This circuit divides your battery voltage through a 1Meg and 470K ratio. So your Moteino analog pin sees [470k/(1000k+470k)] of your battery voltage, or ~0.32 of it. This would allow a max range of 10.31v before the analog pin sees 3.3v and maxes out the 10bit analog range. For instance a hypothetical 5V battery would divide to 1.6v into the analog pin, which will give a reading of about (1024 * 1.6v/3.3v) ~= 496 from analogRead(batteryMonitorPin).

So to get the actual voltage of your battery you have to multiply the analog reading to 0.0032 (=3.3v / 1024, ie 10bits resolution of analog reads). This gives you the voltage at the divider point, or the 1.6v in the hypothetical example of the 5V battery. Then you multiply that result to the inverse of the ratio of the divider (1/0.32), to arrive at the input voltage of 5V:

1.6v =====> 32% (at dividing point)

X =====> 100% (at input point/battery)So solving X = 1.6 * 100/32 = ~5V as expected

The division by 10 in the code shown is just an average of 10 analog samples (to even out any differences in analog reads).

The code you pasted is not using the same ratio since 1/0.32 != 1.42, that was probably more like a 1Meg:2Meg ratio. So each resistor divider ratio will have that third term different, you will have to determine that from your resistors:

`battVolts = analogReading * 0.0032 * dividerRatio;`