Stability and wind effects

[IonAguirre]

A CORRECTION HAS BEEN POSTED BELOW.

 

From now on, stability means "TRANSVERSAL STABILITY"

All units in metrics (I.S.)

KN values for a fixed displacement only.

 

How the ship's stability changes into flood conditions ? What are the effects of wind on the ship's stability ?

 

Everything around these questions starts from the Cross Curves set of each hull, but as I posted a few days ago, here is my proposal for a simpler way, which could avoid having to build tables with curves values and interpolation algorithms.

 

A way I think could be a reasonable approach is using a third degree polinomy instead of real cross curves.

 

A general form could be:

Y=Ax³+Bx²+Cx+D

Where Y is the cross value (KN stability arm) and X is the heeling angle.

 

The key is solving for A,B,C and D for each ship.

 

Four coefficients A,B,C,D require four equations. The system can be solved by using four points from the real KN curve of the ship. Or, as I've done, from three points and the condition of tangency at the origin for the initial stability value GM.

 

Naval building theory says that a straight line, tangent to the stability curve at the origin, will reach a value=GM for x= 1 radian. Where GM= metacentric height

 

Hence, making the first derivative of the polinomy, and forcing x=0, will return that the equation of that line is given by:

Y= Cx

And we know that:

KM= 57C (Where 57 is 1 radian in degrees)

As the whole information has been referenced to the Keel and not to the center of gravity, instead of using GM, we must use KM: KM= KG+GM

 

Now, we have four equations and four unknowns. It can be solved.

 

From KN= Ax³+Bx²+Cx+D, With X as the heeling angle in degrees, KN values will result.

 

 

Lets now take into account the height of the center of gravity ... The righting arm of stability is normally named GZ.

 

For each heeling angle. GZ= KN- KG sin(X)   Where X is the heeling angle again.

 

 

Effect of wind:

Wind works against stability, hence we must compute the heeling arms.

 

The angle from which the wind blows, relative to the ship, will be denoted as W.

The angle at which the yards are turned as V

The angle of attack of the wind on the sails as Aa

 

From simple geometry: Aa= W-V

 

Once the angle of attack is known, its only a matter of having the sails aerodynamic coefficients.

Fortunately, in the very good and clear work presented by William C. Lasher and Logan S. Flaherty about the Suvivability of squared rigged sailing vessels, good approaches to those coefficients are included.

 

For each combination of wind and yards possition, a different angle of attack will result. From the resulting angle of attack, LIFT and DRAG, CL and CD, coefficients of the sails joint can be obtained.

 

But lift and drag mean nothing in a ship, those coeficients must be converted to propulsion coefficient Cx and Heeling coefficient Cy.

 

The conversion is given by:

 

Cx= Cl sin(W) - CD cos(W)

Cy= Cl cos(W) + CD sin(W)

 

I'll not explain what to do with Cx that has nothing to do with heeling and capsize. Lets forget it by now, and lets go with Cy.

 

Cy relates the shape of the sails with the force that wind exerts on them, in such a way, that force will come from:

 

Fy= 0.5*1.24*Cy*Area*WindSpeed²   Where 1.24 is the averaged air density.

 

The point at which this force is applied is called the pressure center, lets say it is at a heigh Z from the keel, and lets call this distance as KZ.

Working against the wind, there is another force, the Lateral resistance of the hull, that can be assumed to be applied at a point half way from the keel to the water surface. Lets call the height of this point from the Keel KL

 

Then, the wind will exert a heeling moment (WM) given by the force Fy and the distance between both points.

 

WM= Fy*(KZ-KL)

 

Lets now see how to obtain a heeling arm for each heeling angle in order to include its value in the final stability calculation.

 

Heeling arms are usually denoted as HA.

With D as the ships displacement in kilograms.

 

HA= [WM* cos(X)^(1.3)]/D   X= heeling angle

 

Then, final righting arm under wind conditions will be given by:

 

GZ= Ax³ +Bx²+ Cx +D - KG sin(x) - [WM cos(x)^(1.3)]/D

 

The value of WM will depend on:

Sails area and height of the resulting pressure center (Two or three fixed values joints could do the work for the game)

Wind speed

Wind angle

Yards angle (The same than for surface and center of pressure)

 

In very simple way, as well, flooding could be included.

 

NOTES:

No aerodynamic sails interaction is taken into account. But for the seek of simplicity its not required.

CL and CD curves can also be approached by polinomials, avoiding the table/interpolation issue.

 

In the graph below there are some calculations for the Frigate DIANA.

Displacement= 1151 mT

KG=2.85m

GMo= 3.53m

Mean draught= 5.2m

Wind speed= 5m/s

Wind angle= 85º

Yards angle= 45º

Sails area= 7470 m² (full rigged)

Pressure center height= 26.38 m

 

Sample points for coefficients calculations:

Heeling=0 ---> KN=0

Heeling=45 --> KN=3.92

Heeling= 90º -> KN= 4.27

Heeling= 57º --> KM= 5.93 (The equation is the first derivative of the polinomial)

 

Results:

A=0: B= -0.0001: C= 0.1034: D=0

 

Resulting curves:

GZ (green) Righting arms from real stability information. No wind.

