Re: [Algorithms] How to derive transformation matrices

[Martin G. <bz...@wi...>]

-----Original Message-----
From: Ron Levine <ro...@do...>
To: gda...@li...
<gda...@li...>
Date: 24 Þëè 2000 ã. 07:11
Subject: Re: [Algorithms] How to derive transformation matrices


Ron, I must admit, your flamethrower is impressive :)

>Here lies the crux of the derivation of the formula for rotation in a
>plane.  Of course, you leave out some essential details--if you start
>out this way, then to expand on your glib "can be derived",  you need
>three trigonometric identities and a mess of algebra.  And how do you
>derive the three trig identities?

We all have learned from elementary math how to express rotations with polar
coordinates and expressing rotation with them. Expanding the cos(a+b) and
sin(a+b) should be no trouble for you, I know ;)


>>As far as I remember (did it too long ago) shearing is derived in the same
way
>>but the equations are different.
>>
>
>You phrases  "the same way", and "the equations are different" provide
>an excellent example of "whistling in the dark".   To have a
>derivation, you would have to start with a definition of "shear".

You obviously missed the words "as far as I remember". I don't know what
Lorrimar ment with 'shear'. But I did a guess 'cause I wanted to be a bit
useful. Shear (as I remember it) is a combination of rotation and scale (no,
I'm not giving an exact mathematical definition, spare me).


>
>>P.S. complex rotation transformations ...
>>
>

>Whatever do you mean by "complex rotation transformations" anyway?
>The fact is that  no rotation is more complex than any other rotation
>(with the exception of the identity, of course).  Every rotation (not
>the identity) has a unique axis and angle.

>
>True, when the axis of the rotation happens to be a coordinate axis in
>the coordinate system you happen to be using, then the matrix of the
>rotation with respect that coordinate system is easy to derive from
>the plane rotation formulas, as you have done above in the case that
>the axis is the y-axis.  But what about when the axis of the rotation
>is not a coordinate axis in the coordinate system you happen to be
>using?  Your putative derivation description

Ok, ok, mea culpa, I had in mind series of rotations around coordinate axes.
Excuse my poor English.
Of course rotating around an arbitrary axis vector does involve more math.

