Sampling with multidimensional transformations

[Cbas]

I am not sure if I have the title right, but here is my problem:
I have a ray which 'should be' shot vertically from a point p, but depending on the situation it can: 1) either be shot in any direction in the hemisphere above p 2) shot with an angle of no more than σ off the vertical 3) shot with an angle of no more than σ off the vertical by with a Gaussian distribution
(See http://imgur.com/BMqWjoQ)
http://imgur.com/BMqWjoQ

First:
I wish to generate a point uniformly distributed on a hemisphere. I did some derivations and I came up with:
θ = acos(R₁)
∅= 2∏R₂
x = sinθcos∅=cos(2∏R₂)sqrt(1-R₁²)
y = sinθsin∅=sin(2∏R₂)sqrt(1-R₁²)
z=cosθ=R₁
I confirmed this wit a textbook, So I am pretty sure its right---
Second:
I want to generate a point uniformly but only within a small solid angle subtended by angle σ
Similar derivation as before but the values for theta and phi are
θ=acos(1-(1-cos(σ)*R₁))
∅= 2∏R₂
Im pretty sure this is also right

Third:
(now this is where I need help)
Instead of using a uniform distribution I would like to use a Gaussian distribution. I know Box Muller is one way of generating random number with a normal distribution (given a set of canonical numbers) but how do I use that now to generate ray directions that are normally distributed?

Thanks for your help

 

[mfb]

Can you do rejection sampling? The "best" implementation would depend on details of the parameters. "Use the second algorithm and reject the point with probablility 1-(value of gaussian)" should give reasonable results if the gaussian is not too narrow.

Alternatively, generate a random number based on the modified distribution f(θ)=sin(θ)*gaussian

 

[Cbas]

Alternatively, generate a random number based on the modified distribution f(θ)=sin(θ)*gaussian

What do you mean by "gaussian"? Is that one of my gaussian random numbers?
I assume sin(theta) is measure of area
Area = ∫^2π₀∫^σ₀sin(θ) Z₁dθd∅

(thanks for your reply)

 

[mfb]

sin(θ) is (proportional to) the length of a circle around the vertical axis. It is just a weight for the gaussian distribution. I don't know if you want that gaussian as function of θ, or the projection on the floor, or whatever, just use what you like there. As function of θ, it gets $$f(\theta)=\sin(\theta)\exp\left(\frac{-\theta^2}{2\sigma^2}\right)$$
Where σ is the width of the gaussian.

 

[blue_raver22]

.I would suggest using a multidimensional transformation to generate points with a Gaussian distribution. This method involves transforming a set of uniformly distributed points into points with a Gaussian distribution.

One way to do this is by using the Box-Muller transform, which involves taking two uniformly distributed points and transforming them into two points with a Gaussian distribution. This can be extended to generate points in higher dimensions.

To generate points within a small solid angle, you can first use the method you described for generating points uniformly within a hemisphere, and then use the Box-Muller transform to transform those points into points within the desired solid angle. This can be done by scaling and shifting the points to fit within the desired range of angles.

Overall, using a multidimensional transformation can provide a more efficient and accurate way to generate points with a Gaussian distribution compared to using a uniform distribution and trying to modify it to fit the desired distribution. Additionally, this method can easily be extended to generate points in higher dimensions if needed.