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I recently developed a sudden interest in flying. I'm wondering whether a pair of small ducted fans could lift a person off the ground.

Lets say the ducted fan is 15cm in diameter. What would be the most thrust a fan like that could produce?

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  • $\begingroup$ Thrust is directly related by flow rate in your setup, which is limited to chocked flow. $\endgroup$ – Sanchises Jun 29 '15 at 8:04
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    $\begingroup$ @sanchises "Choked" flow. "Choked" means restricted by flow area; "chocks" are also used in aviation: they're the wedges you put around wheels to stop vehicles rolling away. $\endgroup$ – David Richerby Jun 29 '15 at 8:38
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    $\begingroup$ @DavidRicherby Oops... I guess chocks can be a limiting factor for most aircraft engines as well... ;) $\endgroup$ – Sanchises Jun 29 '15 at 10:06
  • $\begingroup$ @sanchises Chocks aren't so much a limitation for helicopters, though. :) However, they could still be a problem for whatever (or whoever) gets hit by them as they get blown away by the rotor wash. $\endgroup$ – reirab Jul 1 '15 at 13:44
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    $\begingroup$ I wonder what defines a ducted fan. What's the difference between a ducted fan and a cold jetfan? Fan exit pressure? If you don't limit the fan exit pressure or nozzle exit velocity, you can get quite some thrust out of it. $\endgroup$ – user3528438 Aug 21 '17 at 15:38
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A 15 cm diameter fan could maybe lift 4 kg of mass with a tip speed of 0.6M.

By increasing the power and the blade chord length (in other words, increasing the solidity of the rotor), this may be increased to maybe 6 kg or more, but a single fan would probably never lift more than 10 kg mass.

The thrust of a rotor is $C_T \cdot \rho \cdot \text{Area} \cdot \text{tip speed}^2$.

  • $\rho$ = air density ($1.225~\mathrm{kg~m^{-3}}$ at sea level).
  • $\text{Area}$ = area of rotor disc ($\mathrm{m^2}$)
  • $\text{tip speed}$ = the linear speed at the tip of the rotor = radial speed * radius.
  • $C_T$ = coefficient of thrust. (is usually in the order of 0.02 to 0.05 for small rotors)

Assuming $C_T$ of $0.03$, the calculation shows $40~\mathrm{N}$ ($=4~\mathrm{kg}$) thrust.

Disc area is one of the most important parameters, even a 30 cm disc would generate 4 times the thrust. 60 cm would make almost 65 kg.

Apparently, a 50 cm diameter disc can possibly generate 45 kg thrust on its own. So two of them could carry and maybe even lift-off a person equipped with this backpack (the person + the system must be less than 90 kg).

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  • $\begingroup$ Thank you for being for thorough in your answer. Lets say there are two ducted fans each 20cm in diameter. Utilizing the best technology and engineering could we squeze out over 60kg of thrust out of them? $\endgroup$ – Alexander Yurchenko Jul 2 '15 at 21:12
  • $\begingroup$ I found this NASA's research on magnetically levitated ducted fans here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20070006851.pdf Is it more efficient, thrust power wise, than a regular ducted fan? I wonder if there where two 20cm in diameter versions of this design would that be enough? $\endgroup$ – Alexander Yurchenko Jul 2 '15 at 21:16
  • $\begingroup$ I saw the 0.02 to 0.05 CT values as results of some tests in a rotor design paper. My assumption was 0.03 and with squeezing, that might maybe become 0.06, which is double the numbers I've given. 20cm diameter would not give enough thrust. You could play with the numbers. thanks for accepting this answer. $\endgroup$ – Gürkan Çetin Jul 2 '15 at 21:31
  • $\begingroup$ I'm trying to write out the equation. And not really getting the same results. In the equations where you got 4kg of thrust (with the 15cm diameter) what did you use for your RPM? $\endgroup$ – Alexander Yurchenko Jul 3 '15 at 23:40
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    $\begingroup$ in my sample calculation, the tip speed was M0.65 = 215 m/s. That means a radial speed of 215 / 0.075 = 2866 rad/s. which happens to be more than 27000 RPM. quite a huge number. $\endgroup$ – Gürkan Çetin Jul 4 '15 at 4:48
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Calculating Thrust

One design required sixteen 20-inch (~50 cm) propellers to lift a person.

These were unducted but I doubt that any 15cm ducted fan produces more thrust than eight 50cm propellers can.

