Adding Gravity to Acceleration with Quaternions - c++

I am new on this platform so please let me know if i am doing something wrong before reporting! Also, i already searched for similar problems but i could not find the answer i am looking for.
I have this problem where i need to "add" gravity component to my vector of linear accelerations, and i know this can be done with Quaternions. My vector:
Accelerations = [Xacc, Yacc, Zacc]
and a quaternions that represents my current orientation:
q = [w, x, y, z]
Let's say i start in this position:
q = [1,0,0,0]
Now of course if i move for example my smartwatch i'll get some values in my Accelerations Vector and some values in my Quaternion right?
I thought about converting my Accelerations vector into a Quaternion form:
AccQuat = [0, x, y, z]
And then make a rotation to this quaternion by the opposite of my q, so i can "translate" my Accelerations Quaternion to the Earth frame of reference q = [1,0,0,0], the converting it back into a Vector form, adding the Gravity Vector:
g = [0, 0, -9.81]
Once done, getting the "New" Vector with gravity added back into Quaternion form, rotating it to his native position and then converting it back into his vector form.
Its a little bit tricky but i think it can be done in such a way. Is this a right way to do what im trying to do? Because i have no other information other than an already Normalized Quaternion representing my position and a vector of linear Accelerations.
To multiply a Quaternions i have to apply:
qOut = q*p*q^-1
by doing 2 Quaternions multiplications right?
And if, for example, i have a quaternion representing a 90deg rotation on my Z-Axe:
q = [0.71, 0, 0, 0.71]
if i want to "Rotate back" my Accelerations Quaternion, i need to make
qOut = q*p*q^-1
but as q i need to use my "Reverse" quaternion:
qReverse = [0.71, 0, 0, -0.71]
right?
Im working with C++ to make the code work but my problem now is the mental procedure rather than coding it.
Thanks everyone for the help!
========== EDIT ==================
More informations:
i have a fixed coordinates system called World (0,0,0) always parallel to my floor, my Quaternion rotations are for a system called "0" that refers to this World, they have the same centre in (0,0,0).
Maybe this can explain better:
World is Black, 0 is Blue One.
Lets suppose that my blue axis are my 3 components in the acceleration vector.
I thought about converting them into a Quaternion by simply adding a constant = 0 as w, then rotating them with the Quaternion formula (where q is reverse of my quaternion that i got from the data):
f = q*p*q^-1
So they can coincide with the Black system.
From this add the gravity component from the Z axis as it is // to the floor, then rotating my "New Accelerations quaternion" to his original position with the same formula but inverted q. In the end i want to get my Accelerations vector with Gravity added that's all. I thought it could be done with quaternions. Hope this can explain better, forgive me if im repeating same things but im also new to this argument.

Related

Animations in c++, interpolating keyframes with rotation matrices

In my application, i have a 3D-Model, where i can rotate, scale, translate by setting the values manually..i can also extract the rotation at a given point. So i can store keyframes now, with the 4x4 matrix for each joint.
However, now i want to animate between those frames.
What i want :
interpolate from frame 1 to frame 2 within X seconds. How can i implement this in C++??
I found something like http://www-scf.usc.edu/~amahajan/mocap/mocap4.html ...but i have no clue how to write this in c++
ok i solved this...the Eigen library has already a Quaternion1.slerp(Quaternion2, float T) function. If you dont know what SLERP is: slerp interpolates between one Quaternion and another one, the float T can be considered as the increment, how far much it should interpolate. For example: if you give slerp a unit quaternion, a Y 90° quaternion, and the float T a "0.5" value, the outcome is a Quaternion with Y 45°. For detailed information about slerp you should google it, it is well known.
If you use Eigen, you can simply make Quaternions out of the rotation matrices :
Matrix3f m;
Quaternion<float> q1=m.toRotationMatrix();
With this i simply created a loop that goes from 0.01 to 1.00 with increment 0.01.
