In addition, a representation of the anchor and anchor chain in the sea when external force was applied can be seenin Fig. 80. The catenary forms the suspension part of the anchor chain, from the anchor chain outlet (outside of
hawsepipe) of the bow to the sea bottom.

B: Bell Mouth

T ：External Force

T ：External ForceSea SurfaceyS

L= S +

Cable
Contact

Sea Bottom

Catenary

S ：Catenary length against the external force（m）

L ：Minimum required length ofy ：Water depth + hawsepipe height from sea surface（m）

anchor cable（S ＋）：Minimum required contacted length of the cable（m）

T ：External force（kgf））

Fig. 70

The catenary length of an anchor cable can be calculated with the following calculating formula.

Calculating the Catenary Length of an Anchor ChainS=y2 + 2T
WcyS：Catenary length against the external force（m）y：Water depth + Hawsepipe height from sea surface（m）

Wc' ：Anchor cable weight per m in water（kgs）= 0.87 × Wc（kgs）

T：External force（kgf）

With the above mentioned formula, the more external force ( T ）increases, the longer the catenary is.
Therefore, under the condition that Minimum Required Length of Anchor cable is ﬁxed at a certain level, as thereis the following relationship, if External Force increases while the Anchor cable is ﬁxed, the anchor holding power
decreases, which leads to a vicious circle.

− 48 −

Minimum required length of anchor
B@AKD+HRkWDC@S@BDQS@HMKDUDK+
External force increases

Catenary length (S) also increases
Contacting length of cable (l) becomesshorter accordingly
Anchor holding power decreases because
contacting length of cable is shorter.

Also, it is not easy to manually calculate the catenary length and decrease of anchor holding power along with anincrease in external force. Similar to the calculation used for external force (wind pressure force), the calculation result
will be instantly displayed once the formula is input in the Excel File.
When entering data on holding factor coeﬃcient, observed wind speed, and type of sea bottom coeﬃcient, etc. be on
the safe side when entering the data and do not over rely on it as it is intended to be a guide. Tables 71 and 72 show
catenary length and calculation examples of anchor holding power.

Anchor Holding Power Calculation：Just Reference
1 The formula in this page are to calculate the holding power of your vessel's anchor and anchor cable.
2

In case of External Force<Anchor Holding Power（Wa' × λa），required length of cable is to calculate by following formula.
Required Length of cable = 3 × d + 90 m（Only External Force < Wa’ × λa）

3 Expected total external force should be input by Maximum Impact force. For PCC/CTNR Ship : Wind Pressure on front × 5 〜 6,
Other type of ship : Wind Pressure on front × 3 〜 4 times＊ IE）As per Wind force Cal., Relative Wind Direction θ=0 : 15ton : Max Impact Force 15 × 5 〜 6 for PCC/CTNR= 75 〜 90 tons

Input Data
Expected total external Max. force（MT）：（Ｔｘ
Anchor weight （MT）in Air：
（Ｗａ）
Anchor chain weight in Air （MT/m）：（Ｗｃ）
Kind of anchor（1: JIS, 2: AC14）

Result of Calculation
Impact Force）

85.56

Total height (Bottom to Hawsepipe):

10.5

Catenary length against the external force：
（Ｓ）

174 m

Minimum Required Contacted length of the chain：

150 m

Minimum Required Length of Anchor Chain：Ｌ＝Ｓ＋324m 12ss

0.166 ton
2

Total Length（Shackles）of using Chain on board

12 ss

Water depth（m）：ｄ20.0 mHawsepipe height from the sea surface（m）：ｈ5.0 m

Anchor Holding Factor:（λ a）
Anchor Chain Holding Factor：
（λ c）

25 m

(y)

NoticeKeep Anchor Watch Strictly

AC14 7.0
1.0

Anchor Holding Factor（λ a）：Use Calculation Formula or Input by
mannual.
Anchor Chain Holding Factor（λc ）： 0.75 〜 1.0

Table 71

− 49 −

Calculation Formula in above table◉ Anchor Holding Factor（Subj. to Kind of Sea Bottom）
λ a： 3.0（JIS）：7.0（AC14）λa：Anchor Holding Factor

