BACKGROUND...

If you look at the picture below of the sail rig of Mayfly12 you will see on the sail some (fuzzy) writing (that didn't scan well) that says "55 square feet" to the left of a small circle that represents the center of that area (honest).

The center of that area is often called a "centroid" and you will see it is placed more or less directly above the center of the leeboard's area. That is very important.

As you might imagine a shallow flat hull like this with a deep narrow leeboard wants to pivot around that leeboard. If the forces of the sail, which in a very general way can be centered at the sail's centroid, push sideways forward of the leeboard, the boat will tend to fall off away from the wind. You should be able to hold the boat on course with the rudder but in that case the rudder will have "lee helm" where you have to use the rudder to push the stern of the boat downwind. The load on the rudder will add to the load of the leeboard. Sort of a "two wrongs make a right" situation and generally very bad for performance and safety in that if you release the tiller as you fall overboard the boat will bear off down wind without you.

If the centroid is aft of the leeboard you will have "weather helm", a much better situation. The rudder must be deflected to push the stern towards the wind and the force on it is subtracted from the load on the leeboard. Not only that, but when you release the tiller as you fall overboard the boat should head up into the wind and stall and wait for you if you are lucky. It's a good deal but if you overdo it you can end up with too much load on the rudder.

This balance problem is actually one of the few things about sail rigs that is not arbitrary. The type of rig and its area are pretty arbitrary depending on how fast you want to go, how much you weigh, etc. But balance is quite important and is one of the areas where backyard boaters get into trouble, sometimes changing the boat or rig with no thought of balance. So before you go doing that you should do a little homework. This essay will tell you how to figure sail area and find the centroid.

One last item: the balance situation shown for Mayfly12 is what I have found to be best for this type of boats. Boats with large fin keels don't balance that way - usually the sail centroid is well forward of the keel centroid. That distance is called the "lead". That type of boat is not within my personal experience and I'm not going to get into that. But you still would have to figure the area and centroid.

THREE SIDED SAILS...

This one is really easy. The area is just the base time the height divided by 2. Any side can be the base and the height is aways at a right angle to the base.

So when you lay out the sail you draw it up on thin paper to the same scale as your hull drawing with the leeboard (or daggerboard or centerboard) lowered. Draw a line through the center of the board straight up. Now we're going to locate the scale sail on the boat such that it's centroid falls very close to that line.

Here's how you find the centroid of a triangular sail.

Find the midpoint of each side and and draw a line from that midpoint to the vertex opposite it. The three lines will intersect at the centroid. Actually you only need to find the intersection of two lines but the third line is a good check.

That's it! Now you can take you scale sail drawing and slide it around your hull drawing until the centroid is on that line drawn up from the hull's board. Move it up and down and tilt it until you like the way it looks. But don't cheat much forward or aft of that line.

FOUR SIDED SAILS...

To find the area of a four sided sail you just divide it into two triangles, find the area of each triangle as above, and add the two together.

Now to find the centroid of the four sider. Start by finding the centroids of the two triangles that make up the four sided sail as shown above. Now draw a line from one triangle centroid to the other. The centroid of the four sider is on that line somewhere.

To find exactly where the centroid is on that line, measure the length of that connecting line. You need not use the same scale as is used on the drawing. I prefer to use a millimeter scale for this measurement. Then get out the calculator and work the formula shown in the Figure 4. Let's say for example the length of the connecting line on the scale drawing measures 120 mm (that is measurement L). Let's say the example sail has a lower triangle area of 50 square feet (that is A1). The upper triangle is 35 square feet (that is A2). So the total sail area is 50 + 35 = 85 square feet. The length L1, which will exactly locate the sail's total centroid, is L1 = 120 x 35/85 = 49.4 mm. So you take that millimeter scale and measure up from A1 centroid on the connecting line 49.4 mm and make a tick mark on the connecting line. That is the centroid of the total sail.

Another way to find the centroid, especially of a really odd shaped sail, is to take the scale drawing of the sail and cut it out. Then balance the cutout on a knife edge and mark the balance line, rotate the cutout on the knife edge about 90 degrees and rebalance and mark the new balance line. The centroid lies at the intersection of the two line.

Another way is to dangle the cutout on a pin stuck through a corner and into a wall marked with a vertical line that passes through the pin point. Mark the line that passes through that pivot corner and a vertical. Then rotate the cutout to hang it from another corner, and mark a second line through the second pivot corner and a vertical. The centroid lies at the intersection of those two lines. Back at the missle factory the designers had a favorite place, complete with pivot pin socket hole and vertical line, to hang these cutouts and that place was known as the "weighing wall". Meanwhile the super computer cranked away next door but its answers weren't to be trusted unless they agreed with the cutout hanging at the weighing wall.

