Wetted surface area - approximate formulas
A few notes on wetted surface area (WSA). The notes use only feet, pounds and long tons as I believe readers will be more familiar with them as opposed to metres, kilos and tonnes. I wrote these notes because I have seen on several sites various questions about WSA and the effect on speed, which has more WSA – a catamaran or a monohull, plus several other queries and they are all answered here. I have reviewed many Internet sites and in all reviews of speed, it is assumed that the reader is a designer or a naval architect (NA) who is familiar with calculations and has the background knowledge. I put myself in the position of someone who wants to build a boat perhaps or buy a kit and asks what is WSA all about and how much paint. Simple questions and I had trouble finding anything. If there are postings addressing issues, sort of The Idiot’s Guide to WSA then I apologise to their posters. If you know of anyone who would like some clarification on WSA please point them to this thread.
WSA has three major applications for vessels
There are five ways of finding the area
These are not exact, there are many formulas, and they all disagree to some extent or another. Some specialise in a particular ship type and others are general. Several need some kind of chart showing various factors, others have a set of factors that are used to figure out another factor that is used in the formula, the rest are simple formulas that use the vessel’s dimensions and properties directly.
The formulas are properly called empirical formulas meaning that they are based on experiment or observation or based on known values from ships’ records or file documents. They are not known in the sense of a law such as E = m*c^2 but that they are only approximations.
In the formulas these abbreviations are used throughout
If you know math conventions skip the next two paragraphs
In case you’re not too happy what the symbols in the formulas mean here’s another list
(4*Pi*r^3/3)^(2/3)* a factor call it K = 4*Pi*r^2
skip the math and K = 4.836 - actually it is (4*3^2*Pi)^(1/3)
What it means is that the surface area of a sphere = 4.836*Vol^(2/3)
A cube has a factor of six and all solid shapes have a factor unique to their shape.
The term Vol^(2/3) appears in several formulas as we’ll see later.
Formula examples and application
There are two short formulas that crop up on net on sites dealing with boats
1 WSA = 0.85*LOA*B
2 WSA = L*(B + T). Multiply this by 0.75 for medium displacement yachts and by 0.5 for light displacement yachts.
These are dangerous if you use them for the first time without anything else to compare them with. For example, is it the main hull or does it include bilge keels, rudders and any other appendages to the hull? What is the form or the general shape or the proportions? All these have an effect on the answer. But if you work with only one boat type you will be able to calculate the coefficients using known designs.
To be fair, all the formulas need cross-checking. The general formulas take more account of the possible variations in the hull shape etc but they too still need to be checked. All the following formulas show the wetted surface of only the hull and you must add the appendage areas for the grand total. Rudders, keels, bilge keels are straight-forward but the immersed part of a transom must also be added as the formulas all assume the area curve closes at the waterline ends.
US text-books usually give Taylor’s approximation and maybe a couple of others. For example, the SNAME publication Principles of Naval Arcitecture gives good examples of the difficult secant method etc and Taylor is the only approximation given, see Section 7 of vol 1. Taylor was instrumental in opening up the first US model-testing tank and he published his brilliant Speed and Power of Ships (the full edition) in 1933 based on investigations carried out during the previous 20 years or so. I don’t salute many designers and naval architects but that man is way up there with the best of them.
His formula is based on a warship hull form of the British Royal Navy around 1900, the cruiser HMS Leviathan with a few modifications. Although Taylor made model tests on a large number of models by stretching and squeezing, cut and paste, it is nevertheless limited in the breadth/draft ratio. Before making the expensive models each was drawn to a large scale and the areas were carefully calculated. From these areas Taylor deduced his famous formula. Some designers rarely used it because of limited beam constraints for barge forms and the chore of continually referring to his chart that is shown in the illustration below.