You should do this work in another thread:
for(double i=0.0;i<1.0;i=i+0.01)
{
//let this thread sleep for 10 millisec, this gives us a 100hertz rate
//with this, the whole animation is over in 1 second
boost::this_thread::sleep(boost::posix_time::millisec(10));
//q1= first frame, q2 = second frame
Quaternion<float> q3(q1.slerp(i,q2));
//now apply the quaternion
// for each "bone", or node you need to apply a SLERPed quaternion
m_node->setRotation(q3.toRotationMatrix());
}

3x3 Matrix Rotation in C++

Alright, first off, I know similar questions are all over the web, I have looked at more than I'd care to count, I've been trying to figure it out for almost 3 weeks now (not constantly, just on and off, hoping for a spark of insight).
In the end, what I want to get, is a function where you pass in how much you want to rotate by (currently I'm working in Radian's, but I can go Degrees or Radians) and it returns the rotation matrix, preserving any translations I had.
I understand the formula to rotate on the "Z" axis in a 2D cartesian plane, is:
[cos(radians) -sin(radians) 0]
[sin(radians) cos(radians) 0]
[0 0 1]
I do understand Matrix Maths (Addition, Subtraction, Multiplication and Determinant/Inverse) fairly well, but what I'm not understanding, is how to, step-by-step, make a matrix I can use for rotation, preserving any translation (and whatever else, like scale) that it has.
From what I've gathered from other examples, is to multiply my current Matrix (whatever that may be, let's just use an Identity Matrix for now), by a Matrix like this:
[cos(radians) - sin(radians)]
[sin(radians) + cos(radians)]
[1]
But then my original Matrix would end up as a 3x1 Matrix instead of a 3x3, wouldn't it? I'm not sure what I'm missing, but something just doesn't seem right to me. I'm not necessarily looking for code for someone to write for me, just to understand how to do this properly and then I can write it myself. (not to say I won't look at other's code :) )
(Not sure if it matters to anybody, but just in-case, using Windows 7 64-bit, Visual Studio 2010 Ultimate, and I believe OpenGL, this is for Uni)
While we're at it, can someone double check this for me? Just to make sure it seems right.
A translation Matrix (again, let's use Identity) is something like this:
[1, 0, X translation element]
[0, 1, Y translation element]
[0, 0, 1]
First, You can not have translation 3x3 matrix for 3D space. You have to use homogeneous 4x4 matrices.
After that create a separate matrix for each transformation (translation, rotation, scale) and multiply them to get the final transformation matrix (multiplying 4x4 matrix will give you 4x4 matrix)
Lets clear some points:
Your object consists of 3D points which are basically 3 by 1 matrices.
You need a 3 by 3 rotation matrix to rotate your object: R but if you also add translation terms, transformation matrix will be 4 by 4:
[R11, R12, R13 tx]
[R21, R22, R23 ty]
[R31, R32, R33 tz]
[0, 0, 0, 1]
For R terms you can have look at :http://inside.mines.edu/~gmurray/ArbitraryAxisRotation/, they are dependent on the rotation angles of each axis.
In order to rotate your object, every 3D point is multiplied by this rotation matrix. For every 3 by 1 point you also need to add a 4th term(scale factor) which is 1 assuming fixed scale:
[x y z 1]'
Resulting product vector will be 4 by 1 and the last term is the scale term which is 1 again and can be removed.
Resulting rotated object points are these new 3D product points.
I faced the same problem and found a satisfying formula in this SO question.
Let (cos0, sin0) be respectively the cosine and sine values of your angle, and (x0, y0) the coordinates of the center of your rotation.
To transform a 2d point of coordinates (x,y), you have to multiply its homogeneous 3x1 coordinates (x,y,1) by this 3x3 matrix:
[cos0, -sin0, x0-(cos0*x0 - sin0*y0)]
[sin0, cos0, y0-(sin0*x0 + cos0*y0)]
[ 0, 0, 1 ]
The values on the third column are the amount of translation necessary to apply when you rotation center is not the origin of the system.

