Announcement

Collapse
No announcement yet.

Stability Question (for all you deckies!!)

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Stability Question (for all you deckies!!)

    Is it possible for a ship to have a GM and TCG of the same value?

    Naval Arc question below:

    If, during an Inclining Experiment, a mass of 10t is transversely moved 20m
    over the main deck of a 50,000t Bulk Carrier and produces an angle of heel of 1⁰.
    What is the KG of the vessel as inclined? Take the KM to be 5.78 in this condition.

    I get a:
    TCG of 0.004m
    GM of 0.004m

    and finally KG to be 5.776m

    Can someone please let me know if this correct.
    Many Thanks.

  • #2
    if the TCG is equal to the GM then the angle of list would be 45 degrees, so you've made a mistake somewhere. Tan list = TCG/GM

    Comment


    • #3
      After shifting the GM changes , so intial GM is 0.004 . Calculate final GM & than apply.

      Comment


      • #4
        I really hope I'm not on a vessel with a GM of 0.004m. Might capsize if I sneeze...

        Also, the weight is moved transversely, not vertically, so we can assume KG has not changed, and therefore GM has not changed as we assume M to remain constant at small angles of heel

        Comment


        • #5
          I'm not very hot on inclining experiments, I was at the dentist the day it was covered in class, however I do know

          GMf = Weight x Shieft x length plumb line / Δ x deflection.

          I'm confused as to what you mean by TCG, The centre of gravity is a point in a vessel, therefore it's height must be identified relative to something like the keel?? I hate stability, it's my worst subject at school

          Comment


          • #6
            TCG is Transverse Centre of Gravity. It's how we're taught to calculate small angles of list down here. if you move a weight athwartships then it affects the TCG by (weightxdistance moved)/Displacement. I assume the ship was upright to start with so it's TCG was 0. TCG and GM make a right angle triangle with the angle of list opposite the TCG. If we know the TCG and the angle of list, then we can calculate the GM, and therefore the KG

            Comment


            • #7
              I think i've got it, i'll get my calculator

              Comment


              • #8
                Ok, so GGH = Weight X Shift / Displacement
                = (10x20)/ 50,000
                = 0.004

                GM= GGH/Tan Angle
                = 0.004/Tan 1?
                =0.23m

                KM 5.78
                KG 5.55
                GM 0.23

                KG = 5.55 by my workings.... but like I say, I am terrible at stability Looks like the mention of an inclining experiment is a curve ball.
                Last edited by Martyboy; 15 November 2013, 01:18 PM. Reason: Spelt angle as angel and wanted to avoid the "hot fuzz" references :)

                Comment


                • #9
                  Originally posted by HarmlessWeasel View Post
                  TCG is Transverse Centre of Gravity. It's how we're taught to calculate small angles of list down here. if you move a weight athwartships then it affects the TCG by (weightxdistance moved)/Displacement. I assume the ship was upright to start with so it's TCG was 0. TCG and GM make a right angle triangle with the angle of list opposite the TCG. If we know the TCG and the angle of list, then we can calculate the GM, and therefore the KG
                  I'm not a fan of this TCG being 0 thing tbh.... I would say that G is on the centreline when the vessel is upright, as transverse stability moves vertically as well as horizontally, And if you start referring to G as zero, it may confuzzle you in the future

                  Comment


                  • #10
                    Originally posted by Martyboy View Post
                    Ok, so GGH = Weight X Shift / Displacement
                    = (10x20)/ 50,000
                    = 0.004

                    GM= GGH/Tan Angle
                    = 0.004/Tan 1?
                    =0.23m

                    KM 5.78
                    KG 5.55
                    GM 0.23

                    KG = 5.55 by my workings.... but like I say, I am terrible at stability Looks like the mention of an inclining experiment is a curve ball.
                    Aye Exactly what I got. I was going to let the OP work it out though :P

                    Originally posted by Martyboy View Post
                    I'm not a fan of this TCG being 0 thing tbh.... I would say that G is on the centreline when the vessel is upright, as transverse stability moves vertically as well as horizontally, And if you start referring to G as zero, it may confuzzle you in the future
                    Not sure how transverse stability can move vertically, seeing how it's called transverse... We have KG to describe the vertical component of G, TCG for the transverse, and LCG for the longitudinal. This breaks up the position of G into three vectors so you can deal with each separately. It's how we're being taught at the moment so I'll stick with it for now until I come unstuck later on

                    Comment


                    • #11
                      Yea, i'm up on stability :P

                      Stability is generally broken down into transverse and longitudinal, KG's, GM's KM's etc are classed as part of transverse stability, with GGH and GGV used to describe the vertical and horizontal movement of G, where as LCF, LCG etc are part of longitudinal.

                      Found a definition-

                      "Transverse stability is the measure of a ship’s ability to return to an upright position after being disturbed by a force that rotates it around a longitudinal axis."

                      The MIN with the OOW syllabus for SQA lists all of this under transverse:

                      Transverse Stability

                      a) Calculates shift of G, vertically and horizontally after loading/discharging/shifting a weight
                      b) Calculates final KG or GM by moments about the keel after loading/discharging/shifting
                      weights including appropriate Free Surface Correction
                      c) Calculates distance of G horizontally from the centreline by moments about the centreline
                      after loading/discharging/shifting weights
                      d) Calculates the effect on stability of loading or discharging a weight using ships’ gear
                      e) Calculates the angle of list resulting from 3 a), 3b), 3c) and 3d)
                      f) Explains the difference between list and loll and methods of correction
                      g) Explains the consequences and dangers of a free surface
                      h) Explains that the free surface effect can be expressed as virtual rise of G or as a free
                      surface moment
                      i) Describes the effects on free surface of longitudinal subdivision of a tank


                      Fun times!!!

                      Comment


                      • #12
                        pretty sure I can all of those except f and i, so it sounds like we've been taught different methods to do the same thing. What are GGH and GGV apart from something horizontal and stability?

                        and I obviously didn't meant to suggest you weren't up on stabiltiy :P I'm not about to say I know better than an OOW...

                        Comment


                        • #13
                          GGH and GGV is just the shift of the transverse centre of gravity horizontally and vertically

                          Comment


                          • #14
                            In this case, GGh makes more sense since you can't be certain that the ship's initial G was on the centreline, and so can't definitively say that the TCG is 0.004 m, just that the shift in G ((w*d)/Displacement) is 0.004 m.
                            sigpic
                            Hello! I'm Chris. I'm away a lot so I'm sorry if it takes me a while to reply to messages, but I promise I'll get back to everyone. If it's urgent, please email me directly at [email protected].

                            Need books, Flip Cards or chartwork instruments? Visit SailorShop.co.uk!

                            Comment


                            • #15
                              Originally posted by CharlieDelta View Post
                              In this case, GGh makes more sense since you can't be certain that the ship's initial G was on the centreline, and so can't definitively say that the TCG is 0.004 m, just that the shift in G ((w*d)/Displacement) is 0.004 m.
                              To answer the question don't we need to assume the initial TCG is 0? Otherwise the weight could have been moved from 10m Starboard to 10m Port, giving a GM of 0.11m and so a KG of 5.67

                              Comment

                              Working...
                              X