Lecture One



PROLOGUE: Again this was written over 15 years ago....Time flies when you are having fun (LOL)

Stemming confusion:

Archimedes was a third century BC Greek mathematician. He is credited with formulating the laws of Buoyancy. In these lectures we'll acknowledge his contribution by using the Greek delta triangle symbol to indicate displacement in our formulas beginning with Lecture 2

John La Dage on page one of his classic work states that Stability is the tendency of a vessel to return to its original position after it has been inclined due to external forces.

If you have a copy of professor La Dage's fine book, opening the cover you will see a picture of the "Heavy Lift" ship USNS Pvt. Leonard C Bronston explained as "heeling to a 165 ton test load"

The only exception I have with professor La Dage is in his terminology at times and his excessive use of different types of diagrams that in my estimation tend to confuse. A vessel heels due to wind and sea. A vessel lists due to a man made situation, usually at relatively small angles of inclination. A list is caused due to loading or negligence and can only be relieved by human intervention. My point heeling is due to mother nature and the vessel "will right itself" with positive stability. Human intelligence is required to relieve the list.

I've tried to stay away from definitions and present simple facts, breaking them down, in the most fundimental way I know how, in order to give you a conceptual understanding of stability, so that you find working with the equations relatively simple in solving the Coast Guard's problems.

Rolling : A vessel with a fair size roll, rolls to the right and then left doesn't it?… For example, From a stable position (initial stability) a vessel rolls to the right, it then rolls back to its upright condition, then rolls left, back to upright and then right. I want to make sure that this point is clear. …When it rolls to its full extent right… mark it on your watch till it returns back to the full extent right again. That is termed your roll period. (usually 10 to 20 second range)

Motion: There are Four Lectures on Stability …One Lecture on Trim. Motion as it relates to the stability discipline is always sideways. Motion…left to right..port to starboard. You might encounter the term motion along the transverse plane "all the same thing"...Don't get confused.

We'll also reference motion as vertical motion... measured distances, up and down the centerline, measured from the keel. Distance measured along the vertical plane.

Longitudinal motion (or when the word longitudinal is used) it is reserved for TRIM not STABILITY

First Diagrams:

The two figures that you see above represent the same midsection under different circumstances in order to illustrate several points.

Visualize being astern of the vessel in a launch. The vessel cut in half crosswise. What you view, figuratively, is the vessel's midship cross section seen from your vantage astern.

What I would like you to first be aware of (might seem a small point) is that the centerline is perpendicular to the waterline…or 90 degrees… I point this out because the diagrams were not drawn to mathematical scale as would be necessary in the capacity of a naval architect. The diagrams are drawn only for conceptual understanding. They do however maintain the form of mathematical integrity pointing out the series of triangles associated with trig functions. It's not necessary to make the math association but at least you will be aware of why sine, cosine and tangent angles are used in our equations. This minor point might become more clear in the last diagram that's ok if you don't see it. The last diagram' s intention is to sum up Stability Theory. By that time you should have a good conceptual understanding of the component parts…and where to obtain them from the data given and the data referenced. Whether you see the angular associations or not is not important reserved for a class on naval architecture but they are there nevertheless. A real sectional evaluation would have to come from a lines plan which is where all the hydrostatic properties are determined by the naval architect.

You might have already noticed that the waterline on the left says DWL... the waterline on the right says LWL… "This will be explained shortly"

I'm sure it is fairly obvious, that the portion penciled in blue, is the underwater portion of the vessel's mid section The same section illustrated on the right has an obvious greater volume than the one on the left.

The first point I want to illustrate is a thorough understanding "not a definition" of displacement.

First…Let's visualize an extremely large tank…filled to the very top with sea water…"not another drop could be added to the tank". Now let's take the entire ship represented by the left hand section drawing…pick the vessel up with a crane…"possible of course for illustration purposes". Now let's drop the ship in the tank…What happens is obvious…the water will spill over the sides. The ship has "displaced" the water…. The amount of the water displaced weighs the exact same as the weight of the ship picked up by our gigantic virtual crane. The weight of the vessel equals the displacement of the water weight-wise.

