I am trying to find the intersection point of a curve and a 3D surface with no luck. The surface is in the shape of a cone, and the curve is hyperbolic, as are shown in the figure.
CONE AND THE CURVE
This simulates a ray hits a certain surface. I tried to use bisection method, but it doesn't seem to work. then I tried newton's algorithm, but the results are still not good.
Is there any other good algorithms out there which are suitable for solving this kind of problem?
With the curve given in parametric form
x = fx(t)
y = fy(t)
z = fz(t)
and the surface by one equation of the form
g(x,y,z) = 0
just plug in the curve functions and bisection should work:
g(fx(t), fy(t), fz(t)) = 0
The only problem is to find suitable starting points t1 and t2 where g has opposite sign.
Problem
You are searching for a curve-surface intersection algorithm. Note that both curves and surfaces can be represented in either implicit form or in parametric form. Surface in implicit form is defined by equation F(x, y, z) = 0, which is a quadratic polynomial of x, y, z in case of conic surface. Surface in parametric form is defined by point-valued function S(u, v) of its parameters (e.g. you can use distance along cone axis and polar angle as parameters of conical surface). Curve is usually described only in parametric form, as a function C(t) with parameter t, which could be quadratic for a hyperbolic curve.
Implicit surface
The simplest cases of all is to treat your problem as an intersection of parametric curve against implicit surface. In this case you can write down a single equation q(t) = F(C(t)) = 0 with single variable t. Of course, Newton's iteration is not guaranteed to find all solutions in general case, bisection can only surely find one solution if you find two points with different sign of q(t).
In your case q(t) is a quartic polynomial (after putting quadratic curve parametrization into quadratic surface equation). It can be theoretically solved with Ferrari's analytic formula, but I strongly advise against it, because it is quite unstable numerically. You can apply any popular polynomial solver here, like Jenkins-Traub algorithm or eigenvalues algorithm for companion matrix (also see this question). You can also use methods of interval mathematics: for example, you can recursively subdivide the domain interval of parameter t into smaller pieces, while pruning all the pieces that surely do not contain zeros (interval arithmetic would help you to detect such pieces).
Parametric surface
Now we can move on to the case when both the curve and the surface are represented parametrically. I do not know any solutions that could benefit from the fact the your surface is conical and your curve is hyperbolic, so you have to apply the general curve-surface intersection algorithm. Alternatively, you can fit an implicitly-defined cone into your parametric surface, then use the solution above for quartic polynomial roots.
A lot of reliable general intersection algorithms are based on the subdivision method (which is actually interval mathematics again). The general idea is to continiously divide the curve and the surface into smaller and smaller pieces. The pairs of pieces which surely do not intersect are dropped as soon as possible. At the end you'll have a set of small piece pairs, tightly bounding your intersection points. Yoy might want to run Newton's iteration from them in order to make intersection points precise.
Here is the outline of a sample algorithm:
Start with a single curve piece (the whole input curve) and a single surface
piece (whole surface), and one potentially intersection pair (PIP) of these pieces.
Subdivide each curve piece into two halves (by parameter), subdivide each surface piece into four quadrants (by both parameters).
For each old PIP check all 8 pairs of curve half vs surface quadrant. If they surely do not intersect, forget them. If they can intersect, save them as a new PIP.
Unless all pieces are small enough, repeat from step 2 with new pieces and PIP-s.
For each pair of curve piece and surface piece, you have to check whether they can potentially intersect, which can be easily done by checking their axis-aligned bounding boxes. Also, you can represent your curves and surfaces as NURBS, in which case you can use convex hulls as tighter bounding volumes.
Generally, there are tons of variations and improvements of this algorithms. I advise the following literature for deeper knowledge:
Shape interrogation for computer-aided design and manufacturing.
chapter 4: for root solvers
section 5.7: for curve-surface intersection
PhD of Michael Hohmeyer.
section 4.5: for curve-surface intersection
sections 4.1 and 4.2: for convex hulls intersection (if you are brave enough).
Bottom line
If you are seeking for a simple and working solution, and you are sure that hyperbolas and cones are the only things you have to worry about, then you'd better use implicit definition of cone and solve quartic equation with some standard numerical algorithm from a good library available to you.
Related
I am trying to develop an algorithm that performs the following :
Given a 2D polygon and a 3D polyhedron, determine if the 2D polygon is a projection of the 3D polyhedron (a perspective projection to be precise) without knowing which transformation matrix we may have possibly used for the projection.
input
{2D Polygon}
{3D Polyhedron}
output
{bool} whether or not it's a perspective projection
I am not asking for code, but I would simply like to know if this is feasible in polynomial time.
Any help will be greatly appreciated.
A 3D to 2D perspective projection has 7 degrees of freedom (6 for the relative motion of the scene with respect to the camera, 1 for the focal length).
