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Weighing a ship by bouyancy

  1. Jan 9, 2008 #1
    so your company has built an enormous ship and it is being lowered into a large pool to test if it can float... another company has contacted you to buy your ship and they want to know how much the ship weighs... your company has spent all its money building the ship and you cant afford a scale... if you cannot give the company an accurate weight for the ship they will not buy it and your company will go bankrupt... the only tools you have are a meter that measures the volume of the water in tank, a gun and a basic knowledge of fluids... can you find the weight of the ship and save your company or should you use the gun to off yourself?
     
    Last edited: Jan 9, 2008
  2. jcsd
  3. Jan 9, 2008 #2
    You just need know the water level before and after the ship is lowered into the pool.
    Then multiply this difference (in meters) by the pool's surface (in m2) to get the weight in tons.
     
  4. Jan 9, 2008 #3

    DaveC426913

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    Yeah, I gotta say, Rogerio's method is bound to be simpler. The only thing you need to do is take linear measurements.

    But Rog: what if the pool is amorphous in shape? How would you measure the surface area then?


    shamrock: I don't know what kind of meter will "measure the volume of water in the pool" directly.
     
  5. Jan 9, 2008 #4

    Well Dave, I know my own weight.
    So, I would jump into the pool...
    :smile:
     
  6. Jan 9, 2008 #5
    Heh, assuming you have accurate tools, or the pool has a small surface area! Would be interesting to see about what level of accuracy you could get in that circumstance! Maybe the "volume-meter" is super accurate!

    DaveE
     
  7. Jan 10, 2008 #6
    So, at a rough estimation:

    - Average ratio of length to width of some large ships: 6.7:1
    - Assume pool is rectangular and has 10 feet of extra room around the edges (in reality I don't think they'd use such pools, but if they did they'd probably have more leeway-- but trying for best case)
    - Assume water's density is about 62.3 lbs per cubic foot (it's probably not pure water, and is likely at some temperature between 4C and 20C-- fortunately it doesn't change much)
    - Assuming you can get a 1/2 inch accuracy on your vertical measurement (no clue how realistic that is)

    Ship Length - Expected % error
    100 ft => +/- 5.4 tons
    200 ft => +/- 14.2 tons
    300 ft => +/- 26.9 tons
    400 ft => +/- 43.4 tons
    500 ft => +/- 63.8 tons
    600 ft => +/- 88.1 tons
    700 ft => +/- 116.2 tons
    800 ft => +/- 148.3 tons
    900 ft => +/- 184.1 tons
    1000 ft => +/- 223.9 tons
    1100 ft => +/- 267.5 tons
    1200 ft => +/- 315.0 tons
    1300 ft => +/- 366.4 tons
    1400 ft => +/- 421.6 tons
    1500 ft => +/- 480.8 tons

    I figured that would be too much of a variance (admittedly I haven't checked much), but in comparing a couple actual ship tonnages, it's within 0.2% or so. So that's probably reasonably accurate!

    DaveE
     
  8. Jan 10, 2008 #7

    DaveC426913

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    I'm curious about the gun.
     
  9. Jan 10, 2008 #8
    Well, using the smallest pool in the above, we can know the gun's mass to within 5.4 tons. Will that work?

    DaveE
     
  10. Jan 11, 2008 #9
    answer: well basically the gun had nothing to do with the problem... you can find the weight with laws of buoyancy... a floating object (which has a specific gravity of less than 1) experiences an upward force equal to the weight of the water that it displaces... so you figure out the volume of the water in the tank before the ship is lowered into the pool and then the volume after the ship is floating in the pool... the volume of water which is displaced is the difference in the two values... then with your basic knowledge of how much that volume of water would weigh (roughly 8 pounds per gallon) you can find the weight of the water that was displaced which is equal to the upward force acting on the ship which is equal to the eight of the ship itself!
     
  11. Jan 12, 2008 #10

    Well. it seems the shamrock's reasoning is not related to this problem, since any ship has specific gravity greater than 1...
    :smile:
     
  12. Jan 12, 2008 #11
    Anyway, the correct solution had already been stated (see the second post).
     
    Last edited: Jan 12, 2008
  13. Jan 12, 2008 #12

    RonL

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    Just a little thought to consider.
    Hulls are painted, water lines established, and depth markings are calculated before the boat or ship ever hit the water. -:)
     
  14. Jan 14, 2008 #13
    That part should be re-worded because the volume of the _WATER_ is staying the same, it's obvious what you mean but you should say it.
     
  15. Jan 14, 2008 #14

    DaveC426913

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    OK, I got another water-measurement one.

    You've been tasked with measuring the amount of water in a land-locked lake (no inlets or outlets, i.e its volume does not change over time). It has a highly irregular shape and a highly irregular bottom, the depth of which you don't know. How can you measure the amouint of water?
     
  16. Jan 14, 2008 #15
    Hmm. My first thought was that there might be a solution involving the tides of the lake, but I'm not sure if that would actually work or not. Another thought I had was to put a measurable amount of some sort of traceable chemical.

    Now, stir the lake up :) Use boats, little kids, fish, big spoons-- whatever's handy. Take random samples and test the concentration of the chemical. When your sample sizes start to be more and more consistent (at varying depths, locations, etc), you can stop stirring. At that point, you should be able to get a pretty good guess at how much water there was to start with.

    DaveE
     
  17. Feb 10, 2008 #16
    As the problem is originally stated, read the volume meter in cubic meters. I have no idea how this meter works but I don't care. Lower the boat. The water will rise as the boat is lowered, volume staying the same. Drain the water until the height is back to what it was. Now read off the new volume. Absolute Delta Volume is the mass in metric tons.
     
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