# Coefficient of Scantling

Discussion in 'Boat Design' started by Annode, Apr 3, 2020.

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### TANSLSenior Member

There is not equation. It is just a percentage, 65% or 0.65, of the total weihgt.

2. ### AnnodePrevious Member

1. annoyingly insensitive or slow to understand.
Give it up Gonzo... that troll storm has passed )

>There is not equation. It is just a percentage, 65% or 0.65, of the total weihgt.
>shell and deck plates, 62% of the total

Thank you for these numbers but we are still one step away from the number I am trying to guestimate.
I am looking for the weight of the structure SEPARATE from the weight of the plating, expressed as a fraction of the plating weight which can be easily calculated in CAD from the surface area x steel weight Sq/m.

So of the 62% in this specific example, what percentage is the structure?

Last edited: Apr 11, 2020
3. ### AnnodePrevious Member

Ignacio, I am puzzled. Why have not promoted your software in this thread?
How easy is this to use for someone with zero experience and how quickly could I extract a weight for the entire framework?

4. Joined: Sep 2011
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### TANSLSenior Member

There are various threads in this forum that talk about SCT or, related to it, about ISO 12215-5, which is the standard used to calculate the scantlings of small boats less than 24 m in length.
Like any other software, to use it the user must have a minimum knowledge of calculation of structures and must know something about ISO 12215. Although the program tries to obviate these drawbacks, it is not easy for a neophyte to get good results with the program.
In relation to your boat, I would do the following, which is what I do with my projects:
1. make the 3D model, with Rhino, for example. Model of hull and deck surfaces only. Ask Rhino to give you the square footage of each surface.
2. make a quick calculation with SCT (I can do it for you, if you think it's convenient) of the thickness of 3 or 4 zones of the hull and deck.
3. Calculate the weight of the previous surfaces since you already know their square meters, their thickness and the specific weight of the material to be used. Lets call that weight "Wshell"
4. Assuming that weight is 62% of the total, you get the total weight Tw = Wshell/0.62
5. Add a 10% margin for welding and possible errors. Tw = 1.1 x Tw
6. Frames, stiffeners, brackets,... Wframes = 0.32 x Tw
It is a first approach, valid for a very initial stage of the project but that, as I have already said, gives me very good results

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6. ### AnnodePrevious Member

.32. Ok. Great, now we are finally getting somewhere. So:
Wt = Ws/.62
Wt = Wt * 1.1
Wf = Wt * .32

Solving for the unknowns in those equations: Wf & Wt:
Wf = (Ws/.62) * 1.1 * .32
Wf = Ws * .5677 - Coeff Of Scanlting = .57 (to 2 dec places)
Wt = (Ws + Wf) / .62

ie:
Framing is roughly 60% of the plating weight (weight of the shell)

So:
My hull plating for a 30m example including keel and deck (excluding ballast, rudder, fittings and anything interior)
Ws = 5850 Kg (6mm plate)
Wf = .57 * 5850 = 3334 Kg
Wt = (3334+ 5850) /.62 = 14,800 Kg =15 Tons

Even if 8mm plate is used:
Ws = 7818
Wf = 4456
Wt = 19800 Kg = 20 tons

That seems very light. I was expecting a total weight of around 100 tons.
Did i understand correctly that you mean total weight to be the displacement?

This is the closest thing I could find:
https://sailboatdata.com/sailboat/jongert-21s
Its only 22m but displaces 50,000 Kg (including 9000 Kg of ballast) = 50 tons

I will certainly check out your software when I get to that stage. You should put something in your hadle that alerts newbies like me to this software. I dont know anything about any of you.

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### TANSLSenior Member

I do not understand the first formula and the second is not correct :
Wt = Ws + Wf

File size:
2.4 MB
Views:
14
8. ### AnnodePrevious Member

That is not the formulae I gave )

Wt = (Ws + Wf) / .62 - ie DIVIDED by 0.62 - the number you gave

Wf = Ws * .5677 This is simply your forumla epxressed in terms of the Weight of the Hull plating.

There are three numbers we ar e discussing. I only know ONE of them.. the weight of the hull plating Ws in your nomenclature.

You kmpw the total weight of the boat you are citing. I dont know the total weight of the example boat I am citing. I ONLY KNOW THE WEIGHT OF THE HULL PLATING.

Thank you for the spreadsheet. Do you have one with some numbers in it ))

Can you confirm that you mean displacement when you cite the total weight?

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### TANSLSenior Member

No, is me whos has wrote the correct formula : Wt = Ws + Wf
When I say "total weight" in this context I mean total weight of the structure. The displacement has many other components
The spreadsheet is completely full of numbers. You just have to work as a "developer" (this is one of the Ecxcel header menus) to have full access to them.
I have the numbers and how they are obtained but in Spanish, so I don't think they will be of much help to you. If you want them (I would not like to make them public) give me your email and I will send them to you.

10. ### AnnodePrevious Member

I see. OK. Let me re read your steps and do another round to see if I can get the weight of the framing expressed as a fraction of the weight of the plating..... (why is this so hard?)

