2. SAILOR’S CREED “ I am a United States Sailor.
I will support and defend the Constitution
of the United States of America and I W ill obey the orders of those appointed over me.
I represent the fighting spirit of the Navy and those who have gone before me to defend freedom and democracy around the world.
I proudly serve my country’s Navy combat team with Honor, Courage, and Commitment
I am committed to excellence and fair treatment of all.”
Honor, Courage, and Commitment as we have have just seen were attainable virtues, back then as they are now, that have been demonstrated at the highest level through out our history and are on display right now at every watch station around the world. Our Navy’s history is rich and full of displays of valor, sacrifice, and honorable conduct in dangerous, difficult and even hopeless circumstances. These virtues and values have prevailed.
Possibly one of the leading reason our Sailors today, do not identify with our Core Values more quickly is our shallow sense of our heritage, history and traditions. 1

3. Lesson 4.2 Solid Weight Shifts, Additions and Removals WLB WL G1 Z1 G Z
Tell Story about CG WMEC out of Miami.

4. Example of a SOLID weight addition gone wrong…

5. We must crawl before we can walk!
Lets review Homework #1 and Lesson 4.1

6. References (Are they different?)
NSTM 079 Volume 1
NTTP
Damage Control Book, section II (a)

7. Enabling Objectives to be covered…
Describe movement of reference points.
Describe impact of addition, removal and shift of weight on stability.
Calculate KG1 and GG2.
Calculate Sine and Cosine corrections.
Apply corrections to Static Stability Curve.

8. CLASS TIMELINE
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horiz Weight Additions / Removals 6. Cosine Correction 7. Example Problem

9. Vertical Weight Shifts
GG1 = KG1 - KGo G B M G1 G1 G1 G1
GG1
KG1 G1 G1 G1
KGo
Shifting weight vertically, no matter where onboard it is, will always cause the ship’s center of gravity to move in the same direction as the weight shift.

10. KG1 = (Wo x KGo) ± (w x kg) Wf WHERE; w = Weight Shifted
kg = Distance Shifted
Wo = Original Displacement
KGO = The original height of the ship’s center of gravity (FT)
WO = The ship's displacement prior to shifting weight (LT)
w = The amount of weight shifted (LT)
kg = The vertical distance the weight was shifted (FT)
WF = The ship’s displacement after shifting the weight (LT)
± = When the weight is shifted up use (+)
When the weight is shifted down use (-)
KGo = Original Height of G
Wf = Final Displacement
± = + if shift up/- if shift down

11. KG1 = (Wo x KGo) ± (w x kg) Wf 30 FT 25 LT M Wo = 2000 LT G1 G1 G1 G1B

12. 45 LT
Wo = 3400 LT M
8 FT
33 FT G
15.5 FT B
17 LT

13. WHERE; w1&2 = Weights Shifted
KG1 = (Wo x KGo)±(w1xkg1)±(w2xkg2) Wf
WHERE; w1&2 = Weights Shifted
kg1&2 = Distances Shifted
Wo = Original Displacement
KGo = Original Height of G
Wf = Final Displacement
± = + if shift up/- if shift down

14. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

15. Example of vertical weight addition…

16. Vertical Weight AdditionsM B G

17. Vertical Weight AdditionsM B G G1 M1 B1 M B G G1 M1 B1 M B G G1 M1 B1 M B G G1 M1 B1
GG1
KGo
KG1

18. KG1 = (Wo x KGo) ± (w x kg) Wf WHERE; KGo = Original Height of G
Wo = Original Displacement
± = + if addition/- if removal
w = Weight Added/Removed
kg = Distance Keel to "g" of w
Wf = Final Displacement

19. 16 LT
ADD
42 FT
Wo = 2000 LT
KG1 = ?
KGo = 12 FT

20. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

21. Z CL K B G M B1 CL K B G M B1 CL K B G M B1 CL K B G M B1 CL K B G M
From Lesson 4.01, when the ship's center of gravity is shifted vertically, the length of the righting arm is affected. The Sine Correction is applied to the Righting Arm Curve to reflect the increase/decrease in GZ.

