- hello,

can somebody smarter than me tell me what is the relationship between the

horizontal field of view and the width size in the ptadjust photoshop plugin

from epaperpress when extracting a rectilinear from an equirectangular.

I'd like to extract a 90 degree horizontal view but then have a bit extra

either side please.

regards

mick

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This mail sent through http://www.ukonline.net - crane@... wrote:
> hello,

There has been several discussions on this list as to what should be the

> can somebody smarter than me tell me what is the relationship between the

> horizontal field of view and the width size in the ptadjust photoshop plugin

> from epaperpress when extracting a rectilinear from an equirectangular.

> I'd like to extract a 90 degree horizontal view but then have a bit extra

> either side please.

>

> regards

>

> mick

>

>

optimal dimensions for cube faces when extracting them to make a QTVR.

For 90 deg use width / pi for optimal, to ensure no loss of detail width / 3

What you want to do is keep the same (or a little better) pixels per

degree as the original. For more details see Resolution of Various

Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html

--

Jim Watters

jwatters @ photocreations . ca

http://photocreations.ca - Quoting Jim Watters <jwatters@...>:

> What you want to do is keep the same (or a little better) pixels per

That doesn't explain very clearly, assumes a lot of prior understanding.

> degree as the original. For more details see Resolution of Various

> Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html

eg:

"

The focal length is defined as the distance to the imaging surface at the

center of projection.

For a cubic panorama, this is the distance to the center of one of the faces.

"

I have no idea how to calculate that.

regards

mick

> Jim Watters

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This mail sent through http://www.ukonline.net > Quoting Jim Watters <jwatters@...>:

yes I know i am replying to my own post.

>

> > What you want to do is keep the same (or a little better) pixels per

> > degree as the original. For more details see Resolution of Various

> > Panorama Formats http://www.worldserver.com/turk/quicktimevr/panores.html

>

> That doesn't explain very clearly, assumes a lot of prior understanding.

> eg:

> "

> The focal length is defined as the distance to the imaging surface at the

> center of projection.

>

> For a cubic panorama, this is the distance to the center of one of the faces.

>

> "

> I have no idea how to calculate that.

has anybody explained in terms that are readily understandable what happens to

the arithmetic from real world to sensor image to stitching to projecting onto

the monitor , in QT/ flash etc because I realize I do not have a clue as to

the maths involved and I would like to.

regards

mick

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This mail sent through http://www.ukonline.net- crane wrote:
> has anybody explained in terms that are readily understandable what

If you take fisheye images you have a more or less linear mapping

> happens to

> the arithmetic from real world to sensor image to stitching to

> projecting onto

> the monitor , in QT/ flash etc because I realize I do not have a clue as to

> the maths involved and I would like to.

between distances on the sensor and angles of view in the world. This

is: A certain pixel distance is always a certain angle distance, no

matter where on the sensor you measure.

For equirectangular images it is just the same: The width represents

360° the height 180°, that's why the aspect ratio always is 2:1. How to

calculate this is outlined on

http://wiki.panotools.org/DSLR_spherical_resolution

If it comes to rectilinear it gets a bit more complicated, since the

mapping is a tangent one. But it gets fairly easy if you do it in

geometry. Draw a circle. Now draw a surrounding square. Where the square

touches the circle the pixel density is equal to that one of the circle.

One cube face is the diameter of the circle wide. The diameter of the

circle (and hence the cube face) is circumference / pi.

If the square would not be surrounding but inside the circle, pixel

information would be lost, because the information formerly on the

circle must be compressed to a smaller distance, which means to less

pixels. If you map to equirect again (or map to an arbitrary view in the

viewer) this information can not be re-created.

Hence if you extract a cube face from an equirectangular and you don't

want to loose any information you need a cube face size of equirect

width / pi.

However, as Hans Nyberg demonstrated recently: If you remap from a

maximum size equirect to cube faces for a viewer (be it Flash, QT or

whatever) the best compromise between size and quality is at about 70%

of that size, since any bayer interpolated image can be reduced by about

70% with negligible loss of image quality. Hence you can divide equirect

width by about 4.4 to get the optimum cube face size.

--

Erik Krause

http://www.erik-krause.de