λcType ofAnchorSand

Mud

Dragging
Anchor

JIS

3.5

3.2

1.5

AC14

7.0

10.6

2.0

：Anchor Cable Holding Factor
Holding

Dragging

λc

Sand

Mud

0.75

0.60

0.75 〜 1.0◉ Anchor Chain Holding Factor
λ c：0.75 〜 1.0◉ Total height（Sea Bottom to Hawsepipe）
y ：d ＋ h
Water Depth（d）+ Hawspipe Height from Sea Surface（h）◉ Catenary Length Against the External ForceS ：S =

y² ＋ 2Tx
Wc'

y◉ Anchor Weight in Air（Wa）, Anchor Weight in Water（Wa'）（Wa ）＝ Wa × 0.87
◉ Anchor Cable Weight per m in Air（Wc）, Anchor Cable Weight per m in Water（Wc'）（Wc ）＝ Wc × 0.87
◉ Minimum Required Contacted Length of the Cable：Tx = Wa' ×λ a ＋ Wc' ×λ c ×： =Tx ‑ Wa' ×λ a
Wc' ×λ c◉ Minimum Required Length of Anchor Cable
L：L = S ＋◉ In case of Tx ＜ Wa' ×λ a
：L = 3 × d + 90（m）
Table 72

− 50 −

４−６ Countermeasures to prevent an anchor from dragging instormy weather
Conventionally, there are guidelines, such as those listed below that serve as countermeasures against an anchor
dragging in stormy weather. Consider the following reasons and remarks.

①
Reason②
Reason③
Reason④
Reason⑤
Reason⑥
ReasonMake the deep draft
The shipʼs weight will increase, and the chanceof the vessel horsing will decrease. Also, adecrease in wind pressure area can decrease
the impact of external forces.

Remarks

Consider increasing ballast quantity and hullstrength.

Even keel, or by the head (trim by bow) if possible, is to be applied.
Horsing（yaw and sway) motion can be limited
when wind pressure centre is moved to thestern.

Remarks

Consider increasing ballast quantity and hullstrength. Maintain propeller immersion.

Remarks

It takes 3 minutes heave-in one shackle (longer
time taken during rough seas). The longer theveered anchor cable, the more time it will take.

Veer anchor cable
Because holding factor between the anchor
cable and sea bottom increases and catenary
length increases, it is effective for improvementof anchor holding power and mitigation of
impact force added to cable.

To drop the other anchor as a snubber anchor
It is effective to drop the snubber anchor
because it can halve the amount of horsing(yaw and sway) motion which will decrease
the amount of impact force on the anchor byapproximately 30 to 40%.

Remarks

Consider length needed for a second cable.
One and a half times the depth of water will beneeded. Consider the possibility of cable fouling,
particularly when pitching heavily.

From the outset of anchoring, deploy both anchors
Anchor holding power increases when riding
at two anchors. Two-anchor mooring restricts
horsing (yaw and sway).

Remarks

There is a danger that the anchor cables may
tangle. Two-anchor mooring is not ﬂexible when
the wind direction changes.

Main Engine, steering in combination with bow thruster
It is an effective way to suppress horsing (yawand sway) when an attempt is made to standbow against the wind, using a combination ofdead slow ahead speed and steering.

Remarks

It is necessary to have in-depth meetings withthe engine department. Ensure that the bow
thrusters are kept submerged when the ship is
pitching and rolling. Loosen the anchor chain
temporarily by manoeuvring forward, and notethat any sudden increase in tension may breakout the anchor when the hull falls leeward, which
may lead to anchor dragging.

