#Distance_By_Great_Circle_Sailing ... 👌👇
Your ship departs Yokohama, Japan from position Lat. 35° 27' N; Long. 139° 39' E
bound for San Francisco, California, USA.
At position Lat. 37° 48.5' N, Long. 122° 24' W.
Determine the distance by Great Circle Sailing.
Given:
Lat1 = 35° 27' N
Lat2 = 37° 48.5' N
Long1 = 139° 39' E
Long2 = 122° 24' W
What is asked?
Distance by Great Circle Sailing
Solution:
●1. Solve first for the difference of longitude (Dlong).
Here's the formula in getting Dlong:
If same name, subtract
If different name, plus
Note: Affix name of direction. If Dlo is greater than 180, subtract it to 360 and then affix name of Long1.
Long1 = 139° 39' E
Long2 = 122° 24' W
--------------------------
262° 03'
- 360°
Dlong = 97° 57' E
●2. We can now proceed to solve for Great Circle distance.
Note: Plus (+) when not crossing the equator; Subtract (-) when crossing the equator.
Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlong) + (Sin Lat1 × Sin Lat2)
Cos GCD = ( Cos 35° 27' × Cos 37° 48.5' × Cos 97° 57' ) + ( Sin 35° 27' × Sin 37° 48.5' )
Cos GCD= -0.08902 + 0.35548
GCD = inv Cos 0.26646
GCD = 74.55° × 60 ( to convert to miles)
GCD = 4473 miles ... 👌
====================================
Determining Great Circle Distance & Initial Course ... 👌👇
Determine the Great Circle Distance and initial course
from Lat 24° 52' N; Long 078° 27' W
to Lat 47° 19' N; Long 006° 42' W.
Given:
Lat1 = 24° 52' N
Lat2 = 47° 19' N
Long1 = 078° 27' W
Long2 = 006° 42' W
What is asked?
#Great_Circle_Distance_and_initial_course
Solution: 👇
●1. Solve first for the DLong.
Rule: If same name -Minus
Different name - Plus
And then affix the name of direction.
If the DLong is greater than 180, minus it to 360 then affix the name of Long1.
Long1 = 078° 27' W
Long2 = 006° 42' W
-------------------------
DLong = 071° 45' E
●2. Proceed to the formula for the great circle distance.
Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlong) ± (Sin Lat1 × Sin lat2)
☆Rule:
◇ Plus when not crossing the equator
◇ Minus when crossing the equator
Cos GCD = Cos 24° 52' × Cos 47° 19' × Cos 71° 45') + (Sin 24° 52' × Sin 47° 19')
Cos GCD = 0.50174
GCD = inv Cos 0.50174
GCD = 59° 53'
× 60 to convert it into miles
GCD = 3593 miles
●3. Solving for the Initial Course
Sin IC = Cos Lat2 × Sin Dlong
----------------------------
Sin Distance
Sin IC = Cos 47° 19' × Sin 71° 45'
----------------------------
Sin 59° 53'
Sin IC = 0.74432
IC = inv Sin 0.74432
IC =48°.1 degrees . 👌
=====================================
The position where the Great Circle passes closest to the pole is called #Vertex. At the Vertex the direction of the Great Circle is 090° or 270°. At the Vertex the Great Circle will achieve its highest latitude. There is also a similar Vertex point on the southern hemisphere.
#Latitude_Of_Vertex ... 👌👇
The great circle distance from Lat. 35° 08' S, Long. 019° 26' E to Lat. 33° 16' S; Long. 115° 36' E is 4559 nautical miles and the initial course is 121° T.
Determine the latitude of the vertex.
Given:
Lat1 = 35° 08' S
Lat2 = 33° 16' S
Long1 = 019° 26' E
Long2 = 115° 36' E
Initial Course = 121° T
What is asked?
Latitude of vertex
Solution:
This is the formula to getting the latitude of vertex...
Cos Lat Vertex = Cos Lat1 × Sin I.C.
Cos Lat Vertex = Cos 35° 08' × Sin 121°
Cos Lat Vertex = 0.70100
Lat Vertex = Inv Cos 0.70100
Lat Vertex = 45° 30' S
Note:
Name the Latitude vertex according to the name of Lat1. 👌
Your ship departs Yokohama, Japan from position Lat. 35° 27' N; Long. 139° 39' E
bound for San Francisco, California, USA.
At position Lat. 37° 48.5' N, Long. 122° 24' W.
Determine the distance by Great Circle Sailing.
Given:
Lat1 = 35° 27' N
Lat2 = 37° 48.5' N
Long1 = 139° 39' E
Long2 = 122° 24' W
What is asked?
Distance by Great Circle Sailing
Solution:
●1. Solve first for the difference of longitude (Dlong).
Here's the formula in getting Dlong:
If same name, subtract
If different name, plus
Note: Affix name of direction. If Dlo is greater than 180, subtract it to 360 and then affix name of Long1.
Long1 = 139° 39' E
Long2 = 122° 24' W
--------------------------
262° 03'
- 360°
Dlong = 97° 57' E
●2. We can now proceed to solve for Great Circle distance.
Note: Plus (+) when not crossing the equator; Subtract (-) when crossing the equator.
Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlong) + (Sin Lat1 × Sin Lat2)
Cos GCD = ( Cos 35° 27' × Cos 37° 48.5' × Cos 97° 57' ) + ( Sin 35° 27' × Sin 37° 48.5' )
Cos GCD= -0.08902 + 0.35548
GCD = inv Cos 0.26646
GCD = 74.55° × 60 ( to convert to miles)
GCD = 4473 miles ... 👌
====================================
Determining Great Circle Distance & Initial Course ... 👌👇
Determine the Great Circle Distance and initial course
from Lat 24° 52' N; Long 078° 27' W
to Lat 47° 19' N; Long 006° 42' W.
