# [Matrices] Movement along direction vector

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Hi, let's say I've got a situation like the following, where I have an object in 3d space, whose position and rotation is defined by a 4x4 matrix: I need to move this object accordingly to the direction it's facing, like illustrated here: What is the correct way to manipulate the matrix? Right now I am doing this with spaghetti code that certainly can't be the "elegant" solution. Thanks in advance, a.

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I don't know if the 4x4 transformation matrix is needed for that. That is what it is, right?

If you've got any way to represent the position of the object as a vector, just add the direction vector with it? That's what I'd do..

Mmmm.. spaghetti!

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If the translation vector is known, then a corresponding transformation matrix can be build of it. How this is done depends on whether you use column or else row vectors, and at which index the homogeneous co-ordinate is located. E.g. for column vectors and structure [ x y z w ], a translation by [ x y z ] will be given as
[ 1 0 0 x ][ 0 1 0 y ][ 0 0 1 z ][ 0 0 0 1 ]

To combine this translation matrix with a given transformation matrix, you'll need to multiply them. The correct way if using column vectors is
T * M
where T denotes the new translation and M the transformation matrix so far. While this is the correct mathematical way, some "spaghetti" code can speed up the product, i.e. you exploit the fact that you know that T denotes a translation, and hence know which scalars of the resulting matrix are affected.

Now, if you want to isolate the facing vector from the current transformation matrix, you can use a "pick" vector. Assuming that z is the (local) direction of viewing, then the pick vector would be
p := [ 0 0 1 0 ]T
when using the above example structure. Then the product
M * p
would pick out the 3rd column vector from the matrix. It would obviously be more efficient to read the vector out without performing a matrix vector product. But be aware that the picked vector will contain also some factor if scaling was performed. If so, you need a more elaborate way of so-called "matrix decomposition".

As usual, apply the transpose operation to above formulas if you deal with row vectors instead of column vectors. However, although there is a general way of doing things, exploiting a-priori knowledge can be used to speed up operations. Whether this weights more than elegant code is a matter of the use cases.

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Thanks for the detailed answers. Now it's my time to shed out more details so that you can better understand "where" I am now.. and what do I mean by spaghetti ;)

I am using Direct3d9, coming from OpenGL.
In the engine I am converting, I did the worst thing possible: 3d entities position and rotation where stored as euler angles. Thus my code to "move ahead" a certain entity was and IS this:

FastSinCos( rotation.x*(pi/180), cos_x, sin_x );
FastSinCos( rotation.y*(pi/180), cos_y, sin_y );
FastSinCos( rotation.z*(pi/180), cos_z, sin_z );

g:=cos_x*sin_y*cos_z+sin_x*sin_z;
h:=cos_x*sin_y*sin_z-sin_x*cos_z;
j:=cos_x*cos_y;

position.x:=position.x+G*Spin; <-- Spin is the amount of movement
position.y:=position.y+H*Spin;
position.z:=position.z+J*Spin;

My first rough porting to Matrix math, simply constructs a D3DMatrix out of the position[x,y,z] and rotation[x,y,z] triplets. So maybe I need to take a step back and store position/rotation differently from the beginning? If so, what do you suggest?

Thanks a lot.

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Well, if you have your transformation matrix, the orientation is probably stored like this:
................object position/translation (P)
................|
X-> [ 1 0 0 x ]
Y-> [ 0 1 0 y ]
Z-> [ 0 0 1 z ]
....[ 0 0 0 1 ]

If X,Y,Z hold the orientation, and say, Y is your direction vector then, to move in that direction you'd do:

P = P + speed*Y // P={x,y,z} and Y={0,1,0}

Where 'speed' would be the scalar rate of displacement you want. I guess you can manipulate the matrix in the appropriate places (4th column here), as described in the previous post.

edit: some kind of formatting error(...)

[Edited by - glopen on November 18, 2008 5:58:19 AM]

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