stability and trim
TRANSCRIPT
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Massachusetts Institute of Technology, Subject 2.017
Acknowledgements to Lt. Greg Mitchell for Slides 15-37
STABILITY AND TRIM OF MARINE
VESSELS
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Massachusetts Institute of Technology, Subject 2.017
Concept of Mass Center for a Rigid Body
Centroid – the point about which moments due to gravity are zero:
g mi (xg – xi) = 0
xg = mi xi / mi = mi xi / M
• Calculation applies to all three body axes: x,y,z• x can be referenced to any point, e.g., bow,
waterline, geometric center, etc.• “Enclosed” water has to be included in the mass
if we are talking about inertia
G
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Massachusetts Institute of Technology, Subject 2.017
Center of BuoyancyA similar differential approach with displaced mass:xb = i xi where i is incremental volume,
is total volume
Center of buoyancy is the same as the center of displaced volume: it doesn’t matter what is inside the outer skin, or how it is arranged.
Calculating trim of a flooded vehicle: Use in-water weights of the components, including the water (whose weight is then zero and can be ignored). The calculation gives the center of in-water weight.
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Massachusetts Institute of Technology, Subject 2.017
• For a submerged body, a sufficient condition for stability is that zb is above zg.
g
Righting arm:h =(zb -zg )sin
Righting moment:gh
G
B
Make (zb -zg ) large the “spring” is large and:• Response to an initial heel angle is fast (uncomfortable?)• Wave or loading disturbances don’t cause unacceptably
large motions• But this is also a spring-mass system, that will oscillate unless adequate damping is used, e.g., sails, anti-roll planes, etc.
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Massachusetts Institute of Technology, Subject 2.017
• In most surface vessels, righting stability is provided by the waterplane area.
b
intuition: wedges
B2B1
h
l
RECTANGULAR SECTIONGeometry:d/dx = bh + bl/2 or h = ( ddx – bl/2 ) / b l = b tan
Vertical forces:dFG = -g d no shear)dFB1 = g b h dxdFB2 = g b l dx / 2
K
l/3
b/6
G
M
gy
submerged volume d
d
= Adx
slice thickness dx
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Massachusetts Institute of Technology, Subject 2.017
Moment arms:yG = KG sin yB1 = h sin
/ 2 ; yB2 = (h + l/3) sin
+ b cos
/ 6
Put all this together into a net moment (positive anti-clockwise):
dM/g = -KG d
sin
+ bh2 dx sin
/ 2 + b l dx [ (h+l/3) sin
+ b cos
/ 6] / 2
Linearize (sin
~ tan
~ ), and keep only first-order terms ():
dM g d KG h / 2 b2 / 12 h- KG + A / 2 b + b3 / 12 A
For this rectangular slice, the sum h / 2 + b2 / 12 h must exceed the distance KG for stability. This sum is called KM – the distance from the keel up to the “virtual” buoyancy center M. M is the METACENTER, and it is as if the block is hanging from M!
-KG + KM = GM : the METACENTRIC HEIGHT
(valid until the corner comes out of the water)
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Massachusetts Institute of Technology, Subject 2.The Perfect Storm
G
M
R: righting arm
heel angle
R “down-flood”
How much GM is enough? Around 2-3m in a big boat
Note M does not have to stay on the centerline**!
damage, flooding, ice
Image removed due to copyright restrictions.Please see any publicity images for The Perfect Storm,
such as http://www.imdb.com/media/rm1041734656/tt0177971
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Massachusetts Institute of Technology, Subject 2.017
Considering the Entire Vessel…
Transverse (or roll) stability is calculated using the same moment calculation extended on the length:Total Moment = Integral on Length of dM(x), where (for a vessel with all rectangular cross-sections)dM(x) g [ -KG(x) A(x) + A2(x) / 2 b(x) + b3(x) / 12 dx
First term: Same as –g KG , if
is the ship’s submerged volume, and KG is the value referencing the whole vesselSecond term: Significant if d>b (equivalent to h2 b / 2)Third term: depends only on beam – dominant for most monohulls
Longitudinal (or pitch) stability is similarly calculated, but it is usually secondary, since the waterplane area is very long very high GM
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Massachusetts Institute of Technology, Subject 2.017
Weight Distribution and Trim
• At zero speed, and with no other forces or moments, the vessel has B (submerged) or M (surface) directly above G.
GM
Too bad!
For port-stbd symmetric hulls, keep G on the centerline using a tabulation of component masses and their centroid locations in the hull, i.e.,
mi yi = 0Longitudinal trim should be zero relative to center of waterplane area, in the loaded condition.Pitch trim may be affected by forward motion, but difference is usually only a few degrees.
