How to generate random points on a sphere?











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How do I generate $1000$ points $(x, y, z)$ and make sure they land on a sphere whose center is $(0, 0, 0)$ and its diameter is $20$? Simply, how do I manipulate a point's coordinates so that the point lies on the sphere's "surface"?






geometry computational-geometry spherical-geometry







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  • 4




    Simply Pythagorean theorem?
    – MonkeyKing
    Dec 22 '15 at 20:31






  • 4




    do you want the points to be uniformly random on the surface of the sphere?
    – Alex R.
    Dec 22 '15 at 20:33






  • 8




    Possible duplicate of Picking random points in the volume of sphere with uniform probability
    – leonbloy
    Dec 22 '15 at 20:42






  • 7




    @leonbloy: Not a duplicate, since that one was for the solid sphere.
    – Brian M. Scott
    Dec 22 '15 at 22:39






  • 4




    This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
    – Rahul
    Dec 23 '15 at 1:04

















up vote
36
down vote

favorite
17












How do I generate $1000$ points $(x, y, z)$ and make sure they land on a sphere whose center is $(0, 0, 0)$ and its diameter is $20$? Simply, how do I manipulate a point's coordinates so that the point lies on the sphere's "surface"?






geometry computational-geometry spherical-geometry







share|cite|improve this question




















  • 4




    Simply Pythagorean theorem?
    – MonkeyKing
    Dec 22 '15 at 20:31






  • 4




    do you want the points to be uniformly random on the surface of the sphere?
    – Alex R.
    Dec 22 '15 at 20:33






  • 8




    Possible duplicate of Picking random points in the volume of sphere with uniform probability
    – leonbloy
    Dec 22 '15 at 20:42






  • 7




    @leonbloy: Not a duplicate, since that one was for the solid sphere.
    – Brian M. Scott
    Dec 22 '15 at 22:39






  • 4




    This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
    – Rahul
    Dec 23 '15 at 1:04















up vote
36
down vote

favorite
17









up vote
36
down vote

favorite
17






17





How do I generate $1000$ points $(x, y, z)$ and make sure they land on a sphere whose center is $(0, 0, 0)$ and its diameter is $20$? Simply, how do I manipulate a point's coordinates so that the point lies on the sphere's "surface"?






geometry computational-geometry spherical-geometry







share|cite|improve this question















How do I generate $1000$ points $(x, y, z)$ and make sure they land on a sphere whose center is $(0, 0, 0)$ and its diameter is $20$? Simply, how do I manipulate a point's coordinates so that the point lies on the sphere's "surface"?






geometry computational-geometry spherical-geometry




geometry computational-geometry spherical-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 20:23









Rodrigo de Azevedo

12.8k41853




12.8k41853










asked Dec 22 '15 at 20:29









Filip

184125




184125








  • 4




    Simply Pythagorean theorem?
    – MonkeyKing
    Dec 22 '15 at 20:31






  • 4




    do you want the points to be uniformly random on the surface of the sphere?
    – Alex R.
    Dec 22 '15 at 20:33






  • 8




    Possible duplicate of Picking random points in the volume of sphere with uniform probability
    – leonbloy
    Dec 22 '15 at 20:42






  • 7




    @leonbloy: Not a duplicate, since that one was for the solid sphere.
    – Brian M. Scott
    Dec 22 '15 at 22:39






  • 4




    This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
    – Rahul
    Dec 23 '15 at 1:04
















  • 4




    Simply Pythagorean theorem?
    – MonkeyKing
    Dec 22 '15 at 20:31






  • 4




    do you want the points to be uniformly random on the surface of the sphere?
    – Alex R.
    Dec 22 '15 at 20:33






  • 8




    Possible duplicate of Picking random points in the volume of sphere with uniform probability
    – leonbloy
    Dec 22 '15 at 20:42






  • 7




    @leonbloy: Not a duplicate, since that one was for the solid sphere.
    – Brian M. Scott
    Dec 22 '15 at 22:39






  • 4




    This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
    – Rahul
    Dec 23 '15 at 1:04










