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

Loose spokes after only a few rides

What is this business jet?

What do hard-Brexiteers want with respect to the Irish border?

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

Why are there uneven bright areas in this photo of black hole?

Does adding complexity mean a more secure cipher?

Can you cast a spell on someone in the Ethereal Plane, if you are on the Material Plane and have the True Seeing spell active?

Deal with toxic manager when you can't quit

Correct punctuation for showing a character's confusion

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

Keeping a retro style to sci-fi spaceships?

How did passengers keep warm on sail ships?

Why not take a picture of a closer black hole?

Unitary representations of finite groups over finite fields

I am an eight letter word. What am I?

Inverse Relationship Between Precision and Recall

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

How to obtain a position of last non-zero element

Why was M87 targeted for the Event Horizon Telescope instead of Sagittarius A*?

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

Output the Arecibo Message

Why isn't the circumferential light around the M87 black hole's event horizon symmetric?

Can I have a signal generator on while it's not connected?

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



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












3












$begingroup$


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
    42 mins ago
















3












$begingroup$


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
    42 mins ago














3












3








3


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 23 mins ago









robjohn♦

271k27313642




271k27313642










asked 3 hours ago









The Zach ManThe Zach Man

1057




1057












  • $begingroup$
    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
    42 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
    42 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
42 mins ago




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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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



$mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$



or



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






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 $sqrt{x^2+y^2+z^2}$ 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









      4












      $begingroup$

      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



      $mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$



      or



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






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        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



        $mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$



        or



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






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          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



          $mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$



          or



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






          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



          $mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$



          or



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







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 58 mins 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 $sqrt{x^2+y^2+z^2}$ 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 $sqrt{x^2+y^2+z^2}$ 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 $sqrt{x^2+y^2+z^2}$ 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 $sqrt{x^2+y^2+z^2}$ 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

                      What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...

                      Image
                      What happens when the centripetal force is equal and opposite to the centrifugal force? Why is "la Gestapo" feminine? Nested Dynamic SOQL Query How are passwords stolen from companies if they only store hashes? Exposing a company lying about themselves in a tightly knit industry: Is my career at risk on the long run? Can "few" be used as a subject? If so, what is the rule? Why are there no stars visible in cislunar space? categorizing a variable turns it from insignificant to significant Do I need an EFI partition for each 18.04 ubuntu I have on my HD? What are the consequences of changing the number of hours in ...
                      Read more

                      Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                      Image
                      Tidlegare republikkarSkipingar i 1433Kommunar i SchwyzDistrikt i SchwyzNedleggingar i 1798Stader ved Vierwaldstättersee kommunekantonenSchwyzSveitskm²VierwaldstätterseeLauerz Gersau Kommunevåpenet til Gersau Plassering Styresmakter Land   Sveits Kanton kantonen Schwyz Distrikt Gersau Språk tysk Geografi Flatevidd  - kommune 23,70 km² Innbyggjarar  - kommune (desember 2010)    - folketettleik 2 094   88,4 /km² Koordinatar 46°59′0″N 8°31′0″E  Koordinatar: 46°59′0″N 8°31′0″E  Høgd over havet 435 moh Diverse annan informasjon Postnummer 6442 Nettstad: www.gersau.ch Freistaat und Republik Gersau Fristaten og republikken Gersau By-republikk og alliert protektorat i Det gamle sveitsiske eidsforundet ←   ← 1433–1798 1814–1818 →   → Hovudstad Gersau Styreform Republikk Historisk periode Tidleg moderne tid  - Kjøpte fridom frå ...
                      Read more

                      Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                      Image
                      Frisyrar frisyrehårhårstrikkhårklemmerøyrethesthovudverkeuropeisk motejobbtreningfletterparykkarungdomsopprørmotkulturmusikararmotorsykkelentusiastarpolitietpolitiskforretningverksemdfiskeflette Hestehale sett bakfrå. Hestehale på ein mannleg indianarleiar. Hestehale er ein frisyre der ein samlar det meste eller alt av langt hår [1] i bakhovudet med ein hårstrikk eller liknande. Avhengig av mote kan det òg vere vanleg å sette opp hestehalen med hårklemmer eller å ha hestehalen hengande på den eine sida av hovudet ved å samle håret ved øyret. Hestehale er ein vanleg frisyre for jenter og for gutar med langt hår. Han har fått namnet sitt etter likskapen med halen på ein hest. For stramme hestehaler kan føre til hårtap, [2] i nokre høve også hovudverk. [3] Innhaldsliste 1 Hestehale på kvinner 2 Hestehale på menn 3 Galleri 4 Sjå òg 5 Bakgrunnsstoff 6 Kjelder Hestehale på kvinner | Ungjente med hestehale som held håret unna ansiktet...
                      Read more