Can anyone help with interpolation of 1400 by using this chart. 7000 is the total cargo number. Thanks
use the 1000 row and 2000 row for points 1 and 2
Thanks for helping, what answer r u getting if u dont mind, because im not getting the correct answer.
Yes +1400 interpolation is the answer i need. Thanks so much.
The answer should be 110 but dont know how to get it with interpolation
sorry I transposed values in the table in the first post
I don’t know how they get 110
Do you remember how to interpolate graphically on Loran-A charts? You can do the same thing here.
I know i cant figure it out either, once u have 110 answer u then multiply 90 Ă— the deck cargo 1.70 and u get 153, then + 110 = 263 tons which is final answer
Here’s how you do a linear Interpolation…
1000 = 350
1400 = 110
1500 = 50
1400 is 100 less than 1500, and the total difference between 1500 and 1000 is 500, so (100/500) * 300, which is the total difference between 350, 50…(350-50) = 60
60 + 50 (the tabulated value for 1500) is 110.
This is another thing that can be done very easily with a $12 Casio calculator using the stat function. It’s helpful for compass problems, celestial problems, anything with tabulated values.
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I see the error I made in the table above. Cell D2 is 330 and should be 350.
Ahh the case cracker, thank u
No, you need the stability letter for the vessel.
It’ll tell you the maximum allowable weight of the deck cargo if the vessel has rig water aboard and what the maximum CG of the deck cargo has to be. I think this is an MV Surveyor problem, I’m sure the stability letter is on the internet somewhere.
Once you do find the stability letter, find the moment of the containers:
Moment = W* D
W = (50 * .4 T) = 20 T
D = 3’H/2 = 1.5’
Moment of containers = 30 ft tons
Then, using whatever the stability letter stipulates, calculate the maximum load allowed in tons, the max allowable VCG and multiply to find maximum moments, and subtract 20 tons from the maximum allowable load.
(Max moments - 30ft tons)/(Max cargo - 20 tons) is the answer.
Thanks so much, its the S.S. AMERICAN MARINER, on ever other question it would say refer to the stability data reference book. But it didnt here so i was confused.
Hi,
You’re not working an American Mariner problem. The American Mariner didn’t load rig water it’s a C4 breakbulk ship from WWII.
A lot of stability problems will use the “Stability Data Reference Book” and talk about section 1 (blue pages), section 2 (white pages) and section 3 (salmon pages). The white pages are SS American Mariner Problems a C4 type ship, the salmon pages are tanker problems using an old T-2 tanker, the SS Nordland, and blue pages don’t name a ship but are also for a C4 type ship.
There are also stability questions using a stability letter that is provided with the problem. One is for a MODU called deep driller and another is a supply vessel, the MV Surveyor. The question you are working does not use the stability data reference book. I’m pretty confident it is the M.V. surveyor. Just search for “MV Surveyor” stability letter and you’ll likely find it.
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Ok thats a first, how was someone to know that one without help, cant thank u enough.
Ok it was the surveyor 35 tons of gross cargo is allowed with max rig water at a max vcg of 2ft.
So 35Ă—2.0=70
20Ă—1.5=30
35-20=15
70-30=40
40/15=2.66 answer
Thanks sanyca!!!
You got it, happy to help out.
Check out Capt. Steve Tarrant’s youtube channel “MMADeckPrep” he works a lot of USCG stability problems and is an excellent teacher.
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Been using it for a few days now, thanks. Hey man u better than him, i have a couple more i would like to throw at u . U definitely have alot of knowledge on the subject. Thanks again