Here are some ideas. I could go on. This is not well organized.

As for a strike or other device far away from your power, you can simulate having power out there by using a capacitor.

What you do is feed power through a couple of conductors (just to get decent current flow), and feed a capacitor in the range of 100,000 to 470,000 mfd. It should take a bit of time, maybe a half second or two, to charge this capacitor.

But then you have a MASS of stored electrical power. A control voltage is then sent to a relay at the gate. That relay closes, connecting the charged capacitor to the load.

Now, about the load. If it's to unlock or open a gate, it has to be able to respond and release, or better yet spring open, the gate. A puny amount of current, just enough to open a relay, can thus make several amperes of current flow into a load, a latch release. (In fact this is the definition of a RELAY: it relays the control of and release of a high current using a low current. It's a relay. Use a diode across the coil.

EDIT START
With the large capacitor, you have a HUGE amount of current for a small amount of time. That's why you want the opener to spring the door open; the voltage on the capacitor rapidly lowers to whatever you'd get with the wire all by itself, and the current lowers to whatever the wire would deliver to the load if the capacitor were not there.
EDIT END

About cameras.
The one manufacturer I've spoken with did not even understand my question when I asked them what the minimum voltage was that the camera could function with. So it's up to you to find out.

Meanwhile, when power flows through a wire, the resistance of the wire lowers the voltage that is delivered at the load. The hot and the ground EACH reduce the voltage. I did the math a bit this morning and came up with a few things.

Four 24 gauge wires, wired in parallel, have about the same resistance as one 18 gauge wire. If you had a 100 foot run of 18 gauge wire (which is 200 feet of wire because there's a hundred feet of hot lead and a hundred feet of ground lead), and a camera (PLUS IR LEDs) that draws 500 mA, you'd have a voltage drop of .6385 volts, so a 12 volt supply would deliver 11.3615 volts. Your 200 foot run will have twice the drop, so you'd have 10.723 volts at the load.

To figure voltage drop, determine the current your load requires. Multiply that current by the ohms per foot that the wire has. Putting wires in parallel lowers the resistance. A helpful rule of thumb is never to beat your wife with anything larger in diameter than your thumb. No, wait, that's the origin of the term... rule of thumb that doubling up wires results in a conductor that is nearly equal in resistance to a wire size three numbers smaller. For instance, two 24s in parallel have the resistance of a 21 gauge wire. Two of those 2x24 in parallel have the resistance of an 18 gauge wire.

Once you have the voltage drop, subtract that from the supply voltage. That's the voltage at the load.

Okay, you know that the voltage drops, but what about the amps....
There are some ways that envisioning delivering water through a pipe can help you. If you have some water (electrons) that can be pumped at a particular pressure (voltage) through a pipe that limits the flow (resistance), some things can make sense.

Your regulated supply is your pump. The size of the pipe limits how much water you can push through it. And if you're trying to use the flow and the pressure of the water to do some kind of work at the far end, the flow allowed by the pipe is crucial. The pipe limits the water flow (current) and reduces the push from the pump (the voltage) available to do the work.

The electrical current is the voltage divided by the resistance, so it will simply be what it will be depending on those amounts.

There's sometimes a misunderstanding about current specs of power supplies. A 12 volt 1 am power supply will supply, say, 500 mA to a camera when its LEDs are on. On the other hand, a 12 volt 500 amp power supply will supply 500 mA to the same camera. Yup, that's right -- the same voltage and the same load mean the same current. What's different between those two cases is that the second power supply could power a crapload more cameras, and the second supply could probably start a fire if its output is shorted, since a short would make it deliver 500 amps until protective circuits or devices kick in. But even this follows the same rules about current -- a short circuit is a circuit with zero ohms, so doing the math, the current would be infinite. The power supply can't do that, so it will put out its maximum current of 500 amps.

You end your post asking if something is mathematically correct. Anything you do is mathematically correct because you can't make it mathemaically incorrect. 3+1 is never 5... Will enough power get to the load? That's the question.