# Random Vectors with angle offset

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I'm trying to create a random Vector from an existing Vector and an angle Offset. (for a particle system) i found an old Topic where a user named Kurioes said: You could try this: 1) You have the vector you want to randomize, V. Normalize it. 2) Generate a new random vector, T, with length > epsilon, and normalize the vector. 3) If T and V are parallel (meaning the angle is too far away from 90 degrees, or abs(V dot T ) > 0.5 or something like that) go back to step 1. 4) Define a new vector, S, where S is the cross product of V and T (so is is perpendicular to both V and T); S = V x T 5) T = V x S (T should now be perpendicular to both V and S so that S, T and V are all perpendicular to one another) 6) Generate two random numbers; a between 0 and 2*pi and z between 0 and the maximum angle (in radians) between the input vector and the new randomized vector 7) The randomized V, V# = cos(z) * V + sin(z) * (cos(a)*S + sin(a)*T) -------------------------------------------------------------------------- and here's my code:
float r = rand() / (float)RAND_MAX;
float a = r * 2.0f * (float)M_PI;
r = rand() / (float)RAND_MAX;
Vec3 randSpeed;
if(_emitOffset > 0)
{
float z = r * _emitOffset;
// V# = cos(z) * V + sin(z) * (cos(a)*S + sin(a)*T)
randSpeed =  (_emit+sin(z)) * cos(z) * (_emitS*cos(a) + _emitT*sin(a));
}
else
randSpeed = _emit;


_emit (the emit-vector), _emitS and _emitT are perpendicular to eachother. (atleast 99.9% perpendicular anyways, couldn't get 100% for some reason, perhaps round-off errors) but, when i have _emit(0,-1,0) and _emitOffset = M_PI/4 i get results like {-0.14, 0, 0 } , which isn't right. What's wrong here?

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It might just be a calculations error. If I understand this correctly, you're simply performing two rotations on a basis defined by V x T = S. One rotation has to be an X rotation over [0, 2PI], however the second rotation can be either a Y or Z rotation over [0, maxRadians]. So expanding the follow XZ transformation:
[ 1     0       0  ]   [ cos(z) -sin(z) 0 ]   [ V.x T.x S.x ]   [ V'.x T'.x S'.x ][ 0 cos(x) -sin(x) ] * [ sin(z)  cos(z) 0 ] * [ V.y T.y S.y ] = [ V'.y T'.y S'.y ][ 0 sin(x)  cos(x) ]   [     0       0  1 ]   [ V.z T.z S.z ] = [ V'.z T'.z S'.z ]

Should give you an analytic solution. However as Kurioes mentioned the random solution is only going to be uniform (or whatever distribution your RNG generates) over the input parameters, which are angles in angular space. I wouldn't worry about it though.

Also, if V x T = S, then V x S = -T. Shouldn't matter in the end, but it was bugging me.

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why dont you just use rejection sampling? I don't even think this thing is uniform.

do{
pick 3 random numbers r1,r2,r3 in [0,1]
while(r1^2 + r2^2 + r3^2 > 1)

vector(v1,v2,v3)

or you can use shirley's mapping I can't remember. it's an analitical function with a sin and a cos and a sqrt(maybe not sqrt, can't remember) or something. the advantage is stratification is simple. s,t s+ds,s+dt is equal area for any s,t provided dt and ds are the same of course.

Tim

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timw: by rejection sampling, you mean this, right?

Quote:
 Original post by SiCraneThe simplest to explain method I can think of to do something like this would be:1. generate three random numbers x, y, z each in the range [-1, 1]2. create a vector out of those three numbers3. if the magnitude of the vector is greater than 1, or smaller than some epsilon value, repeat from step 1.4. if the angle between the given vector and the generated vector is greater than the given angle go back to step 1.5. adjust the magntiude of the generated vector to be equal to the magnitude of the given vector.

(taken from the old thread i mentioned)

i COULD use this. However, it is far from optimal, especially when using small offsets.
Let's say i have a particle-emitter with a narrow "cone" (vector with maybe 5 degrees offset). To randomly generate a vector that is offset no more than 5 degrees from a given vector will probably take quite a few attempts.
Considering that emitters might want to emit perticles at rates about 100 particles/second, this doesn't sound very good.

This "Shirley's Mapping" you're talking about could be interesting though. Anyone got any more info on this?

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Quote:
 Original quote by SimonTheSorcererThis "Shirley's Mapping" you're talking about could be interesting though. Anyone got any more info on this?

I too went in search of some more information on this mapping but the best I came across was this, which at least contains some source code to follow. All it does is map a square to a circle (and vice versa) using concentric squares and circles.

