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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 have a three points (A,B,C) that denote objects moving in 2D space. For each node I know its position and its velocity vector. All three objects are moving in the same direction.
I would like to know whether a point C (x3, y3) approximates a "positive" extension to the line formed by points A(x1, y1) and B(x2, y2). That is, I would like to know whether point C is "ahead" of point B (i.e "A->B->C" and not "C->A->B").
I know that checking if points A, B, C are collinear will give me an indication of all three points are lying on the same line, however, i cannot figure out whether point C approximates a positive extension to the line.
Any suggestion would be highly appreciated.
You can calculate the scalar product of the difference vectors AB and BC. If that is positive, then C is what you call 'in front of B. It may be way off to the left or right, though.
The scalar product would be calculated as
(b1-a1)x(c1-b1) + (b2-a2)x(c2-b2).
when A=(a1, a2), B=(b1, b2), C= (c1,c2) - it is the cos of the angle between the two vectors times the lengths of the vectors, and cos is positive for angles less than 90 degree.
Given the points of a line and a quadratic bezier curve, how do you calculate their nearest point?
There exist a scientific paper regarding this question from INRIA: Computing the minimum distance between two Bézier curves (PDF here)
I once wrote a tool to do a similar task. Bezier splines are typically parametric cubic polynomials. To compute the square of the distance between a cubic segment and a line, this is just the square of the distance between two polynomial functions, itself just another polynomial function! Note that I said the square of the distance, not the square root.
Essentially, for any point on a cubic segment, one could compute the square of the distance from that point to the line. This will be a 6th order polynomial. Can we minimize that square of the distance? Yes. The minimum must occur where the derivative of that polynomial is zero. So differentiate, getting a 5th order polynomial. Use your favorite root finding tool that generates all of the roots numerically. Jenkins & Traub, whatever. Choose the correct solution from that set of roots, excluding any solutions that are complex, and only picking a solution if it lies inside the cubic segment in question. Make sure you exclude the points that correspond to local maxima of the distance.
All of this can be efficiently done, and no iterative optimizer besides a polynomial root finder need be used, thus one does not require the use of optimization tools that require starting values, finding only a solution near that starting value.
For example, in the 3-d figure I show a curve generated by a set of points in 3-d (in red), then I took another set of points that lay in a circle outside, I computed the closest point on the inner curve from each, drawing a line down to that curve. These points of minimum distance were generated by the scheme outlined above.
I just wanna give you a few hints, in for the case Q.B.Curve // segment :
to get a fast enough computation, i think you should first think about using a kind of 'bounding box' for your algorithm.
Say P0 is first point of the Q. B. Curve, P2 the second point, P1 the control point, and P3P4 the segment then :
Compute distance from P0, P1, P2 to P3P4
if P0 OR P2 is nearest point --> this is the nearest point of the curve from P3P4. end :=).
if P1 is nearest point, and Pi (i=0 or 1) the second nearest point, the distance beetween PiPC and P3P4 is an estimate of the distance you seek that might be precise enough, depending on your needs.
if you need to be more acurate : compute P1', which is the point on the Q.B.curve the nearest from P1 : you find it applying the BQC formula with t=0.5. --> distance from PiP1' to P3P4 is an even more accurate estimate -but more costly-.
Note that if the line defined by P1P1' intersects P3P4, P1' is the closest point of QBC from P3P4.
if P1P1' does not intersect P3P4, then you're out of luck, you must go the hard way...
Now if (and when) you need precision :
think about using a divide and conquer algorithm on the parameter of the curve :
which is nearest from P3P4 ?? P0P1' or P1'P2 ??? if it is P0P1' --> t is beetween 0 and 0.5 so compute Pm for t=0.25.
Now which is nearest from P3P4?? P0Pm or PmP1' ?? if it is PmP1' --> compute Pm2 for t=0.25+0.125=0.375 then which is nearest ? PmPm2 or Pm2P1' ??? etc
you will come to accurate solution in no time, like 6 iteration and your precision on t is 0.004 !! you might stop the search when distance beetween two points becomes below a given value. (and not difference beetwen two parameters, since for a little change in parameter, points might be far away)
in fact the principle of this algorithm is to approximate the curve with segments more and more precisely each time.
For the curve / curve case i would first 'box' them also to avoid useless computation, so first use segment/segment computation, then (maybe) segment/curve computation, and only if needed curve/curve computation.
For curve/curve, divide and conquer works also, more difficult to explain but you might figure it out. :=)
hope you can find your good balance for speed/accuracy with this :=)
Edit : Think i found for the general case a nice solution :-)
You should iterate on the (inner) bounding triangles of each B.Q.C.
So we have Triangle T1, points A, B, C having 't' parameter tA, tB, tC.
and Triangle T2, points D, E, F, having t parameter tD, tE, tF.
Initially we have tA=0 tB=0.5 tC= 1.0 and same for T2 tD=0, tE=0.5, tF=1.0
The idea is to call a procedure recursivly that will split T1 and/or T2 into smaller rectangles until we are ok with the precision reached.
