If you can help with this issue, thank you in advance. vessel with displacement 19130t KG = 7.01m wants to unload cargo of 170 t located 6.5 sea side from centerline and Kg of cargo 12.4m. Ship is not angled on either side at this time. Wants to unload and load back cargo with heavy lift crane with KG of crane 30 m and 15m from center line. KM for 19130 is 8.47m and for 18960 8.456m. We need all the MG shifts and all the angles before during after loading and unloading of this one cargo. The problem I am facing is that when cargo returns back to original position i do not get angle 0°. Thank you!
Cansomeone help me with this problem A small autonomous sailboat is being tested in the Baltic Sea (𝜌= 1010 [kg/m^3]). It has a length of 4.2 m, max beam of 0.75 m, and a KB of 0.25 m, and its (simplified) loaded waterline is shown, see Figure 1(assume it is wall sided for small changes in draft, and KB remains the same for this problem). The weights and VCG’s of the various components of vessel are shown in Table 1. During testing the sensors get knocked off the keel due to a grounding incident (all external sensors are placed below the keel). The second moment of area for a triangle around its base is 𝐼𝑥=𝑏ℎ3/12𝐼𝑦=ℎ𝑏3/12 Table 1 Component Weight[kg] VCG[m] Lightship 150 0.25 Ballast (lead) 50 0.05 Wing-sail 35 0.8 Internal Electronics 12 0.4 External Sensors 30 -0.25 Calculate the original and the new GM (after the grounding) b)The internal electronics are not attached properly and after hitting big wave they have shifted 0.1 m towards the port side. Determine the angle of heel due to this shift
Sir at 5:42 tan theta = final list moment/final weight*final GM . Is this final value of GM = G1M? Where G1 is the final point of G considering both vertical and horizontal shits
Correct. GM is final taking both factors i.e vertical and horizontal shifts. If there is FSC than GM is also taken by applying that as well. Bottom line.... Final List=> Final GM
If the vessel heeled too much the distance of M will be closer to the G. If that happens we knew we fucked up. chapter 3 section 3.1.2.4 of IMO. It regulates that value of GM must be NOT LESS THAN 0.15metres.
Punit Garg @ 3:28 GG1 ( shift of COG due to loading ) has two components. GGH Horizontal and GGV vertical. If You consider the parallelogram ( or rectangle GGvG1GH ) it is clear GGH= GvG1 . Diagram @ 4:23 might have created confusion due to hand drawing being not perfect. But simultaneously on top right resolution of GG1 is shown in horizontal and vertcal componenets. Generally text books explain the concept with shifting only. Here the effort was made to appreciate the fact list will be affected by horizontal components i.e GGH i.e List moment and at the same time GGV which affects the GM which as also affects the list as per the formula. Trust this clarifies.
The very concept of list is that centre of gravity is off the centre line. Hence COG will change from original position in case of loading, discharging or shiifting. When a vessel is heeled, ship inclines from one side to another, with metacentre at the centre of arc. This metacentre may or may not be on the ship. These are not the mistakes you have pointed out. Kindly let us know any substantial issue which has been missed out.
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Good content
hello, simply discharging 500t from port 2.4 from CL &10.4 kg . do we need to calculate GGv for new GM . and GGH to find the angle of list ,
Wouldn't a vessel have to be grossly over weighted to capsize?
I just try to figure out how this applies to say a 50-foot sailboat.
dablo times d over capital daplo... what these daplos prefer to?
how to get even keel before docking.what is the calculation
sir ,,which video editing software is used for the preparation of this video
camtasia
If you can help with this issue, thank you in advance. vessel with displacement 19130t KG = 7.01m wants to unload cargo of 170 t located 6.5 sea side from centerline and Kg of cargo 12.4m. Ship is not angled on either side at this time. Wants to unload and load back cargo with heavy lift crane with KG of crane 30 m and 15m from center line. KM for 19130 is 8.47m and for 18960 8.456m. We need all the MG shifts and all the angles before during after loading and unloading of this one cargo. The problem I am facing is that when cargo returns back to original position i do not get angle 0°. Thank you!
