Re: Projection and focus sharpness

[bercov@xxxxxxxxxx (John Bercovitz)]

John Dukes writes:
 
> Some few topics ago when the subject of screen brightness came 
> up, I commented on the rather remarkable brightness of the 
> Kodak Ektalite, which has a smooth "silver" surface but which 
> sharply focuses the light to the viewers with a curved surface.  
> (It appears to be a "spherical" curvature, but aspherics would 
> be simple to mold on that scale, I'd suppose.)  John B. raised 
> the question of defocusing at the edges in a projection system 
> designed for a flat screen.
>
> Not wanting to think about the mathematics, there being masters 
> within our midst, I merely measured the depth of the center of 
> the (40 inch by 40 inch) screen compared to a corner. Roughly 2 
> inches.
>
> Don't know if that comes anywhere close to defocusing, given 
> real world projection lenses.  (My personal circle of confusion 
> is unparalleled.)
>
> What thinks thee, John B.?
 
Not sure prezactully what all's being asked of me so I'll flounder 
around a bit, as usual.
 
If you want the radius of curvature given the sagitta and the 
chord, you can solve any number of geometric ways but generally 
you're going to come up with a solution that looks like:
 
 R = (C^2)/(8S) + S/2
 
Since you have a 40x40 screen, the chord (diagonal in this case) 
is 40*2^0.5 = 56.6".  You've given the sag as 2".  Therefore the 
radius is 201" or thereabouts.
 
I've been calling screen manufacturers for literature as I said a 
few weeks ago and one manufacturer (Vutech, Hollywood Florida) 
gave me the radii he uses.  Up to 70x70 he uses a 150" radius.  
The 70x70 screen has a parabolic cross section but has a 150" 
radius at the center.  Larger screens have a 250" radius at the 
center and are also parabolic.  He didn't have further information 
and I felt he was a little shaky on just what a paraboloid is.
 
As I'm sure you recall, a formula for a parabola is 
 
y = (4fx)^0.5          or         x = (y^2)/(4f)
 
where y is half the chord, x is the distance along the optic axis 
from the point where the optic axis pierces the paraboloid, and f 
is the value of x at which the focus lies.  The quantity f is 
half the minimum radius of the paraboloid, that radius being 
found, again, where the optic axis pierces the paraboloid.  From
this formula you can calculate the sagitta which you need (below).
 
As far as the amount of out-of-focus goes (the increase in a 
point's size on the screen due to screen warpage), that's a big 
calculation (for the general case) which I don't have time for 
right now.  So let's cheat and do a first-order calculation.  One 
way to look at it is to transform the warp of the screen into an 
equivalent warp in the slide and see if that warp is significant.  
Longitudinal magnification, for very small changes, is just 
lateral magnification squared.  Your screen is 40" or about 1000 
mm tall and the 5-P slide is 24 mm tall so the lateral 
magnification is 1000/24 = 40 and the longitudinal magnification 
is 40^2 or 1600.  So the 2" warp in your screen equates to a .001" 
warp in a slide (2/1600 = .001).  I would say a .001" warp in a 
slide is doo doo and I wouldn't worry about how much out of focus 
it causes.  How's that for a quickie analysis to prove my fears 
groundless?  8-)  The calculation of the out-of-focus equivalent 
of the Vutech screens is left as an exercise for the student.
 
 
> (And I hope you'll be there Saturday at NSA.  Should you, or 
> someone, speak about this fine group at that time?)
 
I'll be there God willin' and the crik don't rise but I'm no 
public speaker.  Who have we who's good?  Norm?
 
John B
 
PS: Speaking of which, where were all of you when the Oakland 
Camera Club's Stereo Division had its meeting?  Slackers!  Back-
sliders!  8-)  Harold Baize, Eddie Hosey, Dave Spaulding, and 
Allan Woods are excused from this tirade as they were there.  Alex 
Klein, Ryko Prins, Jan Gjessing, Andrew Woods and Steve Spicer are 
also excused because they live so far away.  But the rest of 
you....


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