# [Solved] Quaternions: Extrapolating a rotation based on previous samples

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void CNode::Predict( float fTimeDelta ) // fTimeDelta = time elapsed in seconds since last frame
{
....

// angular estimates.  We need at least 2 items in the list to try this
if( m_spatialHistoryList.size() > 1 )
{
std::list< spatialHistory_t >::const_iterator i = m_spatialHistoryList.begin();

// get the two most recent orientations as confirmed by the server
spatialHistory_t shCurr = (*i);
spatialHistory_t shOlder = (*++i);

// given two quaternions, q1 and q2, a quaternion representing the difference between each is
// quatDiff = q1^-1 q2 where q1^-1 is the inverse.  Use conjugate instead since our quats are unit length.
// Once we have this value, we have the difference for a given time period.
D3DXQUATERNION rotationDelta, oldInverse;
D3DXQuaternionConjugate( &oldInverse, &shOlder.sh.orientation );
D3DXQuaternionMultiply( &rotationDelta, &oldInverse, &shCurr.sh.orientation );

// get the length of time taken to move between these two orientations and convert MS to S
DWORD dwTimeDeltaBetweenUpdates = shCurr.dwServerTimeStamp - shOlder.dwServerTimeStamp;

// We know the rotation difference and the time taken to travel between these two orientations
// so now we need to work out what the equivalent rotation would be for the period fTimeDelta

// what now?

}
}


I'm thinking that I need to work out the frame time as a fraction of the time between the last two updates, then exponentiate by this amount? Any help appreciated. [Edited by - Defrag on December 12, 2006 4:07:35 AM]

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Quote:
 Original post by DefragI'm thinking that I need to work out the frame time as a fraction of the time between the last two updates, then exponentiate by this amount? Any help appreciated.
I don't have much experience with networking, but what you describe sounds reasonable. In pseudocode, it might look something like this:
quaternion last_delta = quaternion_difference(    next_most_recent_sample,    most_recent_sample);quaternion delta = quaternion_exponentiate(    last_delta,    time_step / last_time_step);quaternion new_orientation = quaternion_concatenate(    current_orientation,    delta);
I think this could also be re-arranged to take advantage of the slerp() function:
quaternion last_delta = quaternion_difference(    next_most_recent_sample,    most_recent_sample);quaternion target_orientation = quaternion_concatenate(    current_orientation,    last_delta);quaternion new_orientation = slerp(    current_orientation,    target_orientation,    time_step / last_time_step);
I might have messed something up somewhere, but you might give one or the other of those a try and see what sort of results you get.

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I know you can use slerp (spherical linear interpolation) to smoothly interpolate between two quaternions. With regular lerp on points, when you let 't' go outside the usual [0,1] range you get extrapolated values that form a line, as expected. I'm wondering if the same property holds true for slerp. If it does, then all you'd have to do is slerp between q1 and q2 using the interpolating value (t-t1)/(t2-t1) with the current time being 't'.

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Thanks guys. I'm going to give the first suggestion a try as it more closely mirrors what I had in my head, so it should be easier to implement. Just need to figure out how to do the exponentiation now (I don't think D3DX provides a function to exponentiate) but I have a textbook to get the formula from.

I'll report back when I get it working (or bugger it up).

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Well, that was pretty much painless. It worked first time for my space ship which is controlled by the user. The server runs @ 20 fps, but the client runs as fast as I can render it -- it's smooth :).

My asteroids are really jerky but that's just because of a program logic conflict, I think (the asteroid's own update function is spinning it independently, so I'll have to sort out that later).

I'll move the exponentiation calculation (nabbed from the rather excellent 3D Math Primer for Graphics and Game Development) into its own little function, but here's the code:

// angular estimates.  We need at least 2 items in the list to try this	if( m_spatialHistoryList.size() > 1 )	{					std::list< spatialHistory_t >::const_iterator i = m_spatialHistoryList.begin();				// get the two most recent orientations as confirmed by the server		spatialHistory_t shCurr = (*i);		spatialHistory_t shOlder = (*++i);		        		// given two quaternions, q1 and q2, a quaternion representing the difference between each is		// quatDiff = q1^-1 q2 where q1^-1 is the inverse.  Use conjugate instead since our quats are unit length.		// Once we have this value, we have the difference for a given time period.		D3DXQUATERNION rotationDelta, oldInverse;		D3DXQuaternionConjugate( &oldInverse, &shOlder.sh.orientation );		D3DXQuaternionMultiply( &rotationDelta, &oldInverse, &shCurr.sh.orientation );                		// get the length of time taken to move between these two orientations and convert MS to S		DWORD dwTimeDeltaBetweenUpdates = shCurr.dwServerTimeStamp - shOlder.dwServerTimeStamp;		float fSecsBetweenUpdates = (float)dwTimeDeltaBetweenUpdates * 0.001f;		// We know the rotation difference and the time taken to travel between these two orientations		// so now we need to work out what the equivalent rotation would be for the period fTimeDelta				// Theory: Get the frame time as a fraction of the delta between the two confirmed updates		// then exponentiate the quaternion rotationDelta by this value like so:		// fExpVal = fTimeDelta / fSecsBetweenUpdates		// quaternion finalAngle = quaternionExponentiate( rotationDelta, fExpVal ) 		float fExpVal = fTimeDelta / fSecsBetweenUpdates;		if( fabs( rotationDelta.w ) < .9999f )		{			float fAlpha = acos( rotationDelta.w );			float fNewAlpha = fAlpha * fExpVal;			rotationDelta.w = cos( fNewAlpha );			float fMult = sin( fNewAlpha ) / sin( fAlpha );			rotationDelta.x *= fMult;			rotationDelta.y *= fMult;			rotationDelta.z *= fMult;		}		D3DXQuaternionMultiply( &m_spatial.orientation, &m_spatial.orientation, &rotationDelta );	}

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Quote:
 Original post by DefragJust need to figure out how to do the exponentiation now (I don't think D3DX provides a function to exponentiate)...
You might check out the function D3DXQuaternionToAxisAngle(), since essentially what you want to do is extract the axis and angle of rotation from a quaternion, scale the angle, and then construct a quaternion from the axis and the scaled angle.

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