In the section on the right…We do the same thing…but our senses are telling us… that it's obvious more blue water has been displaced…Exactly right …the vessel's mass has not increased…but obviously something has happened since more water went over the side…correct…The only explanation is that the vessel weighed more on the second lift. Since it was heavier it sank more and displaced more water…Weighing the displaced water on the second lift is the exact same as the weigh of the ship with the obvious added weight…Weight of course is added through loading. The vessel's weight equal's the vessel's displacement.

Archimedes figured this out without a computer… Laws of buoyancy.

What is the difference between DWL and LWL. The obvious difference is that the section designated LWL has a greater draft than the section on the right DWL.

DWL means Designed Waterline…This is the waterline that the naval architect calculated on his lines plan. The light-ship waterline

Light ship means the vessel and it's machinery void of bunkers, stores, crew and cargo.

LWL means Load Waterline…This is the lightship plus bunkers, stores crew and cargo.

Now we have enough conceptual information to open up to our first table in the Blue Sheets.

It's called the Deadweight Scale:..Take a look bottom left…It says Light draft 9-10 Displacement 4521.. This is the light ship DWL in our left diagram. Take a look in the second column…under the heading Deadweight…It says zero..

Now I'm going to give an arbitrary value to our section LWL…the draft in that diagram is 15'. If I look to my right hand column…It tells me my displacement is 7325 "if my eyes are working" It also tells me that my dead weight is approximately 2800 tons MOL…What I deduct is this…My light ship plus cargo etc. equals loaded ship…expressed visually in our drafts. Expressing it another way, the difference in the section on the right from the section on the left is the deadweight…the added cargo… Since the ship weighed more, it displaced more. Weight and displacement are not the same thing… but they mirror each other in weight value..

Looking at deadweight Scale Sheet….we'll get into MTI in the Trim lecture TPI means Tons Per Inch immersion. Every ton of added cargo increases displacement…the figure on the right tells how many added tons necessary for changes in displacement and mean draft ..fairly obvious.

The two columns of values to the right are not so obvious…What's a Block Coefficient: Take a look at the values. on the right…It says 65-75-85. Visualize having a block of wood and a whittling knife with the prospect of whittling out a ship. If you whittled out a tubby looking vessel…you might cut off only 15% of the wood to get your ship…It would probably resemble a barge…If you carved out a sleek looking narrow beamed craft…you might throw out 35 % of the wood.. your block coefficient would be 65%.. get the point…In other words the more the vessel resembles a block, the higher the block coefficient…Most ships run around 73%.

What is a waterplane coefficient…A footprint…a plane in geometry is a flat surface. In our example above we took the vessel out of the water with a giant virtual crane.. Let's say that around the waterline we put a black dye that left an exact footprint. When the vessel was lifted out…the plane surface area within the footprint is the waterplane.."It would of course mirror the shape of the ship at the waterline..

Some of the equations we deal with require that we calculate the value of these two coefficient as part of solving the problem…use "k"..as the decimal value of the coefficient as indicated to the left of the coefficient.

Before we get into the next section of Lecture One, it's worth noting at this time that 95%…maybe more of the problems give you a FWD draft and an AFT draft… This must be converted to a MEAN draft…

Let's review over one of several methods FWD draft 15'- 10" AFT 18'-07". First, convert feet to inches multiplying by 12…Then divide by 2 then divide again by 12… then multiply the decimal value by 12 …EG:: 15 X 12 = 180 +10 =190…. 18 X12 = 216 + 07 = 223 + 190 =413/2 =206.5/12=17.20= 17'…. .20 X12=2.4….ANS 17'-2" Mean Draft…."There are shortcuts…whatever works for you.. but it's imperative that you get it right…No doubt you do this all the time…but as mentioned this is important data necessary to solve the problems you will be confronted with. Any fundamental deficiency needs to be cleared up by Lecture Two when we begin solving the problems.

