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The actual angle that your sail will 'perform' at depends on the entry angle of the foil shape - a rounder shape will have a wider 'angle of attack' than a flatter entry would.

Think of your jib telltales.. they'll 'fly' on both sides when the entry angle evenly splits the apparent wind angle.. a fat, rounded sail shape will have to bear off more to achieve that condition - so it's working at a wider angle of attack vis-a-vis the boat's centerline.

Easing sheets also changes your 'angle of attack' similarly... and your optimum point of sail..

One of the "H"s (Jeff or Rich) can probably give you a more technically correct version

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Not sure if I'm right on this or not - so a grain of salt, please...

In a specific sail, the angle of attack is like Faster explained, but for the boat, like SchockT explained. I think.

In the simplest of terms required for my brain to grasp it, I have always thought of ONLY the very leading edge of a given sail as the best indicator of the angle of attack. As you trim the headsail tighter, for example, the leading edge of the sail tries to become more parallel to the centerline of the boat. I imagine that angle from centerline to the is the 'angle of attack', and that helps me to understand whether I'm pinching or footing, if I'm getting mixed messages from the tell-tales.

Any thoughts on this?

Changing sail shape can change it, but doesn't necessarily. Having a deeper cut jib for instance, loosening the halyard doesn't, but pulling the jib track back does.

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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?

Not sure if I'm right on this or not - so a grain of salt, please...

In a specific sail, the angle of attack is like Faster explained, but for the boat, like SchockT explained. I think.

In the simplest of terms required for my brain to grasp it, I have always thought of ONLY the very leading edge of a given sail as the best indicator of the angle of attack. As you trim the headsail tighter, for example, the leading edge of the sail tries to become more parallel to the centerline of the boat. I imagine that angle from centerline to the is the 'angle of attack', and that helps me to understand whether I'm pinching or footing, if I'm getting mixed messages from the tell-tales.

Any thoughts on this?

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That is geared towards airplane wings, but the principles are the same. The main thing to remember about our sails is that the chord line and the camber line are different at different points along the sail. This is sometimes true for airplane wings also, but not always.

In other words, for any sort of "normal" sail, there isn't just one angle of attack that describes the entire sail. There may be a different angle of attack at each point as you move up the sail.

The suggestion to imagine a huge sheet of plexiglass is a good one for getting the general conceptualization. At the detail level, though, the leech of a sheet of plexiglass wont act like the leech of a fabric sail would. As the sail becomes tighter or looser, the leech moves differently at different points, which is what causes the angle of attack to be different at different points.

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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.

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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.

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Billy.....I am saying f*ck you in a humorous good natured way thatThis 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.

I would respond to a friend, and I hope you take it that way. Good one!!

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Tightening the out haul would not necessarily change the angle of attack very much..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?

What can happen, is that it changes the sail in a way that the AOA gets just to much for the sail shape...

Without moving the boom, the sail would get a flater shape - twist aside now - but the AOA does only increase slightly - the clew is moved further aft and therfor the line from the clew to the cord changes slightly in regard to the apparent wind, which is the AOA as explained above .. It is something that you sometimes do not want to have, since rounder shapes allow for a wider range of AOAs than tighter shapes...

It could happen that you stall a sail just by tightening the out haul...

The same goes for the jip lead... Bringing it back flatens the sail and increases the AOA even more than on the main, bringing it forward does the opposite...

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Flattening the sail decreases the draft of the sail, making it easier to stall, but provides more drive before it stall. However the finer AOA it is shaped for makes it harder to drive the boat inside this narrow band.

As an example, let's say a sails normal designed AOA is 30 degrees. Within a +\-5 degrees. So long as the driver can keep the boat within this 10 degree margin the sail pulls nicely, and everything is good. When we flatten the sail, the designed AOA stays the same 30 degrees, but the margin shrinks to +\- 2 degrees, but provides a better lifting surface inside this margin. However trying to drive this tight groove is very difficult.

Depowering can be acomplished thru inducing extra twist in sails, where you flatten the bottom to twist the top open. Best exemplified by bringing the jib car back. This flatten the bottom third, opens the top third, and power is generated by the middle. This also allows a wider slot, since some part of the sail is always trimmed right, regardless of the driver, but which part of the sail is drive producing may change. The bottom third during an up phase, the top third during a down phase.

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