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Massachusetts Institute of Technology, Subject 2.017
Rotational Dynamics Using the CentroidEquivalent to F = ma in linear case is
T = Jo * d2/ dt2
where T is the sum of acting torques in rollJo is the rotary moment of inertia in roll,
referenced to some location O is roll angle (radians)
J written in terms of incremental masses mi :Jo = mi (yi -yo )2 OR Jg = mi (yi -yg )2
J written in terms of component masses mi and their own moments of inertia Ji (by the parallel axis theorem) :
Jg = mi (yi -yg )2 + Ji
The yi ’s give position of the centroid of each body, and Ji ’s are referenced to those centroids
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Massachusetts Institute of Technology, Subject 2.017
What are the acting torques T ?• Buoyancy righting moment – metacentric height• Dynamic loads on the vessel – e.g., waves, wind, movement of
components, sloshing• Damping due to keel, roll dampers, etc.• Torques due to roll control actuators
An instructive case of damping D, metacentric height GM:J d2 / dt2 = - D d /dt – GM g
OR
J d2 / dt2 + D d /dt + GM g = 0 d2 / dt2 + a d /dt + b = 0 d2 / dt2 + 2 n d /dt + n
2 = 0
A second-order stable system Overdamped or oscillatory response from initial conditions
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Massachusetts Institute of Technology, Subject 2.017
Homogeneous Underdamped Second-Order Systems
x’’ + ax’ + bx = 0; write as x’’ + 2n x’ + n2x = 0
Let x = X est
(s2 + 2n s + n2) est = 0 OR s2 + 2n s + n
2 = 0 s = [-2n +/- sqrt(42n
2 – 4n2)] / 2
= n [-
+/- sqrt2-1)] from quadratic equation
s1 and s2 are complex conjugates if
< 1, in this case:s1 = -n
+ id , s2 = -n
– id where d = n sqrt(1-2)
Recalling er+i
= er ( cos
+ i sin, we havex = e-nt [ (X1
r + iX1i)(cosd t + isind t) +
(X2r + iX2
i)(cosd t – isind t) ] AND
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Massachusetts Institute of Technology, Subject 2.017
x’ = -n x + d e-nt [ (X1r + iX1
i)(-sind t + icosd t) + (X2
r + iX2i)(-sind t – icosd t) ]
Consider initial conditions x’(0) = 0, x(0) = 1:x(t=0) = 1 means X1
r + X2r = 1 (real part) and
X1i + X2
i = 0 (imaginary part)x’(t=0) = 0 means X1
r - X2r = 0 (imaginary part) and
-n + d (X2i – X1
i) = 0 (real part)
Combine these and we find thatX1
r = X2r = ½
X1i = -X2
i = -n / 2 d
Plug into the solution for x and do some trig:
x = e -nt sin(d t + k) / sqrt(1-2), where k = atan(d /n )
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Massachusetts Institute of Technology, Subject 2.017
= 0.0 has fastest rise time but no decay
= 0.2 gives about 50% overshoot
= 0.5 gives about 15% overshoot
= 1.0 gives the fastest response without overshoot
> 1.0 is slower
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STABILITY REFERENCE POINTS
CL
Metacenter
Gravity
B
K
Buoyancy
KeelCourtesy of Greg Mitchell. Used with permission.Massachusetts Institute of Technology, Subject 2.017
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CL
M
G
B
K
GM
KG
BM
KM
LINEAR MEASUREMENTS IN STABILITY
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
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B
WATERLINE
THE CENTER OF BUOYANCY
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
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CENTER OF BUOYANCY
B
WL
G
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
B’
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B1
- The freeboard and reserve buoyancy will also change
Reserve Buoyancy
Draft
CENTER OF BUOYANCY
Freeboard
B0
Courtesy of Greg Mitchell. Used with permission.
B1
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G0
KGo
G MOVES TOWARDS A WEIGHT ADDITION
MOVEMENTS IN THE CENTER OF GRAVITY
G1
KG1
Courtesy of Greg Mitchell. Used with permission.
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G1
KG1KGo
G0
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G MOVES AWAY FROM A WEIGHT REMOVAL
MOVEMENTS IN THE CENTER OF GRAVITY
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G1
G MOVES IN THE DIRECTION OF A WEIGHT SHIFT
MOVEMENTS IN THE CENTER OF GRAVITY
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
G0
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DISPLACEMENT = SHIP’S WIEGHT
20
B
- if it floats, B always equals G
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
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METACENTER
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METACENTER
B SHIFTS
M
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CL
B
M
0o-7/10o
G
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
+GM
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CL
B B20
M
M20
G
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+GM
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CL
M
M20
M45
B
B20 B45
G
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+GM
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G
CL
B
B20B45
M
M20M45
B70
M70
neutral GM
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CL
M20M45
M70
M90
B
B20B45B70
B90
MG
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
-GM
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MOVEMENTS OF THE METACENTER
THE METACENTER WILL CHANGE POSITIONS IN THE VERTICAL PLANE WHEN THE SHIP'S DISPLACEMENT CHANGES
THE METACENTER MOVES IAW THESE TWO RULES:1. WHEN B MOVES UP, M MOVES DOWN.2. WHEN B MOVES DOWN, M MOVES UP.
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
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M
G
B
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
M
G
BG
M
B
M1
B1
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CL
G
M
CL
K
B
G
B1
Z
Massachusetts Institute of Technology, Subject 2.017 Courtesy of Greg Mitchell. Used with permission.
Righting Arm
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RIG
HTI
NG
AR
MS
(FT)
ANGLE OF HEEL (DEGREES)9060300 10 20 40 50 70 80
WL
WL20°
GBZ
WL
40°
G
B
Z
60°
G
B
Z
GZ = 1.4 FT GZ = 2.0 FT GZ = 1 FT
MAXIMUM RIGHTING ARM
ANGLE OF MAXIMUM RIGHTING
ARM
DANGERANGLE
MAXIMUM RANGE OF STABILITY
Righting Arm
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Righting Arm for Various Conditions
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Massachusetts Institute of Technology, Subject 2.017
THINGS TO CONSIDER
• Effects of: – Weight addition/subtraction and movement– Ballasting and loading/unloading operations– Wind, Icing– Damage stability
• result in an adverse movement of G or B• sea-keeping characteristics will change• compensating for flooding (ballast/completely flood a
compartment)• maneuvering for seas/wind
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Massachusetts Institute of Technology, Subject 2.017
References
• NSTM 079 v. I Buoyancy & Stability
• NWP 3-20.31 Ship Survivability
• Ship’s Damage Control Book
• Principles of Naval Architecture v. I
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