RC India

Hi, i was wondering if voltage drop-(1) vs flight time-(1) curve is a linear graph with slope=1
Or,
if voltage drops by 0.5 volts in 5 mins, will it drop by another 0.5 volts in next 5 mins provided throttle, wind speed, and other parameter is constant?

Can we determine the approximate flight time by flying the thing for a few mins?
TIA  :hatsoff:

The slope will not be linear for LiPo and this results in a continuously varying flight time estimate even for a constant load. Coulomb count method i.e. keeping track of the mAh discharged from the battery will be helpful to determine the flying time.

well, i will try to check it this weekend
I think there aren't many who thought over this

It won't be a linear curve as is shown in this plot. (Except the "working range")

Like Santhosh said, a capacity count is the preferred way to estimate flight time.

(http://www.runryder.com/fastphoto/59786/discharge_curves.jpg)

Yeah, not more than a few thousand people...
 :rofl:

I guess it's so obvious that nobody bothered to post it  :giggle: :giggle:

it doesn't answer my question,which was- "is there a way to determine flight time of a setup based on mah consumed in 2 mins, and no other data.

its obvious that keeping track of the mAh discharged from the battery will be helpful to determine the flying time.

few thousand is less  :giggle:

Yes.

You can calculate the average current draw and assuming the same draw over a period, use the 80% rule and calculate estimated flight time.

1. No
2. No
3. Yes

"is there a way to determine flight time of a setup based on mah consumed in 2 mins, and no other data"
Yes.
If flying similarly, the watts consumed in the next 2 minutes will be similar. Since the voltage would be lower by then, the amps drawn would be proportionately more.

Do you have a watt meter?
If so your questions can be answered with just a few trials.

If not, please post whatever data you have, (motor specs, prop, voltage, current draw, mah consumed in 2 minutes etc)
I will try to simulate on test bench and give specific answers.
(However, There are many discharge curves on the net that you can look at)

900kv sunnyskys on 14.8volts and 10x4.5 props
amp draw is 17.6a approx
mAh consumed in 2 mins=.3volts(i guess)
I thought my setup will last for 20 mins or more, maybe 15?

Sir,
Mah consumed has to be in mah, not volts, no?
More tomorrow as battery down  ;D

its .30 volts down, 12:40-12.10v

mah, will tell you asap

@roby,
You mentioned that the amperage was about 17.6A.
At 18amps, consuption is 18,000mah per hour, or 300mah per minute.
Hence using 70% of the capacity of a 3000mah will give you 7 minutes duration.

Actually somewhat less than that, because as the voltage drops, you'll have to increase the current draw (ie, use more throttle) to maintain the same power level

BTW, the voltage drop is not linear. Google 'lipo discharge graph'.

well, that was very insightful :)