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The Matrix and Quaternions FAQ Version 1.21 30th November 2003 ------------------------------- Please mail feedback to matrix_faq@j3d.org with a subject starting with MATRIX-FAQ (otherwise my spam filter will simply kill your message). Any additional suggestions or related questions are welcome. Just send E-mail to the above address. The latest copy of this FAQ can be found at the following web page: http://www.j3d.org/matrix_faq/matrfaq_latest.html Feel free to distribute or copy this FAQ as you please. Contributions ------------- Introduction I1: Steve ????? Correction to Q55 until Q59: Andreas Junghanns Correction to Q50: Morten Ofstad Note to Q39: Tom Nuydens Corrections to Q29 and Q37: Eric Reiss Clarification to Q56: Duncan Murdoch Clarification to Q37: Ron Avitzur Correction to Q1: Mona Wong Corrections to Q36 and Q37: Eric Reiss Improvement to Q34 and Q38: Jon Watte Warning and alternative to Q58 and Q59: Paul Pedriana Correction (and optimization [Lee]) to Q53: Eleanor Groundwater and Lee Morgan Improvement to Q39: jhunpingco Corrections to Q11 and optimization to Q12: Gordon Corrections to Q54 to Q60: Eleanor Groundwater Corrections and improvements to Q23 and Q24: Ben Houston Addition to Q39: Jon Watte Correction to Q61: Adam D. Moss Addition of Q63: Mike Cline Addition of I2: Jacob Marner Correction to Q38 and inception of I2: Armin M?ller Addition of Q60: Pablo Figueroa Correntions and additions to Q14, Q16, Q21 and Q34: Tronster Hartley Correction to Q12 and Q54: Frank DJ Correction to Q34: Robert Funnell History ------- I (Andreas) tried to find "hexapod@(no-spam)netcom.com" who seemed to have maintained this for a while, but the site at netcom.com doesn't exist anymore, emails bounce. Since I (and colleques) wasted quite some time figuring out what was wrong with some of the algorithms given in the earlier versions of this document, I decided to correct it and put it back on the web. The formerly given sites for the location of these documents do not exist anymore: ftp://ftp.netcom.com/pub/he/hexapod/index.html http://www.glue.umd.edu/~rsrodger Versions, dates and links to local copies (so you can compare): matrfaq_1.02.html: Version 1.2 2nd September 1997 matrfaq_1.04.html: Version 1.4 26th December 1998 matrfaq_1.06.html: Version 1.6 30th September 2000 matrfaq_1.07.html: Version 1.7 20th December 2000 matrfaq_1.08.html: Version 1.8 21th December 2000 matrfaq_1.09.html: Version 1.9 16th January 2001 matrfaq_1.10.html: Version 1.10 30th January 2001 matrfaq_1.11.html: Version 1.11 9th February 2001 matrfaq_1.12.html: Version 1.12 26th March 2001 matrfaq_1.13.html: Version 1.13 20th July 2001 matrfaq_1.14.html: Version 1.14 17th August 2001 matrfaq_1.15.html: Version 1.15 20th August 2001 matrfaq_1.16.html: Version 1.16 2nd October 2001 matrfaq_1.17.html: Version 1.17 30th November 2001 matrfaq_1.18.html: Version 1.18 27th January 2002 matrfaq_1.19.html: Version 1.19 20th March 2002 matrfaq_1.20.html: Version 1.20 31st January 2002 matrfaq_1.21.html: Version 1.21 30th November 2003 Please refrain from asking me math questions. I am only maintaining this FAQ and have very little knowledge about the subject. But, if you have a question that is not answered by this FAQ and later happen to find the answer and believe it to be relevant for this FAQ (or its readers), please send all relevant information, hopefully in a pre-digested form, to me to be included here. Thanks! If you prefer to appear as "anonymous" in the contributions list, let me know, otherwise I'll just put you down with whatever name I can gather from your email header. Introduction ------------ I1. Important note relating to OpenGL and this document I2. Important note with respect to normalized inputs 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? Q63. How do I use quaternions to rotate a vector? 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 | I2. Important note with respect to normalized inputs ---------------------------------------------------- Note that most algorithms assume normalized inputs, such as vectors of union length, or matrices with normalized main diagonal etc. It is possible, and often enough the case, that algorithms (and the code snippets provided here) work correctly with arbitrary inputs, but it is usually considered bad practise (and you will pay in debugging time if you fail to observe this suggestion) to rely on this property. 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 01 02 03 | M = | 10 11 12 13 | | 20 21 22 23 | | 30 31 32 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 / ik kj -- k=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 Nx4 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 4x2. 