2B+%28sqr%282%29%2F2%2C+-sqr%282%29%2F2+%29+1.
Note that you don't get your (1,1) vertex mapped the positive Y axis like you wanted, instead you have it mapped to the positive X axis. You might also consider what happened to a vertex that was on the positive Y axis if you applied this transformation. Sure enough, it is transformed to the forward vector.
Therefore it seems like something very fishy is going on with the GLM algorithm. However, I doubt this algorithm is incorrect since it is so popular. What am I missing?
Have a look at GLU source code in Mesa: http://cgit.freedesktop.org/mesa/glu/tree/src/libutil/project.c
First in the implementation of gluPerspective, notice the -1 is using the indices [2][3] and the -2 * zNear * zFar / (zFar - zNear) is using [3][2]. This implies that the indexing is [column][row].
Now in the implementation of gluLookAt, the first row is set to side, the next one to up and the final one to -forward. This gives you the rotation matrix which is post-multiplied by the translation that brings the eye to the origin.
GLM seems to be using the same [column][row] indexing (from the code). And the piece you just posted for lookAt is consistent with the more standard gluLookAt (including the translational part). So at least GLM and GLU agree.
Let's then derive the full construction step by step. Noting C the center position and E the eye position.
Move the whole scene to put the eye position at the origin, i.e. apply a translation of -E.
Rotate the scene to align the axes of the camera with the standard (x, y, z) axes.
2.1 Compute a positive orthonormal basis for the camera:
f = normalize(C - E) (pointing towards the center)
s = normalize(f x u) (pointing to the right side of the eye)
u = s x f (pointing up)
with this, (s, u, -f) is a positive orthonormal basis for the camera.
2.2 Find the rotation matrix R that aligns maps the (s, u, -f) axes to the standard ones (x, y, z). The inverse rotation matrix R^-1 does the opposite and aligns the standard axes to the camera ones, which by definition means that:
(sx ux -fx)
R^-1 = (sy uy -fy)
(sz uz -fz)
Since R^-1 = R^T, we have:
( sx sy sz)
R = ( ux uy uz)
(-fx -fy -fz)
Combine the translation with the rotation. A point M is mapped by the "look at" transform to R (M - E) = R M - R E = R M + t. So the final 4x4 transform matrix for "look at" is indeed:
( sx sy sz tx ) ( sx sy sz -s.E )
L = ( ux uy uz ty ) = ( ux uy uz -u.E )
(-fx -fy -fz tz ) (-fx -fy -fz f.E )
( 0 0 0 1 ) ( 0 0 0 1 )
So when you write:
That is, instead of finding the camera forward being mapped to the -Z
axis, I find the -Z axis being mapped to the camera forward.
it is very surprising, because by construction, the "look at" transform maps the camera forward axis to the -z axis. This "look at" transform should be thought as moving the whole scene to align the camera with the standard origin/axes, it's really what it does.
Using your 2D example:
By using the logic of GLM.lookAt, we should be able to map (1,1) to the Y
axis by using a 2x2 matrix that consists of the forward vector in the
first column and the side vector in the second column.
That's the opposite, following the construction I described, you need a 2x2 matrix with the forward and row vector as rows and not columns to map (1, 1) and the other vector to the y and x axes. To use the definition of the matrix coefficients, you need to have the images of the standard basis vectors by your transform. This gives directly the columns of the matrix. But since what you are looking for is the opposite (mapping your vectors to the standard basis vectors), you have to invert the transformation (transpose, since it's a rotation). And your reference vectors then become rows and not columns.
These guys might give some further insights to your fishy issue:
glm::lookAt vertical camera flips when z <= 0
The answer might be of interest to you?
for a project I need to compute the real world position and orientation of a camera
with respect to a known object.
I have a set of photos, each displays a chessboard from different points of view.
Using CalibrateCamera and solvePnP I am able to reproject Points in 2d, to get a AR-thing.
So my situation is as such:
Intrinsic parameters are known
Distortioncoefficients are known
translation Vector and rotation Vector are known per photo.
I simply cannot figure out how to compute the position of the camera. My guess was:
invert translation vector. (=t')
transform rotation vector to degree (seems to be radian) and invert
use rodriguez on rotation vector
compute RotationMatrix * t'
But the results are somehow totally off...
Basically I want to to compute a ray for each pixel in world coordinates.
If more informations on my problem are needed, I'd be glad to answer quickly.
