Confused by Marchaj

[jam007]

Reading "Aero- Hydrodynamics of sailing" (third edition) I can´t make out the to me conflicting statements on Sweep angle effects (p. 447-460)
On p. 449 he states:
"...a foil with sweep-back behaves similarly to a foil with more taper, while a swept-forward foil act like a foil with less taper."
He goes on at the following pages and illustrates this showing that a swept back foil stalls first att the tip and a swept forward first at the root and gives theoretical reasons for this.

Then at p. 457-458
"[Bad load distribution towards the root or tip leading to larger induced drag due to sweep.] Such a concentration of lift can be alleviated by proper tapering of the foil planform. ... [A figure 2.146 is refered to] For instance, a taper ratio ct/cr of about 0.2 is needed to make the lift distribution of a 30 [degrees] swept-back foil near-elliptical. For the 30 [degrees] swept-forward foil the recommended taper ratio ct/cr is in the order of 1.4, which means that the taper is reversed."

Earlier p 431 he states that a taper ratio of between 0.3 and 0.5 gives near to eliptical loading.

Hm...
the swept-back foil on page 457 should have a smaller ratio than the straight to give it near elliptical loading and the swept-forward higher ratio!? Does this not contradict what is stated at page 449-452

What am I missunderstanding

Anders M

[tspeer]

Sweep loads the aft portion of the foil more than the forward portion of the foil, compared to an unswept foil of the same planform. So the lift coefficient at the end of a swept-back foil is higher than it would be for the unswept foil and the swept-back tip will stall earlier. A swept-forward foil will have a higher lift coefficient at the root, and tend to stall there first.

The spanwise load distribution for a tapered, unswept planform is approximately half-way between being proportional to the taper and the elliptical distribution. A taper ratio in the range of 0.4 - 0.5 results in a lift distribution that most closely approximates the elliptical lift distribution.

As a result, an untapered (rectangular) planform has a lift distribution that is higher at the tip than the elliptical distribution, but not as much as you'd expect from the difference in chord there. So even though the amount of lift is higher near the tip, the lift coefficient is lower. If the planform is more tapered, ie a taper ratio < 0.4 (0 taper ratio is a triangular planform), the lift distribution is lower than elliptical near the tip, but not by as much as the taper would suggest, and the lift coefficient is higher.

Starting from any baseline planform, sweeping aft or adding taper will increase the lift coefficient at the tip. Sweeping forward or adding chord at the tip will decrease the lift coefficient there. So taper and sweep are alike with regard to promoting stall, even though they have opposite effects on the spanwise lift distribution.

BTW, a constant-chord planform with 18 degrees of forward sweep produces a near-elliptical lift distribution for an isolated planar wing with no interference with anything else.

[jam007]

Thanks! Now I understands. I think...
A swept back foil stalls earlier at the tip due to increased lift coefficient and the lift at the tip increases with the lift coefficient as the cord stays the same (or even increases as the flow is diagonal?) to compensate for this far from elliptical lift distribution that results in more induced drag you need to have more taper to shorten the cord.
So. The more you sweep back the more induced drag you get and if you try to decrease the drag by increasing the taper you end up with earlier stall. Hm..

This leads to the question of why the centerboard of my Farrier 25A is swept back. It should be swept forward, shouldn't it?

[Skippy]

Only if you want it to break off when you bash into a shoal.....

Also, I would imagine stalling the tip first keeps the heeling moment down.

[jam007]

Skippy wrote: "Only if you want it to break off when you bash into a shoal....."

Yepp. I assume that is one of the reasons. (But would not the centerboard be self rising if swept 18 degrees forward? )

Stalling the tip to lessen the heeling moment seems also to be a good reason.

A swept forward rectangular foil will be unstable in the sense that vibrations would be enhanced by the oncoming water twisting the tip. Could that be another reason for having vertical or swept back centerboards?

Anders M

[brian eiland]

Marchal kind of jumps around subjects in that edition.

You might have a look at Paul Bogataj's paper "How Do Sails Work"....nice clear explaination and diagrams http://www.northsailsod.com/articles/article6-1.html

I mentioned Paul here on the forum as well, http://boatdesign.net/forums/showpos...5&postcount=59,
"have a look at this article by Paul Bogataj, "How Sails Work", and note particularly this passage about the cord line of the mainsail, "Isolated Sails: A mainsail by itself (cat rig) is tapered, but if the mast is close to vertical it is actually swept forward. Recall that sweep is measured relative to the 25% chord line, which in the case of a tapered sail on an upright mast is angled forward. In this case, the forward sweep would have somewhat of a canceling effect on the increased upwash due to taper. The actual degree of upwash depends on the magnitudes of taper, sweep, and aspect ratio (height/width) of the sail. The sail still operates in the twisted flowfield caused by the boat moving through the earth's boundary layer, so an amount of twist would be appropriate. Raking the mast back increases sweep and will cause additional upwash on the top of the sail, necessitating more twist to the sail. Genoas and jibs are very tapered and swept. Those two features, combined with the already twisted apparent wind, cause significant upwash toward the head of the sail."

[tspeer]

Originally Posted by jam007

...The more you sweep back the more induced drag you get and if you try to decrease the drag by increasing the taper you end up with earlier stall.

