I have a problem. Suppose we have a single cubic bezier curve defined by four control points. Now suppose, the curve is cut from a point and each segment is again represented using cubic bezier curves. So, now if we are given two such beziers B1 and B2, is there a way to know if they can be joined to form another bezier curve B? This is to simplify the geometry by joining two curves and reduce the number of control points.
Some thoughts about this problem.
I suggest there was initial Bezier curve P0-P3 with control points P1 and P2
Let's make two subdivisions at parameters ta and tb.
We have now two subcurves (in yellow) - P0-PA3 and PB0-P3.
Blue interval is lost.
PA1 and PB2 - known control points. We have to find unknown P1 and P2.
Some equations
Initial curve:
C = P0*(1-t)^3+3*P1(1-t)^2*t+3*P2*(1-t)*t^2+P3*t^3
Endpoints:
PA3 = P0*(1-ta)^3+3*P1*(1-ta)^2*ta+3*P2*(1-ta)*ta^2+P3*ta^3
PB0 = P0*(1-tb)^3+3*P1*(1-tb)^2*tb+3*P2*(1-tb)*tb^2+P3*tb^3
Control points of small curves
PA1 = P0*(1-ta)+P1*ta => P1*ta = PA1 – P0*(1-ta)
PB2 = P2*(1-tb)+P3*tb => P2(1-tb) = PB2 – P3*tb
Now substitute unknown points in PA3 equation:
**PA3***(1-tb) = **P0***(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(**PA1** – **P0***(1-ta))+3*(1-ta)*ta^2*( **PB2** – **P3***tb)+**P3***ta^3*(1-tb)
(some multiplication signs have been lost due to SO formatting)
This is vector equation, it contains two scalar equations for two unknowns ta and tb
PA3X*(1-tb) = P0X*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1X – P0X*(1-ta))+3*(1-ta)*ta^2*( PB2X – P3X*tb)+P3X*ta^3*(1-tb)
PA3Y*(1-tb) = P0Y*(1-ta)^3*(1-tb)+3*(1-ta)^2*(1-tb)*(PA1Y – P0Y*(1-ta))+3*(1-ta)*ta^2*( PB2Y – P3Y*tb)+P3Y*ta^3*(1-tb)
This system might be solved both numerically and analytically (indeed Maple solves it with very-very big cubic formula :( )
If we have points with some error, that makes sense to build overdetermined equation system for some points (PA3, PB0, PA2, PB1) and solve it numerically to minimize deviations.
You will find a quite simple solution here: https://math.stackexchange.com/a/879213/65203.
When you split a Bezier, the vectors formed by the last two control points of the first section and the first two control points of the second section are collinear and the ratio of their lengths leads to the value of the parameter at the split. Verifying that the common control point matches that value of the parameter is an easy matter (to avoid the case of accidental collinearity).
Related
Many questions deal with generating normal from depth or depth from normal, but I want to ask about a simple way to generate all the planar surfaces given the depth and normal of an image.
I already have depth and normal of each pixel in the image. For each pixel (ui, vi), assume that we can get its 3D coordinates (xi, yi, zi) with zi as the depth and normal vector (nix, niy, niz). Thus, a unique tangent plane is defined by: nix(x - xi) + niy(y - yi) + niz(z - zi) = 0. Then, for each pixel we can define a unique planar surface by the above equation.
What is a common practice in finding the function f such that f(u, v) = (x, y, z) (from pixel to 3D coordinates)? Is pinhole model (plus the depth data) an effective and accurate one?
How does one generate all the planar surfaces effectively? One way is to iterate through all the pixels in the image and find all the planes, but this seems like an ineffective method.
If its pinhole model
make sure your 3D data is not distorted by projection.
group your points by normal
this is easy or hard depending on the points/normal accuracy. Simply sort the points by normals which leads to O(n.log(n)) where n is number of points.
test/group by planes in single normal group
The idea is to pick 3 points from a group compute plane from it and test which points of the group belongs to it. If too low count you got wrong points picked (not belonging to the same plane) and need to pick different ones. Also if the picked points are too close to each or on the same line you can not get correct plane from it.
