Background
I've been working for some time on attempting to solve the (notoriously painful) Time Difference of Arrival (TDoA) multi-lateration problem, in 3-dimensions and using 4 nodes. If you're unfamiliar with the problem, it is to determine the coordinates of some signal source (X,Y,Z), given the coordinates of n nodes, the time of arrival of the signal at each node, and the velocity of the signal v.
My solution is as follows:
For each node, we write (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 = (v(t_i - T)**2
Where (x_i, y_i, z_i) are the coordinates of the ith node, and T is the time of emission.
We have now 4 equations in 4 unknowns. Four nodes are obviously insufficient. We could try to solve this system directly, however that seems next to impossible given the highly nonlinear nature of the problem (and, indeed, I've tried many direct techniques... and failed). Instead, we simplify this to a linear problem by considering all i/j possibilities, subtracting equation i from equation j. We obtain (n(n-1))/2 =6 equations of the form:
2*(x_j - x_i)*X + 2*(y_j - y_i)*Y + 2*(z_j - z_i)*Z + 2 * v**2 * (t_i - t_j) = v**2 ( t_i**2 - t_j**2) + (x_j**2 + y_j**2 + z_j**2) - (x_i**2 + y_i**2 + z_i**2)
Which look like Xv_1 + Y_v2 + Z_v3 + T_v4 = b. We try now to apply standard linear least squares, where the solution is the matrix vector x in A^T Ax = A^T b. Unfortunately, if you were to try feeding this into any standard linear least squares algorithm, it'll choke up. So, what do we do now?
...
The time of arrival of the signal at node i is given (of course) by:
sqrt( (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 ) / v
This equation implies that the time of arrival, T, is 0. If we have that T = 0, we can drop the T column in matrix A and the problem is greatly simplified. Indeed, NumPy's linalg.lstsq() gives a surprisingly accurate & precise result.
...
So, what I do is normalize the input times by subtracting from each equation the earliest time. All I have to do then is determine the dt that I can add to each time such that the residual of summed squared error for the point found by linear least squares is minimized.
I define the error for some dt to be the squared difference between the arrival time for the point predicted by feeding the input times + dt to the least squares algorithm, minus the input time (normalized), summed over all 4 nodes.
for node, time in nodes, times:
error += ( (sqrt( (X-x_i)**2 + (Y-y_i)**2 + (Z-z_i)**2 ) / v) - time) ** 2
My problem:
I was able to do this somewhat satisfactorily by using brute-force. I started at dt = 0, and moved by some step up to some maximum # of iterations OR until some minimum RSS error is reached, and that was the dt I added to the normalized times to obtain a solution. The resulting solutions were very accurate and precise, but quite slow.
In practice, I'd like to be able to solve this in real time, and therefore a far faster solution will be needed. I began with the assumption that the error function (that is, dt vs error as defined above) would be highly nonlinear-- offhand, this made sense to me.
Since I don't have an actual, mathematical function, I can automatically rule out methods that require differentiation (e.g. Newton-Raphson). The error function will always be positive, so I can rule out bisection, etc. Instead, I try a simple approximation search. Unfortunately, that failed miserably. I then tried Tabu search, followed by a genetic algorithm, and several others. They all failed horribly.
So, I decided to do some investigating. As it turns out the plot of the error function vs dt looks a bit like a square root, only shifted right depending upon the distance from the nodes that the signal source is:
Where dt is on horizontal axis, error on vertical axis
And, in hindsight, of course it does!. I defined the error function to involve square roots so, at least to me, this seems reasonable.
What to do?
So, my issue now is, how do I determine the dt corresponding to the minimum of the error function?
My first (very crude) attempt was to get some points on the error graph (as above), fit it using numpy.polyfit, then feed the results to numpy.root. That root corresponds to the dt. Unfortunately, this failed, too. I tried fitting with various degrees, and also with various points, up to a ridiculous number of points such that I may as well just use brute-force.
How can I determine the dt corresponding to the minimum of this error function?
Since we're dealing with high velocities (radio signals), it's important that the results be precise and accurate, as minor variances in dt can throw off the resulting point.
I'm sure that there's some infinitely simpler approach buried in what I'm doing here however, ignoring everything else, how do I find dt?
My requirements:
Speed is of utmost importance
I have access only to pure Python and NumPy in the environment where this will be run
EDIT:
Here's my code. Admittedly, a bit messy. Here, I'm using the polyfit technique. It will "simulate" a source for you, and compare results:
from numpy import poly1d, linspace, set_printoptions, array, linalg, triu_indices, roots, polyfit
from dataclasses import dataclass
from random import randrange
import math
#dataclass
class Vertexer:
receivers: list
# Defaults
c = 299792
# Receivers:
# [x_1, y_1, z_1]
# [x_2, y_2, z_2]
# [x_3, y_3, z_3]
# Solved:
# [x, y, z]
def error(self, dt, times):
solved = self.linear([time + dt for time in times])
error = 0
for time, receiver in zip(times, self.receivers):
error += ((math.sqrt( (solved[0] - receiver[0])**2 +
(solved[1] - receiver[1])**2 +
(solved[2] - receiver[2])**2 ) / c ) - time)**2
return error
def linear(self, times):
X = array(self.receivers)
t = array(times)
x, y, z = X.T
i, j = triu_indices(len(x), 1)
A = 2 * (X[i] - X[j])
b = self.c**2 * (t[j]**2 - t[i]**2) + (X[i]**2).sum(1) - (X[j]**2).sum(1)
solved, residuals, rank, s = linalg.lstsq(A, b, rcond=None)
return(solved)
def find(self, times):
# Normalize times
times = [time - min(times) for time in times]
# Fit the error function
y = []
x = []
dt = 1E-10
for i in range(50000):
x.append(self.error(dt * i, times))
y.append(dt * i)
p = polyfit(array(x), array(y), 2)
r = roots(p)
return(self.linear([time + r for time in times]))
# SIMPLE CODE FOR SIMULATING A SIGNAL
# Pick nodes to be at random locations
x_1 = randrange(10); y_1 = randrange(10); z_1 = randrange(10)
x_2 = randrange(10); y_2 = randrange(10); z_2 = randrange(10)
x_3 = randrange(10); y_3 = randrange(10); z_3 = randrange(10)
x_4 = randrange(10); y_4 = randrange(10); z_4 = randrange(10)
# Pick source to be at random location
x = randrange(1000); y = randrange(1000); z = randrange(1000)
# Set velocity
c = 299792 # km/ns
# Generate simulated source
t_1 = math.sqrt( (x - x_1)**2 + (y - y_1)**2 + (z - z_1)**2 ) / c
t_2 = math.sqrt( (x - x_2)**2 + (y - y_2)**2 + (z - z_2)**2 ) / c
t_3 = math.sqrt( (x - x_3)**2 + (y - y_3)**2 + (z - z_3)**2 ) / c
t_4 = math.sqrt( (x - x_4)**2 + (y - y_4)**2 + (z - z_4)**2 ) / c
print('Actual:', x, y, z)
myVertexer = Vertexer([[x_1, y_1, z_1],[x_2, y_2, z_2],[x_3, y_3, z_3],[x_4, y_4, z_4]])
solution = myVertexer.find([t_1, t_2, t_3, t_4])
print(solution)
It seems like the Bancroft method applies to this problem? Here's a pure NumPy implementation.
