Drag reduction of a gyrocopter rotor

Jean Claude

Junior Member
The larger the rotor, the less induced drag it will have at a given speed. But this also reduces the auto-rotation Rrpm, and to maintain an acceptable coning in flight, the blades have to be made heavier. A limit is reached due to the bending stress at rest in the roots. When the blades have a constant chord (i,e rectangular), this limit is reached for an aspect ratio of around 20. This is what Igor Bensen chose, and which has just been copied for over 60 years.
However, a tapered shape could further increase the diameter without increasing the bending stress of the roots, and therefore further reduce drag.

Let's start by determining the point of application of the weight, centrifugal force and lift of such a tapered blade.

Position of gravity center (point of application of weight)
The chord of a trapezoidal blade, at a distance r from the hub, can be expressed by the formula
c = C - k (r/R) where C is the root chord, and k a coefficient determining the tip chord.
Thus, k = 0 for a rectangular blade, and 0<k<1 for trapezoidal blades.
Assuming an average density K1 of the material of construction, a blade element located at a distance r from the rotor center and of small width dr will have an elementary mass dm such that :
dm = K1 c² dr
Replacing c by its value gives: dm = K1 [C - k (r/R)]² dr

This means that the total mass of the blade will be :
M blade = ∫ dm = K1 C² ∫[1 + k² (r/R)² - 2 k r/R] dr Hence M = K1 C² (r + ⅓ k² r³ /R² - r² k/R) ,
So between 0 and R the blade mass will be M = K1 C² R (1 + ⅓ k² - k)

On the other hand, the weight of each blade element of mass dm produces a bending moment dΓ at the root, proportional to the distance from the center r, hence dΓ = dm. g r = K1 C² g [r + k² r³/R² - 2 k r²/R]
And the total moment will be: Γ = K1 C² g ∫ (r + k² (r²/R)² - 2 k r²/R) dr
= K1 C² g (½ r² +⅟₄ k² r⁴ /R²- ⅔ k r³/R
From 0 to R: Γ = K1 C² g R² (½ +⅟₄ k² - ⅔ k)

And since this moment Γ is due to the weight M.g of the blade (4) located at a certain distance X from the center, we can therefore say that
X = Γ /M g = K1 C² g R² (½ +⅟₄ k² - ⅔ k) / K1 C² R g (1 + ⅓ k² - k)
Hence X = R (½ +⅟₄ k² - ⅔ k) / (1 + ⅓ k² - k)
for example If k = 0.4 then X = 0.42 R

Position of the lift center

Since the chord c is C (1 - k r/R), then an element of width dr located at a distance r from the center has the elementary surface dS = c. dr = C (1 - k r/R). dr and the lift of this surface being proportional to the square of the velocity/air is :
d(Fp) = K2 r² Ω² dS (assuming a constant angle of attack)
Hence, replacing c by its value: d(Fp) = K2 C Ω² (1 - k r/R) r² dr
and the blade lift is therefore Fp = K2 C Ω² ∫(1 - k r/R) r² dr = K2 C Ω² ∫( r² - k r³/R) dr
= K2 C Ω² (⅓ r³ - ⅟₄ k r⁴/R)
Hence Fp = K2 C Ω² R³ (⅓ - ⅟₄ k)

Lifting moment Mp produced.
The lift of a blade element produces a moment d(Mp) that tends to lift the blade:
d(Fp). r = K2 Ω² C (1 - k r/R) r^3 dr = K2 Ω² C (r³ - k r⁴/R) dr (with K2 = ½ ρ CL)
So the sum is given by : Mp = K2 Ω² C ∫ (r³ - k r⁴/R) dr
= K2 Ω² C (⅟₄ r⁴ - ⅕ k r⁵/R)
Or the total moment from 0 to R: Mp = K2 Ω² C R⁴ (⅟₄ - ⅕ k)

Since this moment is due to the distance to the center Y of the lift (5), it is also equal to Fp .Y and we have
K2 Ω² C R⁴ (⅟₄ - ⅕ k) = K2 C Ω² R³ (⅓ - ⅟₄ k) Y
Hence Y = K2 Ω² C R⁴ (⅟₄ - ⅕ k) / K2 C Ω² R³ (⅓ - ⅟₄ k)
We derive the position of the center of lift: Y = R (⅟₄ - ⅕ k) / (⅓ - ⅟₄ k)