Of the above guidelines, we simulated ③ Veer anchor cable and ⑥ Main Engine, steering in combination with bow
thruster (as per the situation of Vessel C, Case 3-3).
− 51 −

＝ “Increase of anchor holding power in the event of a veered length of the anchor cable” for single anchor mooring ＝
An increase in critical wind speed when the anchor chain was veered due to strong wind, during anchoring with awindlass at 6 shackles (distance of 151 meters from hawsepipe, with Hawsepipe being 25m from sea bottom) of a Pure
Car Carrier laden with 6,000 units was calculated as a simulation under the following conditions:＝ Calculation conditions ＝Anchor weight in air（Wa）：10.5ton

Holding factor（λ a）7.0➡ Anchor weight in water 9.135 tons（Wa'）
Anchor chain weight per m in air（Wc）：0.166 ton/m Holding factor ／（λ c）1.0➡ Anchor weight in water 0.144 ton/m（Wc'）

Water depth + Hawsepipe height from sea surface（y）：25.0mLength of one shackle of anchor cable：27.5mProjected area (front）（A）：800 sqm

The wind force coeﬃcient（CRa）：0.75

Air density（ρ）：0.125kg/sec2/m4① When the anchor cable is constantly fixed
The external force is increased when keeping the anchor cable at 6 shackles and the catenary length of the anchor
cable is enlarged. However, once the external force reaches 63.9 tons (wind speed at 16.9m/s), the anchor cableconsists entirely of catenary which means that the only anchor holding power is by the anchor itself. When calculating
backwards the wind speed to be given a critical wind speed, and dividing the critical wind speed by 1.25 to 1.50 andthen replacing it with the average wind speed, 11.3 to 13.5 m/s will be the anchoring limit.

T :External Force

B:Bell Mouth

63.9t

T :External ForceSea Surface

S′=150.9m

y =25mCatenary

Fig. 73− 52 −

Sea Bottom

（Calculation basis）
Before veering the anchor cable, the ship is moored exclusively with the anchor. This can be compared to theanchoring limit, as the anchor holding power is equal to the external force (impact force). The anchor cable is assumed
to have formed a catenary with no cable lying on the sea bottom. Before determining the critical wind speed, the windforce from ahead must be calculated. It can be demonstrated that the impact force (which, in this case, is the same as
the anchor holding power) should be divided by 6.
Anchor holding power ＝ external force（impact force）：63.9 tonf
➡ Wind force from ahead 10.65 tonf
：150.9 m（5.5ss）

Catenary length（S'）

Calculate critical wind speed against wind force on front (110.65 tonf) from the Hugh’s formula to ﬁnd the criticalwind speed and average wind speed as follows.

Critical wind speed：16.9 m/sec
Average wind speed: 11.3 m/sec 〜 13.5 m/sec＊

The critical wind speed can be converted into the average wind speed which can be solved by dividing thecritical wind speed by1.5 or 1.25.

These average wind speeds are the maximum limits under which the vessel in this example can lay safely atanchor without any more force being exerted on the anchor and cable system than it can withstand.

② Situation after one additional shackle of cable is veered
The critical wind speed was simulated assuming that seven shackles of anchor cable were veered.

B：Bell Mouth

T :External Force 67.3t

Sea Surface

L=S+=178.4m

S=154.8m

y =25m

=23.6mSea Bottom

Cable
Contact

Catenary

Fig. 74− 53 −

When calculating the critical wind speed assuming that one additional shackle of anchor cable is veered, the ship canwithstand an external force of up to 67.3 tons (wind speed at 17.3 m/s). (This external force is balanced assuming that
23.6m of veered anchor cable is contacting the sea bottom. If the external force is greater than this, part of the anchor
cable will be contacting the sea bottom. However, the total anchor holding power will be smaller than the external
force, and the ship will drag anchor.)
Similarly, replacing it with average wind speed, it becomes 11.5m/s to 13.8m/s. Even when compared with 6 shackles
being veered, only an average wind speed of 0.2 to 0.3m/s was applicable.

（Calculation basis）
After a further shackle of cable was veered, the critical wind speed increased. Only part of the longer cable system
will lay along the sea bottom with the remainder forming part of a new catenary, which will now be longer as it will
be extended by the increased external force caused, in turn, by the increased critical wind speed. Veered length of thecable (laid over the sea bottom) ( ｌ ) was calculated by the quadratic equation from the calculation formula as below.

