I need a method to project 3d conics to 2d. None of the articles tell how to do this with rational beziers. Another thing I need a method for is moving 3d or 2d conics to 4d or 3d respectively (as in a reverse projection). I read somewhere that rational beziers can be split by moving them to a higher dimension and splitting the resulting non-rational curve with de Casteljau and then moving back. I seem to recall that perspective projection of conic beziers can be represented exactly with conic beziers, and that it may involve splitting into several curves. I don't understand any of the articles on any site on beziers.
Since there aren't any better answers, here's what I can offer off the top of my head...
Perspective transformation can change parabolas into ellipses or hyperbolas and vice-versa, so even though P0, P1, and P2 can be directly mapped, the weights will change.
Assuming a conic with weights (1,w,1), however, the distance along the line from (P0+P2)/2 to P1 at which it intersects the curve is simply related to the weight w, and that lets you find the new weight as follows:
Map P0, P1, and P2 to P0', P1', P2'
Calculate the midpoint M' = (P1'+P2')/2
Inverse map M' to M, and calculate the intersection point I of the line M-P1 with the original curve.
Map the intersection point I to I', to get the point at which the new curve should intersect M'-P1'
Calculate the new weight w' from the position of the intersection I'. The curve gets to I at t=0.5, so w' = (M'-I')/(P1'-I'). Note that this division makes sense, because the vectors being divided are collinear. You can divide their lengths or just the largest coordinate.
If you expand out all the steps, I'm sure there are ways to simplify this procedure.
Related
Given a point p exterior to an axially aligned, origin centered ellipse E, find the (upto) four unique normals to E passing through p.
This is not a Mathematica question. Direct computation is too slow; I am willing to sacrifice precision and accuracy for speed.
I have searched the web, but all I found involved overly complex calculations which if implemented directly appear to lack the performance I need. Is there a more "programmatical" way to do this, like using matrices or scaling the ellipse into a circle?
Let's assume the ellipse E is in "standard position", center at the origin and axes parallel to the coordinate axes:
(x/a)^2 + (y/b)^2 = 1 where a > b > 0
The boundary cases a=b are circles, where the normal lines are simply ones that pass through the center (origin) and are thus easy to find. So we omit discussion of these cases.
The slope of the tangent to the ellipse at any point (x,y) may be found by implicit differentiation:
dy/dx = -(b^2 x)/(a^2 y)
For the line passing through (x,y) and a specified point p = (u,v) not on the ellipse, that is normal to ellipse E when its slope is the negative reciprocal of dy/dx:
(y-v)/(x-u) * (-b^2 x)/(a^2 y) = -1 (N)
which simplifies to:
(x - (1+g)u) * (y + gv) = -g(1+g)uv where g = b^2/(a^2 - b^2)
In this form we recognize it is the equation for a right rectangular hyperbola. Depending on how many points of intersection there are between the ellipse and the hyperbola (2,3,4), we have that many normals to E passing through p.
By reflected symmetry, if p is assumed exterior to E, we may take p to be in the first quadrant:
(u/a)^2 + (v/b)^2 > 1 (exterior to E)
u,v > 0 (1'st quadrant)
We could have boundary cases where u=0 or v=0, i.e. point p lies on an axis of E, but these cases may be reduced to solving a quadratic, because two normals are the (coinciding) lines through the endpoints of that axis. We defer further discussion of these special cases for the moment.
Here's an illustration with a=u=5,b=v=3 in which only one branch of the hyperbola intersects E, and there will be only two normals:
If the system of two equations in two unknowns (x,y) is reduced to one equation in one unknown, the simplest root-finding method to code is a bisection method, but knowing something about the possible locations of roots/intersections will expedite our search. The intersection in the first quadrant is the nearest point of E to p, and likewise the intersection in the third quadrant is the farthest point of E from p. If the point p were a good bit closer to the upper endpoint of the minor axis, the branches of the hyperbola would shift together enough to create up to two more points of intersection in the fourth quadrant.
