Hmmm...that seems rather long. My HP 1200C scanner can scan an A-size sheet
of paper in much less time than that. I would think there are much faster
CCDs. You can buy specialized digital video cams with frame rates up to
120fps (lower res of course). My real need is to be able to reliably
identify a human size target (1 meter) at 3-5 kilometers. I'm not sure what
pixel density that works out to yet.

Okay Shawn, let's look at the math.
Angular size decreases linearly with distance, or 1/d. Let's consider a
one meter radius circle and a 90 degree field of vision. The circumference
of a one meter radius circle would be about 6.283 meters. One fourth of
that (that 90 degree field) would be about 1.57 meters of arc and a 1,024
pixel wide frame would mean that each pixel was effectively a 1.533
millimeter wide dot on your 1 meter radius circle.
Of course, we tend to view this as a flat field with distance, rather
than a curved section of a circular arc. Nevertheless, we can still use
this method to get an estimate.
If we look at it from the perspective of 90 degrees being 1,024 pixels
wide, then the entire circle would be 4,096 pixels in circumference. And,
each pixel would be 90/1,024 or 0.0879 degrees wide.
Now, let's look at a human target one meter wide at a distance of 5
kilometers. How wide would it appear to us if we had a 90 degree field of
view? Since our 1 meter distance yields a pixel width (for 1K of pixels) of
1.533 millimeters or 0.001533 of our field, and a one meter wide target at
5,000 meters is 0.00005 of our 90 degree field, we will have a target that
is 0.0045 degrees in width, or 0.0512 of a pixel in width.
In other words, nearly 20 people standing side by side would occupy a
single pixel at a distance of 5 kilometers. Time for a zoom lens.

Cheers!

Chip Shults