Given:
Lat1 = 24° 52' N
Lat2 = 47° 19' N
Long1 = 078° 27' W
Long2 = 006° 42' W
What is asked?
#Great_Circle_Distance_and_initial_course
Solution: 👇
●1. Solve first for the DLong.
Rule: If same name -Minus
Different name - Plus
And then affix the name of direction.
If the DLong is greater than 180, minus it to 360 then affix the name of Long1.
Long1 = 078° 27' W
Long2 = 006° 42' W
-------------------------
DLong = 071° 45' E
●2. Proceed to the formula for the great circle distance.
Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlong) ± (Sin Lat1 × Sin lat2)
☆Rule:
◇ Plus when not crossing the equator
◇ Minus when crossing the equator
Cos GCD = Cos 24° 52' × Cos 47° 19' × Cos 71° 45') + (Sin 24° 52' × Sin 47° 19')
Cos GCD = 0.50174
GCD = inv Cos 0.50174
GCD = 59° 53'
× 60 to convert it into miles
GCD = 3593 miles
●3. Solving for the Initial Course
Sin IC = Cos Lat2 × Sin Dlong
----------------------------
Sin Distance
Sin IC = Cos 47° 19' × Sin 71° 45'
----------------------------
Sin 59° 53'
Sin IC = 0.74432
IC = inv Sin 0.74432
IC =48°.1 degrees . 👌
=====================================
The position where the Great Circle passes closest to the pole is called #Vertex. At the Vertex the direction of the Great Circle is 090° or 270°. At the Vertex the Great Circle will achieve its highest latitude. There is also a similar Vertex point on the southern hemisphere.
#Latitude_Of_Vertex ... 👌👇
The great circle distance from Lat. 35° 08' S, Long. 019° 26' E to Lat. 33° 16' S; Long. 115° 36' E is 4559 nautical miles and the initial course is 121° T.
Determine the latitude of the vertex.
Given:
Lat1 = 35° 08' S
Lat2 = 33° 16' S
Long1 = 019° 26' E
Long2 = 115° 36' E
Initial Course = 121° T
What is asked?
Latitude of vertex
Solution:
This is the formula to getting the latitude of vertex...
Cos Lat Vertex = Cos Lat1 × Sin I.C.
Cos Lat Vertex = Cos 35° 08' × Sin 121°
Cos Lat Vertex = 0.70100
Lat Vertex = Inv Cos 0.70100
Lat Vertex = 45° 30' S
Note:
Name the Latitude vertex according to the name of Lat1. 👌
How GMDSS works
The GMDSS enables a ship in distress to send a message in various ways and be virtually certain that it is heard and acted upon. The distress or safety message will be picked up by ships in the area and by shore stations within range if sent on MF and VHF or by shore stations if transmitted using HF, INMARSAT or COSPAS-SARSAT.
While ships can transmit an alert using any or all of the distress equipment available on board, ships will transmit a ship-to ship alert on MF or VHF and ship-to-shore alert in area A3 or A4, as appropriate, by a ship earth station (SES), HF digital selective call (DSC) communications or a satellite EPIRB as appropriate.
Ships sailing in area 2 are at least capable to transmit a ship-to-shore alert on 2187,5 KHz and satellite EPIRB and ship sailing in area A1 are at least capable of transmitting a ship-to-short alert on 156.525 MHz using DSC. The new system means that distress messages will be received much more reliably than ever before. And thanks to the SAR Convention of 1979, it will prove possible for the search and rescue operation that results to be better co-ordinated.
The system established by the SOLAS Convention means that the whole operation will be co-ordinated by a designated rescue co-ordination centre (RCC) which will be informed of the alert either through the INMARSAT or COSPAS-SARSAT systems or from coast stations participating in the GMDSS.
The search itself will also be conducted according to procedures laid down by the SAR Convention and amplified in IAMSAR manual. The GPIDSS, for its part, will provide the means of communications regardless of the area in which the ship is located.
The GMDSS enables a ship in distress to send a message in various ways and be virtually certain that it is heard and acted upon. The distress or safety message will be picked up by ships in the area and by shore stations within range if sent on MF and VHF or by shore stations if transmitted using HF, INMARSAT or COSPAS-SARSAT.
While ships can transmit an alert using any or all of the distress equipment available on board, ships will transmit a ship-to ship alert on MF or VHF and ship-to-shore alert in area A3 or A4, as appropriate, by a ship earth station (SES), HF digital selective call (DSC) communications or a satellite EPIRB as appropriate.
Ships sailing in area 2 are at least capable to transmit a ship-to-shore alert on 2187,5 KHz and satellite EPIRB and ship sailing in area A1 are at least capable of transmitting a ship-to-short alert on 156.525 MHz using DSC. The new system means that distress messages will be received much more reliably than ever before. And thanks to the SAR Convention of 1979, it will prove possible for the search and rescue operation that results to be better co-ordinated.
The system established by the SOLAS Convention means that the whole operation will be co-ordinated by a designated rescue co-ordination centre (RCC) which will be informed of the alert either through the INMARSAT or COSPAS-SARSAT systems or from coast stations participating in the GMDSS.
The search itself will also be conducted according to procedures laid down by the SAR Convention and amplified in IAMSAR manual. The GPIDSS, for its part, will provide the means of communications regardless of the area in which the ship is located.
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