So .32 is what? ........... let me read read very very carefully..... hmmm

So factoring in youe 10% welding materials, your .62 number becomes .5636

Rounding this to .56 because we are just guestimating here

Wt = Ws/.56 (total weight of steel in plating and framing)
so if
Wf = Wt - Ws
then
Wf = .786 * Ws
Now thats a very interesting number. .618 and .786 are fibonacci numbers.
Are you telling me that the coeff of Scantling just happens to be a fib number? Thats too crazy after 5 pages of this...

OK so what is your .32??

Lets take my example:
Ws = 5850 Kg (6mm plate) GIVEN
Wt = 5850/.56 = 10380 Kg approx

So the weight of the framing: Wf = 10380 - 5850 = 4530 Kg

We can calculate that in one step from Ws with:
Wf = .786 * Ws
Wf = .786 * 5850 = 4590 Kg (approx)

But
.32 * 10380 = 3322 Kg ...

11. Joined: Sep 2011
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### TANSLSenior Member

Let's see if I can explain an example so that it can be well understood. Suppose you have a ship that you know the shapes of. This is the process:
- Make the 3D model of the Rhino surfaces, the lining and the cover.
- You ask Rhino to calculate the square meters of those surfaces.
- Through the regulations of a Classification Society, or as you wish, you calculate an average thickness of the plates. If you want a medium thickness for the lining and another medium thickness for the cover.
- You multiply that average thickness by the corresponding surface that Rhino has given you, so you get the volume of material used in the lining and the cover.
- Multiply the volume by the specific weight of the material you want to use. You get the weight of the surfaces (the plates) of the lining and cover: Ws
- known Ws, my figure says that the total weight will be: Wt = Ws / 0.62
- and, as you already know the total weight, Wt, the weight of the frames and reinforcements will be: Wf = 0.38 * Wt
- Now, if you want to be very detailed and a little conservative, you add 10% to each of those pesos.

Ws = 5850 Kg (6mm plate) GIVEN

then Wt = 5 850/0.62 = 9 435.5
and Wf = 0.38*Wt = 0.38*9 435.5 = 3 585.5

Total conservative weight = 1.1*Wt = 1.1 * 9 435.5 = 10 379
plates conservative weight = 1.1* Ws = 1.1 * 5 850 = 6 435
frames conservative weight = 1.1*Wf = 1.1 * 3 585.5 = 3 944

12. ### AnnodePrevious Member

I see. Its the 10% welding factor that is discrepency here. I want to lump it in wih the weight of the framing, you are being more precise and adding it to the plating for the plating. Either way we both end up with the same total weight.

While you are correct in that the welding applies to both the plating and the framing, it is just irrelevant and needless extrax calcualtion to be made to get to the goal of the total weight.

This is all I need:

Wf = .786 * Ws

Coeff Of Scatling (for want of a better name) = 0.786

(I also note that if the welding weight is correctly proportioned to both the plating and the framing, the coeff = .618, another Fib number in the series)

1/.786 = 1.27 (another fib number), so looks like your guestimate was right also in that
Ws = 1.27 * Wf

Its very surprising that you can qoute that number to 2 decimal places. Was that just a lucky guess?

Last edited: Apr 12, 2020
13. ### AnnodePrevious Member

So armed with that information lets continue to do a rough estimate of Displacement.

This is a steel hull sailboat 22m long.
https://sailboatdata.com/sailboat/jongert-21s

Its displacement is 50,000KG
If we just scale it up to 30m by a factor of 1.363, we get a rough displacement of 68,000 Kg
If we scale up the ballast by 1.363 we get 12,365kg

We know the weight of plating and structure is 10,400

So the rest of the boat is going to weight: 68000 - 12365 - 10400 = 45200 KG

ie roughly
hull and framing = 20%
Ballast = 20%

everythng else = 60%

I expected the steel to be a larger % of the total. But what do I know )
Thank you for this number and for your help. You earned my respect )

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### bajansailorMarine Surveyor

Annode, you cannot scale up linearly according to length - you also have to account for the increase in beam and hull draft when going from 22m to 30m in length.

The Jongert's beam is 5.7 m, the displacement is 49.9 tonnes, the LWL is 15.5m and the draft is 3.1m.
For the 22m : the volume of displacement = L x B x T x Block Coefficient
48.678 = 15.5 x 5.7 x 3.1 x BC

Hence BC = 0.178 (which seems very low?)

For a 30 m version, if we assume that the LWL is say 23 m, and keeping the same L/B (2.72) and B/T (1.84) ratios, we have B = 8.46 m and T = 4.6 m.

New volume of displacement = 159.3 cubic metres and the displacement in sea water is 163 tonnes.

So your displacement has gone up by approx 300% - rather a lot more than your assumed value of 36% .

Note also that the cost of building your 30 metre vessel might also be three times greater than the cost of building a 22 metre vessel.

Annode likes this.

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### TANSLSenior Member

That is not right. Since we are talking about volume (or weight), the scale factor should be the cube of the length ratio: 1,363 ^ 3 = 2.53. And still, several other factors would have to be taken into account. But for starters we could consider it as good.

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