|
|
Chandra Aspect Operations |
Version 1.4 26th December 1998
-------------------------------
This FAQ is maintained by "hexapod@netcom.com". Any additional
suggestions or related questions are welcome. Just send E-mail to the above
address.
Feel free to distribute or copy this FAQ as you please.
Contributions
-------------
Introduction I1: steve@mred.bgm.link.com
Introduction
------------
I1. Important note relating to OpenGL and this document
Questions
---------
BASICS
======
Q1. What is a matrix?
Q2. What is the order of a matrix?
Q3. How do I represent a matrix using the C/C++ programming languages?
Q4. What are the advantages of using matrices?
Q5. How do matrices relate to coordinate systems?
ARITHMETIC
==========
Q6. What is the identity matrix?
Q7. What is the major diagonal matrix of a matrix?
Q8. What is the transpose of a matrix?
Q9. How do I add two matrices together?
Q10. How do I subtract two matrices?
Q11. How do I multiply two matrices together?
Q12. How do I square or raise a matrix to a power?
Q13. How do I multiply one or more vectors by a matrix?
DETERMINANTS AND INVERSES
=========================
Q14. What is the determinant of a matrix?
Q15. How do I calculate the determinant of a matrix?
Q16. What are Isotropic and Anisotropic matrices?
Q17. What is the inverse of a matrix?
Q18. How do I calculate the inverse of an arbitary matrix?
Q19. How do I calculate the inverse of an identity matrix?
Q20. How do I calculate the inverse of a rotation matrix?
Q21. How do I calculate the inverse of a matrix using Kramer's rule?
Q22. How do I calculate the inverse of a 2x2 matrix?
Q23. How do I calculate the inverse of a 3x3 matrix?
Q24. How do I calculate the inverse of a 4x4 matrix?
Q25. How do I calculate the inverse of a matrix using linear equations?
TRANSFORMS
==========
Q26. What is a rotation matrix?
Q27. How do rotation matrices relate to coordinate systems?
Q28. How do I generate a rotation matrix in the X-axis?
Q29. How do I generate a rotation matrix in the Y-axis?
Q30. How do I generate a rotation matrix in the Z-axis?
Q31. What are Euler angles?
Q32. What are yaw, roll and pitch?
Q33. How do I combine rotation matrices?
Q34. What is Gimbal Lock?
Q35. What is the correct way to combine rotation matrices?
Q36. How do I generate a rotation matrix from Euler angles?
Q37. How do I generate Euler angles from a rotation matrix?
Q38. How do I generate a rotation matrix for a selected axis and angle?
Q39. How do I generate a rotation matrix to map one vector onto another?
Q40. How do I use matrices to convert between two coordinate systems?
Q41. What is a translation matrix?
Q42. What is a scaling matrix?
Q43. What is a shearing matrix?
Q44. How do I perform linear interpolation between two matrices?
Q45. How do I perform cubic interpolation between four matrices?
Q46. How can I render a matrix?
QUATERNIONS
===========
Q47. What are quaternions?
Q48. How do quaternions relate to 3D animation?
Q49. How do I calculate the conjugate of a quaternion?
Q50. How do I calculate the inverse of a quaternion?
Q51. How do I calculate the magnitude of a quaternion?
Q52. How do I normalise a quaternion?
Q53. How do I multiply two quaternions together?
Q54. How do I convert a quaternion to a rotation matrix?
Q55. How do I convert a rotation matrix to a quaternion?
Q56. How do I convert a rotation axis and angle to a quaternion?
Q57. How do I convert a quaternion to a rotation axis and angle?
Q58. How do I convert spherical rotation angles to a quaternion?
Q59. How do I convert a quaternion to spherical rotation angles?
Q60. How do I convert Euler rotation angles to a quaternion?
Q61. How do I use quaternions to perform linear interpolation between matrices?
Q62. How do I use quaternions to perform cubic interpolation between matrices?
Introduction
------------
I1. Important note relating to OpenGl and this document
-------------------------------------------------------
In this document (as in most math textbooks), all matrices are drawn
in the standard mathematical manner. Unfortunately graphics libraries
like IrisGL, OpenGL and SGI's Performer all represent them with the
rows and columns swapped.
Hence, in this document you will see (for example) a 4x4 Translation
matrix represented as follows:
| 1 0 0 X |
| |
| 0 1 0 Y |
M = | |
| 0 0 1 Z |
| |
| 0 0 0 1 |
In Performer (for example) this would be populated as follows:
M[0][1] = M[0][2] = M[0][3] =
M[1][0] = M[1][2] = M[1][3] =
M[2][0] = M[2][1] = M[2][3] = 0 ;
M[0][0] = M[1][1] = M[2][2] = m[3][3] = 1 ;
M[3][0] = X ;
M[3][1] = Y ;
M[3][2] = Z ;
ie, the matrix is stored like this:
| M[0][0] M[1][0] M[2][0] M[3][0] |
| |
| M[0][1] M[1][1] M[2][1] M[3][1] |
M = | |
| M[0][2] M[1][2] M[2][2] M[3][2] |
| |
| M[0][3] M[1][3] M[2][3] M[3][3] |
OpenGL uses a one-dimensional array to store matrices - but fortunately,
the packing order results in the same layout of bytes in memory - so
taking the address of a pfMatrix and casting it to a float* will allow
you to pass it directly into routines like glLoadMatrixf.
In the code snippets scattered throughout this document, a one-dimensional
array is used to store a matrix. The ordering of the array elements is
transposed with respect to OpenGL.
This Document OpenGL
| 0 1 2 3 | | 0 4 8 12 |
| | | |
| 4 5 6 7 | | 1 5 9 13 |
M = | | M = | |
| 8 9 10 11 | | 2 6 10 14 |
| | | |
| 12 13 14 15 | | 3 7 11 15 |
Answers
-------
BASICS
======
Q1. What is a matrix?
----------------------
A matrix is a two dimensional array of numeric data, where each
row or column consists of one or more numeric values.
Arithmetic operations which can be performed with matrices include
addition, subtraction, multiplication and division.
The size of a matrix is defined in terms of the number of rows
and columns.
A matrix with M rows and N columns is defined as a MxN matrix.
Individual elements of the matrix are referenced using two index
values. Using mathematical notation these are usually assigned the
variables 'i' and 'j'. The order is row first, column second
For example, if a matrix M with order 4x4 exists, then the elements
of the matrix are indexed by the following row:column pairs:
| 00 10 20 30 |
M = | 01 11 21 31 |
| 02 12 22 32 |
| 03 13 23 33 |
The element at the top right of the matrix has i=0 and j=3
This is referenced as follows:
M = M
i,j 0,3
In computer animation, the most commonly used matrices have either
2, 3 or 4 rows and columns. These are referred to as 2x2, 3x3 and 4x4
matrices respectively.
2x2 matrices are used to perform rotations, shears and other types
of image processing. General purpose NxN matrices can be used to
perform image processing functions such as convolution.
3x3 matrices are used to perform low-budget 3D animation. Operations
such as rotation and multiplication can be performed using matrix
operations, but perspective depth projection is performed using
standard optimised into pure divide operations.
4x4 matrices are used to perform high-end 3D animation. Operations
such as multiplication and perspective depth projection can be
performed using matrix mathematics.
Q2. What is the "order" of a matrix?
-------------------------------------
The "order" of a matrix is another name for the size of the matrix.
A matrix with M rows and N columns is said to have order MxN.
Q3. How do I represent a matrix using the C/C++ programming languages?
-----------------------------------------------------------------------
The simplest way of defining a matrix using the C/C++ programming
languages is to make use of the "typedef" keyword. Both 3x3 and 4x4
matrices may be defined in this way ie:
typedef float MATRIX3[9];
typedef float MATRIX4[16];
Since each type of matrix has dimensions 3x3 and 4x4, this requires
9 and 16 data elements respectively.
At first glance, the use of a single linear array of data values may
seem counter-intuitive. The use of two dimensional arrays may seem
more convenient ie.
typedef float MATRIX3[3][3];
typedef float MATRIX4[4][4];
However, the use of two reference systems for each matrix element
very often leads to confusion. With mathemetics, the order is row
first (i), column second (j) ie.
Mij
Using C/C++, this becomes
matrix[j][i]
Using two dimensional arrays also incurs a CPU performance penalty in
that C compilers will often make use of multiplication operations to
resolve array index operations.
