#### Jean Claude

##### Junior Member

- Joined
- Jan 2, 2009

- Messages
- 2,645

- Location
- Centre FRANCE

- Aircraft
- I piloted gliders C800, Bijave, C 310, airplanes Piper J3 , PA 28, Jodel D117, DR 220, Cessna 150, C

- Total Flight Time
- About 500 h (FW + ultra light)

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.

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

for example If k = 0.4 then X = 0.42 R

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)

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:

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:

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

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

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

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 centerPosition 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 forcePoint 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) = K****2****R (⅟₄ - ⅕ k) / K****1****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

**J****rotor****= 2 K****1****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**
Last edited: