so does what you are saying relate to the lift created by an asymmetrical fin only or the lift created by any fin?can you list any references or links that expalin what you are saying a bit more indepth?
As I have understood your posts, you've attempting analyze the difference in hydrostatic pressure between the two sides of an asymmetrical fin using the Bernoulli equation. (By the way, the pressure in the Bernoulli equation refers to normal pressure or hydrostatic pressure. Where here, the term 'normal' is used to refer to a force which is perpendicular to a surface.)
As applying the Bernoulli principle to a fin during a turn would be quite a feat, I've assumed that is not what you've attempted to do, but simply to use the 'drop' down the face to produce some numbers in order to gauge the pressure differential in the Bernoulli equation, see figure 1.
Anyway, if we assume that we know the speed of the incident flow, and we'll take your case of 9.81 m/s. We'll also take you're number for the speed over the 'foiled' portion of 9.91 m/s.
Though the value for the density of salt water is greater than 1000 kg per meter cubed, more like 1020 kg per meter cubed, I've never read, nor heard anyone complain that their tri-fins didn't seem to work as well in fresh water. But then again, I miss a lot.
Using the form of the equation as given in figure 1, we see that there is no need to calculate the pressures directly. That is in order to gauge the pressure differential we need only need the flow speeds.
I'm don't quite understand the reasoning behind your pressure numbers, but as it turns out, there calculation was unnecessary. So, I guess it doesn't matter, at least here.
Here, the left-hand side of the equation (using your numbers) comes to - 985 Pa or - 0.985 Kpa, which is the differential you get when you plug in the numbers. Whether that number is positive or negative will depend on the perspective taken, here the terms with subscript 1 are with respect to the flat side. So, here the curved side is experiencing a hydrostatic pressure drop of 985 Pa relative to the flat side, the direction of lift being towards the foiled side.
In order to calculate the force on the fin which we might attribute to this pressure differential let's use an idealize triangular fin with a base of 12 cm (120 mm) and a height of 15 cm (150 mm). One side of such a fin would have an area of somewhere around 90 cm squared -i.e. one half of base times height, or 0.009 m squared.
Here, a pressure of 985 Pa times an area of 0.009 m squared gives 8.65 N. For those bewildered by SI units, that's about 2 lb of force. The point being that using your numbers, 9 N or 2 lbs is not completely insignificant. Press you finger on a bathroom scale to get a feel for what these numbers represent. It's a little higher if you use 1020 kg per meter cubed for salt water, something around 1000 Pa or 1 KPa. Also, reduce or increase the speeds involved and this number drops or increases. Roughly, 1 meter per second is about 2 miles per hour, so here we are referring to an incident flow approaching 20 miles per hour.
So I guess, assuming that this the simple treatment is not horribly misapplied, that when conditions and the flow is just right, a given individual foil may very well produce such a pressure differential. (The way the differential was calculated here is really crude, but nevertheless possibly enough to gauge a possible effect.)
But toe the fin, or change the angle-of-attack to anything but parallel to the incident flow, and you've got problems. Under such conditions its unlikely that the flow will remain laminar and any simple application of Bernoulli breaks down rapidly, especially if you're dealing with sharp leading edges. Rounding the leading edge a bit, may delay the inevitable, but not for long. And of course there is also that other pesky lateral fin, as well as the possible presence of an additional symmetrically foiled fin -i.e. we are usually dealing with fin systems when asymmetric foils are present.
But let us take a set of asymmetric un-toed twins. Here the pressure differentials would cancel each other, unless your turning. In which case, the outer fin would actually have a higher differential and would be trying to pull you into a bigger turn, that is it would generating a centrifugal (center fleeing) contribution, rather than a centripetal (center seeking) one, as the speed of the flow on the outer fin would be greater. Which would tend to make you 'skid' out? Do asymmetrically un-toed foiled fins make you skid out? I actually never heard such a complaint, but then again I miss a lot. Taking this argument much further than it should, would suggest that you might do better with symmetrically foiled laterals, but there would appear to be testimony which contradicts this.
Of course if you were up on your inside rail, you could argue that as long as that outer fin didn't pop, the net result would be to counteract sinking the inside rail to deep? Maybe that's it? Or maybe not.
… okay, I'll stop, but...
When a single asymmetrically foiled fin is consider in isolation subject to laminar flow conditions, the differentials in hydrostatic pressure between the two sides of the fin is real. When you put together a system with the kind of symmetry generally seen in fin setups involving multiple fins, the net outcome is likely to be somewhat different. Does this mean the consequences of foil goes away in a setup? No. Just that, I believe other effects will tend to dominate the situation.
kc
I would completely agree with this. I was using edge system boxes for a while and changed the fin angles around a lot. Definately made huge differences. And many times the angles were different from one side to the other.
