Tank Vertical Centre of Gravity
For the first subject, I’ve selected nothing regarded to Astronomical Navigation. If the point is to show what different but useful information, I simply couldn’t start there. That’s why the topic today is Tank Vertical Centre of Gravity.
A very important matter, specially to Chief Mate, but also to Officers in charge of any moment of the cargo operation, for it helps in the study of stability, giving the height were the weight is applied on that cargo, that will be added to all other VCG’s onboard and, together, tell the ship’s stability.
According to Brown’s Nautical Almanac, the way to calculate the VCG of a tank, as described by Captain Dr. Ivica Tijardovi?, can be either by a formula, or through a process using heights. The formula can be seen below:
KG(n+1) = [Dn*KGn + w(n+1)*Kg(n+1)]:[Dn + w(n+1)]
where,
KG = “Keel to Gravity”, the height of ship’s Vertical Centre of Gravity (VCG)
D = Ship’s Displacement
w = weights to load and discharge
Kg = heights of Tank’s Vertical Centre of Gravity
n = starter situation
(n+1) = finished situation
About the process, goes as follows: hn=KGn-Kg(n+1) | GnG(n+1)=[w(n+1)*hn]:[Dn+w(n+1)] | KG(n+1)=KGn-GnG(n+1), where h = vertical distance between G and g.
Don’t worry. It seems like a huge formula, but it ain’t that hard. Bear with me.
As an example, a vessel with D=6000t, KG=4m, loading full a tank of 1m height and 400mt of fresh water capacity.
KG1=(6000*4+400*0.5):(6000+400) = 24200:6400 = 3.78125m or
h=4-0.5=3.5m | GG1=(400*3.5):(6000+400) = 1400:6400 = 0.21875m, which finally,
KG1=4-0.21875 = 3.78125m
Always good to remember, that the Tank Capacity Plan gives Kg for the full tank.
That was all for today.