Lifting Spinnakers:does it lift the bow?

Not necessarilly no. Just because the luff is raked down't mean it will produce upward force in any significant quantities. To grossly over simplify, imagine looking at the sail from above. Only the area of cloth you can see from above can be contributing to a vertical force...

 

Absolutely, but there is still some upforce, and none down (if the boat is upright). Therefore the net force is up.

 

Sound thinking, but you're hoist on your own petard. You've got the main's moment, yes, but you then compare it with the asym's force, which is mistaken. The asym's *moment* is bow down, even though it's force reduces the boat's displacement. The two moments add to each other, they are not opposed. Sound impossible? No; imagine a lifting force, *straight up* which is applied at the stern. Force is up (vertical); but moment is bow-down. It's the same with the asym--DRAW THE PICTURE.

Dave

 

It is not necessary to sum the forces at each the attachment point (though you could certainly do it that way--if you had enough money!). It is only necessary to find (or assume) the center of each sail's applied force, and find (or assume) the direction of that force. Do this for each sail; take an origin (I suggest the Center of Buoyancy, but you can use any Origin), and measure the moments around that origin. you can work in just two dimensions if you like, since we are only investigating bow down pitch and not roll or yaw. FWIW, this is precisely how every yacht designer and navel architect designs and investigates forces from his rig.

I say "assume" above, because not very much accuracy is needed to prove the point. All the sails on all sailboats pitch the bow down--except free-flying kites. Period. You'd need a very (very!) long bowsprit and a very short mast to make this any different (I mean, bowsprit to mast base on the order of twice the mast height). Only kites fly sufficiently far from the boat to pitch the bow up.

Please, stop imagining and "feeling" for the answer and just draw the picture. This isn't rocket science. (well, actually, it is, but then rocket science ain't that difficult, either)

Cheers,

Dave

 

Oops! Sorry for the "navel architect" crack, Water Addict. It was unintentional (though one of my favorite jokes--get it? "navel" architect? Sorry, a little design humor. Very little.)

Dave

 

Amazed

Since this discussion originated in the Planing Trimarans thread, I'm kinda amazed that Doug hasn't plugged an email to Ian Farrier to get him to chime in here in support of his lifting bows concept.

I remember a conversation between a friend of mine and Kami Richards at Pineapple sails in the San Francisco Bay area some years ago. My buddy wanted to build a really fast, small trimaran of 15'. He planned to have smallish amas and drive the boat "very actively". He described an assy. spinnaker for the boat to Kami and Kami said that would be fine if he liked sailing submarines.

 

Interesting article on the Cherub UK website:
www.sailingsource.com/cherub/test/doku.php/tech/spinnakerlifts

 

Yes, Jim Champ did that study in response to the 1996 thread on rec.kites. Pretty clearly lays out the whole argument. There are other links to the same essential work, earlier in this thread

Dave

 

I probably didn't make myself clear enough and I think you may have misunderstood me. The asym will produce a forcethat has both a forward and (to a lesser extent) an upward component. The centre of this force acts forward of thr LCF and above the VCF. Therefore the forward component produces a bow down pitching moment whilst the upward component produces a bow up pitching moment. Whichever m,oment is greater will determine whether the bow goes up or down. This depends on the geometry of the specific rig (luff angle, sprit length etc), but for most boats the bow down moment is probably bigger. However, this thread originated from the planing trimaran thread, where it was claimed that the sails produced a downward FORCE. This is not the case (for an upright boat). There is no downward force, just a moment that (may) push the bow down.

 

Yes I know the joke. Been hearing it for a quite few years now. And I've created a couple of my own navels, they're in elementary school and sucking my wallet dry! So I guess I am of the navel and naval arch variety.

