Hi guys -
I'm flying a 651 sq ship with my Misfit (posted in "list you setup")
http://stunthanger.com/smf/index.php/topic,12044.msg346987.html#msg346987
Originally fitted with a 2700 mAh 4-cell pack, flight durations were less than desired - barely making 4 minutes.

I changed to a 5-cell 3700 mAh pack and can get 5 min using only 50% capacity, so I'm good there.

Here's my question.
Using the Castle Link graphing tool, the plots show Watts and another plot of "power out".
While it's obvious to me that current drops as voltage goes up, I would expect the total power draw to stay roughly the same for a given prop/RPM.

In the attached picture, the 4-cell flight (4 min) is on the left, and the 5-cell 5-min flight on the right.
Average wattage (blue plot) is 380 and 428 respectively.  I would assume that the same prop (12x6EP) with the same ship at the same RPM would draw the same power, but apparently not (P=IE).  Seems current drop isn't directly proportional to voltage rise.  I would also expect that, with higher voltage, efficiency would go up and power would go DOWN, not UP.

Also, the Castle data log includes "Power Out" (teal plot) which is up at ~100% on the 4-cell plot but only 70ish with the 5-cell, and it behaves in reverse to Watts .  I can't find a clear description on Castle as to what "Power Out" really means.

Any help?

I agree with what Igor says. You are using a very inefficient set-up. You need to either lower your KV from the 950 where its at to a motor around 700-750 kv and you should be able to use a 5 cell pack of around 2700-3000mah. Or you can keep the same motor and switch back to a 4 cell pack of around 3000-3500 mah. You are also using a bigger motor than you actually need but that's the least of your problems. A 2826 ?(depends on 4 or 5 cell battery) sized motor(brand of your choice) or a 2820 ??(depends on 4 or 5 cell battery) sized motor(brand of your choice) will more than fly that plane and you will drop the weight a few ounces, also making the smaller batteries a better choice .

Your timer sends constant signal in both cases. That is true. Your ESC converts that constant signal to "power out %" which will maintain constnat RPM. As you can see it gives lower % at higher voltage. That is that "throttle" which I wrote in my example. Simply stronger motor needs less throttle.

That throttle % is converted to some Power Width Modulation (PWM). ESC simply switch on current feeding for say 90% and 10% time off at lower voltage. And say 70% on and 30% off at higher volatge. Motor runs at the same rpm. And now comes the trick. Wihile its shaft gives the same power, at the same RPM, also the torque must be the same. Torque is aproximately linear to current in winding and 1/KV. Both are aproximately equivalent (the same motor, same rpm, same power, same torque). But the current is more variable in second case with higher volatge and lower "power out %" because it is feeded only 70% of time while that lower voltage 90%. And while copper thermal loses are R * I^2 also thermal loses in winding are higher. Input power to motor (your blue line) is output power + loses, therefore power comming to motor must be higher regarding that difference of loses.

Simply efficiency of that motor in that conditions is lower, the same power on shaft neeeds more power in.

if the question is, "why did the power increase when the voltage increased", then it's literally because the voltage increased and the load remained the same.  if your motor kv is 950, then at 16.8v you're spinning your 12x6 prop at potentially 15960 (i know it's less, but for ease of explanation). on a five cell it jumps to 19950 rpm. if you're spinning the same prop, the amps go up. you can't swing the same prop faster without drawing more current.

the brown line its the RPM.
It is the same in both cases.

After reading the last few posts (thanks guys!!), I opened up the data file again and measured the average RPM for each of the two different runs (4-cell and 5-cell); 9191 on 4C, 9172 on 5C.
That's using a Hubin FM9 timer set at 9154.  So both runs are within 0.5% of the set-point . 
I was actually amazed, as the Hubin timer has no RPM feedback from the motor - just the fixed 1.401 msec pulse in both cases.

I'm understanding now what "power out %" is and why it's lower with higher input voltage.
I also ran both setups in MotoCalc and sure enough, the higher input voltage results in lower efficiency.  I was always under the impression that, for lower current draw and less I^2R losses, that the efficiency would be higher, but I guess not.

Your motor controller auto senses the new battery and the new "continuous" output current to prevent damage to motor. Review motor power requirements and controller setup.


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