So, it is more efficient to stick with linear arrays. However, one issue
still remains to be resolved. How is an two dimensional matrix mapped
onto a linear array? Since there are only two methods (row first/column
second or column first/row column).
The performance differences between the two are subtle. If all for-next
loops are unravelled, then there is very little difference in the
performance for operations such as matrix-matrix multiplication.
Using the C/C++ programming languages the linear ordering of each
matrix is as follows:
mat[0] = M mat[3] = M
00 03
mat[12] = M mat[15] = M
30 33
| 0 1 2 3 |
| | | 0 1 2 |
| 4 5 6 7 | | |
M = | | M = | 3 4 5 |
| 8 9 10 11 | | |
| | | 6 7 8 |
| 12 13 14 15 |
Q4. What are the advantages of using matrices?
-----------------------------------------------
One of the first questions asked about the use of matrices in computer
animation is why they should be used at all in the first place.
Intuitively, it would appear that the overhead of for-next loops and
matrix multiplication would slow down an application.
Arguments that resolve these objections can be pointed out. These include
the use of CPU registers to handle loop counters on-board data caches
to optimise memory accesses.
Advantages can also be pointed out. By following a mathematical approach
to defining 3D algorithms, it is possible to predict and plan the
design of a 3D animation system. Such mathematical approaches allow
for the implementation of character animation, spline curves and inverse
kinematics.
However, one objection that frequently comes up is that it would be
quicker to just multiply each pair of coordinates by the rotation
coefficients for that axis, rather than perform a full vector-matrix
multiplication.
ie. Rotation in X transforms Y and Z
Rotation in Y transforms X and Z
Rotation in Z transforms X and Y
The argument to this goes as follows:
Given a vertex V = (x,y,z), rotation angles (A,B and C) and translation
(D,E,F). A the algorithm
is defined as follows:
---------------------------
sx = sin(A) // Setup - only done once
cx = cos(A)
sy = sin(B)
cy = cos(B)
sz = sin(C)
cz = cos(C)
x1 = x * cz + y * sz // Rotation of each vertex
y1 = y * cz - x * sz
z1 = z
x2 = x1 * cy + z1 * sy
y2 = z1
z2 = z1 * cy - x1 * sy
x3 = x2
y3 = y2 * cx + z1 * sx
z3 = z2 * cx - x1 * sx
xr = x3 + D // Translation of each vertex
yr = y3 + E
zr = z3 + F
---------------------------
Altogether, this algorithm will use the following amounts of processing
time:
Set-up Per-vertex
------------------------- ------------------------
6 trigonometric functions
6 assignment operations. 12 assignment
12 multiplication
9 addition
------------------------- ------------------------
Assume that the same operations is being performed using matrix
multiplication.
With a 4x4 matrix, the procesing time is used as follows:
Set-up Change Per-vertex Change
-------------------------- ------ ------------------------ ------
6 trigonometric functions 0 0
18 assignment operation -12 3 assignment -9
12 multiplication +12 9 multiplication -3
6 subtraction +6 6 addition -3
-------------------------- ------ ------------------------ ------
Comparing the two tables, it can be seen that setting up a rotation
matrix costs at least 12 multiplication calculations and an extra
18 assignment calls.
However, while this may seem extravagant, the savings come from
processing each vertex. Using matrix multiplication, the savings made
from processing just 4 vertices, will outweigh the additional set-up
cost.
Q5. How do matrices relate to coordinate systems?
--------------------------------------------------
With either 3x3 or 4x4 rotation, translation or shearing matrices, there
is a simple relationship between each matrix and the resulting coordinate
system.
The first three columns of the matrix define the direction vector of the
X, Y and Z axii respectively.
If a 4x4 matrix is defined as:
| A B C D |
M = | E F G H |
| I J K L |
| M N O P |
Then the direction vector for each axis is as follows:
X-axis = [ A E I ]
Y-axis = [ B F J ]
Z-axis = [ C G K ]
ARITHMETIC
==========
Q6. What is the identity matrix?
---------------------------------
The identity matrix is matrix in which has an identical number of rows
and columns. Also, all the elements in which i=j are set one. All others
are set to zero. For example a 4x4 identity matrix is as follows:
| 1 0 0 0 |
M = | 0 1 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
Q7. What is the major diagonal of a matrix?
--------------------------------------------
The major diagonal of a matrix is the set of elements where the
row number is equal to the column number ie.
M where i=j
ij
In the case of the identity matrix, only the elements on the major
diagonal are set to 1, while all others are set to 0.
Q8. What is the transpose of a matrix?
---------------------------------------
The transpose of matrix is the matrix generated when every element in
the matrix is swapped with the opposite relative to the major diagonal
This can be expressed as the mathematical operation:
M' = M
ij ji
However, this can only be performed if a matrix has an equal number
of rows and columns.
If the matrix M is defined as:
| 0.707 -0.866 |
M = | |
| 0.866 0.707 |
Then the transpose is equal to:
| 0.707 0.866 |
T = | |
| -0.866 0.707 |
If the matrix is a rotation matrix, then the transpose is guaranteed
to be the inverse of the matrix.
Q9. How do I add two matrices together?
----------------------------------------
The rule of thumb with adding two matrices together is:
"add row and column to row and column"
This can be expressed mathematically as:
R = M + L
ij ij ij
However, both matrices must be identical in size.
For example, if the 2x2 matrix M is added with the 2x2 matrix L then
the result is as follow:
R = M + L
| A B C | | J K L |
| | | |
= | D E F | + | M N O |
| | | |
| G H I | | P Q R |
| A+J B+K C+L |
| |
= | D+M E+N F+O |
| |
| G+P H+Q I+R |
Q10. How do I subtract two matrices?
-------------------------------------
The rule of thumb with subtracting two matrices is:
"subtract row and column from row and column"
This can be expressed mathematically as:
R = M - L
ij ij ij
However, both matrices must be identical in size.
For example, if the 2x2 matrix L is subtracted from the 2x2 matrix M then
the result is as follows:
R = M - L
| A B C | | J K L |
| | | |
= | D E F | - | M N O |
| | | |
| G H I | | P Q R |
| A-J B-K C-L |
| |
= | D-M E-N F-O |
| |
| G-P H-Q I-R |
Q11. How do I multiply two matrices together?
---------------------------------------------
The rule of thumb with multiplying two matrices together is:
"multiply row into column and sum the result".
This can be expressed mathematically as:
n
--
R = \ M x L
ij / ij ji
--
i=1
If the two matrices to be multiplied together have orders:
M = AxB and L = CxD
then the two values B and C must be identical.
Also, the resulting matrix has an order of AxD
Thus, it is possible to multiply a 4xN matrix with a 4x4 matrix
but not the other way around.
For example, if the 4x4 matrix M is defined as:
| A B C D |
M = | E F G H |
| I J K L |
| M N O P |
and a 4x2 matrix L is defined as:
L = | Q R |
| S T |
| U V |
| W X |
then the size of the resulting matrix is 2x4. The resulting matrix
is defined as:
R = M x L
| A B C D | | Q R |
= | E F G H | x | S T |
| I J K L | | U V |
| M N O P | | W X |
| AQ+BS+CU+DW AR+BT+CV+DX |
= | EQ+FS+GU+HW ER+FT+GV+HX |
| IQ+JS+KU+LW IR+JT+KV+LX |
| MQ+NS+OU+PW MR+NT+OV+PX |
Q12. How do I square or raise a matrix to a power?
--------------------------------------------------
A matrix may be squared or even raised to an integer power. However
there are several restrictions. For all powers, the matrix must be
orthogonal ie. have the same width and height
For example,
-1
M is the inverse of the matrix
0
M generates the identity matrix
1
M leaves the matrix undamaged.
2
M squares the matrix and
3
M generates the cube of the matrix
Raising a matrix to a power greater than one involves multiplying a matrix
by itself a specific number of times.
For example,
2
M = M . M
3
M = M . M . M
and so on.
Raising the identity matrix to any power always generates the identity
matrix ie.
n
I = I
Q13. How do I multiply one or more vectors by a matrix?
-------------------------------------------------------
The best way to perform this task is to treat the list of vectors as
a single matrix, with each vector represented as a column vector.
If N vectors are to be multiplied by a 4x4 matrix, then they can be
treated as a single 4xN matrix:
If the matrix is defined as:
| A B C D |
M = | E F G H |
| I J K L |
| M N O P |
and the list of vectors is defined as:
| x1 x2 x3 x4 x5|
V = | y1 y2 y3 y4 y5|
| z1 z2 z3 z4 z5|
| 1 1 1 1 1|
Note that an additional row of constant terms is added to the vector
list, all of which are set to 1.0. In real life, this row does not
exist. It is simply used to make the orders of the matrix M and the
vector list V match.