Quaternion - Rotate To

I have some object in world space, let's say at (0,0,0) and want to rotate it to face (10,10,10).
How do i do this using quaternions?
This question doesn't quite make sense. You said that you want an object to "face" a specific point, but that doesn't give enough information.
First, what does it mean to face that direction? In OpenGL, it means that the -z axis in the local reference frame is aligned with the specified direction in some external reference frame. In order to make this alignment happen, we need to know what direction the relevant axis of the object is currently "facing".
However, that still doesn't define a unique transformation. Even if you know what direction to make the -z axis point, the object is still free to spin around that axis. This is why the function gluLookAt() requires that you provide an 'at' direction and an 'up' direction.
The next thing that we need to know is what format does the end-result need to be in? The orientation of an object is often stored in quaternion format. However, if you want to graphically rotate the object, then you might need a rotation matrix.
So let's make a few assumptions. I'll assume that your object is centered at the world's point c and has the default alignment. I.e., the object's x, y, and z axes are aligned with the world's x, y, and z axes. This means that the orientation of the object, relative to the world, can be represented as the identity matrix, or the identity quaternion: [1 0 0 0] (using the quaternion convention where w comes first).
If you want the shortest rotation that will align the object's -z axis with point p:=[p.x p.y p.z], then you will rotate by φ around axis a. Now we'll find those values. First we find axis a by normalizing the vector p-c and then taking the cross-product with the unit-length -z vector and then normalizing again:
a = normalize( crossProduct(-z, normalize(p-c) ) );
The shortest angle between those two unit vectors found by taking the inverse cosine of their dot-product:
φ = acos( dotProduct(-z, normalize(p-c) ));
Unfortunately, this is a measure of the absolute value of the angle formed by the two vectors. We need to figure out if it's positive or negative when rotating around a. There must be a more elegant way, but the first way that comes to mind is to find a third axis, perpendicular to both a and -z and then take the sign from its dot-product with our target axis. Vis:
b = crossProduct(a, -z );
if ( dotProduct(b, normalize(p-c) )<0 ) φ = -φ;
Once we have our axis and angle, turning it into a quaternion is easy:
q = [cos(φ/2) sin(φ/2)a];
This new quaternion represents the new orientation of the object. It can be converted into a matrix for rendering purposes, or you can use it to directly rotate the object's vertices, if desired, using the rules of quaternion multiplication.
An example of calculating the Quaternion that represents the rotation between two vectors can be found in the OGRE source code for the Ogre::Vector3 class.
In response to your clarification and to just answer this, I've shamelessly copied a very interesting and neat algorithm for finding the quat between two vectors that looks like I have never seen before from here. Mathematically, it seems valid, and since your question is about the mathematics behind it, I'm sure you'll be able to convert this pseudocode into C++.
quaternion q;
vector3 c = cross(v1,v2);
q.v = c;
if ( vectors are known to be unit length ) {
q.w = 1 + dot(v1,v2);
} else {
q.w = sqrt(v1.length_squared() * v2.length_squared()) + dot(v1,v2);
}
q.normalize();
return q;
Let me know if you need help clarifying any bits of that pseudocode. Should be straightforward though.
dot(a,b) = a1*b1 + a2*b2 + ... + an*bn
and
cross(a,b) = well, the cross product. it's annoying to type out and
can be found anywhere.
You may want to use SLERP (Spherical Linear Interpolation). See this article for reference on how to do it in c++

Implementing a complex rotation-based camera

I am implementing a 3D engine for spatial visualisation, and am writing a camera with the following navigation features:
Rotate the camera (ie, analogous to rotating your head)
Rotate around an arbitrary 3D point (a point in space, which is probably not in the center of the screen; the camera needs to rotate around this keeping the same relative look direction, ie the look direction changes too. This does not look directly at the chosen rotation point)
Pan in the camera's plane (so move up/down or left/right in the plane orthogonal to the camera's look vector)
The camera is not supposed to roll - that is, 'up' remains up. Because of this I represent the camera with a location and two angles, rotations around the X and Y axes (Z would be roll.) The view matrix is then recalculated using the camera location and these two angles. This works great for pan and rotating the eye, but not for rotating around an arbitrary point. Instead I get the following behaviour:
The eye itself apparently moving further up or down than it should
The eye not moving up or down at all when m_dRotationX is 0 or pi. (Gimbal lock? How can I avoid this?)
