OpenGL Roll from view, right, up vectors - 3d space?

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I have a camera defined through three vectors for view, right and up. The camera itself appears to be working fine. However when I try to determine the roll of the camera, compared to the horizon, I am running into some issues. (The reason I am doing trying to do this is to be able to draw a horizon reference line on the display.) I think I have myself pretty confused at the moment, I have a feeling the answer is relatively simple, I just can't see it. My camera vectors are: viewDir, rightVector and upVector; and since I am implementing in OpenGL they initially are set to: viewDir = new Vector3( 0, 0, -1); rightVector = new Vector3( 1, 0, 0); upVector = new Vector3( 0, 1, 0); I'm calculating heading and pitch with the following:
public double GetHeading()
{
return Math.Atan2(viewDir.X, -viewDir.Z);
}

public double GetPitch()
{
return Math.Asin(viewDir.Y);
}

Everything I have tried to calculate the roll relative to the horizon has failed in one way or another. In my current approach (which is probably overcomplicated) I do the following:
public double GetRoll()
{
// initialize temporary vectors
Vector3 tmpView = new Vector3(0,0,-1);
Vector3 tmpRight = new Vector3(1,0,0);
Vector3 tmpUp = new Vector3(0,1,0);

double P = GetPitch();

// rotate around up vector (heading)
tmpView = Vector3.Normalize(
(tmpView * Math.Cos(H)) - (tmpRight * Math.Sin(H));
tmpRight = Vector3.CrossProduct( tmpView, tmpUp);

// rotate around right vector (pitch)
tmpView = Vector3.Normalize(
(tmpView * Math.Cos(P)) + (tmpUp * Math.Sin(P));
tmpUp = -1 * Vector3.CrossProduct( tmpView, tmpRight)

// calculate cross/dot products
Vector3 cross = tmpUp.CrossProduct(upVector);
double  dot = tmpUp.DotProduct(upVector);

double angle = Math.Atan2( cross.Magnitude, dot);

return angle;

}

The problems I have with the above include: 1) Angle is only 'correct' for clockwise rotation (I thought the cross product was supposed to provide me with a direction.) 2) The Pitch and Heading are effecting the Roll angle. For instance when turning 90 degrees right or left from the initial heading, for every 1 degree of pitch up or down, the roll angle is being increased by two degrees. I am sure I have a fundamental problem with my approach, but after 3 days of banging my head against this, and searching google and these forums, i'm no closer to figuring out my mistake. Seems like the only answers I can find are limited to 2D, not 3D. Thanks for any suggestions.

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Ok, I spent the last few hours drawing diagrams trying to understand this better and changed my approach. I changed my the function for calculating the camera 'roll' compared to the horizon with the following:

public double GetRoll(){  double angle = Math.Acos( worldUpV.DotProduct(upVector));  if ( rightVector.Y > 0 ) { angle = -angle; }  return angle;}

However, this does not give me the expected result, once I start changing the pitch of the camera.

For instance, if I roll to 45 deg and pitch to 45 deg, I end up with a calculated angle of 60deg.

Or, if I just pitch the camera up, I get a calculated roll equivalent to my pitch. Which does makes sense to me, as pitching the camera up, moves the upVector away from the world Y axis.

Although pitching is the rotation around the rightVector, while I'm trying to find the rotation around the viewVector/viewDir.

Obviously missing something still - haha.

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Off the top of my head, here's something you might try. Note that this will only work when it's 'meaningful' to compute the roll at all (that is, the computation may fail if the object is headed straight up or down).

Here's what it might look like in pseudocode:
vector3 local_up = object.transform_vector_world_to_local(vector3(0,1,0));local_up.z = 0;local_up.normalize();float horizon_angle = atan2(local_up.y, local_up.x);
That may be totally wrong, but if you're still stuck on this, you might at least try it out and see what you get :)

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Thank you for your suggestion, but I'm not quite sure I understand how the approach is different than what I am currently trying.

Maybe its because I am misunderstanding your intention of the line:

vector3 local_up = object.transform_vector_world_to_local(vector3(0,1,0));

I am assuming that this would be the equivalent of taking the vector(0,1,0)
and applying the heading and pitch rotations.

Equivalent to what I did in my initial post with:
// get Heading and Pitchdouble H = GetAzimuth();double P = GetElevation();            // rotate around up vector (heading)tmpView = Vector3.Normalize((tmpView * Math.Cos(H)) - (tmpRight * Math.Sin(H)));tmpRight = Vector3.CrossProduct( tmpView, tmpUp);            // rotate around right vector (pitch)tmpView = Vector3.Normalize( (tmpView * Math.Cos(P)) + (tmpUp * Math.Sin(P)));tmpUp = -1 * Vector3.CrossProduct(tmpView, tmpRight);

But I think, that difference between what you intended with the 'world_to_local' function and what I do above, is that my 'tmpUp' vector is still described in world coordinates, while your 'local_up' would be described in 'local coordinates'.

