
Sep 16, 2012, 05:18 AM  
Joined Aug 2012
141 Posts

Discussion
Flight time  can somebody explain me why?
So ... an Lipo battery have around 10 000 Mah @3S per 1kg (1000 grams).
So with an 1000 grams hexa multi rotor (Yea I didn't put the engine weight, the frame weight,etc) .... i can fly according to ecald around to 95 minutes: .... http://s16.postimage.org/ptfkoojz9/10k.png Now let's say i wish to fly out more, so i am going to put 10 batteries, each 10 000 mah @3S 900 grams each (those should be really good batteries) http://s16.postimage.org/n0md4njmt/100k.png Why I get only 34 minutes flight time this time? How should I add new batteries, if I hunt for the best flight time I could and to not get this kind of problems? I mean it's kind of weird... you add more batteries and you get less fly time... lol 

Sep 16, 2012, 07:56 AM  

Here's a real world example for you. One of my Hexa's was only getting 4 minutes (with a little buffer room) on a 3S 4500. I wanted to see how much more time I would get by putting two of those in parallel. I only got 6 and half minutes with two. The extra weight and strain on the power system caused it to use more power more quickly. So I was better off flying two 4 minute flights instead of one 6 and half minute flight. Voltage can make a big difference too so my next step was to go to a 4S 4400 and I got 5 and half minutes. More voltage = less amperage for the same power output and longer flight times, so long as the voltage is within spec for the system. I couldn't put a 6S Lipo on it and get more run time because it would blow the system.

Sep 16, 2012, 08:59 AM  
Romania, Dolj, Craiova
Joined Sep 2007
14,013 Posts

Let's illustrate with much numbers the example of Flyboone.
Assume one kg quad, powered from a 3s 2200mAh pack weighing 200 grams, and it's flying 10 minutes. Because 10 minutes is 1/6 of one hour, this means the pack was discharged at 6C rate, thus a current of 2.2 * 6 = 13.2Amps This led to a power consumption of 3(s) * 3.7V * 13.2Amps = 146.5W This is the efficiency, 146.5W / 1.2 kg (1kh frame and 200 grams pack) = 122W/kg, pretty average. The above are real data, rctimer DJI frame with 2830 850kv motors and 10*4.5 gemfan props. Now assume we add a second pack of same capacity and weight. New mass will be 1.4kg, thus new power consumption 1.4 kg * 122 W/kg = 171W New current draw = 171W / 11.1V = 15.4Amps New flight time = double capacity / consumption = 4.4 / 15.4 * 60 minutes an hour = 17.1 minutes, not 20 as expected ! Be aware that in above example I assumed the same efficiency in the both case, but this is not true ! The real flight time is even smaller, a bit over 16 minutes. I just wanted to keep things simple. 
Sep 17, 2012, 06:23 AM  

Let me show you a bit of physics, proving why the maximum flight time for a given frame is obtained with a battery of equal weight (=1/2 of total weight).
Firstly, for a propeller we have: thrust F = kt * p * d^3 * w^2 mechanical input power P = kp * p * d^4 * w^3 where p,d  propeller pitch and diameter, kt and kp  specific constants and w  angular velocity. Playing with the 2 formulas we get: F/P = kt*sqrt(kt)/kp * sqrt(p*d/F) But we have a given frame(p,d  constant) therefore P = k * F * sqrt(F), where k is a constant that shows proportionality (from now on I'll use k to show that, but it'll not be the same k, meaning it will not have the same value), and its value doesn't matter. Note that we have worked out P as mechanical power, we need to find out P(el)  that is electrical power drained by motors from the battery. Now I'm cheating a bit, and assume that P(el) is not propotional to P(mechanical output) for an electric motor, but with P * sqrt(F). Meaning that the more we "push" a motor, the lower its efficiency (I know it's only half true). The term sqrt(F) might not seem to make much sense at this point, but we'll see later on that it drastically simplifies the demonstration. So we write: P(el) = k * F * sqrt(F) * sqrt(F) = k * F^2 Then F must be equal to copter weight (hover), which is (m+M) * g. M is frame mass(constant), m is battery mass and g  gravitational acceleration. Therefore: P(el) = k * (m+M)^2 Now, battery energy is proportional to its mass, E = k * m Flight time is t = E/P(el) Substituting P(el) we get t = k * m/(m+M)^2 After playing with inequalities, we find out that the expression m/(m+M)^2 has a maximum value of 1/(4*M) (remember that M is constant and m is variable), reached if and only if m=M. This means t is maximum when for the given frame, we choose a battery with m=M. Feel free to give examples with real tests or ask questions about the demonstration. 
Sep 17, 2012, 06:54 AM  
Romania, Dolj, Craiova
Joined Sep 2007
14,013 Posts

Wonderful demonstration from a 18 years young, wish I had him as student in the 90s, when I was in high edu...
The confirmation above (true) theory can be seen in the graph below, from mikrokopter site. As you can see, the flight time tops at 14.000 mAh lipo, whose weight is close to the 970 grams of frame. QED. 
Sep 19, 2012, 01:39 PM  
United States, OR, Portland
Joined Aug 2011
459 Posts

@Zen09  excellent description, and @Renatoa  good find. Please crosspost them here.
http://www.rcgroups.com/forums/showthread.php?t=1284741 Jim 
Dec 06, 2012, 05:05 PM  
Joined Nov 2012
1 Posts

Quote:
The formula I found on an MIT course page says that to calculate thrust it is the following: T = kt * rho * n^2 * D^4 where kt is a constant, rho is the air density, n is the angular velocity and D is the propellar diameter. http://web.mit.edu/16.unified/www/SP...es/node86.html 


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