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

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

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edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  47 mins ago
  

  
  
  
  

add a comment | 

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

edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  47 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

$endgroup$

share|cite|improve this question

edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

share|cite|improve this question

edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

edited 28 mins ago

robjohn♦

271k27313642

edited 28 mins ago

robjohn♦

271k27313642

asked 3 hours ago

The Zach ManThe Zach Man

1057

asked 3 hours ago

The Zach ManThe Zach Man

1057

2 Answers
2

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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

$mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$

or

$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$

share|cite|improve this answer

answered 1 hour ago

robjohn♦robjohn

271k27313642

$endgroup$

add a comment | 

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

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

$endgroup$

add a comment | 

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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

$mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$

or

$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$

share|cite|improve this answer

answered 1 hour ago

robjohn♦robjohn

271k27313642

$endgroup$

add a comment | 

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

$mathrmlat=arcsin(2u_1-1),mathrmlon=2pi u_2$

or

$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$

share|cite|improve this answer

answered 1 hour ago

robjohn♦robjohn

271k27313642

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

robjohn♦robjohn

271k27313642

$endgroup$

share|cite|improve this answer

answered 1 hour ago

robjohn♦robjohn

271k27313642

answered 1 hour ago

robjohn♦robjohn

271k27313642

answered 1 hour ago

robjohn♦robjohn

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

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

$endgroup$

add a comment | 

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

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

$endgroup$

add a comment | 

$begingroup$

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

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107

answered 3 hours ago

Misha LavrovMisha Lavrov

49k757107