RIGS WITH MANY SAILS...

Figure 5 shows the rig for Viola22. It has a main gaff sail of 177 square feet, and a mizzen sail of 45square feet. Where is the centroid of the assembly?

It's done exactly as with Figure 4. Draw a line connecting the areas of the two sails. Measure the length of the connecting line. Then run through the same equation as in Figure 4. Nothing to it.

One thing I might point out about the Viola22 rig is that the total centroid falls near the aft edge of the leeboard. By my experience the mizzen is not as efficient as its area suggests so it needs to be a bit oversized by normal rules, fudging the total centroid aft. I think in general the aft sails operate in the scrambled flow of the forward sail, causing loss of force back there.

WHAT IS IT??...

Here is a figure showing AF2 with my guess at the weights of the major components and a first shot at their locations. The object is to find a "center of gravity" location which is sort of an average of all the weights and locations. It would be where all the weights might be concentrated to represent the entire thing in a simple way. For example in the longitudinal direction, the X axis on this figure, you could predict where the hull would balance level while hoisting with a single line. You could figure out where the boat should sit on its trailer and provide the proper tongue weight. You could also figure out a stability curve for the boat by knowing the cg in the Z direction, the vertical direction, and from there tell where the boat will capsize, or how much sail the boat can carry in a certain wind, or how much wind is needed to capsize the design with its given sail area.

In the figure I've shown the reference lines forming the X and Z axes that I used to make the calculations. (The Y axis would be into the page and be a lateral axis.) These reference lines can be drawn anywhere but once you start the calculations they must not be shifted.

VERTICAL CG...

I'm going to start by calculating the vertical cg of the empty boat. The chart below lists the individual items making up the total empty weight, plus the vertical distance from those weights to the reference line, which in this case is a horizontal line drawn through the lowest point of the hull. The third column multiplies the weight times that vertical distance.

A bit of explanation about the weights. The sailing rig bits are roughed in by just figuring the volume of the elements and multiplying by 30 pounds per cubic foot, an approximate density of typical wood. The sail weight is just the sail area converted to yards times maybe 5 ounces per square yard, the typical weight of sailcloth. The hull weight is figured from the weight of the plywood that goes into it, I showed how to do that in an essay about guessing weight a while back. But the element "bottom" is a twist to account for the weight of extra thick planking on the bottom. Here is how I do that. The main hull is 1/4" plywood but the bottom is two plies of 3/8" thick plywood for a total there of 3/4". I look at the hull as a tube of 1/4" plywood that is 36" deep, so its weight is centered 18" above the base line. Then the extra 1/2" thickness of the bottom is accounted for seperately as another 75 pounds about 2" above the baseline to account for some rocker in the bottom. The weights are guesses at this point.

The bottom line totals for the weight W and the moment Wz are simply the sums of those individual columns. The cg height z in the center column is what we are really looking for and is not a simple sum of the center column. To find z of the empty hull you divide the total moment 16530 inch-pounds and divide by the total weight 600 pounds and you get 27.6 inches which is sort of the average location of everything. Remember that this is a first cut approximation and don't be fooled by the decimal precision of the answer.

Actually the boat will never sail in this condition. But with the basic empty hull numbers calculated it is easy to add other items and see the effects. Let's add a 180 pound skipper sitting on a low bench with his own cg 18" above the baseline. This would be the minimum sailing condition. Here it is:

So in this case we've treated the empty boat as a single entity and simply added the skipper. Since the skipper is seated on a very low bench, the effect of his weight is to lower the overall cg height a bit.

And you can go on adding and adding different combinations of people and situations. Here we'll add a crew member to the above figures. The effect is to lower the cg a little bit more since he is sitting so low.

We'll figure one last combination. We will use these numbers next issue when we figure the stability of the different combinations. This last one will be 400 pounds of internal ballast centered 3" above the floor:

So the effect of the 400 pounds of internal ballast is to lower the cg over 6".

LONGITUDINAL CG...

You can also figure the fore-and-aft cg the same way except instead of using the vertical dimension "z" you use the longitudinal dimension "x". The result is the longitudinal cg which effects how your boat will trim fore-and-aft. Here are numbers for the different weight conditions given above but this time the x dimension is used, x being measured from the forward perpendicular of the boat:

First the empty boat longitudinal cg:

Now with skipper:

Now with skipper and crew:

Now with skipper, crewman, and 400# ballast:

Again I've made a lot of assumptions here but in general the empty boat's longitudinal cg is a little forward of center at 117" mostly because of the weight of the mast. When the skipper sits back at the tiller the cg moves aft to 130". Add a crewman sitting aft of the main bulkhead and the cg moves aft again to 132". Add 400 pounds of ballast forward of the main bulkhead and the cg moves forward to 122".