I am familiar with how rig forces are developed and boats are designed. The spin will develop an upward force component. How the total forces from spin are coupled with the hydro and mass forces will determine the total pitch attitude of the boat while sailing. If the spin is forward on the boat, the geometry can be such that the effect of spin forces could be lifting the bow. It depends on the specifics of the boat. For most cases it is probably likely that the chute causes a bow down moment, I'll give you that.

I sent you an email. Care to correspond or do you already have something in the works on this?

 

RHough, you are mistaken about the effect of hoisting the spinny. If, by raisng the kite, the bow down trim decreases, it MUST be due to an opposing moment. If I understand you right, you are saying that because the CofE of the spinny is lower than the main/jib combo, it reduces the overall lever effect. In this you are correct. BUT, the total force has been increased (due to the addition of the spinny), and therefore so has the moment (mmt = lever * force). If the spinny moment were acting in the same direction as the main//jib moment, the bow would trim further down. Take an example. A boat has a main/jib of area = 10sqm, with a CofE 4m above the vertical centre of flotation. Assuming that drive force acts purely horizontal and is proportional to sail area we have a bow down pitching moment proportional to 10*4=40m^3. Now hoist a spinnaker of area 10sqm, with a CofE 3m above the VCF. Its moment is proportional to 30m^3. The total moment is 40+30=70m^3. This is an increase in pitching moment, so the bow MUST go further down. But you have agreed that it does not. The fact that the lever of the combined main/jib/spinny is 70/(10+10)=3.5m (and therefore less than the 4m of the main and jib alone) is immaterial. The moment has increased from 40m^3 to 70m^3. Therefore there must be something else at play. This something else is the fact that the angled luff of the spinny produces an upward force in addition to the forward drive force. Let us say that this upward force acts 4m forward of the longitudinal centre of flotation. Let’s also assume that the upward force is the same magnitude as the forward force (from the spinny). This gives a moment of 4*10=40m^3 acting in the opposite direction. So the total moment is now 40+30-40=30m^3 – which is less than the main/jib combo – hence the bow will rise. This is the only way the bow can rise. If the spinny does not produce a net upward force, the bow can not rise relative to where it is without it.
So, there we have it. There is the maths that us non-believers were asked to produce. Please note, I fully agree that an upward force can produce a ‘downward’ moment. In fact, it definitely does in the case of a sailboat. But it can also produce an upward moment – and it must do if the bow is to rise. But there is no downward force. For those that don’t like maths, surely the upward bend on the end of the bowsprit is all the evidence you need that there is an upward force?

 

Some more numbers…
Imagine a 14ft boat with a 6ft pole for an asymmetric kite. The boat is sailing left to right. When planing, its LCF is 2ft from the stern (as the forward part of the hull is out of the water). Its mast is 20ft long, the main is 100sqft, with a CofE 8ft above the water, the jib is 50sqft (CofE 5ft) and the kite is 150sqft (CofE 6ft). The kite foot goes from the end of the pole to 3ft behind the bow, which means that longitudinally the centre of effort of the kite is about level with the bow (12ft forward of the LCF). In standard displacement mode the LCF is 6ft from the stern. Let’s say that the resultant force from the main and jib are horizontal, whilst the force from the kite acts slightly upwards (I think we’re mostly in agreement on this) at an angle of 30 degrees. Let’s also say that the kite is 1.5 times more efficient (ie force per square foot of sail area) than the main and jib. For now, assume that under main and jib alone the boat does not plane. When the kite goes up, the boat planes. I believe that all these figures and assumptions are quite reasonable for a quick and dirty estimate.
With two sails:
Moment =
100*8 + 50*5
= 1050 units, clockwise

The kite alone:
Moment =
150*6*1.5*cos30 - 150*12*1.5*sin30
= -181 units, clockwise
=181 units anticlockwise

Total moment, sailing with all three sails =
1050-181 = 869 units, clockwise
This means that the bow is still pitching down, but less than if the kite were not being flown. In other words, the kite is lifting the bow, but not enough to overcome the pitching moment. You are better off flying the kite, but you will still nose dive when a gust hits.
However, now lengthen the pole to 10ft. This increases the kite’s distance from the LCF to 16ft and increase the kite’s sail area to 300sqft. Now the kite’s moment is:
300*6*1.5*cos30-300*15*1.5*sin30
= -1262 units, clockwise
=1262 units anticlockwise
The total moment with all three sails is now:
1050-1262 = -212 units, clockwise
=212 units anticlockwise
The net moment is anticlockwise. In other words, the bow is now being lifted in an absolute sense. It is coming out of the water. Nose diving will not occur.
In both cases I would describe the kite as lifting the bow. I can see that there could be some semantic confusion over the first case, where the lift is relative not absolute. In the second case, the situation is clearer. Even if you nit-pick some of my numbers, hopefully you will agree that there is not *always* a strong bow down pitch from the kite as you claim.

 

Boy, I don't know where to start, PI. First, there is only one resultant force from a sail*--devolving that one force into two forces (for instance, "horizontal" and "vertical") is a human construct. Sometimes doing so serves a purpose, but in this case it makes it more complicated than it needs to be. If we can define a resultant force as having both a direction and a magnitude, we don't need to devolve it into vertical versus horizontal, thus we won't become confused about which may be the larger or which will "win" our little tug of war.

Now, in another post you suggest that your "vertical" and "horizontal" force components can be resolved as a 30 degree upwards angle. I dispute this, but am happy to concede it for the argument. And can we agree that this force passes through the aerodynamic center of the sail (not the same as the geometric center, but not too far forward of that)?

Can we further agree to take moments about a common point, and for the sake of this illustration, take that point to be the three-dimensional center of buoyancy of the hull? This is a point on the hull's centerline, somewhere *below* the waterline, where we already have a balance of both buoyancy and gravity forces.

Can we agree that, if we introduce a new force, and a lever arm back to this point, that we will have introduced a moment, therefore a rotation, and that such moment will either, for instance, pitch the bow up or down? Can we also try to stop taking moments about other random points on the boat's structure, and can we stop talking about things like "upwards" and "downward" moments, since moments are torques, not forces, thus cause rotation, not "upwards" or "downwards" movement?

Last, can we agree that any resultant force which passes above the CB will result in a bow-down pitching moment (rotation) and that any resultant force which passes below the CB will result in a bow-up pitching moment (rotation)?

OK, all set (I think): Now, let's look only at the asymmetric spinnaker. Let's leave the main alone, the jib alone, even the mast and bowsprit alone. We agreed that the spi's net resultant force was angled up at 45 degrees (which, again, I maintain is not possible, but surely is a working limit) and that the spi is located somewhere further above the hull's center of buoyancy than forward of same (for instance, the center of effort of this sail may be--on a 14' skiff--perhaps 8' forward of, and 10-12' above the hull's CB. Sound OK? Can we agree that such a spi's resultant force will pass above the hull's CB, thus does--and forevermore will--result in a bow down pitching moment?

The sail would either need to be carried far enough forward to bring the resultant below the CB, or the resultant will need to be "pitched up" enough to increase the resultant angle with the horizontal to achieve the same end, before there is any bow-up pitching moment. There is never "some" bow-up plus "some" bow down from a single sail--we took care of that when we presumed a single resultant force.

Regarding the resultant force from a spi, please do not be confused by the angle of the forestay; a sail's resultant force is *not* perpendicular to the forestay (in profile view); a simple top-view of the situation makes this crystal clear--unless you want to presume the asym sits at 90 degrees to the boat's centerline. Even then, the resultant *cannot* be brought under the hull's CB with any sane combination of sails. The only reason kites can do this (and kites *only*) is that they are flown sufficiently far from the boat that their resultant force can be made to pass below the CB, resulting in bow-up pitch.

Enough?

Dave

*For the real purists, and fans of FEA, there may be as many as an infinite number of "resultant" forces from a sail, but for argument's sake, can we please resolve all of these down into a single force? And, can that force be a vector, with both a direction and a magnitude? Thanks

PS; I find I need to respond to PI's next post, too. Sorry.

 

Sorry, PI, but almost *all* of these presumptions are in error, IMO.

The LCF of a planing hull is near the front of the waterline. A 14'er sometimes gets all but a couple of feet of her hull out, but this is *never* a steady state condition (even though all those photographs might make it look so). LCF is nearer the daggerboard; perhaps 5' forward of the stern, when planing. It's nearer amidships when displacing, ie; 7' forward of the stern. You've got your COE's much to low; the boom itself on a 14'er is at least 3' above the waterline (often 4' as the boat pitches up--*not* due to the spi!); you've got your main's resultant just 5' above the boom.

The Intl 14' specs call for a 25' stick and 200' of working sail, not 150. Also 350 sq ft of spi, not 150 (cf Wikipedia). Your 6' sprit looks a bit short; the photos of 14's I've seen look more like 8'., but a 140 sf main's gonna have its center more like 10' above the boom, thus 13' above water, not 8'. Jib's gonna have its center perhaps 8' above the deck, but the deck is 1.5-2' above water (again due to bow-up position of the boat), so jib CE's gonna be closer to 10' above water, not 5. The spi's gonna reach from the end of the pole to pretty near the end of the boom, not a paltry 3' back of the bow; a 350 sq ft triangle on a 25' stick's gonna have a foot of something like 28', which uses literally all the space there is. then again, the sail is sheeted out, so perhaps your assertion that its center of effort falls near the bow isn't too far off. It's CE is easily 12-14' off the water, however, not 6'--the tip of the bowsprit is 3-4' up, just to begin with.

Second, a mainsail will always have a higher lift coefficient than a spinnaker (you suggest the spi is 1.5 X as powerful as a main, area for area). A main is actually about 25-30% more powerful than the spi, foot for foot--sometimes much more than this, but remember, the main is depowered on a run--it's just too tall not to.

You can re-run your numbers if you wish; the above as a little more than "nit picking" since all your underestimates result in bow-up pitching moments.

Last, you've mis-done your maths, at least once. In your last example (300' kite), you first show anti clockwise-and clockwise- moments from the spi to equal each other= +1282 ft lbs minus 1282 ft lbs (no net rotation from the kite at all--which simply isn't possible, but let's leave that alone), then when you add the bow-down pitch of the main and jib, you get a result in pitch up. Zero plus clockwise cannot result in anti-clockwise.

Here are some photos, for scale:
http://www.yachtsandyachting.com/photos/skiff/2005sanfrancisco14.jpg
http://www.yachtsandyachting.com/photos/skiff/2005sanfrancisco15.jpg
http://www.yachtsandyachting.com/photos/international14/2005eastlothian1.jpg
https://host220.ipowerweb.com/~underthe/undersail/Int'l14ClassWorldChamp2006/Int'l14Worlds06-6066.html
https://host220.ipowerweb.com/~underthe/undersail/Int'l14ClassWorldChamp2006/Int'l14Worlds06-7031.html
https://host220.ipowerweb.com/~underthe/undersail/Int'l14ClassWorldChamp2006/Int'l14Worlds06-8052.html

Dave

 

If the skiff is planing then the sum of vertical forces will be well aft of hydrostatic LCB, perhaps like 2 feet fwd of transom. Assuming x=0 at stem then LCB/F is at x=12. Using your coordinates LCE of spin is at x=0 and if VCE of spin is at z=10 then uplifting moment wins for 45 degree aero force that you list.

The CE on the spin will in reality be much closer to the luff of the sail than the leech.

I see nothing wrong with PIs reasoning. He conceded that in most cases it would likely be a bow down effect. But it is specific to the geometry of whatever the boat is. And also if planing or displacement mode.