## Re: [PanoToolsNG] Re: equirectangular projection

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• ... Santiago, Por favor :-) Later, show us a link to it. abraços AYRTON ... -- ... + 55 21 9982 6313 http://ayrton360.com | http://vrfolio.com |
Message 1 of 8 , May 12, 2009
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On Tue, May 12, 2009 at 7:50 AM, Santiago Ribas <santiago@...>wrote:

> Thanks!!!, I have tested your equations, and they are working great.
> I use both W=4000 and H=8000 and it gives me the result in pixels.
> Today I will build a sample.

Santiago,
Por favor :-)
Later, show us a link to it.

abraços
AYRTON

>
>
> Best regards
>
> Santiago
>
>
> On 09/05/11 19:31, "Erik Krause" <erik.krause@...> wrote:
>
> >
> >
> >
> >
> >
> > Santiago wrote:
> >
> >> > I have a raw 8000x4000 photo, and I know the azimuth and height of the
> >> > sun during the day, and during the year, all I need is a formula to
> >> > convert my solar height (degrees) to equirectangular height in
> pixels.
> >
> > It is very easy, since equirectangular coordinates map directly to
> > height and azimuth (given the horizon is level).
> >
> > In your example given the center of the equirect points south the left
> > and right edge are north, 2000 pixels from left is east, 4000 pixels
> > from left is south and 6000 pixel from left is west.
> >
> > A sun height of 0° is on the horizon and hence in the middle of the
> > panorama, 45° height is half way between the horizon and the upper edge.
> >
> > If in your sun coords south is 180° and you use standard image
> > coordinate system with (0,0) in the upper left corner the formula for
> > (X,Y) equirect coordinates, (w,h) equirect dimensions, (azimuth,height)
> > sun coordinates is:
> >
> > X = azimuth/360° * w
> > Y = (height-90°)/180° * h
> >
> > best regards
>
>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> --
>
>
>
>

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