Thursday, November 29, 2012

PON numerical


PROBLEM   No  25
IF  THE  DISTANCE  BETWEEN  THE  CENTRES  OF  THE  EARTH AND THE MOON  IS  30  TIMES  THE  DIAMETER  OF  THE  EARTH.  CALCULATE  THE  MOON’S  H.P.

IF DIAMETER  IS  2X
RADIUS  IS   X

DISTANCE BETWEEN  CENTRES  60X

SIN  H.P.                    =            X  
                                                60X

H.P.                             =          57.3’                ANS



PROBLEM   No  26
ASSUMING THE EARTH’S RADIUS TO BE 6373 KMS AND THE HORIZONTAL PARALLAX OF MOON TO BE 58.0’.  CALCULATE THE DISTANCE OF THE MOON  FROM  THE  EARTH



SIN  H.P.                    =          AB
                                                AC

SINE  0˚  58’              =          6373
                                                  AC

AC                              =          6373
                                                SIN 0˚ 58’

AC                              =          377755KMS   ANS












PROBLEM  No  34
ASSUMING  THE V OF THE MOON TO BE A CONSTANT 11’,  CALCULTE  THE LENGTH  OF A LUNAR  DAY.


LUNAR  DAY:
TWO SUCCESSIVE  TRANSITS OF  MOON  OVER  THE  SAME  MERIDIAN

GHA  MOON  1HR               14˚  19.0’
CONSTANT V CORRN              11.0’
TOTAL  C. GHA 1 HR          14˚  30.0’

1  HR  ----------------------         14˚  30.0’
1 LUNAR  DAY  ---------       360˚

LUNAR  DAY                       =          360
                                                            14˚  30’

LUNAR  DAY                       =          24H     49M     39S      ANS


Sunday, November 25, 2012

PON NUMERICALS


PROBLEM  No  20
IF  THE  SUN’S  DECLINATION  IS  15º  30’N AND INCREASING,  CALCULATE  THE  SUN’S  SHA  ASSUMING  OBLIQUITY  OF  THE  ECLIPTIC  TO  BE  23º 26.7’.

GIVEN:
SUN’S  DECLINATION  15º  30’N AND INCREASING
OBLIQUITY  OF  THE  ECLIPTIC 23º 26.7’.


SINE  AB       =          TAN  BX  TAN [90-A]
                        =          TAN  15º  30’  TAN  66º  33.3’
AB                  =          39º  45.2’
R. A. OF SUN=          39º  45.2’
SHA OF SUN            =          320º  14.8’                               ANS
                       

PROBLEM  No 21
IF  THE  EQUATION  OF  TIME  WAS  + 04M     06S,  WHEN  RAMS  WAS  14H 32M 15S,  CALCULATE  THE  SUN’S  DECL.

GIVEN:
EQ  OF  TIME            =         04M     06S
R.A.M.S.                      =         14H     32M     15S


EQ  OF  TIME            =          MEAN  SUN  -           TRUE SUN [  WESTERLY]
                                    =          TRUE   SUN  -           MEAN  SUN [ EASTERLY]
                                    =          R.A.T.S.          -           R.A.M.S.
04M     06S                  =          R.A.T.S.          -           14H     32M     15S
R.A.T.S.                      =          14H     36M     21S
            =          219º  05’  15’’



MC                              =          219º  05’  15’’ -           180º
                                    =          39º  05’  15’’

OBLIQUITY              =          23º  26.7’

SINE  MC                   =          TAN  MT  TAN [90-C]
SINE 39º  05’  15’’     =          TAN  MT  TAN [90- 23º  26.7’]
MT                              =          15º  17.2’
SUN’S   DECL           =          15º  17.5’  S                            ANS
                       

PROBLEM  No 22                H.W.
SHAMS  16H 06M     10S      ,  EQUATION  OF  TIME  [-] 02M   48S, OBLIQUITY  OF  THE  ECLIPTIC  TO  BE  23º 26.7’.  CALCULATE  THE  SUN’S  DECLINATION



EQ  OF  TIME            =          MEAN  SUN              -           TRUE SUN [  WESTERLY]
                                    =          SHAMS                      -           SHATS
[-] 02M            48S      =          16H     06M     10S      -           SHATS
SHATS                       =          16H     08M     58S
                                    =          242º  14.5’  -  180º
CE                               =          62º  14.5’        

OBLIQUITY              =          23º  26.7’

SINE  CE                    =          TAN  ET  TAN [90-C]
SINE 62º 14.5’           =          TAN  ET  TAN [90- 23º  26.7’]
                                    =          TAN ET  TAN  66º  33.3’
ET                               =          20º  59.7’
SUN’S   DECL           =          20º  59.7’  N                            ANS


PROBLEM  No 23
GHA  ARIES γ 06º  13’,  GHA  SUN  270º  43’,  SUN’S  DECLINATION  23º  20.9’N.  CALCULATE  OBLIQUITY  OF  ECLIPTIC.