S'+（27.5m ‒）=Wc' ×λ c ×［ Wa' ×λ a ＋Wc'
]× yy2 + 2 ×

S'：Catenary length before one shackle is veered ：150.9m：Veered length of the cable（laid over the sea bottom）（this is to be calculated）
Veered anchor cable contact ：23.6m

Catenary ：3.9m

Total ：

27.5m

③ When the anchor cable is completely veered to 12 shackles
Similar to the above, when calculating the critical wind speed assuming that all additional shackles (12 shackles)of anchor cable have been veered, the ship can withstand up to an external force of 86.3 tons (wind speed at 19.6m/s). However, replacing it with the average wind speed, it becomes 13.1m/s to 15.7m/s. Even compared with theassumption that 6 shackles are veered, only an average wind speed of 1.8 to 2.2 m/s was applicable.

（Calculation basis）（S）Catenary length（ ）Contacted length of the cable（laid over the sea bottom）（L）Total length of anchor cable
The holding power created by the anchor and cable system Total：175.0m：155.0m（+22.4 tonf）：330.0m（12ss）：86.3tonf（Impact force）➡ Wind force from front 14.38tonf

Critical wind ：19.6 m/sec
The average wind speed can then be calculated as13.1 m/sec 〜 15.7 m/sec(This is an increase of 1.8 〜 2.2 m/sec）− 54 −

In summing up; from the viewpoint of the ship commander, even veering anchor cable as a countermeasure ofdragging anchor cannot be considered highly effective for increasing anchor holding power. This is because wind
gusts and does not blow in a constant direction or speed, thus it is also worth considering the necessity of weighing the
dragging anchor in time of an emergency, just in case.

Importantly for those on the bridge, the critical wind speed does not increase
as much as might be expected even if the anchor cable is veered considerably.

＝ Combination use of main engine, steering and bow thruster ＝Countermeasures to be taken under rough sea conditions, using a combination of the main engine, steering and bow
thruster, will be examined.

① Use of the bow thruster
In order to avoid horsing (yaw and sway), it may be necessary to use a bow thruster, if one was outﬁtted on the ship. In
this situation, it is necessary to consider the inﬂuence on the generator as a result of outputting from the bow thruster
or when changing load frequently.
Regarding a pure car carrier laden with 6,000 units, the wind force on front at a force of 16m/s (it will be time at 1.50
= 24m/sec, if under the gusting at storm) is 22 tons. Because 80% of the horsepower (one ton ≒ 100PS) is necessary,
the following amount of power from the bow thruster will be required.

22 tons × 0.8 × 100 ＝ 1,760PS② Combination use with main engine
In theory, it is assumed that ship operating using both the main engine and the steering in order to stand the hull
against the wind can limit horsing (yaw and sway) and reduce tension on the anchor cable. However, from a practicalperspective, a situation that requires the use of an engine will mean that it is getting closer to critical wind speed foranchoring. In addition, according to the usage of the engine, the anchor chain may repeatedly slacken and then becometaut, which may rather increase the impact force and lead to dragging achor.
The result simulated using the case of Vessel C above in 3-3 is introduced in the Maritime Casualty Analysis Report(Vol. 6): “Typhoons and Marine Accidents”. The conditions for simulation are shown in Tables 75.

− 55 −（Simulation method）
Ship's particulars
Loa

224.0 ｍDraft (fore)

8.00m

Length between perpendiculars

215.0 ｍDraft (aft)11.60mTotal width32.2 ｍMean draft

9.80m

Block coeﬃcient of ﬁtness (Cb)

0.821

B

Classiﬁcation of speed force
Speed(full load / ballast)Engine rpm(revolution per minute)

Navigation full speed (Nav. Full)

14.0 knots

75 〜 77 ｒｐｍ

Harbour full speed (S/B Full)

10.3/11.0

56 ｒｐｍHalf ahead eng. speed

8.9/9.6

48 ｒｐｍSlow ahead eng. speed

7.4/8.040 ｒｐｍDead slow ahead eng. speed

5.5/6.030 ｒｐｍClassiﬁcation

Conditor of external force
Wind speed

25m/sWind direction

East-northeast (direction of 67.5 degrees)

Wave height of swell/wavelength

5m/200m

Wave height of swell and wavelength incident angle

Direction of 118 degrees

Table 75

DSP Simulator
Photograph 76 This was not the actual simulator used.

