thanks for helping On Feb 18, 2:22 am, kfj <[email protected]> wrote: > On 16 Feb., 18:56, Dan <[email protected]> wrote: > > > sorry,,i forgot to mention that I was talking about forward transform > > and the image was the source image. and most of time they do inverse > > transform. I just went for forward just to understand. It should work > > atleast (even if it leaves some pixels blank, thats not the concern > > right now), I dont want to go for interpolation. Thing is just it > > should work, quality i will check later on. I was reading the source > > code of libpano and stuck again. Can you tell me what these lines are > > doing in math.c of libpano library ? > > > what is sphere_tp_erect doing? > > sphere_to_erect converts from spherical coordinates to equirectangular > projection, says it's name - and that's about all I know. > > > > > > > > > > > and these lines in it? :: > > phi = x_dest / distanceparam; > > theta = - y_dest / distanceparam + PI / 2; > > if(theta < 0) > > { > > theta = - theta; > > phi += PI; > > } > > if(theta > PI) > > { > > theta = PI - (theta - PI); > > phi += PI; > > } > > > s = sin( theta ); > > v[0] = s * sin( phi ); // y' -> x // sin(theta)*sin(phi) > > v[1] = cos( theta ); // z' -> y // > > cos(theta) > > > r = sqrt( v[1]*v[1] + v[0]*v[0]); > > theta = distanceparam * atan2( r , s * cos( phi ) ); > > > *x_src = theta * v[0] / r; > > *y_src = theta * v[1] / r; > > ... I fear you'll have to figure it out yourself. I suppose it's > reasonably sraightforword trigonometry, but I haven't looked into the > details - I'm more interested in using the transforms than > understanding their inner workings, so I can't really help you there. > > > Where do these formulas come from? Isn't there atleast some > > documentation of this libpano anywhere ??? > > you'll have to do some research. Try > > http://wiki.panotools.org/Projection > > for starters and see where it takes you. It's a complex topic. Trying > to figure out libpano isn't for the fainthearted. If you want it > simple and well-documented, this is probably not where you want to > look. Best of luck > > Kay
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