One approach would be to parameterize E by points of intersection with the x-axis. The lines from p normal to the ellipse must intersect the major axis which is a finite interval [-a,+a]. We can test both the upper and lower points of intersection q=(x,y) of a line passing through p=(u,v) and (z,0) as z sweeps from -a to +a, looking for places where the ellipse and hyperbola intersect.
In more detail:
1. Find the upper and lower points `q` of intersection of E with the
line through `p` and `(z,0)` (amounts to solving a quadratic)
3. Check the sign of a^2 y(x-u) - b^2 x(y-v) at `q=(x,y)`, because it
is zero if and only `q` is a point of normal intersection
Once a subinterval is detected (either for upper or lower portion) where the sign changes, it can be refined to get the desired accuracy. If only modest accuracy is needed, there may be no need to use faster root finding methods, but even if they are needed, having a short subinterval that isolates a root (or root pair in the fourth quadrant) will be useful.
** more to come comparing convergence of various methods **
I had to solve a problem similar to this, for GPS initialization. The question is: what is the latitude of a point interior to the Earth, especially near the center, and is it single-valued? There are lots of methods for converting ECEF cartesian coordinates to geodetic latitude, longitude and altitude (look up "ECEF to Geodetic"). We use a fast one with only one divide and sqrt per iteration, instead of several trig evaluations like most methods, but since I can't find it in the wild, I can't give it to you here. I would start with Lin and Wang's method, since it only uses divisions in its iterations. Here is a plot of the ellipsoid surface normals to points within 100 km of Earth's center (North is up in the diagram, which is really ECEF Z, not Y):
The star-shaped "caustic" in the figure center traces the center of curvature of the WGS-84 ellipsoid as latitude is varied from pole to equator. Note that the center of curvature at the poles is on the opposite side of the equator, due to polar flattening, and that the center of curvature at the equator is nearer to the surface than the axis of rotation.
Wherever lines cross, there is more than one latitude for that cartesian position. The green circle shows where our algorithm was struggling. If you consider that I cut off these normal vectors where they reach the axis, you would have even more normals for a given position for the problem considered in this SO thread. You would have 4 latitudes / normals inside the caustic, and 2 outside.
The problem can be expressed as the solution of a cubic equation which
gives 1, 2, or 3 real roots. For the derivation and closed form
solution see Appendix B of Geodesics on an ellipsoid of revolution. The boundary between 1 and 3 solutions is an astroid.
I am trying to quantize surface normals into let's say 8 bins.
For example, when computing features like HOG to quantize 2D gradients [x,y] into 8 bins we just take the angle with the y plane i.e. arctan(y/x) which will give us an angle between 0-360.
My question is, given a 3D direction [x,y,z], a surface normal in this case, how can we histogram it in a similar way? Do we just project onto one plane and use that angle i.e. the dot product of [x,y,z] and [0,1,0] for example?
Thanks
EDIT
I also read a paper recently where they quantized surface normals by measuring angles between normal and precomputed vectors that which are arranged around a right circular cone shape. I have added a link to this paper in the question (section 3.3.2 last paragraph), is this an effective approach? And if so, how do we compute these vectors?
Quantizing a continuous topological space corresponds to partitioning it and assigning labels to each partition. The straightforward standard approach for this scenario (quantizing normals) is as follows.
Choose your favorite uniform polyhedron:
http://en.wikipedia.org/wiki/Tetrahedron (4 faces)
http://en.wikipedia.org/wiki/Cube (6 faces)
http://en.wikipedia.org/wiki/Octahedron (8 faces)
http://en.wikipedia.org/wiki/Dodecahedron (12 faces)
http://en.wikipedia.org/wiki/Icosahedron (20 faces)
In general: http://en.wikipedia.org/wiki/Schl%C3%A4fli_symbol
Develop a mapping function from a normal on the unit sphere to the face of your chosen polyhedron that the normal intersects.
I would advise doing an argmax across polyhedron faces, taking the dot product of your normal and each polyhedron face normal. The one that gives the highest dot product is the face your normal should be binned into.
Use the face normal for each polyhedron face as the label for that face.
Prefer this approach to the approach suggested by others of mapping to spherical coordinates and then binning those. That approach suffers from too much sensitivity near the poles of the sphere.
