# Passive fin stabilisation of fast catamaran

Discussion in 'Stability' started by groper, Jul 22, 2015.

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### Mr EfficiencySenior Member

Very happy to oblige, you carry on like a petulant space cadet every time someone queries your latest idea. Ta-ta !

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So...what is the estimated top/service speed of your cat?

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### groperSenior Member

Michlet predicted resistance and installed power suggest 25kts cruise. Real world crusing speed may be lowered below 25kts when the sea state is unfavourable \ too uncomfortable. ..

Ddisplaxement full loaded displacement is 4000kgs, installed power is 160kw, lwl is 10.6m.

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OK...so as a start...you can run some very simple calculations.

Since, I assume you know the hydrostatics of your vessel? Assuming you do....you can very quickly work out how much weight on the fore deck would produce a change in trim of 1,degree, 3 degrees, 5 degrees and 10 degrees, for example.

Once you have the weight, this can of course easily be equal to a force..the force from the fins...since to counter any pitching, if the hull pitched up say 3 degrees, you know the magnitude of the force required to restore the vessel to level trim.

So knowing your fwd speed, 25knots...how much area (via your fin shape) do you need to achieve the force required. Then you can see if the fin area is reasonable, or crazy large.

Just run those basic simple numbers first...

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### groperSenior Member

Thanks adhoc - ill get onto working that out shortly, spent all day at the kids soccer carnival

Until i get onto it, ive cut and paste the ship hydrostatics below - keep in mind some of this data will not be sensible at all as the ship weights have not been input into the software model - ive not bothered with that as ive only used this software simply for lofting out the plates etc for the physical building of the boat...

Displacement 3605.244 kg
Volume 3.517 m^3
Draft to Baseline 0.4 m
Immersed depth 0.4 m
Lwl 10.6 m
Beam wl 4.809 m
WSA 22.95 m^2
Max cross sect area 0.462 m^2
Waterplane area 10.494 m^2
Cp 0.718
Cb 0.639
Cm 0.89
Cwp 0.762
LCB from zero pt. (+ve fwd) m -6.595 m
LCF from zero pt. (+ve fwd) m -6.317 m
LCB from zero pt. (+ve fwd) % Lwl -62.213 %
LCF from zero pt. (+ve fwd) % Lwl -59.598 %
KB 0.218 m
KG 0 m
BMt 12.969 m
BMl 20.826 m
GMt 13.187 m
GMl 21.044 m
KMt 13.187 m
KMl 21.044 m

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### groperSenior Member

Ok let me see if i have this right;

First i find moment to trim 1cm ( MTC ) = displacement tonnes X GML / 100LWL
= 3.6 * 21 / 100 * 10.6
= 0.0713

if i assume the forward foil location will be 3.2m forward of the LCF, i have a trim arm (TA) of 3.2m. If i also assume the load (w) to be say 0.5tonnes, then trim change in cm=;
w*TA / MTC
=0.5*3.2 / 0.0713
= 22.44cm change in trim

or we could increase that to say 500kg per hull and call that 44.88cm change in trim for a 1000kg global load.

So that in terms of angular trim gives right on 2 degrees. How much is enough tho?

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OK...you've got the MCT formula correct, but the hydrostatics should give you the MCT anyway too.

So on an Lwl of 10.6, and knowing the LCF location, it is simple trig to find the trim, at 1degree, 3, degree etc. up to 10 degrees. As the similar triangle is the LCF to FP by the total trim/Lwl.

That 'trim'....is your target, since you know the lever to the location, the load is simple to calculate. No need to iterate.

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### groperSenior Member

Yes, but my question was how much trim is enough, ie what's a reasonable trim change to be expected from a typical stabilisation system at full angle of attack?