We used this book at the university: http://www.amazon.com/Introduction-Fluid-Mechanics-Robert-Fox/dp/0471202312
It’s not cheap. But I’ve uploaded the slides our prof used, here: http://users.vtk.be/~s0169770/Fluidum.zip
Almost everything is in the slides, but you would probably need the handbook and a good knowledge of mathematics and physics to understand it. (external flow is in chapter9)
Your lucky that our handbook and slides were in english 
By the way, given the model we're dealing with, in particular the characteristics of the foil involved and that the two speeds are usually not that different. You get can really silly and take this all a bit further, see figure 2.
Which basically restates in more explicit terms what we've already seen above, but now you've got a really simple equation. Pop in the numbers, the result is pretty close to the original, and given the way Bernoulli is being applied here, definitely close enough.
By the way, I don't recommend you run around applying this, at least not in my name. But this does at least suggest that the characteristics of the foil matter.
So, if its all about the pressure differential, then the logical conclusion would be towards thicker foils, within reason, whatever that may be. But why let that stop you (not specifically you hunty, but to the reader, if any, in general.) Why not make and then split a cylinder of diameter 12 cm and height 15 cm and use the halves as laterals. I ain't gonna build them, but I have to admit, out of pure curiosity, I would definitely want to try surfing them. ... I mean, it kind of works for horseshoe crabs, and they've been around longer than some of the retro brands sold in my local shop... I think?
kc
ps
to hans - Path does matter (the assumption here is for steady flow, when streamlines, streaklines and pathlines all coincide). Yes the whole treatment being applied here is crude, nevertheless streamlines are about paths (here.) I inclined to believe, given the quick and dirty nature of huntly’s makeshift continuity argument with respect to the kind of foiling we are addressing, using paths the way he has, isn’t really all that out of line… it’s kind of nice actually.
pss
not necessarily directed towards hans - If you thought that lift/drag referred to an effect rather than a mechanism, you’d be correct. Any force attributed to a moving fluid can be interpreted in such terms. It’s merely a matter of resolving the actual resultant force of the interaction into two orthogonal components. Lift/drag analyses are without a doubt very very useful, but some perspective is usually required. So to say that a flat plate at an angle to a flow will generate lift, does not describe the mechanism itself, only that whatever force is being produced can be resolved in such a way. Bernoulli is about mechanism and it particular its developed using an energy argument, see figure 1 for a reasonable statement of his principle. The forces generated by moving fluids are not limited to changes in hydrostatic pressure, which is what Bernoulli’s principle addresses.
When applied to liquids, such as water, it [Bernoulli’s equation] makes use of conservation of mass, and for incompressible fluids, of which water, in this application, is one, alternatively makes use of conservation of volume. So if you have two equal flows, flow being mass per time, or volume per time, then they can have different speeds if there spatial dimensions are different. How this relates to accompanying differences in hydrostatic pressure relates back to the energetics of the phenomena.
If i may jump in here…
its unfortunate you’ve tried to derive the flow calculations for 2-D shape in an incompressible fluid ( a fin in water) . The NASA webpage clearly identifies the need to use both continuity and momentum equations to solve the flow field.
The NASA page corectly states that the difference in path length between the top and bottom surfaces as Not the true cause of the pressure difference between two surfaces.
however, regardless of its origins the flow on top is faster than the flow on the bottom and bernouli does a pretty good job of determining the pressure distribution surrounding the foil.
If you integrate the resulting pressure distribution around the foil you will end up with a resultant force vector per unit length of the foil.
My point is, dont be side tracked by the fact that bernoulli does not properly explain the flow field around a fin. Once you have calculated the flow field , he does explain the pressure distribution, which does determine resultant lift and drag forces.
more to the point, your calculations do not provide an explanation for the initial question:
what is effect of ‘foiling’ a fin contour ?
the answer is that it provides additional lift.
the combination of camber and/or foil thickness can provide additional lift compared to a symmetric foils at a given angle of attack.
however as greg has already stated, increasing the angle of attack of ANY fin will increase its lift as well. but as you increase AOA you typically increase both lift and drag.
fins that generate their ‘design’ lift at low angles of attack generally have less drag (thus more efficient) than fins generating the same lift at higher AOA.
despite the fact that folks can calculate this. most of the data was derived empirically shortly after the wright bros started flying.
attached is a sample data sheet for one of hundreds of profiles tested by NACA
showing lift and drag at various speeds and angles of attack
(including the effect of flaps)
regards,
-bill
sorry file size limits has made it a bit difficult to read
I totally agree with you!
