magnetic & gyro compass

8/18/2019 Magnetic & Gyro Compass

1/19

Magnetic & Gyro Compass / Chief Mate

Question (1):

At Gibraltar, H= 0.26, Z=+ 0.32, the deiation due to:

(1) !er"anent "a#netis" $as %& ', (2) due to ertial sot iron $as 6& *, (3) due

to horiontal sot iron $as & '. -alulate the deiation due to eah ause at

elbourne $here H= 0.22, Z= / 0.%6, $hat inerene an be dra$n $ith res!et to

(2) aboe $hen loated at or near the "a#neti euator

olution:

1) To calculate the deviation caused by permanent magnetism:

Where δ 1 and 1 related Gibraltar! and δ 2 and " related to Melbourne#

δ 1

δ 2=

H 2

H 1 δ 2=δ 1×

H 1

H 2 δ 2=5×

0.26

0.22

δ 2=5.9⁰W

") To calculate the deviation caused by vertical soft iron since the varies directly asZ

H

Where δ 1 ! 1 and $1 related Gibraltar! and δ 2 !" and $" related to Melbourne#

δ 2=δ 1× H 1

Z 1×

Z 2

H 2 δ 2=6×

0.26

0.32×0.56

0.22 δ 2=12.4⁰W

Question (2):

4 alta (H=0.26) oe5ient !er"anent ( = / 7&). -alulate the deiation on: (1)

036& -, (2) 160& -, (3) 328& -.-alulate also the deiation on 206& - at (a) 9reeto$n (H = 0.30) (b) Halia, ;..

(H= 0.16) and () o4 ai#on (H= 0.


2/19


') δ 2=9×

sin 33

sin 90 δ 2=4.90⁰ E

%") To calculate the deviation heading "( C at: %a) . #'! %b) . #1(! %c) . #

Where δ 1 !1 and θ1 related to the coe+cient! at Malta and , C or "* C

respectively! and δ 2 ! " and θ2 are the re-uired deviation! at re-uired position! and

the heading respectively#

Thus!

δ 1

δ 2=

H 2

H 1× sin θ1

sinθ2 δ 2=δ 1×

H 1

H 2× sin θ2

sinθ1

a) 0reetonδ 2=9×

0.26

0.30× sin 26

sin 90 δ 2=3.42⁰ E

b)alifa2! 3#4δ 2=9×

0.26

0.16×sin 26

sin 90 δ 2=6.41⁰ E

c) 4aigonδ 2=9×

0.26

0.40×sin 26

sin 90 δ 2=2.56⁰ E

Question (3):

he deiation due to !er"anent "a#netis" on a shi! $hen headin# north b>

o"!ass is 8 * at a !osition $here (H = 0.1), (Z = +0.


3/19


δ 1

δ 2=

H 2

H 1× cosθ1

cosθ2 δ 2=δ 1×

H 1

H 2× cosθ2

cosθ1

Where δ 1 !1 and θ1 related to the initial condition! and δ 2 ! " and θ2 related to

5nal condition#

a) δ 2=7× 0.18

0.14× cos 32

cos0 δ 2=7.630W

b)δ 2=7×

0.18

0.14×cos 48

cos0 δ 2=6.020 E

Question (


4/19


") To calculate the deviation on heading "" C:

δ 1

δ 2=

cosθ1

cosθ2 δ 2=δ 1×

cosθ2

cosθ1

Where δ 1 and θ1 related to the values at the 5rst position! and δ 2 and θ2 related

to the values of the second position#

δ 2=δ 1×

cosθ2

cosθ1 δ 2=9×

cos22

cos51 δ 2=13.260 E

') To calculate the deviation due to the changes of course and latitude:

δ 1

δ 2=

H 2

H 1×cosθ1

cosθ2 δ 2=δ 1×

H 1

H 2× cosθ2

cosθ1

Where δ 1 ! 1 and θ1 related to the values at the 5rst position! and δ 2 ! " and

θ2 related to the values of the second position#

δ 2=δ 1× H 1

H 2× cosθ2

cosθ1 δ 2=9×

0.32

0.15×cos73

cos51 δ 2=8.920W

Question (%):

Anal>e the ollo$in# list o deiations into the a!!roi"ate oe5ients A, , -, ?

and * and -alulate the deiation on the ourses 112& - and 272& -.

