Floating Position of a Triangular Pontoon

Discussion in 'Stability' started by Heimfried, Jun 26, 2015.

  1. Heimfried
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    Heimfried Senior Member

    Hello,

    imagin a single pontoon, lenght 4 m. Cross sections square to its length are equilateral triangles (each side of the triangle ist 2 m). Its density ist 700 kg/m³ (43,7 lb/ft³), the density is homogeneous, so the CG is in the geometrical center of the pontoon.

    In which position will it float and why? Edge up, flat side up or other? (Only buoyancy and weight is considered.)
     

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  2. Jamie Kennedy
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    Jamie Kennedy Senior Member

    I'm pretty sure it is other from when I looked at this last. I'll have to find a proof for this. I would find the position of static equilibrium with the greatest metacentric height - height of centre of gravity. I would calculate metacentric height, BM = I/V = (waterlinebeam^3) / (0.7)(3/16)(base^2). I would calculate height of centre of gravity, BG as the distance from the center of the displaced area to the center of the triangle.

    Or if we could just show that the first two are unstable, the answer must be other. This would be the simplest approach if you did not have to determine the stable angle of heel. There has to be at least one stable position because it floats, and it is not circular. For a triangular prism of uniform density if the answer is "other" then there are six such stable positions due to symmetry.
     
  3. Heimfried
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    Heimfried Senior Member

    Thank you, Jamie,

    you are right, it schould be calculated.

    But I'm curiuos if someone would say, the answer is easy to find without any calculation.
     
  4. Jamie Kennedy
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    Jamie Kennedy Senior Member

    I think you could show geometrically that it is unstable in the edge up and flat side up positions, without calculation. With the edge side up the centre of buoyancy is at it's lowest and wants to rise. With flat side up the centre of gravity is at it's highest and wants to fall. In both positions the waterline beam is at a local minima. From this I think one can deduce, even without drawing anything on paper, that both of these positions, or more precisely all six of them are unstable. Next we would state that there must be a stable position somewhere between this unstable positions, and so the answer is "other". It won't just roll forever in neutral stability like a log. It must vary from stable to unstable, 6 times per revolution. You can also show this is true for all specific densities between 0 and 1, but that it becomes stable on the flat at 0 density, and neutral stability in all positions at neutral bouyancy. I think this can all be done as a mental exercise, I just might not be communicating it well. So we will call it a theory and not a proof. ;-)
     
  5. Jamie Kennedy
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    Jamie Kennedy Senior Member

    I believe this is true of all symmetrical uniform density polygons, that they are never stable on their flat or on their corner. Only the circular infinite polygon is neutral. All the finite polygons vary between stable and unstable as you roll them over, and their local unstable maxima are on their edge or on their flat. Which of these is more unstable than the other doesn't really matter but it depends on the density. The point of stable static equilibrium alternates between left canted and right canted, but they are all at the same relative angle from one of the axis of symmetry, but this angle varies with density, and of course the number of sides. So it is true for the equilateral triangle, but you can generalize it for all polygonal prisms of uniform density greater than zero and less than one.
     
  6. Heimfried
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    Heimfried Senior Member

    Your considerations are nearly convincing, but the calculations I did, show partially differences.

    This is a calculation sheet showing that triangle pontoon:

    http://www.bootsphysik.de/rechner/boot_engl2.php

    (A button is named "position of rest / initial state" only "initial state" applys. If it is also the position of rest is to be found.)
     
    Last edited: Jun 26, 2015
  7. Heimfried
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    Heimfried Senior Member

    The way for the test is:

    right side, first input box, "Heel Angle" type 5 (degree) and submit with button "calculate heel / trim" (small disturbance).

    Then click "find equilibrium" (right side, bottom) and be patient.

    (Clicking the buttons marked with just an "i" leads to information.)
     
  8. Jamie Kennedy
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    Jamie Kennedy Senior Member

    So I got a heel angle of 59.9, which is close to 60 but not 60.
     
  9. Heimfried
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    Heimfried Senior Member

    The difference to 60 deg. is because of iteration method.

    After each loop of calculation the programm compares the values of transversal CG and CB. If the righting arm is "nearly" zero, it stops.

    (Some times it is oscillating and dosn't stop until you type Esc-Key.)
     
  10. Heimfried
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    Heimfried Senior Member

    The next attempt ist to lower the density of the pontoon to 400 kg/m³.

    Starting from the initial state type 2080 (kg) in the input box named "Mass to Load/Unload" and click the "unload" button under it.

    The displacement mass is now 2770 kg (volume 6.928 m³).

    And now, like the first try, input heel angle 5°, button "calculate heel / trim" and "find equilibrium".
     

  11. Jamie Kennedy
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    Jamie Kennedy Senior Member

    I was thinking that also, but I'm still not sure it isn't real.
    I will see if I can come up with an analytical solution today.
     
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