D/L units a question

Discussion in 'Boat Design' started by BASIL J WALL, Apr 19, 2012.

?

Does the D/L ratio have units

  1. yes

    0 vote(s)
    0.0%
  2. no

    0 vote(s)
    0.0%
  3. units are cu ft/ cu ft

    0 vote(s)
    0.0%
  4. units are long tons/cu ft

    4 vote(s)
    100.0%
  1. BASIL J WALL
    Joined: Nov 2011
    Posts: 19
    Likes: 0, Points: 0, Legacy Rep: 10
    Location: Canada

    BASIL J WALL designer

    Hi...
    Just a quick little question...
    Do you think the Displacement to length ratio (D/L ratio) used in the British system is a unit less (dimensionless)ratio?
    Thanks
    Basil
     
  2. Ad Hoc
    Joined: Oct 2008
    Posts: 6,760
    Likes: 726, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Not sure what the relevance to the "British" system means.
    ( http://en.wikipedia.org/wiki/Imperial_units )

    Whether you're using metric or feet/pounds (Imperial) system, it is a dimensionless ratio. But the same boat expressed in each system yields totally different answers. For example

    A 40m (131ft) boat that displaces say 140 tonne (137.8 L.Ton):

    L/D = 7.7 metric
    L/D = 50 Imperial units
     
  3. DCockey
    Joined: Oct 2009
    Posts: 4,859
    Likes: 398, Points: 83, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    Dimensionless but the numerial value may depend on unit system used.

    L is a length and is in units of length.

    If "D" is taken as the volume displacement "V" then V will be in units of length cubed. Then the formula "L/D" = L / (V^1/3) will have units of length/length and will be independent of unit system as long as corresponding units of length and volume are used (ft and ft^3, m and m^3, etc).

    A related coefficient is Cv = V / L^3 which will also be independent of unit system as long as corresponding units of length and volume are used (ft and ft^3, m and m^3, etc). Since Cv is typically a small number a modifed version is frequently used Cv = 1000 * V / L^3 This version is also independent of unit system as long as .... The difference is the factor of 1000 for the numerical value.

    It gets more complicated when D is taken as the weight or mass displacement. Then the numerical value can depend on the combination of units used.

    Simple ratios of L / (D^1/3) or D / L^3 have units of length / (cube root of weight or mass) / length cubed and the numerical value will depend on units used.

    A common coefficient used is DLR = D / (0.01 * L)^3 with D in long tonnes (very close to a metric tonne) and L in feet. This has units of long tonnes / feet cubed. The numerical value of DLR will be equal to 28.572 * Cv and will be the same for similar immersed hull shapes of different sizes as long as the specified units are used.

    1 m^3 of fresh water weighs 1 metric ton. So if displacment is taken as the mass displacment expressed in metric tons it will have the same numerical value as the volume displacment in fresh water expressed in m cubed. So
    D / L^3 = V / L^3 as long as L is in m and V is the volume displacement in fresh water.
     
  4. DCockey
    Joined: Oct 2009
    Posts: 4,859
    Likes: 398, Points: 83, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    Dimensionless and unit-less are not the same.

    Dimensionless means the numerical value does not depend on size/dimensions. Unit-less means the numerical value will be the same irregardless of the set of units used for the calculations.
     
  5. daiquiri
    Joined: May 2004
    Posts: 5,373
    Likes: 247, Points: 73, Legacy Rep: 3380
    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Well folks, I have been taught that dimensionless ratios DO NOT depend on units used, because they don't have units. They are pure numbers. However, since english is not my mother tongue, I'll leave that diatribe to you english-native guys.

    Just a comment on D/L ratio. In case of the classic definition the D/L is unit-dependent, because it is calculated as
    D/L = Displacement / ( 0.01 * LWL )^3​
    where displacement is in long tons (2240 lbs) and LWL is in feet. So, it is a Long_tons / feet ratio, which is clearly not a pure number.

    Then we have other ratios, like those DCockey mentioned, but the D/L ratio is traditionally calculated as explained above, so it is a units-dependent, non-dimensionless ratio.

    Cheers

    P.S.
    I have just seen the above clarification by DCockey, so at this point I believe we have a language-related problem here. I agree that unit-less is more appropriate term, instead of dimentionless.
     

  6. Doug Lord
    Joined: May 2009
    Posts: 16,670
    Likes: 337, Points: 93, Legacy Rep: 1362
    Location: Cocoa, Florida

    Doug Lord Flight Ready

    Eric Sponberg did us all a favor a while back and published these design ratios:
    DLR is on page 15.....
     

    Attached Files:

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.