Point of application of centrifugal force

We saw above that dm = K1 c² = K1 C² [1- k (r/R)]² = K1 C² [1 + k² (r/R)² - 2 k r/R] (1)
The elementary centrifugal force is therefore: dFc = dm .Ω² r = K1 Ω² C² [r + k² r³/R² - 2 k r²/R]
Integrating gives the sum: Fc = K1 Ω² C²∫ (r + k² r³/R²-2 k r²/R) dr
Fc = K1 Ω² C²(½ r² + ⅟₄ k² r⁴/ R² - ⅔ k r³/ R)
Let the total force from 0 to R be: Fc = K1 Ω² C² R² (½ + ⅟₄ k² - ⅔ k)

With the blade tip at a height H , each element produces an elementary flattening moment: dℳ = dFc r (H/R) = K1 Ω
With the blade tip at a height H , each element produces an elementary flattening moment: dℳ = dFc r (H/R) = K1 Ω² C² [r² + k² r⁴/R² - 2 k r³/R] (H/R)
whose sum is: ℳ = K1 Ω² C² (H/R) ∫(r² + k² r⁴/R² - 2 k r³/R) dr
ℳ = K1 Ω² C² (H/R) (⅓ r³ + ⅕ k² r⁵/R² - ½ k r⁴ /R)
Let , between 0 and R: ℳ = K1 Ω² C² R³ (H/R) (⅓ + ⅕ k² - ½ k ) (3)

This therefore corresponds to a centrifugal force (2) applied to a height h, such that
h = ℳ / Fc Hence, replacing ℳ and Fc by their respective values (2) and (3):
h = R (⅓+ ⅕ k² - ½ k ) / (½ + ⅟₄ k² - ⅔ k) , a point consequently located at distance d from the center of the blade such that:
d = R (⅓+ ⅕ k² - ½ k ) / (½ + ⅟₄ k² - ⅔ k).

On the other hand, we also know that the cone angle in flight is due to the equality of the moments of lift Mp and flattening ℳ
So, K2 Ω² C R⁴ (⅟₄ - ⅕ k) = K1 Ω² C² R³ (H/R) (⅓ + ⅕ k² - ½ k )
Thus TAN β = (H/R) = K2 R (⅟₄ - ⅕ k) / K1 C (⅓ + ⅕ k² - ½ k )

Rotor inertia
Mass of a blade element dm = K1 c².dr = K1 C² (1 - k r/R)² dr
= K1 C².dr = K1 C² [1 + k² (r/R)² - 2.k r/R] dr
Its inertia relative to the center is dJ = dm.r² = K1 C² [1 + k² (r/R)² - 2 k r/R] r² dr
Hence Irotor = 2 K1 C² ∫ (r² + k² r⁴/R² - 2 k r³/R) dr = 2 K1 C²(⅓ r³ + ⅟5 k² r⁵/R² - ½ k r⁴/R) and
Jrotor = 2 K1 C² R³(⅓ + ⅟5 k² - ½ k)
Using this data, we can see that despite 30% heavier blades, an 8 m diameter rotor with tapered blades of 0.185 m root chord and 0.115 m tip chord (k = 0.38) would not have to undergo any more rest stress than a 6.5 m rotor with rectangular blades of 0.18 m chord.
The taper would be the same, and despite the significant change in diameter,
Detailed comparison below:

Conclusion:
The calculated drag of the 6.5 m x 0.18 m (rectangular) rotor being 382 N at 90 km/h, while the calculated drag of the 8 m x (from 0.185 m to 0.115 m) rotor being 327 N, the tapered blades would allow a gain in Vz at 90 km/h of (382 N - 325 N)* 25 m/s / 2120 = + 0.6 m/s

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Now how about non-linear tapered blades with a twist (in the tip-high direction)?

The left column in the table is for the rectangular blades and vice-versa, right?

Jerry,
If I refer to S. Hoerner's book, " Fluid dynamic drag" he says §7-7 that the extra drag due to the unadequation of a wing twist is
ΔCd min = 4.10-5 α².
Assuming that α is about 2 degrees between the center and the tip, then ΔCd = 16. 10-5, i,e 1.5% of Cd. Which doesn't justify the manufacturing difficulty.

Chuck Beaty experimented with the tapered
blades. I think he was using a Boeing VR-7. They were constructed from fiberglass and wood. The spar was wound fiberglass.