Edit
In the paper you added to your question, the same idea is being used. There, however, the normals are restricted to a hemisphere - the only surfaces directly visible in an image have surface normals no more than 90 degrees away from the vector from the surface to the viewpoint.
The paper wants to quantize these surface normals into 8 values, represented by 8-bit integers with exactly one bit set to 1 and the rest set to 0. The 8 precomputed normals are computed as:
ntx = cos(a)*cos(t)
nty = cos(a)*sin(t)
ntz = sin(a)
where a = pi/4 and t = 0, pi/4, 2*pi/4, 3*pi/4, ..., 7*pi/4.
Notice
[cos(a)*cos(t)]2 + [cos(a)*sin(t)]2 + [sin(a)]2 = cos2(a)[cos2(t) + sin2(t)] + sin2(a) = cos2(a) + sin2(a) = 1
given a 3D direction [x,y,z], a surface normal in this case, how can
we histogram it in a similar way?
In the first case you quantize the polar orientation theta of the gradients. Now you need to quantize the spherical orientations theta and phi in a 2D histogram.
Do we just project onto one plane and use that angle
The binning of the sphere determines how you summarize the information to build a compact yet descriptive histogram.
Projecting the normal is not a good idea, if theta is more important than phi, just use more bins for theta
EDIT
Timothy Shields points in his comment and his answer that a regular binning of theta and phi won't produce a regular binning over the sphere as the bins will be bunched toward the poles.
His answer gives a solution. Alternatively, the non-regular binning described here can be hacked as follows:
Phi is quantized regularly in [0,pi]. For theta rather than quantizing the range [0,pi], the range [-1,1] is quantized instead;
For each quantized value u in [-1,1], theta is computed as
theta = arcsin(sqrt(1 - u * u)) * sign(u)
sign(u) returns -1 if u is negative, 1 otherwise.
The computed theta along with phi produce a regular quantization over the sphere.
To have an idea of the equation given above look at this article. It describes the situation in the context of random sampling though.
EDIT
In the above hack Timothy Shields points out that only the area of the bins is considered. The valence of the vertices (point of intersection of neighboring bins) won't be regular because of the poles singularity.
A hack for the previous hack would be to remesh the bins into a regular quadrilateral mesh and keep the regular area.
A heuristic to optimize this problem with the global constraints of having the same valence and the area can be inspired from Integer-Grid Maps Quad Meshing.
With the two hacks, this answer is too hacky and a little out of context as opposed to Timothy Shields answer.
A 3-dimensional normal cannot be quantized into a 1-D array as easily as for a 2-D normal (e.g., using arctan). I would recommend histogramming it into a 2-d space with a polar angle and an azimuth angle. For example, use spherical coordinates where the r (radius) value is always 1.0 (since your surface normal is normalized, length 1.0). In this case, you can throw away the r-value and just use polar angle θ (theta), and azimuthal angle φ (phi) to quantize the 3D normal.
I'm not sure if this is the right place to ask, but here goes...
Short version: I'm trying to compute the orientation of a triangle on a plane, formed by the intersection of 3 edges, without explicitly computing the intersection points.
Long version: I need to triangulate a PSLG on a triangle in 3D. The vertices of the PSLG are defined by the intersections of line segments with the plane through the triangle, and are guaranteed to lie within the triangle. Assuming I had the intersection points, I could project to 2D and use a point-line-side (or triangle signed area) test to determine the orientation of a triangle between any 3 intersection points.
The problem is I can't explicitly compute the intersection points because of the floating-point error that accumulates when I find the line-plane intersection. To figure out if the line segments strike the triangle in the first place, I'm using some freely available robust geometric predicates, which give the sign of the volume of a tetrahedron, or equivalently which side of a plane a point lies on. I can determine if the line segment endpoints are on opposite sides of the plane through the triangle, then form tetrahedra between the line segment and each edge of the triangle to determine whether the intersection point lies within the triangle.
