I created a 4-point bezier curve. I knew the total bezier curve length using this link. And I knew the length from start point.
I want to know how to get a time value from bezier curve and a point. I found a similar question and divided the bezier curve into 1000 pieces; but it isn't a good solution.
How can I get t value?
Note that for a cubic Bezier curve, there is no "one t value for each coordinate". Cubic Bezier can self-intersect, so you can find multiple t values for a single coordinate. There's two ways to do this: approximately or symbolically.
If you want an approximate answer (like what you're already doing for the length computation), simply construct a lookup table of coordinates-for-t:
buildLUT(a,b,c,d) {
for(t=0; t<=1; t+=0.01) {
LUTx[t*100] = getCoordinate(t, a.x,b.x,c.x,d.x);
LUTy[t*100] = getCoordinate(t, a.y,b.y,c.y,d.y);
}
}
And write an extra function for reverse lookups, or to build the reverse LUTs:
findTforCoordinate(x, y) {
found = []
for(i=0, len=LUTx.length; i<len; i++) {
_x = LUTx[i], _y = LUTy[i]
if(x==_x && y==_y) { found.push(i/len); }
}
return found
}
where a,b,c,d are your curve's control points. Since this is approximate, you're not looking for "t value for coordinate" but "closest t value to coordinate". It won't be perfect.
What WILL be perfect is finding all possible t values for the x and y coordinate components, then finding the one or two t values out of the set of possible six that approach generates that are the same between the x and y solutions. You can do this by using Cardano's approach, which is explain in another stackoverflow question here: Cubic Bezier reverse GetPoint equation: float for Vector <=> Vector for float
Related
I'm currently developing an application that will alert users of incoming rain. To do this I want to check certain area around user location for rainfall (different pixel colours for intensity on rainfall radar image). I would like the checked area to be a circle but I don't know how to do this efficiently.
Let's say I want to check radius of 50km. My current idea is to take subset of image with size 100kmx100km (user+50km west, user+50km east, user+50km north, user+50km south) and then check for each pixel in this subset if it's closer to user than 50km.
My question here is, is there a better solution that is used for this type of problems?
If the occurrence of the event you are searching for (rain or anything) is relatively rare, then there's nothing wrong with scanning a square or pixels and then, only after detecting rain in that square, checking whether that rain is within the desired 50km circle. Note that the key point here is that you don't need to check each pixel of the square for being inside the circle (that would be very inefficient), you have to search for your event (rain) first and only when you found it, check whether it falls into the 50km circle. To implement this efficiently you also have to develop some smart strategy for handling multi-pixel "stains" of rain on your image.
However, since you are scanning a raster image, you can easily implement the well-known Bresenham circle algorithm to find the starting and the ending point of the circle for each scan line. That way you can easily limit your scan to the desired 50km radius.
On the second thought, you don't even need the Bresenham algorithm for that. For each row of pixels in your square, calculate the points of intersection of that row with the 50km circle (using the usual schoolbook formula with square root), and then check all pixels that fall between these intersection points. Process all rows in the same fashion and you are done.
P.S. Unfortunately, the Wikipedia page I linked does not present Bresenham algorithm at all. It has code for Michener circle algorithm instead. Michener algorithm will also work for circle rasterization purposes, but it is less precise than Bresenham algorithm. If you care for precision, find a true Bresenham on somewhere. It is actually surprisingly diffcult to find on the net: most search hits erroneously present Michener as Bresenham.
There is, you can modify the midpoint circle algorithm to give you an array of for each y, the x coordinate where the circle starts (and ends, that's the same thing because of symmetry). This array is easy to compute, pseudocode below.
Then you can just iterate over exactly the right part, without checking anything.
Pseudo code:
data = new int[radius];
int f = 1 - radius, ddF_x = 1;
int ddF_y = -2 * radius;
int x = 0, y = radius;
while (x < y)
{
if (f >= 0)
{
y--;
ddF_y += 2; f += ddF_y;
}
x++;
ddF_x += 2; f += ddF_x;
data[radius - y] = x; data[radius - x] = y;
}
Maybe you can try something that will speed up your algorithm.
In brute force algorithm you will probably use equation:
(x-p)^2 + (y-q)^2 < r^2
(p,q) - center of the circle, user position
r - radius (50km)
If you want to find all pixels (x,y) that satisfy above condition and check them, your algorithm goes to O(n^2)
Instead of scanning all pixels in this circle I will check only only pixels that are on border of the circle.
In that case, you can use some more clever way to define circle.
x = p+r*cos(a)
y = q*r*sin(a)
a - angle measured in radians [0-2pi]
Now you can sample some angles, for example twenty of them, iterate and find all pairs (x,y) that are border for radius 50km. Now check are they on the rain zone and alert user.