Bernouilli give a good idea of the lift, because is the wake is small, the boundary layer (and wake) stay relatively thin so bernouilli comes close to reality. The momentum equation results in where the wake starts, so for a sphere bernouilli is mostly not true because of the big wake, but with an airfoil shape the wake is relatively small. A turbulent boundary layer gives more momentum in the boundary layer and makes the wake appear later so it’s smaller (some scratches on a fin reduces the wake like the dimples in a golfball and make the wake smaller).
My conclusion: the added lift by foiling is very small as hunty calculated. But foiling is still important to reduce the wake and so the drag and to let bernouilli be relatively true! So double foiling appears to be as good as singlefoil, but single foiled fins have a bigger AOA due to the leading edge being closer to the stringer and that’s what you feel. (This is what I think of it)
I feel dumb…
… you’re among friends, amigo. And hopefully it will get even more confusing…
And add to the comments in the figure, ''foil’em if you got em boys… but just remember this is about getting in the way… the right resistance at the right time…"
kc
If i may jump in here…
its unfortunate you’ve
tried to derive the flow calculations for 2-D shape in an
incompressible fluid ( a fin in water) . The NASA webpage clearly
identifies the need to use both continuity and momentum equations to
solve the flow field.
I was aware of the nasa page before i started, the reason i continued with it was that it comes up more often then not when you are talking about asymmetrical side fins that you need them to be asymmetrical due to the lift(or pressure difference) that an asymmetrical fin produces which i assume they get from the bernouilli position.
(also trying to calculate what you are talking about is beyond me)
for example
**Actually lets keep this poor little thread alive for a bit longer. **
The reason you don’t see a bunch of tri fins with two sided foils (foils like the trailing fin) is…
…spinout.
The pressure diferential each side of the fin experiences has a purpose.
http://www2.swaylocks.com/forums/fin-foils
The point i was trying to make was that even if this was a correct way to look at it, the pressure we would be talking about would be very small in the scheme of things.And that other factors would play a more important role imo.
IMO as far as surfing goes, unless you are an epic surfer riding high performance equipment why bother with expensive high performance fins based on performance hype(performance which a fair wack of us wouldnt even notice anyway)?
I say go and buy all the rear fins(from broken fin sets) you can find off ebay for next to nothing and be set for life fin wise.And if you do feel like splashing out, get a beautiful set of hand made fins(which would most likely be right up there performance wise anyway) from guys like huie who have been in the industry long enough to just know through experience what works and what doesnt work.
**
**
wouldnt everything apart from drag cancel out when you are going straight, and in a turn when you are going over on a rail point v into the other side of the fin, so more like this?
So the more perpindicular to the velocity your fin were presented at, would that begin to create more drag than useful lift?in which case would be the advantage of the toe in?trading off some drag going in a straight line for less drag on tight turns?
My sense of it is Yes to all your rhetorical questions. Toe-in also makes turn-in and pumping ([at-]tacking the fin and hull) easier–less resistance if the leading edge is angled into your effort a bit*. Also a toed fin is appropriate given the diagonal flow (in relation to stringer) at basically all times.
which is I think what Mike D said about what concave does at the rail*
**unless your board is too wide, to the point your foot loses leverage trying to sink that rail–too much work too far from your toes
Those are the kind of questions that bother me too. I also agree that it's about trade-offs. Here, with respect to fin systems, but also with respect to a lot of other design elements as well.
With respect to modern fin systems, I think drag is key. It's just not a very romantic notion. But when applied strategically it works wonders to enhance stability and control in general, and allows the surfer to remain in and work the critical region of the wave, with less effort, conscious or otherwise.
Waves are not static things. In fact they are pure acceleration, or if you like acceleration and deceleration. That is they are about changes in motion. As the wave moves towards the beach (starts to break) or over a reef it's decelerating, and the critical region is decelerating faster than the shoulder. Also, as the wave form is decelerating, the flow up its face is accelerating, and the flow up the face in the critical region is accelerating faster than out on the shoulder. Get to far enough ahead or behind in any one of these zones and you loose - that is you need to stay connected or coupled to the wave.
I like to think of a good fin system like a good suspension system and set of tires for a car – just the right amount of damping and friction that allows me to maintain control and connected to the road, for a given set of driving conditions. Fact is, if you've got sufficiently good technique, you could manage with far less. But with the right suspension and tires, maybe you can start getting a little more creative as opposed to being continually concerned about managing the rudiments. I think the modern term for this is 'user friendly'.
Changes in design will often require, or force changes in technique. Often these adjustments to technique are done automatically without much effort, conscious or otherwise, and are quickly forgotten. As you wrote, you may see some benefit during some maneuver, but if fact you need not. It just may be that any negative impact on that maneuver was simply overcome with a slight change in technique, a little more use of rail, a little shift in posture, etc.
So yes, or maybe no... it depends. Gee … that's useful, huh?