FROM  γ  TO  TRUE  SUN
RATS              =          360º     -           270º  43’          +          06º  13’
γTS      =          95º  30’

MT                  =          DECL 23º  20.9’

SINE  MC                   =          TAN  MT  TAN [90-C]
SINE 84º  30’             =          TAN  23º  20.9’ TAN [90- C]
90-C                            =          66º  33.3’
C                                 =          23º  26.7’
OBLIQUITY              =          23º  26.7’                                 ANS







PROBLEM    No  24
GIVEN  DECLINATION OF THE SUN  10º  15’N  AND  DECREASING,  SHAMS  206º  36.8’.   CALCULATE  THE  VALUE  OF  EQUATION  OF  TIME.

SINE  AC                   =          TAN  AB  TAN [90-C]
            =          TAN  10º  15’ TAN [90- 23º  26.7’]
AC                              =          24º  38.6’

SHATS                       =          180º     +          24º  38.6’
                                    =          204º  38.6’

GIVEN  SHAMS       =          206º  36.8’

EQ  OF  TIME            =          MEAN  SUN  -           TRUE SUN [  WESTERLY]
                                    =          SHAMS          -           SHATS
                                    =          206º  36.8’       -           204º  38.6’
                                    =          1º  58.2’
                                    =          [+]  07M          52.8S                           ANS


PROBLEM    No  31
AT A CERTAIN  TIME, THE  SUN’S   RA  WAS  03H  56M  40S  AND  GHA γ  WAS  312º  48.4’.  ASSUMING  OBLIQUITY OF THE ECLIPTIC  23º  27’.  CALCULATE  THE  GP  OF  THE  SUN.


SINE  AB                   =          TAN  BC  TAN [90-A]
SINE 59º  10’             =          TAN  BC   TAN [90- 23º  27’]
            =          TAN  BC   TAN [66º  33’]
BC                               =          20º  25.7’  [LAT OF BODY]             ANS


GHA γ                                    =          312º  48.4’
Gγ [Easterly]               =          47º  11.6’
RA☼                           =          59º  10.0’
LONG                         =          106º  21.6’  E  [LONG  OF  BODY] ANS










PROBLEM  No  2
WHEN  GHA  ARIES γ WAS  212º  14’,  THE  EASTERLY  HOUR  ANGLE  OF  THE TRUE  SUN  WAS  35º  TO  AN  OBSERVER  IN  LONG  35º  WEST.  FIND  THE  R.A.  OF  THE  TRUE  SUN


EHA  IS  MEASURED  EASTWARD  FROM  OBSERVER  TO  TS
R.A. IS MEASURED  EASTWARD  FROM  γ  TO  TS


GHA  γ                       =          RATS
                                    =          212º   14’
                                    =          14H     08M     56S      ANS


PROBLEM  No  4
FIND  THE  TRUE  SUN’S  SHA  AT  THE  INSTANT  WHEN  THE  FIRST  POINT  OF  ARIES  CROSSED  THE  MERIDIAN  OF  85º E,  IF  ON  THAT  DAY  THE  SUN’S  GHA  WAS  60º  12’  WHEN  GHA γ WAS  255º



GIVEN:
GHA  TS                     60º  12’
GHA γ                                    255º
LONG                         85º  E

FROM γ  TO  OBSERVER   =          360º     -           255º     -           85º
                                                =          20º

FOR γ TO  COME  TO  OBSERVER’S  MERIDIAN  IT REQUIRES  TO CROSS  20º
360º---------------          23H  56M  04S  [SIDEREAL  TIME]
20º  ---------------         01H  19M  47S

WHEN  ARIES  COMES  ON  OBSERVERS  MERIDIAN  ,  SUN  MOVES  WESTWARD   BY AN AMOUNT
01H  ---------------        15º  [  SOLAR  TIME ]
       01H  19M  47S ---------------         19º  56.7’

SHA TS          =GHA TS       +LONG+ADDITIONAL AMOUNT TO MERIDIAN
                        =  60º  12’       +85º     +  19º  56.7’
                        =  165º  08.7’              ANS




