Calculating Wind Load on a Katamaran

Discussion in 'Stability' started by Heimfried, Dec 19, 2017.

  1. Heimfried
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    Heimfried Senior Member

    When I calculated the heeling moment of the wind load of "my" houseboat katamaran, I was a bit surprised about the small figures resulting from moderate wind speeds. Because the boat is used only on sheltered inland waterways (small rivers and lakes in europe), I considered wind speed up to Bft 6 (27 kt, 13 m/s) for sailing and Bft 8 (40 kt, 20 m/s) on mooring.

    I would like to ask you if I did a miscalculation.
    I set c(w) = 1 because of nearly vertical house walls, the lateral area above waterline (upright position) is A(L0) = 11.3 m² and the average height of the centroid of the area plus half of the draft is h(0) = 1.02 m, air density rho = 1.25 kg/m³, wind speed v = 20 m/s.

    Heeling Moment is M = c(w) * rho/2 * v² * A(L0) * h(0)
    (I omit the angular correction term (0.25 + 0.75 * cos³phi) because it is virtual 1)

    M = 1 * 0.63 kg/m³ * 400 m²/s² * 11.3 m² * 1.02 m = 2905 Nm

    The righting Moment of this boat at a heel angle of 3.6 deg. is 2936 Nm.
    My conclusion is, this wind load would cause a heel angle of 3.6 deg.

    Hints towards a mistake are welcome.
     
    Last edited: Dec 20, 2017
  2. Ad Hoc
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    Ad Hoc Naval Architect

    Looks correct.
    Perhaps you could show your GZ curve from 0 to 5 degrees?
     
  3. Heimfried
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    Heimfried Senior Member

    Thank you, Ad Hoc.

    GZ curve:

    [​IMG]
     
  4. Ad Hoc
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    Ad Hoc Naval Architect

    Thanks..but just from 0 to 5 as requested...as the rest is not relevant and i can't confirm your values on the scale you've shown. Hence the 0 to 5 degrees only.
     
  5. latestarter
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    latestarter Senior Member

    I can not comment in general however 13 knots would be 6.7 m/s
     
  6. Heimfried
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    Heimfried Senior Member

    Here it is (GZ in mm):

    [​IMG]
     
  7. Heimfried
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    Heimfried Senior Member

    You are right, by mistake I typed the 13 twice, I corrected it. Thank you.
     
  8. Ad Hoc
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    Ad Hoc Naval Architect

    Perfect.

    Now, what is the displacement of the boat?
     
  9. TANSL
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    TANSL Senior Member

  10. Heimfried
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    Heimfried Senior Member

    Yes, DISPM = 1050 kg
     
  11. Ad Hoc
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    Ad Hoc Naval Architect

    Ok...that gives a lever of 282mm...which looks about right on your GZ curve for around 3.6-ish degrees.

    wind heeel.jpg

    The displacement is very light for a house boat, BTW!!
     
  12. Heimfried
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    Heimfried Senior Member

    Yes, originally it is designed to 890 kg, empty weight (wood, epoxi & glass only) 410 kg. But this is not achievable. The first hull (L = 6,3 m, B = 0,60 m), I finished is 85 kg instead of 56 kg.
    The construction is ply (okumé) on frame and it will be not fittet with too much installations.
     
  13. TANSL
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    TANSL Senior Member

    righting arm = (2936/9.81)/1050 = 0,285 m at 3,6 degrees although according to the curve of the graph there seems to be some discrepancy.
     
    Last edited: Dec 20, 2017
  14. Heimfried
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    Heimfried Senior Member

    Yes, but this difference is practical meaningless. If the boat heels a bit, the levers (wind attac and water resistance) are decreasing/increasing the coefficient canges and so on. The value derived from the use of this formula is kind a "better rule of thumb". The further interesting question is how much of the wind force will cause a heel of a light boat with small draft without outer keel or skeg and how much will cause leeway?
     

  15. TANSL
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    TANSL Senior Member

    I have corrected my previous post because the correct term is "righting arm" at 3.6 degrees.
    In fact the difference is very small but it is only about the concepts, which must be correct. Neither inconvenient for doing the calculations well.
    What you want now is easy if you can provide the shapes of your boat and the total weight to which you want to perform the calculations.
     
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