- #1

- 7

- 0

## Main Question or Discussion Point

How do i randomize a position on a sphere? Using a random number between 0 and 360 for longitude and a number between -90 to 90 for latitude would make it more probable to get closer to the poles right?

- Thread starter Abzu
- Start date

- #1

- 7

- 0

How do i randomize a position on a sphere? Using a random number between 0 and 360 for longitude and a number between -90 to 90 for latitude would make it more probable to get closer to the poles right?

- #2

- 97

- 0

picking 3 random numbers between 0 and 1 for the x,y and z coordinates sounds like a better bet...

- #3

Mark44

Mentor

- 33,493

- 5,183

Almost all of the time, this would give you a pointpicking 3 random numbers between 0 and 1 for the x,y and z coordinates sounds like a better bet...

- #4

- 97

- 0

- #5

- 15,393

- 683

A good random number package will provide separate interfaces for drawing from U[0,1), U[0,1], U(0,1), and U(0,1]. A not so good one might only provide just one of these four. There are some very small drawbacks to using not quite the right interface. For example, drawing longitude from U[0,360º] gives a small chance of doubling up on 0 versus 360º. Drawing sine(latitude) from U[-1,1) means you ruled out the North Pole.

- #6

- 7

- 0

Uhm, i have absolutely no idea what this means. I only know very basic math. And in case i was unclear in my first post, i want an even distribution,

A good random number package will provide separate interfaces for drawing from U[0,1), U[0,1], U(0,1), and U(0,1]. A not so good one might only provide just one of these four. There are some very small drawbacks to using not quite the right interface. For example, drawing longitude from U[0,360º] gives a small chance of doubling up on 0 versus 360º. Drawing sine(latitude) from U[-1,1) means you ruled out the North Pole.

- #7

- 7

- 0

- #8

- 15,393

- 683

I'll keep it simple -- after saying what U(a,b) and the related terms mean. Colloquially, U(a,b) means "a random number drawn from between

So, what does this mean? Draw a random number from between -1 and 1. Take the inverse sine of this number and express the result in degrees. This is your latitude. Draw a random number from between 0 and 360º. This is your longitude. If you draw a whole bunch of latitude,longitude pairs this way and plot their locations you will see that this gives a nice uniform spread over the globe. As you have already discovered, drawing latitude directly from between -90º and 90º does not give a nice uniform spread.

- #9

- 7

- 0

=ASIN((RANDBETWEEN(-1000;1000)/1000))/PI()*180

Thanks D H!

- #10

- 15,393

- 683

BTW, =DEGREES(ASIN(RAND()*2-1)) would work quite nicely as well.

My version of Excel doesn't even have a RANDBETWEEN function.

- #11

uart

Science Advisor

- 2,776

- 9

I know this is probably not much help to the OP, but it’s interesting to point out that DH’s method is mathematically equivalent to choosing latitude from a random distribution with a cosine probability density function,

[tex]f(\phi) = 1/2 \cos(\phi) \, : \, -\pi/2 \leq \phi \leq \pi/2[/tex]

Since an element of surface on the sphere is (taking phi as latitude and theta as longitude) given by :

[tex]dS/r^2 = \cos(\phi) \, d\phi \, d\theta[/tex]

and [itex]\cos(\phi) d\phi[/itex] is (for the given pdf) proportional to the probability that latitude is between [itex]\phi[/itex] and [itex]\phi + d\phi[/itex] then it follows that the probability of "hitting" a given element of surface on the sphere is proportional only to the SA of the element, no matter where that element of SA is located.

[tex]f(\phi) = 1/2 \cos(\phi) \, : \, -\pi/2 \leq \phi \leq \pi/2[/tex]

Since an element of surface on the sphere is (taking phi as latitude and theta as longitude) given by :

[tex]dS/r^2 = \cos(\phi) \, d\phi \, d\theta[/tex]

and [itex]\cos(\phi) d\phi[/itex] is (for the given pdf) proportional to the probability that latitude is between [itex]\phi[/itex] and [itex]\phi + d\phi[/itex] then it follows that the probability of "hitting" a given element of surface on the sphere is proportional only to the SA of the element, no matter where that element of SA is located.

Last edited:

- #12

- 1,233

- 17

No polar bias doing it this way, though you need 3 random numbers per point rather than the minimum 2.

- #13

- 15,393

- 683

In the vicinity of the projections onto the sphere of areas in the vicinity of

- The centers of the six faces of the cube: Too few points.
- The twelve edges of the cube: Too many points.
- The eight corners of the cube: Way too many points.

- #14

- 1,233

- 17

Ok I see, I had it very confused. Disregard my first post!

- #15

- 15,393

- 683

Your approach can be made to work. Reject

- Last Post

- Replies
- 2

- Views
- 4K

- Replies
- 18

- Views
- 26K

- Replies
- 41

- Views
- 6K

- Replies
- 11

- Views
- 1K

- Replies
- 8

- Views
- 755

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 4K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 22

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 2K