>
>>...are obtained by expressing them as
>>sequence of simple rotations in XY, XZ or YZ planes.
>
>is another crock of whistling in the dark.  No one has ever given a
>derivation of the general axis rotation in this way.  It would be
>extremely difficult if not impossible. You would first have to give a
>general derivation of an Euler angle decomposition of a rotation with
>given (non-coordinate) axis and angle, (itself a difficult problem
>with many levels of conditional branches needed to get around the
>singularities inherent in Euler angles).  Then algebraically compose
>those Euler rotations, then try somehow to algebraically simplify that
>holy mess in all of its ramified conditional branches.
>
>No, that won't work.  But, there are two ways to derive the matrix for
>a rotation whose axis is not a coordinate axis in the system to which
>the matrix refers.
>
>(1) If C0 is the coordinate system with respect to which you want the
>matrix of the rotation, find a rotation matrix T  that maps, say, the
>y-axis of C0 to the axis of the given rotation.  (Not hard, can be
>done with two Euler rotations, avoiding singularities).  Then the axis
>of your given rotation IS a coordinate axis in the coordinate system
>C1=T C0.  Write down the matrix R for rotation about the y-axis, as
>you have done above, and use the well known similarity transformation
>that we all learned in elementary linear algebra, which tells how to
>get the matrix of a given transformation with respect to a given
>coordinate system from the matrix of the same transformation with
>respect to another coordinate system.  The result for the matrix of
>the general axis rotation  is T R T'  (where T' is the transpose of T)
>or T' R T, depending on whether we are OperatorOnRightists or
>OperatorOnLeftists. Then simplify algebraically.
>
>But, that is still too messy for me, and far too coordinate-dependent.
>I far prefer the following derivation, which does not involve anything
>about similarity transformations, but just vector geometry (which I
>regard as a more basic subject) and only the simplest idea about the
>meaning of a transformation matrix. Namely,
>
>(2) Let A be a unit vector along the axis of rotation and let a be the
>rotation angle.  Let V be ANY given vector.  I am going to write down
>a vector formula that lets you compute the result V' of rotating V
>through angle a about axis A, using just the elementary vector
>operations: addition, multiplication by scalars, dot and cross
>products.   Then, you don't need any matrix whatsover to compute the
>effect of applying the rotation to any vector, you can just use this
>vector formula.  But as a bonus, I'll remind you how to get the matrix
>from this vector formula, anyway.
>
>You get the vector formula by resolving V into two components, one
>parallel to A and one perpendicular to A.  Because the rotation, like
>all linear transformations,  preserves vector sums, we only need to
>compute the effect of the rotation on the two components, then
>recombine the rotated components. Happily, the component parallel to A
>is clearly mapped to itself, and the rotation of the perpendicular
>component is easy to get, because it remains in the plane
>perpendicular to A.   When you carry this out and combine the terms
>appropriately you get
>
> V' = cos(a)V + (A dot V)( 1- cos(a)) A + sin(a) A cross V
>
>(To carry out the details of this derivation, you do need one nice
>vector identity, namely the identity
> A cross (B cross C) = (A dot B) C - (A dot C) B,
>whose derivation is a little messy algebraically, but which you will
>find in the same chapter in your elementary calculus book in which the
>vector cross product is first introduced.)
>
>Note that this formula expresses the rotated vector V' as a linear
>combination of V , A , and the most natural vector that you can
>construct from V and A but is linearly independent of V and A, ,
>namely A cross V.
>
>Actually, this formula is the basis for the formalism of representing
>rotations by quaternion conjugation, but I digress, let's leave
>quaternions out of this.
>
>Now, what about the matrix?  Well, whenever you are able to compute
>the effect of a linear transformation on any given vector, you can get
>the matrix of that transformation with respect to any given coordinate
>system by applying it to the basis vectors of that coordinate system.
>THAT is fundamental to the very  concept of using matrices to
>represent linear transformations.  So, supposing that you have the
>components of A with respect to some coordinate system of which A is
>not necessarily an axis vector,  just apply that formula to the basis
>vectors (1,0,0), (0,1,0), and (0,0,1) and the resulting nine numbers
>are just the elements of the matrix you seek. Again, whether these are
>arranged in rows or columns depends on whether you are
>OperatorOnRightist or OperatorOnLeftist.
>
>What I am trying to stress here is an idea that is foreign to most
>graphics people, but which is extremely important to understanding
>what is going on geometrically, an idea that has been mentioned
>recently in another thread in this list. It is the idea that, however
>much you need them in the end for computing, coordinate systems are an
>enormous hindrance to understanding intrinsic geometry.
>
>God did not attach a canonical coordinate system to the universe for
>us all to use.  And that's good,  because we are free to use whatever
>coordinate system we find most convenient in any problem, and
>frequently we find it convenient to work in several different
>coordinate systems in a single problem!  The only formulas that are an
>aid to understanding, as opposed to  coding, are those that are
>manifestly coordinate invariant.  My vector rotation formula above is
>manfestly coordinate invariant.  Transformation matrices are highly
>coordinate dependent, the opposite of  coordinate invariant.  A given
>transformation, with definite intrinsic geometric meaning, has many
>different matrix representations, one for each different coordinate
>system you might wish to use in any problem.  My manifestly invariant
>vector formula given above contains within it all possible different
>coordinate-dependent matrix representations of the rotation in
>question.
>
>Therefore, all these "tutorials" in graphics API manuals and game
>development web sites,  which present the matrices for coordinate-axis
>rotations and, leaving it at that,  claim to have dealt with
>rotations, are perpetrating a serious crime upon the
>eager newbie who seeks to understand what is a rotation.


Wow, you won't put us all in jail if you were in charge, will you? :)))

>
>Having dealt with rotations, what about shear?  I'll leave that aside
>until someone asks the question using an intrinsic geometric
>definition of "shear" (And it does exist).
>
>P.S. Quaternions, in their role representing rotations, are just as
>coordinate dependent as matrices are.


Ron, if you intend to continue this crash course in expressing
rotations/shearing/etc, I will follow it with close attention. We all can
learn from you, so please be more forgiving with 'newbies' like me.


Martin