# Implementation of the Bancroft method, following
# https://gssc.esa.int/navipedia/index.php/Bancroft_Method
M = np.diag([1, 1, 1, -1])
def lorentz_inner(v, w):
return np.sum(v * (w # M), axis=-1)
B = np.array(
[
[x_1, y_1, z_1, c * t_1],
[x_2, y_2, z_2, c * t_2],
[x_3, y_3, z_3, c * t_3],
[x_4, y_4, z_4, c * t_4],
]
)
one = np.ones(4)
a = 0.5 * lorentz_inner(B, B)
B_inv_one = np.linalg.solve(B, one)
B_inv_a = np.linalg.solve(B, a)
for Lambda in np.roots(
[
lorentz_inner(B_inv_one, B_inv_one),
2 * (lorentz_inner(B_inv_one, B_inv_a) - 1),
lorentz_inner(B_inv_a, B_inv_a),
]
):
x, y, z, c_t = M # np.linalg.solve(B, Lambda * one + a)
print("Candidate:", x, y, z, c_t / c)
My answer might have mistakes (glaring) as I had not heard the TDOA term before this afternoon. Please double check if the method is right.
I could not find solution to your original problem of finding dt corresponding to the minimum error. My answer also deviates from the requirement that other than numpy no third party library had to be used (I used Sympy and largely used the code from here). However I am still posting this thinking that somebody someday might find it useful if all one is interested in ... is to find X,Y,Z of the source emitter. This method also does not take into account real-life situations where white noise or errors might be present or curvature of the earth and other complications.
Your initial test conditions are as below.
from random import randrange
import math
# SIMPLE CODE FOR SIMULATING A SIGNAL
# Pick nodes to be at random locations
x_1 = randrange(10); y_1 = randrange(10); z_1 = randrange(10)
x_2 = randrange(10); y_2 = randrange(10); z_2 = randrange(10)
x_3 = randrange(10); y_3 = randrange(10); z_3 = randrange(10)
x_4 = randrange(10); y_4 = randrange(10); z_4 = randrange(10)
# Pick source to be at random location
x = randrange(1000); y = randrange(1000); z = randrange(1000)
# Set velocity
c = 299792 # km/ns
# Generate simulated source
t_1 = math.sqrt( (x - x_1)**2 + (y - y_1)**2 + (z - z_1)**2 ) / c
t_2 = math.sqrt( (x - x_2)**2 + (y - y_2)**2 + (z - z_2)**2 ) / c
t_3 = math.sqrt( (x - x_3)**2 + (y - y_3)**2 + (z - z_3)**2 ) / c
t_4 = math.sqrt( (x - x_4)**2 + (y - y_4)**2 + (z - z_4)**2 ) / c
print('Actual:', x, y, z)
My solution is as below.
import sympy as sym
X,Y,Z = sym.symbols('X,Y,Z', real=True)
f = sym.Eq((x_1 - X)**2 +(y_1 - Y)**2 + (z_1 - Z)**2 , (c*t_1)**2)
g = sym.Eq((x_2 - X)**2 +(y_2 - Y)**2 + (z_2 - Z)**2 , (c*t_2)**2)
h = sym.Eq((x_3 - X)**2 +(y_3 - Y)**2 + (z_3 - Z)**2 , (c*t_3)**2)
i = sym.Eq((x_4 - X)**2 +(y_4 - Y)**2 + (z_4 - Z)**2 , (c*t_4)**2)
print("Solved coordinates are ", sym.solve([f,g,h,i],X,Y,Z))
print statement from your initial condition gave.
Actual: 111 553 110
and the solution that almost instantly came out was
Solved coordinates are [(111.000000000000, 553.000000000000, 110.000000000000)]
Sorry again if something is totally amiss.
I want to calculate the derivative of points, a few internet posts suggested using np.diff function. However, I tried using np.diff against manually calculated results (chose a random polynomial equation and differentiated it) to see if I end up with the same results. I used the following eq : Y = (X^3) + (X^2) + 7 and the results i ended up with were different. Any ideas why?. Is there any other method to calculate the differential.