4




4




Simply Pythagorean theorem?
– MonkeyKing
Dec 22 '15 at 20:31




Simply Pythagorean theorem?
– MonkeyKing
Dec 22 '15 at 20:31




4




4




do you want the points to be uniformly random on the surface of the sphere?
– Alex R.
Dec 22 '15 at 20:33




do you want the points to be uniformly random on the surface of the sphere?
– Alex R.
Dec 22 '15 at 20:33




8




8




Possible duplicate of Picking random points in the volume of sphere with uniform probability
– leonbloy
Dec 22 '15 at 20:42




Possible duplicate of Picking random points in the volume of sphere with uniform probability
– leonbloy
Dec 22 '15 at 20:42




7




7




@leonbloy: Not a duplicate, since that one was for the solid sphere.
– Brian M. Scott
Dec 22 '15 at 22:39




@leonbloy: Not a duplicate, since that one was for the solid sphere.
– Brian M. Scott
Dec 22 '15 at 22:39




4




4




This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
– Rahul
Dec 23 '15 at 1:04






This is the correct duplicate: How can I pick a random point on the surface of a sphere with equal distribution? But the answers here are better (although they are mostly covered by the MathWorld article I linked above).
– Rahul
Dec 23 '15 at 1:04












6 Answers
6






active

oldest

votes

















up vote
59
down vote













Use the fact that if you cut a sphere of a given radius with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes, not on where they cut the sphere. Thus, you can get a uniform distribution on the surface using two uniformly distributed random variables:




  • a $z$-coordinate, which in your case should be chosen between $-10$ and $10$; and

  • an angle in $[0,2pi)$ corresponding to a longitude.


From those it’s straightforward to generate the $x$- and $y$-coordinates.






share|cite|improve this answer

















  • 12




    This always challenges my intuition.
    – copper.hat
    Dec 22 '15 at 21:17






  • 1




    @copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
    – Brian M. Scott
    Dec 22 '15 at 21:22






  • 1




    So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
    – orangeskid
    Dec 22 '15 at 22:37






  • 3




    This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
    – Brian Tung
    Dec 23 '15 at 0:11






  • 4




    Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
    – Meni Rosenfeld
    Dec 23 '15 at 13:07




















up vote
50
down vote













Using Gaussian distribution for all three coordinates of your point will ensure an uniform distribution on the surface of the sphere. You should proceed as follows




  1. Generate three random numbers $x, y, z$ using Gaussian distribution

  2. Multiply each number by $1/sqrt{x^2+y^2+z^2}$ (a.k.a. Normalise) . You should handle what happens if $x=y=z=0$.

  3. Multiply each number by the radius of your sphere.






share|cite|improve this answer



















  • 1




    Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
    – orangeskid
    Dec 22 '15 at 22:43






  • 21




    Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
    – Robert Israel
    Dec 23 '15 at 0:40


















up vote
14
down vote













Here is a simple but less efficient way:



Generate points uniformly $x in [-10,10]^3$ and reject if $|x| =0 $ (which
should rarely happen) or
$|x| > 10$ (which should happen with probability ${20^3 -{4 over 3} pi 10^3 over 20^3} =1 - {pi over 6} approx 48%$).
Otherwise let $y = {10 over |x|} x$. Then $y$ will be distributed uniformly on the surface of the $10$-sphere.






share|cite|improve this answer























  • The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
    – copper.hat
    Dec 22 '15 at 22:48










  • This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
    – Tryss
    Dec 23 '15 at 21:34










  • @Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
    – heropup
    Dec 23 '15 at 22:11










  • I think for 1000 points the cost is minimal, and the coding fairly simple.
    – copper.hat
    Dec 23 '15 at 22:13










  • @heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
    – Tryss
    Dec 23 '15 at 22:30


















up vote
10
down vote













In addition to Brian Scott's excellent and clever answer, here's another, more straightforward way (in case you want to approach it with a geographical intuition): From two random variables $u_1, u_2$, distributed uniformly on the interval $[0, 1]$, generate (in radians) the latitude



$$
lambda = arccos (2u_1-1)-frac{pi}{2}
$$



and the longitude



$$
phi = 2pi u_2
$$



Then compute the rectangular coordinates accordingly:



$$
x = coslambdacosphi
$$
$$
y = coslambdasinphi
$$
$$
z = sinlambda
$$



ETA (thanks to Tanner Strunk—see comments): This will give coordinates of points on the unit sphere. To have them land on the sphere with diameter $20$ (and therefore radius $10$), simply multiply each by $10$.






