I had asked the following question in another thread and now I think I've found the answer. I'd love feedback from anyone who understands the situation better than I do (probably most people who read this fall in that category). I found two definitions of Exposure Value (EV), one based on luminance and film speed and the other on f-stop and shutter speed. Equating them led me to the following conclusion: ISO=Film Speed (ISO Units) L=Luminance (foot-lamberts) F=f-stop S=Shutter Speed (seconds) L=F^2/(Sx0.32xISO) Or in other words, divide shutter speed times film speed times 0.32 into the square of the f-stop and you'll get luminance in foot-lamberts. Here's some examples: F=f/4, S=1/100 and ISO=200 yields a luminance of 25.1 ft-l F=f/4, S=1/60 and ISO=200 yields a luminance of 14.9 ft-l F=f/8, S=1/100 and ISO=400 yields a luminance of 50.0 ft-l Here's my question. How much do I have to allow for light losses in the lens and camera? These calculations are all designed around the light that gets delivered to my SLR's meter and, presumably, the film. I think using this approach will make my TV output more light that the formula would indicate. For instance, I could set my camera to ISO=200 and f/4 and focus a telephoto lens on an 100 IRE window. If I adjust the "Contrast" setting until the meter says use a shutter speed of 1/100 the formula would say that the TV is producing 25ft-l, right? In reality, maybe 10-20% of the light is lost before it gets to the film/meter so I'd really be driving the TV to something like 28ft-l, wouldn't I? What magnitude should I assume for this effect, assuming no filters on the lens? I'm inclined to calibrate my 36XBR800 to 30ft-l using the formula and figure that ought to be safe for my tube even if it's really putting out closer to 35ft-l or so.