Which means your 'local_up.Z = 0' has a different meaning than if I were to do 'tmpUp.Z = 0;'

This made me think that perhaps what I should do is to subtract the viewDir vector from the tmpUp vector (rather than zero) - but this didn't give the results I am looking for either...

Still confused - heh...

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pitch, yaw & roll are always tricky...

in my vector class, I use this function for pitch/yaw:

// x : LR, z : FB, y : UD
// angle 0, 0, 0 is (0, 0, 1) (straight into screen)
SVector3 SVector3::GetAnglesFromVector()
{
SVector3 vRet(x, y, z);
float fLength = GetLength();
float fForward;
float fYaw, fPitch;

vRet.Normalize();

if (vRet.z == 0 && vRet.x == 0)
{
fYaw = 0;
if (vRet.y > 0)
fPitch = 90.0f;
else
fPitch = 270.0f;
}
else
{
fYaw = (float) (atan2(vRet.x, vRet.z) * RADTODEG);
if (fYaw < 0)
fYaw += 360;

fForward = (float) (sqrt (vRet.x * vRet.x + vRet.z * vRet.z));
fPitch = (float) (atan2(vRet.y, fForward) * RADTODEG);
if (fPitch < 0.0f)
fPitch += 360.0f;
}

vRet.Set(fPitch, fYaw, 0.0f);

return vRet;
}

For the roll, I would take the side vector, Grab the angles and use the pitch

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BuffaloJ, thanks for your suggestion - I tried implementing my method using the approach you suggested - calculating the pitch of the right vector - it looked like it would work, then, I found the same problem when I pitched the camera up and down...

Here is what I implemented:
public double GetRoll(){  double rightLen = Math.Sqrt( (rightVector.X * rightVector.X) + (rightVector.Z * rightVector.Z));  double angle = Math.Atan2( rightVector.Y, rightLen);  if (upVector.Y > 0) { angle = -angle; }  return angle;}

The roll angle changes relative to the pitch angle in that, as the pitch angle approaches 90 degrees, the roll angle approaches zero.

If instance:
Pitch Roll   0    30  15    28.879  30    25.6589  45    20.7048  60    14.4775  75     7.4355  90     0

Which appears as if Roll angle is being multiplied by the Cosine of the pitch angle.

I really feel like I'm missing something completely obvious at this point...

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Quote:
 Original post by awdorrinThank you for your suggestion, but I'm not quite sure I understand how the approach is different than what I am currently trying.Maybe its because I am misunderstanding your intention of the line:vector3 local_up = object.transform_vector_world_to_local(vector3(0,1,0));
Try replacing the above line with something like this:
vector3 local_up;local_up.x = object.side.y;local_up.y = object.up.y;local_up.z = object.forward.y;
And see if that gets you any closer. (If not, post back, and I'll try to post a clearer explanation of what the pseudocode is supposed to do.)

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Actually... I think the last code I posted may be correct. The results I was seeing just were not making sense according to what I expected to see.

Thinking more about it, it does make sense that as I pitch the camera up towards a vertical of 90deg (or down towards -90 degrees) that the roll angle relative to the horizon would change, just in the manner I described (following a circular curve.)

If I am pointing straight up, the roll of the camera, relative to the ground (world X-Z plane) does not effect the 'horizon angle.'

I think the problem I am having in my program now, is not related to the roll angle, but to how I am using that angle to calculate the relative horizon line.
So time to go review that code to see if I can figure out my problems there. ;)

Thanks for the suggestions and pointers - think I might actually be past this mental block!

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After going over this for a few more days, I finally figured it out.

I was getting exactly what I was asking for, the roll relative to the world horizon, however using the Dot Product of the vectors was including the pitch in the angle, in addition to the roll relative to the view vector.

What I settled on was the following code:

public double GetRoll(){  Vector3 relright = viewDir.CrossProduct(worldUpV);  double relativeRoll = rightVector.DotProduct(relright);  double pitchComponent = Math.Cos(this.GetPitch());  double factor = relativeRoll/pitchComponent;  factor = Math.Max(-1, Math.Min(factor, 1));  if (rightVector.Y < 0)  {    return Math.Acos(factor);  }  else  {    return -Math.Acos(factor);  }}

I decided to use the camera viewDir vector and the worldUp vector to calculate a relative right vector and then take the dot product of the camera right vector and the relative right vector, so that the angles would be in the right quadrants.

This gave me the angle that was a combination of the roll and pitch, so I needed to remove the pitch component.

By dividing the dot product by the Cos(pitchAngle) I got the relative roll.

Then, in order to get a full -180 to 180 degrees, I used the direction of the right vector's Y component to negate the angle.

This is finally giving me what I needed to see.

There may be a much easier approach to do this, but for now this appears to do exactly what I was looking to do.

Figured I'd follow up in case any one else would find it useful.

(EDIT: had to make a change to correct for boundary conditions and possible division by zero)

[Edited by - awdorrin on February 12, 2010 12:21:19 PM]

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