LATERAL CG...

The situation with the lateral cg is a bit different for two reasons. One is that most boats are symmetric so the empty boat cg will be on centerline, or have y=0 if you use the centerline as a datum. With AF2 some of the sail rig is not on centerline but I'm going to ignore that for now. Then the only items that would enter into the lateral cg would be the crew assuming they were not sitting on centerline. The effect on heeling can be substantial. The second reason is that I don't know of any free design programs that figure the effect of a cg off the centerline. But in a future issue I'll show you how to figure the effect on heeling moment with hand calculations.

So we'll assume this AF2 is symmetric and that means we can skip the first "empty boat" table.

Now with skipper:

Now with skipper and crew:

Now with skipper, crewman, and 400# ballast:

TO RECAP...

Let's put our cg guesses into one last table showing the four weight conditions we looked at. We'll use this stuff later (trust me).

A few issues ago I described how I like to size the sail of a boat by balancing the sail's ability against the stability of the hull. At that time I made no effort to give any details of the sail other than size and sail area center because those were really the only factors involved. "Efficiency" was of no concern because anything you put up there has the ability to produce the maximum force when hit broad side by a gust. That's when most sailboats capsize, I think,

But now I'll try to get efficiency into the effort. In particular we'll try to go to windward as best we can with a sail and rig that are simple to make. We're not going racing around the world for our corporate sponsor, we're going to keep it all in the "made it myself for next to nothing" catagory that I always design to. I think in the end the direction the essays will take will be to seek simple ways to improve present low tech rigs for sailing to windward.

Why the emphasis on sailing to windward? For one thing, most sail boats will go downwind at about the same speed. You can't outrun the wind without tacking downwind (on some boats) or firing up a motor. But the differences in sailing to windward can be enormous. For example, a boat tacking 60 degrees off the wind at 3 knots will progress to windward 1.5 miles in an hour. That's actually a fair "to windward" pace for a low tech boat in most wind conditions. But anyone could easily walk faster than that or row twice as fast in an ordinary row boat. If the boat will point 10 degrees closer and maintain 3 knots, you get 1.9 knots towards the wind, an improvement of almost 30 percent.

THE PERFECT SAIL....

Here, in a very general way, is what I might use as a starting point as the best sail I could get with low tech materials.

Let me explain. Most of what I'm basing this sail on is tech data from old aero books I have plus some stuff from Marchaj books which I got on library loan because I can't afford them. Using airplane data isn't always appropriate but the aero people from early on did a lot of basic research that applies. I remember seeing a looong time ago at the Air Force Musem in Dayton, Ohio, a wind tunnel built and operated by the Wright brothers. As I recall, in their day the Wrights were on the cutting edge of wind tunnel technology in addition to making the first flying machines that really worked. I also recall that a lot of early aiplane designers were also sailboat designers and they actually did a lot of weird testing back in the first decades of this century, most of which didn't pan out. Indeed, when I see a modern racing yacht (I confess I don't follow yacht racing) it seems that the new ones differ from the old ones mostly in detail and materials. In the same way, modern airplanes still have the same basic layout as those of WW1 although they go through fads which die away each decade as with the T tail and the winglet.

So let's review some thoughts that went into the ideal sail.

First of all it has the normal soft sail. The lines I drew that look like battens really aren't. They are supposed to give you an idea of the cross sectional shape of the soft sail once it is "inflated" by the wind. Really extreme boats have used hard surface sails like the wing of an airplane. I can't see that for a boat that is supposed to be easy to use in a variety of conditions. I know the America's Cup catamaran from a while back had a very successful wing. But the rest of us have to haul our boats on trailers and a full wing is a big and delicate thing, conditions flyers have learned to accept. Then there is the problem of reefing the hard sail. Don't see how it can be done without making the whole project a Rube Goldberg project.

Also, I really don't like battens having been taught by Bolger not to like them. I used to have a Hobie 16 with battens and we got by OK as far as the battens go. The real problem with them was the wear and tear, and the problem in light winds where you can' t look at the sail and see what the wind is doing.

