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• Hi group, can someone tell me, which projection is used by QT, when showing cubic QTs? I m looking for the exact formula to calculate the position of a
Message 1 of 13 , Dec 28 8:56 AM
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Hi group,

can someone tell me, which projection is used by QT, when showing cubic
QTs? I'm looking for the exact formula to calculate the position of a
projected point on the viewers flat area from a given point on the
sphere (given by lon and lat).
Any suggestions?

Regards, H. Hensel
• ... It s simply 6 faces of a cube, each of which is a 90x90 degree rectilinear image.
Message 2 of 13 , Dec 28 10:09 AM
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On Thu, December 28, 2006 8:56 am, Hannes Hensel wrote:
> Hi group,
>
> can someone tell me, which projection is used by QT, when showing cubic
> QTs? I'm looking for the exact formula to calculate the position of a
> projected point on the viewers flat area from a given point on the
> sphere (given by lon and lat).

It's simply 6 faces of a cube, each of which is a 90x90 degree rectilinear
image.
• ... rectilinear ... Forget the cube. I need to get from the imaginary sphere to the flat image shown by the viewer. So, how can I calculate the position of a
Message 3 of 13 , Dec 28 11:31 PM
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--- In PanoToolsNG@yahoogroups.com, "Roger Howard" <rogerhoward@...>
wrote:

> It's simply 6 faces of a cube, each of which is a 90x90 degree
rectilinear
> image.
>
Forget the cube. I need to get from the imaginary sphere to the flat
image shown by the viewer. So, how can I calculate the position of a
given point on the surface of the sphere projected on the viewers flat
rectangular surface?
For example: Lets say I have the point +30Deg Lat and -60Deg Lon on the
sphere. And the QT Viewer is set to +20Deg Tilt, -50Deg Pan and 60Deg
FoW. What I need then, is the formula to calculate the position x, y of
this point in the viewer. Something like x=f(lat,lon,Tilt,Pan,FoW) and
y=f(lat,lon,Tilt,Pan,FoW). Or, if dLat=Tilt-Lat and dLon=Pan-Lon, there
must be x=f(dLat,dLon,Fow), y=f(dLat,dLon,Fow).
You know what I mean? ;-)
• ... I can t, and you shouldn t! ... It s not a sphere, it s a cube. ... First, think of it as a cube. What part of Cubic QuicktimeVR implies a sphere? You
Message 4 of 13 , Dec 29 7:40 AM
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On Dec 28, 2006, at 11:31 PM, Hannes Hensel wrote:

> --- In PanoToolsNG@yahoogroups.com, "Roger Howard" <rogerhoward@...>
> wrote:
>
>> It's simply 6 faces of a cube, each of which is a 90x90 degree
> rectilinear
>> image.
>>
> Forget the cube.

I can't, and you shouldn't!

> I need to get from the imaginary sphere to the flat
> image shown by the viewer.

It's not a sphere, it's a cube.

> So, how can I calculate the position of a
> given point on the surface of the sphere projected on the viewers flat
> rectangular surface?

First, think of it as a cube. What part of Cubic QuicktimeVR implies
a sphere? You asked about what projection it's using, I told you, and
you told me to forget about it!
• ... Ok, so lets ask me for the projection of the cube faces on the viewer area. This is a 2D view of a virtual 3D scene, and I want to know the projectiontype
Message 5 of 13 , Dec 29 11:25 PM
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--- In PanoToolsNG@yahoogroups.com, rogerhoward@... wrote:
> First, think of it as a cube. What part of Cubic QuicktimeVR implies
> a sphere?

Ok, so lets ask me for the projection of the cube faces on the viewer
area. This is a 2D view of a virtual 3D scene, and I want to know the
projectiontype and the formulas (to find x and y for given point with
lon and lat) for it. Can it be, that it is a simple central projection
(as used as perspective view in maya etc.)?

> You asked about what projection it's using, I told you, and
> you told me to forget about it!
You can forget it in this special manner. Because I surley know about
the cubical projection, but the question is for the 3D to 2D conversion
in the QT viewer. Maybe I didn't make myself clear enough here.
• ... Yes, these are simple central projections. To *generate* the cubic image file from spherical coordinates, the sphere is centrally projected onto six
Message 6 of 13 , Dec 30 1:00 AM
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--- In PanoToolsNG@yahoogroups.com, "Hannes Hensel" <h.hensel@...>
wrote:
>
> --- In PanoToolsNG@yahoogroups.com, rogerhoward@ wrote:
> > First, think of it as a cube. What part of Cubic QuicktimeVR
> > implies a sphere?
>
> Ok, so lets ask me for the projection of the cube faces
> on the viewer area. This is a 2D view of a virtual 3D scene,
> and I want to know the projectiontype and the formulas
> (to find x and y for given point with lon and lat) for it.
> Can it be, that it is a simple central projection
> (as used as perspective view in maya etc.)?

