properties more like LDS, and less like blue noise. As stated earlier, one of the greatest challenges of the distributing points evenly on the sphere is that the optimal distribution critically depends on which objective function you use. The blue points are the Poisson points on the sphere. Thus for $N>100$, lattices based on Eqn 3 are better than any geodesic polyhedra. sample_points_on_sphere. But the notation is a little unclear. That distance is equal to the radius. 486 likes. How to distribute points on the surface of a sphere as evenly as possibly is an incredibly important problem in maths, science and computing, and mapping the Fibonacci lattice onto the surface of a sphere via equal-area projection is an extremely fast and effective approximate method to achieve this. Point Sphere. The actual algorithm is very simple. You can then scale the coordinates so that the point is on the sphere of radius r. However, what is also known about this lattice is that the largest Voronoi polygon is at the pole. First, we pick a random point in the unit cube where ​x​, ​y​, and ​z​ all range from -1 to +1. Also, the truncated icosahedron, which is the shape of a $C_{60}$ buckminster fullerene only corresponds to $d^* = 3.125$. For example. We’ll use what is usually the easiest algorithm: a rejection method. Evenly spaced points on a sphere (or other surface) is a common question on this forum that I’ve never paid much attention to as I didn’t need it. I’ve tried delaunay on the spiral result and then remap the mesh to the sphere points but this leaves a hole in the bottom. Agenda Abstract Motivation and Applications Review of Low Discrepancy Sequence Van der Corput sequence on [0; 1] Halton sequence on [0; 1] Halton sequence on [0; 1]n Unit Circle S1 Unit Sphere S2 Sphere Sn and SO(3) Our approach Numerical Experiments Conclusions 3. Required fields are marked *. Every point on the sphere corresponds to a different polarization form. Note that $\varepsilon=\frac{1}{2}$ corresponds to lattice #1, and a value greater than $\frac{1}{2}$ represents a gap near the poles. The behavior of $\epsilon_N$ for varying $N$ can be seen in Figure 3 (blue). 23 (1972), 443–448. Dedicated to the memory of Robert Rankin Abstract Using only basic tools from the theory of modular forms, the rational points of bounded height on the sphere are counted and shown to be uniformly distributed. Given that in our case, our objective constraint is maximizing the minimum pairwise distance separation between points, then it is not guaranteed that such distance-based constraints and relationships will hold after the projection. Some points seem too close together, and some seem too far apart. We now fully generalize $g(n)$ as follows: $$ g(N) = \begin{cases} 3-\phi, & \text{if $k$ is even} \\ A_{j+1}/A_j , &  \text{if $k$ is odd, where $j= (k+7)/2$} \end{cases} \tag{6} $$. However, this does not mean that it is provably optimal. This section shows how to evenly distribute points on a sphere in a manner that optimizes (maximizes)  a more global measure such as the volume of the convex hull. All points are on the unit sphere so should have x j 2 + y j 2 + z j 2 = 1. It's easy to see that the further out on the plane your point p, the closer its projected image on the sphere is to the North pole. For example, should the point corresponding to t_0 be located at the origin in your polar grids? R. Alexander, On the sum of distances betweenN points on a sphere,Ada Math.Acad, Sci. In what sense does the set of points t_i live on the unit square? This is a great answer! Point in Sphere. Let us define $C_N$ as the convex hull of the $N$ points, $$ \epsilon_N = N \left( \frac{4\pi }{3} \; – \textrm{Vol}(C_N) \right)$$. Likewise, I assume eqn (6) should have g(k). Let’s define the following distribution: $$ t_i(\varepsilon) = \left( \frac{i{+ \varepsilon}}{N{-1}{+2\varepsilon}}, \frac{i}{\phi} \right) \quad \textrm{for }\; 0 \leq i \leq N{-1} $$. \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline N & 4-6 & 7-10 & 11-16& 17-26& 27-43& 44-70& 71-114 & 115-184 & 185-300\\ \hline g(n)  &3-\phi & \frac{23}{14} & 3-\phi & \frac{37}{23} & 3-\phi & \frac{60}{37} & 3-\phi & \frac{97}{60} & 3-\phi  \\ \hline \end{array}. It places the points in the midpoints of the intervals (aka the midpoint rule in gausssian quadrature), and so for $n=100$, the values of the first coordinate would simply be: $$ \{ \frac{0.5}{100},\frac{1.5}{100},\frac{2.5}{100},\ldots, \frac{97.5}{100},\frac{98.5}{100},\frac{99.5}{100} \} $$. If the sphere has radius r, the Cartesian coordinates of that point are given by 0 @ x y z 1 A = r 0 @ sin# cos’ sin# sin’ A while back, this post was featured on the front page of Hacker News. This is often referred to as “Tammes’s problem” due to the botanist Tammes, who searched for an explanation of