Despite these and the unusual base ship it gives good results for most ships types with a moderately low Cb and within the constraint 2 <= B <= 5 approximately (fancy mathematical way of saying B is between 2 and 5)
Taylor’s formula: WSA = C*(L*Displ)^(0.5)
C is as shown in the illustration below. The Formula assumes that the Displacement is in salt water so even if your vessel is in fresh water your must correct the displacement. Or you may use the Vol and the formula becomes in either FW or SW
Taylor’s formula revised: WSA = C/5.916*(L*Vol)^(0.5)
Another American formula, that is rarely used, is one by two professors who produced valuable work in relation to the Great Lakes Bulk Freighters – Baier and Bragg. Their formula takes care of dimensional distortions by making use of what is termed the Displacement Length Coefficient. I have often wondered why Taylor did not use something like this for areas since he made extensive use of it in his method of determining the ship resistance. So the formula is;
Baier-Bragg formula: WSA = C*L^2/10
C is as shown in the Baier-Bragg chart below. The source data for this method is a table of C and Displ/(L/100)^3, like Taylor’s the Displ is in long tons of SW The intervals are too big for convenience so I made the chart and fitted a trend-line in Excel. If you are using a pocket calculator use the chart but programming or a spreadsheet use the formula that is (using x for convenience instead of Displ/(L/100)^3);
Baier-Bragg coeff: C = -0.906385E-5*x^2 + 0.954632E-2*x + 0.776457
The formula for the coefficient is obviously too precise but you can tailor it yourself to whatever degree you wish.
The chart has two curves drawn, one is the base values and the other in light red is the trend-line fitted by Excel. The lines closely overlap so you can see the trend-line is a good fit. The formula coefficients for the simple quadratic are shown as is the goddness of fit coefficient R squared.
The next one was deduced by Froude, as a young man he was a junior engineer working for I.K. Brunel and he also worked for him on The Great Eastern doing “marine” calculations. He finally persuaded the British Admiralty to open a test tank to research ships’ resistance. Many people had conducted experiments with models but it is Froude who laid the scientific groundwork for what is done nowadays. He not only explained the frictional drag of a ship he published one of the very early formulas for WSA.
Froude formula: WSA = Vol^(2/3)*(3.4 + L/(2*Vol^(1/3)))
The Admiralty opened a tank at Haslar and the following is an adaptation of Froude’s but still for “warship” forms;
Haslar formula: WSA = Vol^(2/3)*(3.3 + L/(2.09*Vol^(1/3)))
Now we arrive at my own particular favourite, the Denny-Mumford. Mumford worked on powering in the late 19th Century at Denny’s model testing tank in Scotland. This was the period when many of the major Yards in Britain had their own tanks and were experimenting in the early days of model testing in the rush to find more efficient hulls. He produced a formula that is probably the most versatile of all the formulas yet is among the simplest to operate, so simple that it seems it can’t be so accurate. It gives good estimates of the WSA of even the most extremely proportioned forms. The expression is;
Denny-Mumford: WSA = L*(1.7*T + B*Cb)
Another form of this uses the Vol and
Denny-Mumford revised: WSA = 1.7*L*T + Vol/T
I will explain why it is my favourite in a later posting, for now just accept that it is easier to juggle than the others.
Probably one of the most recent formulas is the one developed by Holtrop and Mennen as part of their now classic paper on speed and power. They were researchers at the Dutch testing establishment at Wageningen now known as Marin. The formula is applicable for a wide range of forms and the published limits for powering are;
Holtrop-Mennen formula : WSA=L*(B + 2*T)*Cm^(0.5)*(0.453 + 0.4425*Cb - 0.2862*Cm + 0.003467*B/T + 0.3696*Cw )+ 19.65*ABT/Cb
The last term, 19.65*ABT/Cb, is the bulbous bow wetted surface and it may be added to the WSA for any other formula.
These are examples using the various formulas for a selection of vessels where the results are declared and are verifiable.
All the data has been corrected to a standard length of 27.5’ LWL so you get sense of how the other properties affect the area.
Despite repeated searches I was unable to locate any pleasure craft or yachts that gave enough data to check the formulas against known results.
A table of the final check results etc is shown below in the illustration below. The results by formula closest to the known values are in a different colour and boldly underlined. No firm conclusions should be drawn because such disparate vessels are used. Although only one formula is highlighted in each case some of the others are also quite close and who knows, maybe the “known” answer is wrong.