I dont' get it... somehow the rays are still off. This is my Code btw:
Mat image1CamPos = tvecs[0].clone(); //From calibrateCamera
Mat rot = rvecs[0].clone(); //From calibrateCamera
Rodrigues(rot, rot);
rot = rot.t();
//Position of Camera
Mat pos = rot * image1CamPos;
//Ray-Normal (( (double)mk[i][k].x) are known image-points)
float x = (( (double)mk[i][0].x) / fx) - (cx / fx);
float y = (( (double)mk[i][0].y) / fy) - (cy / fy);
float z = 1;
float mag = sqrt(x*x + y*y + z*z);
x /= mag;
y /= mag;
z /= mag;
Mat unit(3, 1, CV_64F);
unit.at<double>(0, 0) = x;
unit.at<double>(1, 0) = y;
unit.at<double>(2, 0) = z;
//Rotation of Ray
Mat rot = stof1 * unit;
But when plotting this, the rays are off :/
The translation t (3x1 vector) and rotation R (3x3 matrix) of an object with respect to the camera equals the coordinate transformation from object into camera space, which is given by:
v' = R * v + t
The inversion of the rotation matrix is simply the transposed:
R^-1 = R^T
Knowing this, you can easily resolve the transformation (first eq.) to v:
v = R^T * v' - R^T * t
This is the transformation from camera into object space, i.e., the position of the camera with respect to the object (rotation = R^T and translation = -R^T * t).
You can simply get a 4x4 homogeneous transformation matrix from this:
T = ( R^T -R^T * t )
( 0 1 )
If you now have any point in camera coordinates, you can transform it into object coordiantes:
p' = T * (x, y, z, 1)^T
So, if you'd like to project a ray from a pixel with coordinates (a,b) (probably you will need to define the center of the image, i.e. the principal point as reported by CalibrateCamera, as (0,0)) -- let that pixel be P = (a,b)^T. Its 3D coordinates in camera space are then P_3D = (a,b,0)^T. Let's project a ray 100 pixel in positive z-direction, i.e. to the point Q_3D = (a,b,100)^T. All you need to do is transform both 3D coordinates into the object coordinate system using the transformation matrix T and you should be able to draw a line between both points in object space. However, make sure that you don't confuse units: CalibrateCamera will report pixel values while your object coordinate system might be defined in, e.g., cm or mm.
I have a point in 3D space and two angles, I want to calculate the resulting line from this information. I have found how to do this with 2D lines, but not 3D. How can this be calculated?
If it helps: I'm using C++ & OpenGL and have the location of the user's mouse click and the angle of the camera, I want to trace this line for intersections.
In trig terms two angles and a point are required to define a line in 3d space. Converting that to (x,y,z) is just polar coordinates to cartesian coordinates the equations are:
x = r sin(q) cos(f)
y = r sin(q) sin(f)
z = r cos(q)
Where r is the distance from the point P to the origin; the angle q (zenith) between the line OP and the positive polar axis (can be thought of as the z-axis); and the angle f (azimuth) between the initial ray and the projection of OP onto the equatorial plane(usually measured from the x-axis).
Edit:
Okay that was the first part of what you ask. The rest of it, the real question after the updates to the question, is much more complicated than just creating a line from 2 angles and a point in 3d space. This involves using a camera-to-world transformation matrix and was covered in other SO questions. For convenience here's one: How does one convert world coordinates to camera coordinates? The answers cover converting from world-to-camera and camera-to-world.
The line can be fathomed as a point in "time". The equation must be vectorized, or have a direction to make sense, so time is a natural way to think of it. So an equation of a line in 3 dimensions could really be three two dimensional equations of x,y,z related to time, such as:
x = ax*t + cx
y = ay*t + cy
z = az*t + cz
To find that set of equations, assuming the camera is at origin, (0,0,0), and your point is (x1,y1,z1) then
ax = x1 - 0
ay = y1 - 0
az = z1 - 0
cx = cy = cz = 0
so
x = x1*t
y = y1*t
z = z1*t
Note: this also assumes that the "speed" of the line or vector is such that it is at your point (x1,y1,z1) after 1 second.
So to draw that line just fill in the points as fine as you like for as long as required, such as every 1/1000 of a second for 10 seconds or something, might draw a "line", really a series of points that when seen from a distance appear as a line, over 10 seconds worth of distance, determined by the "speed" you choose.
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.