Is there a way to generate a uniformly distributed point on a sphere from a fixed amount of random real...
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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
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add a comment |
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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
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So what you want is a uniform distribution. It would be helpful to state this explicitly.
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– robjohn♦
3 hours ago
2
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Distribute longitude uniformly and the sine of the latitude uniformly. Then the distribution of points on the sphere will be uniform.
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– robjohn♦
3 hours ago
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@robjohn thank you, you're right that I forgot to specify that.
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– The Zach Man
1 hour ago
add a comment |
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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
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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
geometry
edited 1 hour ago
robjohn♦
271k27313642
271k27313642
asked 3 hours ago
The Zach ManThe Zach Man
1107
1107
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So what you want is a uniform distribution. It would be helpful to state this explicitly.
$endgroup$
– robjohn♦
3 hours ago
2
$begingroup$
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
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@robjohn thank you, you're right that I forgot to specify that.
$endgroup$
– The Zach Man
1 hour ago
add a comment |
$begingroup$
So what you want is a uniform distribution. It would be helpful to state this explicitly.
$endgroup$
– robjohn♦
3 hours ago
2
$begingroup$
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
2
2
$begingroup$
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
add a comment |
2 Answers
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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
$z=2u_1-1,x=sqrt{1-z^2}cos(2pi u_2),y=sqrt{1-z^2}sin(2pi u_2)$
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add a comment |
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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.)
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
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
$z=2u_1-1,x=sqrt{1-z^2}cos(2pi u_2),y=sqrt{1-z^2}sin(2pi u_2)$
$endgroup$
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
$mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$
or
$z=2u_1-1,x=sqrt{1-z^2}cos(2pi u_2),y=sqrt{1-z^2}sin(2pi u_2)$
$endgroup$
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
$mathrm{lat}=arcsin(2u_1-1),mathrm{lon}=2pi u_2$
or
$z=2u_1-1,x=sqrt{1-z^2}cos(2pi u_2),y=sqrt{1-z^2}sin(2pi u_2)$
$endgroup$
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
$z=2u_1-1,x=sqrt{1-z^2}cos(2pi u_2),y=sqrt{1-z^2}sin(2pi u_2)$
answered 1 hour ago
robjohn♦robjohn
271k27313642
271k27313642
add a comment |
add a comment |
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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.)
$endgroup$
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 $sqrt{x^2+y^2+z^2}$ 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.)
$endgroup$
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 $sqrt{x^2+y^2+z^2}$ 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.)
$endgroup$
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.)
answered 3 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♦
3 hours ago
2
$begingroup$
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