Interpreting Stability Curves

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

  1. JCD
    Joined: Jul 2006
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    JCD Follow the Bubbles!

    Hello all,

    Been a while since I've been here. I have a question. I have completed another hull for a catamaran and did the calculations for the cross curves and came up with a result that was really wierd and I wanted to know if anybody had an opinion on it. I re-ran the program because I thought it could be a bug, but it came back the same. I wish I knew how to attach these things for better visual, but here are the numbers anyway so that no one lacks information to analyze. I wish I could have attached the lines plan for clearer hull shape understanding.

    My Design Hydrostatics are the following and bear in mind that this is for a single hull shell with the waterline displacement at 4000#'s or half the full designed displacement.

    Design length 34.000 [ft]
    Length over all 34.000 [ft]
    Design beam 6.000 [ft]
    Beam over all 6.000 [ft]
    Design draft 1.516 [ft]
    Midship location 17.000 [ft]
    Water density 64.000 [lbs/ft3]
    Appendage coefficient 1.0000
    Volume properties:
    Displaced volume 62.493 [ft3]
    Displacement 1.786 [tons]
    Total length of submerged body 31.783 [ft]
    Total beam of submerged body 2.872 [ft]
    Block coefficient 0.4515
    Prismatic coefficient 0.6439
    Vert. prismatic coefficient 0.5950
    Wetted surface area 111.49 [ft2]
    Longitudinal center of buoyancy 16.956 [ft]
    Longitudinal center of buoyancy -0.138 [%]
    Vertical center of buoyancy 0.960 [ft]
    Midship properties:
    Midship section area 3.054 [ft2]
    Midship coefficient 0.7012
    Waterplane properties:
    Length on waterline 31.783 [ft]
    Beam on waterline 2.872 [ft]
    Waterplane area 69.284 [ft2]
    Waterplane coefficient 0.7589
    Waterplane center of floatation 15.802 [ft]
    Entrance angle 8.926 [degr.]
    Transverse moment of inertia 35.224 [ft4]
    Longitudinal moment of inertia 4184.2 [ft4]
    Initial stability:
    Transverse metacentric height 1.524 [ft]
    Longitudinal metacentric height 67.915 [ft]
    Lateral plane:
    Lateral area 37.496 [ft2]
    Longitudinal center of effort 18.451 [ft]
    Vertical center of effort 0.866 [ft]

    Center of Gravity X = 16.638 Z= 3.922
    NOTE: All calculated coefficients based on actual dimensions of submerged body.

    Here is the hull hydrostatic for draft:
    LCB 16.956
    VCB 0.960
    Cb 0.4515
    Am 3.054
    Cm 0.7012
    Aw 69.284
    Cw 0.7589
    LCF 15.802
    Cp 0.6439
    S 111.49 sq. ft.
    KMt 1.524
    KMl 67.915

    Here are the cross curve figures for the single hull. Notice that it is pretty standard and maximum righting is at 90 degrees. Seems okay to me.

    KN sin(ø)
    0.0º 0.000
    2.0º 0.053
    5.0º 0.133
    10.0º 0.266
    15.0º 0.400
    20.0º 0.536
    30.0º 0.820
    40.0º 1.173
    50.0º 1.752
    60.0º 2.685
    70.0º 3.866
    80.0º 4.768
    90.0º 4.947
    100.0º 4.796
    110.0º 4.433
    120.0º 3.904
    130.0º 3.248
    140.0º 2.506
    150.0º 1.728
    160.0º 0.996
    170.0º 0.447
    180.0º 0.000

    Now we get to the interesting yet confusing part of this ordeal. Take a look at the cross curve for both hulls at 16 feet centerline to centerline span and total beam overall of 22 feet. The design is now at its full design displacement of 8000 #'s.

    0.0º 0.000
    2.0º 2.514
    5.0º <->
    10.0º <->
    15.0º <->
    20.0º <->
    30.0º <->
    40.0º <->
    50.0º 7.264
    60.0º 6.847
    70.0º 6.412
    80.0º 5.769
    90.0º 4.702
    100.0º 3.236
    110.0º 1.585
    120.0º -0.137
    130.0º -1.855
    140.0º <->
    150.0º <->
    160.0º <->
    170.0º <->
    180.0º 0.000

    Is this weird or what? I'm not sure what happened to the figures within the greater/lesser than brackets because the line is on the graph, but right now it doesn't matter.

    I designed the hulls in such a way that they build up significant bouyancy while they roll and list, but dammmm, if I'm reading this correctly, maximum is at 50 degrees and they are equal and neutral at 119 degrees. Does anyone concurr? I always thought that maximum would be at 10 or so degrees for a catamaran...well, once the windward hull gets unstuck...but the draft is only 1.516 feet and should be out of the water well before 10 degrees.