I extracted a little data from a paper on high speed catamaran active stabilisation, they modelled a catamaran of 54m lwl and 534 tonne displacement, service speed of 40kts. They used front foils of 2.8m^2 upto alpha 12degrees. I calculated the force produced by these to be just over 41 tonnes at 10 degrees alpha. Interceptors were also used in conjunction but calculating the lift from these is above my head.

I'm wondering if simple scaling method might work by simply looking for data on commercial vessels and scaling the system down in terms of fin areas etc for my purposes. Ie, scale down the displacement, or scale off DLR or simply the lwl ???

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It matters not. You need to calculate the weight, or force/lift, required to trim the vessel at 1,3,5, and 10 degrees. See what loads you get. Then from that calculate the fin area required.

If the fin area at 3 degrees is 3m^2..clearly you're going to need some big mother cylinders to move them...if it is just 0.3m^2...it is feasible.

Since most vessels will pitch +/- 5 degrees in a reasonable seastate and even more when in synchronous seas, thus a simple investigation of a series of angles of trim will tell you what you can or can't do.

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### groperSenior Member

well, for 5 degrees we get a required trim change of 1.1m.

that means our triangle needs half that, 55cm.

working backwards; 110cm = w * 3.2(ta) / 0.0713 (mct)
w = (110*0.0713) / 3.2
w= 2.45 tonnes or 1.225t per foil.

I dont think theres a point in working out for 10degrees - clearly we are going to be over the top in terms of loadings.

In order to get 1220kg lift from a foil with an aspect ratio of 2.3, using Dr. Matveev`s foil lift spreadsheet, i need a foil of around 0.3 sqm at 9.8 degrees alpha @ 23kts - i actually get 1270kgs lift to be precise. So that should produce a 5deg change in trim with only 2 bow foils. I can get a greater effect if i can increase my trim arm length, although the physical location of this makes it difficult due to inaccessible WTB compartments up there.

In addition to this, if we used 2 rear foils in conjunction with 2 fwd foils, if i assume an equal trim arm length to the rear foils, then i would only need half as much force to generate the same amount of trim change, so that would be around 610kg per foil, and could be done with foils of only 0.15 sqm area... all this seems very feasible...

Ive started building a 2m LWL model to test all this and make sure the stabilty algorythm is setup correctly so it will only be a simply matter of implementing larger motors to drive the full size system

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Yup..i get the about same. (just slightly different but ball park).

Yes...but don't forget. The change in trim is total, thus accounting for the trim @ the LCF...therefore the 5 degrees is roughly +/- 2.5 degrees in reality from the datum level waterline.

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### groperSenior Member

yes - thats why for the triangle we used half of that @ 55cm. So i used tan 5 degrees * distance to LCF and that gives 55cm.

However the equation requires us to use the required trim change total of 1.1m which is 55cm at each perpendicular - this will give us 5deg from datum level, no?

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The vessel trims about the LCF....thus the total trim, for 5 degrees..is actually roughly half..since trim (aft) and trim(fwd) = total trim. The trim fwd of 550mm is of the "total"....i.e full amplitude. So a trim of 5 degrees giving 1.0m trim (total) ia a full amplitude. Therefore the actual pitch angle is half so +/- 2.5. = 5 degrees total trim.

The vessel does not trim/pitch about the stern.

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### groperSenior Member

Sorry - im not familiar with naval architecture terms

So i think what i actually calculated was for 10degrees total trim in that case.

The triangle was formed with the fwd perpedicular to LCF distance of 6.3m as the "adjacent side" and the "opposite side" is the fwd perpendicular displacement from level.

This was trigged out to give 55cm bow displacement from tan 5 degrees. Are you saying the total is therefore actually 10 degrees as the stern is also displaced 55cm which is +5deg bow and -5deg stern.

Lets look at it practically for a moment - the required force calculated was almost 1/3 of the total vessel displacement for each forward foil - assuming no stern appendages. For 2 fwd foils it was around 60% of the total displacement to give the 5 degree pitch - measured from LCF to level. does this sound reasonable?

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