the surface structure of pollen grains. The second, and from a pragmatic point of view possibly the trickiest to resolve, is that the mapping has a two singularity points at each pole. Random points on a sphere. We also need a way to pick a random point in a unit radius sphere centered at the origin. That is, $$ d_N = \min_{i \neq j} \Vert x_i  – x_j \Vert_2 $$, This value decreases at a rate ~$1/\sqrt{N}$, so it is useful to define the normalized distance, and also the asymptotic limit of the normalized distance as, $$ d^*_N = \sqrt{N} d_N ,\quad  \quad d^* = \lim_{N \rightarrow \infty} d^*_N $$. Very cool. Coxeter demonstrated these arrangements are fundamentally related to the Fibonacci sequence, $F_k =\{1, 1, 2, 3, 5, 8, 13, …\}$ and the golden ratio $\phi = (1+\sqrt{5})/2$. s2point: Single point on the (unit) sphere; s2polygon: Create a s2polygon; s2pp: Spherical point pattern; s2radius: Radius of sphere; s2region: Extract spherical region; s2runif: Generate uniform random points on sphere; spatstat.sphere: Package documentation for spatstat.sphere; sub-.s2pp: Extract a subset of a point pattern on a sphere. The first is that this mapping is area-preserving, not distance preserving. The underlying reason why this works is based on the less well known fact that all numbers $x$ that satisfy the special Mobieus transformation are equivalent to $\phi$ in terms of irrationality. To optimize the point configurations such that the volume of the convex hull is maximized, see my previous post, “Evenly distributing points on a sphere”.) #4. The lattice as per equation 6, is a modification of the canonical Fibonacci lattice that produces a significantly better point distribution as measured by the volume and surface area of the convex hull (Delaunay Mesh). The effects of various polarization instruments are determined by displacements on the sphere. A common variant, which produces a better packing ($d^*=3.09$), is: $$ t_i =   \left(  \frac{i{+0.5}}{N},  \frac{i}{\phi} \right)  \quad \textrm{for }\; 0 \leq i \leq N{-1} \tag{2}$$. Suppose we want to generate uniformly distributed points on a sphere. As per the first reference,  some of the state-of-the-art methods which are typically complex and require recursive and/or dynamic programming are as follows  (the higher the better). You can also use the SAS DATA step to generate the random points on a sphere. Visit here for, Section 2. The original one is strictly only defined for $N$ equal to one of the terms of the Fibonacci sequence, $F_m$ and is very well studied in number theory. “I transform and modernize organizations through innovative data strategies solutions.”. See how the squares get smaller towards the poles? If I try to create a Sphere, I can select Sketch2 as the plane, but as soon as I do that, I look at the Browser and see that I am not really working in Sketch2, but that it has created a temporary Sketch3(it is temporary because once the sphere is created, Sketch3 vanishes). I have the really odd problem when it comes to plotting Bloch spheres using QuTip. A point (x, y, z) is outside the sphere with center (cx, cy, cz) and radius r if. One very elegant solution is modeled after nodes appearing in nature such as the seed distribution on the head of a sunflower or a pine cone, a phenomenon known as spiral phyllotaxis. Thanks. Whilst the second one generalises this to arbitrary $N$, and is used more frequently in computing: $$ t_i =   \left(  \frac{i}{N}, \frac{i }{\phi} \right)  \quad \textrm{for }\; 0 \leq i \leq  N  \tag{1}$$, $$\phi  = \frac{1+\sqrt{5}}{2} = \lim_{n \rightarrow \infty} \left( \frac{F_{n+1}}{F_n} \right)$$. Thus, in trying to maximize $d_N$ by separating the initial polar points in the sequence, actually makes the void at the pole even larger! In science terminology, we could say that need to consider finite-term correction effects. Section 2 will show how we can modify the canonical Fibonacci lattice that produce larger convex hull measures (volume and surface area). Prototype. However 5 points being insufficient is even easier: 5 - 2 < 4 and there is a great circle passing through any two points on a sphere. On the unit square, (and also on the cylindrical projection) they would correspond to two points that are quite distant from each other, and yet when mapped to the surface of the sphere they could be joined be a very small great arc going over the north pole. The closest point on sphere is often used for AI, Physics simuation and special effects. Point patterns that are generated by a uniform process will typically have gaps and clusters. Here a sphere is rolling on a frictional surface but slipping takes place. $$ x = \frac{a\phi+b}{c\phi+d}, \quad \textrm{for all integers} \; \; a,b,c,d \; \textrm{such that } |ad-bc|=1 $$, And thus the connection why $\phi$ and $3-\phi$ work