Since I can't explicitly compute the intersection points, I'm wondering if there is a way to express the same 2D orient calculation in 3D using only the original points. If there are 3 edges striking the triangle that gives me 9 points in total to play with. Assuming what I'm asking is even possible (using only the 3D orient tests), then I'm guessing that I'll need to form some subset of all the possible tetrahedra between those 9 points. I'm having difficultly even visualizing this, let alone distilling it into a formula or code. I can't even google this because I don't know what the industry standard terminology might be for this type of problem.
Any ideas how to proceed with this? Thanks. Perhaps I should ask MathOverflow as well...
EDIT: After reading some of the comments, one thing that occurs to me... Perhaps if I could fit non-overlapping tetrahedra between the 3 line segments, then the orientation of any one of those that crossed the plane would be the answer I'm looking for. Other than when the edges enclose a simple triangular prism, I'm not sure this sub-problem is solvable either.
EDIT: The requested image.
I am answering this on both MO & SO, expanding the comments I made on MO.
My sense is that no computational trick with signed tetrahedra volumes will avoid the precision issues that are your main concern. This is because, if you have tightly twisted segments, the orientation of the triangle depends on the precise positioning of the cutting plane.
[image removed; see below]
In the above example, the upper plane crosses the segments in the order (a,b,c) [ccw from above]: (red,blue,green), while the lower plane crosses in the reverse order (c,b,a): (green,blue,red). The height
of the cutting plane could be determined by your last bit of precision.
Consequently, I think it makes sense to just go ahead and compute the points of intersection in
the cutting plane, using enough precision to make the computation exact. If your segment endpoints coordinates and plane coefficients have L bits of precision, then there is just a small constant-factor increase needed. Although I am not certain of precisely what that factor is, it is small--perhaps 4. You will not need e.g., L2 bits, because the computation is solving linear equations.
So there will not be an explosion in the precision required to compute this exactly.
Good luck!
(I was prevented from posting the clarifying image because I don't have the reputation. See
the MO answer instead.)
Edit: Do see the MO answer, but here's the image:
I would write symbolic vector equations, you know, with dot and cross products, to find the normal of the intersection triangle. Then, the sign of the dot product of this normal with the initial triangle one gives the orientation. So finally you can express this in a form sign(F(p1,...,p9)), where p1 to p9 are your points and F() is an ugly formula including dot and cross products of differences (pi-pj). Don't know if this can be done simpler, but this general approach does the job.
As I understand it, you have three lines intersecting the plane, and you want to calculate the orientation of the triangle formed by the intersection points, without calculating the intersection points themselves?
If so: you have a plane
N·(x - x0) = 0
and six points...
l1a, l1b, l2a, l2b, l3a, l3b
...forming three lines
l1 = l1a + t(l1b - l1a)
l2 = l2a + u(l2b - l2a)
l3 = l3a + v(l3b - l3a)
The intersection points of these lines to the plane occur at specific values of t, u, v, which I'll call ti, ui, vi
N·(l1a + ti(l1b - l1a) - x0) = 0
N·(x0 - l1a)
ti = ----------------
N·(l1b - l1a)
(similarly for ui, vi)
Then the specific points of intersection are
intersect1 = l1a + ti(l1b - l1a)
intersect2 = l2a + ui(l2b - l2a)
intersect3 = l3a + vi(l3b - l3a)
Finally, the orientation of your triangle is
orientation = direction of (intersect2 - intersect1)x(intersect3 - intersect1)
(x is cross-product) Work backwards plugging the values, and you'll have an equation for orientation based only on N, x0, and your six points.
Let's call your triangle vertices T[0], T[1], T[2], and the first line segment's endpoints are L[0] and L[1], the second is L[2] and L[3], and the third is L[4] and L[5]. I imagine you want a function
int Orient(Pt3 T[3], Pt3 L[6]); // index L by L[2*i+j], i=0..2, j=0..1
which returns 1 if the intersections have the same orientation as the triangle, and -1 otherwise.
The result should be symmetric under interchange of j values, antisymmetric under interchange of i values and T indices. As long as you can compute a quantity with these symmetries, that's all you need.
