wwp@yahoogroups.com:
Re: Image width / 96 and height
Dave 360texas.com 2005-Sep-30 14:47:00
Troy - I think I understand what you are saying.
I use a Canon 20d with a Sigma 8 and a portrait orientation.
Image size of 3504h x 2336w and 4 images/360?
Equirectangular stitched image 5000 x 2500 2:1
Angular Resolution 13.9 pixels/degree
Panorama Image Format Perspective (12mm) 1594 pixels cube face
1594x1594=2,540,836 total pixels cube face
2,540,836 x 6 cube faces = 15,245,016 total pixels
So for converting it to a QTVR to achieve a 1:1 I should use 1594.
I suspect that to get a smaller total file size
I might use 1/2 1594 or 797 cube face or TILE SIZE.
Believe it or not... this makes sense. Thank you for taking the time
walk through the exercise.
Regarding your comment about the proper viewer dimensions. I think
the viewer should be about 30% of the image height. 2500h x 30% =
750 pixels high. What viewer height do find provides best rendering?
/s/
Dave [at]360texas [dot]com
--- In #removed#, "thinnairstudios" <#removed#> wrote:
> Hi Markus, and to everyone here :)
>
> Very interesting information at the link. Just the kind of stuff I
> like to geek out on. LOL
>
> I get the "pixels per degree" definition. It is kind of
like "circle
> of confusion" definition as related to the focal plane.
>
> A given lens has a pixels per degree of resolution.
> I use the Nikkor 10.5 full frame fisheye. I used the "Resolution
of a
> Fisheye Lens" calculator.
> 2000 ( width of image the camera takes in vertical orientation)
> 99 deg (FOV in vertical orientation)
> = 20.2 (Angular Resolution(pix/degree))
> So I start with images that have a 20.2 angular resolution
< SNIP FOR SPACE>