# "what is the maximum allowed vcg" How to?

I apologize in advance if this post should be in a different area of the forums. I’m taking my deck safety exam tomorrow, so I’m trying to figure this out asap

How do you solve this problem?

You have 590 tons of below deck tonnage. No liquid mud. 84 tons of cargo above deck with a vcg above deck of 2.7 feet, what is the maximum allowed vcg of the remainder of the deck cargo that is permitted? Illustration d036dg

I know how to use the diagram to determine tonnage, but not vcg. Any help is appreciated! Thanks in advance!

It would be helpful if you provided the problem word for word. Without reading the entire question, which is usually a problem, is the answer 3.12 feet?

I don’t really understand what you are saying there? Yes that is the correct answer though…c. 3.11 feet

Is the solution easily explainable on here? If so thanks in advance!

Information from the problem:

The stability letter for the Hudson says that vertical center of gravity for cargo on deck cannot exceed 3 feet. And, without liquid mud, you cannot have more than 300 long tons of deck cargo. You have 84 tons of deck cargo.

Solution: You can [B]a[[/B]B]dd another 216 tons of deck cargo[/B], as long as the average vertical center of gravity does not exceed 3 feet. Think of it this way: You can put (300 tons x 3 feet average vcg) 900 average foot-tons of deck cargo onboard. You have already put on (84 tons x 2.7’ vcg) 226.8 average vcg foot-tons. Therefore, you can [B]add another 673.2 average foot-tons[/B] (900 - 226.8) of cargo on deck.

673.2 foot-tons divided by 216 tons = 3.117 average vcg

I’m posting from my phone (no computer Internet access here) and studying from the hawsepipe cd so sorry to not post all of the information!

That’s a great, simple way to think of it. Thanks a million for the breakdown and the help!