The first step is to compute distance from T1 from T2, keeping track of with segments were the nearest on each triangle. First 'trick': if on T1 the segment is AC, then stop recursivity on T1, the nearest point on Curve 1 is either A or C. if on T2 the nearest segment is DF, then stop recursivity on T2, the nearest point on Curve2 is either D or F. If we stopped recursivity for both -> return distance = min (AD, AF, CD, CF). then if we have recursivity on T1, and segment AB is nearest, new T1 becomes : A'=A B= point of Curve one with tB=(tA+tC)/2 = 0.25, C=old B. same goes for T2 : apply recursivityif needed and call same algorithm on new T1 and new T2. Stop algorithm when distance found beetween T1 and T2 minus distance found beetween previous T1 and T2 is below a threshold.
the function might look like ComputeDistance(curveParam1, A, C, shouldSplitCurve1, curveParam2, D, F, shouldSplitCurve2, previousDistance) where points store also their t parameters.
note that distance (curve, segment) is just a particular case of this algorithm, and that you should implement distance (triangle, triangle) and distance (segment, triangle) to have it worked. Have fun.
1.Simple bad method - by iteration go by point from first curve and go by point from second curve and get minimum
2.Determine math function of distance between curves and calc limit of this function like:
|Fcur1(t)-Fcur2(t)| ->0
Fs is vector.
I think we can calculate the derivative of this for determine extremums and get nearest and farest points
I think about this some time later, and post full response.
Formulate your problem in terms of standard analysis: You have got a quantity to minimize (distance), so you formulate an equation for this quantity and find the points where the first derivatives are zero. Parameterize with a single parameter by using the curve's parameter p, which is between 0 for the first point and 1 for the last point.
In the line case, the equation is fairly simple: Get the x/y coordinates from the spline's equation and compute the distance to the given line via vector equations (scalar product with the line's normal).
In the curve's case, the analytical solution could get pretty complicated. You might want to use a numerical minimization technique such as Nelder-Mead or, since you have a 1D continuous problem, simple bisection.
In the case of a Bézier curve and a line
There are three candidates for the closest point to the line:
The place on the Bézier curve segment that is parallel to the line (if such a place exists),
One end of the curve segment,
The other end of the curve segment.
Test all three; the shortest distance wins.
In the case of two Bézier curves
Depends if you want the exact analytical result, or if an optimised numerical result is good enough.
Analytical result
Given two Bézier curves A(t) and B(s), you can derive equations for their local orientation A'(t) and B'(s). The point pairs for which A'(t) = B'(s) are candidates, i.e. the (t, s) for which the curves are locally parallel. I haven't checked, but I assume that A'(t) - B'(s) = 0 can be solved analytically. If your curves are anything like those you show in your example, there should be either only one solution or no solution to that equation, but there could be two (or infinitely many in the case where the curves identical but translated -- in which case you can ignore this because the winner will always be one of the curve segment endpoints).
In an approach similar to the curve-line case outline above, test each of these point pairs, plus the curve segment endpoints. The shortest distance wins.
Numerical result
Let's say the points on the two Bézier curves are defined as A(t) and B(s). You want to minimize the distance d( t, s) = |A(t) - B(s)|. It's a simple two-parameter optimization problem: find the s and t that minimize d( t, s) with the constraints 0 ≤ t ≤ 1 and 0 ≤ s ≤ 1.
Since d = SQRT( ( xA - xB)² + (yA - yB)²), you can also just minimize the function f( t, s) = [d( t, s)]² to save a square root calculation.
There are numerous ready-made methods for such optimization problems. Pick and choose.
Note that in both cases above, anything higher-order than quadratic Bézier curves can giver you more than one local minimum, so this is something to watch out for. From the examples you give, it looks like your curves have no inflexion points, so this concern may not apply in your case.
The point where there normals match is their nearest point. I mean u draw a line orthogonal to the line. .if that line is orthogonal to the curve as well then the point of intersection is the nearest point
Given two points, A and B, defined by longitude and latitude I want to determine if another point C is ~between~ A and B. ~between~ is hard for me to define. I don't mean on the line - it almost certainly won't be.
Geometric diagram http://www.freeimagehosting.net/uploads/b5c5ebf480.jpg
In this diagram, point C is ~between~ A and B because it is between the normals of points A and B and the line between them (normals denoted by thin line). Point D is not ~between~ A and B but it is ~between~ B and F.
Another way of saying this is that I want to determine if the triangles ABC and ABD are obtuse or not.
Note that the points will be very close together - within 10s of metres normally.
I'm thinking that the law of haversines may help but I don't know what the inverse of haversine is.
Many thanks for all help.
First, start with translating your points to local tangent plane. We will use the fact that your triangles are much smaller than the earth's radius. (Tangent space is such that equal deltas in each of the two coordinates correspond to equal distances)
This is done by dividing longtitudes by sin(lat):
A_local_x = A_lat_rads;
A_local_y = A_lon_rads/sin(A_lat_rads);
Then,
Compute lengths:
double ABsquared = (A_local_x - B_local_x)*(A_local_x - B_local_x) + (A_local_y - B_local_y)*(A_local_y - B_local_y);
double BCsquared = ..., ACsquared.
Finally:
bool obtuse = (ABsquared+BCsquared < ACsquared) || (ABsquared+ACsquared < BCsquared);
Obtuse means "it is not within the line", as you say. I am not checking whether triangle ABC is obtuse, but whether the angles at B and at A are obtuse. That's it.
note: I haven't tested this code. Please tell me how it works by plugging different points, if there's a bug I will fix it.
If your points are very close—10s of meters could easily qualify—you may be able to approximate it as a 2-d problem, and just calculate the angles CAB, θ and CBA, φ (using dot product). If both θ and φ are less than π/2, you C is "between".
cos(θ) = (AC · AB) / (|AC| |AB|)
If that approximation isn't good enough for you, you will need spherical trigonometry, which is also not too hard.
Note that if I understood your problem correctly, you need to check if the angles CAB and CBA are acute, not that the angle ACB is obtuse or acute.
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.