Let's try
Sign(Product( Orient3D(T[i],T[i+1],L[2*i+0],L[2*i+1]) * -Orient3D(T[i],T[i+1],L[2*i+1],L[2*i+0]) ), i=0..2))
where the product should be taken over cyclic permutations of the indices (modulo 3). I believe this has all the symmetry properties required. Orient3D is Shewchuk's 4-point plane orientation test, which I assume you're using.
I'm using CML to manage the 3D math in an OpenGL-based interface project I'm making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4x4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane.
CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn't necessarily aligned with the z axis of the perspective projection.
I'd like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it?
Thanks!
This seems to be a duplicate of Finding side length of a cross-section of a pyramid frustum/truncated pyramid, if you already have a cross-section of known width a known distance from the apex. If you don't have that and you want to derive the answer yourself you can follow these steps.
Take two adjacent planes and find
their line of intersection L1. You
can use the steps here. Really
what you need is the direction
vector of the line.
Take two more planes, one the same
as in the previous step, and find
their line of intersection L2.
Note that all planes of the form Ax + By + Cz + D = 0 go through the origin, so you know that L1 and L2
intersect.
Draw yourself a picture of the
direction vectors for L1 and L2,
tails at the origin. These form an
angle; call it theta. Find theta
using the formula for the angle
between two vectors, e.g. here.
Draw a bisector of that angle. Draw
a perpendicular to the bisector at
the distance d you want from the
origin (this creates an isosceles
triangle, bisected into two
congruent right triangles). The
length of the perpendicular is your
desired frustum width w. Note that w is
twice the length of one of the bases
of the right triangles.
Let r be the length of the
hypotenuses of the right triangles.
Then rcos(theta/2)=d and
rsin(theta/2)=w/2, so
tan(theta/2)=(w/2)/d which implies
w=2d*tan(theta/2). Since you know d
and theta, you are done.
Note that we have found the length of one side of a cross-section of a frustrum. This will work with any perpendicular cross-section of any frustum. This can be extended to adapt it to a non-perpendicular cross-section.
I'm looking for an algorithm to find the common intersection points between 3 spheres.
Baring a complete algorithm, a thorough/detailed description of the math would be greatly helpful.
This is the only helpful resource I have found so far:
http://mathforum.org/library/drmath/view/63138.html
But neither method described there is detailed enough for me to write an algorithm on.
I would prefer the purely algebraic method described in the second post, but what ever works.
Here is an answer in Python I just ported from the Wikipedia article. There is no need for an algorithm; there is a closed form solution.
import numpy
from numpy import sqrt, dot, cross
from numpy.linalg import norm
# Find the intersection of three spheres
# P1,P2,P3 are the centers, r1,r2,r3 are the radii
# Implementaton based on Wikipedia Trilateration article.
def trilaterate(P1,P2,P3,r1,r2,r3):
temp1 = P2-P1
e_x = temp1/norm(temp1)
temp2 = P3-P1
i = dot(e_x,temp2)
temp3 = temp2 - i*e_x
e_y = temp3/norm(temp3)
e_z = cross(e_x,e_y)
d = norm(P2-P1)
j = dot(e_y,temp2)
x = (r1*r1 - r2*r2 + d*d) / (2*d)
y = (r1*r1 - r3*r3 -2*i*x + i*i + j*j) / (2*j)
temp4 = r1*r1 - x*x - y*y
if temp4<0:
raise Exception("The three spheres do not intersect!");
z = sqrt(temp4)
p_12_a = P1 + x*e_x + y*e_y + z*e_z
p_12_b = P1 + x*e_x + y*e_y - z*e_z
return p_12_a,p_12_b
Probably easier than constructing 3D circles, because working mainly on lines and planes:
For each pair of spheres, get the equation of the plane containing their intersection circle, by subtracting the spheres equations (each of the form X^2+Y^2+Z^2+aX+bY+c*Z+d=0). Then you will have three planes P12 P23 P31.
These planes have a common line L, perpendicular to the plane Q by the three centers of the spheres. The two points you are looking for are on this line. The middle of the points is the intersection H between L and Q.