Yes, these are simple central projections.

To *generate* the cubic image file from spherical coordinates, the
sphere is centrally projected onto six planes, one for each of the
cube faces. In its exact form, this transformation involves lots of
trig for every pixel, to turn sines and cosines of lat/lon into pixel
coordinates. Sometimes computational shortcuts are taken, using
planar triangle approximations instead of doing the trig
independently for every pixel.

To *show* the cubic image file, each visible cube face is projected
to the viewing screen. This transformation involves no trig at all
on a pixel-by-pixel basis. It just projects an oblique plane (one of
the cube faces) onto the viewing plane (the display screen). This
can be implemented as a matrix multiply in homogeneous coordinates,
followed by perspective division.

I forget the exact formulas. You can find the general forms in any
good computer graphics text, but working out the details is always
fiddly.

Can I ask, why are you looking for these formulas?

--Rik
• ... Hey Rik, thanks so far. That goes into the right direction. I m only trying to get a bit deaper in that graphic stuff, to understand what is done there.
Message 7 of 13 , Dec 30 1:49 AM
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> Can I ask, why are you looking for these formulas?
>
> --Rik

Hey Rik, thanks so far. That goes into the right direction. I'm only
trying to get a bit deaper in that graphic stuff, to understand what is
done there. And I wonder, if there is a shortcut, to do those 2 steps
in one (sph > cube and cube > viewer = sph > viewer). Than it should be
a sphere with the vCamera (central point) as center and the viewing
plane inbetween (inner surface of the sphere BEHIND the viewing plane),
right? How is the zoom realized then? Is it only done by bringing the
viewing plane more to the camera (zoom out) or more to the sphere (zoom
in)?
• ... Hannes, Yes, zooming is equivalent to changing the distance between projection plane and center of projection. All of these operations can be combined into
Message 8 of 13 , Dec 30 11:34 AM
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--- In PanoToolsNG@yahoogroups.com, "Hannes Hensel" <h.hensel@...>
wrote:
> How is the zoom realized then? Is it only done
> by bringing the viewing plane more to the camera
> (zoom out) or more to the sphere (zoom in)?

Hannes,

Yes, zooming is equivalent to changing the distance between
projection plane and center of projection.

All of these operations can be combined into one step, for some
definition of "step". They are usually kept separated to simplify
thinking/coding and to move the cost of computations to where they
can be afforded.

A little more discussion...

In my previous reply, I followed your lead and wrote in terms of
lat/lon --> cubic image coordinates --> viewscreen coordinates.

But in fact, the transformations are usually run in the opposite
direction. For each viewscreen pixel, one computes the corresponding
cubic image coordinates. From cubic image coordinates, one computes
the corresponding lat/lon. From lat/lon, one computes the
corresponding coordinates within a "perfected" camera image (after
correction for lens distortion). From those, one computes
coordinates within the camera image -- the one that was fed to
Panorama Tools as a source image.

Pixel value interpolation typically occurs at several places in the
pipeline, once for every actual image that is constructed. Suppose,
for example, that you run Panorama Tools to stitch multiple camera
images into a single equirectangular image, then run PanoCUBE to
generate a cubic file, then view with QuickTime or ptviewer. In this
case there will be at least three separate interpolations of pixel
value: one to go from camera images to equirectangular image, a
second to go from equirectangular image to cubic image, and a third
to go from cubic image to screen image.

In theory, all of this manipulation could be combined into a single
step, generating a screen image directly from camera images with only
a single interpolation of pixel values.

However, that would put the entire cost of the calculation into the
viewing loop. At present, that's not feasible because it would be
too slow. Instead, the expensive calculations are done once, to go
from camera images to cubic image file, so that only very simple
calculations are needed for interactive panning/zooming. The viewing
calculations can be done quickly even in software, and are simple
enough to be done in hardware for even better performance.

It is interesting to notice that if you start with camera images from
a perfect rectilinear lens, then in theory the entire viewing
pipeline can be done without any spherical trig at all! All this
lat/lon stuff is just a convenient way to think about the
calculation. Ultimately, what's being accomplished is only to
project each planar camera image onto the planar viewscreen. That
process requires only matrix multiply and perspective division, using
homogeneous coordinates. I'm not aware of any software that actually
takes this approach, however.

