ISO 12215-7 Global Loads Case 5

When studying the global loads on multihulls, load case 5, the standard proposes a formula to calculate the longitudinal force on the hulls / floats, in which the lower of the values must be chosen, displacement at full load or the weight of a hull / float. In the attached picture you can see the formula.

This must be an error since the mLDC will never be less than the weight of a hull / float. Am I getting it wrong? What is the right thing to do?
Thanks in advance

 

F(LT) is 2 times that of F(LC).
Ergo one hull is now 2 hulls...so, the resultant is the same.

 

WOW, thank you very much, Ad Hoc, you always ready to help and, by the way, brilliant deduction.
I will rephrase the question and put the image larger because, apparently, the previous one was difficult to read.

When studying the global loads on a catamaran, load case 5, the standard proposes a formula to calculate the longitudinal force on the hulls, in which the lower of the values must be chosen, displacement at full load or the weight of a hull. In the attached picture you can see the formula.

This must be an error since the mLDC will never be less than the weight of a hull. Am I getting it wrong? What is the right thing to do?
Thanks in advance

NOTE for those unfamiliar with the ISO standard: mLDC is, for ISO, the displacement at full load in kN.

 

For a trimaran the displacement of an individual hull/float be less than or greater than the weight of the same hull/float.

 

Probably but what I'm asking is, "can that weight, the hull / float, whatever it is, be greater than mLDC?" Because, what I understand of the rule is that the lesser of the two must be chosen. I have no doubt which one is the lesser.
Thank you for your reply.

 

You are thinking only of vessels with a single hull.

The answer is yes for either the outer two or center hull/float of a trimaran.

If you know the answer then why are you asking the question?

 

Well gentlemen, thank you all, really, for your kind responses and for the patience shown with me, but, despite good intentions, you are not able to get me out of my daze. My question is why does the norm tell us to select the lesser of two values when everyone, including myself, knows that one of them, mLDC, is always, always, greater than the other. Thanks again for your help and please try again.

 

The ISO 12215-7 says (End of "Abstract"in a websource) :
"Throughout this document, unless otherwise specified, dimensions are in (m), areas in (m2), masses in (kg), forces in (N), moments in (Nm), Pressures in (kN/m2) (1 kN/m2 = 1 kPa), stresses and elastic modulus in (N/mm2) (1 N/mm2 = 1 MPa)."

mHULL is called the "mass of one catamaran hull", mLDC is also called a mass, not a displacement. The ISO Norm seems to distinguish between masses and forces.

Just my thoughts: If you delete the first number "2,5" in the formula (could be a misprint) there would be left
FLC = min( mLDC; 2,5 mHULL) (kN)
Would this make some sense in the numerical view?
To be "clean" in a physical sense (SI units) it should possibly read
FLC = min( 1,0 mLDC; 2,5 mHULL) (kN) ; 1,0 in kN/t; mLDC in t; 2,5 in kN/t, mHULL in t (t = metric tonnes = 1 000 kg)

Deceleration is said to be 0,25 g (last line in the table). 0,25 g = 0,25 * 9,81 m * s^-2 = about 2,5 m * s^-2 = 2,5 N / kg = 2,5 kN / t
Just an accident?

 

Indeed, that may be the explanation. I'm really grateful to you.
Although, why apply deceleration to the weight of the hull and not to the weight of the entire ship. They both have the same speed.
On the other hand, if the formula says :

FLC = min (mLDC; 2.5 mHULL) (kN)
​

It would be expected that to obtain FLC in kN one would have to use mLDC in kN. Although the standard defines mLDC (my mistake) as the mass in kg of the ship fully loaded. What do you think?
ISOs are full of errors and sometimes difficult to detect and correct. Thanks again.

 

As you will know I'm not a NA nor even an experienced boater, just a first time amateur boat builder with a background of physical knowledge and analytical skills. I never saw one of the the ISO 12215 norms we are talking about, just searched the web yesterday for some fractions of them out of curiosity after I read your OP.

In general I prefer to use "Größengleichungen" (quantity equations) in which every quantity is a (mathematical) product of a measure and a unit (the unit may be "1" as for radiant). If the pictured table would follow this rules an equation like FLC = min(mLDC; 2.5 mHULL) (kN) would mean the input of "mLDC" had to be in kN as you rightly say. But there is an inconsistency because mLDC is a mass not a force.

Another kind of equation we call "zugeschnittene Größengleichung" (equation tailored to particular needs) which only uses numbers (naturally each related to a specific unit) and the numerical result is a number which is matching the given unit. I think, FLC = min(mLDC; 2.5 mHULL) (kN) is this kind of equation. As I don't think, the input of masses in kg would deliver a sensible result, but think the input has to be in t (metric tonnes) there schould be stated an exeption of the general rule near the table or in this section of the norm.
The general statement in the abstract says: "... masses are in (kg) ... unless otherwise specified ... ".

I apologize for stating some very obvious things because I want to prevent misunderstanding.

 

Indeed....

Like all rules, based upon a combination of empirical data and regression analysis, there are caveats and limitations.
Without knowing the background the derivation of the this rule/formulae (and without visibility of this *-7 rule), it is not 100% possible to say.
However, it fits within the norms of caveats for such rule based calculations.

Just look at the preamble of any set of rules, like LR's SSC rules... not "suitable" for vessel below 24m.
Simply because the output, can lead to spurious results that are questionable and require more time/input to arrive at a common consensus.

 

@Heimfried, I thank you very much for your explanations and I believe that, even if you are not NA, at least you have common sense and intention to help, without only trying to show me your enormous wisdom. So thank you, you have helped me a lot and now I know where to look trying to understand that formula.
The formula seems to suggest that the weight of the wet deck +crossbeams + the main deck + superstructure is roughly equal to half a hull. Frankly, I don't think that in general can be said for all catamarans, but I don't know what other interpretation can be given.
From what I have calculated so far, I believe that entering the data in kN gives a logical figure for the longitudinal force FL.
Thanks for your help.

 

I would greatly appreciate if you could explain to me what is the question that I ask, which I answer myself or whose answer I already know.

Sorry, I don't understand what you mean or where you want to go with everything you say. But thanks for your "help"

 

It is a pity you can't leave your ego and emotions at the door rather than constantly bringing them to this forum, and having poor attempts at snide comments...sad.
You may even get assistance with your questions if you weren't so ego driven...

Rules, all rules have caveats.
It is where the limits of their formulation breakdown and are therefore not totally applicable.
So, as noted look at LR SSC rules:



This is simply because the data suggests that vessels of less than 24m, the results are not wholly consistent and can yeild "odd" scantlings.
As such one must make a case for the approval using said rules, to the satisfaction of both parties.

ISO, like any rules, has such caveats.
These caveats are often to prevent, such irregular results, and place maximum or minimum constraints on said rule....rather than spend a significant amount of time trying to obtain additional, yet not much available, data, to cover the extreme ends of the rules. Most rule based societies do not..and simply place a caveat, to "cover all" bases.
That's all.

 

In ISO 12215-7 it is stated " 3.5.6 loaded displacement mass mLDC mass of the boat in the loaded displacement condition NOTE Loaded displacement mass is expressed in kilograms. "
It is also a part of other equations (I counted 8).

Since kg is a force, it can be converted to Newtons or kN force.

I read from the formula minimum (of) 2.5 x mLDC (mono); (or) 2.5 x mHull (of 1 hull of a cat) in kN, Which in my interpretation, 2.5 x "X " whichever is the lesser.

Note this value "longitudinal force" is probably used in other equations in the Rules. I wonder how much simplification I will need to use in order to be consistent with other units.