This concludes the first part, the next posting will be more interesting as it looks at WSA of monohulls versus catamarans. If anyone has any comments or queries – fire away!
If you're going to cover the wetted surface area of multihulls, I hope you plan to include the effects of heel and vertical sail loads, as well as the difference between trimarans and catamarans. The differences are significant - see http://www.steamradio.com/JSYD/Artic...itFertalk.html
Wetted surface area Part 2 - monos or duos?
This posting concerns the wetted surface of a catamaran versus a monohull. All the aspects of stability, suitability, seaworthiness etc are not addressed only the WSA.
All values in the same units as the earlier post – feet and pounds or long tons (2,240#).
I exclude the South Seas praos and the French catimaron although the latter may have inspired the very slow cat houseboats (or inspired by them?).
Monohulls were the standard from the earliest times to the late 19th Century when some catamarans were built as railroad ferries for service in the English Channel between the UK and France. They were largely unsuccessful as the strength of the connections between the hulls and the deck could not be accurately calculated – the old story of a 100 mathematicians taking a 100 years. Because of serious cracking and failures, the ships were pulled out of service and catamarans re-gained a bad reputation.
There had been earlier attempts that were experimental rather than commercial such as those by a medical doctor, Sir William Petty, who discovered that a long narrow hull was faster than a stubby one but less stable. So he attached two hulls, saw the wave patterns, discovered a great deal of useful information but nothing came of it. He built a 20 ft cat and a series of bigger, successful versions. He built his biggest craft in 1684 but it was a failure and he died in 1687, his high hopes unfulfilled and catamarans lay dormant for another 200 years with an un-deservedly bad reputation
Probably the fascination with modern cats began in 1947 in the UK when two Olympic canoeists fastened two hulls together and started to sell “improved versions” built in their father’s boat shop. In the US in 1950, the Hobie appeared and cats never looked back – except at monohulls trying to catch them.
All these catamarans have the same thing in common – speed, and they all follow the same trend in that they are long and skinny compared to a monohull. Too much of an extension and a monohull becomes unstable which is why it is rare to see a B/T ratio less than about 2.25. Instability is often cured on pleasure craft by permanent ballast or lead keels but the principle’s the same – something heavy at a low position in the boat. As boats get bigger and longer that fix becomes impractical as hundreds or thousands of tons would be needed. The only solution is increased beam and the use of lightweight materials topsides.
When any immersed hull increases speed it must overcome two main resistances
At low speed-length ratios, say less than a V/L^(0.5) of 0.8 or so, where V is the speed in knots, the frictional resistance is by far the major component. As the speed-length ratio rises the wave-making resistance increases at a faster rate and soon dominates. At a speed-length ratio of around 1.33 or so the rate of increase is so great that it becomes something like breaking the sound barrier in a plane. Usually only small boats, warships, superfast ferries and racing craft travel at higher speeds.
A speed of just 4 knots on a 16’ boat is the same speed-length ratio as 20 knots on a 400 foot ship or almost 34 knots on a Nimitz class carrier. But the same 34 knots on a typical frigate or destroyer is about a 1.7 speed-length ratio.
The speed at which the wave-making resistance exceeds the frictional resistance depends on the length and shape of the hull. The longer and/or slimmer it is the higher the speed at which it happens. The tendency is therefore to make the cat generally longer than the monohull so it automatically becomes much slimmer. One cat hull is only half the volume of a monohull so it is naturally slimmer but the beam is usually further reduced which decreases the resistance again. The frictional resistance has increased but the wave-making resistance has reduced and the total is less on the two hulls. At very low speeds the frictional resistance will increase by a severe margin but the boat operates at high speeds all the time so the increase is operationally insignificant.
There is a balance that must be achieved in that too long a hull will increase the weight as well as requiring a stronger hull because of higher loadings. As with many other naval architectural problems the best answer is a compromise between the conflicting factors taking account of cost as well as the technicalities.
