Stability formulas - who'll check my algebra?

Discussion in 'Boat Design' started by Stephen Ditmore, Dec 15, 2001.

  1. Stephen Ditmore
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    Stephen Ditmore Senior Member

    I recall you said somewhere that you got an error of about 12.5% comparing my formula's result to one from your program. Question: was the waterterplane of the boat in question truncated at the stern, i.e. did it have an immersed transom, or hard chines and a flat run?
     
  2. Stephen Ditmore
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    Stephen Ditmore Senior Member

    Steve, I see you've addressed some of what I said in your last posting. I'll review it more carefully tonight.

    SD
     
  3. Stephen Ditmore
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    Stephen Ditmore Senior Member

    I didn't do your last posting justice yesterday, Steve. I reviewed it last night, and what I find interesting is that if you multiply your result by the 3/pi factor I've suggested might apply to rounded shapes the result appears to come close to your ProSurf result. I'm still fond of my formulation, but what I might do is to use your "M" to plot M/L against my "n".
     
  4. Steve Hollister
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    Steve Hollister Junior Member

    Hi Steve,

    The hull shape that I tested in ProSurf had a transom stern (cut-off waterplane), a sharp bow, but the mid-portion had some curve to it.

    I think the problem is that with just the input of Awp, there is always going to be some error due to the different shapes that this area can take. What I tried to do with my waterplane of two diamonds and a rectangular mid-body was to try and improve on your basic (but good) correction of (Awp/(L*B))^2 factor. The idea is to use basic shapes, like triangles and rectangles to simulate the shape of the waterplane. That way, you can analytically calculate the moment of inertia from these basic shapes and adjust the shape to match the input Awp.

    I was a little bit surprised that it didn't do better. Perhaps, if I compared the numbers over a wide range of waterplane shapes, this formula would be better on average. The estimate was too high, I think, because it distributed too much area out near the maximum beam. I think I could probably change the shape of this "standard" waterplane to improve the results. However, I think that the only way to test the results of any formula is to methodically compare the results for a wide variety of hull shapes and see what the maximum and average error percentage are.
     
  5. Stephen Ditmore
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    Stephen Ditmore Senior Member

    I'd like to share this contribution, which was e-mailed to me:

    Dear Stephen,

    Your comments in the yacht designer's forum on the 1 degree RM put me in thought, so I went back to my ship design books and tried to find what goes on with the initial hydrostatic approximations. You are correct in the observation that your formula accuracy mainly depends on the approximation of the waterplane's transverse moment of inertia.

    If I'm right, you use the formula
    I = (Cwl)^2 * LB^3 /12, where Cwl = Awp/LB.
    This is a very logical approximation, since it uses the quantity LB^3/12 (moment of inertia of the rectangular area
    with length L and width B) and 'scales' it to the amount the waterplane fills that area (Cwl).

    Looking through literature, I came by an approximation formula for the metacentric radius :
    BM = (0.008 +0.0745 * Cwl^2) * B^2/(Cb*T), where Cb is the block coefficient Cb = (Vol. of Disp.)/LBT.

    From this formula, we can derive an approximation for I :
    I = LB^3/125 + Cwl^2 * LB^3/13.4

    I tried the two formulas on two (totally!) different cases: the YD40 model in Larsson's book, and the 2500tons DWT coaster 'Nadia', which was the ship I designed to obtain my Nav. Arch.& Mar. Eng. degree

    Naming your method 'method 1' and the other one 'method 2', the results I got for the RM 1 degree were ( in ton meters) :

    RM1...........Actual.............Method 1...................Method 2

    YD40..........0.205.........0.203, error 1%.........0.223, error 9%

    Nadia.......42.460.......35.700, error 16%.....39.190, error 8%

    It seems your formula is very accurate in the yacht case. The other formula is not very successful in both cases, but fairs better in the second case - Nadia is a cargo ship with very full lines, so this formula could be useful to the yacht designer only in the case of an old yacht with full form, or in the case of a fishing boat.

    So, I probably haven't told you anything new (!) - just thanks, for providing a good formula and giving me food for thought! I
    have just finished with obtaining my degree from the Nat. Techn. University of Athens(Greece) and doing my military service,
    and use most of my free time in cruising and racing. Yacht design is what I love most, but doesn't constitute a career in Greece
    so I' m looking elsewhere.

    Any discussion or comments are welcome.
    Thanks again

    Iason Chatzakis
     
  6. onelove099
    Joined: Mar 2002
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    onelove099 New Member

    algbera!

    hi everyone.

    does anyone know how to 'simplify' this problems?
    2a(4a²+5a-7)
    y²(2y-5y²+3)

    if anyone knows.. please e-mail me @ onelove099@yahoo.com.

    thank you.
     

  7. Mike D
    Joined: Sep 2002
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    Mike D Senior Member

    Second Moment of Area Approximations

    Hello Stephen

    I am quite new in here so forgive me if I press a wrong button or do something equally silly.

    Let me admit that I am not at all familiar with boat design, the smallest vessel I have ever been involved with was 130 feet long, the longest almost 800 feet so my idea of a boat is anything under 200 feet :)

    I was in charge of a shipyard design and estimating group for many years and a driving interest was finding ways to save time and effort. One of my long-time fascinations was the development of empirical formulas for both design and estimating - my background being naval architecture.

    So I followed this thread watching your frustration grow. Here's some help - I hope. Don't forget these worked well for ships and may need tweaking for yachts etc.

    The abbreviations are L = length, B = beam, T = draft, Cb = block coefficient, Cw = waterplane area coefficient

    I used two approximations for the vertical centre of buoyancy;
    Munro-Smith: KB = T*Cw/(Cb+Cw)
    Normand-Morrish: KB = T*[5/6 - Cb/(3*Cw)]

    The M-S formula usually gives a value about 0.01*T lower than the N-M.

    Transverse 2nd Momt of Area
    I = n*L*B^3 where n is given by

    McCloghrie: n = Cw^2/12 (which you know already)
    Munro-Smith: n = Cw^3/[2*(Cw + 1)*(2*Cw + 1)]
    Hovgaard: n = Cw*(0.0106 + 0.0727*Cw)

    The most reliable is Hovgaard's which gives the highest value of the three. With hollow waterlines the n values can be even greater and the coefficients in the expression could swing to 1/48 and 1/16 respectively. Note that the sum of the two coeffs. is always 1/12.

    Longl. 2nd Momt of Area
    I = m*L^3*B

    Munro-Smith: m = Cw/[12*(3 - 2*Cw)]
    Hovgaard: m = Cw*(.091*Cw - 0.013)

    Below Cw = 0.75 Hovgaard is more reliable, between 0.75 to 0.90 both are about the same and above 0.90 Munro-Smith is better.


    I think your approach is good (same as mine :D , must be good) and I always tried to start off with an approximation as I scoffed at calculations assuming this, that and the next thing but ended by quoting a approximate answer to seven decimals - that was submitted to me by an out-of-town expert, I returned it stating that approximations were inadequate, accuracy was essential. My problem always was that in the initial design stages there was a dearth of information and intelligent guess-work was essential.

    Munro-Smith was my nav. arch. prof many years ago, he instilled a message into his students, "There has to be a better way, your job young men is to find it!" It still amazes me how good some of these approximations can be.

    Keep up the good work, and I do hope I am not too late to help.
    Michael
     
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