Select four vertices in the 2D projection and consider all possible correspondences with polyhedron vertices (there is a polynomial number of such associations). Then form a system of 7 equations in the 7 unknown parameters (unfortunately a nonlinear one; maybe the eighth equation can be useful to select among multiple solutions).
Knowing the parameters, you can check a solution by re-projecting the polyhedron and comparing to the polygon (with further search for correspondences with vertices and edges).
All of this will take polynomial time (quartic if I am right), if one admits that the solver takes bounded time (hence bounded precision).
If the focal length is known, then a better approach is possible. Indeed, with only 6 unknowns, you can find the projection parameters from the projection of just three points. This problem is known to have an analytical solution (actually up to 4 of them), as described at length in "New Algorithms for the Perspective-Three-Point Problem, GAO Xiaoshan & CHEN Hangfei, Vol.16 No.3 J. Comput. Sci. & Technol."
This should lead to an O(N³) exact procedure.
More generally speaking, you form putative correspondences between N pairs of points, solve the corresponding Perspective-N-point problem, and check the hypothesis by reprojecting the polyhedron and comparing to the known projection to validate the hypothesis.
Just an idea for an algorithm:
Take a triangle of the projection made of three points next to each other not on the same line. Iterate through all corresponding triangles of the original. For all possible projections that solve the pair of triangles, check if the rest matches.
I must admit I am not sure right now if there could be infinite solutions for triangles (which would be hard to iterate)? If so, start with four points.
I think it is possible but you have to do a fair amount of reverse engineering. A 2D sketch that represents a 3D object is known as an Orthographic Projection. The link shows you the transformation matrices you need apply to transform the 3D point onto its 2D projection. Now, how do you go the opposite way? Inverse matrices with a mix of some inverse transformations (translation, scaling, rotation...)? I think this is a good lead to follow.
Hi people i have 2 question related to Decasteljau algorithm,they are more of a general questions,but if im right it could help solving many problems.Here it is:
We have some sufrace: Ʃ(i=0,n) Ʃ(j=o,m) Bi,n(U),Bi,m(v) Pi,j analysis that i have found says that first we take some value for one parameter u=uo,then we itterate other parametar v -> 1 get a set of points,then increment u by one etc....for loop inside for loop in code language.My question is can we fix one parameter U=Uo for what ever value,and then just compute points on for parameter v?Because all points that are on one curve are also on the surface,and if distance between curves approaches to zero (which itteration really is) we can apply DeCasteljau algorithm only to one set of curves itterating only one parameter.Or i got something wrong?:)
Second question is i havent really figured out what do we really need DeCasteljau algorithm for,unless we are drawing curves by hand?If we know order of the curve we can easily form Bernstain polynoms for that curve order and compute point for given value of parametar.Because when you unwrap Decasteljau what you get is Bernstain polynom?
So like i said,please help have i got i wrong?
Yes you can fix one parameter (say U) and change the other (V) to generate an iso-U curve.
You can see the things as if you had an NxM array of control points. If you perform a first interpolation on U (actually M interpolations involving N control points), you get M new control points that define a Bezier curve. and by varying U, the curve moves in space.
The De Casteljau's algorithm is used for convenience: it computes the interpolant by using a cascade of linear interpolations between the control points. Direct evaluation of the Bernstein polynomials would require the precomputation of the coefficients, and would not be faster, even when implemented by Horner's scheme, and can be numerically less stable.
The De Casteljau's algorithm is also appreciated for its geometrical interpretation, and for its connection with the subdivision process: if you want to build the control points for just a part of a Bezier curve, De Calsteljau's provides them.
So I'll try to describe the problem in detail, and I'd like some critique on the validity and performance of the process I use to solve it. My main concern is the validity, which I cannot seem to prove.
The idea is that we have a 2D polygon in 3D space (which is described by all its hull points. Each point also has a normalvector which indicates an infinite extrusion. I apologize for using the misleading term 'normal', it's just a normalized vector that represents a normal in a broader context not relevant to this algorithm. To completely clarify, these vectors are neither perpendicular to the plane nor the same on each point. As can be seen in the following image, where the vectors are the red lines:
This means that the sides of this extrusion do not form a flat, but a skewed plane. One can look at its extrusion sides like a Skew polygon (http://en.wikipedia.org/wiki/Skew_polygon).
Anyway, I'd like to see if a point P is inside the extrusion or not. My current procedure is as follows:
Calculate normal N=(A-B)x(B-C) where A,B,C are any three clock-wise points on the 2D polygon. This gives me the 2D polygon plane normal.
Calculate the final plane variable D to get NxX+NyY+Nz*Z+D = 0. Do this by filling in point P in X,Y,Z. This gives the extruded plane.
Extrude the normals at all hull points to get the new hull on the extruded plane. Simply done by finding the intersection of the line they define and the plane.
Here comes the tricky part. Now I have a convex hull on a 3D plane where I know the point is coplanar. I have seen a lot of solutions already on stackoverflow concerning solving linear systems, etc... but they either do not consider 3D, lack a good implementation or look too expensive.
