wwp@yahoogroups.com:
Re: Image width / 96 and height
thinnairstudios 2005-Sep-29 22:42:00
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
I stitch 6+Top+Bottom and output an equirectangular image with
dimensions of 4000x2000. A 2:1 ratio
Next is the resolution of a given panorama
It states the measurement is the "circumference"
Since the image is 4000 wide that is the circumference and this
yields an image at 11.1 angular resolution.
At this point I have lost half of my available angle of resolution
during the output process for an image at 4000x2000
Pixels needed for a given angular resolution.
Inputting the calculated angular resolution of 11.1 yields a value of
3996 for a Equirectangular Spherical image.
This is 4 pixels off from 4000. So this tells me that my 4000x2000
image has an angular resolution of about 11.1 as above. The
calculation checks out. Not sure why there is a 4 pixel difference
but I assume it is from rounding errors??
Now, "if" I wanted to keep the maximum angular resolution my lens can
produce the Equirectangular Spherical image would need to be 7272
wide. Quite large.
So now I have this 4000x2000 Equirectangular Spherical image and I
want to display it at the images angular resolution of 11.1
Using Quicktime the 4000x2000 Equirectangular Spherical image will
be converted into a cube for display. So now I need to know the size
of the cube face to achieve the desired angular resolution of 11.1
Pixels Needed for a Given Angular Resolution:
I now calculate the size of the cube face, which yields 1273 linear
pixels. Since the cube face is square 1273x1273=1,620,529 total
pixels per cube face. 6 cube faces yields a total pixel count of
9,723,174 total pixels for the entire cube.
So at this point, to achieve the maximum angular resolution that the
Equirectangular Spherical image has at 4000x2000 which is 11.1, I
would need to set a cube face of 1273x1273.
This leads me to believe that when choosing a display size of the
panorama using Quicktime I need to set the viewer size at 1273 x
1273, which is about the right size for a full screen display. At
this size I would have an image displaying a 1 to 1 pixel ratio,
which should yield an on screen maximum resolution.
If the image is scaled to browser size then we loose or gain image
resolution due to the monitors ability to display at 72 ? 96 dpi.
But we are assuming a full screen image will be displayed at full
screen any way or else we would pick another display size.
Performing the same calculations again but on a Equirectangular
Spherical image at 2000x1000 we get an angular resolution of 5.6 and
a cube face of 643 linear pixels. As you can see this turns out to
be almost one half of the 4000x2000 image. So everything is working
out nicely so far.
Taking the smaller cube face of 643 tells use to display at 643 x
643.
But what happens to the image when displayed at a 4 to 3 ratio such
as 400x300 (300/400=.75 and .75 of 400 is 300) or in this case 643x482
It seems to me that one should be displaying the image in a 1 to 1
ratio to keep all the angular resolution possible. If you change the
ratio from 1 to 1 to 4 to 3 then the image is being scaled within the
viewer and not utilizing all the angular resolution possible. Or is
it that the viewer is scaling the angular resolution as well and
therefore delivering an equivalent angular resolution at the display
size? This is the part where I am confused. We have a cube face of
643x643. If I display at 643x482 does this do anything to the quality
of the image? Or is the image just being viewed through a "window"
and the image is being rendered at the same angular resolution?
Ok, I "think" I got all this correct. But I am sure I have
misunderstood something. There is no way I could have possibly
figured this out on my own. LOLOLOL.
Fellow camera geeks please sound off.
Troy Ward
--- In #removed#, Markus Altendorff <#removed#> wrote:
> Dave 360texas.com wrote:
> > So what I am gathering.. for a proper image width should
> > be divided by 96, and be a 2:1 ratio.
> >
> > This set width numbers are divisable by 96 and /2 for
> > 2:1. Of course we have to watch the final MOV file size
> > too...
>
> Note that this width rule only applies for Cylinders - if
> you're using a 2:1 ratio, most likely it's not cylindrical,
> but spherical! There's a very thorough explanation of the
> rules and resolution calculations on the homepage of Ken
> Turkowski:
>
> http://www.worldserver.com/turk/quicktimevr/panores.html
>