For more safety I recommend you to use multiple radians (smaller than 50km), because your whole rain cloud can be inside circle, and your app will not recognize him. For example use 3 incircles (r = 5km, 15km, 30km) and do same thing. Efficiency of this algorithm only depends on number of angles and number of incircles.
Pseudocode will be:
checkRainDanger()
p,q <- position
radius[] <- array of radii
for c = 1 to length(radius)
a=0
while(a<2*pi)
x = p + radius[c]*cos(a)
y = q + radius[c]*sin(a)
if rainZone(x,y)
return true
else
a+=pi/10
end_while
end_for
return false //no danger
r2=r*r
for x in range(-r, +r):
max_y=sqrt(r2-x*x)
for y in range(-max_y, +max_y):
# x,y is in range - check for rain
I have two sets of 3D points (original and reconstructed) and correspondence information about pairs - which point from one set represents the second one. I need to find 3D translation and scaling factor which transforms reconstruct set so the sum of square distances would be least (rotation would be nice too, but points are rotated similarly, so this is not main priority and might be omitted in sake of simplicity and speed). And so my question is - is this solved and available somewhere on the Internet? Personally, I would use least square method, but I don't have much time (and although I'm somewhat good at math, I don't use it often, so it would be better for me to avoid it), so I would like to use other's solution if it exists. I prefer solution in C++, for example using OpenCV, but algorithm alone is good enough.
If there is no such solution, I will calculate it by myself, I don't want to bother you so much.
SOLUTION: (from your answers)
For me it's Kabsch alhorithm;
Base info: http://en.wikipedia.org/wiki/Kabsch_algorithm
General solution: http://nghiaho.com/?page_id=671
STILL NOT SOLVED:
I also need scale. Scale values from SVD are not understandable for me; when I need scale about 1-4 for all axises (estimated by me), SVD scale is about [2000, 200, 20], which is not helping at all.
Since you are already using Kabsch algorithm, just have a look at Umeyama's paper which extends it to get scale. All you need to do is to get the standard deviation of your points and calculate scale as:
(1/sigma^2)*trace(D*S)
where D is the diagonal matrix in SVD decomposition in the rotation estimation and S is either identity matrix or [1 1 -1] diagonal matrix, depending on the sign of determinant of UV (which Kabsch uses to correct reflections into proper rotations). So if you have [2000, 200, 20], multiply the last element by +-1 (depending on the sign of determinant of UV), sum them and divide by the standard deviation of your points to get scale.
You can recycle the following code, which is using the Eigen library:
typedef Eigen::Matrix<double, 3, 1, Eigen::DontAlign> Vector3d_U; // microsoft's 32-bit compiler can't put Eigen::Vector3d inside a std::vector. for other compilers or for 64-bit, feel free to replace this by Eigen::Vector3d
/**
* #brief rigidly aligns two sets of poses
*
* This calculates such a relative pose <tt>R, t</tt>, such that:
*
* #code
* _TyVector v_pose = R * r_vertices[i] + t;
* double f_error = (r_tar_vertices[i] - v_pose).squaredNorm();
* #endcode
*
* The sum of squared errors in <tt>f_error</tt> for each <tt>i</tt> is minimized.
*
* #param[in] r_vertices is a set of vertices to be aligned
* #param[in] r_tar_vertices is a set of vertices to align to
*
* #return Returns a relative pose that rigidly aligns the two given sets of poses.
*
* #note This requires the two sets of poses to have the corresponding vertices stored under the same index.