Dick DeGraw approached it an other way. He reduced drag sign significantly by partially powering his rotors. His flies much flatter angle than a normal set of Dragon Wings auto rotation. In my observation and opinion this enables his machines to fly using much smaller displacement engines. The efficiency is demonstrated by the fact that his machines are far more fuel efficient, fly faster at a given power setting, remain one of the quietest machines I’ve ever heard flyby. His twin propeller machine is amazingly fast and amazingly quiet. When he brought it to Mentone, the first time we didn’t realize it had fallen over until it flew by because we never heard it and he was doing about 110 or 115 miles an hour. Just my thoughts of another way to accomplish the same objective using different thoughts and techniques

As I explained, the blade tapering is beneficial because the diameter can be increased without increase the bending stress to the roots at rest.
But without increase of diameter, drag is rather increased, compared with rectangular blades.
Do you know the diameter chosen by Chuck Beaty for his tapered blades?

In my opinion, and according to my spreadsheet calculations, direct drive does indeed reduce rotor drag, but the power saved in propeller thrust is equivalent to that added to the rotor shaft...Unless the rotor pitch is adjusted in flight.

I’m not quite sure. I will ask my father tomorrow when I meet him at Bensen Days. If I had to guess and based on my memory and knowledge of his previous experiments I would say between 22 ft and 24 ft.

Nice analysis JC! It is often under appreciated that the root loads, when the rotor is not rotating, is often the critical condition. Along those lines, any idea what the typical load factors are on typical rotor blades for static (non rotating) root loads. As far as I can tell section t does not specify any negative load factors for the rotor.

As I explained, the blade tapering is beneficial because the diameter can be increased without increase the bending stress to the roots at rest.
But without increase of diameter, drag is rather increased, compared with rectangular blades.
Do you know the diameter chosen by Chuck Beaty for his tapered blades?

In my opinion, and according to my spreadsheet calculations, direct drive does indeed reduce rotor drag, but the power saved in propeller thrust is equivalent to that added to the rotor shaft...Unless the rotor pitch is adjusted in flight.
Jean Claude, Do you think the Prandtl D lift distribution curve is applicable here?
You seem to be suggesting that the efficiency increase only happens with increased span/Diameter.
For Prandtl D there would also have to be washout, or twist with reduced AOA at the tips... is that also applicable to a rotor?

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Jean Claude, Do you think the Prandtl D lift distribution curve is applicable here?
You seem to be suggesting that the efficiency increase only happens with increased span/Diameter.
For Prandtl D there would also have to be washout, or twist with reduced AOA at the tips... is that also applicable to a rotor?
The induced drag of the rotor in forward flight (like a wing) is minimum when the induced flow (downwash) is constant at all points in the rotor disc. Prandl theory in just this as an elleptical wing gives constant cl . Most classical performance calculations on rotors are done assuming constant induced flow and it turns out untwisted blades in the real world are within a couple of percent of this ideal (constant induced flow) . IOW no great benefit to adding twist for a gyro rotor at least from the perspective of induced drag.

On the other hand, twisting the blade (root nose down and tip nose up) can help with delaying stall. This is because the stall develops first at the inboard sections in a gyro rotor. Still the effect as far as I can tell is minimal but others have found some benefit.

The induced drag of the rotor in forward flight (like a wing) is minimum when the induced flow (downwash) is constant at all points in the rotor disc. Prandl theory in just this as an elleptical wing gives constant cl . Most classical performance calculations on rotors are done assuming constant induced flow and it turns out untwisted blades in the real world are within a couple of percent of this ideal (constant induced flow) . IOW no great benefit to adding twist for a gyro rotor at least from the perspective of induced drag.

On the other hand, twisting the blade (root nose down and tip nose up) can help with delaying stall. This is because the stall develops first at the inboard sections in a gyro rotor. Still the effect as far as I can tell is minimal but others have found some benefit.
Prandtl D lift distribution curve is not the elliptical wing planform, it superceded the elliptical wing design and nobody understood it with the exception of the Hortens and Lippish.
Al Bowers rediscovered it and figured it out while working at NASA and built a few flying wing models using Hortens method and Prandtl D.

I was using it in the 90's and early 2000's in flying wing designs, but did not realize that I was using a different methodology, it just happened organically by studying Horten's work in depth. I only had the Ah Ha! moment afte reading Al Bower's paper.....
I did not use enough twist to fully utilize Prandtl D, but as the airplanes get smaller and into low RE #'s I believe the twist and span ratio becomes less important. Mine were all under 14ft. span.

Prandtl D lift distribution curve is not the elliptical wing planform, it superceded the elliptical wing design and nobody understood it with the exception of the Hortens and Lippish.

In the traditional (wing) Prandtl solution the problem is posed as given a fixed (constrained) span what is the best loading for the wing. In this case the answer is indeed elliptical. Another way to pose the problem, as in Prandtl D, is for the span to not be constrained but the bending moment at the root to be constrained. Solving this results in a slightly larger span and bell loading that I suspect is what you are referring to. Al Bowers brought it back in more recent times.