GZs(black) Righting arms from the polinomial expression. No wind.

YR (blue) Tangent line at the center, returning GM for X=57º

GZfs (dashed red) Righting arms under wind conditions.

GZflood(violet) Flooded Righting arms. Water sounding 4 meters.

 

Permanent heel due to wind= 40º

Permanent heel Wind + flood= 45º

 

It looks reasonably well for a game.

 

Any question, dont hersitate asking .........

 

Regards

Edited June 19, 2015 by IonAguirre

[Sharp]

TL DR  Keep G above M

[IonAguirre]

Nop.

Keep M above G

 

GM<0 = capsize

 

The ship in the example, is supposed to have a fully watertight main deck. In the real case, flooding would make the ship sunk.

At the curve, it would be seen as an abrupt curve ending for the flooding heeling angle.

Edited June 18, 2015 by IonAguirre

[maturin]

No time to chew through this on my phone, but does that graph predict that 5 m/s of wind will heel a frigate 40 degrees?

[Sharp]

Nop.

Keep M above G

GM<0 = capsize

The ship in the example, is supposed to have a fully watertight main deck. In the real case, flooding would make the ship sunk.

At the curve, it would be seen as an abrupt curve ending for the flooding heeling angle.

you sir are right, i was trying to remember which was the larger of my two lectures in stability and you are right it was Gina who was bigger, so we don't want her on top...

Edited June 18, 2015 by Sharp

[IonAguirre]

No time to chew through this on my phone, but does that graph predict that 5 m/s of wind will heel a frigate 40 degrees?

Yes it does, 5m/s approx=10 Knots.

But don't forget that the assumption is for the "worst" case, that supposes that each sail will receive the "full" wind, that is not true.

Anyway yes, 10 knots is too much for full rigging, at around 20º the water plane reaches the first battery. This big heel is due to the very high masts. The frigate is not designed for using topgallants under this abeam wind.

[IonAguirre]

Dear Maturin:

 

YOU ARE RIGHT ¡¡¡

 

I made that graph bery quickly and had a very BIG mistake. My Apologizes

 

Heeling and righting moments are not in the same units. While the first is in Kilograms, the second is in Newtons.

 

Here you have the corrected graph with the corresponding source data table. Now results are a lot more logical.

 

Thanks a lot for pointing me to the very big heeling angle. Im a lot shamed for publishing such a nosence.

 

The sails names are in Castilian.

This is a lot more logical. A wind of 11 m/s would heel the ship to the lower battery deck limit.

As you can see, I've set all yards to the same angle.

 

Graphs:

Upper right: Heeling moment due to each sail

Lower left: Righting arms with and without wind.

Lower right: Aerodynamic coefficients aligned with the Ships reference axis. The dotted vertical line corresponds to the averaged angle of attack for all sails.

 

My apologizes again, and thanks for your comments. Im a lot absent-minded.

Edited June 19, 2015 by IonAguirre

[IonAguirre]

FLOODING EFFECT

 

Flooding effect is decreasing the metacentric height, or what is the same, the same effect than loading a weight at a high location.

That "virtual" rise of CG is a function of the water moment of inertia, that depends on volume and shape.

 

For the seek of simplicity and the game requirements, the hull can be assumed to have a constant cross section that can be defined as a parabolla which general equation is:

Y= ax²+ bx +c, where Y is the flood sounding and X is the flood breadth.

 

As we did before coefficents a,b and c can be calculated from three known points of the reference hull cross section. Once those values are known, computing the breadth for any sounding is trivial.

 

Taking into account that we are only looking for a reasonable behaviour and not for a scientific simulator, the flooded volume can be computed as a prism, which lenght is the lenght of the ship (LPP) and which breadth ( is the one computed from the parabollic assumption above.

 

Then, its moment of inertia with respect to the rolling axis (X) is given by:

Ix= (LPP * B³)/12

 

And the "virtual rise" of CG by:

GGv= water density * Ix/Displacement(in Kilograms)

 

The new stability GZ curve, or corrected GZ will result from:

 

GZc= GZ - GGv*sin(X) where X is the heeling angle.

 

Now, with a flooding sounding of 2.7m the ship (DIANA again) would capsize for a heeling of 75º.

But due to the added weight, total flooding will happen before .....new draft would be:

From all assumptions made before (No bulkheads, free flow of water inside the hull, etc .....), or in a more technical term .. Permeability=100%

At the same time, new submerged volume can be assumed as a prism, which base is the inial waterline breath.

Hence, the heigth of such prism which volume equals the flooded one is:

 

h= Flooded volume/waterline breadth

 

And new draft would be:

Final draft= Initial draft+h

Final draft= 5.2+2.56= 7.76m, well above the maindeck ....

 

If a more accurate prediction of this draft is wanted, ....

Assumtion of new volume as the addition of a triangular and a rectangular section along LPP.

Integration of the parabolla between Yo=old draft, and Y=new draft and equaling to the required section area increment, compute Y.

 

Below:

Left:   Parabollic section, with intial water line, blue, and flood, green.

Right: Corrected Static curve of stability

 

Permanent heeling=44º (And now its checked ¡¡) due to flood and wind

Vanishing stability at 75º

Edited June 19, 2015 by IonAguirre