Nailing down our concept of displacement on all levels I feel can be accomplished through successfully answering one question which is this. Does displacement of a vessel increase in fresh water? The answer is NO. The short explanation is the weight of the vessel doesn't change so displacement doesn't change "check with Archimedes." The argument to the contrary is this…draft increases therefore displacement increases. Granting the fact more water is displaced displacement however.is not a characteristic of volume or mass but of weight…Since fresh water weighs less than salt water the increased volume of fresh water weighs the same amount as the weight of the vessel…so displacement remains the same…

Let's get back to our diagram ..figure 3… Conceptually we'll call it the playing field "The players are out on the field"

G & B are the players the reporters want to talk to …they are the stars…All the downward forces… the combined weight of the vessel acts through G. He's sometimes referred to as CG, VCG and on game day KG.

On defense B… He is the opposing force of G…All the upward buoyant forces act through him. G and B are equal as we saw in our discussion on displacement, the scrimmage line is the LWL. He is referred to as CB, VCB, and on game day KB.

K is not a star … but important player nevertheless…K stays at the bottom of the keel and never moves…maybe that's why they call him K.

We've come to regard G & B representing equal but opposing weights; but, when teamed up with K they represent distance…In other words on game day when KB and KG are referenced we are not talking about weight but distance and movement.….The movement is up and down the centerline…that’s the game up and down the centerline… G doesn't go off the centerline…Neither does K "Who doesn't move at all".. B however is a real playmaker. He not only moves up and down the centerline but he can shift laterally. He moves laterally when the loadwater shifts to a new angle.."he's always hanging out around the center of the underportion of the vessel…that's his only legal shift.. If unclear at this point… you'll see on game day… Stability has rules of it's own…

Because B has ability to move laterally as well as vertically, I might point out only along the centerline…" you might regard M as a wide receiver.. plays the position called metacenter... B however gets the big bucks..he plays both sides of the ball. When LWL shifts B shifts and throws a perfect angled pass to M who catches it along the centerline…He's good he's always in position and makes the reception … However M "must" be up field from G when the ball is caught…If G is upfield from M the vessel sinks and you lose… That's called negative stability. When G & M are in the same position you have neutral stability…the vessel doesn't sink..but you don't make it to the playoffs either…you wind up with the angle of loll trophy..not a good thing. " "You'll see on game day."

Since G & B both represent weight and as KG & KB both represent distance. Using football analogy simplifies things; play analysis in the booth is done through a term called moments…weight "W" multiplied by Distance "D" equals moments.. This analysis will be done not on the "telestrator", but by tabulation form. .."KG comes under a lot of analysis"'..for obvious resons..we don't want the vessel to sink.

Since it's so important to keep G downfield from M…their positions are constantly under scrutiny. So much so that the bulk of the questions as it relates to stability are play analysis of locating GM or G's relationship with M..and… KG's. centerline position.

Now that we have a pretty good understanding in the rolls of our players…What affects their movement.

K as mentioned never moves off the bottom of the keel as already pointed out. B on a vessel (with a high block coefficient maybe 85-90% ..a barge type shape) is found at 50% or ½ draft. For the normal shaped ship 53% draft is a more accurate rule or thumb. There's an exact formula, not mentioned here…but I've found the rule of thumb to work just find in determining the correct answer.

The naval architect will determine the initial KG or CG of the vessel…Then it's up to the deck officer through loading cargo and ballasting. The concept is very simple…If the combined weight of the cargo and its respective CG, G or VCG is higher than the G of the vessel…G will move up the centerline…If the cargo is obviously loaded lower than the G of the vessel… G is reduced . For a quick conceptual check… this is particularly true with double bottomed tankers with low cargo loading in its tanks…whereas a container ship with containers four high is another story …. The relationship between G & M are critical on a container ship…M as stated moves only when B moves laterally (through heeling or a list) .but with every added container loaded higher G moves up decreasing our all important GM relationship.

