T3D No. 2 of Dull Double

["Dr. George A. Themelis" <DrT-3d@xxxxxxxxxxxxxxxxxxxxxxxxxxx>]

>(1)window-at-the-screen, and
>(2)infinity-separation-at-standard-interpupillary.  But is there some
>well-known calculation that tells how far to place the projector from
>the screen to achieve (1) and (2) simultaneously?  And what image size
>would typically result? 

Here is how to go about it:  Say that you are using 1.2 mm as your
infinity separation.  You want to project this to 65 mm on the screen.
So you are looking at about 50x magnification.  

The vertical dimension of the 35 mm film is roughtly one inch.  So you 
will project a 50" tall image.  You need 50" tall screen.  

The distance of the projector from the screen will depend on the FL of 
the lenses.  From similar triangles:  
Distance-to-screen = FL x magnification = 50 x FL.  
That's about 17 feet for a TDC with 4" lenses or 21 ft with 5" lenses.

George Themelis


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End of TECH-3D Digest 365
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