Vortices from a gyroplane

Tyger

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I was flying to the Simsbury, CT (4B9) fly-in a couple of years ago, which can get pretty busy, and I had the misfortune (or made the mistake) of landing just behind a pretty big medevac helicopter. He was coming in on a right pattern to runway 3, and everyone else was using left. Anyway, I had a good view of him, he was ahead of me, flying a short right base and final approach, and then he slid immediately over to the helicopter parking area, just west of the approach end of 3 (see diagram here: http://simsburyflyin.com/PROCEDURES)
Well, what I should have done was just go ahead and land long, but what I actually did was try to land just at the approach end, in what turned out to be a big mess of turbulence from his still turning rotor (there was also a pretty good crosswind from the west). Just as I was about to touch down, I suddenly found myself several feet in the air, and then WHAM, I came down fast and hit the runway really hard. Luckily, I was able to keep everything aligned and only my pride was actually damaged (I had a very long look over my landing gear once I had taxied off). Anyway, I consider that a valuable lesson learned about helicopter wakes.
 

kolibri282

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The German company Fraundorfer Aeronautics has developed an aircraft which, they claim, has a rotor efficiency 30% better than existing systems. As you can see in the picture on their webpage the rotor blade is indeed tapered.
https://www.tensor.aero/technology/
The picture below shows, that the specific energy consumption of their Tensor 600 X aircraft is between that of a fixed wing and a helicopter, where you would actually expect a gyroplane to be, whereas most existing gyros have an energy consumption about one third higher than a helo.
I think this basically shows that helicopters over the years have seen a lot of research and improvement whereas autogyros were side lined and have quite some potential for further development still.
signal-2022-10-25-05-49-15-177.jpg
 
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Resasi

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A rotor would simply shed fixed wing style winglets due to centrifugal acceleration, the bending moment created by winglets attached at anything more than the most shallow of angles wrtp. the rotor would just have them fly into the wild blue yonder.
Thank you Kolibri, and again with the post above. Like you I feel that there is more that could be done with gyro rotors to improve their efficiency, but like the old adage, "If you don't see it in aviation it's one of two things, cost or weight.

Autogyros have indeed been a niche backwater that has been neglected. Jay Carter had ideas for aircraft that seemed to almost go back to Fairy Rotordyne proportions. At one time I believe he was working with DARPA on a C-130 scale project but that seems to have come to nothing.

This new concept by Fraundorfer looks very exciting. With the much lower maintenance costs per flying hour than a helicopter, but some of its versatility over fixed wing, I think there are definite commercial possibilities, that today, are not yet available for existing gyros.

With regard to vortices behind gyros. I have for many years dealt with, on a daily basis, and, been very aware of, vortices behind commercial aircraft. As a flight instructor in the sixties operating from an airport that was sandwiched between Miami International and Ft Lauderdale International, and operating into Lauderdale it was inevitable that I would experience encounters with these. They were invisible violent and on one occasion put a briefcase in the back of a 150 completely through the rear wrap around canopy at the back.

Many years later, in a much larger Citation Excel, I was on an Instrument flight plan under radar control and exactly 10 miles in trail of a 747 who had configured for approach going into Riyadh, when we hit his vortices. The aircraft was rolled violently to the left. Luckily, I feel, I was hand flying and managed to stop the roll at what seemed almost a 90 degree bank. We rolled back to straight and level, then, a few seconds later were then rolled as violently to the right. This time already fully charged with adrenaline and alert, this was corrected with much less bank. A terrifying example of just how powerful and lingering these invisible menaces can be.

With these experiences, and knowledge, I was recently flying together with a friend in gyros, and decided to carefully experiment with flying in trail behind him, to see what sort of vortices he was producing. I began quite far back, and moved up and down to see what effect I could find. It was a reasonably still day and no matter where I moved I found little detectable disturbance. In the end I had moved to quite close but still found minimal effect.

This is by no means saying that serious vortices are not produced by gyros, simply, that for a similar sized gyro behind, it seems that they may not be as disruptive. Vortices are produced in proportion to the size and weight of the aircraft producing them, along with, I would expect, wing and, wing tip design. I would expect a heavier two seat gyro to produce more, and higher intensity vortices.
 
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rcflier

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Testosterone maybe...but surely adrenaline....;)
 

Tyger

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Which brings us to today's lesson in the classical origins of common words ;)

ad-renal, above the kidneys (Latin)
epi-nephrine, from above the kidneys (Greek)

And in Latin the word testis means... witness (plural, testes).
 

Resasi

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Testosterone maybe...but surely adrenaline....;)
Which brings us to today's lesson in the classical origins of common words ;)

ad-renal, above the kidneys (Latin)
epi-nephrine, from above the kidneys (Greek)

And in Latin the word testis means... witness (plural, testes).
Thank you both.:)

My "humble' lesson for today. :oops:
 

Aerofoam

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With these experiences, and knowledge, I was recently flying together with a friend in gyros, and decided to carefully experiment with flying in trail behind him, to see what sort of vortices he was producing. I began quite far back, and moved up and down to see what effect I could find. It was a reasonably still day and no matter where I moved I found little detectable disturbance. In the end I had moved to quite close but still found minimal effect.