    How the hell did the maximum stability happen at 50 degrees and point of no return at 119 degrees? Any and all insight would be greatly appreciated.

    I hope I didn't crowd this page out...:eek:

    Thanks
    J
     
  2. Guillermo
    Joined: Mar 2005
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    Guillermo Ingeniero Naval

    Hi JCD,
    It seems to me there's a bug or mistake somewhere, because that KN sin(ø) output for the two hulls makes no sense.
    On the other hand, if you're trying to get the GZ curves, you need to know KG and then apply: GZ = KN - KG sin(ø).
    The question is, what do you need KN sin(ø) for? :confused:
    Cheers
     
  3. JCD
    Joined: Jul 2006
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    JCD Follow the Bubbles!

    Hola,

    No bug or mistake as far as I can see. But then again, I have only 2 braincells left. KN sin(ø) is just the figure that is produced by the data for the cross curves. I think it calculates as a measure of kilo-newton ft. pounds if I interpret the numbers correctly for a lever. For exampe, and correct me if I'm wrong, maximum righting moment is 7.264 and that works out to 26,127 pounds. (1000 x .2248 x 7.264 x 16 where .2248 converts newton to pounds and 16 is the center line distance of the hulls.) Anyway its part of the program. I tested it with other hulls and it looked perfect. I'm not sure why I would need KN sin(ø). The manual from FreeShip reads...

    "15.4 Cross curves.
    Stability calculations are provided in the form of cross curves. For a number of heeling angles and displacements KN sin(ø) is calculated and presented in a graph and table. If only one displacement is provided the KN sin(ø) curve is displayed. If multiple displacements are provided the graph shows the standard cross curves."

    I only used one displacement so I guess that is why KN sin(ø) is denoted.

    Any thoughts on the 50 degrees instead of 10? Take a look at the 119 degrees. Even more unbelievable.

    Thanks
    J
     
  4. Guillermo
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    Guillermo Ingeniero Naval

    There's something I don't understand. Something seems to be wrong. Cross curves do not include the sinus of the heeling angle. They are just a set of curves one for every heeling angle considered, with KN values in the ordinates axis and Displacements on the abcises axis.

    For a given displacement, intersecting the KN curves with a plane perpendicular to the displacements axis and using heeling angles on the new abcises axis (perpendicular to the other two), we get the 'Stability curve' for that displacement.

    [​IMG]

    KN (also named KZ, depending on the notation system) values are never multiplied times the sinus of the angle to get the righting moment RM.
    What we generally use to calculate RM are the GZ values, as per RM(ø) = Disp x GZ(ø). If we bring G to coincide with K (theoretically), then GZ(ø) = KN(ø), and KN becomes the righting arm (values under the 'Stability curves' in the figure).

    Righting moment for a certaing heel angle ø would just be: Displacement x KN(ø), but never Disp x KN x sin(ø)

    Perhaps there's a notation problem and I'm not understanding the thing.
     
  5. Guillermo
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    Guillermo Ingeniero Naval

    As there are missing values and probably the table is wrong, we should not asume maximum righting arm happens at 50º, or 119º is somekind of AVS.
     
  6. JCD
    Joined: Jul 2006
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    JCD Follow the Bubbles!

    Guillermo,

    The way you explain it sure sounds right to me. But as already stated, I have substantial mathematics but some may be way over my head and I stand at over 6 feet.

    If your theoretical exchange in the above calculation can apply, then that would mean that the righting arm is now 7.264 feet? This doesn't sound like an unreasonable height since the CG and CB calculation is made at the center of the span. Solving for Sin50 at approximately 9.5 feet distance from centerline to the ama produces approximately the above 7.264 arm.

    Dammm, that means the windward hull is 15.7 feet above the waterline at 50 degrees. That should cause everyone to stain their shorts. :eek:

    Multiplied by the 8000#'s of displacement then the RM would be at its greatest at 50 degrees at a value of 58,112#'s? This also sounds right.

    Thank you for correcting me in solving the calculations. It was my intention to design the structure to cantilever loads anyway, but this definitely changes how I look at the stability and more importantly the mast.

    Anyway, do you have an opinion on the 50 degrees and 119 degrees and how that could have happened? I tried it again with the hulls at different drafts, displacements and spans and it is still the same result. They get to 2 degrees, spike to fifty, and then hit zero at 119. I'm beginning to think that it may be attributable to the shape of the hull.
    I can't believe it because it would mean a "world accidental precedence". :D Good to dream isn't it?

    Any thoughts would be appreciated.

    Thanks
    J
     
  7. JCD
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    JCD Follow the Bubbles!

    What exactly is an AVS?