PROBLEM  No  6
TO  AN  OBSERVER  ON  A  SHIP  AT  ANCHOR  IN  A  NORTHERN  LATITUDE,  THE  SUN  ROSE  AT  0559  SMT  AND  SET  AT  1806  SMT.  CALCULATE  THE  EQUATION  OF  TIME

SUNRISE                    05H     59M

SUNSET                      18H     06M

TIME FROM SUNSET TO SUNRISE         12H  07M

HALF TIME FROM SUNSET TO SUNRISE         06H  03M  30S

SMT OF MERPASS  =          05H  59M        +          06H  03M  30S
                                    =          12H  02M  30S
LAT  OF  MERPASS =          12H  00M  00S

EQ  OF TIME             =          02M  30S
                                    =          MEAN  TIME            -           APPARENT  TIME
                                    =          12H  02M  30S-          12H  00M  00S
                                    =          [+]       02M  30S        ANS


PROBLEM  No  29
FOR A VESSEL ON A CONSTANT HEADING AT THE SAME POSITION,  THE SUN  ROSE  BEARING  086˚[C]  AND SET  BEARING  292˚[C]. IF THE DEVIATION OF THE COMPASS WAS 2˚E, FIND THE VARIATION

GIVEN;
SUNRISE                    086˚

SUNSET                      292˚

VARIATION             2˚E

SUNRISE                    086˚

SUNSET                      292˚

TIME FROM SUNSET TO SUNRISE         206˚

HALF TIME FROM SUNSET TO SUNRISE         103˚

BRG AT MERPASS  =          086˚     +          103˚
                                    =          189˚
LAT  AT  MERPASS =          189˚

COMPASS  ERROR =          189˚ [C]           -           180˚[T]
                                    =          9˚ W
DEVIATION             =          2˚ E
VARIATION             =          11˚ W              ANS

PROBLEM   No  9
IN  LATITUDE  50˚N,  A STAR WITH  AN  SHA  OF  146˚  10’  HAD  A TRUE  ALTITUDE  OF  51˚  36;,  WHEN  BEARING  TRUE  EAST.  FIND  THE  LOCAL  SIDEREAL  TIME


ZX                   38˚  24’
PZ                   40˚  00’


SIN  PZ           =          TAN [90-P]     TAN  ZX
TAN [90-P]     =          SIN   40˚
                                    TAN  38˚  24’
P                      =          50˚  57.5’
LHA  STAR   =          360      -           P
                        =          309º  02.5’

SINCE  TIME IS  MEASURED  FROM  OBSERVER’S MERIDIAN TO ARIES
LHA STAR    =          LHA γ +          SHA STAR
LHA γ             =          LHA STAR    -           SHA STAR
                        =          309˚  02.5’       -           146˚  10’
                        =          162˚  52.5’      
                        =          10H  49M  43S           ANS

Saturday, July 21, 2012

ECDIS


DEFINITION
  Electronic Chart Display and Information System (ECDIS) means a navigation information system which with adequate back-up arrangements can be accepted as complying with the up-to-date chart required by regulation V/20 of the 1974 SOLAS Convention, by displaying selected information from a system electronic navigational chart (SENC) with positional information from navigation sensors to assist the mariner in route planning and route monitoring, and if required display additional navigation-related information.
PERFORMANCE STANDARD:
  ECDIS should be capable of displaying all chart information necessary for safe and efficient navigation originated by, and distributed on the authority of, government authorized hydrographic offices.
   ECDIS should facilitate simple and reliable updating of the electronic navigational chart.
ECDIS should reduce the navigational workload compared to using the paper chart.
      It should enable the mariner to execute in a convenient and timely manner all route planning, route monitoring and positioning currently performed on paper charts.
   It should be capable of continuously plotting the ship's position.
   ECDIS should have at least the same reliability and availability of presentation as the paper chart published by government authorized hydrographic offices.
  ECDIS should provide appropriate alarms or indications with respect to the information displayed or malfunction of the equipment
  ECDIS should be capable of displaying all SENC information.
  SENC information available for display during route planning and route monitoring
  should be subdivided into the following three categories, Display Base, Standard Display
  and All Other Information (see Appendix 2).
  ECDIS should present the Standard Display at any time by a single operator action.
   When a chart is first displayed on ECDIS, it should provide the Standard Display at the largest scale available in the SENC for the displayed area.
   It should be easy to add or remove information from the ECDIS display. It should not be possible to remove information contained in the Display Base.
  It should be possible for the mariner to select a safety contour from the depth contours provided by the SENC. ECDIS should emphasize the safety contour over other contours on the display.
   It should be possible for the mariner to select a safety depth. ECDIS should emphasize soundings equal to or less than the safety depth whenever spot soundings are selected for display.
  The ENC and all updates to it should be displayed without any degradation of their information content.
   ECDIS should provide a method to ensure that the ENC and all updates to it have been correctly loaded into the SENC.
   The ENC data and updates to it should be clearly distinguishable from other displayed information, such as, for example, that listed in Appendix 3.
  It should not be possible to alter the contents of the ENC.
   Updates should be stored separately from the ENC.
   ECDIS should be capable of accepting official updates to the ENC data provided in conformity with IHO standards. These updates should be automatically applied to the SENC. By whatever means updates are received, the implementation procedure should not interfere with the display in use.
  ECDIS should also be capable of accepting updates to the ENC data entered manually with simple means for verification prior to the final acceptance of the data. They would be distinguishable on the display from ENC information and its official updates and not affect display legibility.
  ECDIS should keep a record of updates including time of application to the SENC.
   ECDIS should allow the mariner to display updates in order to review their contents and to ascertain that they have been included in the SENC.