Then the multiplication is performed as follows:
M . V = V'
| A B C D | | x1 x2 x3 x4 x5 | | A.x1+B.y1+C.z1+D A.x2+B.y2+C.z2+D ... |
| E F G H | . | y1 y2 y3 y4 y5 | = | E.x1+F.y1+G.z1+H E.x2+F.y2+G.z2+H ... |
| I J K L | | z1 z2 y3 y4 z5 | | I.x1+J.y1+K.z1+L I.x2+J.y2+K.z2+L ... |
| M N O P | | 1 1 1 1 1 | | M.x1+N.y1+O.z1+P M.x2+N.y2+O.z2+P ... |
For each vector in the list there will be a total of 12 multiplication
16 addition and 1 division operation (for perspective).
If the matrix is known not to be a rotation or translation matrix then the
division operation can be skipped.
DETERMINANTS AND INVERSES
=========================
Q14. What is the determinant of a matrix?
-----------------------------------------
The determinant of a matrix is a floating point value which is used to
indicate whether the matrix has an inverse or not. If zero, then no
inverse exists. If non-zero, then a determinant exists.
As an example, consider a matrix consisting of a single element:
M = [ 1 ].
For a matrix of this size, the determinant is simply the value of the
single element.
Also, the inverse is simply the reciprocal of this single element:
-1
M = [ 1 / M[0][0] ]
If this single value is non-zero, then an inverse exists. In the case
of the identity matrix, this happens to be 1 / 1 or 1.0
However, if the value of this single element is zero, then the determinant
is also zero.
Attempting to calculate the reciprocal of zero, generates a value of
infinity. This isn't permitted as far a matrices are concerned, so no
inverse of the matrix exists.
For an identity matrix, the determinant is always equal to one.
Any matrix with a determinant of 1.0 is said to be isotropic.
Thus all rotation matrices are said to be isotropic, since the
determinant is always equal to 1.0.
This can be proved as follows:
| A B | | cos X -sin X |
M = | | = | |
| C D | | sin X cos X |
D = AD - BC
D = (cos X . cos X) - (-sin X . sin X)
2 2
D = (cos X ) + (sin X)
2 2
But, cos X + sin X = 1
Therefore,
D = 1
Q15. How do I calculate the determinant of a matrix?
----------------------------------------------------
The determinant of a matrix is calculated using Kramer's rule, where
the value can be calculated by breaking the matrix into smaller
matrices.
For a 2x2 matrix M, the determinant D is calculated as follows:
| A B |
M = | |
| C D |
D = AD - BC
For 3x3 and 4x4 matrices, this is more complicated, but can be solved
by methods such as Kramer's Rule.
Q16. What are Isotropic and Anisotropic matrices?
-------------------------------------------------
An Isotropic matrix is one in which the sum of the squares of all
three rows or columns add up to one.
A matrix in which this is not the case, is said to be Anisotropic.
When 3x3 or 4x4 matrices are used to rotate and scale an object, it
is sometimes necessary to enlarge or shrink one axis more than the
others.
For example, with seismic surveys, it is convenient to enlarge the
Z-axis by a factor or 50 or more, while letting the X and Y axii
remain the same.
Another example is the implementation of "squash" and "stretch"
with character animation. When a character is hit by a heavy object
eg. an anvil, the desired effect is to character stretched out
sideways and squashed vertically:
A suitable matrix would be as follows:
| 2 0 0 0 |
M = | 0 2 0 0 |
| 0 0 0.5 0 |
| 0 0 0 1 |
However, there is problem looming ahead. While this matrix will cause
no problems with the transformation of vertex data, it will cause
problems with gouraud shading using outward normals.
Because the transformation stage is implemented using matrix
multiplication, both vertex data and outward normal data will be
multiplied with this matrix.
While this is not a problem with vertex data (it is the desired effect)
it causes a major headache with the outward normal data.
After raw multiplication, each outward normal will no longer be
normalised and consequently will affect other calculations such as
shading and back-face culling.
Q17. What is the inverse of a matrix?
-------------------------------------
-1
Given a matrix M, then the inverse of that matrix, denoted as M , is
the matrix which satisfies the following expression:
-1
M . M = I
where I is the identity matrix.
Thus, multiplying a matrix with its inverse will generate the identity
matrix. However, several requirements must be satisfied before the
inverse of a matrix can be calculated.
These include that the width and height of the matrix are identical and
that the determinant of the matrix is non-zero.
Calculating the inverse of a matrix is a task often performed in order
to implement inverse kinematics using spline curves.
Q18. How do I calculate the inverse of an arbitary matrix?
----------------------------------------------------------
Depending upon the size of the matrix, the calculation of the inverse
can be trivial or extremely complicated.
For example, the inverse of a 1x1 matrix is simply the reciprocal of
the single element:
ie. M = | x |
Then the inverse is defined as:
-1 | 1 |
M = | - |
| x |
Solving 2x2 matrices and larger can be achieved by using Kramer's Rule
or by solving as a set of simultaneous equations.
However, in certain cases, such as identity or rotation matrices, the
inverse is already known or can be determined from taking the transpose
of the matrix.
Q19. How do I calculate the inverse of an identity matrix?
----------------------------------------------------------
Don't even bother. The inverse of an identity matrix is the identity
matrix. ie.
-1
I . I = I
Any identity matrix will always have a determinant of +1.
Q20. How do I calculate the inverse of a rotation matrix?
---------------------------------------------------------
Since a rotation matrix always generates a determinant of +1,
calculating the inverse is equivalent of calculating the transpose.
Alternatively, if the rotation angle is known, then the rotation
angle can be negated and used to calculate a new rotation matrix.
Q21. How do I calculate the inverse of a matrix using Kramer's rule?
--------------------------------------------------------------------
Given a 3x3 matrix M:
| A B C |
| |
M = | D E F |
| |
| G H I |
Then the determinant is calculated as follows:
n
---
\ i
det M = / M * submat M * -1
--- 0,i 0,i
i=1
where
submat M defines the matrix composed of all rows and columns of M
ij
excluding row i and column j. submat may be called recursively.
ij
If the determinant is non-zero then the inverse of the matrix exists.
In this case, the value of each matrix element is defined by:
-1 1 i+j
M = ----- * det submat M * -1
j,i det M i,j
Q22. How do I calculate the inverse of a 2x2 matrix?
----------------------------------------------------
For a 2x2 matrix, the calculation is slightly harder. If the matrix is
defined as follows:
| A B |
M = | |
| C D |
Then the determinant is defined as:
det = AD - BC
And the inverse is defined as:
-1 1 | D -B |
M = --- | |
det | -C A |
This can be proved using Kramer's rule. Given the matrix M:
| A B |
M = | |
| C D |
Then the determinant is:
0 1
det = M * submat M * -1 + M * submat M * -1
0,0 0,0 0,1 0,1
<=> M * M * 1 + M * M * -1
0,0 1,1 0,1 1,0
<=> A * D + B * C * -1
<=> AD + BC . -1
<=> AD - BC
==============
And the inverse is derived from:
-1 0+0 -1
M = det submat * -1 <=> M = M * 1 <=> D
0,0 0,0 0,0 1,1
-1 1+0 -1
M = det submat * -1 <=> M = M * -1 <=> C * -1
0,1 1,0 0,1 1,0
-1 0+1 -1
M = det submat * -1 <=> M = M * -1 <=> B * -1
1,0 0,1 1,0 0,1
-1 1+1 -1
M = det submat * -1 <=> M = M * 1 <=> A
1,1 1,1 1,1 0,0
Then the inverse matrix is equal to:
-1 1 | D -C |
M = --- | |
det | -B A |
Providing that the determinant is not zero.
Q23. How do I calculate the inverse of a 3x3 matrix?
----------------------------------------------------
For 3x3 matrices and larger, the inverse can be calculated by
either applying Kramer's rule or by solving as a set of linear
equations.