*/
static std::pair<Eigen::Matrix3d, Eigen::Vector3d> t_Align_Points(
const std::vector<Vector3d_U> &r_vertices, const std::vector<Vector3d_U> &r_tar_vertices)
{
_ASSERTE(r_tar_vertices.size() == r_vertices.size());
const size_t n = r_vertices.size();
Eigen::Vector3d v_center_tar3 = Eigen::Vector3d::Zero(), v_center3 = Eigen::Vector3d::Zero();
for(size_t i = 0; i < n; ++ i) {
v_center_tar3 += r_tar_vertices[i];
v_center3 += r_vertices[i];
}
v_center_tar3 /= double(n);
v_center3 /= double(n);
// calculate centers of positions, potentially extend to 3D
double f_sd2_tar = 0, f_sd2 = 0; // only one of those is really needed
Eigen::Matrix3d t_cov = Eigen::Matrix3d::Zero();
for(size_t i = 0; i < n; ++ i) {
Eigen::Vector3d v_vert_i_tar = r_tar_vertices[i] - v_center_tar3;
Eigen::Vector3d v_vert_i = r_vertices[i] - v_center3;
// get both vertices
f_sd2 += v_vert_i.squaredNorm();
f_sd2_tar += v_vert_i_tar.squaredNorm();
// accumulate squared standard deviation (only one of those is really needed)
t_cov.noalias() += v_vert_i * v_vert_i_tar.transpose();
// accumulate covariance
}
// calculate the covariance matrix
Eigen::JacobiSVD<Eigen::Matrix3d> svd(t_cov, Eigen::ComputeFullU | Eigen::ComputeFullV);
// calculate the SVD
Eigen::Matrix3d R = svd.matrixV() * svd.matrixU().transpose();
// compute the rotation
double f_det = R.determinant();
Eigen::Vector3d e(1, 1, (f_det < 0)? -1 : 1);
// calculate determinant of V*U^T to disambiguate rotation sign
if(f_det < 0)
R.noalias() = svd.matrixV() * e.asDiagonal() * svd.matrixU().transpose();
// recompute the rotation part if the determinant was negative
R = Eigen::Quaterniond(R).normalized().toRotationMatrix();
// renormalize the rotation (not needed but gives slightly more orthogonal transformations)
double f_scale = svd.singularValues().dot(e) / f_sd2_tar;
double f_inv_scale = svd.singularValues().dot(e) / f_sd2; // only one of those is needed
// calculate the scale
R *= f_inv_scale;
// apply scale
Eigen::Vector3d t = v_center_tar3 - (R * v_center3); // R needs to contain scale here, otherwise the translation is wrong
// want to align center with ground truth
return std::make_pair(R, t); // or put it in a single 4x4 matrix if you like
}
For 3D points the problem is known as the Absolute Orientation problem. A c++ implementation is available from Eigen http://eigen.tuxfamily.org/dox/group__Geometry__Module.html#gab3f5a82a24490b936f8694cf8fef8e60 and paper http://web.stanford.edu/class/cs273/refs/umeyama.pdf
you can use it via opencv by converting the matrices to eigen with cv::cv2eigen() calls.
Start with translation of both sets of points. So that their centroid coincides with the origin of the coordinate system. Translation vector is just the difference between these centroids.
Now we have two sets of coordinates represented as matrices P and Q. One set of points may be obtained from other one by applying some linear operator (which performs both scaling and rotation). This operator is represented by 3x3 matrix X:
P * X = Q
To find proper scale/rotation we just need to solve this matrix equation, find X, then decompose it into several matrices, each representing some scaling or rotation.
A simple (but probably not numerically stable) way to solve it is to multiply both parts of the equation to the transposed matrix P (to get rid of non-square matrices), then multiply both parts of the equation to the inverted PT * P:
PT * P * X = PT * Q
X = (PT * P)-1 * PT * Q
Applying Singular value decomposition to matrix X gives two rotation matrices and a matrix with scale factors:
X = U * S * V
Here S is a diagonal matrix with scale factors (one scale for each coordinate), U and V are rotation matrices, one properly rotates the points so that they may be scaled along the coordinate axes, other one rotates them once more to align their orientation to second set of points.
Example (2D points are used for simplicity):
P = 1 2 Q = 7.5391 4.3455
2 3 12.9796 5.8897
-2 1 -4.5847 5.3159
-1 -6 -15.9340 -15.5511
After solving the equation:
X = 3.3417 -1.2573
2.0987 2.8014
After SVD decomposition:
U = -0.7317 -0.6816
-0.6816 0.7317
S = 4 0
0 3
V = -0.9689 -0.2474
-0.2474 0.9689
Here SVD has properly reconstructed all manipulations I performed on matrix P to get matrix Q: rotate by the angle 0.75, scale X axis by 4, scale Y axis by 3, rotate by the angle -0.25.
If sets of points are scaled uniformly (scale factor is equal by each axis), this procedure may be significantly simplified.
Just use Kabsch algorithm to get translation/rotation values. Then perform these translation and rotation (centroids should coincide with the origin of the coordinate system). Then for each pair of points (and for each coordinate) estimate Linear regression. Linear regression coefficient is exactly the scale factor.
A good explanation Finding optimal rotation and translation between corresponding 3D points
The code is in matlab but it's trivial to convert to opengl using the cv::SVD function
You might want to try ICP (Iterative closest point).
Given two sets of 3d points, it will tell you the transformation (rotation + translation) to go from the first set to the second one.
If you're interested in a c++ lightweight implementation, try libicp.
Good luck!
