# Forces on transom rudder

Discussion in 'Multihulls' started by MichaelRoberts, Mar 13, 2023.

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### MichaelRobertsArchimedes

This picture shows a kick up foil mounted on the transom of a catamaran I am building
Although I have good data on the likely torque (courtesy of Coursemaster who designed the hydraulics), the lateral forces from a wave, drag and even a broach are difficult to estimate
Thus the maximum shear force on the gudgeon is unknown
Thanks
Michael
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### Robert BieglerSenior Member

If none of the naval architects will respond, I'll relay what I read long ago in an article explaining why a rudder broke.

You want the maximum torque around the longitudinal axis, which is your lever arm multiplied by maximum force. Maximum force you can get from maximum water speed over the rudder multiplied by lift coefficient. The important part of the article I read is that when the angle of attack increases suddenly, flow does not detach immediately, and until it does, the lift coefficient may rise up to 2, far more than a symmetrical foil would give you in a steady state. So take the maximum speed that you expect your boat to reach surfing down a wave, add a bit for the rudder experiencing different flow due to the water moving in the wave (I don't know how much; 10%?), assume a lift coefficient of 2, multiply by the safety factor of your choice.

The rudder that broke was designed for the maximum lift coefficient that profile could achieve in a steady state at 20 knots. Inserting the lift coefficient the rudder could reach for a short time showed that it was only safe up to 12 knots.

Maximum drag would occur when the rudder is at 90 degrees to the flow, so just model that as a flat plate. This is still going to be less force than grounding at speed or hitting a floating object. If the rudder doesn't kick up, you have to decide whether you want rudder, gudgeons or transom to break in that scenario.

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### MichaelRobertsArchimedes

Thank you Robert

Especially your comment that none of the architects will respond - this is a tricky problem

In about an hour I am going to get a whole lot of bollocky advice from two world champion sailors who are coming to lunch
They want a skeg rudder - no room - they want to rotate the engine and do a V drive - no way - the seal would become inaccessible

Catamarans need a kick up rudder - what if we hit a reef - there is no keel - the bottom of the boat could be ripped out
My kick up rudder will have a shear pin - if it goes it goes - still got the other rudder

This cat of mine is one very big exhilarating experiment - and because I am both designer and client I can take the risk.

Keep in touch
Michael

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### John PerrySenior Member

To estimate the maximum shear forces (i.e. sideways forces) on the gudgeons you need to know the vertical spacing between upper and lower gudgeons - I dont see that dimension on your pictures.

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### MichaelRobertsArchimedes

Hello John

The gudgeon measures 359 between supports
The foil centroid is about 706 below the bottom bearing
The pin's max shear load is > 40 kN = 4,000 kg
So max side force on foil is about 2,000 kg

Which leads us to the big question: what is the maximum lateral load from a broadside wave
it's a question of the speed of the broadside wave

Michael

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### John PerrySenior Member

If the hydrodynamic lateral load on the rudder blade is W then the lateral load on the lower gudgeon is W*(1+706/359) which is about 3 times W, not 2 times W.

If the lift coeficient of the foil is Cl then the lateral load on the rudder blade is:
W = Cl *0.5 * fluid_density*foil_area*(vel^2)
i.e. if Cl =1.0 then W = (500*0.559*((20*0.51)^2))/9.81 which is 2964kgf, so in the same ballpark as the calculation in red in your first post.

So for Cl = 1.0, W = 2960kgf and lateral load on lower gudgeon is 8781kgf.
If you take Cl to be 2.0 as Robert sugests then you can double that lateral load.
And if you choose a factor of safety of 2.0 you can double it again!