Probably more noticeable and developed closer to the ground, especially within the range of ground effect.......
 

Abid

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The German company Fraundorfer Aeronautics has developed an aircraft which, they claim, has a rotor efficiency 30% better than existing systems. As you can see in the picture on their webpage the rotor blade is indeed tapered.
https://www.tensor.aero/technology/
The picture below shows, that the specific energy consumption of their Tensor 600 X aircraft is between that of a fixed wing and a helicopter, where you would actually expect a gyroplane to be, whereas most existing gyros have an energy consumption about one third higher than a helo.
I think this basically shows that helicopters over the years have seen a lot of research and improvement whereas autogyros were side lined and have quite some potential for further development still.
View attachment 1156421

I am still trying to understand what problem they are trying to solve. They have 14 foot wings. Obviously that will give them more efficient cruise than a standard gyroplane. So are they saying it is their whole gyroplane or just their rotors in auto-rotation that are more efficient. If it is a claim about their whole gyroplane then have they decided if it is the 14 foot wingspan that is a much bigger contributor in the 30% more efficient number?

Second, so it is not as efficient or as fast as the airplane and cannot hover or lift heavy weights like helicopters. It's cost is as much as a 2 seat R22 at least if not more. So where does it fit? I am looking for a problem that it uniquely solves better at its price point. Do you know Kolibri?
 

kolibri282

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First and foremost I regard this as a tech demonstrator, showing what can be achieved through modern engineering, but with gas prices where they are and quite likely to become higher still a more efficient aircraft should, in my opinion, have a fairly bright future. Perhaps in professional applications like policing, surveillance or camera work rather than for recreational flying but not only every police department but every operator will be glad if they can save on operating costs through higher efficiency. If you clock a lot of operating hours a higher initial investment will, at some point, pay off. I will try to contact them and ask if the would like to comment here.

As for the question of the 30% efficiency increase I think this is for the rotor, since the figure of 0.9 at the best specific efficiency for the conventional gyro in the diagram is more than twice the minimum efficiency of 0.4 for the Fraundorfer and the curve for the conventional gyro increases quite a bit steeper beyond the minimum, so the advantage at higher speeds is even greater.

PS: Quote: "If you don't see it in aviation it's one of two things, cost or weight." /Quote
Thank you Leigh, this is a really nice one, I will add it to my little collection of Aeronautical/Flying quips....;-)
 
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XXavier

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I find the minimum specific energy required of 0,9 Wh/kg-km very high for a gyro, since the corresponding figure given for a helicopter is 0,6 Wh/kg-km...

Is the gyro so inefficient in comparison with a helicopter...?
 

kolibri282

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For a Bensen type gyro, weighing 168 kg, my simulation program gives a power required of 44.88 kW at a flight speed of 96.3 km/h (= 60 mph) .

Plugging these values in gives 43880 W / (168 kg * 96.3 km/h) = 2.715 Wh/(km*kg)
For a Magin VPP M16 I get


47850 kW /(371 kg * 96.3 km/h) = 1.33 Wh/(kg*km)

There is no speed scale in the diagram but it looks to me as if these figures are in the ball park.


For a Cessna 172, weight 2300 lb = 1042kg, cruising at 115 knots, I get

92341 / (2300*0.453*115*1.85) = 92341 kW /( 1042 kg * 213 km/h) = 0.42 Wh/(kg*km)

so the minimum value for a fixed wing of 0.2 seems also Ok
 

XXavier

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For a Bensen type gyro, weighing 168 kg, my simulation program gives a power required of 44.88 kW at a flight speed of 96.3 km/h (= 60 mph) .

Plugging these values in gives 43880 W / (168 kg * 96.3 km/h) = 2.715 Wh/(km*kg)
For a Magin VPP M16 I get


47850 kW /(371 kg * 96.3 km/h) = 1.33 Wh/(kg*km)

There is no speed scale in the diagram but it looks to me as if these figures are in the ball park.


For a Cessna 172, weight 2300 lb = 1042kg, cruising at 115 knots, I get

92341 / (2300*0.453*115*1.85) = 92341 kW /( 1042 kg * 213 km/h) = 0.42 Wh/(kg*km)

so the minimum value for a fixed wing of 0.2 seems also Ok


You're probably using a sophisticated simulation program, but I find your figures too high...