The eye's rotation being inverted (changing the rotation makes it look further up when it should look further down, down when it should look further up) when m_dRotationX is between pi and 2pi.
(a) What is causing this 'drift' in rotation?
This may be gimbal lock. If so, the standard answer to this is 'use quaternions to represent rotation', said many times here on SO (1, 2, 3 for example), but unfortunately without concrete details (example. This is the best answer I've found so far; it's rare.) I've struggled to implemented a camera using quaternions combining the above two types of rotations. I am, in fact, building a quaternion using the two rotations, but a commenter below said there was no reason - it's fine to immediately build the matrix.
This occurs when changing the X and Y rotations (which represent the camera look direction) when rotating around a point, but does not occur simply when directly changing the rotations, i.e. rotating the camera around itself. To me, this doesn't make sense. It's the same values.
(b) Would a different approach (quaternions, for example) be better for this camera? If so, how do I implement all three camera navigation features above?
If a different approach would be better, then please consider providing a concrete implemented example of that approach. (I am using DirectX9 and C++, and the D3DX* library the SDK provides.) In this second case, I will add and award a bounty in a couple of days when I can add one to the question. This might sound like I'm jumping the gun, but I'm low on time and need to implement or solve this quickly (this is a commercial project with a tight deadline.) A detailed answer will also improve the SO archives, because most camera answers I've read so far are light on code.
Thanks for your help :)
Some clarifications
Thanks for the comments and answer so far! I'll try to clarify a few things about the problem:
The view matrix is recalculated from the camera position and the two angles whenever one of those things changes. The matrix itself is never accumulated (i.e. updated) - it is recalculated afresh. However, the camera position and the two angle variables are accumulated (whenever the mouse moves, for example, one or both of the angles will have a small amount added or subtracted, based on the number of pixels the mouse moved up-down and/or left-right onscreen.)
Commenter JCooper states I'm suffering from gimbal lock, and I need to:
add another rotation onto your transform that rotates the eyePos to be
completely in the y-z plane before you apply the transformation, and
then another rotation that moves it back afterward. Rotate around the
y axis by the following angle immediately before and after applying
the yaw-pitch-roll matrix (one of the angles will need to be negated;
trying it out is the fastest way to decide which).
double fixAngle = atan2(oEyeTranslated.z,oEyeTranslated.x);
Unfortunately, when implementing this as described, my eye shoots off above the scene at a very fast rate due to one of the rotations. I'm sure my code is simply a bad implementation of this description, but I still need something more concrete. In general, I find unspecific text descriptions of algorithms are less useful than commented, explained implementations. I am adding a bounty for a concrete, working example that integrates with the code below (i.e. with the other navigation methods, too.) This is because I would like to understand the solution, as well as have something that works, and because I need to implement something that works quickly since I am on a tight deadline.
Please, if you answer with a text description of the algorithm, make sure it is detailed enough to implement ('Rotate around Y, then transform, then rotate back' may make sense to you but lacks the details to know what you mean. Good answers are clear, signposted, will allow others to understand even with a different basis, are 'solid weatherproof information boards.')
In turn, I have tried to be clear describing the problem, and if I can make it clearer please let me know.
