I think I noticed something
 

Has anyone else noticed a resistor that is directly in the path of the negative terminal's conection. If I were to remove it would more power go to the board?

looks like it but that might let too much power go to the board and fry it

you would have to desolder the resistor from the board(leave it connected to the battery wire)...then you would have to use a multimeter (like this )
and test the voltage and amps coming through the resistor, then desolder the resistor completely and test the volts and amps again but without the resistor.....i think these board can only take 8.4 volts max(without FETs)...so if you've got more than that without the resistor then you should leave it in there

I think it's a 5.1v zener diode.

Its a 240k (or 240 ohm I can't read the damn meter) resistor. Its dropping the voltage from 6v to a safe voltage for the board. The resistor can go on either the Positive or negative side, either way it's still dropping the voltage to a safe level

so...... should I try it? cause if it is just dropping the voltage then if we removed it we could use less batteries to power it thus dropping weight.

would probably reduce your runtime if you reduced the number of cells...

It's a zener diode then. There's one on the board, just most people don't figure out where.

dude the cells are wired in a series and each has its own mah and b/c of that it would not reduce run time

what does a zener diode do?

Limits voltage input. Essentially it's a glorified resistor, but there are differences. It only lets a certain amount of current out, no matter how much input there is, until the voltage peak is reached, at which point most (but not necessarily all) of the voltage goes through. Works on the zener effect.

http://www.americanmicrosemi.com/tutorials/zener.htm