Physics of tethered autogyros, autogyro rotors, and windmills at an angle of incidence to the wind

Several topics on this forum touch on autogyros:

Bryan Roberts' Gyromill (Sky WindPower)

Crosswind gyroplane?

Toward the worst AWES

This topic is to collect resources on the physics of tethered autogyros, autogyro rotors, or on the physics of a windmill (not necessarily wind turbine) at an angle of incidence to the wind.

THE AUTOGYRO FOR SHIP PROPULSION.pdf (1983, 8 pages) (4.9 MB)

NASA/CR—2003–212799 | An Overview of Autogyros and The McDonnell XV-1 Convertiplane (2003, 284 pages)

The Autogyro Rotor as a Sail.pdf (1934, 5 pages) (1.5 MB)

This topic is a nice excuse to post this:

Horizon - The Chopper (BBC2 Documentary, 1982)

GYROKITE - Flying a Gyrocopter with No Motor

The wind turns the rotor in autorotation. Once it gets to around 320 rpm, it will spin no faster and up you go. It takes about a 22mph wind in my case at my weight. I think the rotor was 21’9" in diameter with a 7" chord. A standard Bensen rotor.

Nice job there Chris. I’m just a novice, but fascinated by this. Home build I’m assuming. Do you have to change the angle of attack to make the blades rotate. Is this how you attain the controlled lift?

@geepea101able Home built yes. Rotor is fixed in pitch at about 1 1\2 degrees positive pitch. The cyclic allows for tilting the spinning disk but it remains at 1 1\2 positive relative to the rotor hub. The rotor is changing pitch through out its rotation to create the tilt required and in the direction desired by moving the cyclic.

Jim, not easy to answer as no two circumstances will be exactly the same. I can say pretty much without doubt, if you have in full power and unload the rotor for a span of time, the engine torque alone will make the gyro roll over opposite the torque and it may go quickly inverted (unrecoverable). A sustained unloaded rotor can quickly become fatal. First thing in most cases is reduce the throttle to idle (quickly, my helicopter instincts work well here), thus taking away the torque. Momentary unloading in wind gust is normally not an issue.

GYROKITE PT2 - Bob glides over my head

“The rotor was 21’9” “, so about 6.63 m. The tip speed at “320 rpm” is 111 m/s. “Minimum about 22 mph. Best to have 30 and steady”. If the chord is 0.17 m (a little less than 7” which is the mentioned value), the whole blade area is 1.12 m². Assuming the Cl of the profile is 0.7 and air density is 1.2, as an approximation the lift (kg) is:
1/2 x 1.12 x 1.2 x 0.7 x 111²/ (9.81 x 3) = about 197 kg.

Bensen B8.pdf (7.4 MB)

The motorized version (see pdf) has (based on a gross weight of 500 lbs) a Normal Disk Loading of 1.59 {lbs}/{ft^2}

Trying to understand the numbers, in metric units:

Col A Col B Col C
(a) rotor diameter 609.6 cm
(b) rotor chord 17.78 cm
(c) rotor area 10838.7 cm^2
(d) rotor disk area (304.8cm)^2\pi
(e) rotor solidity c/d 0.0371
(f) Normal Gross Weight 226.8kg
(g) Normal Disk Loading f/d 7.77kg/m^2

@PierreB I don’t know what equation you are using.

I like this image:

The above image on hover lift efficiency is taken from figure 3 of this NASA monograph:

The History of the XV-15 Tilt Rotor Research Aircraft: From Concept to Flight (222 pages, many pictures)

One more search result for disk loading+hover, delftAcopter (2016 TU Delft drone):

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In French language: Rz (kg) = 1/2 x Cz du profil x densité de l’air x somme des surfaces des pales (m²) x vitesse d’extrémité des pales (m/s) au carré / (9.81 x 3).

Rough translation, by replacing kg with N:
Lift (N) = 1/2 x CL of the blade profile x rho (air density) x whole blade area (m²) x tip speed (m/s) squared / 3.

This not complete equation works from the blade area, not the disc area. The division by 3 is due to the efficiency of the rotor, the speed of the blades at the root being zero.

Unfortunately I have no reliable source for this equation, as I found it on forums. For example on :

Équation de portance
F portance =1/2 rô V carré S Cz
Équation de traînée
F traînée= 1/2 rô V carré S Cx.

Portance = lift; traînée = drag. On this equation division by 3 is not mentioned.

If the blade area is 1.08 m² (as you indicate) instead of 1.12 m² as I mentioned, the lift (in kg) would be 190 kg.

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I see a constant k of 1/3 in this paper. Instead of Cl they use normalized\ thrust\ (ηT) however, from a quick skim.

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Hundreds of pages of rotary wings in here.

Rotary-Wing Aerodynamics (Dover Books on Aeronautical Engineering)

ISBN-13: 9780486646473

ISBN-10: 0486646475

Authors: W. Z. Stepniewski; C.N. Keys

Edition: Reprint

Binding: Paperback

Publisher: Dover Publications

Published: July 1984

Also I would not hesitate to use to view the details, then perhaps buy a book from the authors if you like what you see


Indeed, on this paper the thrust equation (20) and also the drag equation (28) use respectively normalized thrust (ηT) and normalized drag (ηD), while the lift equation and drag equation I tried to provide, use respectively lift coefficient (CL) and drag coefficient (CD). 1/3 remains a constant in all equations.