My approach is using an assumption that I think is both necessary and sufficient but I cannot prove that last part (the necessary part is obvious). I make the following claim:
For coplanar convex hull consisting of points A->Z and an additional coplanar point P. Then P is within the convex hull if and only if for all clock-wise sequential points A and B and some other third sequential point C the line C-P passes through P before passing through the line A-B.
In the example image above: A'-B' is crossed by D'-P going through P before A'-B', for B'-C' this is achieved by A', for C'-D' also by A' and for D'-A' by B'.
I implement this idea as two 3D parametric lines crossing eachother (A-B and P-C) and looking at the parameter's sign. The parameter in P-C should be non-positive because the crossing point should lie behind P for a parameter going to C.
In essence this would result in a O(n)O(n^2) algorithm with n being the #points of the polygon (which seems to beat all solutions I've seen so far). However, even though I cannot find a counterexample to the above assumption, I also cannot prove it.
Edit: I believe it is possible to prove that a sufficient condition is that for A and B, all other hull points need to satisfy the assumption (and not just C).
Edit2: Clarified claim.
Given a few sample points on a bézier curve, is it possible to work out the set of possible curves these points may lie on?
In my specific application there is a limited set of endpoints the curve may have, so I want to generate the set of possible curves, enumerate all of them and pick out all the ones which may end on a valid end point.
Some people have asked for more detail.
I have a set of points which I know are on a quadratic bezier curve, I want to calculate the formula of the curve and be able to extrapolate new points on the curve.
Bezier curves will always go through starting and ending control points.
If the degree of the curve is equal to the number of sample points then there will be only one curve that will pass through all your points (in a normal case, where all points are different and they don't lie on a bezier curve of a lesser degree).
If the degree of a curve is less then the number of the sample points then, in general case, you will not be able to make the curve go through all the points (in a normal case).
If the degree of a curve is higher then the number of the sample points then, in general case, you will be able to draw infinite number of curves.
In the wiki article you will find references to control points only, but still I believe that I remember the above properties correctly and that they hold for the points on the curves as well.
I think you need to redefine your question and exactly define what type of curves (and of which degree) do you need. Also as Georg pointed out you might be looking for paths - a combination of curves.
EDIT:
First a correction - curve is defined with degree plus one number of control points points (quadratic need three). Control points are not the same as points on the curve - and for three points on the curve and quadratic curve you could have infinite number of solutions (see this for quadratic curve and four points)
As for the solution (but still under assumption that you are looking at a single curve):
For an equation for single quadratic curve you have
B(t) = (1-t)^2*P0 + 2*(1-t)*t*P1 + t^2*P2
Capital letters above are vectors, and P0 corresponds to starting control point (first point), P2 corresponds to ending control point (last point), so you still need to find P1. The variable t is scalar that ranges from 0 to 1.
If working with 2D curves the above vector equation gives two scalar equations for each point on the curve.
Still there is t as an unknown, so you should take 2 more points (4 in total) which will give you 4 unknowns (t for first point, t for second point, x and y of the P1, middle control point) and 4 equation to solve (2 from each sample point).
Solve that with your favourite numerical method and you will get the original curve on which the points came from.
If you still think that you can get more curves and that you will have to choose something then you are not working with bezier curves, but with bezier splines (in a sense of multiple curves joined together). Still the same principle applies and if you work out a way to solve a single curve from the above equations (and if you have enough points) then you can divide the problem into n-segments of actual bezier curves and solve each as outlined above.
If it turns out that you don't have enough points then look at the linked article again - you are probably looking for the smoothest curve and there are some suggestions in the article on how to get there as looking for the exact solution (shortest curve/smoothest curve) seems to be rather complex.
I have problem of packing 2 arbitrary polygons. I.e. we have 2 arbitrary polygons. We are to find such placement of this polygons (we could make rotations and movements), when rectangle, which circumscribes this polygons has minimal area.
I know, that this is a NP-complete problem. I want to choose an efficient algorithm for solving this problem. I' looking for No-Fit-Polygon approach. But I could't find anywhere the simple and clear algorithm for finding the NFP of two arbitrary polygons.
The parameter space does not seem too big and testing it is not too bad either. If you fix one polygon, the other ploygon can be shifted along x-axis by X, and shifted along y-axis by Y and rotated by r.
The interesting region for X and Y can be determined by finding some bounding box for for the polygons. r of course is between and 360 degrees.
So how about you tried a set of a set of equally spaced intervals in the interesting range for X,Y and r. Perhaps, once you found the interesting points in these dimensions, you can do more finer grained search.
If its NP-complete then you need heuristics, not algorithms. I'd try putting each possible pair of sides together and then sliding one against the other to minimise area, constrained by possible overlap if they are concave of course.
There is an implementation of a robust and comprehensive no-fit polygon generation in a C++ library using an orbiting approach: https://github.com/kallaballa/libnfporb
(I am the author of libnfporb)