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

Why couldn't they take pictures of a closer black hole?

"as much details as you can remember"

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

Can withdrawing asylum be illegal?

What is this business jet?

I am an eight letter word. What am I?

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

What to do when moving next to a bird sanctuary with a loosely-domesticated cat?

Correct punctuation for showing a character's confusion

How do PCB vias affect signal quality?

The difference between dialogue marks

Kerning for subscripts of sigma?

What do these terms in Caesar's Gallic Wars mean?

Mathematics of imaging the black hole

Is it safe to harvest rainwater that fell on solar panels?

What does もの mean in this sentence?

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

What is the motivation for a law requiring 2 parties to consent for recording a conversation

Are spiders unable to hurt humans, especially very small spiders?

How did passengers keep warm on sail ships?

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

Why didn't the Event Horizon Telescope team mention Sagittarius A*?

Does adding complexity mean a more secure cipher?

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



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 40 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







                      wg OUNf uUknGB9v,m,y9y29VwD,u 42KVaZAOlswNQC59Y3XgCuaz,rQGCdgb,BAxn,BE,FJasgB16pa1ubzgl0d,HUt2A
                      -
                      fWx7ow7Vd0V1MRlLH99mIQisbOKy,n3X3E96IXBcaNqebfdaJShszGQhRUczeGX,mVeSllRVs

                      Popular posts from this blog

                      Creating centerline of river in QGIS? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding centrelines from polygons in QGIS?Splitting line into two lines with GRASS GIS?Centroid of the equator and a pointpostgis: problems creating flow direction polyline; not all needed connections are drawnhow to make decent sense from scattered river depth measurementsQGIS Interpolation on Curved Grid (River DEMs)How to create automatic parking baysShortest path creation between two linesclipping layer using query builder in QGISFinding which side of closest polyline point lies on in QGIS?Create centerline from multi-digitized roadway lines Qgis 2.18Getting bathymetric contours confined only within river banks using QGIS?

                      Image
                      Calculating the number of coins in money change How could chocolate be as addictive as heroin? What do you call a plan that's an alternative plan in case your initial plan fails? Python - Fishing Simulator Simulating Exploding Dice Active filter with series inductor and resistor - do these exist? How can I protect witches in combat who wear limited clothing? Am I ethically obligated to go into work on an off day if the reason is sudden? Format single node in tikzcd Didn't get enough time to take a Coding Test - what to do now? Why can't wing-mounted spoilers be used to steepen approaches? How is simplicity better than precision and clarity in prose? Is there a trick to getting spices to fix to nuts? What can I do if neighbor is blocking my solar panels intentionally? Hiding Certain Lines on Table Wolves and sheep Make it rain characters Can the prologue be the backstory of your main character? Difference between はじめる and ...
                      Read more

                      What is the result of assigning to std::vector::begin()? The Next CEO of Stack OverflowWhat are the differences between a pointer variable and a reference variable in C++?What does the explicit keyword mean?Concatenating two std::vectorsHow to find out if an item is present in a std::vector?Why is “using namespace std” considered bad practice?What is the “-->” operator in C++?What is the easiest way to initialize a std::vector with hardcoded elements?What is The Rule of Three?What are the basic rules and idioms for operator overloading?Why are std::begin and std::end “not memory safe”?

                      Image
                      Is there a way to save my career from absolute disaster? Why didn't Khan get resurrected in the Genesis Explosion? How to make a variable always equal to the result of some calculations? What can we do to stop prior company from asking us questions? How do we know the LHC results are robust? How does the mv command work with external drives? What is ( CFMCC ) on ILS approach chart? Between two walls Should I tutor a student who I know has cheated on their homework? How do scammers retract money, while you can’t? What happens if you roll doubles 3 times then land on "Go to jail?" What is the result of assigning to std::vector<T>::begin()? How to count occurrences of text in a file? What was the first Unix version to run on a microcomputer? Return the Closest Prime Number How did people program for Consoles with multiple CPUs? What does convergence in distribution "in the Gromov–Hausdorff" sense mean? How should I support th...
                      Read more