To implement this:
compute the equations of P12 P23 P32 (difference of sphere equations)
compute the equation of Q (solve a linear system, or compute a cross product)
compute the coordinates of point H intersection of these four planes. (solve a linear system)
get the normal vector U to Q from its equation (normalize a vector)
compute the distance t between H and a solution X: t^2=R1^2-HC1^2, (C1,R1) are center and radius of the first sphere.
solutions are H+tU and H-tU
A Cabri 3D construction showing the various planes and line L
UPDATE
An implementation of this answer in python complete with an example of usage can be found at this github repo.
It turns out the analytic solution is actually quite nice using this method and can tell you when a solution exists and when it doesn't (it is also possible to have exactly one solution.) There is no reason to use Newton's method.
IMHO, this is far easier to understand and simpler than trilateration given below. However, both techniques give correct answers in my testing.
ORIGINAL ANSWER
Consider the intersection of two spheres. To visualize it, consider the 3D line segment N connecting the two centers of the spheres. Consider this cross section
(source: googlepages.com)
where the red-line is the cross section of the plane with normal N. By symmetry, you can rotate this cross-section from any angle, and the red line segments length can not change. This means that the resulting curve of the intersection of two spheres is a circle, and must lie in a plane with normal N.
That being said, lets get onto finding the intersection. First, we want to describe the resulting circle of the intersection of two spheres. You can not do this with 1 equation, a circle in 3D is essentially a curve in 3D and you cannot describe curves in 3D by 1 eq.
Consider the picture
(source: googlepages.com)
let P be the point of intersection of the blue and red line. Let h be the length of the line segment along the red line from point P upwards. Let the distance between the two centers be denoted by d. Let x be the distance from the small circle center to P. Then we must have
x^2 +h^2 = r1^2
(d-x)^2 +h^2 = r2^2
==> h = sqrt(r1^2 - 1/d^2*(r1^2-r2^2+d^2)^2)
i.e. you can solve for h, which is the radius of the circle of intersection. You can find the center point C of the circle from x, along the line N that joins the 2 circle centers.
Then you can fully describe the circle as (X,C,U,V are all vector)
X = C + (h * cos t) U + (h * sin t) V for t in [0,2*PI)
where U and V are perpendicular vectors that lie in a plane with normal N.
The last part is the easiest. It remains only to find the intersection of this circle with the final sphere. This is simply a plug and chug of the equations (plug in for x,y,z in the last equation the parametric forms of x,y,z for the circle in terms of t and solve for t.)
edit ---
The equation that you will get is actually quite ugly, you will have a whole bunch of sine's and cosine's equal to something. To solve this you can do it 2 ways:
write the cosine's and sine's in terms of exponentials using the equality
e^(it) = cos t + i sin t
then group all the e^(it) terms and you should get a quadratic equations of e^(it)'s
that you can solve for using the quadratic formula, then solve for t. This will give you the exact solution. This method will actually tell you exactly if a solution exists, two exist or one exist depending on how many of the points from the quadratic method are real.
use newton's method to solve for t, this method is not exact but its computationally much easier to understand, and it will work very well for this case.
Basically you need to do this in 3 steps. Let's say you've got three spheres, S1, S2, and S3.
C12 is the circle created by the intersection of S1 and S2.
C23 is the circle created by the intersection of S2 and S3.
P1, P2, are the intersection points of C12 and C13.
The only really hard part in here is the sphere intersection, and thankfully Mathworld has that solved pretty well. In fact, Mathworld also has the solution to the circle intersections.
From this information you should be able to create an algorithm.
after searching the web this is one of the first hits, so i am posting the most clean and easy solution i found after some hours of research here: Trilateration
This wiki site contains a full description of a fast and easy to understand vector approach, so one can code it with little effort.
Here is another interpretation of the picture which Eric posted above:
Let H be the plane spanned by the centers of the three spheres. Let C1,C2,C3 be the intersections of the spheres with H, then C1,C2,C3 are circles. Let Lij be the line connecting the two intersection points of Ci and Cj, then the three lines L12,L23,L13 intersect at one point P. Let M be the line orthogonal to H through P, then your two points of intersection lie on the line M; hence you just need to intersect M with either of the spheres.