Sweeping the board back does not necessarily increase the induced drag - it may reduce it. For example, if you had a highly tapered planform, sweep would load the tip even more, but that would be exactly what you want to get back closer to the ideal lift distribution. It would make the tip stall worse, but it would reduce the drag below the onset of stall.

This leads to the question of why the centerboard of my Farrier 25A is swept back. It should be swept forward, shouldn't it?

The elliptical lift distribution is not actually the optimum for most cases. It is really a by-product that minimizes the induced drag for the special case of an isolated planar lifting surface that has no interaction with anything else. That's not the case for the daggerboard on your F-25. Your daggerboard interacts with the hull and the free surface of the water. The real condition for minimum drag, which is valid for hull and free surface effects, is that induced velocity distribution along the span should be a constant so that the wake comes off like it's a rigid sheet.

If you take into account the free surface interacting with a vertical hydrofoil, the optimum lift distribution not only doesn't look like a semi-ellipse, it doesn't even look like a full ellipse. Instead, it's somewhat egg-shaped, with the maximum value shifted downward. There will be some lift carry-through to the hull, so when you include the hull + free surface, the lift doesn't go to zero at the top of the board. But by sweeping the board aft, you actually get closer to the egg-shaped optimum lift distribution.

Of course the main reason the daggerboard on the F-25 is swept aft is to position the board's center of area in the right place to balance the sailplan, while still putting the board trunk under the mast step to support the compression from the mast. Not everything about keel and board design has to do with hydrodynamics!

[jam007]

Originally Posted by tspeer

If you take into account the free surface interacting with a vertical hydrofoil, the optimum lift distribution not only doesn't look like a semi-ellipse, it doesn't even look like a full ellipse. Instead, it's somewhat egg-shaped, with the maximum value shifted downward. There will be some lift carry-through to the hull, so when you include the hull + free surface, the lift doesn't go to zero at the top of the board. But by sweeping the board aft, you actually get closer to the egg-shaped optimum lift distribution.

Why shifted downwards with a free surface interaction? Is it because the surface will bend downwards on the lowpressure side and uppwards on the high pressure side.

Originally Posted by brian eiland

Recall that sweep is measured relative to the 25% chord line, which in the case of a tapered sail on an upright mast is angled forward. ...

Interesting! I have heard several explanations for the benefits of raking the mast. An aft raking mast would then decrease the induced drag but increase the needed twist and the uppwash?

Anders M

[yokebutt]

Thinking about Brians statement. I recall reading some time ago a statement claiming that lift due to camber is centered at 50% cord, and lift due to angle of attack centers at 25% chord. This may very well do some violence to reality, but I do wonder, isn't a sail a pretty good approximation of a foil being all camber and precious little else? Simply put, where does the center of lift on a sail usually end up chord-wise?

Yoke.

[tspeer]

Originally Posted by jam007

Different coordinate systems can be confusing. Actually often aerofoils have the net force directed forward of the normal of the cord as it is the low pressure area around its forward part that creates the main part of the force....

There were designers in the early days of aviation that didn't appreciate this, and their wings collapsed forward at high angles of attack. If you look at a biplane's structure in a museum, you'll see "drag wires" running diagonally in the cells between the ribs. In both directions.

Originally Posted by yokebutt

Is that also true of the same section without camber? Or is that why cambered foils have their low-drag bucket shifted? ...

Yes, it's true of sections without camber. Another way you see it described, especially when considering sharp vs rounded leading edges, is through "leading edge suction". Negative axial force (total force pointing forward relative to the chord) requires high lift and low drag.

The shifting of the drag bucket with camber is something else. This has to do with the amount of laminar flow on the top and bottom surfaces. And the amount of laminar flow has to do with the slope of the pressure distribution, not the value of the pressures themselves. The easiest way to appreciate this is to consider the NACA a=1 camber line. This results in a uniform pressure difference between the two sides over the whole chord. So when you add camber proportional to the a=1 camber line, you don't change the pressure gradients on either surface at a given angle of attack. But the lift at that angle of attack has changed. So the drag bucket has shifted by the amount of lift added by the a=1 camber line.

Originally Posted by jam007

Why shifted downwards with a free surface interaction? Is it because the surface will bend downwards on the lowpressure side and uppwards on the high pressure side.

The difference in water level is an indication of the difference in pressure between the two sides. But in addition to the difference in level, there will be a difference in vertical velocity between the two sides, especially at high speeds (high chord Froude numbers). Simple theory says that the induced drag is doubled at the surface compared to deeply submerged because of this difference in vertical velocity.

So the optimum planform has its lift distribution shifted downward away from the region where the induced drag due to the same change in spanwise lift distribution is higher.

Originally Posted by jam007

An aft raking mast would then decrease the induced drag but increase the needed twist and the uppwash?

Whether it increases or decreases the induced drag would depend on what the spanwise lift distribution looked like to begin with. If you increase the twist, you may be putting the lift distribution back to where it was (because you're relieving the increased loading at the head due to the added sweep) and not change the induced drag at all.

There are also a whole bunch of other interactions rake has with the rest of the boat, including increased forestay tension, shifting the side load from the keel to the rudder, etc.

It's a lot easier to talk about the effect of sweep on the keel/board because its' never twisted and the crew doesn't change its shape!