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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);

  // get Heading and Pitch
  double H = GetHeading();
  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 Pitch
double 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 awdorrin
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));
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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      creates a special rotation-translation matrix that moves and rotates the grid away from the origin so that when i finally
      normalize all the vertices on my vertex shader i can get a perfect sphere.
      T = glm::translate(glm::dmat4(1.0), glm::dvec3(0.0, 0.0, 1.0)); R = glm::rotate(glm::dmat4(1.0), glm::radians(180.0), glm::dvec3(1.0, 0.0, 0.0)); sides[0] = new TerrainNode(1.0, radius, T * R, glm::dvec2(0.0, 0.0), new TerrainTile(1.0, SIDE_FRONT)); T = glm::translate(glm::dmat4(1.0), glm::dvec3(0.0, 0.0, -1.0)); R = glm::rotate(glm::dmat4(1.0), glm::radians(0.0), glm::dvec3(1.0, 0.0, 0.0)); sides[1] = new TerrainNode(1.0, radius, R * T, glm::dvec2(0.0, 0.0), new TerrainTile(1.0, SIDE_BACK)); // So on and so forth for the rest of the sides As you can see, for the front side grid, i rotate it 180 degrees to make it face the camera and push it towards the eye;
      the back side is handled almost the same way only that i don't need to rotate it but simply push it away from the eye.
      The same technique is applied for the rest of the faces (obviously, with the proper rotations / translations).
      The matrix that result from the multiplication of R and T (in that particular order) is send to my vertex shader as `r_Grid'.
      // spherify vec3 V = normalize((r_Grid * vec4(r_Vertex, 1.0)).xyz); gl_Position = r_ModelViewProjection * vec4(V, 1.0); The `r_ModelViewProjection' matrix is generated on the CPU in this manner.
      // No the most efficient way, but it works. glm::dmat4 Camera::getMatrix() { // Create the view matrix // Roll, Yaw and Pitch are all quaternions. glm::dmat4 View = glm::toMat4(Roll) * glm::toMat4(Pitch) * glm::toMat4(Yaw); // The model matrix is generated by translating in the oposite direction of the camera. glm::dmat4 Model = glm::translate(glm::dmat4(1.0), -Position); // Projection = glm::perspective(fovY, aspect, zNear, zFar); // zNear = 0.1, zFar = 1.0995116e12 return Projection * View * Model; } I managed to get rid of z-fighting by using a technique called Logarithmic Depth Buffer described in this article; it works amazingly well, no z-fighting at all, at least not visible.
      Each frame i'm rendering each node by sending the generated matrices this way.
      // set the r_ModelViewProjection uniform // Sneak in the mRadiusMatrix which is a matrix that contains the radius of my planet. Shader::setUniform(0, Camera::getInstance()->getMatrix() * mRadiusMatrix); // set the r_Grid matrix uniform i created earlier. Shader::setUniform(1, r_Grid); grid->render(); My planet's radius is around 6400000.0 units, absurdly large, but that's what i really want to achieve;
      Everything works well, the node's split and merge as you'd expect, however whenever i get close to the surface
      of the planet the rounding errors start to kick in giving me that lovely stairs effect.
      I've read that if i could render each grid relative to the camera i could get better precision on the surface, effectively
      getting rid of those rounding errors.
       
      My question is how can i achieve this relative to camera rendering in my scenario here?
      I know that i have to do most of the work on the CPU with double, and that's exactly what i'm doing.
      I only use double on the CPU side where i also do most of the matrix multiplications.
      As you can see from my vertex shader i only do the usual r_ModelViewProjection * (some vertex coords).
       
      Thank you for your suggestions!
       
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