Outside of intelligent loading…to keep G at a satisfactory level… another method of lowering G is through ballasting.. as the tanks are lower than the G of the ship..

At this point take another look at the terms below combined with figure 3 above and get used to familiarizing yourself with exactly what… they represent: .KB , KG, KM. BM, GM..

They "ALL" represent distances..I'ts so important to get that straight..because the problems are all referenced to these terms.. The terms found in the reference sheets and or obtained in the question itself, you then plug all relevent data into the equation, then use your calculator to solve the problem. This is the convention used in the subsequent lectures. You will see that every question has a data box prior to resolving the problems. Three stage approach: A Obtain data. B Plug data into equation. C Calculator to solve problem

Before we get to our "game-day play"… as suggested above…We need to visit the Blue Sheets,... Hydrostatic curves… Hydro of course is a derivative of the Greek word for water and the second part is statistics. The Naval Architect's primary concern in the initial stages of a design is calculating these statistics as it relates to the underwater portion of the vessel taken from the lines plan..prior to designing his construction plans.. When the vessel is built, it may undergo a lightship test called the inclining experiment...confirming the architect's calculation for the vessel's all important relationship GM. There are 3 or 4 incline experiment questions in the question bank which I'll show you how to do in Lecture Two ; but , at this time building a good solid conceptual and theoretical base will pay off in calculating the problems.

Hydrostatic Curves

There will be questions which require you to access information from this sheet. Generally the question will give you FWD & AFT drafts…You then determine the mean draft and enter into the table from the left hand column. The first curve TPI we've touched upon. MTI along with longitudinal buoyancy and center of flotation relate more to our discussion on trim. You need not concern yourself with wetted surface or prismatic coefficient.

If you look at the table find the curve " Height of center of Buoyancy". This is your KB…You can also determine displacement next curve . Found in the last third of the table called "HT trans metacenter"… that’s your KM…You will find in the question bank many questions which will give you’re the displacement and drafts…You do a tabulation as mentioned earlier and find KG… You then access KM from this sheet…Km - KG = GM.. obvious but very important… a Coast Guard favorite… We went over Block And Waterplane coefficients…you find the value nearest to your table then go to deadweight scale and use the working decimal value for some reason labeled "k". "you would think with all the other letters in the alphabet"….That's about all you will need to access.. from this sheet

Now I want you to take a brief look at the next sheet call the Cross Curves. Since I have not gone over Righting Arms as yet , we are accessing this sheet for one reason only. You will note the two columns marked Sine's & Cosine's. You might also upon quick glance note that the curves are designed for different displacements when the KG height is 20 feet…That's all I'm going to say about that now.

Now take out your calculator that allows you the ability to do sine, cosine and tangent… If you don't have one..come back when you do..

Calculator Practice:

Going into the problems and fumbling with your calculator.. would be like going on the field without your helmet or pads…So This gives us the opportunity for calculator practice. I trust that everyone knows how to obtain the basic functions of adding, subtracting, multiplying and division on your calculator…That's pretty straight forward stuff.

Looking at the cosine table, we note that the decimal equivalent of the cosine angle of 30 degrees is .87…You must be able to do these calculations. This is how I resolve it on my calculator. I turn it on in the "DEG" mode…I hit the cosine button then 3 and 0 buttons.. then the equals button… giving me the answer.87…Practice that problem along with the other cosine and sine equivalents until it is second nature…15 minutes well spent..now work backwards from the decimal equivalents to degrees until that process is second nature. On my calculator I hit .87… then the equals button… then the button marked 2nd……then the cosine button… then the equals button …which gives me an answer 30 degrees. …a five stage process….Practice that procedure 20 minutes till it becomes second nature. " If you cannot figure out how do it"…contact someone who can show you how…This procedure is a must… The only scientific calculations required if you call it that.

