Primary stability considerations in kayaks

Discussion in 'Stability' started by cthippo, Feb 28, 2012.

  1. gonzo
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    gonzo Senior Member

    We are talking about resistance to capsize. That is what the OP asked about.
     
  2. NoEyeDeer
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    NoEyeDeer Senior Member

    Yes, I realise that stability is not just proportional to the second moment of area of the waterplane. I also realise that, assuming we're ignoring structural considerations for this exercise, the immersed volume will stay the same. The way I see it is this:

    1/ Doubling length gives double I.

    2/ V stays the same.

    3/ From 2/ we can assume that draft is halved. I'm assuming we're going with a basic midship section that has vertical sides, just for simplicity here.

    4/ If I is doubled and V stays the same, then BMT also doubles.

    5/ Presumably, since we're ignoring structural stuff here, we can say CG stays at the same height above the bottom of the hull.

    6/ Net result has to be an increase in GMT on the same displacement.

    7/ Increased GMT will translate directly to an increased righting arm at any smallish angle of heel.

    8/ Since the stability is a product of an increased righting arm and the same displacement, stability must increase. It wont necessarily double, depending on the height of the CG, but it will increase.

    Do you disagree with any of this?

    ETA: Alternatively, if you wish to go with doubling the displacement when you double the length, then in that case BMT would stay exactly the same. That means righting arm would stay exactly the same, but since displacement is doubled the product of the righting arm and displacement is also doubled, and since the product is what determines stability then stability will also double.

    Note that here I'm assuming "stability" means righting moment, not just righting arm. I believe this is the conventional way of measuring stability.
     
  3. Ad Hoc
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    Ad Hoc Naval Architect

    1) Yes
    2) Not unless you reduce the draft accordingly to maintain a constant V
    3) Ok..
    4) Not possible.
    Since if I is doubled, either the V has changed or the draft has changed thus, either V or draft shall effect the final BM.
    5)No we cannot assume that. There is more than just structure in a boat. And to which KG are you referring too..since #4 is not possible.
    6) Please define which parameters have changed..unclear.
    7) See #6
    8) See #6

    Yes, see above.

    You’re mixing up changes that either can or cannot occur and calling it one change only.
     
  4. Ad Hoc
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    Ad Hoc Naval Architect

    So, if beam remains the same yet you double the length, and if you double the displacement the scale is all wrong.

    Since I x L..increase in “I” by a factor L

    V (Lets call it for simplicity LxBxT) , so V new is 2 x LBT.

    Thus BM = I xL/ (LBT x LBT) = 1/[L x(BT)^2)] change which is a decrease.
     
  5. J Feenstra
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    J Feenstra Junior Member

    I guess we all can agree that a Beam increase gives more stability per % then an increase in length. As I recall, Cthippo wants to carry the kayak on the shoulder. With a long kayak, that will give funny situation's when turning in a crowd of people.
     
  6. NoEyeDeer
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    NoEyeDeer Senior Member

    Hold on a second. How can you say 4/ is not possible when you've just said ok to 3/?

    3/ is the halving of draft, so yes "if I is doubled, either the V has changed or the draft has changed" is right, and you have already agreed that for this thought experiment V stays the same.


    Ok, why is 4/ not possible? If you agree to 3/ then 4/ follows logically.


    I have already defined which parameters were changed in my original list. If you want it repeated, I'm assuming that length is doubled and displacement is held constant, which results in draft being halved. No other changes.
     
  7. NoEyeDeer
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    NoEyeDeer Senior Member

    We're just going over basic principles here. Like Gonzo, I wondering why you think that doubling the length of a hull wont increase its stability, since according to what I know it should. I just provided the second scenario (with 2x displacement) as an alternative to the first scenario (constant displacement). In both cases the righting moment will increase, AFAICT.

    Yessssss. But I think this is going to lead to problems with notation later.

    Ok, just above you apparently wanted to use L as a scaling factor, and now you're apparently wanting to use it as an actual length. Your position isn't clear here.

    Ok, hang on. How do you get that formula? Taking the second part of it:

    I xL/ (LBT x LBT)

    If I understand your notation correctly, L is a dimensionless scaling factor for length, B and T are in metres (I'm assuming we're using metric) and I is in metres4. This means the second part of your formula gives a dimensionless output for BM, since (BT x BT) would give units of metres4. That can't be right, since BM is always measured in metres.

    If L is not a dimensionless scaling factor, things get even worse. If you are using L as an actual length, then you don't multiply the original I by L to get the new value of I. You would multiply the original I by Lnew/Loriginal, which in this case is equal to Ix2.

    In any case, why would you want to square beam and draft? As I understand it, BM is equal to the second moment of area of the waterplane (units are metres4) divided by the immersed volume (metres3), which of course gives an answer in metres.

    AFAICT, the problem with the second part of your formula is that by calling L a dimensionless scaling factor, you have effectively removed the actual length in metres from your formula. You then try to re-introduce it somehow but get it wrong. I may not be an NA, but I can certainly handle basic equations. I think you should go over your notation again and re-think things.

    Furthermore, you seem to have changed the formula halfway through your reply:

    I xL/ (LBT x LBT) is most definitely not the same as 1/[L x(BT)^2)]

    If you are using L as a dimensionless scaling factor then the first section is dimensionless, while the second has units of metres-4.

    If you are using L as an actual length in metres, the first section has units of metres-1, while the second has units of metres-5.