Advantage of  ECDIS over paper Chart:

  Position fixing can be done at required interval without manual interference
  Continuous  monitoring of the ships position
  When interfaced with ARPA/RADAR,target can be monitored continuously
  If two position fixing system are available,the discrepancy in two systems can be identified
  Charts can be corrected with help of CD/online
  Passage planning can be done on ECDIS without referring to other publications 
  Various alarm can be set on ECDIS
  Progress of the passage can be monitored in more disciplined manner ,since other navigational data is available on ECDIS
  Various alarm can be activated to draw the attention of OOW
  More accurate ETA can be calculated
  Anchoring can be planned more precisely


ELECTRONIC CHART SYSTEM IS OF TWO TYPES
  Electronic Navigational Charts (ENCs) and
  Raster Navigational Charts (RNCs).
  Electronic Navigational Charts (ENCs) are official vector charts that conform to strict IHO Specifications that have been issued by or on behalf of a national hydrographic authority.
  ENCs consist of digitised data that records all the relevant chart features such as coastlines, buoys, lights, etc. These features and their attributes (such as position,colour, shape) are held in a database -like structure that allows them to be selectively displayed and queried, creating the potential to manipulate the chart image when displayed on screen. Because of their  vector format, ENCs can also be linked to other onboard systems to provide additional automatic features such as warning alarms.
  3 variations of the same ENC, showing minimum, intermediate and maximum data display levels.


RNC
  RNCs use raster data to reproduce paper charts in an electronic format. Their familiar ‘paper chart’ image helps users gain confidence with the use of electronic charts, by providing a direct link between display screen and chart table. RNCs consist of thousands of tiny coloured dots (pixels), that together make a flat digital image. Every pixel is geographically referenced, enabling accurate real-time (continually updated) display of vessel position when your chart display system is linked to GPS. Additional user defined information such as route plans and shoal areas can be overlaid on an RNC to provide automatic links to other  onboard systems (e.g. warning alarms) but unlike ENCs, charted features cannot be selectively displayed or queried















Saturday, July 7, 2012

ECHO SOUNDER


Basic Principle

Short pulses of sound vibrations are transmitted from the bottom of the ship to the seabed. These sound waves are reflected back by the seabed and the time taken from transmission to reception of the reflected sound waves is measured.  Since the speed of sound in water is 1500 m/sec, the depth of the sea bed is calculated which will be half the distance travelled by the sound waves.
The received echoes are converted into electrictal signal by the receiving transducer and after passing through the different stages of the receiver, the current is supplied to stylus which burns out the coating of the thin layer of aluminium powder and produces the black mark on the paper indicating the depth of seabed. 

