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I've read from many sources that Dyson Air Multipliers are more efficient and quieter than normal fans. Now, with the proof of concept, is it possible to use its principles as a propulsion system for, say, a quieter helicopter?

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  • $\begingroup$ Do you mean the bladeless fan, the so-called air multiplier? Rather than hand drier air blade? Though maybe the technology is identical. $\endgroup$ – innisfree Oct 25 '13 at 9:42
  • $\begingroup$ Modern turbojet engines already do this $\endgroup$ – MSalters Oct 25 '13 at 16:22
  • $\begingroup$ @MSalters: Not really. There is bladed fan that moves air, just not through the turbine. $\endgroup$ – user23660 Oct 25 '13 at 18:27
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    $\begingroup$ (I should have said turbofan, as a turbojet proper does not have such a multiplier). @user23660: It indeed uses a somewhat different mechanism, but there's still the multiplier effect, the increased efficiency and reduced noise. Seemed like a sufficient similarity, and with a similar aerospace usage to mention it. $\endgroup$ – MSalters Oct 25 '13 at 19:55
  • $\begingroup$ York - note that helicopters work like airplanes -- because of the shape of the wing. They don't fly by "blowing air down", you know? $\endgroup$ – Fattie Aug 17 '14 at 8:48
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Unlike John Rennie, I think that the problem is not in the efficiency of this system but in the fact, that it will not generate considerable lift. So even if marketing materials are completely true and Dyson Air Multiplier is more energy efficient than conventional fans this efficiency only applies to moving air (which is its intended use) but not to the lifting force.

The principle behind Air Multiplyer (see this video) is creating the flow around the surface of the duct which induces considerably greater flow through the duct. However the resulting flow would be nearly potential and the net force on the duct would be quite small.

Somewhat similar effect does occur in helicopters: Vortex ring state, where under certain conditions increasing the air flow through the rotor does not produce additional lift. In helicopters this is harmful and could even cause the crash, but the Air Multiplier effectively creates similar 'vortex' around the duct for the purpose of moving air.

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This is really a comment, but it got a bit too long for the comment box.

The term efficiency has a very specific meaning in Physics. In the case of a fan we can measure the power the fan consumes simply by measuring what current it draws from the main. We can in principle measure the power the fan produces by measuring the velocity of the air stream produced. The efficiency would then be the ratio of the power produced to the power consumed.

I can't find figures for the efficiency of the Dyson fan, but I would be very surprised if it was anything like as high as a conventional fan. The Dyson fan sucks air in at the base, pushes it along internal tubes to the ring where the air is expelled, under pressure, though orifices in the ring. All of this consumes energy and decreases the efficiency. By contrast a conventional bladed fan has none of these energy losses.

NB you may have seen claims that the Dyson fan moves 15 times as much air as it pumps through the ring. This is not a comment on the efficiency but rather it's the fan equivalent of a gear ratio.

So the bottom line is that I have seen no evidence that the Dyson fan is more efficient than a bladed fan, and therefore better for use in airplanes or helicopters, and would guess the reverse is true.

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  • $\begingroup$ Why do expect the dyson to be less efficient? The mechanical side is probably more efficient in the dyson? Then what other losses are there? Noise, and air displacement in the wrong direction. Both will be higher in a conventional fan. $\endgroup$ – innisfree Oct 25 '13 at 9:55
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    $\begingroup$ When you force air through a small orifice under pressure you will increase viscous losses and some of the work you put in is wasted as heat. The Dyson fan forces air through a small orifice under pressure while a conventional fan does not. $\endgroup$ – John Rennie Oct 25 '13 at 10:01
  • $\begingroup$ @JohnRennie: Building efficient helicopter means producing lift force large enough to keep it in the air with minimal power consumed. So your definition of efficiency is not the main quantity for question at hand. Better use as figures of merit 'thrust-to-weight' and 'thrust-to-power' ratios. $\endgroup$ – user23660 Oct 25 '13 at 14:09
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I honestly thought about it myself before. I think it is to do with the fans power-to-weight ratio.

I couldn't find the fans technical specifications, but adoring to Amazons shipping weight (6.4 lb = 2.9 kg) of a 10 inch (25 cm) Dyson fan, we can estimate it weighs around 2 kg.

This is an awful lot for such a small device! Traditional propellers this size can weigh around 200g (For example the ones used in model aircraft. Of course they are optimised for low mass, yet the difference still seems large.

NB: I know this is a physics forum, and in this post I used mass and weight interchangeably, but I can't think a mass verb like "weigh"

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It seems like the dyson fan uses a regular fan to pull air into its base creating its air multiplying effect. If that's the case a single fan and motor could drive multiple dyson structures at once, which could result in an overall lower net weight. Not sure how well these things would perform outdoors in air turbulence, but they might make for nifty indoor propulsion for a quad if they produced sufficient airflow.

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