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Is there a way to generate a 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
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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
asked 2 hours ago
The Zach ManThe Zach Man
1007
$endgroup$
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
asked 2 hours ago
The Zach ManThe Zach Man
1007
$endgroup$
$begingroup$
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
18 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
asked 2 hours ago
The Zach ManThe Zach Man
1007
$endgroup$
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
asked 2 hours ago
The Zach ManThe Zach Man
1007
asked 2 hours ago
The Zach ManThe Zach Man
1007
asked 2 hours ago
The Zach ManThe Zach Man
1007
asked 2 hours ago
The Zach ManThe Zach Man
1007
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
18 mins ago
add a comment |
$begingroup$
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
18 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
18 mins ago
$begingroup$
@robjohn thank you, you're right that I forgot to specify that.
$endgroup$
– The Zach Man
18 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$
answered 34 mins ago
robjohn♦robjohn
271k27313642
$endgroup$
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.)
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$
answered 34 mins 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
$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$
answered 34 mins 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
$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$
answered 34 mins ago
robjohn♦robjohn
271k27313642
$endgroup$
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
$z=2u_1-1,x=sqrt1-z^2cos(2pi u_2),y=sqrt1-z^2sin(2pi u_2)$
answered 34 mins ago
robjohn♦robjohn
271k27313642
answered 34 mins ago
robjohn♦robjohn
271k27313642
answered 34 mins ago
robjohn♦robjohn
271k27313642
answered 34 mins ago
robjohn♦robjohn
271k27313642
271k27313642
add a comment |
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.)
answered 2 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.)
answered 2 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.)
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
$endgroup$
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.)
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
answered 2 hours ago
Misha LavrovMisha Lavrov
49k757107
49k757107
add a comment |
add a comment |
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$begingroup$
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
18 mins ago