So my idea of the perfect sail is a soft surface sewn to a shape that delivers a good untwisted airforl from top to bottom once inflated by the wind. That cloth fits inside a structure of mast and booms and maybe yards or gaffs. Ideally the structure is rigid to the point where the sail won't lose its ideal shape because deflections in the structure or stre

YOU CAN'T PUSH A ROPE!.....

I don't know how many times I heard that statement when I worked in the aero industry. It is worth repeating. Ropes are pure tension elements, all you can do is pull on them. The same would be true with chains and wires. "But," you say, "I push sideways on ropes all the time..."

POP QUIZ...

Question: Look at figure 1. A rope 20' has a 50# load placed in its center. How much tension must be placed on the rope to make sure there is no deflection at the load?

Answer: It can't be done. The tension would have to be infinite. The deflection itself is the only possible mechanism that can allow the rope to balance the load.

Now look at figure 2A:

Here is a way that the rope could balance the load. Here I have shown the rope deflecting 2' under the load. To do so the rope must stretch on each half from 10' to a length of 10.19'. The rope now slopes at a 2:10 ratio and, since the rope can only develop a force along its own length in tension, it now has a small vertical component that can resist the weight. With the weight in the center as shown, half of the 50 load is resisted by each half of the rope, or 25 pounds per side. But that 25 pounds is just one component of the force in the rope. With a slope of 2:10, the rope must have a another component of 125 pounds to maintain the 2:10 slope. The total tension in the rope as you see must be 127 pounds if the 50 pound load is to be balanced with this geometry.

But as you look at the other examples in figure 2 you will see there are lots of other solutions. The greater the deflection allowed, the less the tension needs to be in the rope to balance the load. We mentioned one extreme in the pop quiz where infinite tension is required for zero deflection. The other extreme case will be when the defection is a full 10' and the load is supported by the two rope halves pulling only straight sideways at 25 pounds each.

So which solution is correct? It depends on the stiffness of the rope. The stiffer the rope, the less it will deflect and the higher will be the tension in the rope.

I dug out a 1995 copy of a famous marine catalog and gleaned from it some strength and stiffness data on 1/4" ropes. Ropes of greater diameter are of course a lot stronger and stiffer.

TYPE ROPE................... STRENGTH...........STRETCH PER 100 POUNDS

3 strand nylon....................2000#............................................7%

braided nylon.......................?..................................................3.5%

3 strand poly......................1350#............................................2%

braided poly.......................1200#............................................?

3 strand dacron..................2000#...........................................2%

braided dacron...................2500#...........................................1%

The 3 strand nylon is actually quite strong but very stretchy which is why it is used a lot for anchor and tow lines. I didn't find the strength of braided nylon but from looking at the other ropes it doesn't look like braiding changes the strength in a big way. Braided rope looks to be about twice as stiff as 3 strand. Braided dacron is hands down winner as far as low stretch. I thought the low stretch of braided poly was a bit of a surprize since it costs a fraction of braided dacron and is available everywhere. It is more prone to sun rot. I have tried cheap braided poly as halyard on my Twixt. It does work well, stiffer than nylon, but is unpleasant to work with. It has a memory and once you tie a knot or bend it around a cleat, it seems to retain the kinks and bends of the action. It is also slick to the touch, not a good quality.

So if we push the nylon braided rope sideways with 50# how far does it stretch? It turns out that if the rope deflects 2.6' sideways (which is 31"), the rope will stretch to 10.33' long (or a stretch of 4" per side over the original length), that produces a load in the nylon braided rope of about 99 pounds and all is in balance. It's the only solution that works with the 1/4" nylon braided rope.

If we push a braided dacron rope sideways with 50#, how far does it stretch. It turns out that if the rope deflects sideways 1.7' (about 20"), the dacron rope stretches to 10.15' long (about 2" per side over the original length), that produces a load in the dacron braided rope of about 145 pounds and all is in balance. It's the only solution that works for this rope.

So the general results of the comparison is that the dacron rope, which is by nature 3-1/2 times as stiff as the nylon rope, reduces the sideways deflection about 30% and the tension in the rope and the reaction loads at the ends of the rope are increased about 45%.

And that is general is the way that rigging works. The stiffer you try to make everything the greater become the tensions and forces in the rig. It's not a linear effect either in that if you try to reduce the rig deflections by a factor of two, for example, the tensions and loads in the rig may increase by a factor of four or six or ??. You may quickly reach a point where the results aren't worth the effort. In extreme racing sailboats, I think there is no rope or wire at all in the standing rigging. The masts are held up with stays made of solid steel rod.

PRETENSION....

One effective way of reducing the flexings and stretchings in a sail rig is to pretension the ropes. For example look at figure 3.