together so well , is that, $$\frac{1}{\phi} = \frac{\phi+1}{2\phi+1}, \quad \quad \frac{1}{3-\phi }= \frac{2\phi+1}{1\phi+1} $$, For the remaining half, we first define an auxiliary sequence $A_N$ that is variant of the Fibonacci sequence, $$ A_1 =1, \; A_2 = 4; \; A_{n+2}= A_{n+1}+A_n \; \textrm{for } n = 1,2,3,… $$, $$ A_N = 1,4,5,9,14,23,37,60,97,157,254,411,…$$. less clusters of high and low density. Perhaps we can drop our requirement for points to be uniformly distributed, but keep them well-distributed. Note that smaller is better. I have used a surface plot (with clear faces) to illustrate the underling sphere in black. Thus, let us generalize equation 1 as follows: $$ t_i =   \left(  \frac{i+1/2}{N},  \frac{i}{g(N)} \right)  \quad \textrm{for }\; 0 \leq i \leq N-1 \tag{4}$$, $$ g(n) = \begin{cases} 3-\phi, & \text{if $k$ is even} \\ \phi, & \text{if $k$ is odd} \end{cases} \tag{5}$$, $$ k = \left\lfloor \textrm{log}_{\phi}(\frac{n}{1.5}) \right\rfloor = \left\lfloor \frac{\ln (n/1.5)}{\ln \phi } \right\rfloor$$. For sake of brevity, this post focuses on just two of these: the minimum packing distance and convex hull / Delaunay mesh measures (volume and area). You can generate each coordinate independently from a normal distribution and use the EUCLID function in Base SAS to compute the Euclidean distance from a point to the origin. \begin{array}{|lr|} \hline \text{} & d^* \\ \hline \text{Coulomb} & 3.37 \\ \text{Log Energy} & 3.37\\ \text{Lattice 3} & 3.34 \\ \text{Zonal Equal Area} & 3.32 \\ \text{Lattice 2} & 3.28 \\ \text{Max Determinant} & 3.19 \\ \text{Lattice 1} & 3.09\\ \hline \end{array}. I find your discussion confusing. This shows that the newly proposed method (green) produces consistently better distribution than the canonical Fibonacci lattice (blue). Random Points on a Sphere Suppose we want to generate uniformly distributed points on a sphere. The radius of the sphere represents the incident irradiance of the light beam (which is usually assumed to be unity). There are two similar definitions of the spherical Fibonacci lattice point set in the literature. The problem of how to evenly distribute points on a sphere has a very long history and is one of the most studied problems in the mathematical literature associated with spherical geometry. Intuitively, it "swaps the inside and outside" of the … Then I need to draw a line passing through these 4 points (thus forming a quadrilateral). I guess we need convex hull, so should probably try weaverbird or something… Cheers, Steven. n is never defined. For example, in this post, we will also find that optimising for packing does not necessarily produce an optimal convex hull, and vice-versa. Samples 3D points on a sphere surface by refining an icosahedron, as in: Hinterstoisser et al., Simultaneous Recognition and Homography Extraction of Local Patches with a Simple Linear Classifier, BMVC 2008 1. As promised, the resulting points are on the circle of radius r=1. where x is uniformly distributed and x âˆˆ [0, 1). Thus for $N=100$, the values of the first coordinate would simply be: $$ \{ \frac{1.5}{103},\frac{2.5}{103},\frac{3.5}{103},\ldots, \frac{98.5}{103},\frac{99.5}{103},\frac{100.5}{103} \} $$. 12 November 2006 With corrections, 18 April 2007 Supervisors: Professor Ian Sloan, Associate Professor Rob Womersley . Convex hulls, Voronoi cells and Delaunay  triangles, http://web.archive.org/web/20120421191837/, http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere, https://perswww.kuleuven.be/~u0017946/publications/Papers97/art97a-Saff-Kuijlaars-MI/Saff-Kuijlaars-MathIntel97.pdf, https://projecteuclid.org/download/pdf_1/euclid.em/1067634731, https://maths-people.anu.edu.au/~leopardi/Leopardi-Sphere-PhD-Thesis.pdf, https://www.irit.fr/~David.Vanderhaeghe/M2IGAI-CO/2016-g1/docs/spherical_fibonacci_mapping.pdf, https://maths-people.anu.edu.au/~leopardi/Macquarie-sphere-talk.pdf. It is of critical importance in many areas of mathematics, physics, chemistry including  numerical analysis, approximation theory, coding theory, crystallography, electrostatics, computer graphics, viral morphology to name just a few. \begin{array}{|c|ccccccc|} \hline N & 20 & 32 & 122 & 272 & 482 & 752 & 1082\\ \hline d^* & 3.19 & 3.63 & 3.16 & 2.99 & 2.90 & 2.84 & 2.81 \\ \hline \end{array}. where $\lfloor x \rfloor$ is the floor function. A point (x, y, z) lies on the sphere with center (cx, cy, cz) and radius r if. One solution is to pick λ âˆˆ [-180°, 180°) as before and then set φ = cos-1(2x - 1), The file names have three components: point … But no point on the plane projects to the North pole itself. My name is Dr Martin Roberts, and I’m a freelance Principal Data Science consultant, who loves working at the intersection of maths and computing. Moving the points farther and farther away from the points, suggests that there is a threshold where it is constructive to place a point … $$t _5/t_4 = 1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{4}}}}$$. \begin{array}{|c|cccccccccc|} \hline N & 12 & 42 & 92 & 162 & 252& 362 & 492 & 642 & 812 & 1002  \\ \hline d^*  & 3.64 & 3.54 & 3.34 & 3.22 & 3.15 & 3.09 & 3.06 & 3.03 & 3.00 & 2.99 \\ \hline \end{array}. 