Subject: Re: Hyperfocal Distances
Thanks for the courteous response, Berg!
The constant selected for the circle of confusion
varies from authority to authority. The FotoSharp
slide rule uses .025mm. Thom Hogan in his Nikon
Field Guide uses F x .001 for the circle of confusion,
where F is the focal length and it is his formula
that I started with. My results coincide with his
table on p98 for focal lengths greater than 50mm.
I suspect that he adjusts his constant for the
circle of confusion term as it becomes more signi-
cant at smaller focal lengths but I do not know
what that adjustment is.
>I realize that you're talking about quick estimates but your figures
>don't look quite right. The hyperfocal distance is normally given by:
> = f^2/(N x c), where
>h = hyperfocal distance
>f = lens focal length
>N = f-number
>c = acceptable circle of confusion diameter (c = 0.03mm for 35mm format)
>Berg
Yes. Now use Thom Hogan's c=F/1000
That F term in the denominator cancels
one of the F x F terms in the numerator. Your correct
formula (using this circle of confusion) is then
H(mm) = 1000 x F/f. Converting the result in mm to feet
yields the H(ft) = 3.3 x F(mm)/f.
Question, Berg: Do your calculations confirm that at
f16 the near focus distance is approximately 1/2 the
hyperfocal distance? By observation of the results on
the FotoSharp slide rule, they do seem to be. Hence the
second simplification that at f16 H= 0.2F and DN= 0.1F.
Thanks for the courteous tone of your response.
....patrick