share|cite|improve this answer























  • This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
    – Level River St
    Dec 24 '15 at 11:06












  • Oops, I think you are right. Thanks for the look out.
    – Brian Tung
    Dec 24 '15 at 15:41






  • 1




    I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
    – Tanner Strunk
    Jan 11 at 17:33






  • 1




    @TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
    – Brian Tung
    Jan 11 at 18:35


















up vote
6
down vote













Same way as on a real sphere, but $(x,y,z) $ multiplied by $i.$






share|cite|improve this answer

















  • 5




    I'm afraid the OP by "imaginary" just meant "mathematical"
    – leonbloy
    Dec 22 '15 at 20:43






  • 3




    That's funny! ${}{}$
    – copper.hat
    Dec 22 '15 at 20:54


















up vote
1
down vote













Wolfram Mathworld provides a methodology for randomly picking a point on a sphere:




To obtain points such that any small area on the sphere is expected to contain the same number of points, choose $u$ and $ν$ to be random variates on $[0,1]$. Then: $$begin{array}{ll}theta=2pi u\
varphi= arccos(2v - 1)end{array}$$ gives the spherical coordinates for a set of points which are uniformly distributed over $mathbb{S}^2$.







share|cite|improve this answer



















  • 1




    Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
    – Ilmari Karonen
    Dec 23 '15 at 18:39











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6 Answers
6






active

oldest

votes








6 Answers
6






active

oldest

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active

oldest

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active

oldest

votes








up vote
59
down vote













Use the fact that if you cut a sphere of a given radius with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes, not on where they cut the sphere. Thus, you can get a uniform distribution on the surface using two uniformly distributed random variables:




  • a $z$-coordinate, which in your case should be chosen between $-10$ and $10$; and

  • an angle in $[0,2pi)$ corresponding to a longitude.


From those it’s straightforward to generate the $x$- and $y$-coordinates.






share|cite|improve this answer

















  • 12




    This always challenges my intuition.
    – copper.hat
    Dec 22 '15 at 21:17






  • 1




    @copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
    – Brian M. Scott
    Dec 22 '15 at 21:22






  • 1




    So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
    – orangeskid
    Dec 22 '15 at 22:37






  • 3




    This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
    – Brian Tung
    Dec 23 '15 at 0:11






  • 4




    Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
    – Meni Rosenfeld
    Dec 23 '15 at 13:07

















up vote
59
down vote













Use the fact that if you cut a sphere of a given radius with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes, not on where they cut the sphere. Thus, you can get a uniform distribution on the surface using two uniformly distributed random variables:




  • a $z$-coordinate, which in your case should be chosen between $-10$ and $10$; and

  • an angle in $[0,2pi)$ corresponding to a longitude.


From those it’s straightforward to generate the $x$- and $y$-coordinates.






share|cite|improve this answer

















  • 12




    This always challenges my intuition.
    – copper.hat
    Dec 22 '15 at 21:17






  • 1




    @copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
    – Brian M. Scott
    Dec 22 '15 at 21:22






  • 1




    So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
    – orangeskid
    Dec 22 '15 at 22:37






  • 3




    This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
    – Brian Tung
    Dec 23 '15 at 0:11






  • 4




    Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
    – Meni Rosenfeld
    Dec 23 '15 at 13:07















up vote
59
down vote










up vote
59
down vote









Use the fact that if you cut a sphere of a given radius with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes, not on where they cut the sphere. Thus, you can get a uniform distribution on the surface using two uniformly distributed random variables:




  • a $z$-coordinate, which in your case should be chosen between $-10$ and $10$; and

  • an angle in $[0,2pi)$ corresponding to a longitude.


From those it’s straightforward to generate the $x$- and $y$-coordinates.






share|cite|improve this answer












Use the fact that if you cut a sphere of a given radius with two parallel planes, the area of the strip of spherical surface between the planes depends only on the distance between the planes, not on where they cut the sphere. Thus, you can get a uniform distribution on the surface using two uniformly distributed random variables:




  • a $z$-coordinate, which in your case should be chosen between $-10$ and $10$; and

  • an angle in $[0,2pi)$ corresponding to a longitude.