My current code
To implement the above three navigation features, in a mouse move event moving based on the pixels the cursor has moved:
// Adjust this to change rotation speed when dragging (units are radians per pixel mouse moves)
// This is both rotating the eye, and rotating around a point
static const double dRotatePixelScale = 0.001;
// Adjust this to change pan speed (units are meters per pixel mouse moves)
static const double dPanPixelScale = 0.15;
switch (m_eCurrentNavigation) {
case ENavigation::eRotatePoint: {
// Rotating around m_oRotateAroundPos
const double dX = (double)(m_oLastMousePos.x - roMousePos.x) * dRotatePixelScale * D3DX_PI;
const double dY = (double)(m_oLastMousePos.y - roMousePos.y) * dRotatePixelScale * D3DX_PI;
// To rotate around the point, translate so the point is at (0,0,0) (this makes the point
// the origin so the eye rotates around the origin), rotate, translate back
// However, the camera is represented as an eye plus two (X and Y) rotation angles
// This needs to keep the same relative rotation.
// Rotate the eye around the point
const D3DXVECTOR3 oEyeTranslated = m_oEyePos - m_oRotateAroundPos;
D3DXMATRIX oRotationMatrix;
D3DXMatrixRotationYawPitchRoll(&oRotationMatrix, dX, dY, 0.0);
D3DXVECTOR4 oEyeRotated;
D3DXVec3Transform(&oEyeRotated, &oEyeTranslated, &oRotationMatrix);
m_oEyePos = D3DXVECTOR3(oEyeRotated.x, oEyeRotated.y, oEyeRotated.z) + m_oRotateAroundPos;
// Increment rotation to keep the same relative look angles
RotateXAxis(dX);
RotateYAxis(dY);
break;
}
case ENavigation::ePanPlane: {
const double dX = (double)(m_oLastMousePos.x - roMousePos.x) * dPanPixelScale;
const double dY = (double)(m_oLastMousePos.y - roMousePos.y) * dPanPixelScale;
m_oEyePos += GetXAxis() * dX; // GetX/YAxis reads from the view matrix, so increments correctly
m_oEyePos += GetYAxis() * -dY; // Inverted compared to screen coords
break;
}
case ENavigation::eRotateEye: {
// Rotate in radians around local (camera not scene space) X and Y axes
const double dX = (double)(m_oLastMousePos.x - roMousePos.x) * dRotatePixelScale * D3DX_PI;
const double dY = (double)(m_oLastMousePos.y - roMousePos.y) * dRotatePixelScale * D3DX_PI;
RotateXAxis(dX);
RotateYAxis(dY);
break;
}
The RotateXAxis and RotateYAxis methods are very simple:
void Camera::RotateXAxis(const double dRadians) {
m_dRotationX += dRadians;
m_dRotationX = fmod(m_dRotationX, 2 * D3DX_PI); // Keep in valid circular range
}
void Camera::RotateYAxis(const double dRadians) {
m_dRotationY += dRadians;
// Limit it so you don't rotate around when looking up and down
m_dRotationY = std::min(m_dRotationY, D3DX_PI * 0.49); // Almost fully up
m_dRotationY = std::max(m_dRotationY, D3DX_PI * -0.49); // Almost fully down
}
And to generate the view matrix from this:
void Camera::UpdateView() const {
const D3DXVECTOR3 oEyePos(GetEyePos());
const D3DXVECTOR3 oUpVector(0.0f, 1.0f, 0.0f); // Keep up "up", always.
// Generate a rotation matrix via a quaternion
D3DXQUATERNION oRotationQuat;
D3DXQuaternionRotationYawPitchRoll(&oRotationQuat, m_dRotationX, m_dRotationY, 0.0);
D3DXMATRIX oRotationMatrix;
D3DXMatrixRotationQuaternion(&oRotationMatrix, &oRotationQuat);
// Generate view matrix by looking at a point 1 unit ahead of the eye (transformed by the above
// rotation)
D3DXVECTOR3 oForward(0.0, 0.0, 1.0);
D3DXVECTOR4 oForward4;
D3DXVec3Transform(&oForward4, &oForward, &oRotationMatrix);
D3DXVECTOR3 oTarget = oEyePos + D3DXVECTOR3(oForward4.x, oForward4.y, oForward4.z); // eye pos + look vector = look target position
D3DXMatrixLookAtLH(&m_oViewMatrix, &oEyePos, &oTarget, &oUpVector);
}
It seems to me that "Roll" shouldn't be possible given the way you form your view matrix. Regardless of all the other code (some of which does look a little funny), the call D3DXMatrixLookAtLH(&m_oViewMatrix, &oEyePos, &oTarget, &oUpVector); should create a matrix without roll when given [0,1,0] as an 'Up' vector unless oTarget-oEyePos happens to be parallel to the up vector. This doesn't seem to be the case since you're restricting m_dRotationY to be within (-.49pi,+.49pi).