--Rik
• ... Yes, that s true, but there is the problem of blendig, then. If the images were perfect, that should be no problem. But if they did not fit 100% or have
Message 9 of 13 , Dec 31 12:20 AM
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--- In PanoToolsNG@yahoogroups.com, "Rik Littlefield"
<rj.littlefield@...> wrote:

> It is interesting to notice that if you start with camera images from
> a perfect rectilinear lens, then in theory the entire viewing
> pipeline can be done without any spherical trig at all!

Yes, that's true, but there is the problem of blendig, then. If the
images were perfect, that should be no problem. But if they did not fit
100% or have some exposure differences, one would notice that in the
viewer. Maybe it would be a good approach, to correct and blend the
images, but to leave their ratio and size. Still it would need some
more calculations, like z-buffering, but that shouldn't be a problem.
The question here is: is there an advantage in having the (almost)
original images available in the viewer or its context? I can already
hear "Protect my images! Copyright!" - which is an important thing...
• ... The fact is there is a guy who stitches panoramas this way. Well not quite but what he does is actually combining 6 images from a fisheye in a 3D software
Message 10 of 13 , Dec 31 2:04 AM
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--- In PanoToolsNG@yahoogroups.com, "Rik Littlefield" <rj.littlefield@...> wrote:

> It is interesting to notice that if you start with camera images from
> a perfect rectilinear lens, then in theory the entire viewing
> pipeline can be done without any spherical trig at all! All this
> lat/lon stuff is just a convenient way to think about the
> calculation. Ultimately, what's being accomplished is only to
> project each planar camera image onto the planar viewscreen. That
> process requires only matrix multiply and perspective division, using
> homogeneous coordinates. I'm not aware of any software that actually
> takes this approach, however.

The fact is there is a guy who stitches panoramas this way.
Well not quite but what he does is actually combining 6 images from a fisheye in a 3D
software (CINEMA 4D). They are converted to rectilinear and cropped to 90x90 degree + a
seam area.
He also use a 15mm fisheye the same way but they have to be combined for each
cubeface.

His Name is Toshio Fuji
http://www.11moon.com/

You would know him if you were on the Quicktime list.

But be prepared if you contact him. He is a very special guy.
He also designs his own panorig and this is what he did to demonstate that his rig was the
best.
http://www.11moon.com/QuickTimeVR/temp/crush_test_s.mov

Hans
www.panoramas.dk
• I use standalone versions of SPi-V and DevalVR to view equirectangular versions of 360x180 panos. I don t know whether or not they build on the fly cubic
Message 11 of 13 , Jan 1, 2007
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I use standalone versions of SPi-V and DevalVR to view equirectangular
versions of 360x180 panos.

I don't know whether or not they build "on the fly" cubic faces...

BTW, the displayed rectilinear image is good and moves are very smooth.
This is very helpful for previews before end of stitcher use, but this
is out of topic!
• Hallo, Hannes ... Search for Euler Angles if you want to know the math behind it... -- Erik Krause Resources, not only for panorama creation:
Message 12 of 13 , Jan 1, 2007
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Hallo, Hannes
on Sat, 30 Dec 2006 09:00:12 -0000 Rik wrote:

> I forget the exact formulas. You can find the general forms in any
> good computer graphics text, but working out the details is always
> fiddly.

Search for "Euler Angles" if you want to know the math behind it...

--
Erik Krause
Resources, not only for panorama creation:
http://www.erik-krause.de/
• My memory of Euler Angles had grown fuzzy with time (hey, I haven t really worried about them since my mechanics classes in physics grad school 20+ years
Message 13 of 13 , Jan 1, 2007
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My memory of Euler Angles had grown fuzzy with time (hey, I haven't
really worried about them since my mechanics classes in physics grad
school 20+ years ago), so I googled them to remind myself. I got a
kick out of the blue-boxed warning at the beginning of the article on
Wikpedia. You know, the same one where they say that the article may
need to be cleaned up or doesn't provide proper references. This one
said: "This article or section may be confusing or unclear for some
readers." I had to laugh 8-) I know I can imaging an unsuspecting
reader wondering what the heck THAT is!

John

John Riley
johnriley@...
jriley@...

On Jan 1, 2007, at 12:06 PM, Erik Krause wrote:
>
>
> Search for "Euler Angles" if you want to know the math behind it...
>
> --
> Erik Krause
> Resources, not only for panorama creation:
> http://www.erik-krause.de/
>
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