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hmm, if I'm getting this, you have an arrow that represents the axis of the emitter-cone and you want to slightly offset in order to shoot your particle in that direction. Check this:

Let v=(v1,v2,v3) be the normalized vector that represents the axis of the emitter. (be careful: not the position, the axis) If it's not normalized you can divide all its components by its magnitude sqrt(v1*v1+v2*v2+v3*v3). This makes it unit vector. Then you can use the following expressions to find it's longitude (east-west orientation) and latitude (north-south of the equator) in a hypothetical spherical coordinates system.

longitude = phi = atan(v2/v1);
latitude = theta = atan( v3/(sqrt(v1*v1+v2*v2)) );

Now you have the orientation (in angles counted in radians) of the emitter's axis. All you need to do now, is add a random offset, i.e. corresponding to +/- 5degrees to these angles and reproject them back to cartesian coordinates using these:

v1' = mag*cos(phi)*cos(theta);
v2' = mag*sin(phi)*cos(theta);
v3' = mag*sin(theta);

where 'mag' should be the desired magnitude of the resulting vector. I suggest you keep this set to 1, since unit vectors are the most useful to work with (when it comes to representing directions)

I hope this has been helpful.

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I guess I should also mention that the longitude 'phi' should be in the range of 0 to 2pi radians, and the latitude 'theta' in the range -pi/2,...,pi/2. I'm mentioning this because there's a bunch of implementations of tan() and atan() out there and they can get a little tricky when you're trying to copy/paste formulas.
I'm also assuming that a right-handed world coord. system is used...

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Quote:
 I hope this has been helpful.

YES! this has been very helpful indeed.
it certainly lets me offset my vectors with ease!

However, if i am to be picky (and i am :P) there's a small flaw here.
Let's say my offset is PI/2 (90 Degrees).
Then i would only have to offset one of theta OR phi by PI/2 to get a new vector that is perpendicular to the original vector.

so i would have to divide the offset-value between the two. (which i'm not doing just yet)

e.g Offset-value is 10 Degrees. Then i might offset phi by [0,3] and phi by [0,7] (totally random ofcourse, however the sum should always be 10)
am i right?

Anyways, here's my source. (as you can see i made a little test in my code)

//TEST_emit = Vec3(0,1,0);//_emit = _emit.Normalize();_emitOffset = 10 * float(DEG_TO_RAD); //Longitudefloat phi = atan2( _emit.GetY() , _emit.GetX() );//Latitudefloat theta = atan( _emit.GetZ()/(sqrt(_emit.GetX()*_emit.GetX()+_emit.GetY()*_emit.GetY())) );	//r is between [0,1]r = rand() / (float)RAND_MAX;//phi is OffSet by [0,_emitOffset]phi += (-_emitOffset + r*2*_emitOffset);//same for thetar = rand() / (float)RAND_MAX;theta += (-_emitOffset + r*2*_emitOffset);					/*//phi is OffSet by M_PI/2 radiansphi += float(M_PI/2);//Theta is not offset at all*/randSpeed.SetX(cos(phi)*cos(theta));randSpeed.SetY((sin(phi)*cos(theta)));randSpeed.SetZ(sin(theta));//float dot = randSpeed.Dot(_emit);

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If you already have a transform for the particle system you can use the following method to emit particles within the specified cone half-angle around the z axis of the particle system:

// m = world transform of particle system// halfangle = half cone angle for spreadvoid RandVec(const mat44& m, float halfangle, vec3& v){   float z = Random(cosf(halfangle), 1.0f);   float t = Random(0.0f, 2.0f*pi);   float r = sqrtf(1.0f - z*z);	   float x = r*cosf(t);   float y = r*sinf(t);	   v = z*m.getZ() + x*m.getX() + y*m.getY();}

For a description of why this works and produces a random distribution have a look at:

http://www.faqs.org/faqs/graphics/algorithms-faq/ (6.08)

There are also a couple of easy optimizations you can make. The cosf(halfangle) can be precomputed and stored for the system and you can take advantage of the fsincos instruction if available to compute the cos and sin. This brings it down to 1 sqrt and 1 fsincos only.

HTH

[edit: missed the m. before getX() and getY() :)]

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Quote:
 Original post by SimonTheSorcererLet's say my offset is PI/2 (90 Degrees).Then i would only have to offset one of theta OR phi by PI/2 to get a new vector that is perpendicular to the original vector.

hmm, you mean you have to offset 'theta XOR phi' by PI/2 to get a perpendicular vector? I'm not quite sure... First glance, I think you'll get perpendicular vectors if you offset either any single one or both of them. I don't think I can proove this within this reply, but there's a very easy way to see if this holds true on not.
2 vectors are perpendicular (in any number of dimensions) if and only if their dot product is zero. So you can feed your algorithm a random vector and try offseting just phi first, then just theta, and then both phi and theta by PI/2 and each time test the dot product of the initial and final vector to see if it's zero.

I guess it wouldn't hurt mentioning that the dot product of two vectors is the sum of the products of their respective components (it's just a number, not a vector)

If you need anything else just shoot, I can provide all the info you want on the theoretical side but I can't help that much on the actual implementation...
Btw, how do you guys post source code snippets in your replies?

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