The above sugests load on the lower gudgeon that is well in excess of the maximum load you have estimated the gudgeon will withstand. But I think this is a pessimistic calculation for the following reasons:

• Are you really going to suddenly swing the helm at a rate that loads the rudder to its maximum possible lift coeficient while you are sailing at 20knots - I think not! And if you do the stern of the boat will start to move laterally as the rudder load increases, this reduces the angle of attack on the rudder and hence the lift coefficient. To calculate this I think you would need the rotational moment of inertia of the boat and make an assumption about how quickly you might be able to turn the rudder if, for example, you suddenly spot an obstruction ahead. Its not an impossible calculation to do, at least approximately, but not something I have looked into.
• The maximum Cl for the part of the foil that is in proximity to the surface is less than for a deeply immersed foil, this is something that I actually started to experiment with about a year ago using a foil attached via load cells to a mounting on the side of a small dinghy powered by an outboard motor. I havent had a chance to get very far with these experiments yet.
I think the problem comes down to knowing what maximum Cl value to allow for, the rest of the calculation is easy. If you google 'rudder stock loads' or something like that you can find papers relating to the design of big ship rudders for which I think that Cl = 1.0 is a typical assumption and on that basis you will need some strong gudgeons (and other related structure).

As for the effect the effect that waves have on the velocity of water flowing past the rudder this is something I did look into a bit since I wanted to calculate how rapidly a control system would need to rotate a lifting hydrofoil to compensate for this. The water in deep water waves is travelling in circular orbits. For water particles at the surface the diameter of the circles is the wave height and the particles make a full rotation in the time it takes the wave to travel the wavelength. As you go deeper the diameter of the circles reduces, the formula I used for this is:
D = D0*e^(-5.97*H/L) where D is the diameter for depth H, D0 is the diameter for water particles at the surface and L is the wave length.

There is a huge amount of information about wave heights and wave periods etc. for UK coastal waters at WaveNet - Cefas (Centre for Environment, Fisheries and Aquaculture Science) https://www.cefas.co.uk/data-and-publications/wavenet/. From this you could estimate how much the wave related water particle velocities add and subtract from the velocity of the boat through the water. I did this for a lifting hydrofoil with 1m wave height and boat speed about 15knots and assuming the boat was sailing to windward at a course 50degrees to the oncoming wave train (that would be nice, wouldn't it!) I think the depth below the mean surface that I considered for the foil was in the same ballpark as the depth of your rudder. My conclusion was that it does not make a huge difference and could be compensated for by a fast acting foil control system. When I tidy up these calculations I will probably write it up for the Amateur Yacht Research Society.

There is lots of data available for measured wave induced water pressures on fixed structures such as breakwaters. I know someone who spent much of his working life studying this, I think he still has a set of pressure sensors mounted on Alderney harbour breakwater. But I think this is different to the effect that a beam sea has on a boat since the whole boat moves laterally in a beam sea rather than being fixed and having to resist the full wave induced water pressure.

I saw the photograph you showed of your catamaran in an earlier post here - it looks to be a huge boat building project that is well on the way to completion - I wish the best for it.

Last edited: Mar 18, 2023
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### MichaelRobertsArchimedes

Great reply, my sincere thanks John

Oops, I made a stupid mistake with the basic beam calculation
Consider a horizontal beam in equilibrium, three units long
Fulcrum is one unit from end
If the down force on long end is 2 tonnes, the balancing down force on short end is 4 tonnes
Total force on fulcrum is 6 tonnes
The force on the lower gudgeon bearing is 6 tonnes.

Dave Gerr's book "Boat Strength" gives 6 tonnes as the shear strength of a 16 mm steel steel bolt - pushing it a bit

Thanks again
Michael

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### MichaelRobertsArchimedes

Hello again John
Forgot to say thanks for encouraging words about the project
It's totally bonkers - 60 feet long - I've just turned 80 - and the only way to get it off our mountain side is a Chinook
Oh well can't stop now - about another year
More soon
Michael

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### John PerrySenior Member

Michael - I did not think my reply was all that helpful since I did not feel able to sugest a lift coefficient to use in designing your rudder fittings.

This standard gives design rudder loads for sailing boats but it is only for monohulls: ISO 12215-8 - Small craft — Hull construction and scantlings — Part 8: Rudders.

I dont have a copy of the above standard but I have an idea that it gives a lift coefficient of 1.5. Given that this is a fairly lightweight craft that can realistically be expected to sail at 20+knots, I suspect that you will have unnecessarily strong and heavy rudder fittings if you design them for that lift coefficient together with a 20knot speed. But who am I to say!