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 square, that is orthogonal and 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 unchanged. 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 One can be bit faster using the following piece of code, note that m and i are both getting changed: m -> matrix to be raised to a power i -> power to raise matrix to a -> matrix that will contain the result I -> identity matrix / -> integer division (ie round down to nearest whole number) % -> integer remainder operation * -> matrix multiplication a=I while(i>0){ if(i%2) a=a*m i=i/2 m=m*m } 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 an inverse 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 make the character look 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 | A B C | M = | D E F | becomes submat = | E F | | G H I | 11 | H I | 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 ); } int m3_inverse( MATRIX3 mr, MATRIX3 ma ) { VFLOAT det = m3_det( ma ); if ( fabs( det ) < 0.0005 ) { m3_identity( mr ); return(0); } 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; return(1); } 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 di, dj, si, sj; // loop through 3x3 submatrix for( di = 0; di < 3; di ++ ) { for( dj = 0; dj < 3; dj ++ ) { // map 3x3 element (destination) to 4x4 element (source) si = di + ( ( di >= i ) ? 1 : 0 ); sj = dj + ( ( dj >= j ) ? 1 : 0 ); // copy element mb[di * 3 + dj] = mr[si * 4 + sj]; } } } 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 ) m4_identity( mr ); 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 impossible to rotate the object in the X-axis. A convenient 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; } /* return only positive angles in [0,360] */ if (angle_x < 0) angle_x += 360; if (angle_y < 0) angle_y += 360; if (angle_z < 0) angle_z += 360; ----------------------------------- Q38. How do I generate a rotation matrix for a selected axis and angle? ----------------------------------------------------------------------- The simplest 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. The following code snipped does most of the work (phi being the rotation angle and (u,v,w) the rotation (axis) vector): rcos = cos(phi); rsin = sin(phi); matrix[0][0] = rcos + u*u*(1-rcos); matrix[1][0] = w * rsin + v*u*(1-rcos); matrix[2][0] = -v * rsin + w*u*(1-rcos); matrix[0][1] = -w * rsin + u*v*(1-rcos); matrix[1][1] = rcos + v*v*(1-rcos); matrix[2][1] = u * rsin + w*v*(1-rcos); matrix[0][2] = v * rsin + u*w*(1-rcos); matrix[1][2] = -u * rsin + v*w*(1-rcos); matrix[2][2] = rcos + w*w*(1-rcos); Don't forget to set the rest of the matrix to 0 (1 at [3][3]) if you are using 4x4 matrices! 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. Note: If Vs and Vf are colinear, the cross product returns (0,0,0). You should test for that case and use any of the 90 degree rotations of either Vs or Vf as a rotation axis, e.g. (y,z,x). If x==y==z, then using one of x,y, or z from the second vector might help. --> Is there a better way? Having the rotation angle and vector, generating the corresponding matrix is easy, see [Q38: How do I generate a rotation matrix for a selected axis and angle?] for details. Don't let the spectre of Gimbal Lock fool you: Euler angles are still a complete representation of any rotation in 3D space; it's just that the actual Euler angles needed to achieve some particular desired rotation may be rather unintuitive. 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, if the quaternion is normalized (or a unit quaternion). In all other cases, the magnitude of the inverse is 1/|q|. 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 = qa->scalar * qb->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->scalar ); v3_add( &va, &va, &vb ); v3_add( &qr->vector, &va, &vc ); quaternion_normalise( qr ); } --------------------------------------------------- An optimization can also be made by rearranging to w = w1w2 - x1x2 - y1y2 - z1z2 x = w1x2 + x1w2 + y1z2 - z1y2 y = w1y2 + y1w2 + z1x2 - x1z2 z = w1z2 + z1w2 + x1y2 - y1x2 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 (Warning: you might have to transpose this matrix if you (do not) follow the OpenGL order!): ? 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; The resulting matrix uses the following positions: ? mat[0] mat[4] mat[ 8] mat[12] ? M = ? mat[1] mat[5] mat[ 9] mat[13] ? ? mat[2] mat[6] mat[10] mat[14] ? ? mat[3] mat[7] mat[11] mat[15] ? 