    J
     
  8. JCD
    Joined: Jul 2006
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    JCD Follow the Bubbles!

    Guillermo,

    The values are missing on the text file but there is a curve on the graph, if two spikes can be called that. If you want, I can zoom into the graph and get those values missing in the text and incorporate them if it will help.

    I tried rolling the hull in the original program "HULLS" and the maximum RM was at 48 degrees. This was of course a single hull without any span, but again, the lightbulb shape of the hull appears to indicate that added boyancy as designed allows her to heel further before RM=HM.

    I know, I know, the minute the hull leaves the water, the cantilever is at maximum, but do you think that the leeward hull submersion increases the RM to a higher angle before it equals the HM because of the shaped design?

    Is it possible that the cantilever can be at max when the hull flies but the leeward hull submersion would increase RM at a greater pace than HM until 50 degrees?

    Looking at the text, the spike goes to 2 degrees almost vertically and then another less vertical spike to 50. Is there a possibility that the hull was on the fly at 2 degrees and then the leewrd hull built bouyancy as it submerged until 50 degrees?

    Just theorizing.:idea:
    Could be theorizing on false data too.:(

    J
     
  9. JCD
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    JCD Follow the Bubbles!

    Anybody?:confused:
    Care to take a guess at this thing? I'm open to any suggestion nomatter how crazy.
    Anybody?:confused:
     
  10. Guillermo
    Joined: Mar 2005
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    Guillermo Ingeniero Naval

    AVS = Angle of Vanishing Stability.
    I insist: something seems (to me) to be wrong with that stability calculation.
    Cheers.
     
  11. verbertus
    Joined: Sep 2007
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    verbertus Junior Member

    Hi JCD,

    Could it be you have leak points in your hulls, not far above the static waterline, that become immersed at 5° heel, coming up above the WL again at 40°, going under water again at 140° and finally surface again at 176-180°? (boy, this is a horrible sentence..)

    You can check for leak points with Tools>check model.
    Freeship and Delftship have many attractive features, but those leak points are a nightmare, whatever the developer says.
    I also miss a normal GZ curve in these programs. I understand cross curves are good for big ship construction, but as a small boat designer I haven't found an easy way to convert them into a more useful stability curve.
    Guillermo's diagram is however a perfect tool for understanding the correlation!

    Best regards
     
  12. JCD
    Joined: Jul 2006
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    JCD Follow the Bubbles!


    Verbertus,

    You are a GENIUS!!!! Had to do that. Double thumbs up. That whole sentence set off all kinds of alarms in my head and made perfect sense.

    Anyway, I did just as you suggested. I always did check model before I mirrored the hull and it came out okay. This time, I did it after and you were 100% correct. Thank you and my hats off to you.

    Here is my next problem. Went through the manual to find out how to correct these things hands on and nomatter what I did, I have the same leaks in the same places as soon as I mirror the hull. When I move the hull to its traverse position, I have twice as many leaks.

    What have you been able to do to correct this damn "leak" so that I can put the hull through its calculation and establish if this is actually the problem that I'm having? Anything? Can the points be tricked?

    Thanks, thanks, thanks,

    J
     
  13. Guillermo
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    Guillermo Ingeniero Naval

    Well, cross curves are essential to calculate GZ curves (which are commonly known as stability curves) both for big ships and small boats. Knowing the Cross Curves for a given trim (And I'm referring to those figuring out the KN values for the several heeling angles) it's only a matter of knowing with as much precision as possible the KG value for a given load condition and then applying the equation I mentioned before:

    GZ = KN - KG sin(ø).

    You easily get the GZ curve for that load condition. asuming the used trim is the correct for that load condition and G is on the longitudinal plan of simmetry, of course.

    You can deduce KG from an stability experiment, or you can estimate it, when in a design or evaluating process.

    And again JCD, just a remark for a humble mortal: KN sin(ø) has no meaning at all (to me) in stability calculations. That's probably a typo in the program, if the notation is the same I'm used to.

    Cheers.

    [​IMG]
     
  14. JCD
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    JCD Follow the Bubbles!

    Guillermo,

    Thanks again for the info. I believe you when you say that KN sin(ø) has no value or meaning in the program as far as you understand the function. It could be a typo, it could be anything. I learned a long time ago to listen to those of greater understanding.

    I did get a possible solution that I will be working on to try to rule it out as a possible problem. Get this...the damn program has me designing leaks! I don't know if I should laugh or be embarrased...it leaks and it ain't even built yet. HA.

    Thanks
    J
     

  15. Guillermo
    Joined: Mar 2005
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    Guillermo Ingeniero Naval

    J,
    I have not worked with Freeship or Delftship programs, but believe me: They are not the only ones to present that leaky problem!
    Cheers.
     
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