Interpreting Stability Curves

Discussion in 'Stability' started by JCD, Sep 27, 2007.

  1. JCD
    Joined: Jul 2006
    Posts: 359
    Likes: 3, Points: 0, Legacy Rep: 36
    Location: Coney

    JCD Follow the Bubbles!

    Yes sir...

    Pulled up the mouse and got busy sealing. It became a chain reaction of fixing and others would pop up, but I got it to 7 leaks way up high and I ran the test. Tee hee, I'm so happy right now. :D :D :D Here are the results.

    KN sin(ø)
    0.0º 0.000
    2.0º 2.632
    5.0º 6.269
    10.0º 8.424
    15.0º 8.448
    20.0º 8.416
    30.0º 8.202
    40.0º 7.835
    50.0º 7.392
    60.0º 6.948

    This is of course all the same conditions as before. Now I have to try to spend some time figuring out what the hell that KN sin(ø) is trying to tell me and then spend some time on the PITA leaks.

    If the above is correct, and I have to multiply the displacement by above, then it sure as hell is higher. It is now 67, 584#'s. I think. Damn, that sounds like a lot.

    Thank you one...thank you all...but I gotta give a special yell to Verbertus. It was exactly as you said!

    J
     
  2. marshmat
    Joined: Apr 2005
    Posts: 4,127
    Likes: 148, Points: 0, Legacy Rep: 2043
    Location: Ontario

    marshmat Senior Member

    I was confused as hell about the KN sin(ø) thing at first.
    Martjin (Freeship's developer) clarified in a thread on here a while back that N is a sort of virtual metacentre valid only at the exact current angle of heel. The diagram he used to explain it is below.
    To get the GZ curve from the KN sin(ø) points: At each point,
    KN sin(ø) - KG sin(ø) = GZ (or so he said...)
    or put another way,
    GZ = GN sin(ø)
    GZ = (KN - KG) sin(ø)
    GZ = KN sin(ø) - KG sin(ø)
    I've never tried Freeship's crosscurve algorithm on a cat. For monos, I have a sheet in my weights/moments spreadsheet set up so that I can paste in the Freeship output and it will make and graph the corrections using the results of the weights/moments table.
     

    Attached Files:

  3. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    Now I realize: there is a notation problem. Freeship notates the "false" Metacenter as "N" while the general understanding is to notate it as "M" (Even if it is not the real metacenter). Now it makes sense.
     
  4. marshmat
    Joined: Apr 2005
    Posts: 4,127
    Likes: 148, Points: 0, Legacy Rep: 2043
    Location: Ontario

    marshmat Senior Member

    Yep, that's more or less it Guillermo. IIRC, the metacentre "M" is calculated for an infinitesimal heel, ie. the boat is almost level. The point "N" is found by the same procedure as "M" is, except that "N" is calculated for an infinitesimal deviation from a given heel angle.
    Sound about right?
     
  5. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    The intersection with the longitudinal plane of simmetry of the vertical coming up from the center of buoyancy B, uses to be noted "M" and no "N". This last is the point at the intersection of such vertical with a perpendicular to it from K (see my attached image in post 13). I think terminology should be universally agreed to, not to confuse people.

    I have been searching internet to post an image of the metacentric curve or evolute, but I've found only a very poor one. Instead, I've found this nice page (A pity images are not working): http://www.1911encyclopedia.org/Shipbuilding
     
  6. JCD
    Joined: Jul 2006
    Posts: 359
    Likes: 3, Points: 0, Legacy Rep: 36
    Location: Coney

    JCD Follow the Bubbles!

    Ahoy there...

    I have a comment. :confused: What?:confused:

    Would this now mean that I don't multiply displacement by the KN to get my figures? Can someone give a detailed example for calculating my RM with my correct curves above? I want to find the answer myself, I just need to know the procedure for getting it. Is this solution method restricted to Freeship only? I'm confused.

    Thanks
    J
     
  7. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    No problem J,
    In your case you must do:
    GZ = KN sin(ø) - KG sin(ø), as marshmat explained. But previously you need to know or estimate KG.
     