Perhaps you can clarify how you know that 'roll' is happening. Do you have a ground plane and the horizon line of that ground plane is departing from horizontal?
As an aside, in UpdateView, the D3DXQuaternionRotationYawPitchRoll seems completely unnecessary since you immediately turn around and change it into a matrix. Just use D3DXMatrixRotationYawPitchRoll as you did in the mouse event. Quaternions are used in cameras because they're a convenient way to accumulate rotations happening in eye coordinates. Since you're only using two axes of rotation in a strict order, your way of accumulating angles should be fine. The vector transformation of (0,0,1) isn't really necessary either. The oRotationMatrix should already have those values in the (_31,_32,_33) entries.
Update
Given that it's not roll, here's the problem: you create a rotation matrix to move the eye in world coordinates, but you want the pitch to happen in camera coordinates. Since roll isn't allowed and yaw is performed last, yaw is always the same in both the world and camera frames of reference. Consider the images below:
Your code works fine for local pitch and yaw because those are accomplished in camera coordinates.
But when you rotate around a reference point, you are creating a rotation matrix that is in world coordinates and using that to rotate the camera center. This works okay if the camera's coordinate system happens to line up with the world's. However, if you don't check to see if you're up against the pitch limit before you rotate the camera position, you will get crazy behavior when you hit that limit. The camera will suddenly start to skate around the world--still 'rotating' around the reference point, but no longer changing orientation.
If the camera's axes don't line up with the world's, strange things will happen. In the extreme case, the camera won't move at all because you're trying to make it roll.
The above is what would normally happen, but since you handle the camera orientation separately, the camera doesn't actually roll.
Instead, it stays upright, but you get strange translation going on.
One way to handle this would be to (1)always put the camera into a canonical position and orientation relative to the reference point, (2)make your rotation, and then (3)put it back when you're done (e.g., similar to the way that you translate the reference point to the origin, apply the Yaw-Pitch rotation, and then translate back). Thinking more about it, however, this probably isn't the best way to go.
Update 2
I think that Generic Human's answer is probably the best. The question remains as to how much pitch should be applied if the rotation is off-axis, but for now, we'll ignore that. Maybe it'll give you acceptable results.
The essence of the answer is this: Before mouse movement, your camera is at c1 = m_oEyePos and being oriented by M1 = D3DXMatrixRotationYawPitchRoll(&M_1,m_dRotationX,m_dRotationY,0). Consider the reference point a = m_oRotateAroundPos. From the point of view of the camera, this point is a'=M1(a-c1).
You want to change the orientation of the camera to M2 = D3DXMatrixRotationYawPitchRoll(&M_2,m_dRotationX+dX,m_dRotationY+dY,0). [Important: Since you won't allow m_dRotationY to fall outside of a specific range, you should make sure that dY doesn't violate that constraint.] As the camera changes orientation, you also want its position to rotate around a to a new point c2. This means that a won't change from the perspective of the camera. I.e., M1(a-c1)==M2(a-c2).
So we solve for c2 (remember that the transpose of a rotation matrix is the same as the inverse):
M2TM1(a-c1)==(a-c2) =>
-M2TM1(a-c1)+a==c2
Now if we look at this as a transformation being applied to c1, then we can see that it is first negated, then translated by a, then rotated by M1, then rotated by M2T, negated again, and then translated by a again. These are transformations that graphics libraries are good at and they can all be squished into a single transformation matrix.
