Does anyone know the correlation between hovering time, flight time, and battery life on a copter? How much of the battery will be used just to hover and how much is used to to fly over a distance?

Thank you!

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well it really all about light waight and efficiency , you need a battery light enough with enough power in amps to deal with the waight of the copter, you also need motors ESC that draw as little amps at full throttle as possible. last pecs is prop size lager prop more lift power,  more lift at lower RPM less AMPs used. 

1. light as possible

2. motors ESC that match your props power amp draw

3. power density of bttery


If I have an all up weight of 2.38 kg on a hexacopter with an 8000mAh 3S1P 30C battery, 12x4.5 propellers, and motors that draw 20 amps at full throttle (so ESCs that draw 30 amps), I'm calculating a pretty low flight time (about 4.5 minutes). Do you have an idea of how that time is proportioned into hover time and actual flying?

Thank you so much!

 lower flight time depend how aggressive you fly but how much power in amps you need to hover equal to total time of hover less than additional power to maneuver equal to x time.

First let's calculate ideal power (I'm converting the 2.4 Kg to Imperial units and assuming 1 pound per thrust unit to make it simpler).

In the following equation, we the thrust of about a pound and divide it by the result of two times the air density in slugs times the area of the prop for induced velocity.  Multiplying this by the thrust gives us the ideal power of the airflow:

12" prop copter: square root of (1 pound/2 * 0.00238 * 0.785sqft.) = 16.4 feet/second^-1 and times our thrust is 16.4 pound feet/sec ^-1.  Dividing by 550 yields an ideal power of 0.0298 horsepower or * 746 = 22.2 watts.

22.2 is the ideal power for each prop, and multiplied by 5, gives you the total hovering power of 111 watts.  Of course this is perfect, and most hobby store props have a figure of merit of about .5, so it will take 222 watts of power at the motor shaft. Assuming a motor efficiency of 70%, you need to supply 317 watts.  Given a nominal Lipoly 3S operating voltage of 10.5 volts, a current of 30.2 amps will be required.  So, with an 8 amp hour pack, I get 15.89 minutes of out-of-ground effect hover without subtracting for losses, etc.  15 minutes sounds about right.

Like the highway mileage for your car, this is under steady-state conditions.  Generally speaking, forward flight would provide some efficiency benefits, perhaps as much as 30% for a typical helicopter.  As they say, maneuvering "around town", your mileage may vary. 

Brad, why multiplied by 5?  She's got a hex, so multiplied by 6 wouldn't it?  I get 133.2W, then 266.4W, then 380W.

But then, the operating voltage is a bit low IMO.  Average voltage over the life of the pack is more like 3.9V.  So that requires 32.5A.  Then only using 80% of the 8000mAH, I get 11.8m hovering. 

Hi Kate,

Brad Hughey is our resident expert on exactly this kind of calculation (although I too would like to know why the multiplier was 5) and his method of calculating is very likely to yield a realistic product.

However, you can check out this kind of thing yourself here:

Just fill in number of rotors, prop size, an appropriate motor and battery  and it will produce a pretty reasonable estimate for hover anyway.

It takes its max flight as full power all the time (can't happen) and it's mixed as a 50 / 50 blend of max throttle and hover throttle (also incredibly unlikely).

Hover time is the most realistic figure and normally flying around will not change that figure by very much, some types of flight can actually increase your total flight time a bit over hover, although most (especially if you are countering gusts or a wind) will reduce it somewhat.

And if you are flying acrobatic high G maneuvers flight time reduces a lot (seems very unlikely with a Hex as you have described though.

Based on my own experience and with Xcopter calc I would say Brad's estimates are very close.

Best Regards,


Hi Kate.

Hovering, especially in calm air, draws the least amount of current. Consequently, battery life decreases proportionally as maneuvers become more complex and aggressive. A copter/battery combo rated for say, 15 minutes of hovering may only yield about 5 minutes of aggressive flying.

If you haven't discovered eCalc yet it's an invaluable tool and can accurately predict expected battery life based on your machine's configuration and flying characteristics:


Hi, Rob.  How've you been?

Yep, hex is 6.  My bad.  But that means that the ideal power at each disk is less, so it would hover longer.  5.3 lbs / 6 =  .88

12" prop copter: square root of (.88 pound/2 * 0.00238 * 0.785sqft.) = 15.3 feet/second^-1 and times our thrust is 13.5 pound feet/sec ^-1.  Dividing by 550 yields an ideal power of 0.0245 horsepower or * 746 = 18.28 watts. 18.28 * 6 = 110 watts (kept the extra .3 lbs in there).  Yep, it would be hard to tell the difference, as ~ 1 lb/sq-ft is really light loading.

I was assuming 3.5V/cell, but let's say 3.7.  That means the current would be less for the same power - 220W for a .5 FM at 70% motor efficiency means 314/11.1 = 28.3 amps.  Discharging to 80% as you say is good practice, or 384 amp minutes = 13.57 minutes.

Suffices to say, about twice as long as Kate's original estimate.  :-)

This is incredibly helpful, thank you for taking the time to calculate this! May I ask where the 550 in the division comes from?


Thank you for sharing Brad's credibility, I will definitely put his advice to use.

I have looked at this tool before and found it overwhelming at first, the selections they offer in the battery/motor/controller fields are not exact to what I am thinking of purchasing, but are close estimates. I think I'm getting used to it now, thanks for clarifying the max/mixed flight times and the variations that affect flight time!



Makes sense. I'm going to try to get better with this tool. Thank you!!


I always say, use a calculator to know.  Understand the math and calculate it yourself to understand.  The whole lift thing is based on mass and acceleration of an air column.  What we end up with is a pound-feet per second of air result, and 550 p-ft/sec is equal to one horsepower.  Rather, we could convert directly from p=ft/sec to watts, i.e. 1 pfs = 1.3558 W.

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