In the problem, I am trying to solve, I have recieved data points of fitted spline function ( not the original data that need to be fitted by spline but the points of the already fitted spline). The x-values are at equal intervals. I only have the points and no equation, what I require is to calculate, the first, second and third derivatives. i.e dy/dx, d2y/dx2, d3y/dx3. Any ideas on how to do this?. Thanks in advance.
xval = [1,2,3,4,5]
yval = []
yval_dashList = []
#selected a polynomial equation
def calc_Y(X):
Y = (X**3) + (X**2) + 7
return(Y)
#calculate y values using equatuion
for i in xval:
yval.append(calc_Y(i))
#output: yval = [9,19,43,87,157]
#manually differentiated the equation or use sympy library (sym.diff(x**3 + x**2 + 7))
def calc_diffY(X):
yval_dash = 3*(X**2) + 2**X
#store differentiated y-values in a list
for i in xval:
yval_dashList.append(yval_dash(i))
#output: yval_dashList = [5,16,35,64,107]
#use numpy diff method on the y values(yval)
numpyDiff = np.diff(yval)
#output: [10,24,44,60]
The values of numpy diff method [10,24,44,60] is different from yval_dashList = [5,16,35,64,107]
The idea behind what you are trying to do is correct, but there are a couple of points to make it work as intended:
There is a typo in calc_diffY(X), the derivative of X**2 is 2*X, not 2**X:
def calc_diffY(X):
yval_dash = 3*(X**2) + 2*X
By doing this you don't obtain much better results:
yval_dash = [5, 16, 33, 56, 85]
numpyDiff = [10. 24. 44. 70.]
To calculate the numerical derivative you should do a "Difference quotient" which is an approximation of a derivative
numpyDiff = np.diff(yval)/np.diff(xval)
The approximation becomes better and better if the values of the points are more dense.
The difference between your points on the x axis is 1, so you end up in this situation (in blue the analytical derivative, in red the numerical):
If you reduce the difference in your x points to 0.1, you get this, which is much better:
Just to add something to this, have a look at this image showing the effect of reducing the distance of the points on which the derivative is numerically calculated, taken from Wikipedia:
I like #lgsp's answer. I will add that you can directly estimate the derivative without having to worry about how much space there is between the values. This just uses the symmetric formula for calculating finite differences, described at this wikipedia page.
Take note, though, of the way delta is specified. I found that when it is too small, higher-order estimates fail. There's probably not a 100% generic value that will always work well!
Also, I simplified your code by taking advantage of numpy broadcasting over arrays to eliminate for loops.
import numpy as np
# selecte a polynomial equation
def f(x):
y = x**3 + x**2 + 7
return y
# manually differentiate the equation
def f_prime(x):
return 3*x**2 + 2*x
# numerically estimate the first three derivatives
def d1(f, x, delta=1e-10):
return (f(x + delta) - f(x - delta)) / (2 * delta)
def d2(f, x, delta=1e-5):
return (d1(f, x + delta, delta) - d1(f, x - delta, delta)) / (2 * delta)
def d3(f, x, delta=1e-2):
return (d2(f, x + delta, delta) - d2(f, x - delta, delta)) / (2 * delta)
# demo output
# note that functions operate in parallel on numpy arrays -- no for loops!
xval = np.array([1,2,3,4,5])
print('y = ', f(xval))
print('y\' = ', f_prime(xval))
print('d1 = ', d1(f, xval))
print('d2 = ', d2(f, xval))
print('d3 = ', d3(f, xval))
And the outputs:
y = [ 9 19 43 87 157]
y' = [ 5 16 33 56 85]
d1 = [ 5.00000041 16.00000132 33.00002049 56.00000463 84.99995374]
d2 = [ 8.0000051 14.00000116 20.00000165 25.99996662 32.00000265]
d3 = [6. 6. 6. 6. 5.99999999]
I'm currently working on a python library to extract pen strokes from TrueType fonts
- Here I'm defining a stroke as a midline running between a test point and its reflected point.
I'm using the term reflected point to refer to the closest point on the opposite side of the 'ink' region which under normal circumstances (apart from say at a serif stem) would also have a tangent in the opposite direction to the test point.
I'm working in python using fontTools and a bezier library that I rolled from the processing code described at http://pomax.github.io/bezierinfo/#extremities .
Where I'm stuck stuck at the moment is on how to find the point on a quadratic bezier curve that has a given tangent, my mathematics skills are pretty rudimentary on a good day with a clear head [which it is not rite now] so I was hoping someone with a sharper mind could point out a birds eye overview on how to achieve this.
At the moment the only thing I can think of is to approach it numerically with something similar to the Newton-Raphson root finding algorithm but evaluating the 1st derivative against the target direction values. I am hoping however there is a symbolic solution as this needs to be run on every other curve for each curve in the glyphs contours.
Using the notation given in http://pomax.github.io/bezierinfo/#extremities, a quadratic Bezier curve is given by:
B(t) = P1*(1-t)**2 + 2*P2*(1-t)*t + P3*t**2
Therefore, (by taking the derivative of B with respect to t) the tangent to the curve is given by
B'(t) = -2*P1*(1-t) + 2*P2*(1-2*t) + 2*P3*t
= 2*(P1 - 2*P2 + P3)*t + 2*(-P1 + P2)
Given some tangent direction V, solve
B'(t) = V
for t. If there is a solution, t = ts, then the point on the Bezier curve which has tangent direction V is given by B(ts).
We essentially want to know if two vectors (B'(t) and V) are parallel or anti-parallel. There is a trick to doing that.
Two vectors, X and Y, are perpendicular if their dot product is zero. If X = (a,b) and Y = (c,d) then the dot product of X and Y is given by
a*c + b*d
So, X and Y are parallel if X and Y_perp are perpendicular, where Y_perp is a vector perpendicular to Y.
In 2-dimensions, if in coordinates Y = (a,b) then Y_perp = (-b, a). (There are two perpendicular vectors possible, but this one will do.) Notice that -- using the formula above -- the dot product of Y and Y_perp is
a*(-b) + b*(a) = 0
So indeed, this jibes with the claim that perpendicular vectors have a dot product equal to 0.
Now, to solve our problem: Let
B'(t) = (a*t+b, c*t+d)
V = (e, f)
Then B'(t) is parallel (or anti-parallel) to V if or when
B'(t) is perpendicular to V_perp, which occurs when
dot product((a*t+b, c*t+d), (-f, e)) = 0
-(a*t+b)*f + (c*t+d)*e = 0
We know a, b, c, d, e and f. Solve for t. If t lies between 0 and 1, then B(t) is part of the Bezier curve segment between P1 and P3.
Thanks to #ubutbu for pointing out the solution for me, figured I would post a working implementation in case someone googles upon this question with a need in the future:
def findParallelT(P, V):
"""finds the t value along a quadratic Bezier such that its tangent (1st derivative) is parallel with the direction vector V.