If Kramer's rule is applied to a matrix M:
| A B C |
M = | D E F |
| G H I |
then the determinant is calculated as follows:
det M = A * (EI - HF) - B * (DI - GF) + C * (DH - GE)
Providing that the determinant is non-zero, then the inverse is
calculated as:
-1 1 | EI-FH -(BI-HC) BF-EC |
M = ----- . | -(DI-FG) AI-GC -(AF-DC) |
det M | DH-GE -(AH-GB) AE-BD |
This can be implemented using a pair of 'C' functions:
---------------------------------
VFLOAT m3_det( MATRIX3 mat )
{
VFLOAT det;
det = mat[0] * ( mat[4]*mat[8] - mat[7]*mat[5] )
- mat[1] * ( mat[3]*mat[8] - mat[6]*mat[5] )
+ mat[2] * ( mat[3]*mat[7] - mat[6]*mat[4] );
return( det );
}
----------------------------------
void m3_inverse( MATRIX3 mr, MATRIX3 ma )
{
VFLOAT det = m3_det( ma );
if ( fabs( det ) < 0.0005 )
{
m3_identity( ma );
return;
}
mr[0] = ma[4]*ma[8] - ma[5]*ma[7] / det;
mr[1] = -( ma[1]*ma[8] - ma[7]*ma[2] ) / det;
mr[2] = ma[1]*ma[5] - ma[4]*ma[2] / det;
mr[3] = -( ma[3]*ma[8] - ma[5]*ma[6] ) / det;
mr[4] = ma[0]*ma[8] - ma[6]*ma[2] / det;
mr[5] = -( ma[0]*ma[5] - ma[3]*ma[2] ) / det;
mr[6] = ma[3]*ma[7] - ma[6]*ma[4] / det;
mr[7] = -( ma[0]*ma[7] - ma[6]*ma[1] ) / det;
mr[8] = ma[0]*ma[4] - ma[1]*ma[3] / det;
}
---------------------------------
Q24. How do I calculate the inverse of a 4x4 matrix?
----------------------------------------------------
As with 3x3 matrices, either Kramer's rule can be applied or the
matrix can be solved as a set of linear equations.
An efficient way is to make use of the existing 'C' functions defined
to calculate the determinant and inverse of a 3x3 matrix.
In order to implement Kramer's rule with 4x4 matrices, it is necessary
to determine individual sub-matrices. This is achieved by the following
routine:
--------------------------
void m4_submat( MATRIX4 mr, MATRIX3 mb, int i, int j )
{
int ti, tj, idst, jdst;
for ( ti = 0; ti < 4; ti++ )
{
if ( ti < i )
idst = ti;
else
if ( ti > i )
idst = ti-1;
for ( tj = 0; tj < 4; tj++ )
{
if ( tj < j )
jdst = tj;
else
if ( tj > j )
jdst = tj-1;
if ( ti != i && tj != j )
mb[idst*3 + jdst] = mr[ti*4 + tj ];
}
}
}
--------------------------
The determinant of a 4x4 matrix can be calculated as follows:
--------------------------
VFLOAT m4_det( MATRIX4 mr )
{
VFLOAT det, result = 0, i = 1;
MATRIX3 msub3;
int n;
for ( n = 0; n < 4; n++, i *= -1 )
{
m4_submat( mr, msub3, 0, n );
det = m3_det( msub3 );
result += mr[n] * det * i;
}
return( result );
}
--------------------------
And the inverse can be calculated as follows:
--------------------------
int m4_inverse( MATRIX4 mr, MATRIX4 ma )
{
VFLOAT mdet = m4_det( ma );
MATRIX3 mtemp;
int i, j, sign;
if ( fabs( mdet ) < 0.0005 )
return( 0 );
for ( i = 0; i < 4; i++ )
for ( j = 0; j < 4; j++ )
{
sign = 1 - ( (i +j) % 2 ) * 2;
m4_submat( ma, mtemp, i, j );
mr[i+j*4] = ( m3_det( mtemp ) * sign ) / mdet;
}
return( 1 );
}
--------------------------
Having a function that can calculate the inverse of any 4x4 matrix is
an incredibly useful tool. Application include being able to calculate
the base matrix for splines, inverse rotations and rearranging matrix
equations.
Q25. How do I calculate the inverse of a matrix using linear equations?
-----------------------------------------------------------------------
If a matrix M exists, such that:
| A B C |
M = | D E F |
| G H I |
then the inverse exists:
| P Q R |
M' = | S T U |
| V W X |
and the following expression is valid:
-1
M . M = I
| A B C | | P Q R | | 1 0 0 |
| D E F | . | S T U | = | 0 1 0 |
| G H I | | V W X | | 0 0 1 |
The inverse can then be calculated through the solution as a set of
linear equations ie.:
| AP + BS + CV | | 1 | Column 0 (X)
| DP + ES + FV | = | 0 |
| GP + HS + IV | | 0 |
| AQ + BT + CW | | 0 | Column 1 (Y)
| DQ + ET + FW | = | 1 |
| GQ + HT + IW | | 0 |
| AR + BU + CX | | 0 | Column 2 (Z)
| DR + EU + FX | = | 0 |
| GR + HU + IX | | 1 |
TRANSFORMS
==========
Q26. What is a rotation matrix?
-------------------------------
A rotation matrix is used to rotate a set of points within a
coordinate system. While the individual points are assigned new
coordinates, their relative distances do not change.
All rotations are defined using the trigonometric "sine" and "cosine"
functions.
For a two-dimensional coordinate system, the rotation matrix is as
follows:
| cos(A) -sin(A) |
| |
| sin(A) cos(A) |
With the rotation angle A set to zero, this generates the identity
matrix:
| 1 0 |
I = | |
| 0 1 |
If the rotation is set to +90 degrees, then the matrix is as follows:
| 0 -1 |
M = | |
| 1 0 |
If the rotation is set to -90 degrees, then the matrix is as follows:
| 0 1 |
M = | |
| -1 0 |
Negating the rotation angle is equivalent to generating the transpose
of the matrix.
If a rotation matrix is multiplied with its transpose, the result is
the identity matrix.
Q27. How do rotation matrices relate to coordinate systems?
------------------------------------------------------
Rotation matrices relate to coordinate systems in the following way.
Mathematical convention requires that a positive rotation angle
generates a clockwise rotation when looking from the origin towards
the positive end of the rotation axis.
Applying this rule, allows for the derivation of three Cartesian
rotation matrices. Consider a right-handed coordinate system. For each
rotation axis, look from the origin towards the positive end of the
selected axis. This generates the following three views:
+----------------------------------------+
| |
| X-axis Y-axis Z-axis |
| |
| |
| ^ Y ^ Z Y ^ |
| | | | |
| | | | |
| | | | |
| | | | |
| O----> Z O----> X X <----O |
| |
+----------------------------------------+
Since a positive rotation angle generates a clockwise rotation, it is
possible to generate a set of coordinate mappings for each rotation.
For simplicity, a rotation of +90 will be considered:
Starting with the X-axis:
( 0, 1, 0 ) -> ( 0, 0, 1 )
( 0, 0, 1 ) -> ( 0,-1, 0 )
( 0,-1, 0 ) -> ( 0, 0,-1 )
( 0, 0,-1 ) -> ( 0, 1, 0 )
These can be simplified to:
X' = X
Y' = -Z
Z' = Y
These can then be placed into a matrix:
| 1 0 0 |
Rx = | 0 cos A -sin A |
| 0 sin A cos A |
Doing the same for the Y-axis:
( 0, 0, 1) -> ( 1,0, 0)
( 1, 0, 0) -> ( 0,0,-1)
( 0, 0,-1) -> (-1,0, 0)
(-1, 0, 0) -> ( 0,0, 1)
These can be simplified to:
X' = Z
Y' = Y
Z' = -X
These can then be placed into a matrix:
| cos A 0 sin A |
Ry = | 0 1 0 |
| -sin A 0 cos A |
And finally for the Z-axis:
( 0, 1, 0 ) -> ( -1, 0, 0 )
(-1, 0, 0 ) -> ( 0, -1, 0 )
( 0,-1, 0 ) -> ( 1, 0, 0 )
( 1, 0, 0 ) -> ( 0, 1, 0 )
These can be simplified to:
X' = -Y
Y' = X
Z' = Z
Placing these into a matrix:
| cos A -sin A 0 |
Rz = | sin A cos A 0 |
| 0 0 1 |
These are the three basic rotation matrices used by OpenGL.
Q28. How do I generate a rotation matrix in the X-axis?
-------------------------------------------------------
Use the 4x4 matrix:
| 1 0 0 0 |
M = | 0 cos(A) -sin(A) 0 |
| 0 sin(A) cos(A) 0 |
| 0 0 0 1 |
Q29. How do I generate a rotation matrix in the Y-axis?