From those it’s straightforward to generate the $x$- and $y$-coordinates.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '15 at 21:05









Brian M. Scott

454k38505906




454k38505906








  • 12




    This always challenges my intuition.
    – copper.hat
    Dec 22 '15 at 21:17






  • 1




    @copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
    – Brian M. Scott
    Dec 22 '15 at 21:22






  • 1




    So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
    – orangeskid
    Dec 22 '15 at 22:37






  • 3




    This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
    – Brian Tung
    Dec 23 '15 at 0:11






  • 4




    Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
    – Meni Rosenfeld
    Dec 23 '15 at 13:07
















  • 12




    This always challenges my intuition.
    – copper.hat
    Dec 22 '15 at 21:17






  • 1




    @copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
    – Brian M. Scott
    Dec 22 '15 at 21:22






  • 1




    So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
    – orangeskid
    Dec 22 '15 at 22:37






  • 3




    This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
    – Brian Tung
    Dec 23 '15 at 0:11






  • 4




    Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
    – Meni Rosenfeld
    Dec 23 '15 at 13:07










12




12




This always challenges my intuition.
– copper.hat
Dec 22 '15 at 21:17




This always challenges my intuition.
– copper.hat
Dec 22 '15 at 21:17




1




1




@copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
– Brian M. Scott
Dec 22 '15 at 21:22




@copper.hat: Yes, I’ve always found it surprising that the flattening and stretching exactly balance the shrinking radius as the planes move towards a pole.
– Brian M. Scott
Dec 22 '15 at 21:22




1




1




So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
– orangeskid
Dec 22 '15 at 22:37




So you should get a uniform distribution on the cylinder and project on the sphere, like Archimedes. Nice!
– orangeskid
Dec 22 '15 at 22:37




3




3




This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
– Brian Tung
Dec 23 '15 at 0:11




This proves, by the way, that the surface area of a sphere's circumscribing cylinder (minus the endcaps) equals that of the sphere itself (and by corollary, the volume of the sphere is one-third the radius times the common surface area).
– Brian Tung
Dec 23 '15 at 0:11




4




4




Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
– Meni Rosenfeld
Dec 23 '15 at 13:07






Does this generalize to higher dimensions? For a 2D circle you just pick $theta$ uniformly. For a 3D sphere you choose a $theta$ and $z$. Is there a similar result for 4D and higher?
– Meni Rosenfeld
Dec 23 '15 at 13:07












up vote
50
down vote













Using Gaussian distribution for all three coordinates of your point will ensure an uniform distribution on the surface of the sphere. You should proceed as follows




  1. Generate three random numbers $x, y, z$ using Gaussian distribution

  2. Multiply each number by $1/sqrt{x^2+y^2+z^2}$ (a.k.a. Normalise) . You should handle what happens if $x=y=z=0$.

  3. Multiply each number by the radius of your sphere.






share|cite|improve this answer



















  • 1




    Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
    – orangeskid
    Dec 22 '15 at 22:43






  • 21




    Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
    – Robert Israel
    Dec 23 '15 at 0:40















up vote
50
down vote













Using Gaussian distribution for all three coordinates of your point will ensure an uniform distribution on the surface of the sphere. You should proceed as follows




  1. Generate three random numbers $x, y, z$ using Gaussian distribution

  2. Multiply each number by $1/sqrt{x^2+y^2+z^2}$ (a.k.a. Normalise) . You should handle what happens if $x=y=z=0$.

  3. Multiply each number by the radius of your sphere.






share|cite|improve this answer



















  • 1




    Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
    – orangeskid
    Dec 22 '15 at 22:43






  • 21




    Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
    – Robert Israel
    Dec 23 '15 at 0:40













up vote
50
down vote










up vote
50
down vote









Using Gaussian distribution for all three coordinates of your point will ensure an uniform distribution on the surface of the sphere. You should proceed as follows




  1. Generate three random numbers $x, y, z$ using Gaussian distribution

  2. Multiply each number by $1/sqrt{x^2+y^2+z^2}$ (a.k.a. Normalise) . You should handle what happens if $x=y=z=0$.

  3. Multiply each number by the radius of your sphere.






share|cite|improve this answer














Using Gaussian distribution for all three coordinates of your point will ensure an uniform distribution on the surface of the sphere. You should proceed as follows




  1. Generate three random numbers $x, y, z$ using Gaussian distribution

  2. Multiply each number by $1/sqrt{x^2+y^2+z^2}$ (a.k.a. Normalise) . You should handle what happens if $x=y=z=0$.

  3. Multiply each number by the radius of your sphere.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 23 '17 at 12:39









Community♦

1




1










answered Dec 22 '15 at 20:40









Henricus V.