#Generic Human deserves credit for the answer, but here's code for it. Of course, you need to implement the function to validate a change in pitch before it's applied, but that's simple. This code probably has a couple typos since I haven't tried to compile:
case ENavigation::eRotatePoint: {
const double dX = (double)(m_oLastMousePos.x - roMousePos.x) * dRotatePixelScale * D3DX_PI;
double dY = (double)(m_oLastMousePos.y - roMousePos.y) * dRotatePixelScale * D3DX_PI;
dY = validatePitch(dY); // dY needs to be kept within bounds so that m_dRotationY is within bounds
D3DXMATRIX oRotationMatrix1; // The camera orientation before mouse-change
D3DXMatrixRotationYawPitchRoll(&oRotationMatrix1, m_dRotationX, m_dRotationY, 0.0);
D3DXMATRIX oRotationMatrix2; // The camera orientation after mouse-change
D3DXMatrixRotationYawPitchRoll(&oRotationMatrix2, m_dRotationX + dX, m_dRotationY + dY, 0.0);
D3DXMATRIX oRotationMatrix2Inv; // The inverse of the orientation
D3DXMatrixTranspose(&oRotationMatrix2Inv,&oRotationMatrix2); // Transpose is the same in this case
D3DXMATRIX oScaleMatrix; // Negative scaling matrix for negating the translation
D3DXMatrixScaling(&oScaleMatrix,-1,-1,-1);
D3DXMATRIX oTranslationMatrix; // Translation by the reference point
D3DXMatrixTranslation(&oTranslationMatrix,
m_oRotateAroundPos.x,m_oRotateAroundPos.y,m_oRotateAroundPos.z);
D3DXMATRIX oTransformMatrix; // The full transform for the eyePos.
// We assume the matrix multiply protects against variable aliasing
D3DXMatrixMultiply(&oTransformMatrix,&oScaleMatrix,&oTranslationMatrix);
D3DXMatrixMultiply(&oTransformMatrix,&oTransformMatrix,&oRotationMatrix1);
D3DXMatrixMultiply(&oTransformMatrix,&oTransformMatrix,&oRotationMatrix2Inv);
D3DXMatrixMultiply(&oTransformMatrix,&oTransformMatrix,&oScaleMatrix);
D3DXMatrixMultiply(&oTransformMatrix,&oTransformMatrix,&oTranslationMatrix);
D3DXVECTOR4 oEyeFinal;
D3DXVec3Transform(&oEyeFinal, &m_oEyePos, &oTransformMatrix);
m_oEyePos = D3DXVECTOR3(oEyeFinal.x, oEyeFinal.y, oEyeFinal.z)
// Increment rotation to keep the same relative look angles
RotateXAxis(dX);
RotateYAxis(dY);
break;
}
I think there is a much simpler solution that lets you sidestep all rotation issues.
Notation: A is the point we want to rotate around, C is the original camera location, M is the original camera rotation matrix that maps global coordinates to the camera's local viewport.
Make a note of the local coordinates of A, which are equal to A' = M × (A - C).
Rotate the camera like you would in normal "eye rotation" mode. Update the view matrix M so that it is modified to M2 and C remains unchanged.
Now we would like to find C2 such that A' = M2 × (A - C2).
This is easily done by the equation C2 = A - M2-1 × A'.
Voilà, the camera has been rotated and because the local coordinates of A are unchanged, A remains at the same location and the same scale and distance.
As an added bonus, the rotation behavior is now consistent between "eye rotation" and "point rotation" mode.
You rotate around the point by repeatedly applying small rotation matrices, this probably cause the drift (small precision errors add up) and I bet you will not really do a perfect circle after some time. Since the angles for the view use simple 1-dimension double, they have much less drift.
