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

I am an eight letter word. What am I?

Is bread bad for ducks?

Unitary representations of finite groups over finite fields

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

Dropping list elements from nested list after evaluation

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

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

Worn-tile Scrabble

What does もの mean in this sentence?

How to display lines in a file like ls displays files in a directory?

Is an up-to-date browser secure on an out-of-date OS?

Are there any other methods to apply to solving simultaneous equations?

Why not take a picture of a closer black hole?

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

Does HR tell a hiring manager about salary negotiations?

Why doesn't UInt have a toDouble()?

"as much details as you can remember"

The difference between dialogue marks

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

Deal with toxic manager when you can't quit

Loose spokes after only a few rides

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

Can there be female White Walkers?

How to type a long/em dash `—`



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












2












$begingroup$


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♦
    2 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♦
    2 hours ago










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
















2












$begingroup$


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♦
    2 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♦
    2 hours ago










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














2












2








2


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 1 min ago









robjohn♦

271k27313642




271k27313642










asked 2 hours ago









The Zach ManThe Zach Man

1007




1007












  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    2 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♦
    2 hours ago










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


















  • $begingroup$
    So what you want is a uniform distribution. It would be helpful to state this explicitly.
    $endgroup$
    – robjohn♦
    2 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♦
    2 hours ago










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
















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




$begingroup$
So what you want is a uniform distribution. It would be helpful to state this explicitly.
$endgroup$
– robjohn♦
2 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♦
2 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♦
2 hours ago












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




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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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









$endgroup$





















    2












    $begingroup$

    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









      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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 36 mins ago









          robjohn♦robjohn

          271k27313642




          271k27313642























              2












              $begingroup$

              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









              $endgroup$


















                2












                $begingroup$

                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









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 2 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

                      “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

                      Image
                      How seriously should I take size and weight limits of hand luggage? Can a virus destroy the BIOS of a modern computer? Can I hook these wires up to find the connection to a dead outlet? How to keep a dark protagonist who wants to keep his humanity dark? How does a dynamic QR code work? My singleton can be called multiple times How to detect integer overflow in C? Can a PhD from a non-TU9 German university become a professor in a TU9 university? How to travel Japan while expressing milk? Is it a bad idea to plug the other end of ESD strap to wall ground? How to show a landlord what we have in s...
                      Read more

                      How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

                      Image
                      Does the US political system, in principle, allow for a no-party system? Is it a Cyclops number? "Nobody" knows! Understanding a proof about a set being closed Rationale to prefer local variables over instance variables? Movie: boy escapes the real world and goes to a fantasy world with big furry trolls Why aren't there more Gauls like Obelix? I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do? Solving Linear Matrix Recurrences What is the generally accepted pronunciation of “topoi”? Is it safe to abruplty remove arduino power ...
                      Read more

                      變成蝙蝠會怎樣？ 參考資料 外部連結 导航菜单Thomas Nagel, "What is it like to be a...

                      Image
                      心靈哲學思想實驗 湯瑪斯·內格爾感質物理心靈解釋鴻溝物理信息蝙蝠物理主義科學還原經驗現象學聲吶超音波神經衝動 成為一隻蝙蝠可能是什麼樣子？ 〈 成為一隻蝙蝠可能是什麼樣子 〉（ What is it like to be a bat? ）是湯瑪斯·內格爾在1974年發表的論文，旨在說明感質在物理與心靈之間的解釋鴻溝（ Explanatory gap （ 英语 ： Explanatory gap ） ）。 根據內格爾的論證，物理信息（ Physical information （ 英语 ： Physical information ） ）無法讓我們知道身為一隻蝙蝠的感覺是什麼，也藉此推知，我們無法知道他人對於顏色、聲音、氣味、疼痛等等的感覺「是什麼感覺」。首先，根據物理主義，世界上所有事物似乎都能由科學來解釋，每個人個別的、主觀的感質，是否也能還原為客觀的物理信息？內格爾認為，「從還原中排除經驗的現象學特徵是不可能的」。 [1] 譬如蝙蝠是由聲吶或回聲來感知世界，藉此得知物體的距離、大小、形狀等等。由於蝙蝠的感受方式與我們有很大的不同，我們無法以自身的內在生活去推知蝙蝠的內在生活，所以我們無法得知身為一隻蝙蝠真正的感覺是什麼。就算以科學方法得知蝙蝠在接收到超音波時，腦中會出現特定的神經衝動，但人類不能知道腦中產生那種神經衝動「是什麼感覺」。只有從蝙蝠的主觀觀點出發，才會知道是蝙蝠是什麼感覺，牠的感覺無法用物理的術語去理解，因為物理術語必定是以許多觀點都能理解為基礎的。 參考資料 ^ 內格爾：〈成為一隻蝙蝠可能是什麼樣子？〉，高新民、儲昭華主編：《心靈哲學》（北京：商務印書館，2002），頁107。 外部連結 Thomas Nagel, "What is it like to be a bat?" 这是一篇與哲學相關的小作品。你可以通过编辑或修订扩充其内容。 查 论 编 This page is only for reference, If you need detailed information, please check here
                      Read more