Immersion

I am (rather rashly!) thinking of buying an Ed Dubois designed Aluminium racing yacht and using it for cruising/racing. Her measurements are 40' (37' lwl) x 12'8 x 8'6 draft. Sexy lines indeed!
She only weighs in at 12000lbs, 5000lbs of that is lead fin keel and ther is another 300 gals of water ballast (I make that 16 crew on the rail - good job I dont have to feed them!). There is a pic of the underbody shape the materials forum
http://boatdesign.net/forums/showthread.php?s=&postid=6403#post6403 - she is about 20 years old.

What I want to roughly calculate is the kg/mm (lb/in?) of immersion. - Is there a rough rule of thumb for this I looked in Principals of Yacht design and got the figures for their YD 40 - but her displacement is a lot more - and to be honest their calculations are a bit beyond me at the moment! I hope to be able to get an estimate of how much weight I can add before she sits very heavily on her lines.

When a displacement figure is given, if it not specified I assume that is "light ship" - what would that include/exclude ?

- I can figure out how to do this if I make the hull in Rhino - but I wont learn anything that way!


Thanks!


Paul

 

Actually Mr Brewer answered my displacement question on his site...

"Most designers figure displacement when half loaded (the boat, not the designer) with stores, liquids and crew."

 

Should have looked harder! just a little further down the page..
(http://www.tedbrewer.com/yachtdesign.html)

"POUNDS PER INCH IMMERSION (PPI): The weight required to sink the yacht one inch. It is calculated by multiplying the LWL area by 5.333 for sea water or 5.2 for fresh. The PPI usually increases as the hull sinks into the water as the LWL area is also increasing due to the shape of the hull above water."


Typical! I spent ages with Larsson and there it was on my favourite designers site all along!

Now all I have to do is take a guess at the WL area

 

waterline area thank's to nico

nico's spreadsheet to calculating hull resistance using the Delft series as given in VPP's and Wave drag... , has an approximation formula for waterline area vis:

=(LWL*BWL)*(1.313*Cp+0.0371*(LWL/Vol^(1/3))-0.0857*Cp*(LWL/Vol^(1/3)))


hope this helps, dionysis

 

Dionysis

The formula can be simplified as
=(LWL*BWL)*(1.313*Cp + LWL/V^(1/3)*(0.0371 - 0.0857*Cp))

There are many empirical formulas published but each is generally specific to a boat/ship type or form. A much simpler one, albeit rough and ready
=(LWL*BWL)*(2*Cb + 1)/3

Polarity
Your favourite designer quotes PPI = 5.333 in salt water. Just in case you don't understand why, here's how.

The waterplane area times one inch/12 gives the change in volume in cubic feet due to one inch sinkage. The volume times 64 is the weight in pounds so Area x 64/12 = lb and guess what 64/12 equals.

If you know the WP area for the YD 40 then just correct it in proportion to the length and beam to get your own boat value. It won’t matter too much that YD 40 is bigger because you are using relative values. So if you had a design 40 ft x 14 ft and you knew the WP area was 390 sq. ft. then a new boat 30 x 10 would be 390 x 300/560 = 208.9 sq. ft. Normally this type of calculation is done using the standard hull form coefficients and the waterplane area coefficient is Cw meaning the waterplane area divided by (length x breadth). It is better to work with the coefficient because if you think your form (the boat, not a beer guzzlin’ you) is a bit less than the base boat it is easier to play with because L x B is constant for the boat. You also get used to a Cw value for a type regardless of the dimensions.

The terms are used in a very loose fashion by most amateurs (and by too many designers). There is a fixed relationship because
Lightweight + Deadweight = Displacement
There is no draft specified nor any operating condition so displacement must have a specified condition to make it make it explicit or definite. “Light ship” is quite clear but too many designers “shade” the numbers.

