I'm trying to find where additional battery capacity becomes worthless in relation to the added weight in terms of a quadcopter. Currently with a 5500 mAh battery, 11.1V, I can get between 12 minutes and 12:30 flight time out of it. My question, then, is this - within the quads lifting capability of course, is there any way to find out where the added weight of a larger battery (or more batteries) cancels out any flight time improvement? Obviously it's not going to be as long as two separate flights, landing and swapping batteries; I'm just trying to maximize my continuous 'in air' time. I'm trying to figure out where the line is (and if I've already crossed it) with tacking bigger batteries onto the quad and seeing diminishing returns. Thanks!

(Again, for now presume that the quad is strong enough to lift whatever you throw at it. With one 5500mAh, ~ 470 grams, my max throttle is about 70%)

For quadcopter designs, you generally want to have around a 2:1 thrust-to-weight ratio. That is, at 100% throttle, you want your combined propeller thrust to be capable of lifting two times the weight of the craft.

You then need to determine the amount of power the motors need to generate that thrust (usually given on the datasheet). When you have that amount of power, you can easily calculate the amount of time your battery will last.

Looking at the data from the motor performance chart, it looks like the absolute maximum weight the motors can support is 560 each, so 2240 grams. This is when the motors are working at 100%, using 95.2 watts, which means about 8.6 amps at 11.1 volts.

Looking at the data from the flight weight vs power and battery life graph, it seems that the helicopter should not exceed 61 ounces (1700 grams). At that weight, 61 watts of power is used, which is 5.5 amps at 11.1 volts.

The graph also shows that one ounce of flight weight would need one watt of power, and the two are linearly correlated. Assuming the craft is 60 ounces, 60 watts of power is needed.

This quote is a bit misleading though. The original author is missing the words "per motor" in their conclusion. 230W/4 motors = 57.5 Watts. That's where they got the "60 ounces needs 60 Watts (per motor)" conclusion. However the general method of plotting performance is beneficial to answering the question.

If you increase the maximum thrust-to-weight ratio, you can keep your throttle lower, use less power, and stay in the air longer. So you can either increase thrust or decrease weight to achieve longer flight times.

To more concisely answer your question, the point of diminishing returns comes when the additional weight added by the battery brings your thrust-to-weight ratio below 1.5:1 on the low end.

Something else to consider is the energy density of your battery technology. How much power per unit weight the battery is capable of. You're probably using Lithium-ion batteries which are probably the best available from a price to performance perspective. But if you're really trying to get longer flight times without increasing weight, you can consider some of the more esoteric (and much more expensive) technologies out there.

And it's probably worth mentioning that even within the Lithium-ion category, not all batteries are created equal.

The batteries should weigh roughly the same as everything else on the airframe combined.

With a given aircraft -- everything known except for the mass of the batteries -- the absolute maximum flight time occurs when the weight of the batteries is twice as much as the weight of everything else on the airframe combined. However, many other characteristics -- time to get up to altitude, responsiveness, avoiding stall, etc. -- work better with less battery weight, so a better rule of thumb is batteries roughly the same (one times) the weight of everything else combined. Reference: Andres 2011. Optimal Battery Capacity. via Heino R. Pull.

This rule of thumb applies to all kinds of battery-powered aircraft -- fixed-wing, quadcopters, helicopters etc.

(Forgive me for recycling my answer to Effectiveness of a Mobile Robot In Relation To Mass ).

EDIT: fixed a self-contradictory sentence -- thanks, Gustav.

• This confuses me, weight of the batteries is twice as much as the weight of everything else. Does not that mean that the batteries should be 2/3 of the vessel? – Gustav Aug 18 '14 at 17:19
• @Gustav: Have you considered maybe looking at Fig. 1 and the last section of the paper by Andres that I linked? When (a) optimizing for absolute maximum flight time, then yes, the weight of the batteries should be roughly two times the total weight of everything else combined. However, (b) many other characteristics -- time to get up to altitude, responsiveness, avoiding stall, etc. -- work better with less battery weight, so a better rule of thumb is batteries roughly the same (one times) the weight of everything else combined. How can I say this more clearly? – David Cary Aug 18 '14 at 19:52
• Thanks! Doing that distinction between those two cases makes it much more clear. You did not write that in your answer. – Gustav Aug 18 '14 at 21:39

have you seen these new zinc air 8cell 10,000mAh 12V drone batteries? http://www.tradekorea.com/product-detail/P00265914/drone_battery_10000mAh.html#.UjpxM2QpY2K saw the chart on different power to weight ratios on the above discussion comparing efficiency to fuel cells (which are still too heavy 1.2kg + fuel^800g+)

was toying round with the idea of a quad fitted with a zinc air 10,000mAh battery, 4xGenesys alternators and 2x super capacitors

Barry Densley

And how about a little experimenting?

I would take the lightest battery i can find and then try adding weights. For each weight I would measure the flight time, make a "flight time / mAh" vs "takeoff weight" plot and fit a curve through the data points. Then I'd go to hobbyking and for every battery pack there I'd calculate flight time using my new magical graph. Find maximum and you're done :-)