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Showing posts with label Ship Stability. Show all posts
Showing posts with label Ship Stability. Show all posts

Question From Readers # 5: Solving for UKC ( under keel clearance)



What would be the approximate keel clearance

 You are on a box-shaped vessel about to enter a river. At the mouth of the river where the relative density is 1.018 ton/m? your draft was 5.80 m. What would be the approximate keel clearance in the river dredged to a depth of 8.50 m if its density 1.005

 Formula for the new draft:

New draft = Old draft x Old density
                     New density

                = 5.80 m x 1.018
                      1.005
                = 5.9044
                     1.005
                = 5.87 m

So we are going now to subtract the depth of the river by our new draft to get the keel clearance.

Keel clearance = Depth of river - ship's new draft
                          = 8.50 m - 5.87 m
                          = 2.63 m

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Question From Readers # 3: Reduction In GM Due To Free Surface

What is the reduction in GM due to free surface

A 7000 ton displacement tank ship carries two slack tanks of alcohol with a SG of 0.8 each tank i 50 ft. long 30 ft. wide. What is the reduction in GM due to free surface with the vessel floating in sea water, SG is 1.026?

Solution:

I will try to answer your question.

 This is the formula we are going to use FSC =           r L B³          
                                                                       420 x Displacement  

1. Divide SG of liquid inside the tank by the density of water where the vessel is floating to get the value of   "r".
                            r =   0.8   = 0.78
                                 1.026

2. Proceeding now to the formula to get the reduction in GM.

                     FSC =           r L B³          
                                420 x Displacement
                           
                             = 0.78 x 50 x 30³
                                 420 x 7000
                             = 0.78 x 50 x 27000
                                 420 x 7000
                             = 1053000 
                                 2940000
                             = 0.36 ft      * as is if there is only one tank. But since there are two slack tanks,
                                  x 2             multiply it by 2.
                                0.72 ft    is our final answer.

I hope this answer to your question. If ever I am wrong in some way, feel free to correct me. And if you may, share this blog to your friends.


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Ship's Squat Formula

Find the approximate calculated squat of your vessel is proceeding to a channel not enclosed with a width of 90 meters deep and dredge surrounding depths of 20 meters. Your vessel's draft is 11 meters and beam of 27 meters, speed 7 knots and block coefficient is 0.8.

Given:

               Speed = 7 knots
Block Coefficient = 0.8

What is asked?

Ship's squat

Solution:

This is the ship's squat formula on open waters and the unit of the answer will be meter.
      
                       Squat = Block Coefficient x Speed²
                                                   100
                                 = 0.8 x 7²
                                        100
                                 = 0.8 x 49
                                        100
                                 = 39.2
                                    100
                       Squat = 0.392 m      

Again, the example above is that we get  the value of the ship's squat in open waters. You might asked then, if there's a formula for squat on open waters, what will be the formula for squat on enclosed water?

To get the value of squat on enclosed water is to multiply block coefficient by the square of speed, and then times 2, then divide the product by 100.

Here is the problem for squat on closed water. Try to solve it yourself.

A container vessel of 12,000 tons displacement is approaching her berth at speed of 4 knots. Its block coefficient is 0.78. Calculate the value of squat.

Answer: 0.25 meters

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Rolling Period Formula In Still Water

Find the still water period of roll for a ship when the radius of gyration is 6 meters and the metacentric height is 0.5 meters.

Given: 

                           Radius = 6 m
Metacentric height (GM) = 0.5 m

What is asked?    

Rolling Period in still water              

Solution:

Still water period of roll = 2π x Radius
                                       √g x GM

                                     = 2 x 3.1416 x 6 m   
                                        √9.81 m/sec² x 0.5

                                     = 37.6992 m       
                                         √4.905 m/sec²

                                     = 17.02 seconds

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Question Fom Readers # 1: How Much More Cargo Can Loaded

How much more cargo can she load in order to be at her summer load-line on reaching salt water?

One of the readers of this blog named MarCo, has submitted a question for you and me to answer his question. This is probably a response from the author's invitation, which is yours truly, who encourage readers to submit questions so that you and me may help solve their problem, so to speak. And this particular question is regarding on cargo loading.

Before anything else, I have to say that I am no expert, so I am open to be corrected. However, I will try to answer your question to the best of my knowledge. Here is the question:

A vessel has a summer draft of 11.184m, TPC 46.99tons and a FWA of 254mm. She is loaded in a berth where the water RD is 1.011 and her present mean draft is 10.98m. How much more cargo can she load in order to be at her summer load-line on reaching salt water?

Analysis of the problem:


If a ship is coming from the sea going to fresh or brackish water, what will happen to the draft is that it will increase. And this will result to decrease to ship's freeboard.

