# Stability Help

Not sure if anyone would be able to help me… I’m taking a Transport Canada Stability and Ship Construction 3 exam soon and have just been self studying to prepare.

I have a copy of an old exam that I’ve been working on, and one question has given me fits. Looking through a couple different stability books, I can’t seem to figure out how to answer it. The question reads :

The motor tug “Sept Iles” is floating in normal operating condition 50% consumables SW. If its FP tank is pumped full with SW, calculate the new metacentric height (GM) corrected for free surfaces (FS) and the vessel’s new longitudinal center of gravity (LCG).

I have the ship stability book available to me, well online http://www.scribd.com/doc/175729282/Sept-Iles-Stability-Book

I’m assuming they want to know GM transverse, although I could be wrong. I’ve used GM L = L / t xGG 1 and can get the longitudinal GM.

For 50% consumables the GM solid is 1.16, GM fluid is 1.12. Would using the GM fluid be all I need to do to “correct for free surfaces”? The FP tank is on the centerline, and they also say it is pumped full, so I can’t see it affecting free surfaces any more. The stability book gives me a Free surface inertia of 36 for the forepeak tank… I cannot find what this means anywhere.

The stability book give me lots of information, I just don’t know how to use it to find the new metacentric height.

I’ve found the LCG just by using the moment of the full forepeak tank, that is fairly easy.

Thanks for any help… figuring out stability is giving me grey hairs…

-Sean

See what I meant before about Canadian license exams testing for a much higher level of competence than the USCG exams? How many of us Master of Towing guys in the US would want to take that exam?

This question is worded piss poorly imo. What on earth is “normal operating condition 50% consumables SW” supposed to mean anyway? SW is not a consumable (unless you have magic engines which happen to run on seawater) and no master would ever allow ALL his fuel oil tanks to be half full!

PROTEST!

[QUOTE=Seancro;132009]Not sure if anyone would be able to help me… I’m taking a Transport Canada Stability and Ship Construction 3 exam soon and have just been self studying to prepare.

I have a copy of an old exam that I’ve been working on, and one question has given me fits. Looking through a couple different stability books, I can’t seem to figure out how to answer it. The question reads :

The motor tug “Sept Iles” is floating in normal operating condition 50% consumables SW. If its FP tank is pumped full with SW, calculate the new metacentric height (GM) corrected for free surfaces (FS) and the vessel’s new longitudinal center of gravity (LCG).

I have the ship stability book available to me, well online http://www.scribd.com/doc/175729282/Sept-Iles-Stability-Book

I’m assuming they want to know GM transverse, although I could be wrong. I’ve used GM L = L / t xGG 1 and can get the longitudinal GM.

For 50% consumables the GM solid is 1.16, GM fluid is 1.12. Would using the GM fluid be all I need to do to “correct for free surfaces”? The FP tank is on the centerline, and they also say it is pumped full, so I can’t see it affecting free surfaces any more. The stability book gives me a Free surface inertia of 36 for the forepeak tank… I cannot find what this means anywhere.

The stability book give me lots of information, I just don’t know how to use it to find the new metacentric height.

I’ve found the LCG just by using the moment of the full forepeak tank, that is fairly easy.

Thanks for any help… figuring out stability is giving me grey hairs…

-Sean[/QUOTE]

Can you find an example of a problem all worked out? That would be the easiest approach.

GM transverse is what you are looking for, usually just called GM.

If the tank is full you do not have to correct for free surface. You don’t need the Free surface inertia.

I am not familiar with GM solid and fluid.

tugsailor - This is only for a 150gt masters certificate…

c.captain - 50% consumables SW (meaning vessel is in SW) just lets me know which figures to get from the stability book. The book provides almost all measurements for Lightweight, 50% consumables, and 100% consumables.

Kennebec Captain - Unfortunately I cannot. The exam is straight from transport canada. The books I’ve been studying from do not have any similar questions.
GM fluid is the GM solid with .04m free surface effect removed for the ship being at 50% consumables, so I think I use that GM Fluid as a starting point.

