Here's what you have:

That's a distributed load ... so the wire-thickness-drop table can look at the distances each current level will be carrying.

After the first light on each leg, you're dealing with less current, so less voltage drop per foot on the subsequent legs.

If you have your split "early"(i.e. close to the source), you can drop the two legs by a wire size or two.

I am a fan of "use one size" instead of playing silly games, *and* "look at the loads" ... do i really care if my lights are losing an extra percent or two? 3% of 12v is 0.36 volts. 5% is 0.6 volts. The last LED won't care that it's a half-volt down.

... you're frequently starting with more than 12v (when the engine's running or they're charging, it's over 13v).

... but if you run them down, it may reach 10 v.

For these dinky loads, i have to admit i'd be sorely tempted to use 16 gauge wire from the first lamp onwards.

(that's a gut feeling, i'm not looking it up since i don't know the various segment lengths involved).

Messing up the "fuse for the smallest wire" dictum is that often these LED pucks come with their own last 6 inches of wire ... and it's invariably far thinner than the wires you're running ... so do you fuse for *those*, or for the main run?

(answer: main run, or maximum expected load ... if you're only feeding these LEDs on that circuit, fuse for 10 amps at most.

7.5 amps if you want quicker fuse-pop.)

Let's look at the raw numbers: 16 gauge wire is 0.00409 ohms per foot.

Your longest length is 18 feet, and (i assume) running two wires (one out, one back) instead of using the Sprinter frame.

So that 36 feet of wire, a total of 36*0.00409 = 0.147 ohms.

Assuming the full 6 amp current through that, we'd see a 6* 0.147 = 0.8837 volt loss.

But we're not running that current the full length. We're only running 6 amps to the split.

Then we're running partial currents from then on.

Let's put the split at 6 feet, with 12 feet going to each final LED.

So now it's 12 wire-feet to the split, or 1/3rd the above full loss: 0.8837/3= 0.295 volt loss

Now it's (on the 5-lamp leg) 3.75 amps along the remaining 24 wire-feet: 3.75*24*0.00409= 0.368 v (3% of 12).

Add those two: 0.295+0.368 = 0.6631 volts.

We're very close to the 5% number, and i'm still vastly over-estimating.

Going to 14 gauge lowers the ohms-per-foot to 0.00258 ... one third of the loss is gone.

That would give you about 0.2 volts loss at the split, and a total of 0.44 volt loss over the full 18 feet on the 5-lamp leg.

--dick