7 Likes. You can contact me through any of these channels. It turns out that local measures such as $d_N^*$ are at times very “unforgiving” inasmuch as a single point in a suboptimal position can catastrophically reduce measure of the entire point distribution. Secondly, in addition to these $N{-2}$ points, the first and final points are defined to be at located each pole. and not strictly blue noise, I bet it has (integration etc!) A point (x, y, z) is inside the sphere with center (cx, cy, cz) and radius r if. Hope you understand now. ( x-cx ) ^2 + (y-cy) ^2 + (z-cz) ^ 2 < r^2. This way of associating to every point in the plane exactly one point on the sphere is called the stereographic projection. Similarly, these point sets can be mapped from the unit square [0, 1]2 which to the sphere by the cylindrical equal area projection: (x, y) → (θ, ϕ): (cos − 1(2x − 1)– π / 2, 2πy) (θ, ϕ) → (x, y, z): (cosθcosϕ, cosθsinϕ, sinθ) Here’s a basic implementation of this in Python. where the normalization factor of $N$ is included, because the absolute discrepancy decreases at a rate $~ 1/N$. The unit sphere S 2 in three-dimensional space R 3 is the set of points (x, y, z) such that x 2 + y 2 + z 2 = 1.Let N = (0, 0, 1) be the "north pole", and let M be the rest of the sphere. Point Sphere. We might start off by picking spherical coordinates (λ, φ) from two uniform distributions, λ ∈ [-180°, 180°) and φ ∈ [-90°, 90°). This mapping suffers from two distinct but inter-related problems. In geometry, inversion in a sphere is a transformation of Euclidean space that fixes the points of a sphere while sending the points inside of the sphere to the outside of the sphere, and vice versa. This post has been superceded by the more recent post: “How to evenly distribute points on a sphere more evenly than the canonical Fibonacci Lattice“. Poisson-disc sampling can of course be applied to other surfaces too e.g. If we want any area on the sphere to contain approximately the same density of points, The most straightforward way to create points on the surface of a sphere are classical spherical coordinates, in which a point is addressed via its two angular coordi-nates, the polar angle # 2 [0;ˇ] and the azimuthal angle ’ 2 [0;2ˇ]. ( x-cx ) ^2 + (y-cy) ^2 + (z-cz) ^ 2 = r^2. Finally it would like to know how many lines did the plane cut through on this sphere. A key insight to further improving on Equation 2, is to realize that the $d^*_N$ always corresponds to the distance between the points $t_0$ and $t_3$, which are at the poles. In our case the sphere has been unwrapped from the side farthest away from us, onto a plane that is in contact with the close side. There are many criteria in use, and they  include: Repeating this point as it is crucial: there is usually no single optimal solution  to this question, because an optimal solution based on one criteria is often not an optimal point distribution for another. The packing criterion asks us to maximize the smallest neighboring distance among the $N$ points. However, we can quickly see that this will result in an uneven distribution, with the density increasing as we get closer to the poles. Lattice 3 (as defined in equation 4) produces a significantly tighter packing than the canonical Fibonacci lattice. Figure 3 shows that this substantially improves the volume discrepancy for half the values of $N$. The key to the closest point on sphere algorithm is to realize that no matter where that point may be, it will always be the same distance from the center of the sphere. A while back, this post was featured on the front page of Hacker News. $$ t_i =   \left(  \frac{i}{F_m},  \frac{i F_{m{-1}}}{F_m} \right)  \quad \textrm{for }\; 0 \leq i \leq  N-1 $$. Hence, the velocity of the bottom-most point is not equal to $0$. As a graphics person, I think it would be neat to explore the differences visually, for path tracing, calculating ambient occlusion, etc! $$ \{ 0; \; \frac{3.5}{106},\frac{4.5}{106},\frac{5.5}{106},\ldots, \frac{99.5}{106},\frac{100.5}{106},\frac{101.5}{106} ; \; 1\} $$. Although the previous section optimized for $d^*_N$, unfortunately these modifications actually make other measures worse, such as the volume of the convex hull (Delaunay mesh). The number of rows is equal to the number of points. Recall that uniformly at random does not mean evenly spaced. Take a look at this image, and see if you can figure out the algorithm to perform the test. In other words, what is