This is a further refinement of the running fix in which the first bearing is taken when the object is at four points (45°) on the bow. When the object is on the beam the range will be the same as the distance run since the first bearing was taken. The disadvantage of the four point bearing is that the range of the single object is not known until it is abeam. This is of little help in passing at a safe distance.
The Transit Bearing
When two charted objects come into line they are said to be in transit. It has already been shown how a transit can be used to check the compass error. A transit can also be used to obtain a fix in conjunction with another position line such as a range (or be used to obtain a fix in conjunction with another position line such as a range (or even a sounding) without use of the compass.
Danger or Clearing Bearings
Many ports have provided leading lights or shapes to guide mariners safely into harbour, avoiding shoals and other dangers.
In places where such aids are not provided, the navigator may still be able to select a leading line provided by the transit of natural features. Thus approaching an anchorage with a coastal hill in transit with a more distant peak may ensure that the vessel clears dangerous rocks.
The advantage of a transit is that the mariner is assured of a safe approach regardless of any compass error
Vertical Sextant Angle
The distance off a light can be found by taking the vertical angle the light subtends at the vessel above sea level.
By Astronomical object
Six common methods are used :
01.Latitude by Polaris (Pole Star)
02.Latitude by meridian altitude
03.Latitude by ex-meridian
04.Longitude by meridian passage of the sun
05.Longitude by chronometer
06.Intercept
Latitude by Polaris (Pole Star)
As the Pole Star is always around the North Pole at radius of 1°, so it is always on or near the meridian passage. The latitude of the observer can be determined. The position line runs in an east-west, or nearly east-west, direction.
Latitude by Meridian Altitude
This method is used to obtain the position line by taking the altitude of the celestial body when it is instantly on the same meridian as the observer’s. In this case, the position line runs in an east-west direction (90°-270°), and coincides with a parallel of latitude.
Latitude by Ex- Meridian Altitude
It is sometimes not possible to obtain the altitude of the celestial body when it is on same observer’s meridian due to cloud, environmental factors, etc. If the altitude of the celestial body can be obtained a few minutes before or after meridian passage, the Ex-Meridian method can be used to reduce the observed altitude to meridian altitude. The latitude of the observer can be determined. The position line runs nearly in an east-west direction.
Longitude by Meridian Passage of the Sun
By knowing that the sun orbits with one completed circle in 24 hours, or 15° for every hour, the observer can determine position at noon by using the chronometer. The advantage of this method is that the DR position is not required.
Longitude by Chronometer
This method is also used to determine the longitude of the observer. The position line runs through the position at DR latitude and observed longitude in a direction perpendicular to the azimuth of the celestial body from the observer.
Intercept
Since it is impractical to draw the large circle of a position circle on the chart, only the part of it in the vicinity of the ship that is perpendicular to the bearing of the body from the ship is drawn. When observing a celestial body, we can obtain its azimuth and altitude. The azimuth is the bearing of the body and the altitude of the body, giving us the zenith distance. As long as the altitude is corrected, the observed zenith distance is the true zenith distance, which is called Observed Zenith Distance or True Zenith Distance. With the D.R. position of the observer at the time of observing, the altitude can be calculated to obtain the zenith distance, which is called Calculated Zenith Distance.
The Transit Bearing
When two charted objects come into line they are said to be in transit. It has already been shown how a transit can be used to check the compass error. A transit can also be used to obtain a fix in conjunction with another position line such as a range (or be used to obtain a fix in conjunction with another position line such as a range (or even a sounding) without use of the compass.
Danger or Clearing Bearings
Many ports have provided leading lights or shapes to guide mariners safely into harbour, avoiding shoals and other dangers.
In places where such aids are not provided, the navigator may still be able to select a leading line provided by the transit of natural features. Thus approaching an anchorage with a coastal hill in transit with a more distant peak may ensure that the vessel clears dangerous rocks.
The advantage of a transit is that the mariner is assured of a safe approach regardless of any compass error
Vertical Sextant Angle
The distance off a light can be found by taking the vertical angle the light subtends at the vessel above sea level.
By Astronomical object
Six common methods are used :
01.Latitude by Polaris (Pole Star)
02.Latitude by meridian altitude
03.Latitude by ex-meridian
04.Longitude by meridian passage of the sun
05.Longitude by chronometer
06.Intercept
Latitude by Polaris (Pole Star)
As the Pole Star is always around the North Pole at radius of 1°, so it is always on or near the meridian passage. The latitude of the observer can be determined. The position line runs in an east-west, or nearly east-west, direction.
Latitude by Meridian Altitude
This method is used to obtain the position line by taking the altitude of the celestial body when it is instantly on the same meridian as the observer’s. In this case, the position line runs in an east-west direction (90°-270°), and coincides with a parallel of latitude.
Latitude by Ex- Meridian Altitude
It is sometimes not possible to obtain the altitude of the celestial body when it is on same observer’s meridian due to cloud, environmental factors, etc. If the altitude of the celestial body can be obtained a few minutes before or after meridian passage, the Ex-Meridian method can be used to reduce the observed altitude to meridian altitude. The latitude of the observer can be determined. The position line runs nearly in an east-west direction.
Longitude by Meridian Passage of the Sun
By knowing that the sun orbits with one completed circle in 24 hours, or 15° for every hour, the observer can determine position at noon by using the chronometer. The advantage of this method is that the DR position is not required.
Longitude by Chronometer
This method is also used to determine the longitude of the observer. The position line runs through the position at DR latitude and observed longitude in a direction perpendicular to the azimuth of the celestial body from the observer.
Intercept
Since it is impractical to draw the large circle of a position circle on the chart, only the part of it in the vicinity of the ship that is perpendicular to the bearing of the body from the ship is drawn. When observing a celestial body, we can obtain its azimuth and altitude. The azimuth is the bearing of the body and the altitude of the body, giving us the zenith distance. As long as the altitude is corrected, the observed zenith distance is the true zenith distance, which is called Observed Zenith Distance or True Zenith Distance. With the D.R. position of the observer at the time of observing, the altitude can be calculated to obtain the zenith distance, which is called Calculated Zenith Distance.
The difference between the observer or true zenith distance and the calculated zenith distance is the intercept.
If the true zenith distance is smaller than the calculated zenith distance, then the observer is nearer toward the geographical position of the celestial body compared with the DR position, and the intercept is called TOWARD.
If the true zenith distance is greater than calculated zenith distance, then the observer is further away from the geographical position of the celestial body compared with the DR position, and the intercept is called AWAY.
If the true zenith distance is smaller than the calculated zenith distance, then the observer is nearer toward the geographical position of the celestial body compared with the DR position, and the intercept is called TOWARD.
If the true zenith distance is greater than calculated zenith distance, then the observer is further away from the geographical position of the celestial body compared with the DR position, and the intercept is called AWAY.
#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. 👌