Random insideUnitSphere but outside other UnitSphere?
I have a question, sorry for my bad English. I know and use Random.insideUnitSphere. Now I would like to create a random point inside the sphere, but outside another smaller sphere (with the same center).
Like:
Well the easy but not good solution would be to use a loop in that you create a random point in the outer sphere, check if it is inside the small excluded space and then create a new one if it is. That should do for most problems but can take a few iterations until you find a suitable point and it is not really a nice solution.
The better solution would probably be to create a random point within the radius of the outer sphere - radius of the inner sphere. Then you translate the point by a vector of length of the inner sphere radius into the negative direction towards the sphere center.
Answer by _dns_ · Mar 10, 2017 at 03:48 PM
EDIT: this answer does not work, my mistake, thanks @Bunny83 for pointing it out!
randomValue = Random.insideUnitSphere ((1.0-smallSphereRadius/bigSphereRadius) + (smallSphereRadius/bigSphereRadius)) bigSphereRadius;
To simplify/Optimize if needed
TESTED but still imperfect answer: So, if I recall, I was trying to find a solution without normalizing because it still costs some CPU.
Solution with normalizing (corresponding to OP's solution) :
float delta = BigRadius - SmallRadius;
float length = SmallRadius + delta * Random.value;
Vector3 position = Random.insideUnitSphere.normalized * length;
Solution without normalizing (Notice it's using onUnitSphere)
float delta = BigRadius - SmallRadius;
float length = delta * Random.value;
Vector3 pos = Random.onUnitSphere * (BigRadius - length);
I've tested both and they seem to work the same.
I've timed both methods with a release build on an i7: method without normalizing is around 2.4 time faster (repeating a million time the last 2 lines, results might differ with IL2CPP). I could not see the internal code Unity uses because it's in C++
That being said, visually, it seems that both methods generate more points in the center than in the outer sphere (obvious when Comparing with a simple Vector3 pos = Random.insideUnitSphere * BigRadius
It shows that the distribution is not good with both methods (that rely on the same principle because the normalization is equivalent as taking a point on the sphere). There must be more math involved to have a uniform distribution (though it seems correct to discard values from Random.insideUnitSphere * BigRadius that are inside SmallRadius, not CPU efficient but uniform, as far as I understand)
EDIT2 better solution: Ok, so I've recalled what I was trying to do first: I was trying to "compress" the random points inside the sphere = imagine a sphere with random points in it, then you slowly inflate the center until it reaches SmallRadius size, compressing all the points but not ejecting them out of the sphere. I think this conserves the uniform distribution, but I'm not mathematically sure it does. Visually, it works well and looks uniform, even with extreme values of radius (SmallRadius = 0 and SmallRadius = BigRadius - epsilon)
Vector3 posInSphere = Random.insideUnitSphere;
float length = posInSphere.magnitude;
float ratioRadius = SmallRadius/BigRadius;
Vector3 pos = (((1.0f-ratioRadius)*length + ratioRadius) / length) * BigRadius * posInSphere;
It involves computing a vector's magnitude so it's close to the cost of a normalization but it's better distributed anyway :-)
Just because a similar question has just been asked over here I stumbled on this question. I would like to say that this is not a solution and does not simplify anything. What your code actually does is this:
randomValue = Random.insideUnitSphere * bigSphereRadius;
That's because your bracket term:
((1.0-smallSphereRadius/bigSphereRadius) + (smallSphereRadius/bigSphereRadius))
is actually just "1". $$anonymous$$aybe it becomes more obvious when you substitude the ratio between the small and large radius with a variable
float ratio = smallSphereRadius / bigSphereRadius;
((1.0 - ratio) + (ratio)) ==
(1.0 - ratio + ratio) ==
(1.0)
It doesn't ensure a $$anonymous$$imum distance at all. Note that the solution posted by the OP does work as he uses a normalized direction vector and calculates a length that has a $$anonymous$$imum length + some random "addition". However the "outterRadius$$anonymous$$ultiplier" doesn't specify the outer radius but the max addition to the inner radius. So the max radius is the inner radius + the outer radius.
Answer by WILEz1975 · Mar 10, 2017 at 08:48 PM
Solved:
The result in editor at runtime:
Work fine!