Catamaran Wave Interference

Discussion in 'Boat Design' started by fallguy, Sep 6, 2019.

  1. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    Yes.

    I suppose this is why the cabin weight data is not in there?
     
    Last edited: Sep 9, 2019
  2. rxcomposite
    Joined: Jan 2005
    Posts: 1,925
    Likes: 151, Points: 63, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    AH- Here is the data and formula comparison. Stability Formula_image.jpg
    I made a scaled test model of the central hull, distributed the dummy load to scale, placed it in my water tank. The hull flipped over in either way. It was unstable.

    I placed all the scaled load (modeling clay) inside the shell deep down, placed it again in the tank. The hull became stable and relatively difficult to tip.

    I was just testing the central hull against the formula.
     

    Attached Files:

  3. Ad Hoc
    Joined: Oct 2008
    Posts: 6,037
    Likes: 338, Points: 83, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    These numbers do not make any sense.


    KB = 0.184m

    BMt = 0.12m


    Therefore KM is roughly 0.30m


    If the GMt is – 0.43m, this gives the KG at roughly 0.73m. Which also means with a negative GM she is unstable.


    Yet confusingly the tables says VCG of hull at 0.365m…this KG value is usually an estimate for the vessel in the condition she is floating at, thus, confused by the major difference. But even so, it is still negative??


    We can work out the I of the WPA from knowing the BM

    I = BM x V so, roughly 0.12 x 2 = 0.24m^4


    But a rough and ready check….I = Ib^3/12 = (9.5 x 0.84^3)/ 12 = 0.47 x 0.80 (to account for shape as it is not a box) = 0.38m^4.

    To be the same as the I from the BM the shape factor would need to be 0.63. Which doesn’t make much sense – it suggests a canoe shape fwd and aft. If I assume 2 isosceles triangles even this gives me a shape factor of 0.5, which suggest the hull is not far off 2 long pointy triangles end to end at their base. Is this the shape of your hull?


    But now looking at the WPA, the actual value of the area, the shape factor = 6.25/(9.5 x 0.84) = 0.78 (or same as Cw).

    This is not 0.63.


    So the numbers are not even close.


    Then looking at BM(long) = 12.206m

    That gives KM(long) as roughly = KB + BM = 12.40m.


    If the KG is roughly at 0.73m, this give GM(Long) roughly as 11.7m.

    Yet the hydros state 12.60m


    But as before lets work out the I from the MB -> I(long) = 12.026 x 2 = 24.4m^4


    Rough check of I = (9.5^3x 0.84)/12 = 60.0 x 0.8 = 48m^4.

    This double.


    If we use the same shape factor as calculated 60 x 0.63 = 37.8m^4, again nowhere close.


    If we assume the WPA is correct = 6.25m^2…how far apart do we need to place these hulls to get a positive GM, lets say around 1.5m based upon the KG as given


    KM = KG + GM


    So KM = 0.7 + 0.5 = 2.20m


    With KB = 0.2 -> The BM = 2.0m


    So I = BM x V = 2.0 x 2.0 = 4.0m^4 - this is 16 times greater than the single hull.


    I = 2 x a h^2


    So h = SQR( 4.0/ 2 x 6.25) = 0.32m


    Which if the hull beam is 0.84, the distance apart is less than half the beam? i.e. not much, or the fact that as soon as you provide a separation between the hulls you GM will be positive anyway.
     
    Last edited: Sep 10, 2019
    rxcomposite likes this.
  4. rxcomposite
    Joined: Jan 2005
    Posts: 1,925
    Likes: 151, Points: 63, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    Thank AH for post #32. Always the forum mentor.

    I will add this long method of computation. I use the RM method to find the weight (buoyancy of outer hull). It is tedious plugging in the numbers for every degree of heel.
     
  5. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    The waterplane is square stern canoe-like.

    I estimate the waterplane at 6; so the numbers are good; perhaps a tad high for our minute beam reduction.
     
  6. Ad Hoc
    Joined: Oct 2008
    Posts: 6,037
    Likes: 338, Points: 83, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    So....do you have what you need now?
     
  7. Dolfiman
    Joined: Aug 2017
    Posts: 890
    Likes: 399, Points: 63
    Location: NICE (France)

    Dolfiman Senior Member

    Attached my tentative estimation of the residuary drag of the project catamaran, also including the frictional drag of the hulls, + the spreadsheet (under open office) with the data and figures used for this estimation, compiled from various model test series (the sources are mentioned) and put in the same adimensional format for an easy use. The approach is then simple :
    ** At first, the residuray drag of slender monohull (when Lw/Bw > 7) , in adimensional Dr/Mg (%) in function of Froude Fn and with the length-displacement ratio Lw/D^(1/3) as unique graduation in the given Figure 1.
    ** Then, the amplification factor (1+K) due to the hulls interference in catamaran configuration :
    Dr/Mg (cata) = (1+K) Dr/Mg (monohull)
    K being both function of the space ratio S/Lw (S the transversal space between hulls axis) and of the hull ratios (Lw/D^(1/3), Lw/Bw, Bw/Tc (by order of presumed importance). K can be estimated through either proximity or slight interpolation from the given 10 adimensional Figures directly derived from the model test series.
     