P : a 3x2 matrix containing the control points of B(t).
V : a pair of values representing the direction of interest (magnitude is ignored).
returns 0.0 <= t <= 1.0 or None
Note the result may be in the same direction or flipped 180 degrees from V"""
#refer to answer given by 'unutbu' on the stackoverflow question:
#http://stackoverflow.com/questions/20825173/how-to-find-a-point-if-any-on-quadratic-bezier-with-a-given-tangent-direction
#for explanation.
# also the 'Rearrange It' app at http://www.wolframalpha.com/widgets/view.jsp?id=4be4308d0f9d17d1da68eea39de9b2ce was invaluable.
assert len(P)==3 and len(P[0])==2 and len(P[1])==2 and len(P[2])==2
assert len(V)==2
P1=P[0]
P2=P[1]
P3=P[2]
# B(t) = P1 * (1-t)**2 + 2*P2*(1-t)*t + P3*t**2
# B'(t) = 2 * (1-t) * (P2 - P1) + 2*t*(P3-P2)
# B'(t) = (a*t + b, c*t + d), V = (e, f)
a = -2 * (P2[0] - P1[0]) + 2 * (P3[0]-P2[0])
b = 2 * (P2[0] - P1[0])
c = -2 * (P2[1] - P1[1]) + 2 * (P3[1]-P2[1])
d = 2 * (P2[1] - P1[1])
e = V[0]
f = V[1]
# -(a*t+b)*f + (c*t+d)*e = 0 for parallel... t = (d*e - b*f) / (a*f - c*e)
num = float(d * e - b * f)
den = float(a * f - c * e)
if den == 0:
return None
else:
t = num / den
return t if 0.0 <= t <= 1.0 else None
if __name__ == "__main__":
assert findParallelT([[0.0,0.0], [0.0,25.0], [25.0,25.0]], [25.0,25.0]) == 0.5
assert findParallelT([[0.0,0.0], [0.0,25.0], [25.0,25.0]], [25.0,-25.0]) == None
assert findParallelT([[200.0,200.0], [10.0, 20.0], [300.0, 0.0]], [10.0,0.0]) == None
assert findParallelT([[407.5, 376.5],[321.0,463.0],[321.0,586.0]], [-246.0,0.0] ) == None
assert findParallelT([[617.0, 882.0],[740.0, 882.0], [826.5, 795.5]], [-245.0,0.0]) == 0.0
shadertoy.com has a search function and "bezier" gets you to analytic solutions for calculation (over 2 domains)
The shortest distance of any planar point to a planar quadratic-bezier, by calculating 2 roots of a cubic function
(quadratic Bezier will only have 2 real roots) (euclidean distance increases exponent of polynomial by +1).
3 points on the Bezier that equate to the 3 local extrema of the cubic function, one for each root of the quadratic.
so you just calculate which of the 3 points is closest.
I need an algorithm that can give me positions around a sphere for N points (less than 20, probably) that vaguely spreads them out. There's no need for "perfection", but I just need it so none of them are bunched together.
This question provided good code, but I couldn't find a way to make this uniform, as this seemed 100% randomized.
This blog post recommended had two ways allowing input of number of points on the sphere, but the Saff and Kuijlaars algorithm is exactly in psuedocode I could transcribe, and the code example I found contained "node[k]", which I couldn't see explained and ruined that possibility. The second blog example was the Golden Section Spiral, which gave me strange, bunched up results, with no clear way to define a constant radius.
This algorithm from this question seems like it could possibly work, but I can't piece together what's on that page into psuedocode or anything.
A few other question threads I came across spoke of randomized uniform distribution, which adds a level of complexity I'm not concerned about. I apologize that this is such a silly question, but I wanted to show that I've truly looked hard and still come up short.
So, what I'm looking for is simple pseudocode to evenly distribute N points around a unit sphere, that either returns in spherical or Cartesian coordinates. Even better if it can even distribute with a bit of randomization (think planets around a star, decently spread out, but with room for leeway).
The Fibonacci sphere algorithm is great for this. It is fast and gives results that at a glance will easily fool the human eye. You can see an example done with processing which will show the result over time as points are added. Here's another great interactive example made by #gman. And here's a simple implementation in python.
import math
def fibonacci_sphere(samples=1000):
points = []
phi = math.pi * (3. - math.sqrt(5.)) # golden angle in radians
for i in range(samples):
y = 1 - (i / float(samples - 1)) * 2 # y goes from 1 to -1
radius = math.sqrt(1 - y * y) # radius at y
theta = phi * i # golden angle increment
x = math.cos(theta) * radius
z = math.sin(theta) * radius
points.append((x, y, z))
return points
1000 samples gives you this:
The golden spiral method
You said you couldn’t get the golden spiral method to work and that’s a shame because it’s really, really good. I would like to give you a complete understanding of it so that maybe you can understand how to keep this away from being “bunched up.”
So here’s a fast, non-random way to create a lattice that is approximately correct; as discussed above, no lattice will be perfect, but this may be good enough. It is compared to other methods e.g. at BendWavy.org but it just has a nice and pretty look as well as a guarantee about even spacing in the limit.
Primer: sunflower spirals on the unit disk
To understand this algorithm, I first invite you to look at the 2D sunflower spiral algorithm. This is based on the fact that the most irrational number is the golden ratio (1 + sqrt(5))/2 and if one emits points by the approach “stand at the center, turn a golden ratio of whole turns, then emit another point in that direction,” one naturally constructs a spiral which, as you get to higher and higher numbers of points, nevertheless refuses to have well-defined ‘bars’ that the points line up on.(Note 1.)
The algorithm for even spacing on a disk is,
from numpy import pi, cos, sin, sqrt, arange
import matplotlib.pyplot as pp
num_pts = 100
indices = arange(0, num_pts, dtype=float) + 0.5
r = sqrt(indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
pp.scatter(r*cos(theta), r*sin(theta))
pp.show()
and it produces results that look like (n=100 and n=1000):
Spacing the points radially
The key strange thing is the formula r = sqrt(indices / num_pts); how did I come to that one? (Note 2.)