− 56 −（Simulation method）
By rotating the propeller during horsing (yaw and sway) motion, in order to supply the need for holding power
against the force, propeller thrust power applicable to the ship speed in ﬂat water of 6 to 14 knots was given. Then, anevaluation was conducted using the following calculation, in the event of giving the propeller thrust power.
Maximum of cable tension（ Tmax ）Maximum of holding power（ FAmax ）Minimum of holding power（ FAmin ）The method of evaluation will be described below. A summary of it is shown in Table 77.
When Tmax does not exceed FAmin ：（ Tmax ＜ FAmin ） ⇒ “Non-dragging of anchor”
When Tmax exceeded FAmin：（ Tmax ＞ FAmin ） ⇒ “Dragging anchor”

Anchorage in heavy weather using engine Simulation results
Anchor cable
VeeredEquiv. to 6kt(D.Slow)
Dead slow aheadeng. speed

Equiv. to 8kt(Slow)
Slow aheadeng. speed

Equiv. to 10kt
(Half)
Half aheadeng. speed

Equiv. to 11kt
(S/B Full)
Harbour full
speed

Equiv. to 14kt
(Nav. Full)
Navigation fullspeed

(tons）

51.1

48.8

50.0

0.0

0.0

FAmin (tons）

16.2

48.8

48.8

64.2

64.2

FAmax (tons）

52.6

56.5

64.2

64.2

64.2

Judgement

Dragging
anchorNon-dragging
of anchor

Excess thrust

Excess thrust

Excess thrust(tons）

51.9

48.8

58.8

0.0

0.0

FAmin (tons）

22.1

48.8

21.4

70.2

70.2

Propulsion /judgement

Tmax

6 shackles

Tmax

8 shackles

FAmax (tons）

58.6

62.5

70.2

70.2

70.2

Judgement

Dragging
anchorNon-dragging
of anchor

Excess thrust

Excess thrust

Excess thrust(tons）

48.8

48.8

61.2

0.0

0.0

FAmin (tons）

13.6

48.8

48.8

82.2

82.2

FAmax (tons）

70.6

74.6

82.2

82.2

82.2

Judgement

Dragging
anchorNon-dragging
of anchor

Excess thrust

Excess thrust

Excess thrust

Tmax

12 shackles

Table 77

− 57 −

Regarding Vessel C case, the Master started to use the engine before anchor dragging started. Under the 25m/s windspeed condition, it was possible to prevent anchor dragging with engine thrust power at the slow ahead engine speed(Slow Ahead), but there was a shortage of thrust power when it was set to Dead slow ahead engine speed (D. Slow
Ahead). However, the thrust power was too large at more than Half Ahead engine speed (Half Ahead) which caused
the anchor to drag as a result.
Vessel C attempted to control the ship’s posture by setting the engine to full speed, however, she was exposed to thewind from the Unmanouvrable area during dragging anchor caused by wind pressure, which will be referred to below,
and as a result, she was presumably grounded without the Master being able to control the ship’s posture.
In addition, because the wind does not blow in the same direction and speed, it is necessary to fully understand thatwrong usage of an engine may rather lead to greater impact force.
In the event that large vessels can evacuate to the open sea, it would be ideal to refuge at a point away from thetyphoon by heaving up anchor rather than staying anchored while using the engine.

４−７

Diﬃculty in maintaining manoeuvrability after dragging
anchor has started

If the ship swings broad on to the wind and begins to move to leeward, considerable time and space may be required
before effective control can be restored by using the main engine, rudder and bow thruster. In addition, when thepropeller is working the eﬀect of the bow thruster will be decreased by about 20% per 1 knot of ahead speed. (Ahead
speed at about 5 knots, turning round yields no aﬀect even when setting to full power.) It is important to understandthe Unmanouvrable area of the ship.
12
Wind velocity/ ship speed ratio

10

Unmanouvrable area

Unmanouvrable area

8
6
Maneuvering Area
4
2
0

60

120

180

240

300

360

Relative wind direction from bow (degrees）
A pure car carrier laden with 4,500 units: h（water depth）/d（draft）= 1.3
Graph 78