    Either way, this is obviously incorrect for an equation, unless I am very much mistaken. If I am mistaken, please show the derivation.

    My understanding is that it would go like this:

    L = actual waterline length in metres.

    B and T we agree on.

    Therefore, after doubling the length the formula would be BM = (2 x I)/(2 x L)(B x T), meaning BM would stay exactly the same as the initial (half as long) hull. However, since displacement is doubled and righting arm is the same, righting moment should double.
     
  8. gonzo
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    gonzo Senior Member

    Basic kayak design dictates that if you make it longer it can also be flatter. Also, it can have less rocker. I used the example of two kayaks end to end as an illustration.
     
  9. Ad Hoc
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    Ad Hoc Naval Architect

    Wrote #33 after 2 glasses of wine with dinner and now just woke and not thinking straight (hence deleted incorrect previous post) ..ok, here goes again, I’ll double check my maths this time…haha

    Doubling length = L + L or 2L

    New I becomes (L+L)xB^3/12 =2LxB^3/12 = I x 2 or 2I, quite right

    Let Displacement is LxBxT (ignoring CB and density for simplicity), for ease = V

    Doubling the displacement is 2 x( LxBxT), BUT, if the beam and draft remain the same, it cannot be 2x(LxBxT), since B & T do not scale with L, since your premise is keeping the draft and beam the same, yes?...otherwise BM shall decrease.

    So, assuming displacement doubles, where V = XSA x L = (BxT) x L
    Thus new V= 2L x (BxT) = 2V

    So new I = Ix2/2V = same no change.

    I think I’ve had enough strong tea to wake me up now and think clearly
     
  10. Ad Hoc
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    Ad Hoc Naval Architect

    Well on the face of it, seems plausible. But, doubling the length of a boat does not imply the KG remains the same. I have never known it to be the same; KG always increases. Thus GM is less. (See my bit about the Series 64 hulls, good example, same result).

    So, for small angles, one could argue that it is better, but is the range of stability the same, with an increase in KG....unlikely. Which then comes down to what is your criteria to measure "stability"...whole range, amount of stored energy....max angle of heel when a moment is applied, angle of vanishing stability and so on.

    I should not drink wine when i am posting replies...hahaha :D:eek:
     
  11. gonzo
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    gonzo Senior Member

    For taking photos only the first degrees of initial stability are really important. On kayaks, the body of the person is the larger percentage of the displacement. They are very different from ships, because they have movable ballast that is probably 70% of the displacement.
     
  12. Ad Hoc
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    Ad Hoc Naval Architect

    I did a stability report for a simple box like hull car ferry several years ago. I dug it out and it is an easy task, once in, to double the length and keep the draft the same and double the displacement, as it was ostensibly a box.

    Below is a plot of the GZ and you can see the effect.

    effect of double L & displ.jpg

    I have made an allowance for the increase in Lightship and for ease and erring on the pessimistic side, kept the KG the same (but usually increases). The total full load is double and the KG new full load is much lower. Yet as you can see the differences is very very minor. So close one could call them the same.

    Principal particulars here (output from program):

    Displacement 90.000 tonnes
    Longitudinal Centre of Gravity 0.000 metres
    Vertical Centre of Gravity 2.500 metres
    Transverse Centre of Gravity 0.000 metres
    Equilibrium GM 13.406 metres
    Equilibrium Heel Angle 0.000 degrees
    Equilibrium Draught 0.483 metres Equilibrium Trim Between Marks 0.000 metres by the bow
    Angle of Vanishing Stability 55.2 degrees to stbd -55.2 degrees to port
    Maximum GZ 1.948 metres to stbd 1.948 metres to port
    Maximum GZ Angle 13.4 degrees to stbd -13.4 degrees to port

    New hull (double L and displacement):

    2 L displacment
    Displacement 179.860 tonnes
    Longitudinal Centre of Gravity 0.000 metres
    Vertical Centre of Gravity 2.230 metres
    Transverse Centre of Gravity 0.000 metres
    Equilibrium GM 13.688 metres
    Equilibrium Heel Angle 0.000 degrees
    Equilibrium Draught 0.483 metres Equilibrium Trim Between Marks 0.000 metres
    Angle of Vanishing Stability 59.3 degrees to stbd 59.3 degrees to port
    Maximum GZ 2.014 metres to stbd 2.014 metres to port
    Maximum GZ Angle 13.9 degrees to stbd 13.9 degrees to port
     
  13. gonzo
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    gonzo Senior Member

    Kayaks are completely different animals though
     
  14. Petros
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    Petros Senior Member

    so it added 4 degrees to max stablity angle.

    making it longer than will increase stablity, a little.

    I am sure that Chippo's eyes have glazed over long before now, how about we get back to making a recommendation for him on what he should build.

    I like the stability of a simple greenland type hull, for your size I would recommend about 23 inch beam, fairly flat bottom, 18 ft long, about 4 inches of rocker. I personally would make the bow and stern more vertical to increase the water line length. and to improve the stablity you can also make the bow and sturn more "full" at the gunwales rather than the slim pointy hollow bow and stern (this gives you more volume inside the hull as well).

    If you stick to the basic traditional size and shape, you will end up with a kayak that paddles lightly, behaves well, and is pleasing to the eye.
     

  15. NoEyeDeer
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    NoEyeDeer Senior Member

    One thing you missed. The displacement is now double the original, which means that even though I and righting arm are the same, the righting moment is also doubled, which means stability is doubled.
     
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