H?G ?*@ A - ? *; 6 7 86 86 86;* 1" 7 81" 81"*

* 1* W 91* 91* 1" W 91" 91" 91"

' * 7 8* 8*' 1" 7 81" 81" 81"

;' 7 8 8

Coff . A=δ ( N + NE+ E+SE+S+SW +W + NW )

8 Coff . A=+7.5 E

Coff . B=δ ( E−W )

2 Coff .B=−6W


5/19


Coff .C =δ ( N −S)

2 Coff .C =+10 E

Coff . D=δ ( NE+SW )−( NW +SE)

4 Coff .D=+7.75 E

Coff . E=δ ( N +S)−( E+W )

4 Coff .E=−4W


6/19


1) ;t course 11"

δ = A+B sinco .+C cos co .+ D sin2co .+ E cos2co .

δ =(7.5)+(−6)sin1120+(10)cos1120+(7.75)sin 2240+(−4 )cos224⁰

δ =−4.3W

") ;t course ","

δ = A+B sin co .+C cos co .+ D sin 2co .+ E cos2co .

δ =(7.5)+(−6)sin 2920+(10)cos2920+(7.75)sin2240+(−4)cos224⁰

δ =+14.3 E

Question (6):

9ro" >our stud> o "a#neti o"!ass e!lain the ollo$in#:

1) the nature o oe5ient .

2) -ause o oe5ient .

3) *4et o han#in# latitude and ourse.


7/19


b)


8/19


Question (8):

'hen headin#


9/19


3ote: ;t position %") $ is negative so that the polarity in the ship=s vertical soft iron is

reversed ith the change from 3orthern to 4outhern hemisphere and the sign of induced A is

also reversed#

;t >osition %") total coe+cient A . >ermanent A 8


10/19


Question (7):

E the deiation due to hard iron in a shi! is 10 ', $hen headin# 070 (-), at a

!osition (1) $here H= 20 a"!eres !er "eter (AC"), Dnd the deiation due to the

sa"e iron:

(a) $hen on the sa"e headin# at a !osition (2) $here H= 16 AC".

(b) $hen headin# 31% (-) at a !osition $here H= 30 AC".

The deviation due to this iron comprise coe+cient permanent A and since Westerly deviationis caused on an 7asterly course it is named negative# The deviation varies inversely as hard

directly as the 4in Co#

olution:

a) Comparing the deviation at position %1) & %")

δ 2

δ 1=

H 1

H 2× sin θ2

sinθ1 δ 2

−10=

20

16× sin 90

sin 90 δ 2=−10×

20

16× 1

1

δ 2=−12.5 ° or %1"#D W)

a) Comparing the deviation at position %1) & %')

δ 3

δ 1=

H 1

H 3× sinθ3

sinθ1 δ 3

−10=

20

30×sin315

sin 90

δ 3=−10×20

30×−0.707

1

δ 3=4.7 ° or %#*D 7)

1


11/19


Question (10):

9ro" belo$ table obtain a list o deiations, (1) anal>se the" into the a!!roi"ate

oe5ients and, i the s$in# tooF !lae on the "a#neti euator, (2) state ho$ the

arious orretors should be adusted. Et $as disoered that the ai"uth "irror

$ith $hih the bearin#s $ere taFen introdued oe5ient a!!arent A=/1.

Head ; ;* * * ' ' ;'

rue earin# 107 107.% 110 111 111 111.% 112 11<-o"!ass

earin#110.% 112.% 111.% 111.% 118.% 12


12/19


%E) The soft iron spheres are over compensating and should be moved further from the

compass to e-ual distances each side in order to change the deviation 'D on any one of the

four intercardinal compass heading#

%7) Aecause coe+cient 7 is small and unliKely to embarrass the navigator! it is! in this case!

usti5ably ignored#

1"


13/19


Question (11):

*!lain ?a"!in# *rror

This error is present in a compass hich is damped in tilt# ; compass damped in tilt alays

settles east of the meridian and above the horiHon in 3! and vice9versa#


14/19


Question (1 ontrol in #>roso!e $ith sFethin# the e!lanation

The gyro spin a2is can be made meridian9seeKing by the use of a pendulum acting under the

inNuence of earth gravity# The pendulum causes a force to act upon the gyro assembly

causing it to process# >recession! the second fundamental property of a gyroscope! enables

the instrument to become north9seeKing# ;s the pendulum sings toards the center of

gravity! a donard force is applied to the heel a2le! hich causes horiHontal precession to

occur#

Question (1%):'hat is the e4et o o! hea> in #>roso!e and used orGyrocompasses to be e@ectively top9eighted and use an anticlocKise spinning rotor# Autadding a eight to the top of the rotor casing produces a number of Ondesirable e@ects#