-------------------------------------------------------
Use the 4x4 matrix:
| cos(A) 0 -sin(A) 0 |
M = | 0 1 0 0 |
| sin(A) 0 cos(A) 0 |
| 0 0 0 1 |
Q30. How do I generate a rotation matrix in the Z-axis?
-------------------------------------------------------
Use the 4x4 matrix:
| cos(A) -sin(A) 0 0 |
M = | sin(A) cos(A) 0 0 |
| 0 0 1 0 |
| 0 0 0 1 |
Q31. What are Euler angles?
---------------------------
Euler angles are the name given to the set of rotation angles which
specify the rotation in each of the X, Y and Z rotation axii.
These are specfied in vector format eg. |x y z| and can be stored
as a VECTOR data structure.
For example, the set
| 0 0 0 | will always generate the identity matrix.
Other angles are represented as follows:
| 90 0 0 | is a rotation of +90 degrees in the X-axis.
| 0 90 0 | is a rotation of +90 degrees in the Y-axis and
| 0 0 90 | is a rotation of +90 degrees in the Z-axis.
Euler angles can be represented using a single vector data structure.
Q32. What are Yaw, Roll and Pitch?
----------------------------------
Yaw, Roll and Pitch are aeronautical terms for rotation using the
Euclidean coordinate system (Euler angles), relative to the local
coordinate system of an aeroplane.
Imagine you are viewing an aeroplane from above and from directly behind.
The Z-axis is lined up with the tail and nose of the aeroplane.
The X-axis runs from the tip of the left wing to the tip of the right
wing.
The Y axis points straight up from the ground.
Pitch then becomes rotation in the X-axis, Yaw becomes rotation in the
Y-axis and Roll becomes rotation in the Z-axis.
Q33. How do I combine rotation matrices?
----------------------------------------
Rotation matrices are combined together using matrix multiplication.
As a result, the order of multiplication is very important.
Q34. What is Gimbal lock?
-------------------------
Gimbal lock is the name given to a problem that occurs with the use of
Euler angles. Because the final rotation matrix depends on the order
of multiplication, it is sometimes the case that the rotation in one
axis will be mapped onto another rotation axis.
Even worse, it may become impossible to rotate an object in a desired
axis. This is called Gimbal lock.
For example, assume that an object is being rotated in the order Z,Y,X
and that the rotation in the Y-axis is 90 degrees.
In this case, rotation in the Z-axis is performed first and therefore
correctly. The Y-axis is also rotated correctly. However, after
rotation in the Y axis, the X-axis is rotated onto the Z-axis.
Thus, any rotation in the X-axis actually rotates the object in the
Z-axis. Even worse, it becomes to rotate the object in the X-axis.
The only solution to this problem is to make use of Quaternions.
Q35. What is the correct way to combine rotation matrices?
----------------------------------------------------------
Really, there is no "correct way" of combining rotation matrices.
However, in order to be able to predict the result of combining
matrices together, some organisation is required. This is also
necessary if a full 3D matrix library is to be built.
The simplest way to rotate an object is to multiply the matrices
using the order:
M = X.Y.Z
where M is the final rotation matrix, and X,Y,Z are the individual
rotation matrices. This defines a rotation in the X-axis (pitch) first,
followed by the Y-axis (yaw) and a final rotation in the Z-axis (roll).
However, whenever the view from the camera viewpoint is being
evaluated, then the order and signs of the rotation is reversed.
For example, if you are standing up, and turn to your left, everything
in your field of view appears to move towards the right.
However, someone else facing you will say that you turned towards their
right.
Thus the view from the camera is modelled using the order:
M = -Z.-Y.-X
This is the inverse (or transpose) of the rotation matrix generated
if the camera were being rendered as another object.
Q36. How do I generate a rotation matrix from Euler angles?
-----------------------------------------------------------
At first glance, the most obvious method to generate a rotation matrix
from a set of Euler angles is to generate each matrix individually and
multiply all three together ie.
m3_rotx( mat_x, vec -> angle_x );
m3_roty( mat_y, vec -> angle_y );
m3_rotz( mat_z, vec -> angle_z );
m3_mult( mat_tmp, mat_z, mat_y );
m3_mult( mat_final, mat_tmp, mat_x );
This set of calls could be placed in a separate routine eg.
m3_fromeuler( MATRIX *mat_final, VECTOR3 *euler )
However, to perform this sequence of calls is very wasteful in terms
of processing time. Given that each 4x4 rotation matrix is guaranteed
to have 10 elements with value zero (0), 2 elements with value one (1)
and four others of arbitary value, over 75% of every matrix operation
is wasted. This does not include the set up and initialisation of each
matrix.
Altogether, over 75% of all matrix operations are spent processing
arithmetic expressions which lead to either zero or one.
A more efficient way must be found. Fortunately, there is another way
of determining the final resulting matrix.
If all three matrices are combined in algebraic format, the following
expression is defined:
M = X.Y.Z
where M is the final matrix,
X is the rotation matrix for the X-axis,
Y is the rotation matrix for the Y-axis,
Z is the rotation matrix for the Z-axis.
Expanding into rotation matrices in algebraic format gives:
| 1 0 0 |
X = | 0 A -B |
| 0 B A |
| C 0 -D |
Y = | 0 1 0 |
| D 0 C |
| E -F 0 |
Z = | F E 0 |
| 0 0 1 |
where A,B are the cosine and sine of the X-axis rotation axis,
C,D are the cosine and sine of the Y-axis rotation axis,
E,F are the cosine and sine of the Z-axis rotation axis.
Then the expression:
M = X.Y.Z
can be split into two matrix multiplications:
M' = X.Y
M = M'.Z
Evaluating M' first:
M' = X.Y
| 1 0 0 | | C 0 -D |
M' = | 0 A -B | . | 0 1 0 |
| 0 B A | | D 0 C |
| 1.C + 0.0 + 0.D 1.0 + 0.1 + 0.0 1.-D + 0.0 + 0.C |
M' = | 0.C + A.0 + -B.D 0.0 + A.1 + -B.0 0.-D + A.0 + -B.C |
| 0.C + B.0 + A.D 0.0 + B.1 + A.0 0.-D + B.0 + A.C |
Simplifying M' gives:
| C 0 -D |
M' = | -B.D A -B.C |
| A.D B A.C |
Evaluating M gives:
M = M'.Z
| C 0 -D | | E -F 0 |
M = | -BD A -BC | . | F E 0 |
| AD B AC | | 0 0 1 |
| C.E + 0.F + -D.0 C.-F + 0.E + -D.0 C.0 + 0.0 + -D.1 |
M = | -BD.E + A.F + -BC.0 -BD.-F + A.E + -BC.0 -BD.0 + A.0 + -BC.1 |
| AD.E + B.F + AC.0 AD.-F + B.E + AC.0 AD.0 + 0.0 + AC.1 |
Simplifying M gives a 3x3 matrix:
| CE -CF -D |
M = | -BDE+AF -BDF+AE -BC |
| ADE+BF -ADF+BE AC |
This is the final rotation matrix. As a 4x4 matrix this is:
| CE -CF -D 0 |
M = | -BDE+AF BDF+AE -BC 0 |
| ADE+BF -ADF+BE AC 0 |
| 0 0 0 1 |
The individual values of A,B,C,D,E and F are evaluated first. Also, the
values of BD and AD are also evaluated since they occur more than once.
Thus, the final algorithm is as follows:
-----------------------
A = cos(angle_x);
B = sin(angle_x);
C = cos(angle_y);
D = sin(angle_y);
E = cos(angle_z);
F = sin(angle_z);
AD = A * D;
BD = B * D;
mat[0] = C * E;
mat[1] = -C * F;
mat[2] = -D;
mat[4] = -BD * E + A * F;
mat[5] = BD * F + A * E;
mat[6] = -B * C;
mat[8] = AD * E + B * F;
mat[9] = -AD * F + B * E;
mat[10] = A * C;
mat[3] = mat[7] = mat[11] = mat[12] = mat[13] = mat[14] = 0;
mat[15] = 1;
-----------------------
Using basic matrix calculations, the operation count would reach
128 multiplications, 96 additions and 80 assignments operations.
Using the optimised algorithm, only 12 multiplications, 6 subtractions
and 18 assignment operations are required.
So, it is obvious that by using the optimised algorithm, a performance
achievement of 1000% is achieved!