Well, I am using the square root here because I want these to have even-area spacing around the disk. That is the same as saying that in the limit of large N I want a little region R ∈ (r, r + dr), Θ ∈ (θ, θ + dθ) to contain a number of points proportional to its area, which is r dr dθ. Now if we pretend that we are talking about a random variable here, this has a straightforward interpretation as saying that the joint probability density for (R, Θ) is just c r for some constant c. Normalization on the unit disk would then force c = 1/π.
Now let me introduce a trick. It comes from probability theory where it’s known as sampling the inverse CDF: suppose you wanted to generate a random variable with a probability density f(z) and you have a random variable U ~ Uniform(0, 1), just like comes out of random() in most programming languages. How do you do this?
First, turn your density into a cumulative distribution function or CDF, which we will call F(z). A CDF, remember, increases monotonically from 0 to 1 with derivative f(z).
Then calculate the CDF’s inverse function F-1(z).
You will find that Z = F-1(U) is distributed according to the target density. (Note 3).
Now the golden-ratio spiral trick spaces the points out in a nicely even pattern for θ so let’s integrate that out; for the unit disk we are left with F(r) = r2. So the inverse function is F-1(u) = u1/2, and therefore we would generate random points on the disk in polar coordinates with r = sqrt(random()); theta = 2 * pi * random().
Now instead of randomly sampling this inverse function we’re uniformly sampling it, and the nice thing about uniform sampling is that our results about how points are spread out in the limit of large N will behave as if we had randomly sampled it. This combination is the trick. Instead of random() we use (arange(0, num_pts, dtype=float) + 0.5)/num_pts, so that, say, if we want to sample 10 points they are r = 0.05, 0.15, 0.25, ... 0.95. We uniformly sample r to get equal-area spacing, and we use the sunflower increment to avoid awful “bars” of points in the output.
Now doing the sunflower on a sphere
The changes that we need to make to dot the sphere with points merely involve switching out the polar coordinates for spherical coordinates. The radial coordinate of course doesn't enter into this because we're on a unit sphere. To keep things a little more consistent here, even though I was trained as a physicist I'll use mathematicians' coordinates where 0 ≤ φ ≤ π is latitude coming down from the pole and 0 ≤ θ ≤ 2π is longitude. So the difference from above is that we are basically replacing the variable r with φ.
Our area element, which was r dr dθ, now becomes the not-much-more-complicated sin(φ) dφ dθ. So our joint density for uniform spacing is sin(φ)/4π. Integrating out θ, we find f(φ) = sin(φ)/2, thus F(φ) = (1 − cos(φ))/2. Inverting this we can see that a uniform random variable would look like acos(1 - 2 u), but we sample uniformly instead of randomly, so we instead use φk = acos(1 − 2 (k + 0.5)/N). And the rest of the algorithm is just projecting this onto the x, y, and z coordinates:
from numpy import pi, cos, sin, arccos, arange
import mpl_toolkits.mplot3d
import matplotlib.pyplot as pp
num_pts = 1000
indices = arange(0, num_pts, dtype=float) + 0.5
phi = arccos(1 - 2*indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
x, y, z = cos(theta) * sin(phi), sin(theta) * sin(phi), cos(phi);
pp.figure().add_subplot(111, projection='3d').scatter(x, y, z);
pp.show()
Again for n=100 and n=1000 the results look like:
Further research
I wanted to give a shout out to Martin Roberts’s blog. Note that above I created an offset of my indices by adding 0.5 to each index. This was just visually appealing to me, but it turns out that the choice of offset matters a lot and is not constant over the interval and can mean getting as much as 8% better accuracy in packing if chosen correctly. There should also be a way to get his R2 sequence to cover a sphere and it would be interesting to see if this also produced a nice even covering, perhaps as-is but perhaps needing to be, say, taken from only a half of the unit square cut diagonally or so and stretched around to get a circle.
Notes
Those “bars” are formed by rational approximations to a number, and the best rational approximations to a number come from its continued fraction expression, z + 1/(n_1 + 1/(n_2 + 1/(n_3 + ...))) where z is an integer and n_1, n_2, n_3, ... is either a finite or infinite sequence of positive integers:
def continued_fraction(r):
while r != 0:
n = floor(r)
yield n
r = 1/(r - n)
Since the fraction part 1/(...) is always between zero and one, a large integer in the continued fraction allows for a particularly good rational approximation: “one divided by something between 100 and 101” is better than “one divided by something between 1 and 2.” The most irrational number is therefore the one which is 1 + 1/(1 + 1/(1 + ...)) and has no particularly good rational approximations; one can solve φ = 1 + 1/φ by multiplying through by φ to get the formula for the golden ratio.
For folks who are not so familiar with NumPy -- all of the functions are “vectorized,” so that sqrt(array) is the same as what other languages might write map(sqrt, array). So this is a component-by-component sqrt application. The same also holds for division by a scalar or addition with scalars -- those apply to all components in parallel.
The proof is simple once you know that this is the result. If you ask what's the probability that z < Z < z + dz, this is the same as asking what's the probability that z < F-1(U) < z + dz, apply F to all three expressions noting that it is a monotonically increasing function, hence F(z) < U < F(z + dz), expand the right hand side out to find F(z) + f(z) dz, and since U is uniform this probability is just f(z) dz as promised.
This is known as packing points on a sphere, and there is no (known) general, perfect solution. However, there are plenty of imperfect solutions. The three most popular seem to be:
Create a simulation. Treat each point as an electron constrained to a sphere, then run a simulation for a certain number of steps. The electrons' repulsion will naturally tend the system to a more stable state, where the points are about as far away from each other as they can get.
Hypercube rejection. This fancy-sounding method is actually really simple: you uniformly choose points (much more than n of them) inside of the cube surrounding the sphere, then reject the points outside of the sphere. Treat the remaining points as vectors, and normalize them. These are your "samples" - choose n of them using some method (randomly, greedy, etc).
Spiral approximations. You trace a spiral around a sphere, and evenly-distribute the points around the spiral. Because of the mathematics involved, these are more complicated to understand than the simulation, but much faster (and probably involving less code). The most popular seems to be by Saff, et al.