Here we have the same solution for braided nylon 1/4" rope that we had before except someone has yanked on the ends of the rope and pulled them outward with a force of 150 pounds. The rope stretches out in length from10.33' to 10.53' per side, the end supports pull apart a bit such that there is no longer 20' between them, but 20.78'. And the 50 pound load which used to cause a deflection of 2.6' is drawn inward so that the deflection is now 1.7', a 35% reduction. I've also figured the case where the rope is loaded to 500 pounds of tension. Deflection of the 50 pound load is reduced to 1.2', a reduction of 54% over the original case. There is a price of course. Internal loading in the structure is now 500 pounds where the original was 99 pounds!

Now look at the effect of preloading a halyard. Sailor 1 hoists his yard to the top of his 10' mast and cleats it off with no preload. The wind comes by and applies a 50 pound load to the yard which transfers it to the halyard. The 10' halyard stretches 1.75% or 2" under the load and a gap of 2" will appear between yard and masthead.

Sailor 2 hoists his yard and yanks on the halyard while cleating it with a 50 pound force. In doing so he forces the yard against the masthead with a 50 pound force and that prestretches the halyard 2". As with Sailor 1 the wind pipes up and applies the same 50# load. Now how much does the pretensioned halyard stretch? Answer: Until the yard load exceeds 50 pounds the halyard won't stretch at all beyond its original prestretch and no gap will appear between yard and masthead. What happens is that the preload force between the yard and the masthead will reduce to zero and the force in the halyard stays at 50 pounds. As the yard load exceeds 50 pounds, say to 100 pounds, the halyard load will now increase from 50 pounds to 100 pounds and a gap between yard and mast will open up to 2" where with a similar 100 pound load the gap of sailor 1's rig will open to 4".

Actually almost any type of sail rig will benefit from lots of preload in the halyard. If you have a simple rig made of stout materials you can hardly overdo it. You can often preload a sail's halyard most easily by pulling down on the tack line since that line is usually quite short and can be made a multi part tackle. In the extreme you can preload he rig to the point of destruction even with no wind load on the sail!

Here is another thought. Clearly in all these cases the total amount of rope stretch is about directly related to the length of the line. So one way of reducing rig deflections is simply to keep the ropes short. Thus some Bolger boats like Birdwatcher have their sail tops tied directly to the top of the mast, there is no halyard. Similarly some racing boats have gizmos on their mastheads where you hook the halyard after hoisting sail, effectively eliminating the halyard stretch. Both of these schemes have the added benefit of eliminating half of the preload compression in the mast. And one might reconsider plans to lead lines that set the sail long ways into the cockpit for convenience. All the extra lengths add to the sail rig deflections.

EFFECTS OF HIKING....

In the last issues I looked at the effect of ballast on the boat's stability but limited things to having the CG on centerline. But the effect of a crew's being off the center of the boat, "hiking", can be significant even for mid sized boats like Jukebox and Jewelbox. Let's look at a couple of examples based on the new boat Frolic2 shown down in the featured boat section.

Take a look at Figure 1 which shows why hiking works so well. If the boat's overall CG is offset from centerline, that offset is added, to a certain degree, directly to the righting arm calculated by the stability program. I say "to a degree" because the added effect is actually the offset multipiled by the cosine of the heel angle. That would mean it is quite effective at small heel angles, drecreasing as heeling increases to the point where hiking has no benefit at 90 degrees of heel. (Unlike traditional ballast which is most effective at 90 degrees of heel.)

So here is how you can do the numbers. First you need to figure the CG. But now we also need to figure the lateral CG where in previous examples we figured only the vertical CG.

So the chart shows the Frolic2 with a 200 pound crew sitting on a side bench has a total weight of 640 pounds, it's total vertical CG is 20.5" above the inside bottom (which I used as a reference point), and 7.5" off centerline to the side the crewman is sitting on.

With this information, and knowing the lines of the boat, we can fire up Hullform6S and crank out a few stability calculations to get an overall picture of the boats stability. But there's a problem for now: Hullform6S won't figure CG's that are not on centerline. So we run the stability calculations on centerline and adjust them afterward by hand (or write a simple spreadsheet if you must).

In detail here is how I do it. Hullform gives the righting moment (in foot pounds) and I divide those numbers by the weight (in pounds) to get the righting arm (in feet). Then I take the CG offset (in feet it equals .625'), multiply by the cosine of the heel angle, and add that to the Hullform righting arm to get the new righting arm (in feet). That gets multiplied by the weight (in pounds) to get the "hiking" righting moment.