A rough estimate follows: for a Bensen gyro with a mass of 168 kg, (weight 1646 N) if the L/D is probably 3,5 [so, the glide ratio is 3,5:1] at 96,3 km/h = 26,75 m/s, the sink speed in a glide is 26,75/3,5 = 7,64 m/s, and the ‘gravitational power’ associated with that glide is 1646 · 7,64 = 12575 W. If, instead of a glide, we want s/l flight at the same speed, the power required would be the same. Thus, for a s/l flight of one hour at 96,3 km/h, the mechanical energy per kg and km to be supplied to the prop, (assuming 100% prop efficiency) would be 12575 W · 1 h/(168 kg · 96,3 km) = 0,78 Wh/kg-km. In the real world, the propeller efficiency would be probably 70 %, so the real figure would be 0,78/0,7 = 1,11 Wh/kg-km...
 

kolibri282

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The value for the Bensen is probably a bit to high but 1.11 is probably a bit to low. I get 1.3 for a Euro-Tub kind of gyro and the diagram seems to depict values for one as well (judging from the aircraft there), a Bensen with those rough and ready wood blades that beats a fully faired aircraft for efficiency seems a bit odd. On the other hand the value for the 172 seems quite Ok, so we both seem to agree, that today's gyros a pretty dirty beasts....;-)
 

Jean Claude

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I find the minimum specific energy required of 0,9 Wh/kg-km very high for a gyro, since the corresponding figure given for a helicopter is 0,6 Wh/kg-km...
Is the gyro so inefficient in comparison with a helicopter...?
Sans titre.png
Several things surprise me about Fraundorfer's graph.
1) Why, at zero speed, does the energy consumed per km travelled not tend towards infinity, since the time of flight then becomes infini?
2) Why, when it comes to quantifying the cost of transportation, is energy related to the total mass, instead of the payload?

For a Magni M16 of 475 kg at 125 km/h I find 1.1 wh/kg/km
For a Robinson R22 at 185 km/h , I find 0.75 wh/kg/km
For a Cessna 150 at 185 km/h, I find 0.4 wh/kg/km

I think that the values announced for the Fraundorfer's autogyro are largely underestimated:
First, I don't see that tapering the blades or twisting them would gain more than 2% efficiency (?) on the rotor.
I assume that the 30% improvement is the decrease in rotor drag, due to the unloading by the fixed wing.
And in this case, the additional drag and weight of the auxiliary wing obviously moderates this gain.
Taking this into account, even with careful aerodynamic work on the parasitic drag, the transport energy does not go below 0.75 wh/kg/km, far from the 0.4 mentioned.

Nota:
Relatively to the payload, I find 3.3 wh/kg/kmfor the Magni M16, 3 wh/kg/km for the Robinson R22, and 1.75 wh/kg/km for the Cessna 150
 
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kolibri282

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Note that only the graph of the helicopter goes all the way to zero, the other ones stop somewhere before that, which, I think, makes sense because at zero speed these aircraft don't stay aloft.

Power is Force * Velocity, thus the coefficient

c = P/(m*V) is also equal to c= (F*V/(m*V) = F/m

so for zero speed of the helicopter you might cancel the velocity in numerator and denominator and get the force required to hover, which does not modify the dimensions of the coefficient. Just an idea.

I am not sure, what you mean by "relative to the payload", it might be easier to see what you are doing if you write out all the numbers you have plugged in to get your results.

I had tried to contact the company but did so far not get any reply.
 

Jean Claude

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My calculation is the same as yours, Juergen
Helicopter R22, for example:
To fly at 185 km/h, it needs 115 hp (ie 86000 W) that makes an energy of 86000 Wh for 185 km.
If it weighs 620 kg, then the specific energy is
86 000/ (185*620) =0.75 Wh/kg/km

So, any non-zero power needed to fly will require infinite energy when the transport speed tends to 0
Thus, I wonder how Fraundorfer reasons to arrive at a finite value when V=0
I am not sure, what you mean by "relative to the payload", it might be easier to see what you are doing if you write out all the numbers you have plugged in to get your results.
Since the payload is only the two passengers (say 156 kg) then the transport energy (per km and per kg paying) is rather:
86000/(185*156)= 3 Wh/kg/km
 
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Jean Claude

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Being now a closed gyrocopter well faired for which the parasitic S.Cd would be 0.3 m2 (instead of 0.7 for a Magni M16), weighing 475 kg.
Its parasitic drag at 200 km/h is then 565 N and its rotor drag is 430 N. This requires a power of 75 kW on the propeller shaft and the specific power would be: 75000/(200*475) =0.8 Wh/kg/km

Let us add a wing of 3.5 m2 with an aspect ratio of 10. carrying 142 kg (i.e 30%), that means that its Cl is 0.210 and its Cdi = Cl^2 /Pi.A = 0.0014
If Cd0 wing = 0.006, then its drag is 0.5* rho* S Cd* V^2 = 50 N approximately, while the drag of the rotor is only 430 N * 0.7 = 300 N
The total drag then requires a power of :
(300 N + 50 N + 565 N) * 55.5 m/s /0.73= 70 kW
and the specific energy would be : 70000/(200*475) =0.74Wh/kg/km
 
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