    Attached Files:

  8. Dolfiman
    Joined: Aug 2017
    Posts: 890
    Likes: 399, Points: 63
    Location: NICE (France)

    Dolfiman Senior Member

    …. Oups , with the right speadsheet, in the previous one I forgot to mention the value of Lw/D^(1/3) on some Figures for K.
     

    Attached Files:

  9. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    Here are the numbers I ran. (keep in mind this is amateur hour on my end)

    lwl -> 9.504M
    bwl -> 0.841M
    cw -> 0.78, based on accepted waterplane area
    I -> cw^2*LWL*BWL^3 or 3.46M^4
    Volume of Hull -> 1.96 M^3
    BM -> I/V or 1.77M
    KM = BM + KB
    KM = 1.77 + (0.451-0.184) = 2.033
    GM = 2.033 - VCG = 1.67

    Does this mean I will never be good at math (it is wrong) or does it mean the GM will never be 2? Took me an extra day because I had a damn divide by 12 in the math to get to feet!!@@#?@@!..

    Also, what effect does width between hull centers have on this calculation? I don't get that aspect of it.
     
  10. Ad Hoc
    Joined: Oct 2008
    Posts: 6,037
    Likes: 338, Points: 83, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Nearly there. You went wrong with the I

    If this is for just one hull, then I = L.B^3/12

    You missed the 1/12.

    But, to work out the I for single hull, is not so easy without more data. As it is the "second moment of area".
    Strictly speaking the I is a bit more 'complex', but I made a simple assumption.

    Why?...well if we take a simple box shape, L x B is the area and the I = L.B^3/12.
    But the hull is not a box shape.

    And by example, if we take a triangle, its I = L.B^3/36
    I used a triangle as an example because it can also be 'bond' by a box shape the very same as the simple box to define its Length, L, and its width/beam, B. But the shapes are totally different.

    Hence why is "assumed" a certain shape function from the data.
     
    rxcomposite likes this.
  11. Dolfiman
    Joined: Aug 2017
    Posts: 890
    Likes: 399, Points: 63
    Location: NICE (France)

    Dolfiman Senior Member

    I have adjusted a numerical hull with Gene-Hull in order to address the stability issue inc. this aspect, i.e. the influence of the space S on the transversal GM.
    The numerical hull hydrostatics and ratios are as close as possible to the project one, and among output there is a very good concordance for :
    - VCG : 0,370 m / 0,365 m , which is (in Gene-Hull) the height of the center of gravity when assuming a uniform hull skin weight/sq meter.
    - VCB : -0,184 m / -0,185 m

    For the influence of the space S : is varying from 0 (= 2 hulls without space, which is equivalent to each hull stand alone) up to 3,807 m (the project cata space for S/Lw = 0,4). Results is :
    Space = 0 >>> transversal GM = - 0,40 m , the hull alone is unstable (not a surprise).
    Space = 3,807 m >> GM = 11,33 m

    The influence of the height of the VCG, just linear : GM 11,7 m with VCG = 0 to GM = 9,7 m with VCG = 2 m

    The righting arm GZ and moment RM with heel angle, done at constant space 3,807 m, weight 4054 kg and VCG is assumed at 1 m after completion of the project with a cabin >>> the maximum moment (GZ = 1,58 m ; RM = 63 kN.m) is reached for an heel angle of 12°, which coincide with the off watering of one hull.

    On the other hand, I am intriguing by your longitudinal values of LCB (5,978 m) and Longt. center of floatation (5,939 m) : seems very aft (aft of the bow very end or aft of the fore perpendicular ?) and it is not usual that this two values are the same, usually Lcenter of floatation is behind the LCB by few % Lw (5,77 % Lw for my numerical hull) . Also, the LCB is very different of LCG "skin hull" (LCB 5,978 m / LCG 4,869 m) , not usual too.
     

    Attached Files:

    rxcomposite and fallguy like this.
  12. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    The bows are knife like. It is a semi-displacement cat.

    I expect the boat may squat a bit without ballasting. More so that I went up with engines 100# and 20hp each for better alternators.

    Thanks. I will check out those numbers Dolfiman.

    Thank you as well everyone; AH.

    I spent a lot of time reviewing these against Brian Tenhaile's published work on Understanding Ship and Boat Stability.

    I gathered late last nite the errant 12 was related to feet..not meters. Alas, no.
     
  13. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    The hulls are also hardchine. Not sure if this nit matters much; the drawing is otherwise quite close.
    Your work is so cool. Thank you a bunch. So the gmt of a single hull is negative. This means that I have been telling everyone she'd flip over on her own and that is correct, ftmp.

    I am sure when the dust settles, Woods will like this paper as well.

    He posted a picture of the skoota beat up by Dorian from a drone closeup and it looks like the cabin is sort of gone, but hard to tell if the hulls and beams are intact. That is, could he limp her out of Abaco or is he insured. I don't know.. it was ground tied with a bunch of others that survived a bit
     
  14. rxcomposite
    Joined: Jan 2005
    Posts: 1,925
    Likes: 151, Points: 63, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    cat or tri hulls by itself will most likely flip over. It is a long slim canoe body. Try standing on a canoe.
     

  15. fallguy
    Joined: Dec 2016
    Posts: 1,902
    Likes: 105, Points: 63, Legacy Rep: 10
    Location: usa

    fallguy Senior Member

    I have a Gil Gilpatrick Laker canoe shortened 3" and I can't flip it.

    But it is a bear to paddle.
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.