22. Z Z1 Z1 Z Z Z1 Z Z1 Z1 Z Z1 Z Z1 Z Z G G1 G1 G M G G1 G G1 G1 G G1 GB1
When gravity rises from position G to position G1, the righting arm is reduced from length GZ to length G1Z1. The distance the righting arm was reduced is shown by length GT. K CL

23. Z1 Z G1 G M T

24. G1 G T Sin  = OPP HYP WHERE; OPP = GT  HYP = GG1 Sin  = GT GG1
REDUCTION IN
RIGHTING ARM
The value of length GT varies by the sine function of the angle of inclination. To correct the righting arm curve, the Sine Correction curve is plotted using the equation:
GT = GG1 x Sin 

25. THINK VERTICAL DIRECTION ONLY!!!
“SINE correction is vertical only… I use KG1 to find SINE Correction”

26. KG1 = 19.8 FT GG1 = 0.8 FT GT = GG1 x Sin O1 -1 2 3 4 5 10 20 30 40 50 60 70 80 90
ANGLE OF INCLINATION - DEGREES
RIGHTING ARMS (FT)
KGA = 19 FT
KG1 = 19.8 FT
GG1 = 0.8 FT
GT = GG1 x Sin O
Sin 0° = 0 Sin 30°= 0.5 Sin 90°= 1.0
GT0°= .8FT x 0 = 0 FT
GT30°= .8FT x .5 = .4 FT
GT90°= .8FT x 1 = .8 FT
FINAL CURVE X
.8 FT X
.4 FT
The Sine Correction curve is applied directly to the righting arm curve, and indicates exactly how much righting arm has been lost due to the higher position of G throughout the range of stability. X

27. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

28. Horizontal Weight ShiftsM G G2 G2 G G2 G G G2 G2 G2 G2 G2
GG2
Shifting weight horizontally, no matter where onboard it is, will always cause the ship’s center of gravity to move in the same direction as the weight shift. NOTE: A weight shift causing the ship’s center of gravity to move off centerline will always reduce the stability of the ship. B

29. GG2 = w x d Wf WHERE; w = Weight Shifted d = Distance shifted
Wf = Final Displacement
w = The amount of weight shifted (LT)
d = The horizontal distance the weight is shifted (FT)
WF = The ship’s displacement after the weight is shifted (LT)

30. Horizontal Weight Shifts
GG2 = w x d Wf M
Wo = 3000 LT G G2 G2 G2 G G2 G G G2 G2 G2 G2
GG2 B
34 LT
24 FT

31. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

32. Horizontal Weight Additions and Removals
GG2 = w x d Wf B M G G2 G2 G2 G2 G2
GG2
When an off-center weight is added or removed to/from a ship, the ship’s center of gravity will move off centerline, the ship will develop a list.

33. GG2 = w x d Wf WHERE; w = Weight Added or Removed
d = Distance Added/Removed from Centerline
w = The amount of weight added/removed (LT)
d = The distance from the center of gravity of the weight to the ship’s centerline (FT)
WF = The ship’s displacement after the weight is shifted (LT)
Wf = Final Displacement

34. Horizontal Weight Additions and Removals
GG2 = w x d Wf B M G
Wo = 4500 LT
50 LT G2 G2 G2 G2 G2
GG2
28 FT

35. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

36. Cosine Correction B B1 M G G2 B B1 M G G2 B B1 M G G2 B B1 M G G2 B B1Z G G2 G2 G G G2 G G2
When the ship's center of gravity is shifted off centerline, the length of the righting arm is affected. The Cosine Correction is applied to the Righting Arm Curve to reflect the increase/decrease in GZ. B

37. Z Z2 M G2 G
When gravity moves from position G to position G2, the righting arm is reduced from length GZ to length G2Z2. The distance the righting arm was reduced is shown by length GP.

