center of lateral resistance

I can see how using the area of the hull to estimate the CLR would work on a traditional long keel hull, however if you were talking about something with a deep fin keel or centerboards then it would be a different matter as the effect from a foil is more dependent on its shape than its size.

Are you talking about working out the contribution of the hull then adding the effect of the appendages to it?

 

No leeway = no lift....
No angle of attac means no lift....

 

The hull can be (I think) set up with lee helm. The rudder is used to make the hull foil assymetric. There is plenty of sail trim available with 24 sails on a fully-rigged ship. Steering is still possible with the helm at - say - 5 degrees to leeward.

I remember aspects from Colin Mudie's "Sailing Ships", and this is my interpretation of his comments. I feel that our ancestors were better at getting ships to windward than our "just start the engine" modern crews. Its not easy, but it can be done. With the death of cheap oil, it may be an advantage to re-learn such tricks.

 

I rest my case
If you beleive that is more effiecient than a modern foil....

 

Imagine looking at a sailboat from directly above. as a boat heels, the center of effort of the sails moves further and further away from the center of resistance of the hull. The further it goes, the more torque you get forcing the boat into the wind.

A good example is sailing downwind in quartering seas with a non-planing boat. When a wave lifts the stern quarter, you get massive weather helm, as the wave moves along, and the stern quarter sinks into the trough behind it, you get massive lee helm.

Ergo, planing boats are a lot easier on the nerves, and a lot more fun.

Yoke.

 

Lateral area vs Leeway?

If lateral area is used to balance sail forces, how much leeway is needed?

For example:
The total lateral area of a hull, daggerboard and rudder at 15 degrees of heel is 25.7 ft^2.

The Sail Carrying Power is 291lbs at 15 degrees of heel.

For 25.7ft^2 to generate 291lbs of force it must be moving laterally at 2.38fps. That's assuming that the total lateral area has the same drag as a flat plate (Cd=2.0).

The upwind speed is 5 knots or 8.44fps.

Is a lateral speed (leeway) of 28% of boat speed correct?

The dagger board is 8.16ft^2

For Cl=0.5 and 8.44 fps the daggerboard produces 290 lbs of lift (lateral resistance).

The lateral hull and rudder area have only to generate 1 lb of lateral resistance.

To get 1 lb of lateral resistance only 0.14fps of leeway is needed (assuming Cd=2.0). This is 1.7% of forward speed.

Using the total lateral plane area to determine a "balance point" for Sail Area CE lead can be very wrong. It would only apply if the CLP of the hull, rudder and keel is in line with the 1/4 chord of the keel.

The lateral resistance centre is the resolved sums of the hull, rudder, and keel forces. It may or may not fall close to the lateral center of area.

Rudder load depends on the distance between the boats CG and the centre of lateral resistance. If the CLR is forward of the CG it imparts a weather helm moment to the hull. The rudder must balance this, either with leeway velocity and drag or with lift. The total lateral force must act through the boat's CG. As the keel is moved further forward of the CG the rudder load increases. When the 1/4 chord of the keel is moved aft towards the longitudinal CG the rudder load decreases.

Just make things a bit more complicated ...

When the keel is lifting it also creates a yaw moment due to induced drag. Thus a keel placed at the LCG of the hull will require a load on the rudder to balance the yaw induced by the keel's lift.

To set the "lead" of the sail plan. The magnitude and distance from the CG are resolved to find the moment at the CG. There is a pitch down moment due to distance above the CG. There is a weather going yaw moment due to heel and a lateral force.

By moving the sail's centre of effort forward (increasing distance to the CG) it creates a yaw moment about the CG that opposes the yaw due to heel.

This is why a boat with lead tends to have lee helm at small heel angles and weather helm at large heel angles. The longitudinal distance between the sail effort and CG remains constant while the horizontal distance changes. In other words the lee helm arm is fixed while the weather helm arm changes.

If the boat is to have the same "balance" or feel at all normal heel angles, the CE must move forward with heel or the CLP most move aft with heel.

The rudder's force can be changed to move the CLP (not CLA). Indeed, increasing rudder angle with heel is exactly what most boats require to maintain course as heel increases.

 

You get a higher lift coeffiecient than that of a flat plate dragged sideways.
I clip from http://www.grc.nasa.gov/WWW/K-12/airplane/kitelift.html:

For a thin flat plate at a low angle of attack , the lift coefficient Clo is equal to 2.0 times pi (3.14159) times the angle a expressed in radians (180 degrees equals pi radians): Clo = 2 * pi * a
We use Clo for the lift coefficient because there is another aerodynamic effect present on most kites. If we think of a kite as an aircraft wing, and use the terminology associated with aircraft wings, most kites have a low wing span (length from side to side) relative to the surface area. Most kites therefore have a low aspect ratio AR which is defined to be span s squared divided by the area A. AR = s^2 / A

Near the tips of a wing the flow spills from the underside to the top side because of the difference in pressure. This creates a downwash which changes the effective angle of attack of the flow over a portion of the wing. For low aspect ratio wings, the portion of the wing affected by the downwash is greater than for high aspect ratio wings. Since most kites have a low aspect ratio AR, we have to include the effect of the downwash on the lift coefficient. The equation for the correction is given at the bottom right of the slide: Cl = Clo / (1 + Clo / (pi * AR) )

 

Its good because it's all a matter of degree. If the CE is too far ahead of CLP, you will have lee helm (boat tending to point down wind).

If it's not far enough forward of CLP, you will have unacceptable (pull your arms out of their sockets) weather helm (boat tending to point upwind).

Experienced designers develope a feel for this sort of thing. How much is enough and how much is either too much or too little.

Bob

 

RHough:

Except for a few too many acrynoms, I was able to read and totally comprehend your post. I am not able to verify your calculations nor comment whether there are any fundamental flaws in your logic; but, you could write "Leeway For Dummies" and I would buy it.

 

SuperPiper, I think that using the Cd=2.0 gives to little lift.
With 5 degrees leway you have a lift coefficient of approx 2.0.
The lift is dependent on the speed forward and the angle of attac, not only the sideways component of the speed. If you sail fast you will have less leway.

 

And Raggi_Thor:

That explains exactly why my fat little boat does not sail very well to windward.

 

Yes
If you sail at 3 knots your lateral area (or your keel) should be two and a half time as large as if you sail in 5 knots. 3x3=9 while 5x5=25

 

What section gives Cl=2.0 at 5 degrees AOA?

Most keel sections give Cl=.5 at 5 degrees.

 

I just used the formula for a flat plate, Clo = 2 * pi * a
5 degrees = PI * 5/180 ~0.17 Rad
0.17 * 2 * pi = 2.2
Maybe This is not applicable for that angle of attac...

 

Thanks.

The graph shows a positive Cl at 0 AOA ... can't be done with a flat plate or uncambered foil.

At 0 AOA the lift line of a plate passes through the origin.

IIRC correctly the upper limit of Cl for a symmetric foil is around 1.5 or so and the lift line is between .10 and .11 per degree for NACA sections.

I thought maybe you had section that I hadn't discovered.