I am referring to the airplane shown here.

The battery is said to have 640 Whr capacity (actually this is the discharge rate), the load being a 2700 W motor. Then the endurance is giving as 110 minutes. Neglecting any other effects and focusing on the battery-engine, how is the battery time calculated?

I guess one can't know without knowing the battery's capacity in Coulomb.

  • \$\begingroup\$ Wh is a measure of energy, not current. \$\endgroup\$ Apr 13, 2014 at 4:24

3 Answers 3


Specifications here: http://www.uavfactory.com/product/69

  • Energy = power x time (Wh)

  • Duration = Energy / power hours or seconds

The battery is said to have 640 Whr capacity
(actually this is the discharge rate)

No. Capacity is capacity and is correctly stated in Wh = Watt-hour
Capacity = Energy available = Power x time = (here) Watts x hours.

the load being a 2700 W motor.

The missing link is that they are claiming a MAXIMUM (implied) motor power of 2700 Watts.

Your overall problem is not appreciating the various units used and how they relate.

Power = instantaneous rate of doing work. Unit = Watts.

Energy = Sum of Power x time products. Unit = Watt.seconds = Joule, or watt hours.

A motor that runs at a POWER level of 100 Watts for one hour (3600 seconds)
uses 3600s x 100W
= 360,000 Watt seconds of energy
= 100 Watt hours.

So if
the available ENERGY is 640 Wh
and the duration = 110 minutes
= 110/60 ~= 1.83... hours

then IF the figures are correct at the power level involved then
MEAN power = Energy/time_operating
= 640 Wh / 1.83h ~= 350 Watts.

The actual Watts will vary across the operating cycle and
the battery capacity will vary with discharge rate
and load is affected by windspeed and air density and flight profile and ...
and some of these vary with time of day and location and ...
so such figures are an approximation.

  • \$\begingroup\$ Aha...So they are assuming the motor is operating at way less than its maximum value, but they, for marketing maybe, won't explicitly mention that! \$\endgroup\$
    – student1
    Apr 13, 2014 at 15:04
  • \$\begingroup\$ It MAY be called on to produce maximum power during takeoff run and climbout but in level flight it will use very much less. FWIW: Power taken by aerodynamic drag ~~~= 0.5 x Da x Cd x A x V^3. In SI units Da = air density = 1.2 kg/m^43 at sea level, Cd = drag coefficient = say 0.5 to start, A = frontal projected area and V = velocity in m/s. That will ABOUT equal cruise Watts. Very useful formula - works for all sorts of things. \$\endgroup\$
    – Russell McMahon
    Apr 14, 2014 at 20:36

Now as Ignaci pointed out, the "640Whr" is a meassure of energy not current.

Now ask yourself, what is the voltage output of these batteries? Therefore; Capacity in Ah(Discharge rate per say) =640Whr / voltage

Say voltage = 7.4v Then the batteries discharge rate = 86.48 Usually the discharge rate for good cbatteries is 30C for RC batteries. But it depends, you will have to check for the discharge rate. Say it's 30C. Then your capacity = 86.48/30 = 2.8Ahr = 2800mAhr

Now you can easily find out the amout of current withdrwal by the motor by simply I = P/V And devinding the 2800mA / Motor's current in mA = The amount of time you can use your batteries.

  • \$\begingroup\$ So with no information about the voltage/discharge rate I cannot tell for sure, right? \$\endgroup\$
    – student1
    Apr 13, 2014 at 5:02
  • \$\begingroup\$ As for the voltage? you can easily measure it using a multimeter. Your main problem is with discharge rate. \$\endgroup\$
    – Adel Bibi
    Apr 13, 2014 at 5:38

640 Whr/2700W=0.237 hrs, regardless of anything else.

0.237 hrs x 60 mins/hrs= 14.2 mins

Given 110 mins duration (=1.833 hrs), 640 Whrs / 1.833Hrs gives you 350 W maximum absorbed power from the plane.

  • \$\begingroup\$ Thanks, they should have explicitly mention that. From 2700 to 250 is a HUGE difference! \$\endgroup\$
    – student1
    Apr 13, 2014 at 15:06

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