- Hans Nyberg wrote, On 01.11.2008 10:41 Uhr:
> While doing experiments with Pano2VR I discovered that the last Mac version defaults to Pi

This is chosen by design, and there are two reason:

> conversion of cubefaces in the convert input option but when you do movies and flash the

> default size is width/4.

>

> This might be very confusing for people who are unaware of this.

* The tile size width/Pi is practical for converting between formats to

avoid losing quality but for outputting a panorama width/4 is enough

because then the optimal resolution is in the middle between center and

the border and this is imho a good compromise. With width/Pi it is not

possible to produce a sharp edge near the cube face corners. To produce

the best possible output the equirectangular would need to have a width

of Pi*sqrt(2) ~ 4.443 the tile size. Of course it would loose quality in

the center but then it would be possible to produce a sharp line near

the cube face edge. So from the view as a demand to the panorama source

image as "best output possible" the correct setting would be width/4.443.

* Pano2VR has a lot of unexperienced/new panographers as users and they

don't understand the concept of tile size. They reduce the input image

size instead (for example in Photoshop) to produce smaller file sizes.

So a width/4 gives good results and is easier to understand then

width/Pi or width/(Pi*sqrt(2)).

--

MfG,

Thomas - Hi Thomas,

But, if the default value is width/4 you are using this figure:

http://www.devalvr.com/fiero/equicube2.jpg , loosing a lot of

information in the conversion.

If you take the output equirect from PTGui, all values superior to

3.14 in the divisor are loosing information in the conversion.

regards!

fiero

--- In PanoToolsNG@yahoogroups.com, Thomas Rauscher <yahoo@...> wrote:

>

> fierodeval wrote, On 01.11.2008 21:59 Uhr:

> > Yes, I understand. But when you speak about the "sqrt(2)", you are

> > speaking about the figure in the right of this image

> > http://www.devalvr.com/fiero/equicube.jpg . Te correct figure is in

> > the left and this figure is valid in the two directions.

> >

> > Then, if I understand correctly, you say that the figure for

> > equirect->cube conversion is the left image and for cube->equirect is

> > the right image?

>

> Yes. correct. Because the cube faces *could* contain more

information in

> the corners.

>

> If the source was a equirectangular with 1/Pi as factor for the cube

> faces size then it is safe to just use the left type of conversion

again

> because the pixels in corners are stretched by a factor of sqrt(2).

>

> If the source images are cube faces, for example of a rendering from a

> 3D software, then you need to use the conversion on the right side to

> avoid losing resolution/information.

>

> The factor Pi*sqrt(2) would also be true if the equi->cube was executed

> with a factor smaller then 1/Pi*sqrt(2) to get the maximum out of the

> cube faces. Another case would be if you recover a equirectangular

from

> a QTVR where you don't know the original resolution.

>

> --

> MfG,

> Thomas

>