I think Richard Woods said something like sailing at x knots is similarly scary to driving a car at about one tenth of x knots. So 20knots is like a formula 1 car at 200mph. But maybe for a 60 foot catamaran its not quite that bad, perhaps the factor could be one fith rather than one tenth. So I am thinking that loading the rudders to such a lift coefficient while sailing at 20knots would be like suddenly applying full steering lock to a car while driving at 100mph - would anyone other than perhaps a stunt driver do such a thing? And if you did the equivalent with a sailboat, broken rudder fittings might not be your only problem.

I have checked your figure of 6 tons for a 16mm bolt against the table provided by the British Stainless Steel Association. That table gives 4.8 tons for a property class 70 16mm 316 SS bolt in single shear but I think they are considering the shear area of the threaded part, I assume you will be using a pin without thread so that probably explains the difference. But most rudder fittings on larger craft are designed to place the rudder pivot pin in double shear, i.e. the pin is supported above and below a fitting attached to the rudder itself. You can allow twice the load for double shear. And you can use a stronger material than 316 stainless - since you only need a few inches of plain bar stock you may be able to find a bit of 17-4PH stainless steel, Nitronic 50 or grade 5 titanium for example. And of course, the structure that carries the pin must have strength compatible with the pin.

Your project seems amazing, if bonkers - are you really going to have a Chinook to lift it?

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### MichaelRobertsArchimedes

I am in your debt John
The double shear idea has inspired me to use a longer gudgeon pin as part of the kick up safety fuse
And it will be in double shear so I don't have to rebuild the box that holds the foil
Haven't quite worked it all out yet, hope to have the details sorted in a day or so
Michael

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### Robert BieglerSenior Member

I agree that the rudders of a boat as large as this are not easy to turn quickly enough to see the transient effects I read about. The increase in coefficient of lift may happen anyway while broaching. The rudder at 90 degrees to the flow can happen when sailing backwards after missing a tack, though then the speed should be less.

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### MichaelRobertsArchimedes

Thanks for your excellent guidance gentlemen
I've been pondering the reality of six tonnes pressing sideways on the foil holder - three Land Rovers!
Something will break - and it probably won't be the gudgeon pin
I've just finished the difficult task of installing the hydraulic steering system - circuit attached - about 50 metres of 12 mm copper tube.
This means I can now do some real world torque and lateral load testing
The torque will be tested with a 1 metre lever clamped onto the foil and a 300 kg scale
Then lateral load will be applied with the help of a chain block, a nearby tree and a webbing sling around the foil.
This test will take a week or so to set up - then I'll report back
Michael

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### MichaelRobertsArchimedes

Maybe this is a new insight into a flaw in hydraulic steering systems

If this boat had a tiller the problem of lateral forces from a rogue wave would go away - you don't fight the tiller
(Chichester had to push on Gypsy Moth's tiller with both feet - until he got to Australia and they did some work on the keel)

Because a hydraulic system has check valves the rogue pressures can't be released
Pressure builds up on the hydraulic rams

Consider a rogue lateral wave that applies force on the foil's centroid.
This centroid is about 200 mm behind the axis (aft of the centre of lift)
That lateral force generates torque, it is resisted by pressure in the hydraulic system

IF THERE WERE A PRESSURE RELIEF VALVE THE LATERAL FORCE WOULD BE LIMITED

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### John PerrySenior Member

Michael - pressure relief valve sounds sensible - better than your previous idea to test your rudders to destruction before the boat is even afloat!

Assumes rudders are not too well balanced - our 10m trimaran has single rudder directly attached to tiller and with only slight weather helm and partly balanced rudder it can be steered with only finger force on tiller so even if it had hydraulic steering a relief valve as you describe would probably not work very well.

Be aware that the centre of lift of a rudder that is working as a foil (i.e. not stalled) is not at the centroid of the immersed area! - it is close to the quarter cord line. So you need the pivot axis forward of that - from the pictures you showed earlier it looks as though you should be ok. In any case, with 'kickup' style rudders you can probably find a way to adjust rudder blade rake if you need to increase or reduce the balance effect.

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