Whats wrong with my calculations?

What is the problem?

One of the values we gathered from the towing test was at 1.85m/s, our boat experienced a 3.5N drag. Using this piece of information, i attempt to break down the 3.5N into frictional and residuary resistance using Rf = FSV^n for frictional, subtracting it from 3.5N would yield residuary.

For residuary resistance, i believe i could use

Residuary of actual boat/Residuary of model = (length of boat/length of model)^3

During towing, my model's displacement would be 330g(weight of the model without anything) + 480g(weight of loaded weights) at a draft of 3cm, the wetted surface area was 818cm^2. I then estimated the WSA of the actual size boat to be the square of the boat: model ratio which is 20:1.

Ultimately, using FSV^n (actual boat) + (Residuary resistance model x 20^3) sheould give me a close estimate of what the actual size boat resistance would be.However, the frictional resistance i calculat
ed for the model is too small to be true.

L=length of boat=0.54m
P=density of water=1000kg/m^3
F = (P/1000)(0.1392 + (0.2581/2.68+L))

Rf = FSV^n
= 0.219 x 0.0818 x 1.85^1.825
=0.055N

0.055 is way to small...

 

What is Your boat's particulars?

 

Power catamaran model

Length=0.54m
Beam=0.25m
depth=0.06m
draft=0.03m
Ratio to actual size boat:1:20

 

depthofit,

personally I'd use a different approach and use the ITTC 57 prodedure to calculate the true scale resistance from model test results.

I. Total resistance of scale model is broken down to a resistance coefficient:

cT,M = RT,M / (uM² * ρ * SM * 0.5), with

cT,M = Total Resistance Coefficient of Model
RT,M = Total Resistance Force of Model
um = Model Towing Velocity
ρ = Water Density
SM = Wetted Surface of Model

cR,T,m = 3.5 / (1.85² * 1000 * 0.0818 * 0.5) = 0.025


II. Residual Resistance Coefficient of the model is

cR,M = cT,M - cF,M

with cF,M being the Friction Resistance Coefficient calculated by the ITTC 57 formula:

cF,M = 0.075 / (log Rn - 2)²

Rn = Reynolds Number = (L * u) / ν with

L = LWL and

ν = Kinematic Viscosity (1.14E-6 m²/s for fresh water @ 15°C)

-> RnM = (0.54 * 1.85) / 1.14E-6 = 8.76E5

Thus cF,M = 0.075 / (log 8.76E5 - 2)² = 0.0048

and cR,M = 0.025 - 0.0048 = 0.0202 accordingly.


III. When sailing at equal Froude Number, cR,M = cR,S = cR (Index "S" means the ship in full scale size).

Fn = u / √(g * LWL) = 1.85 / √(9.81 * 0.54) = 0.804

This means the full scale ship's speed will be uS = Fn * √(g * LWLS) = 0.804 * √(9.81 * 10.8) = 8.276 m/s

The full scale ship's Total Resistance Coefficient can now be calculated as

cT,S = cR + cF,S + cA

with cA being the Model Correlation Coefficient which - in short - takes several scale effects like skin roughness, towing tank inaccuracies and the like into account. For simplicity I'll use the ITTC standard value of cA = 0.0004

RnS = (LWLS * uS) / ν = (10.8 * 8.276) / 1.19E-6 = 7.51E7 (using ν = 1.19E-6 m²/s for salt water with 3.5% salinity @ said 15°C)

cF,S = 0.075 / (log RnS - 2)² = 0.0022

cT,S = 0.0202 + 0.0022 + 0.0004 = 0.0228

IV. Now it's possible to calculate the full scale overall resistance with

RT,S = CT,S * uS² * ρ * 0.5 * SS = 8.276² * 0.0228 * 1025.9 * 0.5 * 32.72 = 26209.83N = 26.2 kN


However, If I understand correctly you just want to know the friction resistance portions of both model and actual boat (and my drivel above was unnecessary):

RF,M = CF,M * uM² * ρ * 0.5 * SM = 1.85² * 0.0048 * 1000 * 0.5 * 0.0818 = 0.672 N

RF,S = CF,S * uS² * ρ * 0.5 * SS = 8.276² * 0.0022 * 1025.9 * 0.5 * 32.72 = 2529 N = 25.3 kN


Hopefully I didn't mix up my numbers while trying to be a clever Dick...

 

Eh Olav, fist time I see German in writing. What does it say in English ? All I understand is 'dick'

 

Germaths

Well, I used to believe maths and its symbols to be some kind of an "international" language - in addition I now fear my parents fooled me when they claimed my native language they taught me as a child to be German...

 

ROTFL!!
Fanie, I was sipping down some tea when I saw this and it nearly cost me my life!

 

Wow olav that was really some fabulous math u have there, thanks alot. However, is it really possible to have only 0.672N coming from fricitonal resistance at that speed? With my model being a displacement boat, that speed i mention there is definitely above hullspeed, does that explains for the low frictional resistance?
Are the values i provided even credible or logical? 3.5N at 1.85m/s for a 0.54m cat?

 

you need to separate the resistance calculated for laminar and turbulent flow. Do you know where the pins were located and hence the WSA for each?

 

Ad hoc,

you are right that there should be some tubulation generators near the bow to assure the laminar flow turning into turbulent at the right position (i.e. where it gets turbulent on the actual boat). However, it seems to me that depthofit did some DIY tank testing without such devices which makes the measured resistance force a little bit uncertain (no offence here, depthofit!). This may also explain the low overall resistance of the model although it's within the realistic ballpark imho and probably the best result one can achieve without access to professional towing tank equipment.

 

Thanks.

 

Hi, one thing i don't utnderstand is that for Pe = Rt x V. The Rt is the total resistance at what speed? If Rt is the total resistance at a particular speed, den wad is the V for?

Having calculated the total resistance at 8.276m/s, how do i apply the effective power formula?

 

It's actually just a local dialect by the costs of Nort Sea and Baltic

 

depthofit,

"V" in your formula is the boatspeed that "belongs" to the particular resistance force (I called it "u" in my calculation but it's actually the same). Thus your Effective Power at 8.276 m/s is PE = 26.2 kN * 8.276 m/s = 216.83 kW.

The other thing you wanted to know is about the scientific notation (interestingly I got a notification by e-mail that you asked this question but that post isn't visible here for some reason... ). For example, 2E6 means 2*10^6 ("2 to the power of 6") = 2,000,000 or 2E-6 means 2*10^(-6) = 0.000002.

 

Haha ya, i wanted to ask you about the E but den a fren of mine enlighten me so i quickly change the post into a thnks. Thanks anyway.