In the first post I said that the Denny-Mumford formula is my favourite and this is why. All the terms are known in the very basic design, no calculation of other hull form coefficients or characteristics is necessary and no coefficients need be looked up. If you play with a hull’s proportions making it wider with less draft and so on you will find that the area calculated also changes but it doesn’t change using most other formulas. Taylor’s formula means continually checking the diagram for any significant shifts but the diagram has dead positions and others are lively. All my searching seemed to be on the lively areas so I was forced to use the diagram.
Denny-Mumford has one advantage which is missing in the others, you can find the B and T values that give the minimum WSA
Keeping the length, Cb and displacement constant means that B*T must also be a constant, call it K so the formula can be rewritten as; (If you don’t like calculus go to the last line of this paragraph)
WSA = L*(1.7*K/B + B*Cb). Differentiate this expression with respect to B therefore
dWSA/dB = L*(-1.7*K/B^2 + Cb). L is common, divide by L and transpose so that
B^2 = 1.7*K/Cb = 1.7*B*T/Cb then
B = 1.7*T/Cb or T = B*Cb/1.7 for minimum WSA.
WARNING: these are the standard formulas for displacement hull ships and may give false results for other vessels. Take “your” hull type and work out the coefficient in place of 1.7 and maybe the second term in brackets should also have a coefficient instead of 1 (one). It is straight-forward in spreadsheets because they all have routines to find the best fit of a formula to suit the data – in Excel this is Solver. If you are unhappy about this then post a message and I’ll write an Excel spreadsheet and post it for you. I’ll save it in the oldest version of Excel that supports it so it should be understood by Lotus or Quattro
Starting with the basic D-M, these last two formulas and a little algebra it is an easy matter to see what happens to the WSA when we begin slicing and dicing.
N is the number of new geometrically similar hulls (geosims), so N is 2 for a cat. The basic monohull has the dimensions L, B etc and the new hulls have L1, B1 etc. If the cat hulls are the same shape as the monohull ie geosims then from the first post we know that a coefficient times Vol^(2/3) gives the WSA and the constant is the same for the cats and the monohull because they have the identical shape, only smaller. This means we can forget it because it is constant and we need only look at proportions. (The derived forms that are explained are similar to each other, which is true for a catamaran but not for a trimaran and up. If you need a trimaran set-up it is easy to figure out from the following how to do it.)
Vol1 = Vol/N so the WSA ratios are
(Vol/N)^(2/3)*N compared to Vol^(2/3)
divide the first expression by the second and it results in
(1/N)^(2/3)*N = N(1/3) = 1.26 for a catamaran and 1.44 for a trimaran
In other words, a proportional shrinkage all round means the two cat hulls’ WSA is 26% greater than the monohull, 44% greater for a trimaran. This is when the volume of displacement remains constant and it is unlikely due to the additional weight but this would only worsen the situation. To keep it simple it is always assumed that the volume is constant. So what size is the cat hull compared to the monohull?
We know that N*L1*B1*T1*Cb=L*B*T*Cb but we kept the proportions the same, Cb is equal so this simplifies to L1^3=L^3/N and L1 = L/N^(1/3). This proportion (1/N)^(1/3) is 0.794 for the cat and 0.693 for the trimaran. So a monohull 20 feet long x 10 wide x 4 deep would shrink to 15.88 x 7.94 x 3.176 for the cat.