A lot more information about this problem can be found here
In this example code node[k] is just the kth node. You are generating an array N points and node[k] is the kth (from 0 to N-1). If that is all that is confusing you, hopefully you can use that now.
(in other words, k is an array of size N that is defined before the code fragment starts, and which contains a list of the points).
Alternatively, building on the other answer here (and using Python):
> cat ll.py
from math import asin
nx = 4; ny = 5
for x in range(nx):
lon = 360 * ((x+0.5) / nx)
for y in range(ny):
midpt = (y+0.5) / ny
lat = 180 * asin(2*((y+0.5)/ny-0.5))
print lon,lat
> python2.7 ll.py
45.0 -166.91313924
45.0 -74.0730322921
45.0 0.0
45.0 74.0730322921
45.0 166.91313924
135.0 -166.91313924
135.0 -74.0730322921
135.0 0.0
135.0 74.0730322921
135.0 166.91313924
225.0 -166.91313924
225.0 -74.0730322921
225.0 0.0
225.0 74.0730322921
225.0 166.91313924
315.0 -166.91313924
315.0 -74.0730322921
315.0 0.0
315.0 74.0730322921
315.0 166.91313924
If you plot that, you'll see that the vertical spacing is larger near the poles so that each point is situated in about the same total area of space (near the poles there's less space "horizontally", so it gives more "vertically").
This isn't the same as all points having about the same distance to their neighbours (which is what I think your links are talking about), but it may be sufficient for what you want and improves on simply making a uniform lat/lon grid.
What you are looking for is called a spherical covering. The spherical covering problem is very hard and solutions are unknown except for small numbers of points. One thing that is known for sure is that given n points on a sphere, there always exist two points of distance d = (4-csc^2(\pi n/6(n-2)))^(1/2) or closer.
If you want a probabilistic method for generating points uniformly distributed on a sphere, it's easy: generate points in space uniformly by Gaussian distribution (it's built into Java, not hard to find the code for other languages). So in 3-dimensional space, you need something like
Random r = new Random();
double[] p = { r.nextGaussian(), r.nextGaussian(), r.nextGaussian() };
Then project the point onto the sphere by normalizing its distance from the origin
double norm = Math.sqrt( (p[0])^2 + (p[1])^2 + (p[2])^2 );
double[] sphereRandomPoint = { p[0]/norm, p[1]/norm, p[2]/norm };
The Gaussian distribution in n dimensions is spherically symmetric so the projection onto the sphere is uniform.
Of course, there's no guarantee that the distance between any two points in a collection of uniformly generated points will be bounded below, so you can use rejection to enforce any such conditions that you might have: probably it's best to generate the whole collection and then reject the whole collection if necessary. (Or use "early rejection" to reject the whole collection you've generated so far; just don't keep some points and drop others.) You can use the formula for d given above, minus some slack, to determine the min distance between points below which you will reject a set of points. You'll have to calculate n choose 2 distances, and the probability of rejection will depend on the slack; it's hard to say how, so run a simulation to get a feel for the relevant statistics.
This answer is based on the same 'theory' that is outlined well by this answer
I'm adding this answer as:
-- None of the other options fit the 'uniformity' need 'spot-on' (or not obviously-clearly so). (Noting to get the planet like distribution looking behavior particurally wanted in the original ask, you just reject from the finite list of the k uniformly created points at random (random wrt the index count in the k items back).)
--The closest other impl forced you to decide the 'N' by 'angular axis', vs. just 'one value of N' across both angular axis values ( which at low counts of N is very tricky to know what may, or may not matter (e.g. you want '5' points -- have fun ) )
--Furthermore, it's very hard to 'grok' how to differentiate between the other options without any imagery, so here's what this option looks like (below), and the ready-to-run implementation that goes with it.
with N at 20:
and then N at 80:
here's the ready-to-run python3 code, where the emulation is that same source: " http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere " found by others. ( The plotting I've included, that fires when run as 'main,' is taken from: http://www.scipy.org/Cookbook/Matplotlib/mplot3D )
from math import cos, sin, pi, sqrt
def GetPointsEquiAngularlyDistancedOnSphere(numberOfPoints=45):
""" each point you get will be of form 'x, y, z'; in cartesian coordinates
eg. the 'l2 distance' from the origion [0., 0., 0.] for each point will be 1.0
------------
converted from: http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere )
"""
dlong = pi*(3.0-sqrt(5.0)) # ~2.39996323
dz = 2.0/numberOfPoints
long = 0.0
z = 1.0 - dz/2.0
ptsOnSphere =[]
for k in range( 0, numberOfPoints):
r = sqrt(1.0-z*z)
ptNew = (cos(long)*r, sin(long)*r, z)
ptsOnSphere.append( ptNew )
z = z - dz
long = long + dlong
return ptsOnSphere
if __name__ == '__main__':
ptsOnSphere = GetPointsEquiAngularlyDistancedOnSphere( 80)
#toggle True/False to print them
if( True ):
for pt in ptsOnSphere: print( pt)
#toggle True/False to plot them
if(True):
from numpy import *
import pylab as p
import mpl_toolkits.mplot3d.axes3d as p3
fig=p.figure()
ax = p3.Axes3D(fig)
x_s=[];y_s=[]; z_s=[]
for pt in ptsOnSphere:
x_s.append( pt[0]); y_s.append( pt[1]); z_s.append( pt[2])
ax.scatter3D( array( x_s), array( y_s), array( z_s) )
ax.set_xlabel('X'); ax.set_ylabel('Y'); ax.set_zlabel('Z')
p.show()
#end
tested at low counts (N in 2, 5, 7, 13, etc) and seems to work 'nice'
Try:
function sphere ( N:float,k:int):Vector3 {
var inc = Mathf.PI * (3 - Mathf.Sqrt(5));
var off = 2 / N;
var y = k * off - 1 + (off / 2);
var r = Mathf.Sqrt(1 - y*y);
var phi = k * inc;
return Vector3((Mathf.Cos(phi)*r), y, Mathf.Sin(phi)*r);
};
The above function should run in loop with N loop total and k loop current iteration.