... but I really did enjoy your application of Bernoulli's principle... that was fun. Throwing the word around is one thing, but getting your hands dirty... hey, that takes a set.
kc
Hi kcasey, thanks for your replys.you’ve definately put up some interesting info and im still trying to digest it.Thankfully hans posted up some links to reference material and i found a second hand book on amazon at a great price so i should really be able to sink my teeth into this a bit more.
I can really say ive learnt a thing or two from this thread, which i am pleased about.I think what i was wanting to know about an asymmetrical fin opposed to a symmetrical fin is answered in my mind.That was the goal, as everything i have been getting up until now never quite made sense to me.
sorry for the long winded non-answer.
i did not intend to be critical of your math skills.
most of my explanation was excerpts from college fluid-mechanics courses (some 40 yrs ago)
its interesting that lift in non-viscous conditions is quite accurately quantified by bernoulli type calculations ( conservation of mass). even better, the lift calculated IS equal to change in momentum of the fluid ( the other INCORRECT theory offered by NASA) But all the examples from college days were simple enough that the equations could be boiled down to exact solutions. For the vast majority of flow conditions we used empiracle test results or waited for the advent of finite element solutions on fast computers ( fancy term for an infinite number guesses until one happens to work).
but back on topic.
the answer to your question is in the lift/drag coefficient data sheets for the NACA fin in the attachment above.
lift= .5 * cl * density* velocity * velocity *area
at 0 degrees angle of attack the lift coefficiect (cl) = .4
assuming the board velocity is 20 ft/sec ( 6.1 m/sec, sorry i hate metric)
lift force= .5(.4) (1.98) (2020) A =127 lbs per square ft of fin surface.
not insignificant, even though the area of a thruster fin is probabaly on the order of .1-.2 ft^2 .
13-25 lbs lift
AND by definition, a symmetric fin generates zero lift at 0 degrees AOA.
This is not to try and talk you out of any opinion, i’m just trying to help quantify the choices.
regards,
-bill
OK that is interesting, so it seems that bournilli only works in 0 degree angle of attack?so still the asymmetrical fin is not of much use to us because when we want lift is when we are up on a rail in a turn is it not?does this mean that what we are mainly concerned with is newtonian type lift in this situation?Also when you say not insignificant can you give a figure for comparison of lift produced in a turn?
Maybe they are helpful in making it harder to get the board over on the rail due to the lift at 0 degree AOA.Is that why you see alot of the traditional fish with wide tails having symetrical twins, because the extra width gives this lift more leverage being further out making the board that little bit harder to get on the rail?.
Actually given that the side fins have toe in therefore never at 0 degree AOA are they not producing this type of lift at all?
Hunty,
Fins perform two distinct and diametrically opposed functions. The first and what you are refering to in directional control. Angle of attack and minimizing leading edge turbulence have alot to do with how well fins perform this function and you have covered alot of the theoretical reason well in the posts above.
The second function is kinetic propulsion. Pumping your fins to get through an area of the wave that is not providing enough power on it's own to get you where you want to go. This is where concave and flat sided assymetrical fins perform better than convex 80/20 or 50/50 symetrical fins. The inside surface of a flat sided or concave fin grabs and redirects more water rather than shedding it like a convex or symetrical fin. This is where leading edge drag is a benifit and why I say diametrically opposed. The traits that make one category of fins perform better in weaker conditions limit them in more powerful conditions.
That’s why options are a good thing.
I really agree here. A fin lives in a very dynamic world of contantly changing parameters. But the ability to create speed through drive is certainly one aspect that isn’t nessasarily related to the standard concept of a less dynamic use of a wing. Thanks Tom for bringing up a different perspective.
not really,
what i meant was that neither of us can really calculate the flow field around any particular fin section
without the benefit of a 2-d potential flow code that solves the appropriate continuity equations.
However, if you did know the velocities everywhere on the profile, you would then be able to calculate the pressure distribution and the resultant lift and drag forces. The velocities around the fin change quite dramatically and you need to integrate the entire distribution to calculate the net lift and drag forces. [once you have the flow field, the assumption that foil surface becomes one of the streamlines is acceptable. At this point, bernoulli’s theorem which identifies a streamline as path of constant pressure (potential) is still valid and conveniently defines the reltionship between static pressure and velocity]
…or you could just look up the data for your particular foil profile
using the same example above.
lift coefficient increases linearly from 0 degrees to approx 16 deg AOA
at 16 degrees the lift coefficient is 1.6 ( a 4 fold increase)
after 16 degrees the foil stalls and lift will crumble while the drag coefficient jumps dramatically
real numbers for a specific yet extremely “1-dimentional” example.
you still need to make the data fit real life applications.
-bill
See: birds
; )
etc