Is there a simple equation for finding GM with one added weight? GM is 1.12m, the VCG of the FP tank is 6.73 and is 30.44 m.t. when full. Does the metacenter even move when a weight is added, or is it just KG that is affected? If the latter, then I can simply solve for the new KG by using the moment of the FP tank, and then find the GM…

[QUOTE=Seancro;132020]Does the metacenter even move when a weight is added, or is it just KG that is affected? If the latter, then I can simply solve for the new KG by using the moment of the FP tank, and then find the GM…[/QUOTE]

KM always changes with a change in displacement (even a small change)…that is what the crosscurves of stability tell you

[QUOTE=Seancro;132020]
Is there a simple equation for finding GM with one added weight? GM is 1.12m, the VCG of the FP tank is 6.73 and is 30.44 m.t. when full. Does the metacenter even move when a weight is added, or is it just KG that is affected? If the latter, then I can simply solve for the new KG by using the moment of the FP tank, and then find the GM…[/QUOTE]

You can use moments to find the new KG. Multiply weights by VCG and get the moments then divide total moments by total weights.

weight 1 X VCG 1 = Moment 1 (moment before weight is added)
weigh 2 X VCG 2 = moment 2 ( moment added weight)

Divide total moments (moment 1 + moment 2) by total weight (weight 1 + weight 2) and you will get new KG

[QUOTE=tugsailor;132011]See what I meant before about Canadian license exams testing for a much higher level of competence than the USCG exams? How many of us Master of Towing guys in the US would want to take that exam?[/QUOTE]

Son, this is ‘merica! Let me explain how we get’ r done. USCG writes a question, USCG posts the answer to said question, don’t matter how bad the question is as long as you can memorize the answers… That’s what makes better sailors… A good memory.

[QUOTE=Kennebec Captain;132024]You can use moments to find the new KG. Multiply weights by VCG and get the moments then divide total moments by total weights.

weight 1 X VCG 1 = Moment 1 (moment before weight is added)
weigh 2 X VCG 2 = moment 2 ( moment added weight)

Divide total moments (moment 1 + moment 2) by total weight (weight 1 + weight 2) and you will get new KG[/QUOTE]

Right, I figured that out a different way (moment of the added weight from the vessel VCG / Displacement With added weight) but glad to see your method gave me the same answer. So I’ve solved for KG, but I still need to solve for GM.

gm = km - kg

[QUOTE=Kennebec Captain;132036]gm = km - kg[/QUOTE]

Correct, but I don’t have KM either. I have the initial GM before adding weight 1.12m, and I now have the new KG 4.98m with the added weight. I need to find GM or KM now with the added weight.

``  Would M increase the same amount as G when adding the weight?     I wouldn't think so, but I'm running out of ideas...``

[QUOTE=Seancro;132037]Correct, but I don’t have KM either. I have the initial GM before adding weight 1.12m, and I now have the new KG 4.98m with the added weight. I need to find GM or KM now with the added weight.

``````  Would M increase the same amount as G when adding the weight?     I wouldn't think so, but I'm running out of ideas...[/QUOTE]
``````

KM has to be given, there should be some curves somewhere showing KM at different drafts, or it will be given in the problem, you can’t figure it on your own.

There is a very good explanation with diagrams here.

[QUOTE=Seancro;132037]Correct, but I don’t have KM either. I have the initial GM before adding weight 1.12m, and I now have the new KG 4.98m with the added weight. I need to find GM or KM now with the added weight.

``````  Would M increase the same amount as G when adding the weight?     I wouldn't think so, but I'm running out of ideas...[/QUOTE]
``````

How have got to be missing something in the booklet…any change in displacement will create a change in KM but not related to the change in G. Regardless of the height to which weight is added, B goes up slightly because the immersed hull area is larger and thus M goes up as well. G can go up or down depending on where the weight is added, M always goes higher for increased displacement. KB is the centroid of the immersed area and BM is the radius to the point which centers on the arc followed by B as the ship inclines to about 10deg P&S.