Now the question is how does this apply to rotors ?

Well first due the way the blades carry the lift (thrust), tension is the primary loading and not bending moment at the root, like in a traditional wing. So in other words Prandtl D is not relevant. However, bending moment (due to weight of blade) in the static (non rotating case) case at the root is relevant. Though in this case the rotor is not lifiting. This is what JC addressed

Like Prandtl D, JC posed the following question: What if you constrain the downward bending moment (at rest) then does the increase rotor diameter by linearly tapering the chord result in lower rotor drag ? The answer he showed was that it indeed reduced rotor drag.

To generalize and take JCs work a step further would be to ask what is the optimal distribution of rotor chord based on a constrained root bending moment due to blade weight. I am sure this has been solved in the literature ..... but could be a fun to derive....

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Jean Claude, Do you think the Prandtl D lift distribution curve is applicable here?
You seem to be suggesting that the efficiency increase only happens with increased span/Diameter.
For Prandtl D there would also have to be washout, or twist with reduced AOA at the tips... is that also applicable to a rotor?

Just my answer applicable to autorotating rotors. In short not even worth bothering about. Too small a gain to go do all the work and expense.

Probably correct, but the taper and increased span would accomplish part of the design anyway.
The other big question mark for me is how much does the centrifugal pumping along a rotary wing contribute, or increase spanwise flow?
Part of the increased efficiency of Prandtl D is greatly reduced spanwise flow and the resultant tip vortices that follow....

On another note in the same song, could downward angled tiplets re-direct the spanwise flow into a lift component thus increasing efficiency?
At some point the forward speed would potentially be generating a downward force on the LE of the entire disk (And maybe a separation on the lower surface?), but would it be an issue at typical low gyro speeds?
Or could the angle be the exact opposite of the coning angle, so it presented it's low side at the LE?

Probably correct, but the taper and increased span would accomplish part of the design anyway.
The other big question mark for me is how much does the centrifugal pumping along a rotary wing contribute, or increase spanwise flow?
Part of the increased efficiency of Prandtl D is greatly reduced spanwise flow and the resultant tip vortices that follow....

On another note in the same song, could downward angled tiplets re-direct the spanwise flow into a lift component thus increasing efficiency?
At some point the forward speed would potentially be generating a downward force on the LE of the entire disk (And maybe a separation on the lower surface?), but would it be an issue at typical low gyro speeds?
Or could the angle be the exact opposite of the coning angle, so it presented it's low side at the LE?

Making tapered rotors is much more expensive than simpler straight ones. Increasing the span is the best bang for the buck easily.
I don’t know what centrifugal pumping means.
Instead of doing triplets it’s much simpler and safer to do a small wing on the fuselage that is set so that at cruise speeds it starts to create lift without much of a pitching moment like using UI-1720 airfoil.

Prandtl D is irrelevant with rotors as mentioned. Improving span efficiency for a rotor has very little impact- span efficiency is already very high and further given the large profile losses of the rotor, any benefit from higher span efficiency is very minimal.

Prandtl D is irrelevant with rotors as mentioned. Improving span efficiency for a rotor has very little impact- span efficiency is already very high and further given the large profile losses of the rotor, any benefit from higher span efficiency is very minimal.
That is a blanket statement that I would reject for lack of proof.
My initial point was that some of Prandtl D could be inferred by Jean Claude's increased efficiency with span and taper, maybe with the twist to counter spanwise flow.
This is all hypothetical, but seems to have some real confirmation with the generalization of increased span increases efficiency with a tapered section.

As far as tapered being harder to make, with composites, the mold does not care......

That is a blanket statement that I would reject for lack of proof.
My initial point was that some of Prandtl D could be inferred by Jean Claude's increased efficiency with span and taper, maybe with the twist to counter spanwise flow.
This is all hypothetical, but seems to have some real confirmation with the generalization of increased span increases efficiency with a tapered section.

As far as tapered being harder to make, with composites, the mold does not care......

You are right the mold does not care but the taper and washout maybe gets you 1.5% overall in the aircraft. Barely noticeable. Easier and better to make the rotor 6” longer and get noticeable performance difference.

The mold may not care, but the (frequently metal) spar does. It is inexpensive to squirt out extruded spars, once you've invested in the die.

The mold may not care, but the (frequently metal) spar does. It is inexpensive to squirt out extruded spars, once you've invested in the die.
The spar or LE extrusion can remain linear, in fact it could more than likely be the same extrusion that was being used in aluminum blades.
All the taper/twist/ tip shape modifications would happen in the mold with the composite layup.

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