“Full and Down” describes the desirable but rare condition in which a vessel has all her cargo space filled and is loaded down to her Plimsoll marks. Achieving the state of “full and down” requires some arithmetic. The old examination bank of questions has three “full and down” problems, two of which require elementary algebra. The new examination bank has dropped two of these questions leaving only one relatively simple calculation problem.

Before getting to the questions, we need to define some terms. “SF” is the stowage factor of a cargo. It represents the number of cubic feet one (long) ton of a product occupies. For example, if a cargo has a Stowage Factor of 72, then 2240 pounds (one long ton) of the product will take up 72 cubic feet. The smaller the Stowage Factor, the denser the cargo. Feathers have a large stowage factor while lead has a relatively small stowage factor. A ton of lead occupies much less space than a ton of feathers.

Your vessel has a deadweight capacity of 5000 tons and a cubic capacity of 300,000 cu. ft. You are to load lead, with a stowage factor of 18, and cotton, with a stowage factor of 80. If you load full and down, how much cotton should you load? |

A. 1613 tons |

B. 2190 tons |

C. 2819 tons |

D. 3387 tons |

Answer: D |

### Solution:

Maximum Weight: 5000 tons

Available Space: 300,000 cubic feet

Let “x” represent lead

Let “y” represent cotton

x + y = 5000 (tons)

18x + 80y = 300,000 cubic feet

Translate x into y:

x = 5000 – y

Substitute this value of x into second formula:

18 (5000 – y) + 80y = 300000

90000 – 18y + 80y = 300000

90000 + 62y = 300000

62y = 210000

**y = 3387 tons of cotton
**

Your vessel's available bale cubic capacity is 625,000 and her available cargo capacity is 10,000 deadweight tons. Disregarding broken stowage, how many tons of pyrite (stowage factor - 13) and how many tons of cork (stowage factor - 150) must be loaded to be full and down? |

A. 6387 tons pyrite, 3613 tons cork |

B. 6721 tons pyrite, 3270 tons cork |

C. 7500 tons pyrite, 2500 tons cork |

D. 9133 tons pyrite, 867 tons cork |

Answer: A |

### Solution:

Maximum Weight: 10000 tons

Available Space: 625,000 cubic feet

Let “x” represent pyrite

Let “y” represent cork

x + y = 10000 (tons)

13x + 150y = 625,000 cubic feet

Translate x into y:

x = 10000 – y

Substitute this value of x into the second formula:

13 (10000 – y) + 150y = 625000

130000 – 13y + 150y = 625000

130000 + 137y = 625000

137y = 495000

**y = 3613 tons of cork
**

**10000 – 3613 = 6387 tons of pyrite**

### Solution:

In order to satisfy both the “full” in terms of capacity and the “down” in terms of weight, we are looking for the cargo that fills the space and meets the weight requirements. The space available is 500000 cubic feet. The weight is 9000 tons (10,500 tons less the 1500 tons for fuel, water and stores).

With a stowage factor of 7.1, zinc is too heavy. 500000 cubic feet of zinc weighs 70422 tons, far beyond the allotted 9000 tons. On the other hand, 500000 cubic feet of barbed wire weighs roughly 9000 tons. A good fit.

Both tallow (at 7485 tons per 50000 cubic feet) and rubber (3566 tons) fail the “down” requirement.