This would not happen as the length would be increased so we’ll call the ratio of the lengths X i.e X*L = L1 and for the moment keep the B/T ratio the same i.e. B/T = B1/T1
We know that N*X*L*B1*T1*Cb = L*B*T*Cb therefore
N*X*B1*T1 = B*T and so B1*T1 = B*T/(N*X)
But we kept these the same proportion so then
B1 = B/(N*X)^(0.5) and T1 = T/(N*X)^(0.5)
This time we’ll keep the cat length the same as the monohull so X = 1 and both B1 and T1 change in the proportion 1/N^(0.5) = 1/2^(0.5) = 1/1.414 = 0.707. As check the volume of displacement of the cat hulls is 2*L*0.707*B*0.707*T*Cb = L*B*T*Cb. The WSA is 2*L*(1.7*0.707*T + 0.707*B*Cb) which is the basic monohull times 2*0.707 equals 1.414 meaning a 41.4% increase in WSA. This is a consistent pattern, some slight improvements can be had by optimising B1/T1 but it is quite small and the cat hulls are becoming much heavier which would worsen the picture
In the last paragraph the WSA was tailored for a catamaran and the general formula is
WSA = L*(N*X)^(0.5)*(T + B*Cb)
Remember that it is deduced from a monohull using the monohull dimensions and block coefficient. If you use a calculator be sure to enter a whole number for N, you could enter 2.5 if you wish and you’ll get an answer but is has no meaning. What is half a hull?
Using the last formula we can see the effect of changing the length of the hull by any reasonable proportion we wish, if a cat hull was half the length of a monohull we would end up with the same WSA but a strange looking form. Also remember that we have retained the original Cb and B/T ratio but there would be changed by the designer as the design progresses. A full rigorous design would start with good guesses then evaluate the speed, stability etc then make changes as a result and do it all again until the guessed input equals the required output. Incidentally there is a paper by Stephen M. Hollister in the Boatbuilding Articles entitled The Design Spiral for Computer-Aided Boat Design. It describes the design spiral, this round and round process until you get the right answer, trial and error, call it what you wish. It is one of the Must Read papers in this site for those engaged in designing a boat. You may find it heavy going at times as it looks into everything that a Buyer would consider if he were laying out a very large amount of money. So skim through some of the parts but read the rest carefully. But note how you start the whole thing – define the purpose and then start the design.
Some very approximate calculations have been made of the resistance for a small boat and catamaran derivatives. You must be familiar with the terms WAG (Wild assed guess) and SWAG (Scientific wag), I favour referring to these calculations as DOG estimates because they are ruff, ruff, ruff.
The base monohull is a displacement type 26 feet L x 7 feet B x 2.25 feet T with a block coefficient of 0.525, totally made up without reference to anything. The total volume of displacement, approximately 215 cu. ft., is constant on all the derivatives and its WSA is 195 sq. ft.
The three derivatives are
If anyone wishes to run a few calculations or compare hulls etc this is how I derived the power. The method used is a combination of two superseded approaches, frictional resistance by Froude and wake-making by Taylor. The Froude method gives results that are very close to the modern Cf method. In fact the modern method was modified, some claim, to more closely match the Froude values. The resistance is given as
Resistance in lb = f*WSA*V^1.825
V = speed in knots
f = 0.00871 + 0.053/(8.8 + L)
The wave-making resistance is much more difficult so I used a formula that was published by D.W. Taylor in the early days of his researches before the publication of the final edition of the The Speed and Power of Ships. His formula was put forward as being reasonably accurate at most speeds and especially good at a speed-length ratio up to about 1.1 making it unsuitable for the higher speeds in this quick review. It has the merit of ease of use and that is good enough since its results conform to the typical patterns expected. The formula is
Resistance in lb = 12.5*Cb*V^4/L^2*Displ in long tons
Horse power = Total Resistance *V/163 delivered to the propeller, approximately.
This allows a 50% loss for the propeller etc.
Three diagrams are shown below which clearly show the rapid growth in resistance and the associated power. When a vessel moves through water a series of waves is created along the hull, very slow gives many and very fast means less than one, the fewer the waves the higher the resistance. There is a time when there is only one wave in the length and the vessel sort of sits in a hole in the water, this occurs at the “magic number” when the speed-length ratio is about 1.35, the actual value depends on the shape and fullness of the hull.
It is difficult for the vessel to get out of the hole and it takes a lot of power but when it succeeds, there is a reduction in the required power even though it is moving faster. This up and down pattern is repeated but it is not shown in the very simple approach taken here. The results are only approximate and simply a general guide.
The third diagram shows what happens at low speeds, say something like a houseboat that will not go cruising and only moves slowly to another location to moor there for a while for a change of scenery. But it is clear that the power for the monohull is less than any catamaran and the WSA is important as the wave-making resistance has not yet passed the frictional resistance.