These e@ects become pronounced hen a ship is subected to severe movement in heavyeather# To counteract unanted e@ects! an PapparentP top eighting of the compass isachieved by the use of a mercury Nuid ballistic contained in to reservoirs or ballistic pots#

Question (16):

'hat is the use o botto" hea> in #>ro so!e ("aFe >our e!lanations b>sFethin#)

This action causes the north end of the spin a2is! of a gravity9controlled undamped gyro! todescribe an ellipse about the meridian# Aecause it is un damped! the gyro ill not settle onthe meridian# 4hos this action for a gyro ith a clocKise rotating spinner# The ellipseproduced ill be anticlocKise due to the constant e2ternal inNuences acting upon the gyro#

The e2tent of the ellipse ill! hoever! vary depending upon the initial displacement of thegyro spin a2is from the meridian and from the earthPs horiHontal# The term Pnorth9seeKingP isgiven to the undamped gravity controlled gyro mechanism because the northeast end of thespin a2is describes an ellipse around the 3orth >ole but never settles# Bbviously such a gyrois not suitable for use as a precise north reference compass aid

Question (18):*!lain $ith sFethin# in brie the !rourer to "aFe #>roso!e north seeFin#Ay top or bottom heavy# Aottom9heavy control and a clocKise rotating gyro# top heavysystem ith an anticlocKise rotating spinner#With bottom9heavy control! tilting upards of the south end produces a donard force onthe other end! hich! for this direction of spinner rotation! produces a precession of the northend to the est#


15/19


1


16/19


Question (1):*!lain the da"!in# settlin# error in north I outh he"is!here b> sFethin# onl>

Question (17):

*!lain the ballisti deJetion error and auses o it

This condition arising from changes in speed or course hereby an acceleration or

deceleration force is set up hich alays acts on the pendulous mass of the gyro as ell as

on the damping Nuid# This force no matter ho small! acting on the pendulous mass! ill

cause the gyro to precess#

?esults from changes in shipPs south north component of velocity and speed changes#

Question (20):

9ro" >our stud> o #>roso!e desribe the ollo$in#:

1# #>roso!e inertia. (Ki#idit> in the s!ae:( it is the ability of a free gyroscope to

remain ith its spin a2is pointing in the same 52ed direction in space regardless of ho

the gimbal support system may"# ?rit: 9 is the direction in hich spin a2is points relative to the true north#'# ilt: /is the angle of elevation or de9prerecession of the spin a2is above or belo the

horiHontal## Breession: 9 is the movement at right angles as a reaction of applying a tor-ue to a

spin a2is at right angle#


17/19


Question (21):'hat do >ou understand ro" the ite"s

• Tilting: 9 is the rate of change of the tilt of the spin a2is# Tg . 1 cos lat# Q sin ;H#• Erifting is the rate of change of the drift %;H#) of the spin a2is# Eg . 1 sin lat#

Question (22):

*!lain ho$ a ree #>roso!e an be "ade north seeFin# b> the use o #rait>

ontrol

Ay placing top/heavy eight on top on the rotor casing! such that hen the spin a2is is

horiHontal the vertical through the center of gravity of the eight passes through the center

of the rotor#

g This achieved by means of gravity control in all lats# %394)! the north end of the gyro spin

a2is ill tilt up if it is to east of the meridian and fall if it is to the est of the meridian this tiltcauses the precession in the re-uired direction by top eight# %the south end rotating

anticlocKise! otherise ill be bottom eight)

1*


18/19


Question (23):

tate the e4et o da"!in# to "aFe #>roso!e as #>ro o"!ass $ith sFeth as

need to e!lain

Eamping! in order to respond to the drift! tilt and precession hich maKes it north seeKing!

the suspension of gyroscope must be virtually frictionless and a gravity controlled gyroscope!

ould then oscillate inde5nitely on either side of the meridian#

Eamping in tilt means that hen the north seeKing end of the gyroscope a2is is tilted a

damping tor-ue is applied in a horiHontal plane#

Question (2ro o"!ass $hen steerin#

160 () at s!eed 21.2 Fts. En lat


19/19


. 9 1#*' o

To calculate hen altering course

To calculate hen change speed

Bn 4outherly courses error is o // Bn 3ortherly courses error is high

Question (2%):

En $hat lat. 'ill the ourse, lat and s!eed error not eeed % $hen stea"in# at 20

Lts.

Cos 16 or . one

1,

magnetic & gyro compass

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