While a ship from fresh or brackish water going to the sea will decrease her draft, thus increasing ship's freeboard as in the case of this problem.

So now, we have to get the dock water allowance for our ship's summer draft. Here's the formula for getting dock water allowance:

         Dock Water Allowance = 254 (1.025 - 1.011)
                                                             25
                                              = 254 x 0.014   
                                                        25
                                              = 3.556
                                                    25
         Dock Water Allowance = 0.14224 meters

So with our dock water allowance above, it means that with regards to the density of water where our ship is berthed and loaded, we can add 0.14224 m to the summer draft of 11.184 m.

            So, 11.184 m
               +   0.14224 m
                  11.32624 m     Now our mean draft is 10.98 m. So how much more tons shall we load?
               - 10.98 m           We have to subtract to get the difference.
                   0.34624 m 
               x         100         To convert it to centimeter so that we can apply the TPC.
                   34.624 cm      Now since tons per centimeter immersion is 46.99 tons, so multiply it by
               x   46.99          
                   1626.98 tons is the additional cargo to be loaded.

I hope this answer to the question. But again, I am open to correction. So if my answer is wrong, correct me.

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Solving For The Righting Arm

A bulk carrier has a displacement of 40,000 tons and the KG is 9 meters. The KN at 10° degrees shheel is 1.87 meters. What is the righting arm at this heel?

Given:

                KG = 9 m
                KN = 1.87 m
Angle of heel = 10 degrees
Displacement = 40,000 tons

What is asked:

Righting arm at 10 degrees heel

Solution:

GZ is the common symbol of the righting arm.

GZ = KN - (KG x Sin θ)
      = 1.87 m - (9 m x Sin 10°)
      = 1.87 m - 1.56283 m
GZ = 0.307 m


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Solving For The Reduction In GM

A liquid mud tank measures 30 ft. long, 15 ft. wide, 6 ft. deep. The vessel displaces 968 tons. The specific gravity of the mud is 1.8. What is the reduction in GM if two of this tank are slack?

Given:

        Lenght = 30 ft
Width/beam = 15 ft
S.G. of mud = 1.8
displacement = 968 tons

What is asked:

Reduction in GM

Solution:

Free surface correction formula:

FSC =     r L B3                        
         (420 x displacement)

1. First we have to solve for relative density. "r" can be solve by dividing the specific gravity of the fluid inside tank by the density of water where the vessel floats. Since the problem does not specify, let us assume that it is sea water.

         r = 1.8     = 1.76
              1.025

2. We can now solve for the reduction in GM.

  FSC =     r L B3                        
            (420 x displacement)
          = 1.76 x 30 x (15)3
                (420 x 968)
          = 1.76 x 30 x 3375
              (420 x 968)
          =  178200 
              406560
          = 0.43831
                 x    2    (since there two slack tanks)
            0.87     is the final answer


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Find The Height Of The Metacenter Above Keel

When a ship of 12,000 tons displacement is heeled 5° 25' the moment of statical stability is 300 tons-meters, KG = 7.5 m. Find the height of the metacenter above the keel.

Given:

Angle of heel = 5° 25'
Displacement = 12,000 tons
              MSS = 300 tons-meters
                KG = 7.5 meters

What is asked:

Height of the metacenter above the keel.

Analysis:

Height of the metacenter is also called KM. While metacentric height is also called GM. Keep in mind that the height of the metacenter and the metacentric height are not the same thing.

To find for the height of the metacenter, we have to add GM and KG. So the formula is KM = KG + GM. In the problem we are given a value of KG, so we have to solve for the GM first, and then add it to the KG to get the value of KM.

Solution:
 
1. Solving for metacentric height (GM)

GM =  Moment of Statical Stability 
              Displacement x Sin Ɵ

       =       300 tons-meters      
          12,000 tons x Sin 5° 25'
       
       = 300  tons -meters   
          1132.77 tons 


GM = 0.26 meters

2. Solving for the height of the metacenter (KM)

KM = KG + GM

      = 7.5 m + 0.26 m
      = 7.76 m is the height of the metacenter above keel  



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Change Of Trim

A weight of 350 tons is loaded on your vessel 85 feet forward of the tipping center. The vessels MTI is 1150 foot-tons. What is the total change of trim?

Given:

Distance = 85 ft
  Weight = 350 tons
       MTI = 1150 ft/tons

What is asked?

Total change of trim

Solution:

Change of trim is equal to the product of weight times distance over MTI.

Change of trim =     weight × distance    
                           Moment to trim 1 inch
               COT =    350 tons × 85 ft   
                              1150 ft-tons/inch
               COT =   29750 ft-tons     
                            1150 ft-tons/inch
               COT = 25.86 inches

Change of Trim

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