Q37. How do I convert a rotation matrix to Euler angles?
--------------------------------------------------------
This operation is the exact opposite to the one answered in the question
above. Given that the rotation matrix is:
| CE -CF -D 0 |
M = | -BDE+AF BDF+AE -BC 0 |
| ADE+BF -ADF+BE AC 0 |
| 0 0 0 1 |
where A,B are the cosine and sine of the X-axis rotation axis,
C,D are the cosine and sine of the Y-axis rotation axis,
E,F are the cosine and sine of the Z-axis rotation axis.
Using the C data structure for a 4x4 matrix, the index values are
as follows:
| 0 1 2 3 |
M = | 4 5 6 7 |
| 8 9 10 11 |
| 12 13 14 15 |
From looking at these two tables, it can be see that array element
[2] has the value of -D or -sin(Y). Then the rotation angle in the
Y-axis can be calculated from a call to to the inverse-sine function.
Passing this value to the cosine function then gives the value of
C.
If C is not zero, then the rotation angles in each of the X and Z
axii, can be derived from the terms on the third column and first row
respectively. These are as follows:
X-axis: M[6] = -BC
M[10] = AC
Z-axis: M[0] = CE
M[1] = -CF
The actual rotation angles can be derived by taking each pair of values
dividing by C and passing the results to the inverse tangent function.
If C is zero, then these calculations are not possible. In this case
the rotation angle in the Y-axis will be either -90 or +90. Thus D
will either have the value of 1 or -1.
In this case, Gimbal Lock will have occurred. Rotations in both the
X and Z axii will appear to be in the same axis. This can be seen
through the evaluation of the rotation axis.
| 0.E -0.F 1 0 |
M = | -B.1.E+AF B.1.F+AE -B.0 0 |
| A.1.E+BF -A.1.F+BE A.0 0 |
| 0 0 0 1 |
Multiplying out each term gives:
| 0 0 1 0 |
M = | -BE+AF BF+AE 0 0 |
| AE+BF -AF+BE 0 0 |
| 0 0 0 1 |
Rearranging these terms gives:
| 0 0 1 0 |
M = | -BE+AF AE+BF 0 0 |
| AE+BF BE-AF 0 0 |
| 0 0 0 1 |
Then it can be seen that the matrix is really of the form:
| 0 0 -1 0 |
M = | -V W 0 0 |
| W V 0 0 |
| 0 0 0 1 |
Where V has the value of BE-AF and
W has the value of AE+BF
These two values can be considered to be the sin and cosine of a
single rotation axis.
The final algorithm is then as follows:
-----------------------------------
angle_y = D = -asin( mat[2]); /* Calculate Y-axis angle */
C = cos( angle_y );
angle_y *= RADIANS;
if ( fabs( C ) > 0.005 ) /* Gimball lock? */
{
trx = mat[10] / C; /* No, so get X-axis angle */
try = -mat[6] / C;
angle_x = atan2( try, trx ) * RADIANS;
trx = mat[0] / C; /* Get Z-axis angle */
try = -mat[1] / C;
angle_z = atan2( try, trx ) * RADIANS;
}
else /* Gimball lock has occurred */
{
angle_x = 0; /* Set X-axis angle to zero */
trx = mat[5]; /* And calculate Z-axis angle */
try = mat[4];
angle_z = atan2( try, trx ) * RADIANS;
}
angle_x = clamp( angle_x, 0, 360 ); /* Clamp all angles to range */
angle_y = clamp( angle_y, 0, 360 );
angle_z = clamp( angle_z, 0, 360 );
-----------------------------------
Q38. How do I generate a rotation matrix for a selected axis and angle?
-----------------------------------------------------------------------
The only way to generate this type of rotation matrix is through the
use of quaternion mathematics.
See question [Q53. How do I convert a quaternion to a rotation matrix?]
for further details.
Q39. How do I generate a rotation matrix to map one vector onto another?
------------------------------------------------------------------------
When developing animation software, a common requirement is to find
a rotation matrix that will map one direction vector onto another.
This problem may be visualised by considering the two direction
vectors to be attached at their starting points. Then the entire
rotation space forms a unit sphere.
In theory, there are an infinite number of rotation axii and angles
that will map one vector onto the other. All of these axii lie on the
plane where all of the points are the exact same distance from both
vectors.
However, only one solution is of practical interest. This is the path
which covers the shortest angular distance between the two vectors.
The rotation axis to this path is calculated by taking the cross
product between the two vectors:
Vaxis = Vs x Vf
The rotation angle is calculated by taking the dot product between the
two vectors:
-1
Vangle = cos ( Vs . Vf )
One practical application of the solution to this problem is finding
the shortest flight path between two cities. In this case, each city
is represented as a direction vector generated from spherical
coordinates. Since planet Earth is spherical, the desired flight path
is the shortest angular rotation between the two cities.
Q40. How do I use matrices to convert one coordinate system to another?
-----------------------------------------------------------------------
Similar to the previous problem, the requirement is to map one
coordinate system onto another. However, instead of just trying to
map one coordinate axis onto another, all three axii have to be matched.
Both coordinate systems are therefore represented as either 3x3 or
4x4 matrices.
The problem is therefore to find the rotation matrix that will map one
matrix onto another. This can be expressed mathematically:
Mfinal = Mrot . Morig
where Mfinal is the final coordinate system matrix,
Morig is the original coordinate system and
Mrot is the desired rotation matrix.
The goal is then to find the matrix Mrot. This can be achieved by
rearranging the equation to give:
-1
Mfinal . Morig = Mrot
-1
Mrot = Mfinal . Morig
Thus, the desired rotation matrix can be by calculatng the inverse of
the original coordinate system and multiplying it with the final
rotation matrix.
As a check, consider the cases when either the original or final
rotation matrices are the identity matrix. In each case, the rotation
matrix should match the final matrix and the inverse of the final
matrix respectively.
Once calculated, the rotation matrix may be converted into a
quaternion.
Q41. What is a translation matrix?
----------------------------------
A translation matrix is used to position an object within 3D space
without rotating in any way. Translation operations using matrix
multiplication can only be performed using 4x4 matrices.
If the translation is defined by the vector [X Y Z ], then the 4x4
matrix to implement translation is as follows:
| 1 0 0 X |
| |
| 0 1 0 Y |
M = | |
| 0 0 1 Z |
| |
| 0 0 0 1 |
If the vector is [0 0 0] then the vertex list will remain as before.
Q42. What is a scaling matrix?
------------------------------
A scaling matrix is used to enlarge or shrink the size of a 3D
model.
If the scaling vector is [X Y Z] then the matrix to perform this is
as follows:
| X 0 0 0 |
| |
| 0 Y 0 0 |
M = | |
| 0 0 Z 0 |
| |
| 0 0 0 1 |
If the scaling vector is [1 1 1], then this generates the identity
matrix and vertex geometry will remain unchanged.
Q43. What is a shearing matrix?
-------------------------------
A shearing matrix is used to make a 3D model appear to slant sideways.
For example, "italic" text requires each character to slant towards the
right.
In three dimensions six possible shearing directions exist:
o shear X by Y
o shear X by Z
o shear Y by X
o shear Y by Z
o shear Z by X
o shear Z by Y
All six shearing directions may be combined into a single matrix:
| 1 Syx Szx 0 |
| |
| Sxy 1 Szy 0 |
M = | |
| Sxz Syz 1 0 |
| |
| 0 0 0 1 |
| |
Where Sij implements a shear of I by J
Thus, Sxy shears X by Y
In theory, rotation in three dimensions may be considered a combination
of six shearing directions.
Q44. How do I perform linear interpolation between two matrices?
----------------------------------------------------------------
Given two rotation matrices, the problem is to find a way of
determining intermediate positions specified by a parametric
variable t, where t ranges from 0.0 to 1.0
This can be achieved by converting the two matrices into either
Euler angles or Spherical rotation angles (via quaternions) and
a translation vector.
In either case, each matrix is converted into a pair of 3D vectors.
Interpolation between these two vectors can then be performed
through the use of the standard linear interpolation equation:
Vr = Va + t .(Vb - Va )
where Vr is the resulting vector
Va is the start position vector
Vb is the final position vector
This equation may be applied to both translation and rotation
vectors.
Once determined, the resulting translation and rotation are then
converted back into the desired intermediate matrix.
Q45. How do I perform cubic interpolation between four matrices?
