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

Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?

Button changing its text & action. Good or terrible?

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

How can I add encounters in the Lost Mine of Phandelver campaign without giving PCs too much XP?

Match Roman Numerals

Falsification in Math vs Science

Does adding complexity mean a more secure cipher?

Kerning for subscripts of sigma?

Correct punctuation for showing a character's confusion

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

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

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Geography at the pixel level

Is Cinnamon a desktop environment or a window manager? (Or both?)

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

Finding the area between two curves with Integrate

Pokemon Turn Based battle (Python)

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

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

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()?

Unitary representations of finite groups over finite fields

Worn-tile Scrabble

How do PCB vias affect signal quality?

The difference between dialogue marks



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
    50 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
    50 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 31 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
    50 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
    50 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
50 mins ago




$begingroup$
@robjohn thank you, you're right that I forgot to specify that.
$endgroup$
– The Zach Man
50 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



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



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









      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



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



      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









      $endgroup$


















        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



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



        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









        $endgroup$
















          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



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



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

                      Fairchild Swearingen Metro Inhaltsverzeichnis Geschichte | Innenausstattung | Nutzung | Zwischenfälle...

                      Image
                      ZubringerflugzeugTransportflugzeugZweimotoriges FlugzeugFairchildMilitärluftfahrzeug (Belgien)Militärluftfahrzeug (Schweden)Militärluftfahrzeug (Südafrika)Erstflug 1965 Vereinigte StaatenIATA-CodeTurbopropflugzeugUS-amerikanischenFlugzeugherstellersSwearingenFairchildSan AntonioEd SwearingenGeschäftsreiseflugzeugsSwearingen Merlin IIFairchildUSASkywest AirlinesAir WisconsinHorizon AirlinesComairWings WestAmerican Eagle AirlinesFairchild DornierAerocondeutschsprachigen LändernC-1C-2VC-3VC-4C-5C-6C-7C-8C-9KC-10C-11C-12YC-14YC-15C-17C-18C-19C-20C-21C-22C-23C-24VC-25C-26C-27C-28C-29C-31C-32C-33UC-35C-37C-38C-39C-40C-41KC-45KC-46C-143HC-144C-145C-146KC-767UC-880 Fairchild Swearingen Metro Swearingen Metroliner der Ostfriesische Lufttransport auf dem Flughafen Bremen Typ: Turbopropflugzeug Entwurfsland: Vereinigte Staaten  Vereinigte Staaten Hersteller: Fairchild/Swearingen Erstflug: 1965 Indienststellung: 1971 Produktionszeit:
                      Read more

                      Pilgersdorf Inhaltsverzeichnis Geografie | Geschichte | Bevölkerungsentwicklung | Politik | Kultur...

                      Image
                      Gemeinde im BurgenlandOrt im Bezirk OberpullendorfPilgersdorfErsterwähnung 1225 ungarischkroatischBurgenlandBezirk OberpullendorfÖsterreichZöbernbachKirchschlag in der Buckligen WeltLockenhausKatastralgemeindenNoricumkeltischen Höhensiedlung BurgBurgbergPannoniaLudwigs des DeutschenDominicusRatpotsRihherisMagyarenGüssinger FehdeAlbrechtFranziskanernKanizsayKirchschlagKrumbachTürkenProtestantenSteinbachHaidukenAugustiner-EremitenKuruzenPestVolkszählungGemeinderatBürgermeisterdirektwahlErstenZweiten WeltkriegsDeutschkreutzDraßmarktFrankenau-UnterpullendorfGroßwarasdorfHoritschonKaisersdorfKobersdorfLackenbachLackendorfLockenhausLutzmannsburgMannersdorf an der RabnitzMarkt Sankt MartinNeckenmarktNeutalNikitschOberloisdorfOberpullendorfPilgersdorfPiringsdorfRaidingRitzingSteinberg-DörflStoobUnterfrauenhaidUnterrabnitz-SchwendgrabenWeingrabenWeppersdorf Dieser Artikel erläutert die burgenländische Gemeinde. Siehe auch Pilgersdorf (Gemeinde Lasberg), Oberösterreich bzw. Pilgersdorf,
                      Read more

                      Marineschifffahrtleitung Inhaltsverzeichnis Geschichte | Heutige Organisation der NATO | Nationale und...

                      Image
                      MarineMinenkriegsführung (See)Geleit (Schifffahrt)NATO-Organisation SeestreitkräftenHandelsschifffahrtHandelsschiffenZweiten WeltkriegKalten KriegesGeleitKriegsschiffeU-BootenErsten Weltkrieguneingeschränkten U-Boot-Kriegerhebliche Anstrengungenweit reichenden SchutzmaßnahmenSowjetunionNATOAllied Maritime Command Headquarters NorthwoodLondonOfficer in Tactical CommandKapitänePartnerschaft für den FriedenReedereienFischereiKonvoisMinenbedrohungMinenabwehrfahrzeugenReserveoffizierenautische PatenteKonvoisInternationale Seeschifffahrts-Organisationengl.SonderorganisationVereinten NationenLondonInternational Ship and Port Facility Security CodeInternationalen HandelskammerLondonPiraterieKuala LumpurOperation AtalantaAustralienNeuseelandBundesmarineEmdenFlottenkommandoBefehlshaber der Seestreitkräfte der NordseeMarinekommandoRostockHandelsschiffsoffizierenReserveoffizierePiraterie vor der Küste SomaliasOperation Allied ProtectorOperation Ocean ShieldGolfs von AdenInternational Recommended
                      Read more