# Rudder Moment Calculation

Discussion in 'Hydrodynamics and Aerodynamics' started by eXXonWaldez, Aug 15, 2013.

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### eXXonWaldezNew Member

Hi,

Firstly, hi all, a am new to forum. I have searched similar threads but i could not find, so if there is please forgive me.

My question is, how you can calculate a rudder moment in zero speed with a certain of turning time, lets say 70 degree (port to starboard) in 20 seconds?

If anyone can help, i'd really appreciated.

Thank You.

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### PetrosSenior Member

Welcome to the forum.

not sure I understand the question. what do you mean "zero speed"?

when I worked in aerospace determining control loads and forces on the control surfaces was a very difficult and complex exercise. I do not think I have ever seen it done for a boat rudder, I am sure the companies that design large vessels do it. I am sure that most designers more likely they use tables developed from rudder moment forces actually measured on a large data base on existing vessels.

Please explain what exactly you are looking for. the moment on the hull for turning? Or the moment forces on the rudder/rudder shaft or control linkage?

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### Ad HocNaval Architect

Hi ExxonW

Welcome to the forum.

Your question is indicating one of two questions or I should say solutions.

1) The amount of “moment” generated by a vessel and from its rudder
2) A vessels tactical diameter of turn.

1) This related to the rudders characteristics in terms of “how much” but how much is then related to what is called Straight Line Stability. For a vessel to continue on a straight course requires several conditions to occur. There are 4 “cases” that define such straight line stability.

Case 1, a vessel experiences a disturbance and its final path is no longer the same as the original in direction, but still in a straight line, that is to say at “some angle” from the original direction.

Case 2, this is where after a disturbance it oscillates before the vessel continues in the same direction and a straight line, but not necessarily on exactly the same path as before the disturbance, i.e. offset by some transverse distance..

Case 3, this is the same as Case 2, but no oscillation after the disturbance.

Case 4, After the disturbance the vessel continues in a straight line and on the same path as the original prior to the disturbance, this is “positive stability”.

Reason for exampling this is that it is related to what is called the Stability Index. This relates to how a vessel turns, and the important part of this direction control is called the Neutral Point. This is where an applied force on the “point’s” location does not cause the vessel to deviate from a constant heading likes cases 1-4 above. (Sometimes it can be advantages to have a slightly directionally unstable vessel.)

So this point can be affected by larges rudders or large skeg etc, since this moves the neutral point from a poor location to a good location relative to the LCG of the vessel by the “moment” is produces. Once the directional stability has been assessed as being either in either cases 1 -4. So skeg that is too large or a rudder that is too large, may assist in making a vessel directionally stable, but, it may then render a moment required to “turn” the vessel greater than that is available from the rudder!

You need to conduct tank test experiment to determine the location of the neutral point

2) A vessels turning diameter, or turning circle, is related to the displacement the GM its speed, angle of heel, and the centre of lateral resistance.

The total force is the hull force minus the rudder force = displacement x speed^2/Radius. This is applied at a location “G” on the centreline of the vessel.

So knowing the basic parameters the turning circle can be approximated to:

R.gSin(theta)/V^2=GE/GM

R = turning radius
G=9.81
Sin.(theta) is angle of heel during a turn
GE= distance from centre of later resistance to the applied moment of hull/rudder.

In most cases the central of lateral resistance and the rudder force are very close, at small angles of heel. A quick approximation is that half the draft = this value.

So getting back to your question, you need to define what is you require a bit better. Are you trying to calculate how much force/moment is applied by a rudder?..how much force is required to produce a turn…what turning radius is produced/required etc etc? As the question is a bit vague.

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### DCockeySenior Member

Another possibility is the OP is asking about the moment required to move the rudder from 35 deg to port to 35 deg to starboard in 20 seconds when the vessel is stationary, ie zero speed. No idea what the reason for that question would be though.

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### Ad HocNaval Architect

Rudders general only go from 30 to 30 degrees P-S.