The general transformation, as well the scale can be retrieved via Procrustes Analysis. It works by superimposing the objects on top of each other and tries to estimate the transformation from that setting. It has been used in the context of ICP, many times. In fact, your preference, Kabash algorithm is a special case of this.
Moreover, Horn's alignment algorithm (based on quaternions) also finds a very good solution, while being quite efficient. A Matlab implementation is also available.
Scale can be inferred without SVD, if your points are uniformly scaled in all directions (I could not make sense of SVD-s scale matrix either). Here is how I solved the same problem:
Measure distances of each point to other points in the point cloud to get a 2d table of distances, where entry at (i,j) is norm(point_i-point_j). Do the same thing for the other point cloud, so you get two tables -- one for original and the other for reconstructed points.
Divide all values in one table by the corresponding values in the other table. Because the points correspond to each other, the distances do too. Ideally, the resulting table has all values being equal to each other, and this is the scale.
The median value of the divisions should be pretty close to the scale you are looking for. The mean value is also close, but I chose median just to exclude outliers.
Now you can use the scale value to scale all the reconstructed points and then proceed to estimating the rotation.
Tip: If there are too many points in the point clouds to find distances between all of them, then a smaller subset of distances will work, too, as long as it is the same subset for both point clouds. Ideally, just one distance pair would work if there is no measurement noise, e.g when one point cloud is directly derived from the other by just rotating it.
you can also use ScaleRatio ICP proposed by BaoweiLin
The code can be found in github
I am trying to implement Bezier Curves for an assignment. I am trying to move a ball (using bezier curves) by giving my function an array of key frames. The function should give me all the frames in between the key frames ... or control points ... but although I'm using the formula found on wikipedia... it is not really working :s
her's my code:
private void interpolate(){
float x,y,b, t = 0;
frames = new Frame[keyFrames.length];
for(int i =0;i<keyFrames.length;++i){
t+=0.001;
b = Bint(i,keyFrames.length,t);
x = b*keyFrames[i].x;
y = b*keyFrames[i].y;
frames[i] = new Frame(x,y);
}
}
private float Bint(int i, int n, float t){
float Cni = fact(n)/(fact(i) * fact(n-i));
return Cni * pow(1-t,n-i) * pow(t,i);
}
Also I've noticed that the frames[] array should be much bigger but I can't find any other text which is more programmer friendly
Thanks in advance.
There are lots of things that don't look quite right here.
Doing it this way, your interpolation will pass exactly through the first and last control points, but not through the others. Is that what you want?
If you have lots of key frames, you're using a very-high-degree polynomial for your interpolation. Polynomials of high degree are notoriously badly-behaved, you may get your position oscillating wildly in between the key frame positions. (This is one reason why the answer to question 1 should probably be no.)
Assuming for the sake of argument that you really do want to do this, your value of t should go from 0 at the start to 1 at the end. Do you happen to have exactly 1001 of these key frames? If not, you'll be doing the wrong thing.
Evaluating these polynomials with lots of calls to fact and pow is likely to be inefficient, especially if n is large.
I'm reluctant to go into much detail about what you should do without knowing more about the scope of your assignment -- it will do no one any good for Stack Overflow to do your homework for you! What have you already been told about Bezier curves? What exactly does your assignment ask you to do?
EDITED to add:
The simplest way to do interpolation using Bezier curves is probably this. Have one (cubic) Bezier curve between each pair of key-points. The endpoints (first and last control points) of each Bezier curve are those keypoints. You need two more control points. For motion to be smooth as you move through a given keypoint, you need (keypoint minus previous control point) = (next control point minus keypoint). So you're choosing a single vector at each keypoint, which will determine where the previous and subsequent control points go. As you move through each keypoint, you'll be moving in the direction of that vector, and the longer the vector is the faster you'll be moving. (If the vector is zero then your cubic Bezier degenerates into a simple straight-line path.)
Choosing that vector so that everything looks nice is highly nontrivial, but you probably aren't really being asked to do that at this stage. So something pretty simple will probably be good enough. You might, e.g., take the vector to be proportional to (next keypoint minus previous keypoint). You'll need to do something a bit different at the start and end of your path if you do that.