the mapping? The key to improving the volume discrepancy is to note that although the use of $\phi$, the golden ratio intuitively makes sense as $N \rightarrow \infty$, it does not necessarily follow that it is the best value for finite $N$. Something I think worth pointing out is that this “The packing criterion asks us to maximize the smallest neighboring distance among the points” is exactly what Mitchell’s best candidate algorithm does, so this seems to be blue noise distributed sampling points on a sphere, but without the need to do point generation in advance. You can also use the SAS DATA step to generate the random points on a sphere. from qutip import * import numpy as np import matplotlib.pyplot as plt b = Bloch() s = Qobj([[1],[0]]) b.add_states(s) b.show() But when I try to plot points on the Bloch sphere … Thus, to improve $d_N$ the points near the poles should be positioned farther apart. This produces spectral characteristics similar to blue noise, i.e. The algorithm. We recall that a 3D unit-sphere (and hence a direction) is parametrized only by two variables; elevation \theta \in [0; \pi] and azimuth \varphi \in [0; 2\,\pi] which can be converted to Cartesian coordinates as The more difficult case of points with a given height is also treated. Firstly, it uses $\varepsilon =\frac{7}{2}$ for $1 \leq  i \leq N{-2}$. 483 likes. And here are the corresponding values of $d^*$ for various $N$-point geodesic domes based on the dodecahedron. We reject this point and try … Your email address will not be published. How to evenly distribute points on a sphere more evenly than the canonical Fibonacci Lattice, A Comparison of poopular point confugrations on S^2, https://www.linkedin.com/in/martinroberts/, Uniform Random sampling on an $n$-sphere an in an $n$-ball, A new method to construct isotropic blue noise point sets with uniform projections, Maximal Posison disc sampling: an improved version of Bridson’s algorithm, New top story on Hacker News: Evenly distributing points on a sphere – New Content, New top story on Hacker News: Evenly distributing points on a sphere – World Best News, New top story on Hacker News: Evenly distributing points on a sphere – News about world, New top story on Hacker News: Evenly distributing points on a sphere – Latest news, New top story on Hacker News: Evenly distributing points on a sphere – Tech + Hckr News. Please reference this newer link wherever possible. An example of these Fibonacci Grids is shown below. Plotting state vectors works perfectly fine e.g. As the mapping from the unit square to the surface of the sphere is done via an area-preserving projection, one can expect that if the original points are evenly distributed then they will also quite well evenly distributed on the sphere’s surface. Thus, we present an alternative to lattice 2 which is generally more preferable, as it does not exhibit such a large void near the poles. How to generate random points on a sphere This question often pops up, when you need a random direction vector to place things in 3D or you want to do a particle simulation. Optimising the Convex hull (Delaunay mesh). Should eqn (4) have g(N) instead of g(n)? . This is shown in figure 5. These points sets can be transformed to a well-known Fibonacci spirals via the transformation. Steven: how to mesh these points. Furthermore, the solution to this problem is critically dependent on the criteria used to judge the uniformity. Or even g(i), perhaps? We might start off by picking spherical coordinates (λ, Ï†) from two uniform distributions, λ âˆˆ [-180°, 180°) and φ âˆˆ [-90°, 90°). Although we’ve successfully generated uniformly distributed points on a sphere, it feels messy. Point-Sphere is proud to be an all-in-one curated stop that offers sophisticated, unique and prime quality products. It is almost identical to lattice 2 but with two differences. It would help to label the sources of the grid points in a simple case. That is, $$ t_0 = (0,0);  \; \; t_i =   \left(  \frac{i{+\frac{7}{2}}}{N{+6}},  \frac{i}{\phi} \right); \;\; t_{N{-1}} = (0,1);  \quad \textrm{for }\; 1 \leq i \leq N{-2} $$. it is often used on 2D planes. They should have all been $N$, but sometimes I had inconsistently written $n$. Quote:Original post by defferDo you mean: random point on a sphere?Or do you want to generate a sequence of points of a sphere in some uniform manner?the latter, yes Please help For each point set, the text file has four items per row: the x j, y j, and z j cartesian coordinates in [-1, 1], and the cubature weight w j for that point. If the distance from the point to the center of the sphere is less than the radius of the sphere, the sphere contains the point! The discrepancy between the volume of the convex hull of points and the volume of a unit sphere. Sampling with Halton Points on n-Sphere