I assume he means a turn from an original course then bearing away by 70 degrees, from that course, in a turn.

Other than curiosity how to calculate such an event, ditto.

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### eXXonWaldezNew Member

Hi again thank you all for responses. when a ship has a speed of 0 knots, stationary. Maybe for an example, a fin stabilizer, in zero speed generates force to keep the ship upright. But there should be a certain amount of moment to turn the fin in certain amount of time. What i need to calculte is how much moment should i apply to the shaft. Maybe now, i explain my question better.

Thank you

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### PetrosSenior Member

sorry, that does not help.

if you have zero speed, how would a fin generate any force? Usually the ship is kept upright by the shape of the hull, except for a deep keel sailboat? Is this what you mean?

Please list the size of the boat, lenght and displace, you are asking about.

Please list the type of the ship, sailboat, pleasure boat, ski boat, cargo hauler, what?

Is it powered by sail, motor/prop, what?

What do you mean by fin stablizier? a fin going straight down under the hull? is it weighted? how large is the fin?

what shaft are you applying a moment to? prop shaft? rudder shaft? fin shaft?

Perhaps you need to have someone help you with your terminology if you expect to get some help. We can not help you if we can not understand what you want to know.

Can you draw a picture, scan it and post what you are talking about?

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### eXXonWaldezNew Member

Hi Petros,

There are stabilization systems using hydrofoil shaped fins. There are two types of these systems.
Firstone called "active stabilization": which generates force by changing angle of attack of fins. Which works same manner of rudder
Second is called "zero speed" or "at anchor" stabilizer: This is also an hydrofoil which stabilizes the boat active(e.g. 12 knots,10 knots), also at zero speed (at anchor).
(Sure there is a shape difference between active fins and zero speed fins.)
If you may check this website, that might help,

http://www.quantumstabilizers.com/onanchor-large.htm

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### BMcFSenior Member

Generating sufficient forces for zero speed control using fins or trim tabs requires high actuator slew rates and a very non-linear form of control algorithm to drive them.

Essentially, the movement of the fins (or tabs) is "bang-bang" in nature, moving at a very high rate from one limit position to the other in response to angular rate of the vessel (in roll) exceeding a certain threshold value. The control forces, actually impulses in nature, result from the added mass and viscous resistance of the water as the fin is rotated moved rapidly through it. "Balanced" fins (those with lift center on or very near the shaft centerline) make poor zero-speed control devices because their center of area is fairly close to the shaft. The farther away from the shaft the center of area is, in the direction of the trailing edge, the larger the zero-speed control forces it can develop. Fins with extendable/retractable trailing edges have been developed to cover the requirements of conventional underway roll stabilization and zero speed stabilization. Where trim tabs are employed for active roll and trim stabilization underway, they often make excellent zero-speed control effectors because they are, inherently, as unbalanced as can be achieved with respect to having center of area as far as possible from the axis of rotation.

The actuation and bearing support design (and peak hydraulic power requirements) are typically much more robust for a device used for zero-speed control because of the need to reliably achieve such high actuation rates coupled with the higher torques required to operate an unbalanced fin. I've not seen zero speed roll stabilization implemented using rudders yet.

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### Murat124Junior Member

There is a calculation method formula at Turkish LLoyd Rules, if you will change speed in to 0 in this formula you can calculate easily.

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### Murat124Junior Member

Calculation v = 0 for v< 10 knots in given formula
http://www.iacs.org.uk/document/public/Publications/Resolution_changes/PDF/UR_S10_Rev2_pdf1330.pdf

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### TANSLSenior Member

All regulations of classification societies have chapters devoted to the calculation of scantlings of rudders. But the question is:
and I fear that some rudder data are needed to answer this question.
The moment is equal to force times distance. The force depends on the rudder´s area and distance on its geometry.

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### Murat124Junior Member

) Yess sure you need a rudder to calculate its moment

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