I hope this has answered the general question about the WSA of a monohull and a catamaran and what happens.
But why did I not optimize the B/T ratio? Because there was only a marginal improvement of a good (or lucky?) selection in the first place. But the main reason was that I wanted you to pick some dimensions yourself and see what happens. There is no doubt that the proportions would change on fast designs making a slimmer, go-fast hull.
Any queries? Just ask.
i got this simple rough formula for hullspeed: WL^0,5*4.5=S.
square root of the WaterLine times 4.5 equals Speed i am told and as far as i know this formula also applies on cat and trimarans hulls adding those lenghts up in that sum, so multihulls go faster!? forexample where a single -waterreplacing- hull 13 meter boat would do max 16 km a tri-hull can do 3*13m. =39m. hull^0.5*4.5=max 27 km. all considering wetted surface, an ideal 10 to 1 hull ratio etc. for a pleasure cruiser/houseboat that size and speed (and fuel economy) increase is attractive it seems to me. maybe you math wonders can give me some ideas if this is 3/4 correct? maybe even know some rules on spacing between hulls? the effect of a wider middle beam ( to keep the profile down), other interesting facts? hope the metrics are ok.
i found the same formula in ft. and speed in knots: Hull Speed = 1.34 x sqrt(WL in feet) and yes that converts ok. but can i double the WL for a cat and triple WL for a trimaran in this form.???
yipster anyone knows what this formula is based on?
(any other readings on this?) [IMG]http://[/IMG]
Last edited by yipster : 10-27-2002 at 01:02 PM.
My apologies for not getting back to you earlier.
The formula goes back to the early days of using scale models of ships in model testing tanks to predict the power needed to propel ships. In the late 19th Century Froude in England introduced the “modern” methods of comparison and prediction.
At that time all ships and boats were displacement type meaning that boats did not go fast enough to plane although semi-planing was just possible with lightweight steam reciprocating engines and boilers. But Froude began his experiments for the British Admiralty researching warship hull forms and power so they were all displacement types.
Froude published his famous Law of Comparison in 1876 from which the formula V/Rt(L) is derived. This was the speed in UK knots divided by the square root of the length in feet on the waterline. The difference between the old UK knot and the SI knot now in common usage is almost negligible and can usually be neglected (1 UK knot = 1.000639 metric knots). So models were tested at the so-called comparative speeds of the full-sized sized ships. Suppose a ship was 324 feet long and the design speed was 22.5 knots giving V/Rt(L)=1.25 then a model 16 feet long would be towed at 5 knots.
When the ship power is plotted against V/Rt(L) it rises fairly rapidly but at about V/Rt(L) = 1.33 (or slightly more) there is a very sudden rise and speeds are usually uneconomical. This is the history of the formula and it is still used by naval architects as a guide as it is better to compare hulls of different sizes that way. In order to get away from knots and feet testing tanks use the Froude Number which is V/(g x L)^0.5 where V is the speed in feet per second (or metres per second), g is the acceleration due to gravity in feet per second per second (or metres per second per second) and L is the length in feet (or metres). The advantage is that you get the same number whichever system you use.
When a vessel is moving in water it must overcome the frictional resistance or the drag due to the water and also the wave-making resistance. At low speeds friction resistance is the greater but as speed increases the wave-making resistance increases much faster and becomes greater than frictional resistance. So the wetted surface area is very important at low speeds whereas at high speeds it becomes more important to change the hull shape to something more suitable having reduced wave-making resistance. This explains why hull forms at low design speeds are usually wide but high-speed forms are usually long and skinny.
Nowadays catamarans etc are extremely popular but you must consider them as a set of separate hulls and look at each hull individually. For example; say a simple single hull 16 feet long travelling at 6 knots is V/Rt(L) 1.5, a catamaran with two hulls 16 feet long is the same. A Trimaran with say a centre hull of 16 feet and two outer hulls each 9 feet long has two ratios. The centre hull is still at 1.5 but the outer hulls are at 2.