See Autonomous human transport for details of how the designer calculated thrust.

He used Thrust (pounds) = R2D4Tc where Tc is an empirically measured constant for which he had a value of 2.7734 x 10-12. R is RPM, D is diameter (inches).

I imagine max RPM might be limited by the need to keep the fan tips subsonic (e.g. < M0.5).

Note that thrust is shown as depending on the fourth power of diameter, sixteen 50cm propellers will therfore produce about 1000 x the thrust of two 15 cm propellers of the same design at the same RPM.

Ducted vs Free

It seems you need to be careful when comparing ducted fans with propellers. Using higher RPM to compensate for smaller diameters results in lower efficiencies (you need bigger motors).

Small diameter, high disk loading ducted fans are often conceived to allow the use of a high rpm engine running a direct drive propeller. While these highly loaded fans (if properly designed) will be more efficient than a free propeller of the same diameter, they typically won’t match the efficiency of a larger free propeller (of much lower disk loading)

From Duct Myths, Duct Physics

Twin ducted-fan backpack

The $150000 Martin Jetpack uses two ducted fans powered by a 2-litre two-stroke engine of 200hp (~150000 watts?). The fan diameter looks much larger than your 15cm. The width of the machine is given as over 2 meters so I'd estimate the fan diameter is close to 80 cm.

The company website doesn't say how they calculated thrust. From their use of larger diameter fans I'd guess there are reasons that 15cm fans are unsuitable.

enter image description here

Related

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  • $\begingroup$ There's also the Gravity jet pack, which uses two jet engines strapped to each arm and a fifth one strapped to the back for flight. $\endgroup$ – nick012000 Dec 8 '20 at 1:01
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An Apache AH-64 has a rotor that's about 100x times the diameter, so the swept area is 10000x larger. It can lift around 10000 kg, which means your ducted fan would lift about 1 kg. You'd need 2 fans with approximately one meter diameter.

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    $\begingroup$ It's not just about size. Plus the author is specifically asking for ducted fans. $\endgroup$ – Antzi Jul 1 '15 at 16:42
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    $\begingroup$ @Antzi: True, but that's a major factor. See RedGrittyBrick's answer, in particular the two fans of about 80 centimeters. "Ducted" is more about personal safety and noise than about thrust. $\endgroup$ – MSalters Jul 2 '15 at 20:43
  • $\begingroup$ The interesting thing is that this extremely simple example hits on the correct order of magnitude, according to the other answers. This shows how much back of the envelope calculations can tell... $\endgroup$ – vidarlo Dec 7 '20 at 7:32
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Here are technical data for existing 15 cm EDF from Schübeler Technologies GmbH https://www.schuebeler-jets.de/en/products/hst-en Technical Data DS-130-DIA HST® with DSM 7857-470:

Inner shroud diameter: 152 mm Fan swept area: 130 cm² Weight incl. motor, wires, connectors and Secure Fan Fix: 1750 g Static Thrust Range: 135 – 175 N Thrust range: 92 – 105 m/s Exhaust speed range: 17.500 – 20.000 rpm Input Power: 8,0 – 12,0 kW Allowed battery: 12 – 14S 14000 mAh Overall efficiency:76 – 74%

Regards blue

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Thrust (in lbs) = 9.35(horsepower x diameter of ducted fan in feet)2/3 [power of 2/3]

This is the formula I recall from the book.

Assuming you design your fan for high static thrust:

for a 6 inch diameter to produce say 300lbs of thrust (lifting man and 100lb machine) you will need a 360hp powerplant.

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    $\begingroup$ From what book? $\endgroup$ – fooot♦ Apr 14 '17 at 15:12
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    $\begingroup$ You can't keep throwing power at a fan and expect an unlimited thrust increase. The blades will stall at a certain point. Every lifting surface has a maximum $C_L$ $\endgroup$ – Koyovis Aug 21 '17 at 14:21
  • $\begingroup$ Indeed, but a multi-stage set of rotors and stators has no such limitation (c.f. jet engines). You can do the math - given a standard atmosphere's density and pressure, you can calculate the mass flow into a 15 cm diameter hole that has zero pressure behind it. Given that mass flow, you can calculate the velocity it needs to be expelled to achieve the desired thrust. Multiply that velocity times the thrust and you get the power needed. Of course, that assumes zero losses from rotor and stator drag, motor, circuitry, etc. so the real life would be even worse. $\endgroup$ – JimHorn Aug 31 '20 at 20:19

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