Lightweight is the weight (strictly speaking mass) of the vessel in the lightship condition and could be in pounds, kilos or tonnes depending generally on the size (and do you prefer the metric system). It is easier to think of lightweight in the sense of what it does not include. It is the weight of the empty boat/ship with its permanent inventory but excludes cargo, fuel, lubricants, ballast, stores, people and their possessions. Ballast here is water ballast that is used for trim or deeper draft in bad weather etc. Fixed and permanent ballast are in lightweight so that lead keels and such would also be in lightweight. A ship has engines in lightweight so a sailboat has sails, rigging, blocks and so on in lightweight

It is clearer for a ship than a boat because the ship is built using the Class Rules etc and so anchors and cables are required and there are formulas for the size and lengths etc. So too, there are defined machinery and equipment spare parts that must be onboard and these are part of the lightweight. If the Owner wants more spares these would be also be in the lightweight but there would be a signed Change Order (C.O.) or equivalent between him and the Builder covering such changes. The C.O. would give the agreed contractual effect and cover the extra cost, weight and effect on schedule.

A boat is more “free” in scope it seems. When you buy a boat does it include chairs and bedding for example? Theoretically the weights should be in lightweight but has the designer and builder allowed for them. I have checked out a few designers’ sites and some do point out that the declared weight uses their specified equipment and so will change depending on the buyer’s choices. So that even though lightweight has a known definition and extent it can still vary widely in scope.

Displacement is even more complicated. Displacement only means the weight of the water displaced by a floating body when floating free. This means that lightship has a displacement, put anything onboard or take anything off and the displacement changes.

Ships are normally designed to carry a specified deadweight at a specified draft. Usually the draft is determined by calculating the Freeboard which is more or less a function of hull size and fullness with changes as a result of ********* on the main deck. Sea-going ships are designed to operate in salt-water with a specific gravity of 1.025 and the freeboard is in summer – the allowable draft changes with the season, the Plimsoll marks on the hull show them. OK, so the ship specification will refer to the deadweight as so many tonnes on the summer draft and this draft is usually the design draft. Sometimes the design draft may be a bit less to give a margin.

Boats usually, except for luxury yachts and such, don’t have a range in the design draft, it is usually a fixed value and typically is the maximum draft in accordance with the designer’s codes and other regulatory standards where applicable. I’ve seen this referred to as the design draft, maximum draft and a few other terms so I’ll just call it the design draft. This then, automatically gives the design displacement, subtract the lightweight and you have the deadweight. This is sometimes called the load, or variable load, or charge, or capacity.

I have seen some sites where there is no definition of draft other than “Draft = 5’ 6” for example. So is it the mean draft amidships, the stern draft or what? Some sites quote both ends so it is quite clear. Some give the max displacement while others give a half-loaded value. A few sites give enough values so that they all tell you something, quoting light drafts with a displacement or weight and the max or design drafts with a displacement say it all.

What was your question again, Paul?

I would doubt that an unspecified displacement = lightweight, I would have guessed the half-loaded displacement but your question sure does highlight the loose terminology.


Michael

 

Thanks Mike and dyonysis

:idea: I get it now!

So without looking it up, in the metric system to get kg/mm ... WPm2 * .001m * 1000 L per m3 * 1.025 for saltwater. Or quite simply the WP area in m2*1.025 = kg/mm in saltwater. Ahh the wonders of metric!

OK so the YD40 has a Aw (their symbol for waterplane area) of 22.61m2 with a Lwl of 10.02m and Bwl of 3.17m (half load) so the Cw would be 0.71 .She's about the same hullform

So for my boat Aw = 11.28m Lwl * 3.28m Bwl (I dont know this exactly so I used the ratio Bmax:Bwl for the YD40) * .71 = 26.26m2 * 1.025kg/L for Saltwater= 26.91kg/mm - Thanks !

So my next question is .. whats the "usual" definition of Half loaded? Half full tanks (excl. water ballast) and... half a toothbrush? ..half a loaf of bread?...

Paul

 

paul

Suppose your boat is designed to carry 5 people, food, water, provisions, beer and booze for 5 days. 5 toothbrushes, toiletries, plus fuel, 5 books, 5 magazines and 5 sets of whatever you feel appropriate and it all totals say 3,307 lb or 1,500 kg

So then 2.5 people, .....5 days, 2.5 t...., t..... plus f, 2.5 b.., 2.5 mags........... = 750 kg

But if you tossed the half body overboard and topped up on beer and booze, replaced the fuel with liquid refreshment, a band, a buffet and broads and somehow you ended up with 750 kg then the draft would still be the same.