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 ) = 1 + mat[0] + mat[5] + mat[10] If the trace of the matrix is greater than zero, then perform an "instant" calculation. Important note wrt. rouning errors: Test if ( T > 0.00000001 ) to avoid large distortions! S = sqrt(T) * 2; X = ( mat[9] - mat[6] ) / S; Y = ( mat[2] - mat[8] ) / S; Z = ( mat[4] - mat[1] ) / S; W = 0.25 * S; If the trace of the matrix is equal to zero then identify which major diagonal element has the greatest value. Depending on this, calculate the following: if ( mat[0] > mat[5] && mat[0] > mat[10] ) { // Column 0: S = sqrt( 1.0 + mat[0] - mat[5] - mat[10] ) * 2; X = 0.25 * S; Y = (mat[4] + mat[1] ) / S; Z = (mat[2] + mat[8] ) / S; W = (mat[9] - mat[6] ) / S; } else if ( mat[5] > mat[10] ) { // Column 1: S = sqrt( 1.0 + mat[5] - mat[0] - mat[10] ) * 2; X = (mat[4] + mat[1] ) / S; Y = 0.25 * S; Z = (mat[9] + mat[6] ) / S; W = (mat[2] - mat[8] ) / S; } else { // Column 2: S = sqrt( 1.0 + mat[10] - mat[0] - mat[5] ) * 2; X = (mat[2] + mat[8] ) / S; Y = (mat[9] + mat[6] ) / S; Z = 0.25 * S; W = (mat[4] - mat[1] ) / S; } The quaternion is then defined as: Q = | X Y Z W | Q56. How do I convert a rotation axis and angle to a quaternion? ---------------------------------------------------------------- Given a rotation axis and angle, the following algorithm may be used to generate a quaternion: vector_normalize(axis); sin_a = sin( angle / 2 ); cos_a = cos( angle / 2 ); X = axis -> x * sin_a; Y = axis -> y * sin_a; Z = axis -> z * sin_a; W = cos_a; 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( |X,Y,Z,W| ); cos_a = W; angle = acos( cos_a ) * 2; sin_a = sqrt( 1.0 - cos_a * cos_a ); if ( fabs( sin_a ) < 0.0005 ) sin_a = 1; axis -> x = X / sin_a; axis -> y = Y / sin_a; axis -> z = Z / sin_a; 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 ) X = sin_a * cos_lat * sin_long Y = sin_a * sin_lat Z = sin_a * sin_lat * cos_long W = cos_a WARNING: There might be a problem in this code. An alternative is the code snipped given in [Q60: How do I convert Euler rotation angles to a quaternion?"]. 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_a = W; sin_a = sqrt( 1.0 - cos_a * cos_a ); angle = acos( cos_a ) * 2; if ( fabs( sin_angle ) < 0.0005 ) sin_a = 1; tx = X / sin_a; ty = Y / sin_a; tz = Z / sin_a; latitude = -asin( ty ); if ( tx * tx + tz * tz < 0.0005 ) longitude = 0; else longitude = atan2( tx, tz ); if ( longitude < 0 ) longitude += 360.0; WARNING: In this code might still be a problem. Please let me know what it is and how to fix this. 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 ); } The following more or less comes from: http://vered.rose.utoronto.ca/people/david_dir/GEMS/GEMS.html //Pitch->X axis, Yaw->Y axis, Roll->Z axis Quaternion::Quaternion(float fPitch, float fYaw, float fRoll) { const float fSinPitch(sin(fPitch*0.5F)); const float fCosPitch(cos(fPitch*0.5F)); const float fSinYaw(sin(fYaw*0.5F)); const float fCosYaw(cos(fYaw*0.5F)); const float fSinRoll(sin(fRoll*0.5F)); const float fCosRoll(cos(fRoll*0.5F)); const float fCosPitchCosYaw(fCosPitch*fCosYaw); const float fSinPitchSinYaw(fSinPitch*fSinYaw); X = fSinRoll * fCosPitchCosYaw - fCosRoll * fSinPitchSinYaw; Y = fCosRoll * fSinPitch * fCosYaw + fSinRoll * fCosPitch * fSinYaw; Z = fCosRoll * fCosPitch * fSinYaw - fSinRoll * fSinPitch * fCosYaw; W = fCosRoll * fCosPitchCosYaw + fSinRoll * fSinPitchSinYaw; } The following might also work: QUATERNION qx = { cos(pitch/2f), sin(pitch/2f), 0, 0 } QUATERNION qy = { cos(yaw/2f), 0, sin(yaw/2f), 0 } QUATERNION qz = { cos(roll/2f), 0, 0, sin(roll/2f) } quaternion_multiply( &qt, &qx, &qy ); quaternion_multiply( &q, &qt, &qz ); In Java looks like this: public static void setQ( Quat4f q, float pitch, float yaw, float roll ) { Quat4f qx = new Quat4f((float) Math.cos(pitch/2f), (float) Math.sin(pitch/2f), 0, 0); Quat4f qy = new Quat4f((float) Math.cos(yaw/2f), 0, (float) Math.sin(yaw/2f),0); Quat4f qz = new Quat4f((float) Math.cos(roll/2f), 0, 0, (float) Math.sin(roll/2f) ); Quat4f qt = new Quat4f(); qt.set( qx ); qt.mul( qy ); qt.mul( qz ); q.set( qt ); } 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, MF, 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 */ Q63. How do I use quaternions to rotate a vector? ------------------------------------------------------------------------------ A rather elegant way to rotate a vector using a quaternion directly is the following (qr being the rotation quaternion): -1 v' = qr * v * qr This can easily be realised and is most likely faster then the transformation using a rotation matrix. Discuss this article in the forums
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