# Forces involved analysis …

### We try to do a little analysis of what are the forces acting on the section of a blade of our generator, faithful to the principle explained in this post.

#### Static analysis of the distribution of lift and drag forces.

We start from an analysis of the stationary machine: It assumed null the angular velocity of the blade around the main axis ω = 0.
The lift and drag values are dependent by incidence of the blade section relative to the wind direction, for simplicity in the analysis of the breakdown we will write only P and R to indicate the functions P (i) and R (i) with i = angle of incidence.

P(i)=Cp(i) ρ S V^2
R(i)=Cr(i) ρ S V^2

#### Resistance component

We define: b the distance of the blade section from the main axis, α the angle that describes the position of the blade and V the speed of the true wind. We calculate the length of the arm in function of α:

b = 2 r abs ((sin (α / 2))).

We calculate the resistance of the blade along the tangent of the circle that represents the direction in which the blade is free to move: R ‘= R cos (α), the other orthogonal component is nulled by the constraint .

We calculate the moment (M) respect to the main axis:

M = b*f = b*R’ = – 2 r R cos(α) ∣sen(α/2)∣

From this formula we obtain:

•   α = 0° we have b = 0 , then M = 0;
•  α = 90° we have R’ = 0,  R is perpendicular to the direction of the blade , then M = 0;
•  α = 180° =>  b = 2r , R =R’ and R has max value.
•  α = 270° => R’ = 0 , R is perpendicular to the direction of the blade , then M = 0;
• -90°<α<90° the resistance opposed to the motion of the blades and vanishes for α = 0° .

We can observe that in the starting phase, when angular velocity is null, the wind resistance contributes negatively to the displacement of the blade for a half cycle (-90 °> α <90 °) in which the arm is, however, smaller and the resistance the same is lower due to the transverse position to the wind of the blade.

Contrary with 90 °> α <225 ° there is a positive contribution of the resistive component on the half cycle which has longer arm.
Assuming a linear relationship between the coefficient of resistance and angle of incidence with a maximum value of R = 1, setting b (max) = 1 and S = 1 we have the trend as in the figure:

#### Lift component

We try to do a similar analysis to the lift force generated by the blade: We define

β = α – 90 °

as the angle between the lift vector and the direction of the blade ( tangent line on the α position).

we can write the P component of the lift as P ‘= P cos (β).
then:

• on α = 0° =>  b = 0 and  M = 0;
• α = 45° => M is approximately = 0.268 * 2r* P(i) ==>so it contributes to the motion
• α = 90° => M is approximately = 0.707 * 2r* P(i) ==> contributes to the motion
• α = 180° => M = 0 , P is perpendicular to the direction of the blade .
• α = 225° => M is approximately = -0,65 * 2r *P(i) , P reverses the direction and still has that lift contributes to the motion of the blade in the desired direction.

We can therefore say that: the contribution of the lift to the motion of the blade is always positive except for α = 0 ° and α = 180 °, respectively, in which they cancel the arm length and lift force.
Here, respectively, the graphs relating to the lift contribution (| cos (β) | * | sin (α / 2) |) depending on the angle α and, approximately, the value of the coefficient of lift force (Cp (i)) based on the angle of incidence (α / 2) for a flat surface:  By product of the two graphs you get the trend of the force transmitted to the machine from the lift force.

The peak of lift occurs with an angle of incidence of about 15 ° (flat surface) which corresponds to α = 30 ° where the contribution of the lift blade movement is about 70 ~ 80%.

#### Dynamic operation In the above image we try to establish how  the apparent wind vector (v ‘) is positioned  respect to the (section of)  blade when the tangential velocity  is close ( or equal) to the true wind speed (v).

We can see as, approaching the tangential velocity to the real wind velocity, the vector v’ tends to form a smaller angle respect to the profile of the blade (i => 0 °), and in consequence of this:

The resistance tends to decrease considerably and its vector tends to rotate counterclockwise. The lift force occupies a major role in generating force to be transferred to the machine and improves the efficiency with the approach to an angle of incidence of 15 ° (maximum lift of the yield, see image on the left) that is obtained with a tangential velocity equal to ~% 96 true wind speed.

The machine behaves as resistive generators (eg Savonius) with vertical axis with resistance operation at startup, this also leads to a lower start-up wind (speed lower cut).

As the speed increases, the behavior is like to wind turbines with horizontal axis, where the lift force becomes the main component in the generation of the force that actively contributes to the motion of the blade.

In next post an detailed analysis on blade profile…stay tuned.

p.s. : Sorry for my bad english.