Here is what is looks like:

Well, you can see that the effect of the skipper hiking (or in this case just sitting on the upwind side of the boat), about doubles the righting moment of the boat. I'm a bit surprised that the boat still shows a positive righting moment up to 70 degrees of heel. But the Hullform analysis indicated that the rail is going under at 60 degrees. If it really works out that way it ain't so bad. It certainly ought to give the crew a good indication that it's time to back off the sheet! Actually, if the oarports are left uncovered there will be some flooding even before then.

On the chart I've also shown the effect of a second 200 pounder on the upwind seat. Quite an effect. Righting moments are way up again, although the program showed the boat would now flood over the rail at 50 degrees of heel. I want to show this because the crew you sail with will certainly have an effect on the impression a boat leaves you. Sometimes I've admired how some folks seem totally unfazed by stiff winds only to find they are always sailing with heavy crews. (My old Jinni capsized on me twice while solo, but never with two crew even though there were some wild sailing days.) One similar problem is the case where a swimmer tries to reenter a light boat. If solo, the boat might easily capsize as he loads his weight on the rail to pull himself into the boat. He may not think too much about that if he never sails solo and the extra crew weight in always in there to stabilize things as he climbs back in over the rail.

Another thing to consider is the case of the solo skipper who is seated upwind with the boat clipping along at maximum righting moment. He has to lean to the center or lee side to clear a line or something. The force on the sail is not affected by his actions, but the boats righting abililty is cut in half, or less. The boat capsizes! He needs to depower the sail by slacking the sheets before leaning over to the lee side. I suspect this may one of the most common reasons folks capsize small boats.

CONVERTING RIGHTING MOMENTS TO SAIL FORCES....

This is easy as pie! Look at Figure 3:

Once you've figured the righting moment, all you need to do to balance the boat is to divide the righting moment by the distance between the center of area of the centerboard (or daggerboard of leeboard or keel, whatever you are using to prevent leeway) and the center of area of the sail. On Frolic2 the two area centers are about 12' apart. So if the maximum righting moment is 900 foot pounds, the maximum sail force (and centerboard force, on a normal boat the sail force and centerboard forces are equal and opposite) is 900/12=75 pounds.

What does that mean? We'll return to this subject in a while and relate the righting moment curve and the resulting sail forces to the wind's speed over the sail and the water's speed over the board. Given the righting moment curve and a certain sail we can figure the wind needed to capsize the boat. Or given a wind speed figure the maximum sail we can carry.

In the 1 April issue I presented a way of determining the maximum sail force your boat could handle and not capsize. To review a bit, first you need to figure a weight and CG for your boat, model your boat on a program like Hullforms 6S complete with that weight and CG, work up a stability curve of righting moment versus heel angle. Then you find the maximum righting moment and divide that by the distance between the center of the sail area and the center of the underwater board area to find the maximum allowable sail force (which is usually also the maximum underwater board force). If the wind force on the sail exceeds that value your hull will not have the ability to handle it and your boat will capsize.

Here is the righting moment curve we worked up for Frolic2.

So for Frolic2 the maximum righting moments are 1550 ft pounds for the boat with two crew members sitting to windward, 900 ft pounds for a solo crew sitting to windward, and 500 ft pounds for a solo crew centered. The full sail area of Frolic2 is 12 feet above the underwater board area. So the first crew can have a max sail force of 1550/12=129 pounds. The second crew can handle 75 pounds of sail force and the solo centered crew can only take 42 pounds. Quite a difference!

THE FORCE OF THE WIND....

How to relate the wind speed and sail area to sail force. The equation for figuring the wind's force on a panel (sail or wing or whatever) is .0034 x S x C x V x V in normal air at sea level.

S is the area of the sail. For Frolic2 it is 114 square feet.

C is a force coefficient. It's value is subject to endless debate but tends to peak at about 1.5 for a reasonable sail. I got most of this info from Marchaj's excellent books. What was interesting about his data was that a square rigger running downwind has sails that also operating at a C of about 1.5. Really super wings operating to windward might peak at a C of about 2 but that takes very high technology, quite unlike anything I would design.

V is the wind speed in knots. But note that the value V is actually "square" , that is, multiplied in there twice. As a result, the wind speed usually becomes the most important factor in the equation.

So, for the Frolic2 example, the equation boils down to F=.0034 x 114 x 1.5 x V x V. That would be the maximum force since I'm using the full sail and the full value of C. It multiplies out to .58 x V x V. If we put in a few values for V we might get a table like this:

V(knots)...F(pounds)

2................2.3

4................9.3

6...............21

8................37

10.............58

12.............84

14.............114

16.............148

18.............188

20.............232

25.............362

30.............522

Looking back to the Frolic2 rightning moment data, we'd predict that with the double crew to windward she could take a max wind of about 15 knots, the solo crew to windward could take about 11 knots wind, and the solo centered crew only about 9 knots.