  8. JCD
    Joined: Jul 2006
    Posts: 359
    Likes: 3, Points: 0, Legacy Rep: 36
    Location: Coney

    JCD Follow the Bubbles!

    Guillermo,

    I dont see KG sin(ø) in any of the program calculations. If I had to estimate KG prior to this, do I now do that again or will I find it in the program? Should it be in there as part of the solution? This little jam right here takes it to a different program level.

    Thanks
    J
     
  9. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    The program will never give you KG, unless it has a module to calculate it from an inclining experiment. If your program does not have that module (which it seems not to have), or it has it but you are in a designing process (not working with a real boat to which perform an inclining experiment), you have to estimate KG by a careful weights and centers of gravity calculations, or then take a value close to a known one from a similar boat or series of boats (less precise). This last is what uses to be done in the early stages of designing.
    So, you should estimate KG by the best way you can, and then build up the GZ curve using the formula marshmat posted. An spreadsheet will do the job easily.
    Cheers.
     
  10. JCD
    Joined: Jul 2006
    Posts: 359
    Likes: 3, Points: 0, Legacy Rep: 36
    Location: Coney

    JCD Follow the Bubbles!

    Guillermo, Okay.

    What I was doing is taking estimated measurements from the Hulls program before I exported the file to the FreeShip program because that program did the actual inclinitation of the hull.

    Would it be more accurate to solve for the trigonometry knowing the KN sin(ø) and the Vertical Center of Bouyancy and the Vertical Center of Gravity at those angles?

    I know it's trying to take a shortcut, but I'm trying to see if I can use known values in the program to solve more accurately.

    Thanks
    J
     
  11. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    I'm somewhat lost with your questions.:confused:

    Knowing KN sin (ø) (in Freeship's notation), you need the Center of Buoyancy position for nothing. Further on, the position of the center of gravity (for this discussion's purposes just what we have been calling KG), does not depend of the angle of heel (free surfaces in tanks effect aside).

    You must estimate KG and then use that value to calculate the GZ curve using the simple formula given. Probably the program will obtain the curve by itself once the KG has been input, but if needed to be done 'by hand' it's also a very, very simple calculation. I'm afraid there are not shortcuts for this.

    Probably I'm not explaing things properly or I'm not understanding you. Excuse me if that's so.

    Cheers.
     
  12. JCD
    Joined: Jul 2006
    Posts: 359
    Likes: 3, Points: 0, Legacy Rep: 36
    Location: Coney

    JCD Follow the Bubbles!

    Hola Guillermo,

    No, no, no. You explained it correctly and clearly sir. I put bouyancy in there by accident and it appears I may have written it right once and wrong once. Sorry. I didn't even check my own post. I have actually already performed calculations. Here they are.

    I know:
    VCG (KG)= 3.33 feet.
    KN sin(ø) = 6.25 @ 20 Degrees @ 2.612 Tonns or 5850#'s.

    Again I'm not sure why it happens at 20 degrees instead of 10, but that is the maximum KN sin(ø) for the maximum displacement so I'll solve for that.

    GZ = KN sin(ø) - KG sin(ø)
    = 6.25 - (3.33 x .342)
    = 6.25 - 1.14
    = 5.11

    RM = 5.11 x 5850#'s
    = 29,894#'s

    Now I'm to the point where I have begun to calculate the required mast sections modulus. Here it is.

    29894#'s x 12 = 358728 in #’s
    Aluminum: yield stress 60,000psi,
    358728 / (60,000/3) = 17.9364^3 = Required Section Modulus

    Please feel welcome to correct me on any of the above if I missed something or if you believe the program may still be deceiving me.

    Thanks
    J
     

  13. Guillermo
    Joined: Mar 2005
    Posts: 3,644
    Likes: 185, Points: 63, Legacy Rep: 2247
    Location: Pontevedra, Spain

    Guillermo Ingeniero Naval

    Hola J,
    For mast section and other rig calculations I recommend the reading of Principles of Yacht Design, or the like.

    Cheers and good luck!
     
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.