Finally got What I needed! Here's what I did:
private void interpolate() {
float t = 0;
float x,y,b;
for(int f =0;f<frames.length;f++) {
x=0;
y=0;
for(int i = 0; i<keyFrames.length; i++) {
b = Bint(i,keyFrames.length-1,map(t,0,time,0,1));
x += b*keyFrames[i].x;
y += b*keyFrames[i].y;
}
frames[f] = new Frame(x,y);
t+=partialTime;
}
}
private void createInterpolationData() {
time = keyFrames[keyFrames.length-1].time -
keyFrames[0].time;
noOfFrames = 60*time;
partialTime = time/noOfFrames;
frames = new Frame[ceil(noOfFrames)];
}
I have a set of data points:
(x1, y1) (x2, y2) (x3, y3) ... (xn, yn)
The number of sample points can be thousands. I want to represent the same curve as accurately as possible with minimal (lets suppose 30) set of points. I want to capture as many inflection points as possible. However, I have a hard limit on the number of allowed points to represent the data.
What is the best algorithm to achieve the same? Is there any free software library that can help?
PS: I have tried to implement relative slope difference based point elimination, but this does not always result in the best possible data representation.
You are searching for an interpolation algorithm. Is your set of points a function in a mathematical sense (all x values are disjunct from each other) then you can go for a polynomial interpolation, or are they distributed over the 2d plane, then you could use bezier curves.
Late answer after years:
Have a look at the Douglas-Peucker algorithm:
function DouglasPeucker(PointList[], epsilon)
// Find the point with the maximum distance
dmax = 0
index = 0
end = length(PointList)
for i = 2 to ( end - 1) {
d = perpendicularDistance(PointList[i], Line(PointList[1], PointList[end]))
if ( d > dmax ) {
index = i
dmax = d
}
}
// If max distance is greater than epsilon, recursively simplify
if ( dmax > epsilon ) {
// Recursive call
recResults1[] = DouglasPeucker(PointList[1...index], epsilon)
recResults2[] = DouglasPeucker(PointList[index...end], epsilon)
// Build the result list
ResultList[] = {recResults1[1...length(recResults1)-1], recResults2[1...length(recResults2)]}
} else {
ResultList[] = {PointList[1], PointList[end]}
}
// Return the result
return ResultList[]
end
It is frequently used to simplify GPS tracks and reduce the number of waypoints. As a preparation, you may have to sort your points to store neighbour points adjacent in your list or array.
it depends on must your curve intersect each point or it is approximation. Try:
Take points
Apply any interpolation (http://en.wikipedia.org/wiki/Polynomial_interpolation) to get equation of curve
Then take sample points with specific step.
I would like to get some code in AS2 to interpolate a quadratic bezier curve. the nodes are meant to be at constant distance away from each other. Basically it is to animate a ball at constant speed along a non-hyperbolic quadratic bezier curve defined by 3 pts.
Thanks!
The Bezier curve math is really quite simple, so I'll help you out with that and you can translate it into ActionScript.
A 2D quadratic Bezier curve is defined by three (x,y) coordinates. I will refer to these as P0 = (x0,y0), P1 = (x1,y1) and P2 = (x2,y2). Additionally a parameter value t, which ranges from 0 to 1, is used to indicate any position along the curve. All x, y and t variables are real-valued (floating point).
The equation for a quadratic Bezier curve is:
P(t) = P0*(1-t)^2 + P1*2*(1-t)*t + P2*t^2
So, using pseudocode, we can smoothly trace out the Bezier curve like so:
for i = 0 to step_count
t = i / step_count
u = 1 - t
P = P0*u*u + P1*2*u*t + P2*t*t
draw_ball_at_position( P )
This assumes that you have already defined the points P0, P1 and P2 as above. If you space the control points evenly then you should get nice even steps along the curve. Just define step_count to be the number of steps along the curve that you would like to see.
Please note that the expression can be done much more efficient mathematically.
P(t) = P0*(1-t)^2 + P1*2*(1-t)*t + P2*t^2
and
P = P0*u*u + P1*2*u*t + P2*t*t
both hold t multiplications which can be simplified.
For example:
C = A*t + B(1-t) = A*t + B - B*t = t*(A-B) + B = You saved one multiplication = Double performance.
The solution proposed by Naaff, that is P(t) = P0*(1-t)^2 + P1*2*(1-t)*t + P2*t^2, will get you the correct "shape", but selecting evenly-spaced t in the [0:1] interval will not produce evenly-spaced P(t). In other words, the speed is not constant (you can differentiate the previous equation with respect to t to see see it).
Usually, a common method to traverse a parametric curve at constant-speed is to reparametrize by arc-length. This means expressing P as P(s) where s is the length traversed along the curve. Obviously, s varies from zero to the total length of the curve. In the case of a quadratic bezier curve, there's a closed-form solution for the arc-length as a function of t, but it's a bit complicated. Computationally, it's often faster to just integrate numerically using your favorite method. Notice however that the idea is to compute the inverse relation, that is, t(s), so as to express P as P(t(s)). Then, choosing evenly-spaced s will produce evenly-space P.