14.9k21947




14.9k21947








  • 1




    Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
    – orangeskid
    Dec 22 '15 at 22:43






  • 21




    Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
    – Robert Israel
    Dec 23 '15 at 0:40














  • 1




    Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
    – orangeskid
    Dec 22 '15 at 22:43






  • 21




    Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
    – Robert Israel
    Dec 23 '15 at 0:40








1




1




Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
– orangeskid
Dec 22 '15 at 22:43




Excellent! This works in any dimension and seems not to be widely known, although I've seen some older paper (late $'40$'s) where the authors claim that they've learned it from Harald Cramér.
– orangeskid
Dec 22 '15 at 22:43




21




21




Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
– Robert Israel
Dec 23 '15 at 0:40




Not widely known? Every probabilist should know that the multivariate standard normal distribution is spherically symmetric, from which this follows immediately.
– Robert Israel
Dec 23 '15 at 0:40










up vote
14
down vote













Here is a simple but less efficient way:



Generate points uniformly $x in [-10,10]^3$ and reject if $|x| =0 $ (which
should rarely happen) or
$|x| > 10$ (which should happen with probability ${20^3 -{4 over 3} pi 10^3 over 20^3} =1 - {pi over 6} approx 48%$).
Otherwise let $y = {10 over |x|} x$. Then $y$ will be distributed uniformly on the surface of the $10$-sphere.






share|cite|improve this answer























  • The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
    – copper.hat
    Dec 22 '15 at 22:48










  • This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
    – Tryss
    Dec 23 '15 at 21:34










  • @Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
    – heropup
    Dec 23 '15 at 22:11










  • I think for 1000 points the cost is minimal, and the coding fairly simple.
    – copper.hat
    Dec 23 '15 at 22:13










  • @heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
    – Tryss
    Dec 23 '15 at 22:30















up vote
14
down vote













Here is a simple but less efficient way:



Generate points uniformly $x in [-10,10]^3$ and reject if $|x| =0 $ (which
should rarely happen) or
$|x| > 10$ (which should happen with probability ${20^3 -{4 over 3} pi 10^3 over 20^3} =1 - {pi over 6} approx 48%$).
Otherwise let $y = {10 over |x|} x$. Then $y$ will be distributed uniformly on the surface of the $10$-sphere.






share|cite|improve this answer























  • The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
    – copper.hat
    Dec 22 '15 at 22:48










  • This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
    – Tryss
    Dec 23 '15 at 21:34










  • @Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
    – heropup
    Dec 23 '15 at 22:11










  • I think for 1000 points the cost is minimal, and the coding fairly simple.
    – copper.hat
    Dec 23 '15 at 22:13










  • @heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
    – Tryss
    Dec 23 '15 at 22:30













up vote
14
down vote










up vote
14
down vote









Here is a simple but less efficient way:



Generate points uniformly $x in [-10,10]^3$ and reject if $|x| =0 $ (which
should rarely happen) or
$|x| > 10$ (which should happen with probability ${20^3 -{4 over 3} pi 10^3 over 20^3} =1 - {pi over 6} approx 48%$).
Otherwise let $y = {10 over |x|} x$. Then $y$ will be distributed uniformly on the surface of the $10$-sphere.






share|cite|improve this answer














Here is a simple but less efficient way:



Generate points uniformly $x in [-10,10]^3$ and reject if $|x| =0 $ (which
should rarely happen) or
$|x| > 10$ (which should happen with probability ${20^3 -{4 over 3} pi 10^3 over 20^3} =1 - {pi over 6} approx 48%$).
Otherwise let $y = {10 over |x|} x$. Then $y$ will be distributed uniformly on the surface of the $10$-sphere.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '15 at 21:14

























answered Dec 22 '15 at 21:01









copper.hat

125k559159




125k559159












  • The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
    – copper.hat
    Dec 22 '15 at 22:48










