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Overshoot

Overshoot

Figure 10.3 Zig-zag manoeuvre the initial response of the ship to the rudder being put over that can be vital in trying to avoid a collision. This inidal response is studied in the zig-zag manoeuvre. In it the ship proceeds on a straight course at a steady speed, a rudder angle of 20° is applied and held until the ship's head has changed by 20° and then the rudder is reversed to 20° the other way and held until the ship's head has changed 20° in the opposite direction. The manoeuvre is repeated for different speeds, rudder angles and heading changes.

The important measurements from the manoeuvre, Figure 10.3, are:

(1) the overshoot angle. This is the amount the heading increases by after the rudder is reversed. Large angles would represent a ship in which the helmsman would have difficulty in deciding when to take rudder off to check a turn. Values of 5.5 and 8.5° would be reasonable aims for ships at 8 and 16 knots respectively, varying roughly with speed. The angle does not depend upon ship length.

(2) the times to the first rudder reversal and the first maximum heading change. It has been suggested1 that for reasonable designs, times to change heading by 20° would be of the order of 80 to 30 seconds for a 150 metre ship over the range 6 to 20 knots. The time would be roughly proportional to length.

(3) the steady overshoot angle and the period of the cycle once a steady condition is reached.

If the ship is stable there will be a unique rate of turn for each rudder angle. If the ship is unstable the plot has two 'arms' for the smaller rudder angles, depending upon whether the rudder angle is approached from above or below the value. Within the rudder angles for which there is no unique response it is impossible to predict which way the ship will turn, let alone the turn rate, as this will depend upon other disturbing factors present in the ocean. The manoeuvre does not give a direct measure of the degree of stability, although the range of rudder angles over which response is indeterminate is a rough guide. To know the minimum rudder angle needed to ensure the ship turns in the desired direction is very useful.

### The pull-out manoeuvre

This manoeuvre1 is also related to the directional stability of the ship. The rudder is put over to a certain angle and held until the ship is turning at a steady rate. The rudder is returned to amidships and the change in the turn rate with time is noted. For a stable ship the turn rate will reduce to zero and the ship takes up a new steady straight line course. A plot of the log of the rate of turn against time is a straight line after a short transition period. If the ship is unstable the turn rate will not reduce to zero but there will remain some steady rate of turn. The area under the plot of turn rate against time gives the total heading change after the rudder angle is taken off. The smaller this is the more stable the ship.

If the ship is conducting turning trials it will be in a state of steady turning at the end of the run. If the rudder is centred the pull-out manoeuvre can be carried out immediately for that speed and rudder angle.

MANOEUVRING DEVICES Rudder forces and torques

### Rudder forces

Rudders are streamlined to produce high lift with minimum drag. They are symmetrical to produce the same lift characteristics whichever way they are turned. The force on the rudder, F, depends upon the cross-sectional shape, area A, the velocity Vthrough the water and the angle of attack a.

The constant depends upon the cross section and the rudder profile, in particular the ratio of die rudder depth to its chord length and the degree of rounding off on the lower corners. The lift is also sensitive to the clearance between the upper rudder surface and the hull. If this is very small the lift is augmented by the mirror image of the rudder in the hull. f{a) increases roughly linearly with a up to the stall angle which is typically about 35°. f{a) will then decrease.

Various approximate formulae have been proposed for calculating F. An early one was:

F = 577 AV2 sin a newtons

In this an allowance was made for the effect of the propeller race by multiplying Vby 1.3 for a rudder immediately behind a propeller and by 1.2 for a centreline rudder behind twin screws. Other formulations based on the true speed of the ship are:

F = 21.1 AV2 a newtons for ahead motion F - 19.1 AV2 a newtons for astern motion F = 18.0 AV2 a newtons

The first two were proposed for twin rudders behind twin screws and the third for a centreline rudder behind a single screw.

If wind or water tunnel data is available for the rudder cross section this should be used to calculate the lift and the centre of pressure position.

Typically the rudder area in merchant ships is between ^ and % of the product of length and draught.

### Rudder torques

To establish the torque needed to turn a rudder it is necessary to find the position on the rudder at which the rudder force acts. That position is the centre of pressure. For a rectangular flat plate of breadth B at angle of attack a, this can be taken as (0.195 + 0.305 sin a) B aft of the leading edge. For a typical rudder section it has been suggested2 that the centre of pressure for a rectangular rudder can be taken at K X (chord length) aft of the leading edge, where:

K = 0.35 for a rudder aft of a fin or skeg, the ship going ahead. = 0.31 for a rudder in open water.

The open water figure is used for both configurations for a ship going astern.

For a non-rectangular rudder an approximation to the centre of pressure position can be obtained by dividing the rudder into a number of rectangular sections and integrating the individual forces and moments over the total area. This method can also be used to estimate the vertical location of the centre of pressure, which dictates the bending moment on the rudder stock or forces on the supporting pintles.

### Example 10.1

A rudder with an area of 20 m2 when turned to 35° has the centre of pressure 1.2 m from the stock centreline. If the ship speed is 15 knots, and the rudder is located aft of the single propeller, calculate the diameter of the stock able to take this torque, assuming an allowable stress of 70MN/m2.

Solution

Using the simple formula from above to calculate the rudder force and a factor of 1.3 to allow for the screw race:

= 577 X 20 X (15 X 1.3 X 0.5144)2 X sin 35° = 0.666 MN

This can be equated to qj/r where r is the stock radius, q is the allowable stress, and/is the second moment of area about a polar axis equal to 7rr4/2. Hence

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