You must not add the hull lengths, you cannot say that the catamaran is 2 x 16 = 32 so that V/RtL is 6/32^0.5. If you take two or three hulls and make them in line with no space between them you can add the lengths to make one complete hull. This is common with tugs for example. On the Mississippi River in the USA you can see many barges strapped together with a pusher tug at the stern. Or there is the design of one large hull with a notch in the stern and a tug that fits in the notch. Both these arrangements get the advantage that you describe because they are in line with each other not abreast or at the side of each other. If you do a search on the net for tug push or tug ATB you’ll find dozens of examples. ATB means articulated tug barge - a tug fitting in a stern notch of a barge or hull and linked with arms or pins which allow rotation.
I read somewhere about houseboat proportions but I forget where. I remember that it said the hulls are about 10 units long with a beam of 1 unit. The space between is 8 units making the overall width 10 units and the house is about 7.5 units long. You should not use more than two hulls unless you wish to go very fast and that would raise more problems due to spray etc under the house. There are a few sites that discuss houseboats just search the net.
Good luck with your project
thanks for the reply, again educative. not you, but i must apologize for having the wrong idea that multihull lenghts can be added up as in my example. where i got it from i dont recall, might be this same i-net! the advantage of your formule that accepts imperial as well as metric looks very handy to me but i still have to sit and try it. i understand now multihulls must be seen as a set of individual hulls. i like to learn a bit more on this but allready see that it will get quit complicated. fact remains however that multihulls are simply faster than monohulls (i sailed small cats and felt it, in 1914 the sea sled trimaran allready broke records, round the world champs now are in sail and power multi(tri)hulls). is it optimum designed WSA and the long slim hull's that breake wave making resistance what makes them faster? looking on the net at specifics by cats the power, lenght, speed, figers dont calculate. smaller engines, shorter boats but remarkable faster! are there not more good reasons? i guess you mentioned the very slow pontoon house boats in your original article becouse they had catched your eye before but not as example. yes, i like to live on the water again and i am designing a boat, more boat than house, so a little spray is ok. but why would i be going faster (how much faster?) on a 3e hull? gee, i did read this also on the net! many thanks for correcting my wrong formula, but i keep wondering. maybe someone else here on the forum has knowledge on cat and tri hulls and can explain a little more? meanwhile i'll be carefully surfin...
Wave making formula
Thanks for the WSA. I now incorporate choices for WSA calculation in my program. No two comes out the same so it places me in good balpark figure and know where i stand.
I wrote a program that analyzes Frictional resistance against Wavemaking resistance. The original formula is too complex. I am going to use the formula you published. Is there any other formula wavemaking formula that is being used?
It seems that even the skin friction formula has many variations. I have heard of the newest. Grigsons algorithym.
Would you write something about this?
Thats very sad news. His post was really informative. I had it printed and now belongs to my library of Boat Design.
What is really what you want to integrate?, Frames developments, water line lengths, ... ?
These integrations are possible but the results are not very approximate and in ships with transom, knuckles, bulbous bow, can be very misleading.
It is best to divide the hull in small triangular surfaces and add their indiviual areas.
The calculation of the WSA is one of the most complicated in naval architecture. The quality of software for naval architecture can be evaluated, to some extent, by the method used to calculate WSA and the degree of accuracy of the calculations.
Using CAD programs such as AutoCAD, for example, you can perform a very accurate calculation, starting from the water lines.
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TANSL - thanks. I was trying to figure out a way to establish if there is a was to integrate using a matrix comprised of the delta values between a square grid of points along the surface of the hull from LWL to bilge line, bow to stern. It seems to me that there is an integral function for finding the length of the curve from bow to stern at any given level. as well integral function for finding the length of the curve of the hull perpendicular to the longitudinal axis from LWL to bilge line along each station. I was wondering if the two integrals could be combined via a matrix to sum the cross products for each grid section - is this possible? How do you do it without the CAD software? Thank you !!
and then updating it in the light of boundary layer data that was not
available when Grigson developed his formula. See:
and the references in the biblio.
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