  • This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
    – Tryss
    Dec 23 '15 at 21:34










  • @Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
    – heropup
    Dec 23 '15 at 22:11










  • I think for 1000 points the cost is minimal, and the coding fairly simple.
    – copper.hat
    Dec 23 '15 at 22:13










  • @heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
    – Tryss
    Dec 23 '15 at 22:30


















  • The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
    – copper.hat
    Dec 22 '15 at 22:48










  • This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
    – Tryss
    Dec 23 '15 at 21:34










  • @Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
    – heropup
    Dec 23 '15 at 22:11










  • I think for 1000 points the cost is minimal, and the coding fairly simple.
    – copper.hat
    Dec 23 '15 at 22:13










  • @heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
    – Tryss
    Dec 23 '15 at 22:30
















The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
– copper.hat
Dec 22 '15 at 22:48




The points that are not rejected are uniformly distributed in the sphere, and projecting such points to the surface preserves uniformity.
– copper.hat
Dec 22 '15 at 22:48












This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
– Tryss
Dec 23 '15 at 21:34




This method is probably faster than the others from a computationnal point of view. The only complex operations are a division and a square root, and both are quite fast
– Tryss
Dec 23 '15 at 21:34












@Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
– heropup
Dec 23 '15 at 22:11




@Tryss Computationally, yes, but probabilistically the rejection rate is not to be ignored, especially in high dimensions (cf. curse of dimensionality). There are also quite fast implementations of standard normal random variates. It may be worth doing some simulations although I would be surprised if no one has written a paper about it.
– heropup
Dec 23 '15 at 22:11












I think for 1000 points the cost is minimal, and the coding fairly simple.
– copper.hat
Dec 23 '15 at 22:13




I think for 1000 points the cost is minimal, and the coding fairly simple.
– copper.hat
Dec 23 '15 at 22:13












@heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
– Tryss
Dec 23 '15 at 22:30




@heropup : yes, in high dimensions the rejection rate is big, and the gaussian method scale lineary contrary to this method. But in low dimensions, it works very well
– Tryss
Dec 23 '15 at 22:30










up vote
10
down vote













In addition to Brian Scott's excellent and clever answer, here's another, more straightforward way (in case you want to approach it with a geographical intuition): From two random variables $u_1, u_2$, distributed uniformly on the interval $[0, 1]$, generate (in radians) the latitude



$$
lambda = arccos (2u_1-1)-frac{pi}{2}
$$



and the longitude



$$
phi = 2pi u_2
$$



Then compute the rectangular coordinates accordingly:



$$
x = coslambdacosphi
$$
$$
y = coslambdasinphi
$$
$$
z = sinlambda
$$



ETA (thanks to Tanner Strunk—see comments): This will give coordinates of points on the unit sphere. To have them land on the sphere with diameter $20$ (and therefore radius $10$), simply multiply each by $10$.






share|cite|improve this answer























  • This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
    – Level River St
    Dec 24 '15 at 11:06












  • Oops, I think you are right. Thanks for the look out.
    – Brian Tung
    Dec 24 '15 at 15:41






  • 1




    I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
    – Tanner Strunk
    Jan 11 at 17:33






  • 1




    @TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
    – Brian Tung
    Jan 11 at 18:35















up vote
10
down vote













In addition to Brian Scott's excellent and clever answer, here's another, more straightforward way (in case you want to approach it with a geographical intuition): From two random variables $u_1, u_2$, distributed uniformly on the interval $[0, 1]$, generate (in radians) the latitude



$$
lambda = arccos (2u_1-1)-frac{pi}{2}
$$



and the longitude



$$
phi = 2pi u_2
$$



Then compute the rectangular coordinates accordingly:



$$
x = coslambdacosphi
$$
$$
y = coslambdasinphi
$$
$$
z = sinlambda
$$



ETA (thanks to Tanner Strunk—see comments): This will give coordinates of points on the unit sphere. To have them land on the sphere with diameter $20$ (and therefore radius $10$), simply multiply each by $10$.






share|cite|improve this answer























  • This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
    – Level River St
    Dec 24 '15 at 11:06












  • Oops, I think you are right. Thanks for the look out.
    – Brian Tung
    Dec 24 '15 at 15:41