Before we get into the "game-day problem" that ties up the "Entire Argument" relating to stability theory in one diagram… I would like you to take a look at Table Four in the white sheets Titled "Free Surface Correction and Tank Capacities." You will access information here that generally will be applied to your tabulation (which I'll get more specific about in a minute) Look at the headings tank capacities… a fuel oil tank is full at 97%…in our problems… If it doesn't have 97%… it's considered slack…and you use the corresponding values…Again for our purposes a tank is either pressed up to (97% or slack nothing in between..use the values indicated accordingly… salt water tanks…100% or they are slack)…You will of course access your VCG when indicated or implied in a problem…which as you have already learned is the Center of Gravity of the cargo in the tank.

Tabulation :

Free surface problems and KG problems are solved by setting up a table which I call a tabulation The headings are indicated: W X D = Moments.. you therefore have three columns You enter your weights in column one…"that includes the displacement of the vessel".. for example the vessel displaces 7000 tons with a KG of 31…7000 first column 31 second column you loaded 500 tons of cargo in # 3 lower hold the VCG of the cargo 6 feet above the keel…500 goes under 7000..six feet goes under 31.." Different terminology is being used to describe the same thing. A cargo with a VCG 6 feet above the keel has a KG of 6…The problem will go on ..keep on filling in your tabulation…add your total weights then add your total moments… divide moments by the weights and you have a new KG for the vessel.. Knowing what the new weight of the vessel is…you now can go into your tables determine a new mean draft..then obtain M as I explained above found in the Hydrostatic Tables..now that you have KM and you calculated KG through tabulation..you subtract KG from KM and get GM… That procedure is a favorite question in the pool of questions. If you are familiar with those concepts, you know the rolls of the players very well…and we are ready for the big game.

Game Day:

Looking at the diagram in front of you, note that it is similar to the last diagram that you saw. In fact it is the same diagram marked fig 3 on the lower left hand corner. I took the other diagram and went back to my drawing board. . You will note that I drew in an inclined waterline marked WL1.. You will also note the waterline angle as indicated is 30 degrees. This graphically shows the vessel is heeled 30 degrees to the starboard side.. You do remember that I said B has the ability to shift laterally. That's exactly what it did. If you look on the left hand side situated between WL1 and LWL… what looks like sloppy erasing was done on purpose. This was to graphically point out that portion that was under water with the vessel at rest is now out of the water with the vessel healing at 30 degrees. Obvious is that portion which was out of the water is now under water on the right under WL1.. B merely shifted with the new distribution of displacement basically in the same relative position. . You do remember that I stressed the importance of the centerline being perpendicular to the LWL…. Graphically the key to the diagram is the line drawn from B1 PERPENDICULAR through WL 1 intersecting the original centerline. M has shifted downward . If the vessel inclined another 10 degrees.. i.e. if I drew in WL 2 indicating an angle of 40 degrees M would obviously shift even further downward. Diagrams are to visually tell stories… and as mentioned earlier when M could graphically posture itself lower than G the vessel would sink. Graphically if you look where the section ends…and the main deck would begin 15 more degrees of heel indicated by waterline 3 (if drawn in) ..would suggest deck immersion and the vessel would capsize. The capsizing forces would quickly exceed the righting forces