The math function for plane is:
x*nx + y*ny + z*nz + d = 0
where (nx,ny,nz) is your normal of the group (unit vector) and (x,y,z) is your point position. So you just compute d from a known point (one of the picked ones (x0,y0,z0) ) ...
d = -x0*nx -y0*ny -z0*nz
and then just test which points are sattisfying this condition:
threshod=1e-20; // just accuracy margin
fabs(x*nx + y*ny + z*nz + d) <= threshod
now remove matched points from the group (move them into found plane object) and apply this bullet again on the remaining points until they count is low or no valid plane is found...
then test another group until no groups are left...
I think RANSAC can speed things up to avoid brute force in this case but never used it myself so google ...
A possible approach for the planes is to consider the set of normal vectors and perform clustering on them (for instance by k-means). Then every cluster can correspond to several parallel surfaces. By evaluating the distance from the origin (a scalar function), you can form sub-clusters which will separate those surfaces. Finally, points at constant distance can belong to different coplanar patches, which you can separate by connected component labelling.
It is likely that clustering on the normal vectors and distance simultaneously (hence in a 4D space) will yield better results and be simpler. Be sure to normalize the vectors. Another option is to represent the vectors by just two parameters (such as spherical angles), but this will lead to a quite non-uniform mapping, and create phase wrapping issues.
I start this thread asking for your help in Excel.
The main goal is to determine the coordinates of the intersection point P=(x,y) between two curves (curve A, curve B) modeled by points.
The curves are non-linear and each defining point is determined using complex equations (equations are dependent by a lot of parameters chosen by user, as well as user will choose the number of points which will define the accuracy of the curves). That is to say that each curve (curve A and curve B) is always changing in the plane XY (Z coordinate is always zero, we are working on the XY plane) according to the input parameters and the number of the defining points is also depending by the user choice.
My first attempt was to determine the intersection point through the trend equations of each curve (I used the LINEST function to determine the coefficients of the polynomial equation) and by solving the solution putting them into a system. The problem is that Excel is not interpolating very well the curves because they are too wide, then the intersection point (the solution of the system) is very far from the real solution.
Then, what I want to do is to shorten the ranges of points to be able to find two defining trend equations for the curves, cutting away the portion of curves where cannot exist the intersection.
Today, in order to find the solution, I plot the curves on Siemens NX cad using multi-segment splines with order 3 and then I can easily find the coordinates of the intersection point. Please notice that I am using the multi-segment splines to be more precise with the approximation of the functions curve A and curve B.
Since I want to avoid the CAD tool and stay always on Excel, is there a way to select a shorter range of the defining points close to the intersection point in order to better approximate curve A and curve B with trend equations (Linest function with 4 points and 3rd order spline) and then find the solution?
I attach a picture to give you an example of Curve A and Curve B on the plane:
https://postimg.cc/MfnKYqtk
At the following link you can find the Excel file with the coordinate points and the curve plot:
https://www.mediafire.com/file/jqph8jrnin0i7g1/intersection.xlsx/file
I hope to solve this problem with your help, thank you in advance!
kalo86
Your question gave me some days of thinking and research.
With the help of https://pomax.github.io/bezierinfo/
§ 27 - Intersections (Line-line intersections)
and
§ 28 - Curve/curve intersection
your problem can be solved in Excel.
About the mystery of Excel smoothed lines you find details here:
https://blog.splitwise.com/2012/01/31/mystery-solved-the-secret-of-excel-curved-line-interpolation/
The author of this fit is Dr. Brian T. Murphy, PhD, PE from www.xlrotor.com. You find details here:
https://www.xlrotor.com/index.php/our-company/about-dr-murphy
https://www.xlrotor.com/index.php/knowledge-center/files
=>see Smooth_curve_bezier_example_file.xls
https://www.xlrotor.com/smooth_curve_bezier_example_file.zip
These knitted together you get the following results for the intersection of your given curves:
for the straight line intersection:
(x = -1,02914127711195 / y = 23,2340949174492)
for the smooth line intersection:
(x = -1,02947493047196 / y = 23,2370611219553)
For a full automation of your task you would need to add more details regarding the needed accuracy and what details you need for further processing (and this is actually not the scope of this website ;-).