  • 1




    I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
    – Tanner Strunk
    Jan 11 at 17:33






  • 1




    @TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
    – Brian Tung
    Jan 11 at 18:35













up vote
10
down vote










up vote
10
down vote









In addition to Brian Scott's excellent and clever answer, here's another, more straightforward way (in case you want to approach it with a geographical intuition): From two random variables $u_1, u_2$, distributed uniformly on the interval $[0, 1]$, generate (in radians) the latitude



$$
lambda = arccos (2u_1-1)-frac{pi}{2}
$$



and the longitude



$$
phi = 2pi u_2
$$



Then compute the rectangular coordinates accordingly:



$$
x = coslambdacosphi
$$
$$
y = coslambdasinphi
$$
$$
z = sinlambda
$$



ETA (thanks to Tanner Strunk—see comments): This will give coordinates of points on the unit sphere. To have them land on the sphere with diameter $20$ (and therefore radius $10$), simply multiply each by $10$.






share|cite|improve this answer














In addition to Brian Scott's excellent and clever answer, here's another, more straightforward way (in case you want to approach it with a geographical intuition): From two random variables $u_1, u_2$, distributed uniformly on the interval $[0, 1]$, generate (in radians) the latitude



$$
lambda = arccos (2u_1-1)-frac{pi}{2}
$$



and the longitude



$$
phi = 2pi u_2
$$



Then compute the rectangular coordinates accordingly:



$$
x = coslambdacosphi
$$
$$
y = coslambdasinphi
$$
$$
z = sinlambda
$$



ETA (thanks to Tanner Strunk—see comments): This will give coordinates of points on the unit sphere. To have them land on the sphere with diameter $20$ (and therefore radius $10$), simply multiply each by $10$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 18:36

























answered Dec 23 '15 at 0:14









Brian Tung

25.7k32553




25.7k32553












  • This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
    – Level River St
    Dec 24 '15 at 11:06












  • Oops, I think you are right. Thanks for the look out.
    – Brian Tung
    Dec 24 '15 at 15:41






  • 1




    I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
    – Tanner Strunk
    Jan 11 at 17:33






  • 1




    @TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
    – Brian Tung
    Jan 11 at 18:35


















  • This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
    – Level River St
    Dec 24 '15 at 11:06












  • Oops, I think you are right. Thanks for the look out.
    – Brian Tung
    Dec 24 '15 at 15:41






  • 1




    I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
    – Tanner Strunk
    Jan 11 at 17:33






  • 1




    @TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
    – Brian Tung
    Jan 11 at 18:35
















This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
– Level River St
Dec 24 '15 at 11:06






This is the best way. Some points: 1. arccos(-1..1) normally gives the range 0..pi, so I would be inclined to subtract pi/2 instead of pi, to get a number in the range -pi/2 .. pi/2 which is how a geographer would present the angle. Actually I think this change is required, because the current answer will always give negative z. 2. Having done that, I see nothing wrong with z = 2u -1 and cos lamba = sqrt(1-z^2) 3. phi works just as well with 2*npiu2, which may be convenient in some computer implementations.
– Level River St
Dec 24 '15 at 11:06














Oops, I think you are right. Thanks for the look out.
– Brian Tung
Dec 24 '15 at 15:41




Oops, I think you are right. Thanks for the look out.
– Brian Tung
Dec 24 '15 at 15:41




1




1




I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
– Tanner Strunk
Jan 11 at 17:33




I'm a bit late in coming to this question (ha), but you should be scaling those x,y,z expressions by 10 to land on the sphere, right? (Tiny, nit-picky detail, but I figure it should be noted for future readers.) Nice avatar by the way, Brian.
– Tanner Strunk
Jan 11 at 17:33




1




1




@TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
– Brian Tung
Jan 11 at 18:35




@TannerStrunk: Yes, you're quite right, thanks! I'll add an edit.
– Brian Tung
Jan 11 at 18:35










up vote
6
down vote













Same way as on a real sphere, but $(x,y,z) $ multiplied by $i.$






share|cite|improve this answer

















  • 5




    I'm afraid the OP by "imaginary" just meant "mathematical"
    – leonbloy
    Dec 22 '15 at 20:43