It is based on a sunflower seeds pattern, except the sunflower seeds are curved around into a half dome, and again into a sphere.
Here is a picture, except I put the camera half way inside the sphere so it looks 2d instead of 3d because the camera is same distance from all points.
http://3.bp.blogspot.com/-9lbPHLccQHA/USXf88_bvVI/AAAAAAAAADY/j7qhQsSZsA8/s640/sphere.jpg
Healpix solves a closely related problem (pixelating the sphere with equal area pixels):
http://healpix.sourceforge.net/
It's probably overkill, but maybe after looking at it you'll realize some of it's other nice properties are interesting to you. It's way more than just a function that outputs a point cloud.
I landed here trying to find it again; the name "healpix" doesn't exactly evoke spheres...
edit: This does not answer the question the OP meant to ask, leaving it here in case people find it useful somehow.
We use the multiplication rule of probability, combined with infinitessimals. This results in 2 lines of code to achieve your desired result:
longitude: φ = uniform([0,2pi))
azimuth: θ = -arcsin(1 - 2*uniform([0,1]))
(defined in the following coordinate system:)
Your language typically has a uniform random number primitive. For example in python you can use random.random() to return a number in the range [0,1). You can multiply this number by k to get a random number in the range [0,k). Thus in python, uniform([0,2pi)) would mean random.random()*2*math.pi.
Proof
Now we can't assign θ uniformly, otherwise we'd get clumping at the poles. We wish to assign probabilities proportional to the surface area of the spherical wedge (the θ in this diagram is actually φ):
An angular displacement dφ at the equator will result in a displacement of dφ*r. What will that displacement be at an arbitrary azimuth θ? Well, the radius from the z-axis is r*sin(θ), so the arclength of that "latitude" intersecting the wedge is dφ * r*sin(θ). Thus we calculate the cumulative distribution of the area to sample from it, by integrating the area of the slice from the south pole to the north pole.
(where stuff=dφ*r)
We will now attempt to get the inverse of the CDF to sample from it: http://en.wikipedia.org/wiki/Inverse_transform_sampling
First we normalize by dividing our almost-CDF by its maximum value. This has the side-effect of cancelling out the dφ and r.
azimuthalCDF: cumProb = (sin(θ)+1)/2 from -pi/2 to pi/2
inverseCDF: θ = -sin^(-1)(1 - 2*cumProb)
Thus:
let x by a random float in range [0,1]
θ = -arcsin(1-2*x)
with small numbers of points you could run a simulation:
from random import random,randint
r = 10
n = 20
best_closest_d = 0
best_points = []
points = [(r,0,0) for i in range(n)]
for simulation in range(10000):
x = random()*r
y = random()*r
z = r-(x**2+y**2)**0.5
if randint(0,1):
x = -x
if randint(0,1):
y = -y
if randint(0,1):
z = -z
closest_dist = (2*r)**2
closest_index = None
for i in range(n):
for j in range(n):
if i==j:
continue
p1,p2 = points[i],points[j]
x1,y1,z1 = p1
x2,y2,z2 = p2
d = (x1-x2)**2+(y1-y2)**2+(z1-z2)**2
if d < closest_dist:
closest_dist = d
closest_index = i
if simulation % 100 == 0:
print simulation,closest_dist
if closest_dist > best_closest_d:
best_closest_d = closest_dist
best_points = points[:]
points[closest_index]=(x,y,z)
print best_points
>>> best_points
[(9.921692138442777, -9.930808529773849, 4.037839326088124),
(5.141893371460546, 1.7274947332807744, -4.575674650522637),
(-4.917695758662436, -1.090127967097737, -4.9629263893193745),
(3.6164803265540666, 7.004158551438312, -2.1172868271109184),
(-9.550655088997003, -9.580386054762917, 3.5277052594769422),
(-0.062238110294250415, 6.803105171979587, 3.1966101417463655),
(-9.600996012203195, 9.488067284474834, -3.498242301168819),
(-8.601522086624803, 4.519484132245867, -0.2834204048792728),
(-1.1198210500791472, -2.2916581379035694, 7.44937337008726),
(7.981831370440529, 8.539378431788634, 1.6889099589074377),
(0.513546008372332, -2.974333486904779, -6.981657873262494),
(-4.13615438946178, -6.707488383678717, 2.1197605651446807),
(2.2859494919024326, -8.14336582650039, 1.5418694699275672),
(-7.241410895247996, 9.907335206038226, 2.271647103735541),
(-9.433349952523232, -7.999106443463781, -2.3682575660694347),
(3.704772125650199, 1.0526567864085812, 6.148581714099761),
(-3.5710511242327048, 5.512552040316693, -3.4318468250897647),
(-7.483466337225052, -1.506434920354559, 2.36641535124918),
(7.73363824231576, -8.460241422163824, -1.4623228616326003),
(10, 0, 0)]
Take the two largest factors of your N, if N==20 then the two largest factors are {5,4}, or, more generally {a,b}. Calculate
dlat = 180/(a+1)
dlong = 360/(b+1})
Put your first point at {90-dlat/2,(dlong/2)-180}, your second at {90-dlat/2,(3*dlong/2)-180}, your 3rd at {90-dlat/2,(5*dlong/2)-180}, until you've tripped round the world once, by which time you've got to about {75,150} when you go next to {90-3*dlat/2,(dlong/2)-180}.
Obviously I'm working this in degrees on the surface of the spherical earth, with the usual conventions for translating +/- to N/S or E/W. And obviously this gives you a completely non-random distribution, but it is uniform and the points are not bunched together.
To add some degree of randomness, you could generate 2 normally-distributed (with mean 0 and std dev of {dlat/3, dlong/3} as appropriate) and add them to your uniformly distributed points.
OR... to place 20 points, compute the centers of the icosahedronal faces. For 12 points, find the vertices of the icosahedron. For 30 points, the mid point of the edges of the icosahedron. you can do the same thing with the tetrahedron, cube, dodecahedron and octahedrons: one set of points is on the vertices, another on the center of the face and another on the center of the edges. They cannot be mixed, however.
