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How to implement LookRotation


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#1 ryt   Members   -  Reputation: 294

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Posted 12 October 2013 - 01:33 PM

I would like to implement the function LookRotation() using quaternions, the same one seen in Unity 3D Quaternion.LookRotation(Vector3 forward).

This article almost explains it, at the end is a whole solution but I don't understand the quaternion stuff. I would like to implement it with Quaternion.AngleAxis(). So we would need an angle and rotation vector. But how we can find the rotation vector since LookRotation() only accepts one parameter ?

 

We could take the cross between forward and Vector3.up to get the right vector. And then cross forward and right to get localUp. But this way we would end up always with a up vector. But we need an arbitrary rotation vector because we need to rotate in all directions not only one (with up vector we can rotate only in one direction depending on left or right hand system).



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#2 Doublefris   Members   -  Reputation: 316

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Posted 16 October 2013 - 09:20 AM

I tried doing this problem myself, but in the end I ended up more or less doing a quaternion - matrix conversion.

What you can do if you're only rotating between directions on the x-z plane ( the 'ground) is use a simple cross and dot product with the halfway vector to get the cos(angle/2) and sin(angle/2).

However when you need a fixed up vector it becomes more complicated. There is no 'fixed' way to rotate between two vectors so you need to take care of that. I believe unity actually uses a fixed up vector { 0, 1, 0).

 

here is some code you can use :

It should actually be quite fast.

quaternion<T>::lookAt(const vecT& trgt, const vecT& eye, const vecT& upVec){
	const vecT f	= (eye - trgt).normalize();
	const vecT s	= (upVec.cross(f)).normalize();
	const vecT u	= f.cross(s);

	T z = 1 + s.x + u.y + f.z;
	T fd = 2 * sqrt(z);

	quaternion result;

	if(z > T(0.000001)){
		result.w = T(0.25) * fd;
		result.x = (f.y-u.z) / fd;
		result.y = (s.z-f.x) / fd;
		result.z = (u.x-s.y) / fd;
	}
	else if(s.x>u.y && s.x>f.z)	{
		fd = 2 * sqrt(1 + s.x - u.y - f.z);
		result.w = (f.y-u.z) / fd;
		result.x = T(0.25)		* fd;
		result.y = (u.x+s.y) / fd;
		result.z = (s.z+f.x) / fd;
	}
	else if(u.y > f.z){ 
		fd = 2 * sqrt(1 + u.y - s.x - f.z);	
		result.w = (s.z-f.x) / fd;
		result.x = (u.x+s.y) / fd;	
		result.y = T(0.25) * fd;
		result.z = (f.y+u.z) / fd;		
	}
	else {
		fd = 2 * sqrt(1 + f.z - s.x - u.y);
		result.w = (u.x-s.y) / fd;
		result.x = (s.z+f.x) / fd;
		result.y = (f.y+u.z) / fd;	
		result.z = T(0.25)		* fd;	
	}

	return result;
}

Edited by Doublefris, 16 October 2013 - 09:26 AM.


#3 ryt   Members   -  Reputation: 294

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Posted 23 October 2013 - 10:18 AM

I don't quite really understand the quaternion stuff that you do.

Could that be written in angle axis format ?

 

Before when I used to calculate with quaternions I used to do it with Quaternion(cos(a/2), sin(a/2)*n.x, sin(a/2)*n.y, sin(a/2)*n.z) where a is angle and n is a unit vector.

Could maybe your function be rewritten in this format ?


Edited by ryt, 23 October 2013 - 10:19 AM.


#4 Doublefris   Members   -  Reputation: 316

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Posted 29 November 2013 - 06:19 PM

I came up with an another solution, not using matrix conversion.

I added commentary to the code.

This is using a right handed coordinate system.

quat quat::CreateLook(vec3& o, vec3& t, vec3& up)
{
	// forward and size vectors of the coordinate frame 
	const vec3 f = normalize(o - t);
	const vec3 side = normalize(cross(up, f));

	// cross product of bisection and [0, 0, -1] gives you the 
	// half-sine values required to orientate [0, 0, -1] to f
	// the dot product gives you half the cosine
	vec3 b = bisect(f, vec3(0, 0, -1));
	const quat p = quat(dot(b, f), cross(b, f));

	// now we need an additional rotation around the f vector
	// to orientate the side vector.
	vec3 r = vec3(p.s*p.s + p.i*p.i - p.j*p.j - p.k*p.k,
	(2*p.i * p.j) - (2*p.s * p.k),
	2*p.i * p.k + 2*p.s * p.j);

	b = bisect(side, r);
	quat q = quat(dot(b, side), cross(side, b));
	// now we can take the product of q and p

	return p * q;
}


where bisect = a + b / || a + b || or normalize(a+b) 

 

I am reaally wondering if anybody could come up with a more elegant solution. (also : FASTER solution)


Edited by Doublefris, 29 November 2013 - 06:27 PM.





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