Graph 78 indicates the Unmanouvrable area of a pure car carrier laden with 4,500 units. This illustrates the limitations
on manoeuvrability imposed by various wind strengths compared to the vessel’s speed. Numbers entered in the− 58 −

vertical axis are wind speed per ship’s speed and the wind force angle is entered along the horizontal axis. Except forwhen being exposed to tailwind from the stern direction, it is an Unmanouvrable area area if the wind speed is morethan four times that of the ship’s speed.
For example, in the event of turning round to the starboard side (windward) by being exposed to wind from the
starboard side, it is impossible to change course at more than an 8 of wind velocity/ship speed ratio. On the other hand,in the event of turning round to the port side (leeward) by being exposed to wind from the port side, it is impossible tochange course at a 4 of wind velocity/ship speed ratio. The cause is the transient change due to the mutual inﬂuencesof wind pressure moment, water resistance moment and steering moment.
With a wind speed of 20m/s, the ship’s speed would have to be more than 5 knots (2.5m/s), in order to stand a chanceof stemming the wind. If it is more than 5 knots, the bow thruster will cease to be eﬀective.

Note:

A difference in a ship’s speed will permit as to whether the vessel can maintain a given course.
This should not be confused with limit for course keeping. The limit for course keeping is an area
whereby it is not possible to keep on course if the wind velocity/ship speed ratio is large during
the voyage. It is different from limitation of manoeuvring by rudder.

PCC loaded with 6000 vehicles
Curve of limit of ability to maintain course (rudder angle of 30°)

The graph on the right shows
the limit for course keeping,
the area whereby it is or is not
the rudder angle at 30 degrees.
Once the wind velocity/ship
speed ratio exceeds 3.7, it will
be in an area whereby it will not
be possible to keep on course
because of the relative winddirection from bow. This should
not be confused with limitationof manoeuvring by rudder.

Wind speed (Va)/Vessel Speed(Vs)possible to keep on course at

Region in whichcourse cannot be
maintained

Region in whichcourse can be
maintained
°

°

°

°

°

°

°

Relative wind direction
Graph 79

４−８ Safe distance from other ships, shallows and other facilities
There is no definite criteria regarding a safe distance to keep from other ships, shallows and other facilities. The
reasons are as follows.
The following items are to be considered, when focusing on the area of sea one is to use for restoring the ship’s postureusing a combination of the main engine, steering and bow thruster promptly after discovering a dragging anchor.

− 59 −

①

A radius of swinging circle

A circle with a radius of minimum required length of anchor cable + theship’s Loa②

S p e e d of d r a g g i n g a n c h o r
under wind pressure force:

approximately 3 - 4 knots③

Required time to weigh anchor:

In general, an anchor cable will be retrieved at a rate of about 9 m/min andto retrieve 1 shackle takes about 3 minutes.
It may not be possible to heave in a taut cable continuously, thus prolonging
the operation of weighing anchor.

④

Time needed for preparing themain engine:

Early S/B Eng.

⑤

Time required to obtain speed ahead needed to restore the ship to full manoeuvrable condition after being
exposed to wind from the leeward

Compared with the situation of weighing the anchor during dragging anchor with yaw and sway (the ﬁrst stage) andthen regaining control of the hull, with weighing the anchor after the anchor has dragged caused by wind pressure, thesimulation result indicates that approximately 3.5 times the amount of water area on the leeward side will be required.

Simulation of the time full control is achieved over the ship’s manoeuvrability
following dragging anchor: Case study
Turning Circle

