[Adonthell-commits] CVS: adonthell/doc/devel map.dxt,1.1.2.1,1.1.2.2

[Alexandre Courbot <address@hidden>]

Update of /cvsroot/adonthell/adonthell/doc/devel
In directory subversions:/tmp/cvs-serv9639/doc/devel

Modified Files:
      Tag: Branch_road_to_0-4
        map.dxt 
Log Message:
UPDATED map engine specifications
ADDED Python callback functionality to the input system
RE-ENABLED (at least a little bit) the Python wrappers


Index: map.dxt
===================================================================
RCS file: /cvsroot/adonthell/adonthell/doc/devel/map.dxt,v
retrieving revision 1.1.2.1
retrieving revision 1.1.2.2
diff -C2 -r1.1.2.1 -r1.1.2.2
*** map.dxt     10 Aug 2002 20:46:33 -0000      1.1.2.1
--- map.dxt     11 Aug 2002 20:57:44 -0000      1.1.2.2
***************
*** 80,87 ****
--- 80,91 ----
  same terms that are used in the implementation:
  
+ <b>Points</b>
+ 
  The most basic thing that will be used is the point. I won't make the
  offence to explain what is a point - we will simply
  describe them with their X, Y (and Z) positions in the space.
  
+ <b>Vectors</b>
+ 
  The first real thing to deal with are vectors. Vectors are widely used
  in the map engine, as they represent every displacement of %objects,
***************
*** 91,95 ****
  (it's \e norm, noted \f$||u||\f$), is equal to
  
! \f$||u|| = \sqrt{ux^2 + uy^2 + uz^2}\f$
    
  In most operations however, we don't need the exact norm of a vector
--- 95,99 ----
  (it's \e norm, noted \f$||u||\f$), is equal to
  
! \f$||u|| = \sqrt{u_x^2 + u_y^2 + u_z^2}\f$
    
  In most operations however, we don't need the exact norm of a vector
***************
*** 107,114 ****
  \image html vectors1.png "Here, v2 and v3 are pointing in the same direction 
than v1 (the invert-clockwise angle between them in < 180°)"
  
! \image latex vectors1.eps "But here, v2 and v3 are pointing in the opposite 
direction of v1 (the invert-clockwise angle between them in >= 180°)" width=10cm
! \image html vectors1.png "But here, v2 and v3 are pointing in the opposite 
direction of v1 (the invert-clockwise angle between them in >= 180°)"
  
  
  
  Lines, as we hear them, are simply a vector which is given a starting
--- 111,146 ----
  \image html vectors1.png "Here, v2 and v3 are pointing in the same direction 
than v1 (the invert-clockwise angle between them in < 180°)"
  
! \image latex vectors2.eps "But here, v2 and v3 are pointing in the opposite 
direction of v1 (the invert-clockwise angle between them in >= 180°)" width=10cm
! \image html vectors2.png "But here, v2 and v3 are pointing in the opposite 
direction of v1 (the invert-clockwise angle between them in >= 180°)"
! 
! In easier words, <i>v points in the same direction than u</i> can be
! interpreted as <i>when having the same starting point, v is on u's
! left</i>, and <i>v points in the opposite direction of u</i> can be
! heard as <i>when having the same starting point, v is on u's
! right</i>.
! 
! The cross product of two vectors \e u and \e v (noted \f$||u.v||)\f$
! can help us knowing if two vectors point in the same
! direction. Indeed:
! 
! \f$||u.v|| = ||u|| * ||v|| * \sin(\theta)\f$
! 
! \f$||u.v|| = u_x * v_y - u_y * v_x\f$
! 
! Therefore:
! 
! \f$\sin(\theta) = \frac{u_x * v_y - u_y * v_x}{||u|| * ||v||}\f$
! 
! The sinus of an invert-clockwise angle will be negative if the angle
! is between 0 and 180 degrees (the vectors are pointing in the same
! direction), positive otherwise (the vectors are pointing in opposite
! directions).
  
+ For the engine, we will only need the sign of
+ \f$\sin(\theta)\f$. Therefore, we can get rid of the division, and
+ directly consider the sign of the cross-product to know if two vectors
+ point in the same direction or not.
  
+ <b>Lines</b>
  
  Lines, as we hear them, are simply a vector which is given a starting
***************
*** 119,125 ****
  check wether AB and CD intersect. If they do, then:
  
! \f$Ax + r(Bx - Ax) = Cx + s(Dx - Cx)\f$, and
  
! \f$Ay + r(By - Ay) = Cy + s(Dy - Cy)\f$
  
  for \e r and \e s between 0 and 1.
--- 151,157 ----
  check wether AB and CD intersect. If they do, then:
  
! \f$A_x + r(B_x - A_x) = C_x + s(D_x - C_x)\f$, and
  
! \f$A_y + r(B_y - A_y) = C_y + s(D_y - C_y)\f$
  
  for \e r and \e s between 0 and 1.
***************
*** 127,136 ****
  Finding \e r and \e s is as simple as solving:
  
! \f$r = \frac{(Ay - Cy)(Dx - Cx) - (Ax - Cx)(Dy - Cy)}{(Bx - Ax)(Dy - Cy) - 
(By - Ay)(Dx - Cx)}\f$ and
  
! \f$r = \frac{(Ay - Cy)(Bx - Ax) - (Ax - Cx)(By - Ay)}{(Bx - Ax)(Dy - Cy) - 
(By - Ay)(Dx - Cx)}\f$.
  
  Note that the denominator is the same - we can save one division by
! setting a variable to \f$frac{1}{denom}\f$ and using multiplications
  instead of divisions while calculating \e r and \e s. Moreover,
  pre-calculating it is mandatory anyway, as if it is egal to 0 the
--- 159,168 ----
  Finding \e r and \e s is as simple as solving:
  
! \f$r = \frac{(A_y - C_y)(D_x - C_x) - (A_x - C_x)(D_y - C_y)}{(B_x - A_x)(D_y 
- C_y) - (B_y - A_y)(D_x - C_x)}\f$ and
  
! \f$r = \frac{(A_y - C_y)(B_x - A_x) - (A_x - C_x)(B_y - A_y)}{(B_x - A_x)(D_y 
- C_y) - (B_y - A_y)(D_x - C_x)}\f$.
  
  Note that the denominator is the same - we can save one division by
! setting a variable to \f$\frac{1}{denom}\f$ and using multiplications
  instead of divisions while calculating \e r and \e s. Moreover,
  pre-calculating it is mandatory anyway, as if it is egal to 0 the
***************
*** 144,149 ****
  point \e P with:
  
! \f$Px = Ax + r(Bx - Ax)\f$
! \f$Py = Ay + r(By - Ay)\f$
  
  So if used as if, this algorithm will require one float division and four
--- 176,181 ----
  point \e P with:
  
! \f$P_x = A_x + r(B_x - A_x)\f$
! \f$P_y = A_y + r(B_y - A_y)\f$
  
  So if used as if, this algorithm will require one float division and four