D/L units a question

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

 

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

 

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.

 

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.

 

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.

 

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