From: Richard Knoppow (dickburk@ix.netcom.com)
Date: 01/11/03-11:46:02 PM Z
----- Original Message -----
From: "Michael Healy" <mjhealy@kcnet.com>
To: <alt-photo-process-l@sask.usask.ca>
Sent: Saturday, January 11, 2003 10:55 AM
> I posted this inquiry to Pure Silver this AM, but thought
there might be
> some stray wonks here that don't find their way over to
that neck of the
> woods.
>
> I am working with Alan Greene's "Primitive Photography" in
designing a 7x17
> folding camera. Trying to, anyhow. I find myself stuck on
his discussion of
> the way to determine the angle of view of a lens. Is there
someone here who
> might be able to lend me a
> bit of expertise on this? I know this is going to sound a
bit esoteric. It's
> just that my focal length determines all sorts of other
measures that figure
> in the conversion of his 10x12 blueprint to 7x17.
>
> I will quote Greene at some length (all from pp. 88-89).
He is using
> Clarence Woodman's table. "Finding the angle of view for a
given len and
> format combination is very easy," he says. "The following
method,
> established by Clarence Woodman, is recommended:
>
> 1. Determine the diagonal of the format... For example,
the diagonal of a
> 210mm x 270mm rectangle is 342mm.
>
> 2. Determine the quotient of the lens and format being
used by dividing the
> lens focal length into the diagonal. For example, with a
250mm focal length
> lens used with the 342mm format diagonal, 342 is divided
by 25 to arrive at
> 0.8047.
>
> 3. Taking the quotient from step 2, refer to Table 3.1 to
determine the
> angle of view for the lens. For example, using the
quotient 0.8047 from step
> 2, a 400mm lens used with a 210 x 270mm format negative is
found to have an
> angle of view measuring approximately 44 degrees. <end
quote>
>
> I have two problems with this. By my calculations, it
would seem that Greene
> means to divide diag 342mm by foc 250mm, **not** 25. IE,
it looks to me like
> the book has a typo. However, 342 divided by 250 is not
0.0847, either, it's
> 1.368, which (in Woodman) doesn't come to an angle of 44
degrees, but an
> angle of 68 degrees. This (I think) sounds plausible in
this example, since
> a 250mm lens on 210x270 (8.5x10.5???) would be somewhat
wide (wouldn't it?).
> But I do NOT see how his arithmatic comes even close to
reaching 0.8047 (an
> angle of 44 degrees).
>
> My next puzzler is how Greene manages to get from step 2
to step 3. Suddenly
> he's on another example, a long lens; but instead of
recalculating, he's
> applying the 250mm lens's (miscalculated?) angle of view
to a 400mm lens. I
> don't see how this follows from his contention that the
angle of view is
> "FOR a given lens and format combination".
>
> Anyone able to shed light on this? Am I just missing
something because it's
> the size of a barndoor? Oh, one more thing, case anyone on
this list should
> happen -- ahem -- to possess the largest library of
photographic materials
> in the history of the common man... Greene's Table 3.1 is
titled "Clarence
> Woodman's Table for determining the angle of view (found
by dividing the
> diagonal of the format by the focal length of the lens)".
It is said to have
> been adapted from Tho[ma]s Bolas and George E. Brown, "The
Lens: a practical
> guide to the choice, use, and testing of photographic
objectives" (London:
> Dawbarn and Ward, 1902), 28. I've been unsuccessful in
finding Woodman on
> the internet, or I might have tried independently to
verify Greene's method.
>
> Mike
>
I answered this question in the Pure Silver list but might
add a couple of things because I think the above is a long
way around.
The angle of view is proportional to the tangent of the
ratio of the focal length to the dimension of interest. I
use that term rather than diagonal since it is sometimes
desired to know the horizontal or vertical angle as well as
the diagonal.
The calculation can be done on any hand calculator or on
the calculator in Windows or Mac operating systems.
Where
A = viewing angle
D = dimension
f = focal length
A = 2*arctan{(D/f)/2}
Where the focal length of a lens giving the desired viewing
angle is wanted.
f = D*{tan(A/2)}/2
There is probably a better way of expressing this in ASCII
As an example of the second equasion, say a 90degree view
is wanted from a 4x5 negative. The diagonal is 150mm (film
is a little smaller than the "nominal size").
f= 150 x (tan 45) /2 or 150 x 1/2 or 75mm.
Where the focal length of the lens equals the diagonal the
angle is about 53degrees. For a 300mm lens its about
28degrees.
--- Richard Knoppow Los Angeles, CA, USA dickburk@ix.netcom.com
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