Wow, there's a lot to say. Sorry.
capn, "just use the calculator" will work if we can somehow find out the actual amount of tolerable voltage drop.
Example 12v 1 amp and you want the same at 300 ft.
"The same" means no voltage drop. "No voltage drop" requires zero resistance, which is impossible. But let's figure out what will happen with your example, using 12/2. In practice, 12/2 seems ridiculously thick for camera power, but the numbers might surprise us.
12/2's resistance is 1.588 ohms per thousand feet. 300 feet has a resistance of 0.529 ohms. There's a hot lead and a ground lead, so resistance is twice that, or 1.059 ohms. At one ampere, the voltage drop is 1.059 volts, so the delivered voltage is 10.941 volts.
We now have the question, will the camera work properly with 10.941 volts? Have you ever seen a "minimum applied power" voltage specification? And we're still making the assumption that the connection points will have zero resistance. (Just for reference, a 300 foot run of 16/2 would deliver 9.59 volts.)
Keeps you from hoping the voltage you feel is high enough at the end of 16/2 is enough
You've still got hoping and feeling in there, instead of knowing and measuring. I'll bet no camera manufacturer specifies the ACTUAL lowest operating voltage of its products, so guessing and hoping are still part of the situation.
The worst situation as an installer would be replacing a non-IR camera at the end of a long run with a camera that has IRs.. It could work fine during the day, but when the sun goes down and the LEDs attempt to turn on, the voltage drop could be too much. The camera could switch on and off repeatedly, or reach some stasis where it's not quite on. I'm not sure.