Intersection of the straight lines:
Intersection of the smoothed lines:
comparison charts:
solution,
Thank you very much for the anwer, you perfectly centered my goal.
Your solution (for the smoothed lines) is very very close to what I determine in Siemens NX.
I'm going to read the documentation at the provided link https://pomax.github.io/bezierinfo/ in order to better understand the math behind this argument.
Then, to resume my request, you have been able to find the coordinates (x,y) of the intersection point between two curves without passing through an advanced CAD system with a very good precision.
I am starting to study now, best regards!
kalo86
I'm trying to code the Ritter's bounding sphere algorithm in arbitrary dimensions, and I'm stuck on the part of creating a sphere which would have 3 given points on it's edge, or in other words, a sphere which would be defined by 3 points in N-dimensional space.
That sphere's center would be the minimal-distance equidistant point from the (defining) 3 points.
I know how to solve it in 2-D (circumcenter of a triangle defined by 3 points), and I've seen some vector calculations for 3D, but I don't know what the best method would be for N-D, and if it's even possible.
(I'd also appreciate any other advice about the smallest bounding sphere calculations in ND, in case I'm going in the wrong direction.)
so if I get it right:
Wanted point p is intersection between 3 hyper-spheres of the same radius r where the centers of hyper-spheres are your points p0,p1,p2 and radius r is minimum of all possible solutions. In n-D is arbitrary point defined as (x1,x2,x3,...xn)
so solve following equations:
|p-p0|=r
|p-p1|=r
|p-p2|=r
where p,r are unknowns and p0,p1,p2 are knowns. This lead to 3*n equations and n+1 unknowns. So get all the nonzero r solutions and select the minimal. To compute correctly chose some non trivial equation (0=r) from each sphere to form system of n+1 =equations and n+1 unknowns and solve it.
[notes]
To ease up the processing you can have your equations in this form:
(p.xi-p0.xi)^2=r^2
and use sqrt(r^2) only after solution is found (ignoring negative radius).
there is also another simpler approach possible:
You can compute the plane in which the points p0,p1,p2 lies so just find u,v coordinates of these points inside this plane. Then solve your problem in 2D on (u,v) coordinates and after that convert found solution form (u,v) back to your n-D space.
n=(p1-p0)x(p2-p0); // x is cross product
u=(p1-p0); u/=|u|;
v=u x n; v/=|v|; // x is cross product
if memory of mine serves me well then conversion n-D -> u,v is done like this:
P0=(0,0);
P1=(|p1-p0|,0);
P2=(dot(p2-p0,u),dot(p2-p0,v));
where P0,P1,P2 are 2D points in (u,v) coordinate system of the plane corresponding to points p0,p1,p2 in n-D space.
conversion back is done like this:
p=(P.u*u)+(P.v*v);
My Bounding Sphere algorithm only calculates a near-optimal sphere, in 3 dimensions.
Fischer has an exact, minimal bounding hyper-sphere (N dimensions.) See his paper: http://people.inf.ethz.ch/gaertner/texts/own_work/seb.pdf.
His (C++/Java)code: https://github.com/hbf/miniball.
Jack Ritter
jack#houseofwords.com
I have two rings in 3 space each represented by a normal vector, center coordinate, and a radius.
How can I determine if the rings are linked. I.e. a point on one circle lies in the other disk.
This will be implemented in a tight loop so I am concerned about performance. I feel like there should be a closed form solution. So far I've only thought iterative solutions.
Any help?
Outline of the algorithm:
Handle the cases where the planes of the circles are parallel or concurrent.
Find the line of intersection of the planes of the circles.
Find the intersections of the circles with this line.
All the circle intersections with the line are on both planes. We can now do distance checks to see if the circles are linked.
Details:
I'll assume that the circles C1 and C2 are given by a center point (p1, p2), unit normal axis vector (n1, n2) and radii (r1, r2).
If n1 == k n2 for some scalar k, then the planes are parallel or concurrent. If they're concurrent this becomes the trivial 2d problem: check whether dist(p1, p2) < r1+r2.
Otherwise, the planes intersect at a line L. If the circles are linked, then they must both intersect L since linking implies that they mutually pass through each other's plane of definition. This gives you your first trivial rejection test.