----------------------------------------------------------------
Given four rotation or translation matrices, the problem is to
find a way of determining intermediate positions specified by a
parametric variable t.
This can be achieved by making use of cubic interpolation. As with
linear interpolation, the four matrices are converted into their
corresponding translation and rotation vectors (Again, either Euler
angles or spherical rotation angles).
Each set of four vectors is then converted into a single geometry
vector G. Through the use of spline mathematics, this geometry vector
is converted into an interpolation matrix M.
If the geometry vector is defined as:
| x1 x2 x3 x4 |
G = | y1 y2 y3 y4 |
| z1 z2 z3 z4 |
Then multiplication by the base matrix:
| -4.5 9.0 -5.5 1.0 |
Mb = | 13.5 -22.5 9.0 0.0 |
| -13.5 18.0 -4.5 0.0 |
| 4.5 -4.5 1.0 0.0 |
will generate the 3x4 interpolation matrix Mi:
Mi = G .Mb
This can be implemented through a standard matrix-vector multiplication.
Interpolation can then be performed by the use of the parametric
variable t:
R = Mi . t
|t^3|
| xr | | A B C D | |t^2|
| yr | = | E F G H | . |t |
| zr | | I J K L | |1 |
The result vector can then be converted back into a rotation or
translation matrix.
It should be noted that the rotation paths that are generated may
occasionally become rather loopy. This is normal, as the algorithm
is trying to find the path with the least amount of rotation between
all four vectors.
Of the two methods, spherical rotation angles will usually be seen to
provide the cleanest interpolation paths for rotation.
Q46. How can I render a matrix?
-------------------------------
When using a graphics window for 3D animation, it is convenient to be
able to view a rotation matrix concurrently with the animation.
However, displaying a rotation matrix as an array of numeric values
does not provide a very meaningful context.
An alternative to rendering numeric data is to make use of graphical
display methods such as bar-graphs.
Much like a graphic equalizer on a stereo, a rotation matrix may be
displayed in a bar graph format. Each element of the rotation matrix
is rendered as an individual bar-graph in the range -1 to +1.
A 3x3 matrix would look like the following:
+--+ +--+ +--+
|##| | | | |
+--+ +--+ +--+
| | | | | |
+--+ +--+ +--+
+--+ +--+ +--+
| | |##| | |
+--+ +--+ +--+
| | | | | |
+--+ +--+ +--+
+--+ +--+ +--+
| | | | |##|
+--+ +--+ +--+
| | | | | |
+--+ +--+ +--+
In this case, the rotation matrix is the identity matrix, since each
element in the major diagonal is +1, and all others are zero.
For added visual clarity, parameters which are negative may shaded
in a different colour than those which are positive.
QUATERNIONS
===========
Q47. What are quaternions?
--------------------------
Quaternions extend the concept of rotation in three dimensions to
rotation in four dimensions. This avoids the problem of "gimbal-lock"
and allows for the implementation of smooth and continuous rotation.
In effect, they may be considered to add a additional rotation angle
to spherical coordinates ie. Longitude, Latitude and Rotation angles
A Quaternion is defined using four floating point values |x y z w|.
These are calculated from the combination of the three coordinates
of the rotation axis and the rotation angle.
Q48. How do quaternions relate to 3D animation?
-----------------------------------------------
As mentioned before, Euler angles have the disadvantage of being
susceptible to "Gimbal lock" where attempts to rotate an
object fail to appear as expected, due to the order in which the
rotations are performed.
Quaternions are a solution to this problem. Instead of rotating an
object through a series of successive rotations, quaternions allow
the programmer to rotate an object through an arbitary rotation axis
and angle.
The rotation is still performed using matrix mathematics. However,
instead of multiplying matrices together, quaternions representing
the axii of rotation are multiplied together. The final resulting
quaternion is then converted to the desired rotation matrix.
Because the rotation axis is specifed as a unit direction vector,
it may also be calculated through vector mathematics or from
spherical coordinates ie (longitude/latitude).
Quaternions offer another advantage in that they be interpolated.
This allows for smooth and predictable rotation effects.
Q49. How do I calculate the conjugate of a quaternion?
------------------------------------------------------
This can be achieved by reversing the polarity (or negating) the
vector part of the quaternion, ie:
--
Qr = ( Qr.scalar, -Qr.vector )
----------------------------------------------------------------
quaternion_conjugate( QUAT *qr, QUAT *qa )
{
qr -> qw = qa -> qw;
qr -> qx = -qa -> qx;
qr -> qy = -qa -> qy;
qr -> qz = -qa -> qz;
}
--------------------------------------
Q50. How do I calculate the inverse of a quaternion?
----------------------------------------------------
This is equivalent to calculating the conjugate of the quaternion.
See Q49. How do I calculate the conjugate of a quaternion?
Q51. How do I calculate the magnitude of a quaternion?
------------------------------------------------------
The magnitude of a quaternion is calculated by multiplying the
quaternion with its conjugate ie:
------------
/ --
|Qr| = \/ Qr.Qr
This can be implemented as the following code sequence:
-------------------------------------------------------------------
QFLOAT quaternion_magnitude( QUAT *qa )
{
return( sqrt(qa->qw*qa->qw+
qa->qx*qa->qx+ qa->qy*qa->qy+qa->qz*qa->qz) )
}
-------------------------------------------------------------------
Q52. How do I normalise a quaternion?
-------------------------------------
A quaternion can be normalised in a way similar to vectors. The
magnitude of the quaternion is calculated first. Then both the
scalar and vector part of the quaternion are divided by this value.
A unit quaternion will always have a magnitude of 1.0
Q53. How do I multiply two quaternions together?
------------------------------------------------
Given two quaternions Q1 and Q2, the goal is to calculate the
combined rotation Qr:
Qr = Q1.Q2
This is achieved through the expression:
Qr = Q1.Q2 = ( w1.w2 - v1.v2, w1.v2 + w2.v1 + v1 x v2 )
where v1 = (x,y,z) of Q1
w1 = (w) of Q1
v2 = (x,y,z) of Q2
w2 = (w) of Q2
and both . and x are the standard vector dot and cross products.
This can be implemented using the following code segment:
---------------------------------------------------
quaternion_multiply( QUAT *qr, QUAT *qa, QUAT *qb )
{
qr.scalar = v3_dot( &qa->vector, &qb->vector );
v3_cross( &va, &qa->vector, &qb->vector );
v3_scalef( &vb, &qa->vector, &qb->scalar );
v3_scalef( &vc, &qb->vector, &qa->scaler );
v3_add( &va, &va, &vb );
v3_add( &qr->vector, &va, &vc );
quaternion_normalise( qr );
}
---------------------------------------------------
Q54. How do I convert a quaternion to a rotation matrix?
--------------------------------------------------------
Assuming that a quaternion has been created in the form:
Q = |X Y Z W|
Then the quaternion can then be converted into a 4x4 rotation
matrix using the following expression:
| 2 2 |
| 1 - 2Y - 2Z 2XY - 2ZW 2XZ + 2YW |
| |
| 2 2 |
M = | 2XY + 2ZW 1 - 2X - 2Z 2YZ - 2XW |
| |
| 2 2 |
| 2XZ - 2YW 2YZ + 2XW 1 - 2X - 2Y |
| |
If a 4x4 matrix is required, then the bottom row and right-most column
may be added.
The matrix may be generated using the following expression:
----------------
xx = X * X;
xy = X * Y;
xz = X * Z;
xw = X * W;
yy = Y * Y;
yz = Y * Z;
yw = Y * W;
zz = Z * Z;
zw = Z * W;
mat[0] = 1 - 2 * ( yy + zz );
mat[1] = 2 * ( xy - zw );
mat[2] = 2 * ( xz + yw );
mat[4] = 2 * ( xy + zw );
mat[5] = 1 - 2 * ( xx + zz );
mat[6] = 2 * ( yz - xw );
mat[8] = 2 * ( xz - yw );
mat[9] = 2 * ( yz + xw );
mat[10] = 1 - 2 * ( xx + yy );
mat[3] = mat[7] = mat[11 = mat[12] = mat[13] = mat[14] = 0;
mat[15] = 1;
----------------
Q55. How do I convert a rotation matrix to a quaternion?
--------------------------------------------------------
A rotation may be converted back to a quaternion through the use of
the following algorithm:
The process is performed in the following stages, which are as follows:
Calculate the trace of the matrix T from the equation:
2 2 2
T = 4 - 4x - 4y - 4z
2 2 2
= 4( 1 -x - y - z )
= mat[0] + mat[5] + mat[10] + 1
If the trace of the matrix is greater than zero, then
perform an "instant" calculation.