Wai-Shing Luk1 1School of Microelectronics Fudan University April 6, 2014 2. This video details the challenges of spacing n points equally on a sphere and some possible methods of doing it. Here’s a basic implementation of this in Python. Visit here for discussion. AFAIK blue noise is nice for visual patterns (perceptual error), but LDS is better for integration in general. Great write-up. Now fixed. Now I pick 4 random points on this unit sphere. A more æsthetically pleasing pattern can be generated using Poisson-disc sampling, where no points are less than a minimum distance apart. The convergents of this sequence all have elegant continued fractions, and in the limit converge to $\phi$. LinkedIn: https://www.linkedin.com/in/martinroberts/, Twitter: @TechSparx  https://twitter.com/TechSparx, email: Martin (at) RobertsAnalytics (dot) com. Distributing points on the sphere: Partitions, separation, quadrature and energy Paul Leopardi Thesis submitted in fulfilment of requirements for Doctor of Philosophy in Mathematics at the University of New South Wales. Oh, and don’t forget to drag the spheres. Later if this is successful I would like to draw a plane passing through the center of the sphere. Article Google Scholar Generate a number of candidate samples and pick the furthest from all previous samples. Hungar. The spherical fibonacci spiral produced from equation 1, results in a value of $d_N^*$ for all $N$, and so $d^* =  2$. Testing if a point is inside a sphere is one of the simplest 3D tests. Your email address will not be published. Unfortunately, with the exception of a few special cases (namely the platonic solids) it is not possible to exactly equally distribute points on the sphere. Mitchell’s best-candidate algorithm is a straightforward approximation. The following table is a summary of the value of $g(N)$ for various $N$. This post has been kept as an archival record. Even though spherical Fibonacci point sets are not the globally best distribution of samples on a sphere, (because their solutions do not coincide with the platonic solids for $n=4,6,8,12,20$), they yield excellent sampling properties and are extremely simple to construct in contrast to other more sophisticated spherical sampling schemes. Here are the corresponding values of $d^*$ for various $N$-point geodesic domes based on the icosahedron. Figure 3 shows that, in relation to convex hull volume, this new distribution is better than the canonical lattice for all values of $n$. Point-Sphere is proud to be an all-in-one curated stop that offers sophisticated, unique and prime quality products. Points that are near the plane suffer less distortion and points that are near the hole get very stretched. In our case, regardless of how large $N$ is, the $d_N^*$ is typically determined by the four points closest to each pole, especially $t_0$ and $t_3$. It is this particular issue that makes many of the spiral mappings sub-optimal. This post has been kept as an archival record. Devoting most of an entire day searching and failing to get anywhere close to my objective is quite annoying!! For large $N$ this value of $d^*=3.35$ compares extremely well compared to other methods, such as geodesic domes, which are based on triangulated projections from the faces of platonic solids to the surface of the sphere. corellaman (_corellaman) July 24, 2020, 7:36pm #3. Hi Nadav, You’re right. However, since it’s a low discrepancy sequence (golden ratio!) One way to explain this is to look at the how the area of a given “square” of width Δλ and height Δφ varies with φ. Please reference this newer link wherever possible. there are a number of solutions. Based on this model, it can be shown that  for $N\geq 20 $, compared to the canonical spherical Fibonacci lattice,  the following simple expression corresponding to $\varepsilon=\frac{3}{2}$ produces a even better packing ($d^* = 3.27$): $$ t_i =   \left(  \frac{i{+1.5}}{N{+3}},  \frac{i}{\phi} \right)  \quad \textrm{for }\; 0 \leq i \leq N{-1} \tag{3} $$. Figure 3. here is one INVALID way: image 2800×1839 844 KB. sample points uniformly on a sphere by dividing an icosahedron(the largest convex regular polyhedron) Theory. Now, I want to create a Sphere at this point in Sketch2 and Join it with the existing body. Rational points on the sphere W. Duke ∗ DepartmentofMathematics, UniversityofCalifornia, LosAngeles, CA98888. $$ t_0=(0,0); \; t_{N{-1}} = (1,0); \quad t_i =   \left(  \frac{i{+3.5}}{N{+6}},  \frac{i}{\phi} \right)  \quad \textrm{for }\; 1 \leq i \leq N{-2} \tag{4} $$ This lattice results in $d^*=3.313$, which for $N>90$ is even better than lattice #2. My name is Dr Martin Roberts, and I’m a freelance Principal Data Science consultant, who loves working at the intersection of maths and computing. That is. Of interest, this distribution also slightly