There is no good formula for calculating KB or BM which a mariner can use in a rough fashion to the best of my knowledge.

[QUOTE=Kennebec Captain;132038]KM has to be given, there should be some curves somewhere showing KM at different drafts, or it will be given in the problem, you can’t figure it on your own.

Well I feel a lil slow… Your right, on the Hydrostatic Curves Graph in the stability book they have (in blurry letters) KMT… which I can read off that and figure out and subtract KG and voila,= GM. Here I was thinking it would be more equations but the answer is right there in the stability book… Well thanks a ton for your help and patience!

M - Metacenter: As the ship is inclined through small angles of heel, the lines of buoyant force intersect at a point called the metacenter.

As the ship is inclined, the center of buoyancy moves in an arc as it continues to seek the geometric center of the underwater hull body. This arc describes the metacentric radius.

The position M has to do with the shape of the hull.

ok, theres another (simpler) question that I’ve lost confidence in over time. If anyone is up for some stability trivia…

When a vessel is trimmed by stern and has its center of flotation 3m abaft amidships, fuel oil is transferred from forward to aft. The mean draft will:
A. not change
B. increase
C. decrease
D. increase or decrease

My first thought was A. Not change. The trim will change, but the mean draft will not because no weight has been added, just shifted.

But then I read a chapter in a book about True Mean Draft and Arithmetic Mean Draft, saying that if the LCF is not amidships and a weight is shifted fore to aft, the AMD will decrease.

I hope it is a simple question and A is the answer, but I’d like to know if I’m missing something…?

[QUOTE=Seancro;132064]ok, theres another (simpler) question that I’ve lost confidence in over time. If anyone is up for some stability trivia…

When a vessel is trimmed by stern and has its center of flotation 3m abaft amidships, fuel oil is transferred from forward to aft. The mean draft will:
A. not change
B. increase
C. decrease
D. increase or decrease

My first thought was A. Not change. The trim will change, but the mean draft will not because no weight has been added, just shifted.

But then I read a chapter in a book about True Mean Draft and Arithmetic Mean Draft, saying that if the LCF is not amidships and a weight is shifted fore to aft, the AMD will decrease.

I hope it is a simple question and A is the answer, but I’d like to know if I’m missing something…?[/QUOTE]

Not sure, I think C because the ass if fat and the bow is skinny. When you shift weight fwd the bow will go down more then the ass will come up.

[QUOTE=Kennebec Captain;132065]Not sure, I think C because the ass if fat and the bow is skinny. When you shift weight fwd the bow will go down more then the ass will come up.[/QUOTE]

since mean draft is measured at midships but the point of trimming is aft of that point then increasing draft aft will cause the midships draft measurement to decrease.

[QUOTE=c.captain;132067]since mean draft is measured at midships but the point of trimming is aft of that point then increasing draft aft will cause the midships draft measurement to decrease.[/QUOTE]

That makes sense, that the draft at mid ships would decrease, but to confuse things of course, my Ship Stability for Masters and Mates book defines mean draft as : Draft measured from waterline to keel at a position immediately under the LCF.

I guess it would depend on what Transport Canada’s definition for Mean Draft is…

[QUOTE=Seancro;132069]That makes sense, that the draft at mid ships would decrease, but to confuse things of course, my Ship Stability for Masters and Mates book defines mean draft as : Draft measured from waterline to keel at a position immediately under the LCF.

I guess it would depend on what Transport Canada’s definition for Mean Draft is… [/quote]

I have never read that one but if that is what Transport Canada goes by then mean draft would not change with shift in weights since LCF would not change. Is that book one of the standard references for exams? If so, then go by that and if you got the question wrong then you could protest it.

I will say in that researching these questions, I am surprised at how much is online concerning ship stability…isn’t the internet a wonderful invention? Pictures of ships and nekid wimen both!