RE: New Lipo battery problem
Responding to Terry....
I'll give this a shot. Hopefully I'm not boring you with stuff you know.
Let's assume that the motor was propped in the two different setups so that you got constant watts out of each one. Since the higher voltage setup will lead to a higher RPM, you would need to reduce pitch or diameter or both to achieve this condition.
OK, so this would mean that Watts(Low Volt setup, 514W) = Watts(High Volt setup,386W), or 14.8v * 34 amps = 18.5V * X amps. Solving for X we get 27.2 amps. You say you get 28.1 A, on this 18.5V setup, which gives 18.5V * 28.1 A = 519.8W. But you also say your measured watts is only 386W. 386 does not equal 519, so even if you left the same prop on the two setups ... something is wrong.
Since you measured watts and amps in the high volt setup, we should be able to calculate battery voltage with some confidence, 386W / 28.1 A = 13.7V. Whew! That is a long way from 18.5v! 18.5 - 13.7 = 4.7v ... how old did you say that 11.1v battery is?
I would recommend putting a voltmeter on the battery side with the system loaded, and see what voltages you are getting out of each battery. Or, put a whatmeter in series with the battery, and record voltage, current, watts. If the watts and amps indications you provided for the high volt system are correct, then the voltage should be much lower than your nominal 18.5v. W = V * A.