I don't know if that is right, but it's what I was thinking. Saying the chord of the sail relative to the wind doesn't take into consideration the shape (draft) of the sail. If the sail is flat then maybe the chord would determine the angle. Let me ask this, without moving the boom, would tightening the out haul change the angle of attack. And by the same token, would moving the jib lead change the angle of attack for the head sail?
OH, yes, I was just going to say that.This is all you really need to know:
The airfoil is modeled as a thin lifting mean-line (camber line). The mean-line, y(x), is considered to produce a distribution of vorticity \gamma (s) along the line, s. By the Kutta condition, the vorticity is zero at the trailing edge. Since the airfoil is thin, x (chord position) can be used instead of s, and all angles can be approximated as small.
From the Biot–Savart law, this vorticity produces a flow field w(x) where
w(x) = \frac{1} **(2 \pi)} \int_{0}^{c} \frac **\gamma (x')}**(x-x')} dx'
where x is the location where induced velocity is produced, x' is the location of the vortex element producing the velocity and c is the chord length of the airfoil.
Since there is no flow normal to the curved surface of the airfoil, w(x) balances that from the component of main flow V, which is locally normal to the plate—the main flow is locally inclined to the plate by an angle \alpha - dy/dx. That is:
V \; (\alpha - dy/dx) = w(x) = \frac{1} **(2 \pi)} \int_{0}^{c} \frac **\gamma (x')}**(x-x')} dx'
This integral equation can by solved for \gamma(x), after replacing x by
\ x = c(1 - \cos (\theta ))/2 ,
as a Fourier series in A_n \sin(n \theta) with a modified lead term A_0 (1 + \cos (\theta)) / \sin(\theta)
That is
\frac{\gamma(\theta)} **(2V)} = A_0 \frac **(1 + \cos(\theta))} **\sin(\theta)} + \sum A_n \; \sin (n \theta))
(These terms are known as the Glauert integral).
The coefficients are given by
A_0 = \alpha - \frac {1}**\pi} \int_{0}^**\pi} (dy/dx) \; d\theta
and
A_n = \frac {2}**\pi} \int_{0}^**\pi} \cos (n \theta) (dy/dx) \; d\theta
By the Kutta–Joukowski theorem, the total lift force F is proportional to
\rho V \int_{0}^{c} \gamma (x) \; dx
and its moment M about the leading edge to
\rho V \int_{0}^{c} x \; \gamma (x) \; dx
The calculated Lift coefficient depends only on the first two terms of the Fourier series, as
\ C_L = 2 \pi (A_0 + A_1/2)
The moment M about the leading edge depends only on A_0, A_1 and A_2 , as
\ C_M = - 0.5 \pi (A_0+A_1-A_2/2)
The moment about the 1/4 chord point will thus be,
\ C_M(1/4c) = - \pi /4 (A_1 - A_2) .
From this it follows that the center of pressure is aft of the 'quarter-chord' point 0.25 c, by
\ \Delta x /c = \pi /4 ((A_1-A_2)/C_L)
The aerodynamic center, AC, is at the quarter-chord point. The AC is where the pitching moment M' does not vary with angle of attack, i.e.,
\frac ** \partial (C_{M'}) }** \partial (C_L)} = 0
Got that? Good!
Next question......
Ops....forgot to credit Wiki for the above.
This is all you really need to know:
The airfoil is modeled as a thin lifting mean-line (camber line). The mean-line, y(x), is considered to produce a distribution of vorticity \gamma (s) along the line, s. By the Kutta condition, the vorticity is zero at the trailing edge. Since the airfoil is thin, x (chord position) can be used instead of s, and all angles can be approximated as small.
From the Biot–Savart law, this vorticity produces a flow field w(x) where
w(x) = \frac{1} **(2 \pi)} \int_{0}^{c} \frac **\gamma (x')}**(x-x')} dx'
where x is the location where induced velocity is produced, x' is the location of the vortex element producing the velocity and c is the chord length of the airfoil.
Since there is no flow normal to the curved surface of the airfoil, w(x) balances that from the component of main flow V, which is locally normal to the plate—the main flow is locally inclined to the plate by an angle \alpha - dy/dx. That is:
V \; (\alpha - dy/dx) = w(x) = \frac{1} **(2 \pi)} \int_{0}^{c} \frac **\gamma (x')}**(x-x')} dx'
This integral equation can by solved for \gamma(x), after replacing x by
\ x = c(1 - \cos (\theta ))/2 ,
as a Fourier series in A_n \sin(n \theta) with a modified lead term A_0 (1 + \cos (\theta)) / \sin(\theta)
That is
\frac{\gamma(\theta)} **(2V)} = A_0 \frac **(1 + \cos(\theta))} **\sin(\theta)} + \sum A_n \; \sin (n \theta))
(These terms are known as the Glauert integral).
The coefficients are given by
A_0 = \alpha - \frac {1}**\pi} \int_{0}^**\pi} (dy/dx) \; d\theta
and
A_n = \frac {2}**\pi} \int_{0}^**\pi} \cos (n \theta) (dy/dx) \; d\theta
By the Kutta–Joukowski theorem, the total lift force F is proportional to
\rho V \int_{0}^{c} \gamma (x) \; dx
and its moment M about the leading edge to
\rho V \int_{0}^{c} x \; \gamma (x) \; dx
The calculated Lift coefficient depends only on the first two terms of the Fourier series, as
\ C_L = 2 \pi (A_0 + A_1/2)
The moment M about the leading edge depends only on A_0, A_1 and A_2 , as
\ C_M = - 0.5 \pi (A_0+A_1-A_2/2)
The moment about the 1/4 chord point will thus be,
\ C_M(1/4c) = - \pi /4 (A_1 - A_2) .
From this it follows that the center of pressure is aft of the 'quarter-chord' point 0.25 c, by
\ \Delta x /c = \pi /4 ((A_1-A_2)/C_L)
The aerodynamic center, AC, is at the quarter-chord point. The AC is where the pitching moment M' does not vary with angle of attack, i.e.,
\frac ** \partial (C_{M'}) }** \partial (C_L)} = 0
Got that? Good!
Next question......
Ops....forgot to credit Wiki for the above.
Tightening the out haul would not necessarily change the angle of attack very much..
