# Boundary-layer trip

Discussion in 'Hydrodynamics and Aerodynamics' started by Remmlinger, Oct 20, 2013.

1. Joined: Jan 2011
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### Remmlingerengineer

Mikko, I am sorry, but there was an error in my paper. The diagram showed drag area, not friction coefficient (the wetted surface was missing in the denominator).
Also in my first shot I took the position of the sand strips from a drawing in a paper. Jasper den Ouden wrote me now, that on the actual Sysser 72 model in his archive the roughness strips are missing and he does not know why. For all other models he gave me the exact position. Obviously I had bad luck with my initial choice.
In my updated paper http://www.remmlinger.com/B-L-trip.pdf I use Sysser 23 as an example. The numerical values are for the model size.
If I wanted to split the total resistance into 1.) the viscous resistance at the right dynamic sinkage and trim and 2.) the wave resistance, then you were right. This seems to be too difficult. My current approach is to make the split between 1.) the viscous resistance of the flat water, untrimmed hull and 2.) the residuary resistance. This is the common approach, introduced by W. Froude. The residuary resistance includes the change in wetted surface and any change of the separated flow at the stern. It exhibits therefore a slight dependency on the Reynolds number. The ITTC invented therefore the "correlation allowance".

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### Mikko BrummerSenior Member

OK, thanks, that helps.

Looking at the Fig 12, up to Fn 0,15 or so, the drag should be practically only viscous. So should not your drag line lie a little bit higher up, somewhere along the Delft method line at Fn 0,15 or 0,20? This would bring the residuary curve close to zero at Fn 0,10, and point it towards zero at Fn 0, like it should?

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3. Joined: Jan 2011
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### Remmlingerengineer

The force coefficient is not necessarily zero at Fn = 0.

I have attached the diagram for the resistance forces. It is no problem to extend the curves down to zero at Fn = 0.
At these low Fn values the residuary resistance is equal to Rr = Constant*Fn^n i.e. it is proportional to the nth power of Fn. The value of "n" is not known a priori. It depends on the hull-form, on the hight of the towing point etc...
The force coefficient is defined as:
Cr = drag-area / Awet = Rr / (rho/2 * V^2 * Awet)
from this definition it follows:
If "n" is larger than 2, then Cr will go down to zero at Fn = 0
If "n" is equal to 2, then Cr will remain constant down to Fn = 0
If "n" is smaller than 2, then Cr will go to infinity at Fn = 0

The force coefficient (or drag area) is a much more sensitive tool for the comparison of test results and predictions than the force itself. It magnifies possible differences. May be this is the reason that most people compare their results on the basis of the forces.

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4. Joined: May 2006
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### Mikko BrummerSenior Member

Thank you Uli. Always so nice to learn something new,

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