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Revision 3ea19d1c

Added by Leszek Koltunski about 8 years ago

rearrange comments

View differences:

src/main/res/raw/main_vertex_shader.glsl
199 199
//    Here we have to have X'(1) = 2(mx-flx)-vx which is positive <==> vx<2(mx-flx). We also have to
200 200
//    have vy<2(my-fly) so that Y'(t)>0 (this is a must otherwise we have local loops!)
201 201
//    Similarly for the Fr--->M' part of the curve we have the same equation except for the fact that
202
//    this time we have to have X'(1)<0 so now we have to have vx>2(mx-flx).
202
//    this time we have to have X'(1)<0 so now we have to have vx>2(mx-frx).
203 203
//
204 204
//    If we are stretching the left or right edge of the bitmap then the only difference is that we
205 205
//    have to have (X'(1),Y'(1)) = (0,+-c) with + or - c depending on which part of the curve
......
210 210
//
211 211
//    If we are dragging the top edge:    
212 212
//
213
//    Now point (x,u_objD.x) on the top edge will move by vector (X(t),Y(t)) where those functions
214
//    are given by (*) and t =  x < dSx ? (u_objD.x+x)/(u_objD.x+dSx) : (u_objD.x-x)/(u_objD.x-dSx)
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//    Then point (x,h/2) on the top edge will move by vector (X(t),Y(t)) where those functions are
214
//    given by (*) and t =  x < dSx ? (w/2+x)/(w/2+dSx) : (w/2-x)/(w/2-dSx)    (-w/2 < x < +w/2 !)
215 215
//    (this is 'vec2 time' below in the code).
216
//    Any point (x,y) will move by vector (a*X(t),a*Y(t)) where a is (y+u_objD.y)/(2*u_objD.y)
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//    Any point (x,y) will move by vector (a*X(t),a*Y(t)) where a is (y+h/2)/h
217 217
  
218 218
void deform(in int effect, inout vec4 v)
219 219
  {

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