:idea: Half load = design load/2 measured in lb or kg regardless of the component parts.

Michael

 

5 people??... well that's 4 too many for a start!

She's designed as a single/short handed, long distance, ocean race boat (I'm pretty antisocial!). Don't know what the design load is, all I have is "12000 lbs displacement!"- 8'6" draft
I would assume her design load would include 2 person x 1 Round Britain or 1 x person x 1 Atlantic crossing....

Time to start drilling holes in the toothbrush...

Paul

 

Then you should conduct a "lightweight survey" which is the term meaning to determine the weight and the longitudinal centre of gravity of the empty boat.

It can be done with the boat in the water or on dry land. Then you make a calculation of things that are missing and must be added and a calculation of things that are onboard and must be removed. Add everything up and the result is lightweight.

Knowing the loaded condition and having hydrostaic data you can now do all the loading conditions.


Michael

 

mmm Right.

I did a "lightweight survey", all my friends agreed that I was one - and indeed half loaded at the time.
My stability on dry land was not so good either... I thought it best to avoid the water.


Seriously, thanks Mike. I have the kg/mm and an idea of what might be included in the 12000 lbs, so I have a rough idea that I can add another 1.3 T and only be 5cm off the dwl. - I and I learnt something! - (I need a bigger set of scales!)

Paul

 

Paul

Why do you need a bigger set of scales? It sounds fishy to me!

Factual and serious:
Usually the calculation of the volume of the hull is conducted longitudinally but it could also be done vertically.

So if you had the PPI or kg/mm and you put them through Simpson's Rule then you'd get the displacement within the range of draft that you used.

A short example. Using drafts of 2.4 m, 2.5 m and 2.6 m the corresponding kg/mm are 25, 25.1 and 25.25 and Simpson's #1 Rule is Common Interval x 1/3 x (25 + 4x25.1 + 25.25)

We must use a spacing of mm as the sinkage coeff, is kg/mm and the answer will be in kg. So then 100/3 x (25+100.4+25.25) = 5021.67 kg. If we took the mid-range value we would get 200 mm x 25.1 kg/mm = 5020 kg. Very close and good enough unless we have a substantial change in the hull form (hence the waterplane area) and we need a precise answer.

Michael

 

Paul

I submitted instead of previewing -

Sometimes it may be quicker to do this, it just depends on the numbers and how quick are you at mental arithmetic.

• Take the mean of the lower and upper kg/mm values: (25 + 25.25)/2 = 25.125
• From the result subtract the mid value: 25.125 - 25.1 = 0.025
• Add one third of this to the mid value: 25.1 + 0.025/3 = 25.10833.......
• Multiply this by the range of draft: 25.10833 x 200 = 5021.67

Or it might be just as quick (and without a calculator) to do it by parts

• Multiply the mid value by the range = 25.1 x 200 = 5020
• Take the mean of the lower and upper kg/mm values: (25 + 25.25)/2 = 25.125
• From the result subtract the mid value: 25.125 - 25.1 = 0.025
• Multiply the result by 200 and divide by 3: 200 x 0.025/3 = 5/3 = 1.67
• Add the two partial results: 5020 + 1.67 = 5021.67

Easy when you know how and only with a pencil and paper.

Michael

 

Hi Mike

1 question..

"Take the mean of the lower and upper kg/mm values"...

I only got one answer , as I only have one set of numbers .. what I miss?

PAul

 

Paul

What don't you understand?

Michael

 

Mike

Sorry to be dense here but I only got one value.......

"So for my boat Aw = 11.28m Lwl * 3.28m Bwl (I dont know this exactly so I used the ratio Bmax:Bwl for the YD40) * .71 = 26.26m2 * 1.025kg/L for Saltwater= 26.91kg/mm"

So I dont understand where the other values came from or what I'm taking the mean of!


PAul