Stability Issues

If I understand correctly, the COG should be directly above the Center of Flotation, right?

What is the difference between the Center of Flotation and the Center of Bouyancy?

 

The CB is the center of underwater volume. The CF is the center of area of the waterline plane. The boat will trim untill the CG is over the CB. Or else capsize. Whichever comes first.

 

Thanks for the quick and very informative response! On my model, the COG is about 9 or 10 inches forward of the COB, but it is only the Area COG, without the deck, bulkheads, motor, etc. The deck and bulkheads will move the COG rearward a little, as will the motor.

What I need to know is if there is a way to place a weight to represent the motor on the transom in Rhino, or will I need to calculate the adjusted COG by hand?

 

I do not know Rhino ... and I do not know your model. But I normaly do a weight calculation at side and not in the design program. Reason is simple, you have to add persons, tanks, effects etc ... and of course this all will effect your LCG as well as VCG. For tanks you have also the effect of free surface moments. Depending of your model you should think about that .... specially for VCG.

 

What do you mean by LCG and VCG?

 

LCG = longitudinal center of gravity .... should be above LCB(oyuncie) otherwise trim forward or aft
VCG = vertical center of gravity .... main point to discuss stability calculations.

 

How much tolerance are you allowed as far as locations of the COG and the COB? Right now, my COG is about 3.5" behind my COB. I can adjust fuel location to bring it closer, but if it is not absolutely necessary, I would rather leave it where it is.

 

LCG position not LOB position = trim for or aft. As far as I understood your posting mrbrown, you would get aft trim.
To state how much trim, it`s impossible with known datas. I.e. is your model a 60m motoryacht ... forget the distance. Is is a 3m motorboat ... forget the distance, if helmsman is not on your calculated place, the effect of trim will be much bigger.
Be a little bit carefull by shifting tanks, sometimes they are full, sometimes empty. I normaly calculate with full tanks, 50% tanks and 10% tanks, of course 3 different VCG`s and perhaps, depending on the form of the tank, also different LCG of the tank.

 

Thanks for the information. The boat is an 18 foot center console bay boat, so the effects should be negligible, correct?
drdmowe

 

How do you calculate the "moment to trim 1 inch?" I pieced my way through it, but don't think I did it right. I just rotated the hull 1" aftward at the midship point, took the new center of bouyancy location and multiplied the distance from it to the midship point by the original displacement. I then solved an equilibrium equation for the weight required at a point 10" behind the transom (approximate location of the motor), and came up with about 297 lbs. required to trim the transom down 1 inch.

I am sure I messed up somewhere, not having anything to go by.

 

The hull doesn't trim about the midship point - you should use the geometrical center of the waterplane (LCF - the longitudinal center of floatation) instead!

And remember to consider if the trimming weight is added, subtracted or just moved. If it is added or subtracted, the boat will float on another waterplane (with another LCF).

As something to go by, may I suggest Larsson & Eliasson: "Principles of Yacht Design", where all these calculations are described.

 

MCT = (D x GM (L)) / (100 x L) (in metric)

where:

MCT = Moment to Change Trim One Centimetre
D = Displacement in tonnes
GM(L) = Longitudinal Metacentric Height in metres
L = Length on Waterline in metres

You seem to be hell-bent on pounding this stability stuff into your head, MrBrown, so you should arm yourself with the proper tool for head-bashing. I would suggest that you purchase a copy of "Ship Stability for Master and Mates" by D.R.Derrett. It is readily available worldwide, and in the USA from Hoffman Services in Lafayette, LA at http://www.maritimeusa.com/metconts/section59.html and listed as item _BK465

 

Thanks for the information gentlemen. It is greatly appreciated! I can change the formula that you gave over to lbs. and inches, right? (I assume that you were using metric tons). Also, is Longitudinal Metacentric Height the depth of the waterline at the centroid of its surface area?

Inquisitive feller ain't I?

Thanks

 

GM(L) is the distance between the CoG and the longitudinal metacenter (M(L)).

GM(L) = BM(L) - BG

where:

BM(L) = I(L)/Displ (The longitudinal moment of inertia of the waterplane (taken about the centre of floatation) divided by the displacement)
BG = The distance between the centre of buoyancy and the CoG

I strongly urge you to read at least one of the books mentioned above. Taking a boat design course in this forum will take too long...

Good luck!

 

LOL yep, it would take a long time. I have a book on order, and will order more soon. I am just being impatient, I guess. Thanks for your help!