[photo-3d] Re: Algorithm Wanted, aligning any stereo pair

["Abram Klooswyk" <abram.klooswyk@xxxxxx>]

It has been somewhat long ago since Bruce came up with a 
seemingly useless but interesting problem :-). 
In those cases I often have the disadvantage of trying to find 
something in my memory, and than indeed finding something, but
that tends to prevent further original thinking.

George Themelis 29 Aug 2000: 
>You can think of the stereoscopic deviations as vectors 
>which, in a properly recorded and mounted stereo pair, lie in 
>the horizontal direction. 

and Aug 31, 2000:
> You position one pair randomly "over the other", select 
> a number of homologous points and compute the displacement 
> vectors. You sum up all the vectors together. 

In the past there have been times when only _mentioning_ a 
term like "vector" would have caused expelling to Tech-3D, but 
now George makes this a Math-3D list :-).

To avoid speaking of vectors I have sometimes used the thought 
experiment to take a stereo picture of a huge horizontal 
wooden arrow, pointing in a median plane from far away to 
some 2 meter from the camera. (A median plane is a vertical 
plane extending from your nose, or from a point midway between 
your camera lenses, to "infinity".)

On the mounted slide the arrow will show as a horizontal 
arrow, pointing to the right on the left chip, and to the left 
on the right chip. Drawing such arrows on two blank pieces of 
film, or just transparent pieces of plastic (marked L and R), 
is instructive to see how reversing, rotating and transposing 
affect the arrow directions. The only correct way is: pointing 
to each other. Tilted arrows will not be good of course.
(The arrows are the deviation directions.)

Bruce Springsteen 30 Aug 2000:
>I'm looking for the most efficient algorithm - a set of steps -
>for locating and aligning those vectors in an un-registered 
>pair of images. 

This has been done, and even published, for X-rays, over 
thirty years ago. Quoting from memory. 

It is easy to see that any two X-rays of the same object, 
unless identical, will form a stereopair, and this is true for 
ordinary photographs too, only sometimes the differences can 
be too large for viewing (but then sometimes can be used for
calculatios). 

Suppose someone who has a bullet fragment somewhere inside his 
skull. Two X-rays have been made to form a stereopair, but due 
to the hectic circumstances in the E.R. it is not clear how 
patient and/or X-ray tube and/or film cassette have moved. In 
any case, movement is not parallel  to the film edge, and 
there is some unknown rotation of object and film between both 
exposures. How to align them for stereo viewing? 

Put the films one above the other on a viewing box with 
strong light. 
Superimpose two homologous points. Rotate the upper film about 
the superimposed points as axis, until a second pair of 
homologous points (not too close by) also superimpose.
Then find a third pair of homologous points, preferably having 
a large mutual distance. Draw (on the upper film, using a 
glass pencil and a ruler) a line joining these points, and 
extend the line over some length. Interchange the films and 
draw a similar line on the second film. 

Then the two lines are the _directions of the deviation_. 
Putting the two films side by side with the two lines aligned 
horizontally and on the same height will result in the correct 
viewing position. 

Transparencies from 35 mm cameras can be aligned in a similar 
way on a bright viewing box, using a strong magnifier 
(monocular), or a pair of head mounted magnifiers as used by 
watch makers. You don't want markings on the images, but
scratches on the edges of the film, along the extensions of 
the line that runs over the third homologous pair, will work 
(again using some ruler).

In practice I mostly prefer a trial and error method, but 
using mounting reference grid with multiple horizontal lines, 
and several vertical lines (so  not just a few horizontal and 
three vertical lines as mounting jigs often have). 
For paper prints the grid should placed over the pictures.

Bruce: 
>I have no "right" solution to offer, just my own ideas.

Please, tell the next time your own ideas first, before 
putting us to the test :-) 

And speaking about Math, isn't this problem covered at length 
in photogrammetry textbooks?

Abram Klooswyk