Notice how the force goes totally bonkers at the higher wind speeds! So if you are sailing in 10 knot winds and get a 30 knot gust you don't have a prayer, right? All you can do is to feather the sail, letting out the sheet and allow it to weathervane into the wind and hope the wind direction is steady enough that you don't get caught by the full force. Marchaj's data showed that the max sail force comes at an "angle of attack" of about 20 degrees for a typical sail so if you can keep the sail feathered less than that angle you can hope to keep the value of C well below the max value.

Actually the force on the wind can be much more subtle than that. If you were sailing the Frolic2 alone with full sail in 11 knot winds, having a great time at near max power, heeled close to 20 degrees, you might get lazy and tie off the sheet. But a gust to 14 knots, an increase you might not even feel on your face, increases the sail force by 50%! If you don't act fast you might go over.

By the way, you might place in your memory that a 14 knot wind will produce about 1 pound per square foot on a typical sail.

SIZING THAT SAIL

There's another way to maneuver that equation. You can assume a certain wind strength and figure the sail area needed to produce the desired sail force.

For example, take the Frolic2. Assume you want to sail in 15 knot winds. You are sailing solo and think you are good enough to keep your weight to windward. How much sail should you carry?

Going back to the righting moment curve, it takes 75 pounds of sail force to capsize the boat with that crew situation. So you might write the equation as 75=.0034 x S x 1.5 x 15 x 15. Solve for S, the sail area, and you get S=75/(.0034 x 1.5 x 15 x15)=65 square feet. You might measure the sail area with different reef points and make a mental note of what wind speed that would correspond to. (Actually you might find you can carry slightly more than 65 square feet since most sails reefed set lower on the mast than when full. Thus, instead of dividing the maximum righting moment by 12 you might end up dividing by 10.)

I've found this approach to be pretty accurate, given that you never really know how strong the wind is blowing.

Here's a chart of pressure on the sail versus wind speed. I've shown three values of C in case you question the value of 1.5 that I prefer.

This issue will tell you how to figure the size for the "underwater board" that is used to balance the side loads produced by the sail.

By "underwater board" I mean a leeboard or daggerboard or centerboard or fixed fin or keel. I don't see how the exact type makes much difference as far a abililty to counteract side load goes. There are other practical differences of course.

As in previous discussions I'll assume the underwater board will counteract all the lateral force of the sail. "But," you might say, "the immersed hull itself provides some lateral resistance." True, but that is usually quite small in proportion to that of a good fin because of aspect ratio considerations that we'll go over in the next issue. You might also ask, "Hey, if you make all the sail's force side force, what's to drive the boat forward?" Correct again, but when close hauled a sail produces mostly lateral force with only a small of forward force left to push the boat forward. It's a vector thing and vectors don't add up like regular numbers. For example if you 100 pounds of sail force directed 60 degrees off centerline, you would have 50 pounds of force pushing the boat forward and 86 pounds pushing to the side. Assuming all of the sail's force is side force doesn't introduce a huge error.

I'm going to use Frolic2 as an example again. Remember in previous issues we worked up a righting moment curve for the boat using the Hullforms6S program we got as freeware at Blue Peter Marine's web page. From there we found that the maximum righting moment for the boat with two big men sitting to windward was 1550 ft pounds. The sail (114 square feet) and leeboard areas are 12 feet apart on Frolic2 so the maximum sail force we can stand without capsize is 129 pounds (which happens in about 15 knots of wind).

The force on the leeboard is assumed to be the same as the lateral force of the sail at that time, so the maximum force on the board would be 129 pounds. If it exceeds that the boat will capsize with that crew configuration. How do we size the leeboard such that we can be pretty sure it will produce that 129 pounds of lateral force?

THE FORCE OF THE WATER ON THE LEEBOARD....

The leeboard, or any underwater board, "flies" through the water in the same way that an airplane wing flies through the air. If the board is pointed dead ahead into the flow of water, it produces no lateral force, only drag. If a small "angle of attack" is introduced, a large amount of lateral force can be produced. To produce that angle of attack , skipper need only point the boat slightly upwind of his desired course. So the boat points one way and goes slightly downwind of where it is pointed. You might view this as "leeway".

The equation for the lift, or lateral force, developed by the board is F=2.86 x S x C x V x V. It's a lot like the equation used for figuring sail force in the wind. In fact it's exactly the same equation with an allowance that the density of water is about 900 times the desity of air. But there are some other differences.

The S in this equation is the area of the underwater board in square feet. Remember it is only the area that is immersed in the water flow.

In the case of the underwater board, the value of C is, I would think, less that the C of 1.5 that I recommended for sails. The reason is that the underwater boards cannot have camber, as a soft sail can have, and function through tacking left and right. It must be symmetrical. (OK, boats with two leeboards can have cambered boards.) Thin airfoils with no camber seldom have maximum lift coeffecients much greater than 1. So I recommend the value used for C in this equation be 1. Then the equation reduces to F=2.86 x S x V x V.

(Below is shown a chart of lift coefficients for various fins of various "aspect ratios". I'll explain what aspect ratio is next issue. This chart was gleaned from info from Marchaj. Looks like his test data indicated the C maximum was 1.2 but the chart seems highly idealized, which is OK. Besides the note of C max, note that the general trend is for each fin to gain C directly proportional to the angle of attack until the maximum is reached. Then it levels off.)

V in this case is the boat's speed through the water, not the wind speed. How fast is your boat going to go? Boy, is that a tough question. The speed you want to use in the equation in not the "hull speed", the usual assumed maximum speed of the boat The hull speed (in knots) of a displacement boat is often shown as about 1.3 times the square root of the waterline length (in feet). So the Frolic2, with a waterline length of about 18' would have expect to have a top speed of about 5.5 knots. But the speed we want to use in the equation is about half of that maximum speed. Why? Because for the underwater board, the worst condition is when beating to windward at low speed. At that time the load on the sail, and thus the load on the leeboard, is maximum, but the boat's speed is well below maximum as it beats through the waves. Let's call this something like "beating speed". So for Frolic2 we might take the beating speed to be 2.75 knots. (It's still pretty fast for a sailboat going to windward.) So now for Frolic2 the equation becomes F=2.86 x S x 2.75 x 2.75. which is equal to F = S x 21.6. If we think the maximum force on the board is going to be 129 pounds as per our stability analysis, we can solve for S = 129/21.6 =6 square feet of leeboard area.

(Here is a chart that you can use to figure the pressure that water exerts on an underwater board at various boat speeds. I've shown two values of the force coefficient, C=1 that I recommend, and C=1.5 for you optomists.)

That is actually a pretty large board for Frolic2. As designed it has about 4.4 square feet of leeboard immersed. What does that mean? That means that with two big boys sailing on the rail in 15 knot winds, they need to maintain a speed that will produce about 129/4.4 = 29 psf on the leeboard. If you look at figure 2, you will see that is 3.2 knots. If they don't do that here is what happens. The crew will slide off to leeward. Rudder movements intended to increase the board's angle of attack and thus its lift will do no good. About the only things they can do to regain complete control would be to head downwind a bit (to increase the boat's speed), or let out the sheet a bit to feather the sail somewhat and thus reduce the sail load to something the board can handle. If they try to "pinch" the boat closer to windward, their hull speed will diminish more, they will be stuck, bouncing up and down in the waves, probably going one foot sideways for each foot forward. I've been there many times! How about you? (Another solution is to reef to reduce the sail force.)

Now if you don't like doing math, I'll tell you where I probably got the value of 4.4 square feet when I designed Frolic2. It turns out that if you have a pretty normal boat and make the board are 4% of the sail area, you will probably have something that works quite well. Frolic2 has 114 square feet of sail, so 114 x .04 = 4.6 square feet. I guess that's how I figured it.

I do think it is possible to make the fin too big. Certainly you can add a lot of drag by making it huge. The reason I say this is that when I made my Bolger Jinni a long time ago, I used a leeboard that was about 5 feet long. The next season I chopped about 12" off its bottom. Easier to handle. And I can't say I ever really noticed any difference in its performance going to windward. As a contrast to that, I enlarged the leeboard on my Piccup Pram after a couple of seasons and noticed an enjoyable improvement in its ability to sail to windward, especially with the larger sails had I started to use.

One interesting thing about the way the board behaves is that if you have two boats with the same rig in the same wind, and one is faster than the other, the faster boat can get by with a smaller board. Why? Because its faster speed through the water produces more pressure on the board than the slow boat. But I don't think you can say something like, "I'm going to increase sail area. That will increase my speed and then I can reduce my board area." It might happen that way if you are lucky. But when you increased the sail area you increased the sail force. And you will need increased board force to balance it. Whether the boat speed increase will sufficicently increase the board force is hard to say.