Re: Large Binocular Telescope

[T3D John Bercovitz <bercov@xxxxxxxxxx>]

> Am I correct, that telephoto lenses in a stereo rig do not require 
> an enlarged stereo base, compared to standard lenses? 
 
That's correct.  There is a series in the technical directory in the 
photo-3d directory at bobcat that shows the geometric basis for this.  
Actually, there's no difference between this and binoculars or 
telescopes so you can understand it intuitively from looking through 
binoculars or a telescope to see how things are flattened in depth.  
And looking through, oh what's that thing called?, the thing that is 
two periscopes that increases your stereobase.  Disnomia strikes 
again.  Anyway, that thing will quickly show you that increasing the 
stereobase makes a scene look like a miniature scale view, perfectly 
scaled down in all three dimensions.  So while magnification flattens 
only the depth dimension, increased stereobase scales all three 
dimensions downward equally.  So you cannot balance one change with 
the other (contrary to the infamous "PePax" theory).
 
Oops!  I think you just got the standard John B diatribe free for 
nothing.  What you're _really_ asking is, "Don't worry about 
distortions, using the base and magnification of the LBT, would you be 
able to see any of the depth dimension on the moon?"  OK.  The maximum 
stereoacuity recorded is around 3" of arc, I think, but typical is 
more like 10 or 15" of arc.  This is for high-contrast objects under 
the very best of test conditions.  So let's say the moon can somehow 
give us these conditions.  I think for resolved (non-blank) 
magnification you're allowed 60X per inch of objective up to 4" which 
is the diameter of a turbulence cell.  So the LBT can go to 240X.  
Peter's the telescopist, not I.  If I'm wrong, please change it.  15" 
is 7.3E-5 rad or one part in 14 000.  If we can go 240X, then our new 
stereoacuity is 7.3E-5 rad/240 = 3.1E-7 rad or one part in 3.3E6.  The 
stereobase of the LBT is 14 m.  The moon is 406,680 m distant.  That 
sounds a bit too close.  238 000 mi * 5280 ft/mi / 3.2808 ft/m = 
3.83E8 m  Sound better?  
 
>From convscar in the SL3D directory in the photo-3d directory:
 
              S
Z = {----------------------- } - D
                   S
     tan [(arctan ---  - A]
                   D
Where:
A is the stereoacuity (3.1E-7 rad at 240X)
S is the stereobase  (14 m for the LBT)
D is the distance to the object. (3.83E8 m in our case)
Z is the minimum distance at which a second object must lie behind 
the first object for the second object to be observably behind the 
first object.
 
Since our angles are so small, we can get away with dumping the trig 
functions.
 
          S                       14
Z = {-----------} - D = {------------------} - 3.83E8
        S                   14
       ---  - A           ------  -  3.1E-7
        D                 3.83E8
 
I get a negative answer.  This means the moon is so far away already 
that it can't be separated even from the infinitely-distant background 
stars.  No joy.  How close would the moon have to be to be able to 
separate it from infinity?  For that, essentially we would have to set 
the denominator to zero:
 
S
- = A    so D = S/A = 14/3.1E-7 = 4.5E7 meters
D
 
So the moon is about 10 times too far away to see stereoscopically 
with this instrument.
 
Now with Bill's depth sensing method (out-of-focus near and in-focus 
far objects), you'd have a different problem which you'd no doubt be 
better at calculating than I since you're into telescopes.
 
John B


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