L is given by a point q and a vector v. Finding v is easy: It's just the cross product of n1 and n2. q is a little trickier, but we can choose a point of nearest approach of the lines
l1(s) = p1 + s (v X n1)
l2(t) = p2 + t (v X n2)
These lines are perpendicular to v, n1 and n2, and are on the planes of C1 and C2. Their points of nearest approach must be on L.
You can refer to this post to see how to find the points of nearest approach.
Finally, the only way the circles can intersect is if they both have points on L. If they do, then consider the intersection points of C1 and L as they relate to C2. The circles are linked if there are two intersection points q11 and q12 of C1 and L and exactly one of them are inside of C2. Since L is on the plane of C2, this becomes a planar point-in-circle test:
IF dist(p1, q11) < r1 THEN
linked = (dist(p1, q12) > r1)
ELSE
linked = (dist(p1, q11) < r1)
ENDIF
Of course this pseudo-code is a little sloppy about handling the case that the circles actually intersect, but how that should be handled depends on your application.
A straightforward solution that is relatively easy to code: compute the line of intersection of the two planes and rotate and translate everything (in my case the six points defining the two circles) together so that the line becomes the Z axis (x=y=0). Then the planes can be rotated separately around Z so that the first circle, C1, lies on the XZ plane and C2 lies on the YZ plane. Now the centers and radii are easy to find (in my situation I don't initially have them). These rotations do not change the link/no link property and the link or lack of it can be easily determined from the order of the four intersections of the circles with the Z axis. (If either of the circles does not intersect x=y=0, there is no link.) (I may have confused the axes but I think this will work.)
Thank you.
Is this the 3D circle-disk intersection problem? Or, is it the circle-circle intersection problem?
If it's the former, there is a fast, closed form algorithm. First, weed out the circles-too-distant case: dist(CIR1.c, CIR2.c) > CIR1.r + CIR2.r
Handle the edge case: coplanar circles.
Extrude the disk into its plane, then intersect the circle with this plane. If there are 1 or 2 intersection points, test to see if they meet pointInDisk(p, CIR) logic. Report out any surviving intersection points.
A computational geometry textbook might have a good solution!
But I don't have a computation geometry textbook handy at the moment. If I were going to roll my own right now, based on my own (very limited) intuition, I think I'd start with 3d axis-aligned bounding boxes.
e.g., create an "just inside the circle" bounding box, as well as a "just outside the circle" bounding box. I believe it's trivial to figure out if two axis-aligned boxes overlap (section 2.1 of a homework assignment I once did talks about a related problem in the 2D situation --- finding the nearest rectangle to a point: http://juliusdavies.ca/uvic/report.html ).
I suspect the following assertions would hold (but I could be wrong):
If the inner-boxes overlap, then the rings are definitely connected. (I now think this is wrong...)
If the outer-boxes DO NOT overlap, then they are definitely not connected --- I am very confident this assertion holds! :-) :-)
If the outer-boxes overlap, but the inner-boxes do not, then they might be connected.
That's where I'd start, if I were rolling my own.
Conclusion: uhhh, I hope you have better luck with the other answers. :-)
I'm trying to find/make an algorithm to compute the intersection (a new filled object) of two arbitrary filled 2D objects. The objects are defined using either lines or cubic beziers and may have holes or self-intersect. I'm aware of several existing algorithms doing the same with polygons, listed here. However, I'd like to support beziers without subdividing them into polygons, and the output should have roughly the same control points as the input in areas where there are no intersections.
This is for an interactive program to do some CSG but the clipping doesn't need to be real-time. I've searched for a while but haven't found good starting points.
I found the following publication to be the best of information regarding Bezier Clipping:
T. W. Sederberg, BYU, Computer Aided Geometric Design Course Notes
Chapter 7 that talks about Curve Intersection is available online. It outlines 4 different approaches to find intersections and describes Bezier Clipping in detail:
https://scholarsarchive.byu.edu/cgi/viewcontent.cgi?article=1000&context=facpub
I know I'm at risk of being redundant, but I was investigating the same issue and found a solution that I'd read in academic papers but hadn't found a working solution for.
