I have been researching the Lilium aircraft and can’t find information about thrust. An early version of this aircraft was 2 seats with a loaded weight of 1,400 lbs, 36 ducted fans and 320 kW peak power. They are testing a five seater also for vertical takeoff. Either the standard calculations for thrust are wrong in ducted fans or something else is going on. Any guesses?


1 Answer 1


The highest thrust is normally needed during hovering. During hovering thrust equals weight so the thrust generated by each fan should be simply:


Since in these designs the total failure of some fans is always taken into account, $nr_{fan}$ is the minimum number of them which can sustain the flight (let's say some 80% of the total number). For the lilium, I suppose a bit more of thrust is needed during the transition to forward flight so the previous value has to be increased, let's say of some 10%. Obviously this final value doesn't take into account any aerodynamic interference


Just out of curiosity I did some basic calculation. The thrust generated by an ideal not-ducted rotor is:

$ T = \sqrt[3]{2 \rho A P^2} $

For a ducted rotor this increases of some 30%, the power being equal:

$ T = 1.3 \sqrt[3]{2 \rho A P^2} $

From the pictures of the aircraft I would say that each rotor has some 20cm diameter and therefore $A$ becomes:

$ T = 1.3 \sqrt[3]{2 \rho (0.8*36 \pi 0.1^2) P^2} $

where the number of rotors (36) has been multiplied by 0.8 to take into account the possibly failed rotors.

As said the max needed power could be some 10% higher than in hover to cope with the transition from hover to forward flight:

$ T = 1.3 \sqrt[3]{2 \rho (0.8*36 \pi 0.1^2)*(P/1.1)^2} $

In hover thrust equals weight $T = W$. Using the ISA value for $\rho$ and the official 320kW for the total engines power we get some:

$ T = W = 737kg = 1620lb $

which correlates quite well with the given max takeoff weight of $1400lb$


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