this one will...
Cats then stepped v-hulls then non-stepped v-hulls'''
Wetted surface area II: Cat hulls or Monos?
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Hello all
This posting concerns the wetted surface of a catamaran versus a monohull. None the aspects of stability, suitability, seaworthiness etc are addressed only the WSA and the effect on speed.
All values in the same units as the earlier post – feet and pounds or long tons (2,240#).
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 gained a bad reputation.
There were 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, deduced the effects and 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 undeservedly bad repuation
There is a balance that must be achieved in that too long a hull will increase the weight as well as requiring extensive strengthening 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 a 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 always obliged 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: this are standard formulas for displacement hulls and may give false results for other types. 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, good for short formulas. 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 these 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 and the designer's aim of providing it at the least cost. 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 the following 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
Geosims – identical proportions of the monohull and same Cb. Dimensions 20.64’ x 5.56’ x 1.75’ and WSA 245.7
Equal L – the same length as the monohull, same Cb. Dimensions 26 x 4.95 x 1.59 and WSA 277.8
25% on L – the cat hull is 25% longer than the monohull but has same Cb. Dimensions 32.5 x 4.43 x 1.23 and WSA 337.8
The method is oversimplified as I had no wish to spend too much time running calculations but the general trends and patterns are OK and typical of what actually happens.
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 protected 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. The ultra-simple approach I took on the calculations would not have shown any improvement and a more accurate method is necessary.
Any queries? Just ask.
Based on this we have stepped v-bottoms that equal parameters of wetted surfaces equal or better than Cats...criteria? Find a 38'- 42' stepped boat that will do 70mph with a total of 800hp....85 mph with a total of 1000hp and 150mph with 2000hp and you have a v-bottom equal to a cat....there are a few out there that can do this....interesting stuff....we did mockups on cad-cam...only 3 Manufacturers have boats the public can buy out of the box and get these results with....and the interesting thing is that NOW all 3 manufacturers boats can do this and take a good pounding in the seas and do just fine....
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