Basic question about displacement and power

Discussion in 'Hydrodynamics and Aerodynamics' started by mcarling, Jan 4, 2016.

  1. mcarling
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    mcarling Junior Member

    Apologies if this question is terribly elementary. I think I understand but would like a sanity check.

    Consider two hulls of equal length with a high L/B ratio (over 20:1). Hull A has twice the displacement of Hull B and each has a shape optimized to minimize drag at 6 knots (well below hull speed, which anyway is probably immaterial with a high L/B ratio). If I understand correctly, the power needed to cruise at 6 knots will be approximately but not exactly double for Hull A as for Hull B i.e. approximately proportional to displacement. Correct?

    Obviously, there will be some secondary effects, for example, Hull B will have a higher L/B ratio than Hull A, which is why I don't expect the difference to be exactly double. Are there any other non-negligble secondary effects?

    Is there a formula into which I can insert displacement, length, speed, beam, draft, prismatic coefficient, etc. and get an approximation of the power needed to overcome drag?
  2. philSweet
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    philSweet Senior Member

    For narrow hulls well below hull speed, friction is the overriding concern. The wetted area of the heavy hull will be 2^2/3 = 1.587 for an exact geometric scaling (different length). For a fixed length, but still very skinny boat, the area ratio will be 1.414. This is the ratio of drag since wave drag would be less than 10% of the total.

    Also, your question sort of contradicts itself. A 20:1 L/B isn't going to the optimal shape well below hull speed. But if we just take the length and speed as given, then it's ok and we optimize the cross sections for each case. We can assume the midsection shape and prismatic coefficient will not change by much. The last assumption is that the changes in the ratios L/B and L/D don't effect the wetted surface coefficient of a really skinny boat when you scale B and D.
  3. Joakim
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    Joakim Senior Member

    That ratio assumes that the hull sections are sections of a circle up to half a circle. It's a different story, if the displacements are not low enough to enable this at around 20:1 L/B. At high displacement and constant L/B draft will almost double and area ratio will be close to 2.
  4. Ad Hoc
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    Ad Hoc Naval Architect


    The resistance, simplistically, is broken down into friction and wave making, or residuary. The residuary can be broken down further, but it shall just complicate matters for now.

    Thus each hull has each part of the resistance broken down.

    The length to displacement ratio...affects the residuary. If the hulls are identical in shape, it may be considered as a pro-rata of hull A to hull B. But how much?..that is the unknown, since hull B's total resistance will be different. As its WSA will be different and thus its total resistance will be different. Therefore a simple pro-rata of the LD ratio cannot be assured, thus the whole resistance is not so simplistic as a doubling either.
  5. philSweet
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    philSweet Senior Member

    Not really, it only assumes that the cross sections are similar, meaning beam and draft are scaled by an equal amount, and by the same amount along the entire length. They can be any shape at all, but since the OP did specify an optimized hull, it will be quite like a semicircle.

  6. rubenova
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    rubenova Junior Member

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