38. Z Z2 M G2 G
The distance the righting arm was reduced is shown by length GP. P

39. P G G2  Cos  = ADJ WHERE; HYP ADJ = GP HYP = GG2 Cos  = GP GG2
REDUCTION IN RIGHTING ARM G2 G P 
Cos  = ADJ
HYP
WHERE;
ADJ = GP
HYP = GG2
Cos  = GP
GG2
GP = GG2 x Cos 

40. THINK TRANSVERSE DIRECTION ONLY!!!
“COSINE correction is transverse (Port/Stbd) only… I use GG2 to find COSINE Correction”

41. KGA = 19 FT KG1 = 19.0 FT GG2 = .9 FT GP = GG2 x Cos O1 -1 2 3 4 5 10 20 30 40 50 60 70 80 90
ANGLE OF INCLINATION - DEGREES
RIGHTING ARMS (FT)
KGA = 19 FT
KG1 = 19.0 FT
GG2 = .9 FT
GP = GG2 x Cos O
Cos 0° = 1.0 Cos 60°= 0.5 Cos 90°= 0
GP0°= .9FT x 1 = 0.9 FT
GP60°= .9FT x .5 = .45 FT
GP90°= .9FT x 0 = 0 FT
FINAL CURVE X
.9 FT X
.45 FT
The Cosine Correction curve is applied directly to the righting arm curve, and indicates exactly how much righting arm has been lost due to G being located off centerline, throughout the range of stability.
Connect the three points with a straight edge. The region between the Cosine Correction curve and the original curve (known as the uncorrected curve) represents the remaining righting arms.
The righting arm curve can be corrected with the Cosine Correction curve using either of the methods listed for correcting the sine curve.
Whenever weight is added or removed to/from a ship, the ship’s center of gravity rarely moves in only one direction. Fortunately, the effects are cumulative. First, calculate the vertical shift in the center of gravity and correct the statical stability curve using the sine correction. Next, calculate the horizontal shift in the center of gravity and correct the previously sine corrected statical stability curve with the cosine correction. The result is the final statical stability curve. X
Angle of List

42. Ok… Lets see both corrections applied to the same curve…
i.e. The “BIG PICTURE”

43. ORIGINAL CURVE 1 -1 2 3 4 5 10 20 30 40 50 60 70 80 90
ANGLE OF INCLINATION - DEGREES
RIGHTING ARMS (FT)
CORRECT FOR KG1
FINAL CURVE
SINE CORRECTION
COSINE CORRECTION
With the combination of sine and cosine are cumulative on stability and result in a overall lose of stability. MAKE SURE YOU APPLY WHICHEVER CORRECTION IS DONE SECOND TO THE RIGHTING ARM CURVE OF THE FIRST CORRECTION!!!
Note: Apply second correction to the first corrected curve… doesn’t matter which one you do first.

44. CLASS TOPICS
1. Vertical Weight Shifts 2. Vertical Weight Additions/Removals 3. Sine Correction 4. Horizontal Weight Shifts 5. Horizontal Weight Additions/Removals 6. Cosine Correction 7. Example Problem

45. EXAMPLE PROBLEM

46. V = 67 LT * 43 FT3/LT
V = 2881 FT3
wH2O = 2881 FT3 / 35 FT3/LT
wH2O = LT
wadded = LT – 67 LT = LT

47. This is what you plot… 2 1 3 0.62 0.16 3673.31 19.62 652.79 19.2 **
HELICOPTER
C/L
+ 11 32
352
BOAT P
- 3 54 31
162 93
AMMO
PS 13 25 27
325
351
TANK S
2881 9 10
137.79
153.1
814.79
162
597.1
STBD 1 3
3650
19.62
652.79
0.62
19.2 **
19.26
597.1
0.36
0.16
0.36
19.0
This is what you plot…

48. Review of Enabling Objectives
Describe movement of reference points.
Describe impact of addition, removal and shift of weight on stability.
Calculate KG1 and GG2.
Calculate Sine and Cosine corrections.
Apply corrections to Static Stability Curve.

49. Quiz…
If we have a weight addition (w) the notation in the KG1 equation is…?
ANS: +
What are the SINE values for 0, 30 and 90 degrees?
ANS: 0 = 0, 30 = 0.5, 90 = 1
What are the COSINE values for 0, 60 and 90 degrees?
ANS: 0 = 1, 60 = 0.5, 90 = 0

50. Instructor will now…
Assign Homework for lesson 4.2 (Stability Problems #1, #2, #3)
Read Student Guide!!