S = 0.5 / sqrt(T)
W = 0.25 / S
X = ( mat[9] - mat[6] ) * S
Y = ( mat[2] - mat[8] ) * S
Z = ( mat[4] - mat[1] ) * S
If the trace of the matrix is less than or equal to zero
then identify which major diagonal element has the greatest
value.
Depending on this value, calculate the following:
Column 0:
S = sqrt( 1.0 + mr[0] - mr[5] - mr[10] ) * 2;
Qx = 0.5 / S;
Qy = (mr[1] + mr[4] ) / S;
Qz = (mr[2] + mr[8] ) / S;
Qw = (mr[6] + mr[9] ) / S;
Column 1:
S = sqrt( 1.0 + mr[5] - mr[0] - mr[10] ) * 2;
Qx = (mr[1] + mr[4] ) / S;
Qy = 0.5 / S;
Qz = (mr[6] + mr[9] ) / S;
Qw = (mr[2] + mr[8] ) / S;
Column 2:
S = sqrt( 1.0 + mr[10] - mr[0] - mr[5] ) * 2;
Qx = (mr[2] + mr[8] ) / S;
Qy = (mr[6] + mr[9] ) / S;
Qz = 0.5 / S;
Qw = (mr[1] + mr[4] ) / S;
The quaternion is then defined as:
Q = | Qx Qy Qz Qw |
Q56. How do I convert a rotation axis and angle to a quaternion?
----------------------------------------------------------------
Given the rotation axis and angle, the following algorithm may be
used to generate a quaternion:
------------------------------------------------
sin_a = sin( angle / 2 )
cos_a = cos( angle / 2 )
q -> x = axis -> x * sin_a
q -> y = axis -> y * sin_a
q -> z = axis -> z * sin_a
q -> w = cos_a
quaternion_normalise( q );
------------------------------------------------
It is necessary to normalise the quaternion in case any values are
very close to zero.
Q57. How do I convert a quaternion to a rotation axis and angle?
----------------------------------------------------------------
A quaternion can be converted back to a rotation axis and angle
using the following algorithm:
---------------------------------------------------
quaternion_normalise( qr );
cos_angle = qr -> qw;
angle = acos( cos_angle ) * 2 * RADIANS;
sin_angle = sqrt( 1.0 - cos_angle * cos_angle );
if ( fabs( sin_angle ) < 0.0005 )
sa = 1;
axis -> vx = qr -> qx / sa;
axis -> vy = qr -> qy / sa;
axis -> vz = qr -> qz / sa;
---------------------------------------------------
Q58. How do I convert spherical rotation angles to a quaternion?
----------------------------------------------------------------
A rotation axis itself may be defined using spherical coordinates
(latitude and longitude) and a rotation angle
In this case, the quaternion can be calculated as follows:
-----------------------
sin_a = sin( angle / 2 )
cos_a = cos( angle / 2 )
sin_lat = sin( latitude )
cos_lat = cos( latitude )
sin_long = sin( longitude )
cos_long = cos( longitude )
qx = sin_a * cos_lat * sin_long
qy = sin_a * sin_lat
qz = sin_a * sin_lat * cos_long
qw = cos_a
-----------------------
Q59. How do I convert a quaternion to spherical rotation angles?
----------------------------------------------------------------
A quaternion can be converted to spherical coordinates by extending
the conversion process:
-----------------------
cos_angle = q -> qw;
sin_angle = sqrt( 1.0 - cos_angle * cos_angle );
angle = acos( cos_angle ) * 2 * RADIANS;
if ( fabs( sin_angle ) < 0.0005 )
sa = 1;
tx = q -> qx / sa;
ty = q -> qy / sa;
tz = q -> qz / sa;
latitude = -asin( ty );
if ( tx * tx + tz * tz < 0.0005 )
longitude = 0;
else
longitude = atan2( tx, tz ) * RADIANS;
if ( longitude < 0 )
longitude += 360.0;
-----------------------
Q60. How do I convert Euler rotation angles to a quaternion?
-------------------------------------------------------------------
Converting Euler rotation angles to quaterions can be achieved through
the use of quaternion multiplication. Each rotation angle is converted
to an axis-angle pair, with the axis corresponding to one of the
Euclidean axii. The axis-angle pairs are converted to quaternions and
multiplied together. The final quaternion is the desired result.
The following code segment demonstrates this:
---------------------------------------------
quaternion_from_euler( QUATERNION *q, VFLOAT ax, VFLOAT ay, VFLOAT az )
{
VECTOR3 vx = { 1, 0, 0 }, vy = { 0, 1, 0 }, vz = { 0, 0, 1 };
QUATERNION qx, qy, qz, qt;
quaternion_from_axisangle( qx, &vx, rx );
quaternion_from_axisangle( qy, &vy, ry );
quaternion_from_axisangle( qz, &vz, rz );
quaternion_multiply( &qt, &qx, &qy );
quaternion_multiply( &q, &qt, &qz );
}
---------------------------------------------
Q61. How do I use quaternions to perform linear interpolation between matrices?
-------------------------------------------------------------------------------
For many animation applications, it is necessary to interpolate
between two rotation positions of a given object. These positions may
have been specified using keyframe animation or inverse kinematics.
Using either method, at least two rotation matrices must be known, and
the desired goal is to interpolate between them. The two matrices are
referred to as the starting and finish matrices( MS and MF).
Using linear interpolation, the interpolated rotation matrix is
generated using a blending equation with the parameter T, which
ranges from 0.0 to 1.0.
At T=0, the interpolated matrix is equal to the starting matrix.
At T=1, the interpolated matrix is equal to the finishing matrix.
Then the interpolated rotation matrix (MI) is specified as:
MI = F( MS, MI, T )
where F is a blending function.
The first stage in interpolating between the two matrices is to
determine the rotation matrix that will convert MS to MF.
This is achieved using the following expression:
-1
T = Ms . Mf
where Ms is the start matrix,
Mf is the finish matrix,
and T is the intermediate matrix.
The next stage is to convert this matrix into a rotation axis and
angle. This is achieved by converting the matrix into a quaternion
and finally into the required rotation axis and angle.
In order to generate the interpolated rotation matrix, it is only
necessary to scale the rotation angle and convert this angle and
the rotation axis back into a rotation matrix.
Using a 4x4 matrix library, this is as follows:
m4_transpose( mt, ms ); /* Inverse */
m4_mult( ms, mt, mb ); /* Rotation matrix */
m4_to_axisangle( ms, axis, angle ); /* Rotation axis/angle */
for ( t = 0; t < 1.0; t += 0.05 )
{
m4_from_axisangle( mi, axis, angle * t ); /* Final interpolation */
... whatever ...
}
where t is the interpolation factor ranging from 0.0 to 1.0
Q62. How do I use quaternions to perform cubic interpolation between matrices?
------------------------------------------------------------------------------
For some applications, it may not be convenient or possible to use linear
interpolation for animation purposes. In this case, cubic interpolation
is another alternative.
In order to use cubic interpolation, at least four rotation matrices must
be known.
Each of these is then converted into a set of spherical rotations
via quaternions and spherical rotation angles (ie. longitude, latitude
and rotation angle).
These are then multiplied with the base matrix for a Cardinal spline
curve. This interpolation matrix can then be used to determine the
intermediate spherical rotation angles.
Once the interpolated coordinates are known (latitude, longitude and
rotation angle), the interpolated rotation matrix can then be generated
through the conversion to quaternions.
Using a 4x4 matrix library, the algorithm is as follows:
----------------------------------------------------------------------
for ( n = 0; n < 4; n++ )
m4_to_spherical( mat[n], &v_sph[n] ); /* Spherical coordinates */
m4_multspline( m_cardinal, v_sph, v_interp ); /* Interpolation vector */
...
v3_cubic( v_pos, v_interp, t ); /* Interpolation */
m4_from_spherical( m_rot, v_pos ); /* Back to a matrix */
----------------------------------------------------------------------
Last modified:04/22/09
 |
The Chandra X-Ray
Center (CXC) is operated for NASA by the Smithsonian Astrophysical Observatory. 60 Garden Street, Cambridge, MA 02138 USA. Email: aspect_help@head.cfa.harvard.edu Smithsonian Institution, Copyright © 1998-2004. All rights reserved. |