Voltage Reduction
HI John
As usual Dave is correct in all his advice but what its worth I will add a bit more.
I learnt the hard way see my bit in previous forum subject " Fireboat c of g ".
You do not say what capacity your batteries have or your motor type so cannot estimate the sort of current your system will draw.
Basically a drive system using brushed motors is a series connected circuit I.e. batteries, wiring and speed controller/motor are all connected in a ring and consequently are subjected to the same current ( amps ) flowing through all of them. Dependent on the Impedance ( resistance ) of each item, some of the voltage available from the battery is dropped across each element including the battery, this has internal resistance of its own so as the current rises the battery terminal volts fall. in the motor, most, you hope, of the voltage drop is converted into mechanical power, however in all the rest of the components virtually all the drop is converted into heat.
My experience of the " Blue " esc is that they work but mine results in a very large voltage drop across it, that reduces the voltage at the motor and consequently the power available, it also means that the electronics gets very hot and can cause its death very easily, as many on this site can testify. So be very careful in how long you run your motor and check that the esc is not getting hot quickly, if you have a dc voltmeter measure the voltage drop across the motor leads of the esc.
In general I would not have thought that the motor control part of the esc would be that sensitive to the output available from a 6 cell I.e. 12 volt lead/acid battery, but if the controller has a bec output connector for the r/c system, this may be a bit more delicate. it may be sensible not to use this until you have some confidence in the rest of the system.
It would also be advisable to incorporate a fuse in the system to limit the current that the esc/motor can draw.
Another safety item worth considering is adding a dropping resistor in the circuit of about 0.1 ohms this at 10 amps will lower the voltage available to the rest of the circuit by 1 volt but will at this current result in 10 watts of heating. At other currents watts can calculated by the equation.
Watts =Ohms times Amps squared.
Once you have some confidence in the system this can be removed.
But in the final outcome, as Dave advises, it is your decision what to do but be careful and proceed slowly and if running with the prop connected make sure it is always clear of obstruction and stalling the motor.
nasraf