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

Multi tool use
Multi tool use

Straighten subgroup lattice

"as much details as you can remember"

Why doesn't shell automatically fix "useless use of cat"?

Mathematics of imaging the black hole

Pokemon Turn Based battle (Python)

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

The difference between dialogue marks

Is it a good practice to use a static variable in a Test Class and use that in the actual class instead of Test.isRunningTest()?

How to support a colleague who finds meetings extremely tiring?

Ubuntu Server install with full GUI

How to translate "being like"?

What is this sharp, curved notch on my knife for?

How to charge AirPods to keep battery healthy?

Falsification in Math vs Science

How do you keep chess fun when your opponent constantly beats you?

Unitary representations of finite groups over finite fields

Is bread bad for ducks?

What is the meaning of Triage in Cybersec world?

Can a flute soloist sit?

Does adding complexity mean a more secure cipher?

Getting crown tickets for Statue of Liberty

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

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

Match Roman Numerals



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
    51 mins 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
    51 mins 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 32 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
    51 mins 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
    51 mins 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
51 mins ago




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







                      WlZPQ 9R0c83,FR3zzkC dv6nAD,k2 QIH,gLn0CQVg6doALs0M,ucFm3e9Y
                      -
                      CGOjb1V,0fQ

                      Popular posts from this blog

                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                      Image
                      Flight departed from the gate 5 min before scheduled departure time. Refund options How to show a density matrix is in a pure/mixed state? Can we cancel the equality mark here? How to ask rejected full-time candidates to apply to teach individual courses? Are there any AGPL-style licences that require source code modifications to be public? Why did Israel vote against lifting the American embargo on Cuba? What should one know about term logic before studying propositional and predicate logic? Why is Rajasthan pro BJP in the LS elections but not in the state elections? New Order #6: Easter Egg Question on Gÿongy' lemma proof How to resize main filesystem Is it OK if I do not take the receipt in Germany? My mentor says to set image to Fine instead of RAW — how is this different from JPG? Can I feed enough spin up electron to a black hole to affect it's angular momentum? How do I find my Spellcasting Ability for my D&D character? Is my guitar...
                      Read more

                      2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                      Image
                      Objetos do disco dispersoObjetos transnetunianos objeto transnetunianodisco dispersoSistema Solarcorpo celesteressonância orbitalplanetaplanetamagnitude absolutadiâmetro9 de abril2013Outer Solar System Origins Surveyórbitaexcentricidadesemieixo maiorUAperiélioSolafélio 2013 GY136 Data da descoberta 9 de abril de 2013 Descoberto por Outer Solar System Origins Survey Categoria Transnetuniano Objeto do disco disperso Elementos orbitais Semieixo maior 55,961 UA Perélio 32,595 UA Afélio 79,327 UA Excentricidade 0,418 Inclinação 10,9° Características físicas Dimensões 133 km Magnitude absoluta 7,6 ver 2013 GY 136 , também escrito como 2013 GY136 , é um objeto transnetuniano que está localizado no disco disperso, uma região do Sistema Solar. Este corpo celeste está em uma ressonância orbital de 2:5 com o planeta planeta. Ele possui uma magnitude absoluta de 7,6 [ 1 ] e tem um diâmetro estimad...
                      Read more

                      Metrô de Los Teques Índice Linhas | Estações | Ver também | Referências Ligações externas | Menu de navegação«INSTITUCIÓN»«Mapa de rutas»originalMetrô de Los TequesC.A. Metro Los Teques |Alcaldía de Guaicaipuro – Sitio OficialGobernacion de Mirandaeeeeeee

                      Image
                      Metrô de Los TequesFundações na Venezuela em 2006 metropolitanoLos TequesGuaicaipuroMirandaGoverno da VenezuelaLinha 1Linha 2estaçõesLinha 1Linha 2IndependenciaSan AntonioLinha 3GuaicaipuroLibertadorEstação Las AdjuntasMetrô de Caracasalgarismocorestaçõesnomecorestaçõesestações Metrô de Los Teques Uma das composições que operam no sistema. Informações Proprietário Governo da Venezuela Local [[ Los Teques, Miranda]] País variant = Tipo de transporte Metropolitano Número de linhas 2 Número de estações 5 Website [1] Funcionamento Início de funcionamento 3 de novembro de 2006 (12 anos) [ 1 ] Operadora(s) C. A. Metro Los Teques Dados técnicos Extensão do sistema 11,2 km Bitola Bitola padrão (1435 mm) Velocidade máxima 80 km/h ...
                      Read more