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Daily Question for Feb. 18, 2018

Your vessel's draft is 16'-00" fwd. and 18'-00" aft. The MT1 is 500 ft-tons. How many tons of water must be shifted from the after peak to the forepeak, a distance of 250 feet, to bring her to an even draft forward and aft?

A. 52 tons

B. 50 tons

C. 48 tons

D. 24 tons

Comments

  1. UnknownFebruary 18, 2018 at 2:27 PM

    Heck if I know.

    ReplyDelete
  2. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERFebruary 18, 2018 at 3:43 PM

    The first thing you are being tested on in this question is if you can calculate the vessel's trim given its forward and after drafts. So you need to know that trim is simply the difference between the forward and after drafts. In this case it is 2 feet or 24 inches by the stern or aft because the after draft is deeper than the forward draft.

    The second thing you need to know is what does "even draft forward and aft" mean. It means no trim or zero trim.

    The third thing is to determine the Change of Trim = 2 feet - 0 feet = 2 feet or 24 inches.
    (This can be confusing because the vessel's initial "trim" and the "change of trim" required are the same.)

    In previous question we have discussed "Change of Trim = Trim Moment / MT1". In this question we are given a value of MT1 = 500 ft-tons/inch, but notice they left out the "/ inch" in the question. (I guess this is another test of what you are suppose to know.)

    I have also discuss Moment before but let me refresh you that a Moment is simply something time a Distance in this case the Trimming Moment will be equal to the weight we need to find multiplied by the distance of 250 feet between the FPT and APT.

    So let us put it all together and find the answer to this question.

    Change of Trim = Trim Moment / MT1
    Where:
    MT1 = Moment to Trim one Inch = 500 ft-tons/inch
    Change of Trim = 24 inches Note: you must use inches to keep the units correct.
    Trim Moment = the weight we need to find multiplied by the distance of 250 feet

    Change of Trim = Weight x distance / MT1
    Solving for Weight
    Weight = Change of trim x MT1 / distance
    Weight = (24 inches x 500 ft-tons/inch) / 250 feet = 48 tons
    Please note how all the units cancel out and the answer is in tons!
    The correct answer is C. This is a relatively simple problem but there is a lot knowledge being tested in addition to being able to solve an equation.

    ReplyDelete
  3. UnknownFebruary 18, 2018 at 4:40 PM

    Thanks again George, I was over thinking this one.

    ReplyDelete
  4. UnknownFebruary 24, 2018 at 5:59 AM

    This one makes sense
    ( change in trim needed * MT1) / distance or (24*500)/250=48

    Hold on here.
    OK I have a question for the commissioner.

    Mr. George.

    I have run into something loading water on a supply boat. The stern was higher because the stern tanks where empty. I needed to bring her flat by loading water aft only.

    so lets play with this one.
    Lets say we started at 18' fwd and 16 ft aft and desire to bring her even keel by loading water in the aft tank only. I'm thinking one would need almost double the weight on one end since we are not shifting the weight from one end to the other.

    ReplyDelete
  5. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERFebruary 24, 2018 at 9:32 AM

    If you are adding weight not shifting it I would look at it this way:

    1. Parallel sinkage = weight / TPI

    2. Change of Trim = weight x distance from LCF / MT1

    ReplyDelete

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