but consistently reduces the discrepancy between the surface area of the convex hull and the surface area of the unit sphere. Section 1 will show how we can modify the canonical Fibonacci lattice to consistently produce a tighter packing configuration. Note that the points are no longer independent of each other, hence they are no longer uniformly distributed. Try picking any square on the graticule (the spherical grid). grasshopper3d.com Equal spacing of points on a sphere . This is great stuff, thank you for sharing! First published: 10 August, 2018 Last modified: 7 June , 2020. Consider two points very close to the pole but 180 degrees different in longitude. With two differences, CA98888, lattices based on the front page of point on sphere.. Via the transformation using QuTip perform the test at random does not mean that it is particular. For half the values of $ N $ -point geodesic domes based on front... 2020, 7:36pm # 3 be applied to other surfaces too e.g this image, and ​z​ range! X j 2 = r^2 Fibonacci lattice point set in the plane cut through this! Hence they are no longer uniformly distributed points on a sphere are the corresponding values $! $ the points near the hole get very stretched where $ \lfloor x \rfloor $ is the function. Improves the volume of the light beam ( which is usually the easiest algorithm: a rejection.! You for sharing this video details the challenges of spacing N points equally on a sphere Suppose we want area... Associating to every point on the sphere corresponds to a well-known Fibonacci via... All range from -1 to +1 area on the icosahedron patterns ( error! And some possible methods of doing it better for integration in general sphere to. Lattice 2 but with two differences algorithm: a rejection method the velocity of the convex,! The discrepancy between the volume of the value of $ g ( N ) $ for varying N. We could say that need to consider finite-term correction effects published: 10,..., and see if you can figure out the algorithm to perform the test ( 4 ) consistently..., I want to generate uniformly distributed, but sometimes I had inconsistently written N. \Lfloor x \rfloor $ is the floor function one point on the icosahedron summary of the sphere distances points. Points t_i live on the unit sphere so should probably try weaverbird or something… Cheers Steven... This sphere incident irradiance of the grid points in a simple case 1 will show how we can modify canonical. Which is usually the easiest algorithm: a rejection method in the plane cut through on this.!, Steven of rows is equal to $ \phi $ want any area on the sphere the near... Data step to generate uniformly distributed, but sometimes I had inconsistently written $ N $, but I. The newly proposed method ( green ) produces consistently better distribution than the canonical lattice. Located at the pole but 180 degrees different in longitude are better than geodesic. Need convex hull of points, there are a number of candidate and! Process will typically have gaps and clusters transform and modernize organizations through innovative strategies... Improves the volume of the sphere is called the stereographic projection and the volume discrepancy for the. Among the $ N $ -point geodesic domes based on eqn 3 are better than any geodesic polyhedra see the! Correction effects all points are on the dodecahedron July 24, 2020, 7:36pm # 3 7:36pm! The value of $ g ( N ) sequence all have elegant continued fractions, and less like blue is... Volume and surface area ) are less than a minimum distance apart g ( N ) -point domes... A random point in the literature, because the absolute discrepancy decreases at rate. Underling sphere in black the canonical Fibonacci lattice to consistently produce a tighter packing configuration squares get towards... Case of points t_i live on the sphere plane passing through these 4 points ( thus forming a )! In general have x j 2 + z j 2 + z j 2 + y 2... Been kept as an archival record be applied to other surfaces too e.g front page of News... Density of points, there are two similar definitions of the sphere to! N ) points t_i live on the icosahedron random does not mean evenly.! Is quite annoying! like blue noise, I want to create a sphere at this point in and... Minimum distance apart behavior of $ g ( N ) instead of g ( N ) $ for various N... Sometimes I had inconsistently written $ N $ -point geodesic domes based on the criteria used judge! Stop that offers sophisticated, unique and prime quality products smallest neighboring among. Used for AI, Physics simuation and special effects y j 2 + y j =. S a low discrepancy sequence ( golden ratio! to $ 0.. Bloch spheres using QuTip our requirement for points to be an all-in-one curated stop that offers sophisticated, and. The packing criterion asks us to maximize the smallest neighboring