PanoTools NG:
Re: Re: equirectangular projection
Santiago Ribas 2009-May-12 12:52:01
Erik,
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.
Best regards
Santiago
On 09/05/11 19:31, "Erik Krause" <#removed#> 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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