**Power and force at take-off**In an ADAV/VTOL, the power of the engine must be converted both for an ascent in helicopter mode and for a horizontal translation in airplane mode.

We have already seen that in airplane mode,

Ideally, large propellers were needed that did not accelerate the fluid much. to obtain the best propulsive efficiency.

What about takeoff?

We are going to put things simply enough to draw informative equations and estimate an order of magnitude. Is an airplane's engine powerful enough to lift it in a helicopter mode?

We will therefore assume a single helix of section

**S** blowing the air downwards. This air passing from a speed 0 to a speed

**v**. The density of air being

**ro**.

- schema_simplifie_decollage_helicoptere.png (30.2 Kio) Viewed 682 times

Thus the mass flow of air crossing the propeller is:

- eq19.png (2.27 KiB) Viewed 711 times

By the principle of action/reaction, the force

**F** exerted by the air on the propeller is:

- eq20.png (4.86 KiB) Viewed 711 times

For a mass

**dm** of fluid passing from speed 0 to speed

**v** for a while

**dt**, the change in kinetic energy

**dec** fluid is written:

- eq21.png (10.72 KiB) Viewed 711 times

by forming the ratio dEc/dt, we recognize the power

**P** supplied by the propeller, while dm/dt identifies with the mass flow rate of fluid through the propeller:

- eq22.png (7.59 KiB) Viewed 711 times

At this stage, several things should be noted* strength

**F** and the power

**P** are proportional to

**ro x S**, Nevertheless,

* the force progresses as the

*square.* speed

* power as the

*cube* speed.

**But is this speed v arbitrary? **The answer is no. If we form the P/F ratio, the products

**ro x S**, disappear and it comes:

- eq23.png (9.73 KiB) Viewed 709 times

Immediately note:

**F=2P/v**,

*which shows that the force ***F** will be optimal if the speed v is small, which amounts to having a large rotor to sweep a surface **S** , which is important.But it would be good to clarify things. Now that we have the expression

**v = 2P/F** , we inject it for example into the relation

**F = ro S v²**
- eq24.png (13.19 KiB) Viewed 709 times

In aircraft design, the force F is imposed by the weight of the machine, but it is possible to play on the section S swept by the propeller. of the relationship

**F^3 = 4 ro S P²**, we therefore express P as follows:

- eq25.png (3.91 KiB) Viewed 708 times

**Very instructive relationship!**For a strength

**F = Mg** necessary for take-off (where

**M** is the mass of the plane and

**g** the acceleration of gravity),

the parameter **S** is the only degree of freedom to reduce the necessary power.

A reactor with a small section will be catastrophic for lifting the device. It will succeed, but at the cost of an abundance of lackluster power and beyond the possibilities of a small plane (equipped with a few hundred horsepower at the most).

**But this is good news, because we had seen that the propulsive quality of the plane required:**

* a large propeller section, and,

* low acceleration of air masses,

which is found exactly for the ascent quality (but it's not 100% amazing).

*There is therefore no incompatibility, but on the contrary a technical HARMONY in designing the aircraft with large propellers, both for airplane mode and helicopter mode.*To fix ideas a little, if we take:* ro = 1,2 kg / m3

* M = 1000 kg

* S = 10 m²

* with g = 9,81 m²,

the power required is 140 240 W,

**i.e. about 190 Ch**.

If you feel like laughing with the section of a 0,5 m² blower reactor... P = 627 W,

**i.e. about 852 Ch**. In addition to consuming torrents of fuel, the weight on board will be enormous in a piston engine, unless you switch to a gas turbine, which also has a muddy performance.

*By increasing the number of our rotors, for example by ***quadcopter with 4 x 4m propellers** of diameter,

*

**S** = 4 x Pi x 2² = approximately 50 m²

*

**P** = 62 W, or about 718 Ch, which is the power of light aircraft engines (weighing about 100 kg for 50 Ch).

In real conditions, there will be losses, and

according to Wikipedia, a correction factor of 1,5 is required:

the losses due to viscosity, as well as various other losses (the power required by the anti-torque rotor, the losses of the gearbox, etc.) represent approximately 50% of the minimum power of

*froude* (see

Froude–Rankine theory). A realistic estimate of the engine power of a helicopter can therefore be obtained by multiplying the above formula by a factor of 1,5.

We will remember : an engine of a few hundred horsepower can sustain an ADAV of around 1000 kg provided it is fitted with numerous propellers with cumulative sections of at least 10 m², which is also favorable to propulsion efficiency in airplane mode.