COMPONENTS
Basically an echo sounder has following components:
Transducer – to generate the sound vibrations and also receive the reflected sound vibration.
Pulse generator – to produce electrical oscillations for the transmitting transducer.
Amplifier – to amplify the weak electrical oscillations that has been generated by the receiving transducer on reception of the reflected sound vibration. 
Recorder  - for measuring and indicating depth. 
CONTROLS
An echo sounder will normally have the following controls:
Range Switch – to select the range between which the depth is be checked e.g.  0- 50 m, 1 – 100 m, 100 – 200 m  etc.  Always check the lowest range first before shifting to a higher range.
Unit selector switch – to select the unit feet, fathoms or meter as required.
Gain switch – to be adjusted such that the clearest echo line is recorded on the paper.
Paper speed control – to select the speed of the paper – usually two speeds available.
Zero Adjustment or Draught setting control – the echo sounder will normally display the depth below the keel.  This switch can be used to feed the ship’s draught such that the echo sounder will display the total sea depth.  This switch is also used to adjust the start of the transmission of the sound pulse to be in line with the zero of the scale in use.
Fix or event marker  - this button is used to draw a line on the paper as a mark to indicate certain time e.g. passing a navigational mark, when a position is plotted on the chart etc.
Transducer changeover switch – in case vessel has more than one switch e.g. forward and aft transducer.
Dimmer – to illuminate the display as required.  
Pulse Length
The pulse length is the duration between the leading edge and the trailing edge.The pulse length determine the minimum distance that can be measured by the echo sounder.The minimum measurable distance will be equal to the half of the pulse length.for the shallow water short pulse is used while for the deeper water long pulse is used.
Pulse repetition frequency
This is the nos of pulse transmitted per second.This determines the maximum range that can be measured by the echo sounder.The PRF is normally automatically selected and changes as the range scale is changed.for lower range,High PRF is used whereas for the higher range ,low PRF is used.
TRANSDUCER
Electrostrictive transducer
This type makes use of the special properties of crystals (e.g. crystals of barium-titanate and lead zirconate). If an alternating voltage is applied to the opposite faces of a flat piece of one of the above materials, the crystal will expand and contract, and hence vibrate creating sound waves for  as long as the vibrations continue. The process is reversible, i.e. when varying pressure from a returning echo, is applied to the opposite faces, an alternating voltage is generated across the faces and the same can be further amplified and used to activate an indicator.
Magnetostrictive transducer
In this type, the use is made of the magneto-striction effect which is a phenomenon whereby magnetization of ferromagnetic materials produce a small change in their dimensions, and conversely the application of mechanical stresses such as weak pressure vibrations, as from an echo to them, produce magnetic changes in them; e.g. a nickel bar when placed in the direction of or strength of the magnetic field. If the nickel bar is placed in a coil with an alternating current flowing through it (a solenoid), the varying current and magnetic field will cause the ends of the bar to vibrate and hence create a sound wave. This is what happens when the transducer is transmitting.
Echo sounder
SITING OF TRANSDUCER
Factors affecting the siting of transducer:
AIR BUBBLE & CROSS NOISE: The transducer should be installed in a position where there is very less chance of formation of the air bubbles.Air bubbles will act as large reflectors of transmitted energy if lot of air bubbles are formed close the transducer.This will also create the cross noise.
  There are various locations on the ship where formation of air bubble is less e.g.
a) On large ,fast,deep draft ships-1/8 to ¼ L of the ship from forward
On medium speed ships- forward most portion of the ship.
On slow cargo ships-1/4 L from aft
On oil tanker –normally forward end of the E/Room bulkhead.
Ranging
In echo sounder the stylus is moving with certain constant speed and transmission takes place when the stylus passes the zero marks.When the higher range is selected the speed of the stylus is reduced as stylus has to paper for the longer duration.This system is called the ranging.
PHASING
In phasing the speed of the stylus motor remains constant.In stead of changing the speed of the stylus,the transmission point is advanced.
The sensors are positioned around the stylus belt.The magnet generates the pulse when it passes the sensors which in turns activate the transmitter.
PHASING
ERRORS OF ECHO SOUNDER
1.Velocity of propagation in water:
      The velocity taken for the calculation of the is 15oom/sec.The velocity of the sound wave is changing due to the change of the salinity and temperature of the sea water. As velocity is varying hence depth recorded will be erroneous.
2. STYLUS SPEED ERROR:The speed of the stylus is such that the time taken by the stylus to travel from top to bottom on chart is same as the time taken by sound wave to travel twice the range selected.
 but due to fluctuation in voltage supplied to stylus motor ,will cause error in the recorded depth.
3. PYTHAGORAS ERROR:
    This error is found when two transducer are used one for transmission and one for reception.This error is calculated using the Pythagoras principle.
4.Multiple ECHO:The echo may be reflected  no of times from the bottom of the sea bed,hence providing the multiple depth marks on paper.
5.The thermal and density layers:
     The density of the water varies with temperature and salinity ,which all tends to form different layers.The sound wave may be reflected from these layers .
  6.Zero line adjustment error:
    If the zero is not adjusted properly,it will give error in reading
CROSS NOISE:
    If sensitivity of the amplifier is high,just after zero marking a narrow line alongwith the several irregular dots and dashes appear and this is called cross noise.The main reasons for the cross noise are aeration and picking up the transmitted pulse.If intensity of cross noise is high,it will completely mask the shallow water depths.This is controlled by swept gain control circuit.
 
AERATION:
  When the sound wave is reflected from the reflected from the air bubbles,it will appear as dots,this is known as aeration.