A possible fix would be to store a dedicated yaw/pitch and relative position from the point when you enter that view mode, and using those to do the math. This requires a bit more bookkeeping, since you need to update those when moving the camera. Note that it will also make the camera move if the point move, which I think is an improvement.
If I understand correctly, you are satisfied with the rotation component in the final matrix (save for inverted rotation controls in the problem #3), but not with the translation part, is that so?
The problem seems to come from the fact that you treating them differently: you are recalculating the rotation part from scratch every time, but accumulate the translation part (m_oEyePos). Other comments mention precision problems, but it's actually more significant than just FP precision: accumulating rotations from small yaw/pitch values is simply not the same---mathematically---as making one big rotation from the accumulated yaw/pitch. Hence the rotation/translation discrepancy. To fix this, try recalculating eye position from scratch simultaneously with the rotation part, similarly to how you find "oTarget = oEyePos + ...":
oEyePos = m_oRotateAroundPos - dist * D3DXVECTOR3(oForward4.x, oForward4.y, oForward4.z)
dist can be fixed or calculated from the old eye position. That will keep the rotation point in the screen center; in the more general case (which you are interested in), -dist * oForward here should be replaced by the old/initial m_oEyePos - m_oRotateAroundPos multiplied by the old/initial camera rotation to bring it to the camera space (finding a constant offset vector in camera's coordinate system), then multiplied by the inverted new camera rotation to get the new direction in the world.
This will, of course, be subject to gimbal lock when the pitch is straight up or down. You'll need to define precisely what behavior you expect in these cases to solve this part. On the other hand, locking at m_dRotationX=0 or =pi is rather strange (this is yaw, not pitch, right?) and might be related to the above.

Direct3D - How do I calculate Roll from View Matrix?

This one has been eating up my entire night, and I'm finally throwing up my hands for some assistance. Basically, it's fairly straightforward to calculate the Pitch and Yaw from the View Matrix right after you do a camera update:
D3DXMatrixLookAtLH(&m_View, &sCam.pos, &vLookAt, &sCam.up);
pDev->SetTransform(D3DTS_VIEW, &m_View);
// Set the camera axes from the view matrix
sCam.right.x = m_View._11;
sCam.right.y = m_View._21;
sCam.right.z = m_View._31;
sCam.up.x = m_View._12;
sCam.up.y = m_View._22;
sCam.up.z = m_View._32;
sCam.look.x = m_View._13;
sCam.look.y = m_View._23;
sCam.look.z = m_View._33;
// Calculate yaw and pitch and roll
float lookLengthOnXZ = sqrtf( sCam.look.z^2 + sCam.look.x^2 );
fPitch = atan2f( sCam.look.y, lookLengthOnXZ );
fYaw = atan2f( sCam.look.x, sCam.look.z );
So my problem is: there must be some way to obtain the Roll in radians similar to how the Pitch and Yaw are being obtained at the end of the code there. I've tried several dozen algorithms that seemed to make sense to me, but none of them gave quite the desired result. I realize that many developers don't track these values, however I am writing some re-centering routines and setting clipping based on the values, and manual tracking breaks down as you apply mixed rotations to the View Matrix. So, does anyone have the formula for getting the Roll in radians (or degrees, I don't care) from the View Matrix?
Woohoo, got it! Thank you stacker for the link to the Euler Angles entry on Wikipedia, the gamma formula at the end was indeed the correct solution! So, to calculate Roll in radians from a "vertical plane" I'm using the following line of C++ code:
fRoll = atan2f( sCam.up.y, sCam.right.y ) - D3DX_PI/2;
As Euler pointed out, you're looking for the arc tangent of the Y (up) matrix's Z value against the X (right) matrix's Z value - though in this case I tried the Z values and they did not yield the desired result so I tried the Y values on a whim and I got a value which was off by +PI/2 radians (right angle to the desired value), which is easily compensated for. Finally I have a gryscopically accurate and stable 3D camera platform with self-correcting balance. =)
It seems the Wikipedia article about Euler angles contains the formula you're lookin for at the end.