# Mass Moment of Inertia

Discussion in 'Boat Design' started by netjaws, Mar 23, 2009.

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### netjawsJunior Member

Hello,

I have calculated the transverse moment of inertia of a ship to be 59,823.3 ft^4 at the DWL.

How do I go about finding the mass moment of inertia, in slug*ft^2?

Thanks

p.s.- at the DWL, the ship displaces 8,203,242.4 ft^3, or approximately 234,378.4 LT of saltwater

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### Eric SponbergSenior Member

You have to do a very detailed weight estimate of the boat first, with the weight and center of gravity of all components determined and a weight moment calculation made to determine the overall center of gravity of the boat. The mass moment of inertia is then the moment calculation done one more time, but with respect to the center of gravity. That is:

Imass = weight/g x (arm to center of gravity)^2

Where g = acceleration of gravity

Your "transverse moment of inertia" that you show is a function of some area, either of the hull cross-section or the waterplane, as indicated by the units, ft^4. Mass moment of inertia is a function of the vessel's overall mass, which is a function, of course, of weight. That is why you need to do the weight estimate before you can do the mass moment of inertia calculation.

Eric

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### netjawsJunior Member

Thank you, that clarifies - I figured my calculated Ix of the waterplane would have something to do with this, guess not-

To be clear, I'm trying to find the "virtual" mass moment of inertia in roll, which equals mass moment of inertia + "added" mass moment of inertia. To my understanding, the latter should be about 20% of the former.

But the units of each - and the whole - is slug*ft^2 (or kg*m^2) - so, as you say, there is an "arm" associated with this, which must be squared. The question is - the arm from the center of gravity to what?

p.s.- I have vcg, lcg - but I think the assumption with a ship this big is that tcg=0

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### TeddyDiverGollywobbler

To cg of each individual part of the ship..

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### netjawsJunior Member

red hell and bloody death.

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### netjawsJunior Member

well is there an accepted average for specific types/sizes of ships? maybe a simple "formula" to give a rough approximation?

I'm searching and searching, and not finding

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### TeddyDiverGollywobbler

For smth "simple" like fully loaded supertanker or a log the "arm to center of gravity" comes close to LOA/4, which is simple way to "approximate". An uniform part of any vessel can be judged like a single object and then ad the effects of "not uniform parts" like bow, superstructure, stern...

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### jehardimanSenior Member

Netjaws,

Eric is right, the only way to get the true answer is a full calculation from the complete weight report. That said, Teddy is correct in his approximation.

Think of the ship as a hollow shell of steel. Ixx, Iyy, and Izz can be approximated as m*k^2 where k is ~ .35B to 0.4B for Ixx and ~.25LOA for Iyy and Izz.

Now if you have a longitudinal weight estimate you can refine that number, and a shell expansion can also help to spread machinery weights like peanut butter over the body. In my experience, these estimates have been +/-10% of the actual numbers.

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### netjawsJunior Member

Alright, thanks

My guess was even more conservative, 0.25B for Ixx

Thanks for all the help

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### jehardimanSenior Member

Remember the corners...check the radius of gyration (k) for hollow square.

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### Eric SponbergSenior Member

Imagine your X, Y, and Z reference lines going through the vessel CG. The mass moment of inertia will be the sum of second order moments (mass x moment arm^2) of each item of mass where arm is the distance from each center of mass to the reference axes. So you'll have three mass moments of inertia in the three orthogonal directions.

Added mass is another matter, and that relates to the mass of entrained water that moves with the vessel as it rolls. That is probably impossible to calculate accurately. Probably the only way to positively determine that is to measure it in a model test or full-scale test. There may be published guidelines to make estimates, but much depends on the actual geometry of the hull, and it is very hard to generalize hull geometries, particularly with, say, a sailboat.

Eric

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### netjawsJunior Member

ha-ha, I think you are correct

on that note - I never understood why people so commonly truncate 'added mass moment of inertia' to, simply, 'added mass' - mass and mass moment of inertia are two very different things, and someone just getting into this, imo, can be misled terribly

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### jehardimanSenior Member

It gets worse.

"Added mass" is actually not a mass at all. It is the intergral of the differnetial pressure over the surface of the wetted boundary caused by the ridgid body motion. It is called "added mass" because when the first experiments were done in fluid dynamics with models, they were found not to agree with F=m*a....i.e. there was "additional" mass needed to make the measured a agree with the measured F. This is why the "added mass" coefficient, Cm, for a sphere in translation is 0.5, while the "added mass" is 1.5 because added mass = m*(1+Cm). BTW, the Cm for a sphere in rotation is effecteively 0.

FWIW, for a quick look, you should be calculating Cm directly using strip theory which would work quite well for a VLCC/ULCC. Go see Chapter 4 in Dynamics of Marine Vehicles by Bhattacharyya for a cookbook example.

OOOps.... Cm for a sphere is 0.5, not 1.0 as I recalled from memory. 1.0 is for a cylinder transverse to it's long axis...too many years working with submarines...

Last edited: Mar 25, 2009

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### netjawsJunior Member

interestingly, bhattacharyya is the only literature I've come across where he goes to the trouble of spelling out "added mass moment of inertia"

he's great.

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