Case 2

Steering takes eﬀect

0.23 nautical miles

1.0 nautical miles
5kts
Weigh the anchor

Case 1

2.4 nautical miles

Weigh the anchor

1.0 nautical miles

5kts
0.3 nautical miles（3L）

0.625 nautical miles
Fig. 80− 60 −

Case1 Weighing anchor after dragging anchor is detected
In the case of

Fig. 80

The ship’s LOA is 200m.
She is lying at anchor with 8 shackles.
Time to weigh anchor is 1.5 times longer than is usually required.
During that period, the ship is being brought to leeward at 4 knots.
After weighing anchor, it takes 15 minutes until the ship’s speed reaches 5 knots.
In the interval, the ship’s side has been pressed by the wind.
The main engine and bow thruster are ready prior to dragging anchor.
Whilst weighing anchor, both are available. The wind speed was at 20m/sec.
It has been clarified that accidents may be caused, if there is no navigable water area with 3.63 nautical miles to the leeward
direction and 0.925 nautical miles in the horizontal direction. It would be difficult in many cases to secure this size of waterarea when needing to shelter from a typhoon.
1 Turning Radius of the Circle is:
8 shackles (220 m) ＋ 200 m ＝ 420 m ＝ 0.227 nautical miles
2 Time needed for weighing the anchor:
After dragging anchor caused by wind pressure starts, it is no longer possible to manoeuvre the ship. The anchor cable
recovery rate in moderate weather is known to be 9m per minute. Therefore, it would take about 24 minutes to weigh anchor
in moderate weather. However, in this situation, it would take 1.5 times as long or about 36 minutes.
In this interval, the distance which the ship is pressed to leeward = 4 knots x 0.6 hours = 2.400 nautical miles (distance coveredin 36 minutes at 4 knots).
3 After the anchor is aweigh, a further 15 minutes elapse before the vessel’s speed reaches 5 knots. During this period,
the vessel will drift about 1 mile to leeward (distance covered in 15 minutes at 4 knots).
0.25 hours × 4 knots ＝ 1.000 nautical mile
Leeward direction:

3.627 nautical miles (6,717m) is necessary.

4 The distance sliding transverse
Advance at 5 knots is (3L)

＝ 0.300 nautical miles（0 knot ＋ 5knots）÷2 × 0.25 hours＝ 0.625 nautical miles

Horizontal direction: 0.925 nautical miles (1,713m) are required

Case2 Started to weigh the anchor during dragging anchor with yaw and sway
In the case of

Fig. 80

If weighing the anchor at the early stage of dragging anchor with yaw and sway, it may be possible to weigh the anchor
almost right away using the engine for operation to weigh the anchor accordingly. Even if the vessel has entered into an
unmanouvrable area and, even worse, exposed to wind from the side, as long as the following water area both for leeward
direction and sliding transverse are secured, it is possible to achieve full control over the ship’s manoeuvrability.

Leeward direction:

1.230 nautical miles (2,278m)

Horizontal direction: 0.925 nautical miles (1,713m)
Most accidents such as collisions and groundings involving ships occur when an anchor drags because an insuﬃcient amount of sea
area was not secured.

− 61 −

§5

Conclusion

Three diﬀerent cases and mechanisms of dragging anchor were introduced. We are sometimes forced to anchor to
shelter from stormy weather such as from large swells and wave heights exceeding 5 meters.
However, as can be seen from the introduced cases, even though there were opportunities to evacuate from the sea
area away from the typhoon, a large number of accidents were caused due to a lack of weather information.
In the case of large vessels that had a method of riding at two anchors and two-anchor mooring, it is concluded thatmainly single anchor mooring was decided on because of the diﬃculty of ship handling.
Therefore, it is a requirement that as much weather information be available as possible in order better ensure safety.
Also, the Master should not be alone in making a judgement, it is also necessary for operators and ship managementcompanies to support the Master. Even if dragging anchor is already under way, the ship’s posture may be adequately
controlled if it is detected at an early stage.

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References
・

Maritime Casualty Analysis Report (Vol. 6):
Typhoons and marine accidents issued by Marine Accident Inquiry Agency (MAIA) in 2006

・

Marine Accident Inquiry and Safety Investigation Association, “Determination record by Marine Accidents
Inquiry Agency”

・Theory and Practice of Ship Handling, by Dr. Kinzo Inoue・Japan Captains’ Association, DVD “Safety Maneuvering in Restricted Water Area”

・

Ship Handling, by Dr. Keinosuke Honda

・

Ship Handling, by Dr. Satoshi Iwai・

Study of Ship Anchoring Supporting system (Kobe University)

Reference documents were provided by:・Japan Captains’ Association

・

Nippon Yusen Kabushiki Kaisha・Japan Marine Science Inc.

− 63 −

Capt. Takuzo Okada
Master Mariner
General Manager
Loss Prevention and Ship Inspection Dept.
The Japan Ship Ownersʼ Mutual Protection & Indemnity Association

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Translated by : YUZEN Translation LLC− 64 −