• Welcome to ScubaBoard

  1. Welcome to ScubaBoard, the world's largest scuba diving community. Registration is not required to read the forums, but we encourage you to join. Joining has its benefits and enables you to participate in the discussions.

    Benefits of registering include

    • Ability to post and comment on topics and discussions.
    • A Free photo gallery to share your dive photos with the world.
    • You can make this box go away

    Joining is quick and easy. Login or Register now by clicking on the button

Bouyancy Mystifies Me

Discussion in 'Basic Scuba Discussions' started by pauldw, Oct 19, 2019.

  1. tarponchik

    tarponchik Loggerhead Turtle

    # of Dives: 500 - 999
    Location: USA
    They do the same mistake/ As I mentioned, for any uniform object, the centre of buoyancy can not be located above the object's mass centre.
  2. Jcp2

    Jcp2 Literally virtually diving ScubaBoard Supporter

    I reclaim both links as I believe that the arrows represent force vectors rather than the true centers of buoyancy and mass, but am willing to listen to an alternate opinion rather than being socratically beaten into submission by a pedant.
  3. tarponchik

    tarponchik Loggerhead Turtle

    # of Dives: 500 - 999
    Location: USA
    OK, here is the answer. The lowest potential energy would be not when the center of gravity is as low as possible, but when the distance between the centre of mass of the object and centre of mass of the displaced water is the shortest.

    So, going back to the log, assume the ratio of the log length (L) to diameter is 10. In the vertical position, the distance between the two centres would be 1/4 of L for a log with density of 0.5, and in the horizontal position the distance would be something like 1/60 of L. So the log floats horizontally.

    Now let's make our log square in cut. We already know that it will float horizontally for the same reason as the log above, but will it turn face up or edge up? Do your math and you'll see that it will float edge up. The mass centre of the square log will be located at the water level in both cases; however, the mass centre of the displaced water will be positioned higher when the edge is up, and this makes a difference.

    Finally, lets shorten our square-cut log until it becomes a cube. Following the same math as above, we conclude that the uniform cube will float vertex up.
    eleniel, peterak and BlueTrin like this.
  4. tarponchik

    tarponchik Loggerhead Turtle

    # of Dives: 500 - 999
    Location: USA
    Sorry if that's what I sound to you.
  5. BlueTrin

    BlueTrin ScubaBoard Supporter ScubaBoard Supporter

    # of Dives: 100 - 199
    Location: London
    This link is interesting as well for this problem.
  6. EFX

    EFX ScubaBoard Supporter ScubaBoard Supporter

    # of Dives: 200 - 499
    Location: North Central Florida
    Now I'm going to ruin everyone's fun with some math and an example problem:

    If the dive kit doesn't change we only need to find the change in buoyancy due to changes in body volume and weight. We know the dry weight changed from 210 lbs to 190 lbs. If we can find the volume we can calculate the buoyancy.

    B = VD - W

    According to the equation the buoyant force B is the difference between the weight of the water displaced by the object (VD) and the dry weight (W) of the object. If B is zero the object is neutrally buoyant and will neither sink nor float. If B is negative (-) the object will sink and if B is positive (+) the object will float.

    Let's calculate the buoyancy (B) for both dry weights and see what the difference is. We can find the approximate volume by dividing the weight by the density of the human body. The density is 1 g/cm3 (from http://www.ehow.com/how_7853815_calculate-volume-person.html). The procedure to find volume is:

    1. Measure the weight in lbs.
    2. Convert the weight to grams (W lbs x 453.6 g/lb).
    3. Get volume (in cm3) by dividing weight by density (V = W/D).
    4. Convert volume in cm3 to ft3 using 28,316.8 cm3/ft3.

    W[210] = 210 lbs x 453.6 g/lb = 95,256 g
    W[190] = 190 lbs x 453.6 g/lb = 86,184 g
    V[210] = 95,256 g / 1 g/cm3 = 95,256 cm3 x 1 ft3 / 28,317 cm3 = 3.36 ft3
    V[190] = 86,184 g / 1 g/cm3 = 86,184 cm3 x 1 ft3 / 28,317 cm3 = 3.04 ft3

    Using the equation B = VD - W and 64 lbs/ft3 as the density of salt water the B for each weight is:

    B[210] = 3.36(64) - 210 = 5.04 lbs.
    B[190] = 3.04(64) - 190 = 4.56 lbs.

    Dropping 20 lbs in body weight lowers your buoyancy by 0.48 lbs (a 9.5% change). This suggests reducing your lead weight by only 1 lb. If you are interested in the buoyancy in fresh water use 62 lbs/ft3 for the density of water. (Just for comparison B[210] in fresh water = -1.68 lbs).

Share This Page