You can rewrite the bezier curves as a set of two bi-variate cubic equations like this:
∆x = ax₁*t₁^3 + bx₁*t₁^2 + cx₁*t₁ + dx₁ - ax₂*t₂^3 - bx₂*t₂^2 - cx₂*t₂ - dx₂
∆y = ay₁*t₁^3 + by₁*t₁^2 + cy₁*t₁ + dy₁ - ay₂*t₂^3 - by₂*t₂^2 - cy₂*t₂ - dy₂
Obviously, the curves intersect for values of (t₁,t₂) where ∆x = ∆y = 0. Unfortunately, it's complicated by the fact that it is difficult to find roots in two dimensions, and approximate approaches tend to (as another writer put it) blow up.
But if you're using integers or rational numbers for your control points, then you can use Groebner bases to rewrite your bi-variate order-3 polynomials into a (possibly-up-to-order-9-thus-your-nine-possible-intersections) monovariate polynomial. After that you just need to find your roots (for, say t₂) in one dimension, and plug your results back into one of your original equations to find the other dimension.
Burchburger has a layman-friendly introduction to Groebner Bases called "Gröbner Bases: A Short Introduction for Systems Theorists" that was very helpful for me. Google it. The other paper that was helpful was one called "Fast, precise flattening of cubic Bézier path and offset curves" by TF Hain, which has lots of utility equations for bezier curves, including how to find the polynomial coefficients for the x and y equations.
As for whether the Bezier clipping will help with this particular method, I doubt it, but bezier clipping is a method for narrowing down where intersections might be, not for finding a final (though possibly approximate) answer of where it is. A lot of time with this method will be spent in finding the mono-variate equation, and that task doesn't get any easier with clipping. Finding the roots is by comparison trivial.
However, one of the advantages of this method is that it doesn't depend on recursively subdividing the curve, and the problem becomes a simple one-dimensional root-finding problem, which is not easy, but well documented. The major disadvantage is that computing Groebner bases is costly and becomes very unwieldy if you're dealing with floating point polynomials or using higher order Bezier curves.
If you want some finished code in Haskell to find the intersections, let me know.
I wrote code to do this a long, long time ago. The project I was working on defined 2D objects using piecewise Bezier boundaries that were generated as PostScipt paths.
The approach I used was:
Let curves p, q, be defined by Bezier control points. Do they intersect?
Compute the bounding boxes of the control points. If they don't overlap, then the two curves don't intersect. Otherwise:
p.x(t), p.y(t), q.x(u), q.y(u) are cubic polynomials on 0 <= t,u <= 1.0.
The distance squared (p.x - q.x) ** 2 + (p.y - q.y) ** 2 is a polynomial on (t,u).
Use Newton-Raphson to try and solve that for zero. Discard any solutions outside 0 <= t,u <= 1.0
N-R may or may not converge. The curves might not intersect, or N-R can just blow up when the two curves are nearly parallel. So cut off N-R if it's not converging after after some arbitrary number of iterations. This can be a fairly small number.
If N-R doesn't converge on a solution, split one curve (say, p) into two curves pa, pb at t = 0.5. This is easy, it's just computing midpoints, as the linked article shows. Then recursively test (q, pa) and (q, pb) for intersections. (Note that in the next layer of recursion that q has become p, so that p and q are alternately split on each ply of the recursion.)
Most of the recursive calls return quickly because the bounding boxes are non-overlapping.
You will have to cut off the recursion at some arbitrary depth, to handle weird cases where the two curves are parallel and don't quite touch, but the distance between them is arbitrarily small -- perhaps only 1 ULP of difference.
When you do find an intersection, you're not done, because cubic curves can have multiple crossings. So you have to split each curve at the intersecting point, and recursively check for more interections between (pa, qa), (pa, qb), (pb, qa), (pb, qb).
There are a number of academic research papers on doing bezier clipping:
http://www.andrew.cmu.edu/user/sowen/abstracts/Se306.html
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.61.6669
http://www.dm.unibo.it/~casciola/html/research_rr.html
I recommend the interval methods because as you describe, you don't have to divide down to polygons, and you can get guaranteed results as well as define your own arbitrary precision for the resultset. For more information on interval rendering, you may also refer to http://www.sunfishstudio.com