Based on fnord's answer, here is a Unity3D version with added ranges :
Code :
// golden angle in radians
static float Phi = Mathf.PI * ( 3f - Mathf.Sqrt( 5f ) );
static float Pi2 = Mathf.PI * 2;
public static Vector3 Point( float radius , int index , int total , float min = 0f, float max = 1f , float angleStartDeg = 0f, float angleRangeDeg = 360 )
{
// y goes from min (-) to max (+)
var y = ( ( index / ( total - 1f ) ) * ( max - min ) + min ) * 2f - 1f;
// golden angle increment
var theta = Phi * index ;
if( angleStartDeg != 0 || angleRangeDeg != 360 )
{
theta = ( theta % ( Pi2 ) ) ;
theta = theta < 0 ? theta + Pi2 : theta ;
var a1 = angleStartDeg * Mathf.Deg2Rad;
var a2 = angleRangeDeg * Mathf.Deg2Rad;
theta = theta * a2 / Pi2 + a1;
}
// https://stackoverflow.com/a/26127012/2496170
// radius at y
var rY = Mathf.Sqrt( 1 - y * y );
var x = Mathf.Cos( theta ) * rY;
var z = Mathf.Sin( theta ) * rY;
return new Vector3( x, y, z ) * radius;
}
Gist : https://gist.github.com/nukadelic/7449f0872f708065bc1afeb19df666f7/edit
Preview:
# create uniform spiral grid
numOfPoints = varargin[0]
vxyz = zeros((numOfPoints,3),dtype=float)
sq0 = 0.00033333333**2
sq2 = 0.9999998**2
sumsq = 2*sq0 + sq2
vxyz[numOfPoints -1] = array([(sqrt(sq0/sumsq)),
(sqrt(sq0/sumsq)),
(-sqrt(sq2/sumsq))])
vxyz[0] = -vxyz[numOfPoints -1]
phi2 = sqrt(5)*0.5 + 2.5
rootCnt = sqrt(numOfPoints)
prevLongitude = 0
for index in arange(1, (numOfPoints -1), 1, dtype=float):
zInc = (2*index)/(numOfPoints) -1
radius = sqrt(1-zInc**2)
longitude = phi2/(rootCnt*radius)
longitude = longitude + prevLongitude
while (longitude > 2*pi):
longitude = longitude - 2*pi
prevLongitude = longitude
if (longitude > pi):
longitude = longitude - 2*pi
latitude = arccos(zInc) - pi/2
vxyz[index] = array([ (cos(latitude) * cos(longitude)) ,
(cos(latitude) * sin(longitude)),
sin(latitude)])
#robert king It's a really nice solution but has some sloppy bugs in it. I know it helped me a lot though, so never mind the sloppiness. :)
Here is a cleaned up version....
from math import pi, asin, sin, degrees
halfpi, twopi = .5 * pi, 2 * pi
sphere_area = lambda R=1.0: 4 * pi * R ** 2
lat_dist = lambda lat, R=1.0: R*(1-sin(lat))
#A = 2*pi*R^2(1-sin(lat))
def sphere_latarea(lat, R=1.0):
if -halfpi > lat or lat > halfpi:
raise ValueError("lat must be between -halfpi and halfpi")
return 2 * pi * R ** 2 * (1-sin(lat))
sphere_lonarea = lambda lon, R=1.0: \
4 * pi * R ** 2 * lon / twopi
#A = 2*pi*R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|/360
# = (pi/180)R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|
sphere_rectarea = lambda lat0, lat1, lon0, lon1, R=1.0: \
(sphere_latarea(lat0, R)-sphere_latarea(lat1, R)) * (lon1-lon0) / twopi
def test_sphere(n_lats=10, n_lons=19, radius=540.0):
total_area = 0.0
for i_lons in range(n_lons):
lon0 = twopi * float(i_lons) / n_lons
lon1 = twopi * float(i_lons+1) / n_lons
for i_lats in range(n_lats):
lat0 = asin(2 * float(i_lats) / n_lats - 1)
lat1 = asin(2 * float(i_lats+1)/n_lats - 1)
area = sphere_rectarea(lat0, lat1, lon0, lon1, radius)
print("{:} {:}: {:9.4f} to {:9.4f}, {:9.4f} to {:9.4f} => area {:10.4f}"
.format(i_lats, i_lons
, degrees(lat0), degrees(lat1)
, degrees(lon0), degrees(lon1)
, area))
total_area += area
print("total_area = {:10.4f} (difference of {:10.4f})"
.format(total_area, abs(total_area) - sphere_area(radius)))
test_sphere()
This works and it's deadly simple. As many points as you want:
private function moveTweets():void {
var newScale:Number=Scale(meshes.length,50,500,6,2);
trace("new scale:"+newScale);
var l:Number=this.meshes.length;
var tweetMeshInstance:TweetMesh;
var destx:Number;
var desty:Number;
var destz:Number;
for (var i:Number=0;i<this.meshes.length;i++){
tweetMeshInstance=meshes[i];
var phi:Number = Math.acos( -1 + ( 2 * i ) / l );
var theta:Number = Math.sqrt( l * Math.PI ) * phi;
tweetMeshInstance.origX = (sphereRadius+5) * Math.cos( theta ) * Math.sin( phi );
tweetMeshInstance.origY= (sphereRadius+5) * Math.sin( theta ) * Math.sin( phi );
tweetMeshInstance.origZ = (sphereRadius+5) * Math.cos( phi );
destx=sphereRadius * Math.cos( theta ) * Math.sin( phi );
desty=sphereRadius * Math.sin( theta ) * Math.sin( phi );
destz=sphereRadius * Math.cos( phi );
tweetMeshInstance.lookAt(new Vector3D());
TweenMax.to(tweetMeshInstance, 1, {scaleX:newScale,scaleY:newScale,x:destx,y:desty,z:destz,onUpdate:onLookAtTween, onUpdateParams:[tweetMeshInstance]});
}
}
private function onLookAtTween(theMesh:TweetMesh):void {
theMesh.lookAt(new Vector3D());
}