1) Should I have a negative bus near the battery ground to make it easier?
Not needed. The ground on the fuse block will be fine.
2) The inverter is a 35A max unit, which is 5A over the fuse panels single circuit max. Should it be ran directly to the battery?
You have to figure the load at whatever the inverter's lowest operating voltage is. So let's say it shuts down at a voltage of 10.5v. That would be 400w / 10.5v = 38a. If it shuts down at 11v - 400w / 11v = 36a. Shutdown at 11.5v - 400 w/ 11.5v = 35a.
So okay, say it's a 35a max draw.
So first, you have to find out what is the minimum wire size for a 35a load (gonna answer your question #4 here). For that, check a handy AWG (American Wire Gauge) chart:
http://www.engineeringtoolbox.com/wire-gauges-d_419.html
There you see that a 10 gauge (or what an electrician would call #10) could, at 52a, theoretically handle the load - IF it's a solid core wire. But almost no one uses a solid core wire above #12 size, and in vehicle use, almost no one uses ANY solid core wire - it's all stranded. So almost everyone just uses the 1-3 cores rating, which for #10 would be 30a, which is not enough.
So looking at the 1-3 core stranded wire rating, we see that we need a minimum of #8 (and a 40a fuse to protect it) to handle the load of the inverter, no matter if it draws a max of 35a or a max of 38a.
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Next thing to figure is voltage drop. There are 4 variables needed to figure voltage drop - voltage, load, wire size and distance.
We know the voltage...sort of...for now we'll just say, "12v". We know the load (say 35a), we know the wire size (say #8) and now we need the distance. So let's just try one on for size. We'll say a 10' straight line distance, which would be loop of 20' of wire. Now we gotta check out a voltage drop calculator:
http://www.calculator.net/voltage-d...ance=10&distanceunit=feet&eres=35&x=49&y=5
So we plug in our numbers to the voltage drop calculator, and we see that using those numbers, we'll end up with:
Voltage drop: 0.44
Voltage drop percentage: 3.67%
Voltage at the end: 11.56
But that's a bit deceptive. The fully charged battery will be resting at 12.8v, and if the inverter disconnects at say 10.5v, then we have to do TWO voltage drop calculations to know what's really going on - one at 12.8v, and another at 10.5v:
Plugging in 12.8v we get:
Voltage drop: 0.44
Voltage drop percentage: 3.44%
Voltage at the end: 12.36
Plugging in 10.5v we get:
Voltage drop: 0.44
Voltage drop percentage: 4.19%
Voltage at the end: 10.06
Okay, so we see that in all three cases, we're talking about roughly a 1/2 volt drop with that load, wire size and distance. What does that mean? It means, that the actual voltage on the loop is going to be .5v lower than whatever the battery voltage is. So if the inverter is set to shut down at 10.5v, it will see that on the loop and shut down - even though the battery is actually at 11v.
That might not be a bad thing - better for the battery to do it that way. But if you were designing for an off-grid solar powered cabin, and were trying to squeeze out every percent of efficiency, you might decide to run a larger wire size. Not because you need it to handle the load, but because you want to minimize the voltage drop.
In your case, it's a little more complicated, because you're going to have part of the run (from the battery to the fuse block) using a lot bigger wire. Add in more complexity because that bigger wire is also feeding other loads. Blah...the math gets to be a real pain in a hurry.
So, I'd just keep it simple. You know you need a minimum of #8 stranded wire (and a 40a fuse to protect it) to handle the load, and over a loop of 20' with a 35a load, you'll get around .5v drop in the voltage so the inverter will shut down when the battery is at 11v. Works for me.
ALSO, that's all figured on the "worst case" - 35a load. But how often are you really going to draw the full 400w out of that inverter? Probably not that often.
So for a 10' straight line run, I'd go with #8. For a longer run, I'd up-size the wire to the next size, #6.
And to answer your question about running directly to the battery - no need. You'll be running a lot bigger wire to feed the fuse block anyway. Your only issue is the 30a limit on fuse size. So what's that work out to? 30a * 10.5v = 315w. So your "400w" inverter is actually only a "300w" inverter - but ONLY IF the battery is nearly dead AND you are trying to run the inverter at full load. Otherwise, for normal day to day use, you'll be very unlikely to blow a 30a fuse.
Of course, if you only use a 30a fuse, then you could use #10 wire. I wouldn't because it would make the voltage drop a lot higher (try it in the calculator). So even with a 30a fuse, I'd still use at least #8 to supply that inverter.
Now for the pop quiz to see if you got the idea...
I deliberately introduced an error into the voltage drop calculations I just did. It's not enough to invalidate the results, but it's off a bit. Not *quite* right.
Can you spot the error?
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Now let's go back to the wire size chart and see what size wire you need for the big stuff.
Say you use a 90a fuse or breaker. What is the smallest wire size you can use?
Using the 1-3 core stranded numbers, you'll need at least #2 wire.
3) A breaker is the same as a fuse, just can be reset or disconnected correct?
Yup.
4) Wire size is confusing, is there a chart that breaks it down?
Already covered.
5) Positive and negative should be the same size?
Always.
Never make the negative (or neutral in AC wiring) smaller than the positive (or hot in AC).
A circuit is a loop, and the entire loop will carry the same load.
So if you run a #2 to feed from the battery to the fuse block positive, also use a #2 for the negative from the fuse block to the battery.
The only nit I can see to pick with your diagram is that you're missing a fuse between the solar array and the input side of the charge controller.
Cheers.
