tOM,
Yes . . the FOV (in degrees) does shrink as the film plane is shifted
off-center from the lens. I was still struggling with a pile of equations
working it geometrically using the principle that the three interior angles
of a triangle always add to 180 degrees (acute or obtuse). Then I realized
it could be shown via constructive geometry and all became very clear . . .
that my brain had run so fast the first time that I had actually visualized
it, but only for a very fleeting moment before leaping so quickly to its
conclusion that the path that got me there was lost.
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!].
Now copy the length of one of the legs of the smaller center isosceles
triangle and mark its length down the leg of the larger one we first drew
starting at the apex. It will fall short of reaching the base. Connect
that end point with the nearest trisection point on the base of the large
triangle (distance "a" from the end of the base). We now have another
isosceles triangle with the same leg length of the center one. It should
be obvious now that this new isosceles triangle's base is shorter than the
base of of the one with length "a" in the middle. Why? If you look past
its base, there is a length "a" at an angle to it that must be, by
necessity, longer! Because it has the same length of leg with a shorter
base, its apex angle *must* be less. Thus, the apex angles of the two
obtuse triangles with base length "a" are smaller than the apex angle of
the isosceles in the middle with length "a" base. By intuitive logic this
special case can be generalized.
Quo Erat Demonstrandum:
The Field of View shrinks as the film gate of fixed dimensions is shifted
off-axis in the image circle.
An additional note. If you think about it, the FOV as defined by the fixed
film gate dimensions must shrink because its center is being moved farther
from the apex of the image cone.
Thankyou tOM for challenging me on this . . . seriously . . . it forced me
to walk through it in detail to demonstrate it with certainty. I make too
many quick and dirty mental analyses and it frightens me at times that I
haven't thought it through thoroughly enough.
-- John Lind
[who must now explain the sudden TP shortage and stubby Crayolas to the
rest of the household.]
At 02:29 AM 7/6/04, tOM wrote:
>Ackshully, I think I was dead wrong. The angle viewed is less at the
>outside of the image circle than at the middle, for a given negative
>width.
>
>tOM
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