ship hydrostatics pls
sir,please explain about wall sided formulae
Cansomeone help me with this problem
A small autonomous sailboat is being tested in the Baltic Sea (𝜌= 1010 [kg/m^3]). It has a length of 4.2 m, max beam of 0.75 m, and a KB of 0.25 m, and its (simplified) loaded waterline is shown, see Figure 1(assume it is wall sided for small changes in draft, and KB remains the same for this problem). The weights and VCG’s of the various components of vessel are shown in Table 1. During testing the sensors get knocked off the keel due to a grounding incident (all external sensors are placed below the keel). The second moment of area for a triangle around its base is 𝐼𝑥=𝑏ℎ3/12𝐼𝑦=ℎ𝑏3/12
Table 1 Component Weight[kg] VCG[m]
Lightship 150 0.25
Ballast (lead) 50 0.05
Wing-sail 35 0.8
Internal Electronics 12 0.4
External Sensors 30 -0.25
Calculate the original and the new GM (after the grounding)
b)The internal electronics are not attached properly and after hitting big wave they have shifted 0.1 m towards the port side. Determine the angle of heel due to this shift
inclining experiment ship you have any formtion about this
and thankx you
+mohammad khader
Request noted. Will revert soon
How is GVG1 = GGH?
But its physically can't be rectangle, because as more down from the corner as more GG' is
Sir at 5:42 tan theta = final list moment/final weight*final GM . Is this final value of GM = G1M? Where G1 is the final point of G considering both vertical and horizontal shits
Correct. GM is final taking both factors i.e vertical and horizontal shifts. If there is FSC than GM is also taken by applying that as well. Bottom line.... Final List=> Final GM
@@sailorstube3259 thank you sir
Much the metacentric is low more stable is a ship ?
Contrary
If the vessel heeled too much the distance of M will be closer to the G. If that happens we knew we fucked up. chapter 3 section 3.1.2.4 of IMO. It regulates that value of GM must be NOT LESS THAN 0.15metres.
3:16
GGh ≠ GvG1
Your Gv G1 is not equal to G Gh, don't you see that
Refer reply to Punit Garg , one year ago. Please see to that...First Comment.
They can be equal if MB and MB1 is parallel
Please how can we use the tabular form to solve this problem
You mean list problems. There are videos on this channel.Kindly have a look.
Ship sunk. GG.
Tan theta= P/B
But you took B/P.
How?
Where???
@@sailorstube3259 i got confused...everything is okay
3:20 It is a stupid mistake. They can be equal if MB and MB1 is parallel
yes he made mistake haha
Even I second that. Error in tutor
how gg1 = ggh
Punit Garg
@ 3:28 GG1 ( shift of COG due to loading ) has two components. GGH Horizontal and GGV vertical. If You consider the parallelogram ( or rectangle GGvG1GH ) it is clear GGH= GvG1 . Diagram @ 4:23 might have created confusion due to hand drawing being not perfect. But simultaneously on top right resolution of GG1 is shown in horizontal and vertcal componenets. Generally text books explain the concept with shifting only. Here the effort was made to appreciate the fact list will be affected by horizontal components i.e GGH i.e List moment and at the same time GGV which affects the GM which as also affects the list as per the formula. Trust this clarifies.
But its physically can't be rectangle, because as more down from the corner as more GG' is
@@sailorstube3259, you are wrong, Vjaceslavs Borzovs
are right
You mull the parallelogram and a trapezoid
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the center of gravity does not change. So many mistakes, and the ship is not rotating around pivot point M.
The very concept of list is that centre of gravity is off the centre line. Hence COG will change from original position in case of loading, discharging or shiifting. When a vessel is heeled, ship inclines from one side to another, with metacentre at the centre of arc. This metacentre may or may not be on the ship. These are not the mistakes you have pointed out. Kindly let us know any substantial issue which has been missed out.