Last word -- I promise -- on this topic.
I think the confusion began with the same term, "field of view," being used in
two different senses. I believe the way Piers uses it, and not incorrectly, I
suppose, it refers to the entire image the lens "sees" and projects. The way I
use it, it refers to whatever portion of the scene the lens "sees" that makes
it onto the film.
In other words, in one sense, field of view would be a constant; for me it's a
variable. For instance, my baby Graphics and XLRF cameras accept
interchangeable backs. I have four different formats: 4.5x6, 6x6, 6x7 and 6x9.
Using my definition, the field of view for any particular lens is different
for each format.
John's calculations and explanation brought back my headache, but it did remind
me of a story that might appeal to those who may have understood what he was
talking about:
There once was an Indian chief who had three wives. He kept one wife in a
teepee made of deer hide, one in a teepee made of buffalo hide, and the third
wife lived in a teepee made of hippopotamus hide. The wives in the deer and
buffalo hide teepees each gave birth to a single child every other year, while
the third wife had twins every year. Which means -- anyone? anyone? -- the
product of the squaw of the hippopotamus hide is equal to twice the product of
the squaws of the other two hides.
Now I think I'd better go hide.
Walt
-------------- Original message from "John A. Lind" : --------------
[snip]
> I will describe it with an example. Draw an isosceles triangle with a
> height of arbitrary length "b" and a base with length of three times some
> arbitrary length "a" (this trisects the base from the get-go). Draw two
> lines from the apex to the two points that trisect the base. We now have
> three triangles: one isosceles in the middle bounded by two identical
> obtuse triangles on each side . . . all having a horizontal base with
> length "a". The length of the leg of this center isosceles equals the
> squareroot(.25*a^2 + b^2) [Thank You Pythagoras!].
>
[snip]
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