  • 3




    That's funny! ${}{}$
    – copper.hat
    Dec 22 '15 at 20:54















up vote
6
down vote













Same way as on a real sphere, but $(x,y,z) $ multiplied by $i.$






share|cite|improve this answer

















  • 5




    I'm afraid the OP by "imaginary" just meant "mathematical"
    – leonbloy
    Dec 22 '15 at 20:43






  • 3




    That's funny! ${}{}$
    – copper.hat
    Dec 22 '15 at 20:54













up vote
6
down vote










up vote
6
down vote









Same way as on a real sphere, but $(x,y,z) $ multiplied by $i.$






share|cite|improve this answer












Same way as on a real sphere, but $(x,y,z) $ multiplied by $i.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '15 at 20:35









Narasimham

20.6k52158




20.6k52158








  • 5




    I'm afraid the OP by "imaginary" just meant "mathematical"
    – leonbloy
    Dec 22 '15 at 20:43






  • 3




    That's funny! ${}{}$
    – copper.hat
    Dec 22 '15 at 20:54














  • 5




    I'm afraid the OP by "imaginary" just meant "mathematical"
    – leonbloy
    Dec 22 '15 at 20:43






  • 3




    That's funny! ${}{}$
    – copper.hat
    Dec 22 '15 at 20:54








5




5




I'm afraid the OP by "imaginary" just meant "mathematical"
– leonbloy
Dec 22 '15 at 20:43




I'm afraid the OP by "imaginary" just meant "mathematical"
– leonbloy
Dec 22 '15 at 20:43




3




3




That's funny! ${}{}$
– copper.hat
Dec 22 '15 at 20:54




That's funny! ${}{}$
– copper.hat
Dec 22 '15 at 20:54










up vote
1
down vote













Wolfram Mathworld provides a methodology for randomly picking a point on a sphere:




To obtain points such that any small area on the sphere is expected to contain the same number of points, choose $u$ and $ν$ to be random variates on $[0,1]$. Then: $$begin{array}{ll}theta=2pi u\
varphi= arccos(2v - 1)end{array}$$ gives the spherical coordinates for a set of points which are uniformly distributed over $mathbb{S}^2$.







share|cite|improve this answer



















  • 1




    Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
    – Ilmari Karonen
    Dec 23 '15 at 18:39















up vote
1
down vote













Wolfram Mathworld provides a methodology for randomly picking a point on a sphere:




To obtain points such that any small area on the sphere is expected to contain the same number of points, choose $u$ and $ν$ to be random variates on $[0,1]$. Then: $$begin{array}{ll}theta=2pi u\
varphi= arccos(2v - 1)end{array}$$ gives the spherical coordinates for a set of points which are uniformly distributed over $mathbb{S}^2$.







share|cite|improve this answer



















  • 1




    Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
    – Ilmari Karonen
    Dec 23 '15 at 18:39













up vote
1
down vote










up vote
1
down vote









Wolfram Mathworld provides a methodology for randomly picking a point on a sphere:




To obtain points such that any small area on the sphere is expected to contain the same number of points, choose $u$ and $ν$ to be random variates on $[0,1]$. Then: $$begin{array}{ll}theta=2pi u\
varphi= arccos(2v - 1)end{array}$$ gives the spherical coordinates for a set of points which are uniformly distributed over $mathbb{S}^2$.







share|cite|improve this answer














Wolfram Mathworld provides a methodology for randomly picking a point on a sphere:




To obtain points such that any small area on the sphere is expected to contain the same number of points, choose $u$ and $ν$ to be random variates on $[0,1]$. Then: $$begin{array}{ll}theta=2pi u\
varphi= arccos(2v - 1)end{array}$$ gives the spherical coordinates for a set of points which are uniformly distributed over $mathbb{S}^2$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '15 at 13:05









Daniel Fischer♦

173k16159281




173k16159281










answered Dec 23 '15 at 12:06









varunk

1192




1192








  • 1




    Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
    – Ilmari Karonen
    Dec 23 '15 at 18:39














  • 1




    Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
    – Ilmari Karonen
    Dec 23 '15 at 18:39








1




1




Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
– Ilmari Karonen
Dec 23 '15 at 18:39




Except for the link, this seems to be exactly the same as Brian Tung's earlier answer.
– Ilmari Karonen
Dec 23 '15 at 18:39


















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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