Thanks very much, Richard, for the extensive explanation. I managed to follow almost all of it, and the excerpt above is where I have a problem still. I tried the DoF calculator you left a link for, and it pretty well confirmed what everyone has been saying: for equivalent field of view, the depth of field is nearly proportional to the f-number divided by the sensor width. That worked out with every example I tried.However both the 80mm lens on the 5D and the 50mm lens on the 7D were set at f/4 the same shutter speed would get the same exposure results. However, if you wanted to use an 80mm lens on the 5D and a 50mm on the 7D and if you demanded the same DOF, you would need to open the 50mm lens approximately one stop to f/2.8 which would require a shutter speed of 1/2 the duration of your exposure at f/4.
In the part I quoted above, you use this principle to obtain equivalent fields of view at the same subject distance by using a 50mm lens on the 7D and an 80mm lens on the 5D, then getting equal DoF by using f/2.8 on the 5D and f/4 on the 7D. Hunky-dory so far. But why would you need twice the shutter speed on the 7D under those circumstances? The physical aperture is essentially the same size on both (and would be exactly the same if the f-stop on the 5D had been 4/1.6 in order to produce exactly the same depth of field. So the same amount of light enters the shutter and the same amount of light impinges on both sensors. So why do you need to change the shutter speed?
My conclusion would be that the 5D will have the same depth of field as the 7D given the pair of lenses that yield the same field of view (1.6 ratio of focal lengths) and setting the aperture of the 5D to be the same physical size as that of the 7D (i.e., f# 1.6 times as large), with no need to increase ISO sensitivity or use a longer exposure time. But no one else seems to think so.
As I mentioned, I really want to understand this, because I'm contemplating buying a camera, but I don't want to go large-format if it means sacrificing DoF.