P3D Re: Projector question!

[bercov@xxxxxxxxxxx (John Bercovitz)]

> I was just wondering:  In order to fill my 50" screen with my VM 
> projector, I have to have the projector quite a distance away from 
> the screen.  Could anyone tell me how many feet away a TDC 116 or 
> 716 has to be to fill a 50" screen?  

The distance to the screen from the projection lens* is the ratio of 
the screen image size to the slide size times the focal length of the 
projection lens.  

Let's look at some specific calculations.  Did you say you were 
projecting 5P or 7P?  Is your screen square?  I will assume it is.

If 5P, the height will usually be greater than the width for this 
screen size as the height of the slide can be up to 24 mm which is 
24/25.4 or .945 inches.  50" screen / .945 inch slide gives 52.9 for 
a magnification.  So the distance from the lens to the screen is 
52.9*4 inch projection lens = 212 inches or 17.6 feet.  If you had a 
5" lens, the magnification from slide to screen would of course be the 
same but the focal length would be 25% greater so now we have 
(50/.945)5 = 265" or 22 feet.

If a 7P slide, it will be the width at 28 mm (1.1 inches) which is 
greater than the height.  So now the calculation for the 4" lens is 
(50/1.1)*4 = 181" or 15.1'.  Note that it is less than the result for 
the 5P slide.  For 5" lenses, the calc would be (50/1.1)*5 = 18.9 feet.

John B

*A Newtonian magnification equation is m = x'/f which can be rearranged 
to be x' = f*m.  x' is distance from front focus of lens to the screen 
and it is the distance I talk about when I say "distance to the screen 
from the projection lens" above.  f is the focal length of the projection 
lens.  The position of the front focus of the lens can be located by 
pointing the back of the lens at an infinitely distant object and then 
noting where it comes to a focus in front of the front part of the lens.  
This focus is the front focus of the lens.  Measure from it to the screen.


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