But getting back to what we have in front of us is this… the vessel shows a reasonable amount of positive stability at a 30 degree angle of inclination to the starboard side. The vessel heeled 30 degrees due to wind and sea will have a tendency to move back to its original position. You will notice we picked up a new player called Z. If I had the means to draw this section to scale from the mid ship section of the lines plan, I'd take my architectural ruler and measure to scale the lateral line GZ. That distance multiplied by the displacement represents the righting force. ..My displacement would represent the tipping force. Obviously if my GZ arm is 2' I would have a righting force twice the tipping force. I want you now to turn your attention to G. We said that G can only move along the centerline.. perfectly true.. If from the time of our preceding diagram I had loaded my vessel in such a way to lower "G" This would suggest a greater distance in the GZ arm which would translate mathematically into a greater righting force viz. the tipping force which of course would remain the same given the same 30 degree angle of inclination…So what we see in GZ is an indicator of stability. The longer the GZ arm the more stability…to a point. What I mean by this is.. GZ "on paper" will get…larger and larger long after the ship has capsized and sunk.. other critical forces will take hold which I'll touch upon

We should conclude that the position of G is a very important factor in determining the stability or righting force of the vessel. A vessel inclined by wind and sea will indeed right itself provided KM is greater than KG. The distinction between list and heel, comes in handy in resolving the stability load problems…You'll see how this plays out in some of the CG problems.

Now let's take a quick look at Sheet 3 & 4 . The Cross Curves and the Statical Stability Curves. You should have no problem using these sheets… If you are a bit uncertain about KG displacement , righting arms…review. Sheet three tells me that given a vessel with a KG of 20 feet with a displacement of 10,000 tons at an angle of inclination of 30 degrees… my GZ arm is 4 feet..

I know there is some Chief Mate on a container ship out there who looking at the cross curves in light of our discussion… which suggests a 10,000 ton vessel with an angle of inclination of 75 degrees, has a righting arm of 4.6 feet is insane… Very true…reality doesn't necessarily modify theory. On a container ship change of underwear may be necessitated after undergoing repeated rolls of 15 degrees.

Take a look at the little diagram that I stuck with throughout this presentation.

The waterline has been inclined 40 degrees (WL2).The diagram also suggests that the starboard side of the waterline is at the edge of deck immersion.... B-2 of course has shifted as the center of displacement by definition, but what is more graphically relevent is that we see M2 and KG at the same point...We not only have neutral stability...but the diagram aptly points out the non existence of a GZ arm, translating into a very important point..Righting force equals tipping force. The vessel really has no ambition to right itself at this point...with waterline 2 at the edge, deck immersion will ensue, tipping force will increase, and vessel in imminent danger of capsizing. As a naval architect, if this section was taken off the lines plan and a couple of adjustments were made to validate my waterline as it related to displacement, I could conclude that the permanent angle of list to be 40 degrees... One design remedy would be of course to increase freeboard. The central point of the diagram is this, when the GZ arm for varying angles of heel, in the form of a curve is zero (as you will see in lecture three), positive stability is non existent.

The Coast Guard considers the danger angle to be ½ the permanent angle or 22.5 degrees more or less. Permanent angle being a little less than 45 degrees…"It will be touched upon in the problems."

What I've tried to accomplish in Lecture One is to give you a good fundamental understanding of Stability theory and practice, combined with the terms faced in the problems found in Lectures Two, Three and Four. As already mentioned..If you can do the problems outlined in these next three lectures, you are capable of answering all the questions in the Coast Guard data bank as it relates to stability.

If you have an idea of what has ben presented and can follow the logic of the following problem, you are on the right track.

Q A vessel with a GM of 6 feet is inclined 30 degrees. What is the value of the Righting Arm ? GZ equals GM multiplied by the sine angle…. 6 times sin of 30 degrees equals 3 ANS. (This is what I have meant relating to the triangular references suggested in the above diagram....note the succession of right triangles.. and angular relationships...this is what is actually being resolved by sine ,cosine and tangent.)

Please Note: Lecture One along with the Reference Library has been composed in HTML format allowing easier composition and insertion of our diagrams along with the necessary hyperlinks in order to access the required data sheets. All the other Lectures with the exception of Part One, Damage Stability have been composed in document format making it possible mathematical practice.

Before tackling the other lectures, might I suggest you go to the sites Reference Library for some further points on navigating the link.


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