distance among the $ N $ is the floor.! So should probably try weaverbird or something… Cheers, Steven also known this! That produce larger convex hull, so should have x j 2 1! While back, this does not mean evenly spaced applied to other surfaces too.... On this unit sphere than a minimum distance apart previous samples discrepancy (! Points uniformly on a sphere Suppose we want to generate uniformly distributed points on a sphere this! The plane projects to the pole a uniform process will typically have gaps clusters! The hole get very stretched projects to the North pole itself course be applied to other surfaces e.g... Of various polarization instruments are determined by displacements on the icosahedron transform and modernize organizations through innovative DATA solutions.... Makes many of the spiral mappings sub-optimal all been $ N $ quite annoying!. The closest point on the dodecahedron all previous samples consider two points very close to my objective quite! + y j 2 + z j 2 + z j 2 r^2! These channels has ( integration etc! error ), but keep them well-distributed one. Plane passing through the center of the spiral mappings sub-optimal, hence they are no longer of! X j 2 + y j 2 + z j 2 + y j 2 z! Hull, so should have all been $ N $ -point geodesic domes based on the.. Betweenn points on the unit sphere $ \epsilon_N $ for various $ N $ Alexander on. To this problem is critically dependent on the dodecahedron identical to lattice 2 but with two differences a different form. Plane suffer less distortion and points that are generated by a uniform process will have. Very close to the North pole itself with a given height is also known this. Judge the uniformity error ), but keep them well-distributed challenges of N... Really odd problem when it comes to plotting Bloch spheres using QuTip I want to create a is. The volume of a point on sphere sphere so should probably try weaverbird or something…,. If you can also use the SAS DATA step to generate uniformly distributed points on a sphere it!, lattices based on the front page of Hacker News for visual patterns ( perceptual )... $ points distributed, but keep them well-distributed spherical grid ) corresponds to a polarization! Continued fractions, and less like blue noise is nice for visual patterns ( perceptual error,. I need to draw a line passing through the center of the value of $ $... Summary of the grid points in a simple case often used for AI, simuation! $, but LDS is better for integration in general back, this post featured... Use what is also treated not distance preserving finally it would help to label the sources of the of. 180 degrees different in longitude: a rejection method points to be an all-in-one curated that! The velocity of the bottom-most point is inside a sphere Rob Womersley the point. See how the squares get smaller towards the poles \phi $ typically have gaps and clusters minimum distance.... Front page of Hacker News be applied to other surfaces too e.g, it feels messy successfully generated distributed. Thus for $ N $ LosAngeles, CA98888 ( which is usually the easiest:. A minimum distance apart modernize organizations through innovative DATA strategies solutions. ” all previous samples neighboring distance among the N. \Phi $ for example, should the point corresponding to t_0 be at! Generated point on sphere Poisson-disc sampling can of course be applied to other surfaces too e.g draw plane. Applied to other surfaces too e.g the transformation, i.e slipping takes place asks us to maximize the smallest distance. ( which is usually assumed to be unity ) can also use the DATA. Even better Physics simuation and special effects drag the spheres first, we could that! A line passing through these 4 points ( thus forming a quadrilateral ) the absolute discrepancy at! Lattices based on eqn 3 are better than any geodesic polyhedra transformed to a different polarization form help to the. Basic implementation of this sequence all have elegant continued fractions, and ​z​ all range from -1 to.... Point in Sketch2 point on sphere Join it with the existing body the set of points with a height... Are better than any geodesic polyhedra sphere Suppose we want to create a sphere, it feels messy 2 y. Math.Acad, Sci two similar definitions of the spherical Fibonacci lattice that larger! That it is provably optimal error ), but sometimes I had inconsistently written $ N $ what is known! Afaik blue noise, I want to generate the random points on this unit sphere so should have (. That the largest convex regular polyhedron ) Theory are on the criteria to... ) July 24, 2020, 7:36pm # 3 behavior of $ d^ * for... Is also treated DATA strategies solutions. ” sets can be seen in figure 3 ( defined... ^2 + ( y-cy ) ^2 + ( y-cy ) ^2 + ( y-cy ^2.

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