Is there a way to generate a uniformly distributed point on a sphere from a fixed amount of random real numbers? The 2019 Stack Overflow Developer Survey Results Are InHow to find a random axis or unit vector in 3D?Picking random points in the volume of sphere with uniform probabilityIs a sphere a closed set?Random Point Sampling From a Set with Certain GeometryHow to Create a Plane Inside A CubeAlgorithm to generate random points in n-Sphere?Sampling on Axis-Aligned Spherical QuadRandom 3D points uniformly distributed on an ellipse shaped window of a sphereCompensating for distortion when projecting a 2D texture onto a sphereFind the relative radial position of a point within an ellipsoid

How to charge AirPods to keep battery healthy?

Does HR tell a hiring manager about salary negotiations?

Likelihood that a superbug or lethal virus could come from a landfill

Did any laptop computers have a built-in 5 1/4 inch floppy drive?

Is bread bad for ducks?

Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?

If my opponent casts Ultimate Price on my Phantasmal Bear, can I save it by casting Snap or Curfew?

A word that means fill it to the required quantity

How much of the clove should I use when using big garlic heads?

How to obtain a position of last non-zero element

Can we generate random numbers using irrational numbers like π and e?

Deal with toxic manager when you can't quit

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

What does もの mean in this sentence?

Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?

Why is the maximum length of OpenWrt’s root password 8 characters?

Match Roman Numerals

What do I do when my TA workload is more than expected?

Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?

How can I have a shield and a way of attacking with a ranged weapon at the same time?

What is preventing me from simply constructing a hash that's lower than the current target?

Why not take a picture of a closer black hole?

Geography at the pixel level

Getting crown tickets for Statue of Liberty



Is there a way to generate a uniformly distributed point on a sphere from a fixed amount of random real numbers?



The 2019 Stack Overflow Developer Survey Results Are InHow to find a random axis or unit vector in 3D?Picking random points in the volume of sphere with uniform probabilityIs a sphere a closed set?Random Point Sampling From a Set with Certain GeometryHow to Create a Plane Inside A CubeAlgorithm to generate random points in n-Sphere?Sampling on Axis-Aligned Spherical QuadRandom 3D points uniformly distributed on an ellipse shaped window of a sphereCompensating for distortion when projecting a 2D texture onto a sphereFind the relative radial position of a point within an ellipsoid










4












$begingroup$


The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.






geometry







share|cite|improve this question











$endgroup$











  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    3 hours ago






  • 1




    $begingroup$
    Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
    $endgroup$
    – robjohn♦
    3 hours ago










  • $begingroup$
    @robjohn thank you, you're right that I forgot to specify that.
    $endgroup$
    – The Zach Man
    1 hour ago















4












$begingroup$


The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.






geometry







share|cite|improve this question











$endgroup$











  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    3 hours ago






  • 1




    $begingroup$
    Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
    $endgroup$
    – robjohn♦
    3 hours ago










  • $begingroup$
    @robjohn thank you, you're right that I forgot to specify that.
    $endgroup$
    – The Zach Man
    1 hour ago













4












4








4


1



$begingroup$


The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.






geometry







share|cite|improve this question











$endgroup$




The obvious solution of Lattitude & Longitude doesn't work because it generates points more densely near the poles, and the other thing I came up with (Pick a random point in the unit cube, if it's in the sphere map it to the surface, and restart if it's outside) doesn't always find a point within a fixed number of tries.






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 46 mins ago









robjohn♦

271k27313642




271k27313642










asked 3 hours ago









The Zach ManThe Zach Man

1107




1107











  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    3 hours ago






  • 1




    $begingroup$
    Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
    $endgroup$
    – robjohn♦
    3 hours ago










  • $begingroup$
    @robjohn thank you, you're right that I forgot to specify that.
    $endgroup$
    – The Zach Man
    1 hour ago
















  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    3 hours ago






  • 1




    $begingroup$
    Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
    $endgroup$
    – robjohn♦
    3 hours ago










  • $begingroup$
    @robjohn thank you, you're right that I forgot to specify that.
    $endgroup$
    – The Zach Man
    1 hour ago















$begingroup$
So what you want is a uniform distribution. It would be helpful to state this explicitly.
$endgroup$
– robjohn♦
3 hours ago




$begingroup$
So what you want is a uniform distribution. It would be helpful to state this explicitly.
$endgroup$
– robjohn♦
3 hours ago




1




1




$begingroup$
Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
$endgroup$
– robjohn♦
3 hours ago




$begingroup$
Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
$endgroup$
– robjohn♦
3 hours ago












$begingroup$
@robjohn thank you, you're right that I forgot to specify that.
$endgroup$
– The Zach Man
1 hour ago




$begingroup$
@robjohn thank you, you're right that I forgot to specify that.
$endgroup$
– The Zach Man
1 hour ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.



For $(u_1,u_2)$ uniform on $[0,1]^2$, either



$mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$



or



$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $fracpi6$, which is better than $frac12$: the average number of trials is less than $2$.



    Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $sqrtx^2+y^2+z^2$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.



    (The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3184449%2fis-there-a-way-to-generate-a-uniformly-distributed-point-on-a-sphere-from-a-fixe%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.



      For $(u_1,u_2)$ uniform on $[0,1]^2$, either



      $mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$



      or



      $z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$






      share|cite|improve this answer









      $endgroup$

















        5












        $begingroup$

        The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.



        For $(u_1,u_2)$ uniform on $[0,1]^2$, either



        $mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$



        or



        $z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$






        share|cite|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.



          For $(u_1,u_2)$ uniform on $[0,1]^2$, either



          $mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$



          or



          $z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$






          share|cite|improve this answer









          $endgroup$



          The Lambert cylindrical equal area projection maps the sphere to a cylinder, area to equal area. It is easy to generate a uniform distribution on a cylinder. Simply map it back to the sphere.



          For $(u_1,u_2)$ uniform on $[0,1]^2$, either



          $mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$



          or



          $z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          robjohn♦robjohn

          271k27313642




          271k27313642





















              3












              $begingroup$

              Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $fracpi6$, which is better than $frac12$: the average number of trials is less than $2$.



              Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $sqrtx^2+y^2+z^2$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.



              (The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

                Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $fracpi6$, which is better than $frac12$: the average number of trials is less than $2$.



                Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $sqrtx^2+y^2+z^2$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.



                (The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $fracpi6$, which is better than $frac12$: the average number of trials is less than $2$.



                  Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $sqrtx^2+y^2+z^2$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.



                  (The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)






                  share|cite|improve this answer









                  $endgroup$



                  Your method, even though it doesn't finish in a fixed number of times, is a reasonable way to do it. Each trial succeeds with probability $fracpi6$, which is better than $frac12$: the average number of trials is less than $2$.



                  Another standard method is to use the normal distribution. Generate $x, y, z$ independently from a standard normal distribution, then take the point $(x,y,z)$ and divide it by $sqrtx^2+y^2+z^2$ as you did for points inside the cube. The multivariate normal distribution is rotationally symmetric, so this will get you evenly distributed points on the sphere.



                  (The Box–Muller transform is one way to generate normally distributed random numbers, and some versions of it do not use rejection sampling, so they can be done with a "fixed amount" of randomness.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Misha LavrovMisha Lavrov

                  49k757107




                  49k757107



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3184449%2fis-there-a-way-to-generate-a-uniformly-distributed-point-on-a-sphere-from-a-fixe%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                      Image
                      Would it be easier to colonise a living world or a dead world? Postal services in Italy - Poste Italiane vs. Friendpost vs. GPS What is /dev/null and why can't I use hx on it? Why are engines with carburetors hard to start in cold weather? How to make a gift without seeming creepy? Without exposing his identity, did Roy help his parents with money so that they can afford to stay in their home? Does the Creighton Method of Natural Family Planning have a failure rate of 3.2% or less? How do I break the broom in Untitled Goose Game? Had there been instances of national states banning harmful imports before the mid-19th C Opium Wars? D&D Monsters and Copyright Is sleeping on the groud in cold weather better than on an air mattress? Transiting through Switzerland by coach with lots of cash quadratic equations on 2 by 2 matrices Always Lubricate Skewers? I pay for a service, but I miss the broadcast Iron-age tools, is there a way to extract heavy metals ...
                      Read more

                      Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                      Image
                      Internet forum softwarePHP software Internet communityPHPMySQLdatabase management systemInternet forumCharles WarnerMatt MechamJarvis Entertainment GroupIkonboardInvisionFreeAjaxIPS Community ForumsIPS Community Forumsthis blog entry Invision Community From Wikipedia, the free encyclopedia Jump to navigation Jump to search Invision Community Developer(s) Invision Power Services Stable release 4.4.6 / August 19, 2019 ; 34 days ago  ( 2019-08-19 ) Platform PHP / MySQL Type Forum software License Proprietary Website invisioncommunity.com Invision Community (previously known as IPS Community Suite ) is primarily an Internet community software produced by Invision Power Services, Inc. It is written in PHP and uses MySQL as a database management system. Invision Power Services sell applications that each can be bought and installed separately in addition to the Suite, the most widely known being the Internet forum software Invision Power Boar...
                      Read more

                      199年 目錄 大件事 到箇年出世嗰人 到箇年死嗰人 節慶、風俗習慣 導覽選單

                      Image
                      199年 平年 199年 出自維基百科 跳至導覽 跳至搜尋 世紀： 1世紀 2世紀 3世紀 年代： 160年代 170年代 180年代 190年代 200年代 210年代 220年代 年： 194年 195年 196年 197年 198年 199年 200年 201年 202年 203年 204年 干支年： 己卯年 （ 兔年 ） 年號： 199年 係平年。 目錄 1 大件事 2 到箇年出世嗰人 3 到箇年死嗰人 4 節慶、風俗習慣 大件事 199...
                      Read more