GREAT CIRCLE DISTANCE
GREAT CIRCLE DISTANCE
The
The Great-CircleGreat-Circle distance ordistance or OrthodromicOrthodromic distance is the shortest distance between any twodistance is the shortest distance between any two points on the surface of the earth measured along a path on its surface, assuming earth to be a perfect points on the surface of the earth measured along a path on its surface, assuming earth to be a perfect sphere. (Figure 1)
sphere. (Figure 1)
Figure 1: Great Circle Distance between two points on the Figure 1: Great Circle Distance between two points on the earthearth
GREAT CIRCLE GREAT CIRCLE If a sphere
If a sphere is cut by a plane at is cut by a plane at any arbitrary distance and any arbitrary angle, the any arbitrary distance and any arbitrary angle, the section is alwayssection is always a perfect circle. The diameter of this circle will be less than or equal to the diameter of the earth.
The largest diameter of the section circle is achieved when the intercepting plane passes through The largest diameter of the section circle is achieved when the intercepting plane passes through the center of the earth, in which case the diameter of the circle is maximum and equal to the diameter of the center of the earth, in which case the diameter of the circle is maximum and equal to the diameter of the earth (Figure 2). Such circles are called Great Circles. It is also to be noted that the Great Circles will the earth (Figure 2). Such circles are called Great Circles. It is also to be noted that the Great Circles will always have their centers exactly at the center of the earth.
always have their centers exactly at the center of the earth.
SHORTEST PATH AND GREAT-CIRCLE SHORTEST PATH AND GREAT-CIRCLE
The shortest path between any two points on the earth surface is the path along the Great Circle The shortest path between any two points on the earth surface is the path along the Great Circle that passes through both the points. For
that passes through both the points. For any two points on the earth any two points on the earth there could be only one there could be only one unique Great-unique Great-Circle passing through both of them.
Circle passing through both of them. Note:
Note: The exception to this are when both the points are exactly opposite sides of the sphere and whenThe exception to this are when both the points are exactly opposite sides of the sphere and when both the points are at exact same location. Both these cases have infinite number of Great-Circles passing both the points are at exact same location. Both these cases have infinite number of Great-Circles passing through them.
through them.
CALCULATION OF THE GREAT CIRCLE DISTANCE CALCULATION OF THE GREAT CIRCLE DISTANCE
Let us considering the earth at the center of the co-ordinate system with the North-Pole on the Let us considering the earth at the center of the co-ordinate system with the North-Pole on the positive
positive z-axis z-axis andand0˚0˚longitude (Greenwich Meridian) in the direction of longitude (Greenwich Meridian) in the direction of x-axis x-axis. (Figure 3). (Figure 3)
Now consider two points
Now consider two points AA andand BB on the earth surface. on the earth surface. Let the co-latitudLet the co-latitude and longitude of poe and longitude of pointint A
A bebeδδAA andandλ λ AA respectively. And the co-latitude and respectively. And the co-latitude and longitude of pointlongitude of point BB bebeδδBB andandλ λ BB respectively.respectively.
Note: The co-latitude is the angle measured from the
Note: The co-latitude is the angle measured from the z-axis z-axis(North Pole). It is related to (North Pole). It is related to the actual latitudethe actual latitude
σ σ byby
--- (1) --- (1)
Let us now draw a line from the origin to point
Let us now draw a line from the origin to point AA. This line forms the position vector. This line forms the position vector r r A A for thefor the
point
point AA in the co-ordinate system. Similarly construct the position vectorin the co-ordinate system. Similarly construct the position vector r r B B for the pointfor the point BB. Since we. Since we
know that the points
know that the points AA andand BB are located on the surface are located on the surface of the earth, the of the earth, the magnitudes of the vectorsmagnitudes of the vectors r r A A andand
r
r B B will be equal to the radius of the earth.will be equal to the radius of the earth.
Where Where R
R ≈ 6371 km ≈ 6371 km ≈ 3959 miles ≈ 3959 miles (the Mean Radius of Earth)(the Mean Radius of Earth)
Figure 3: Spherical co-ordinate system for
Figure 3: Spherical co-ordinate system for the Earththe Earth
For mathematical convenience let us now resolve the position vectors
For mathematical convenience let us now resolve the position vectors r r A Aandand r r B B into rectangularinto rectangular
co-ordinate system (Cartesian coordinate system). To do this we need to know the
co-ordinate system (Cartesian coordinate system). To do this we need to know the x x,, y y andand z z componentcomponent of the points
of the points AA andand BB.. If
If AA’’the projection of the pointthe projection of the point AA on theon the x-y x-yplaneplane,, then from Figure 4;then from Figure 4; OA’
OA’== R Rsin (sin (δδAA) ) --- --- (3)(3)
z
z A A == R Rcos (cos (δδAA) ) --- --- (4)(4)
x
x A A == OA’OA’ cos (cos (λ λ AA ) ) == R R sin (sin (δδAA) cos () cos ( λ λ AA ) ) --- --- (5)(5)
y
Figure 4: Resolving Point A
Figure 4: Resolving Point A into Rectangular Coordinatesinto Rectangular Coordinates
Hence; Hence; r
r A A == R Rsin (sin (δδAA) ) cos cos ((λ λ AA )) x x ++ R Rsin (sin (δδAA) sin () sin (λ λ AA )) y y ++ R R cos (cos (δδAA)) z z --- (7)--- (7)
Similarly for the point Similarly for the point BB;; r
r B B == R Rsin (sin (δδBB) ) cos cos ((λ λ BB)) x x ++ R Rsin (sin (δδBB) sin () sin ( λ λ BB )) y y ++ R R cos (cos (δδBB)) z z --- (8)--- (8)
Since the shortest path from
Since the shortest path from AA toto BB on the surface of the earth is along a Great Circle, the radiuson the surface of the earth is along a Great Circle, the radius of this arc (path) will be
of this arc (path) will be R R, the radius of the earth. Hence the length of this arc , the radius of the earth. Hence the length of this arc could be found by knowingcould be found by knowing the angle
Figure 5: The Great-Circle through the points A and B Figure 5: The Great-Circle through the points A and B
From equation (7) and (8),
From equation (7) and (8), the dot product of the the dot product of the vectorsvectors r r A A andand r r B B isis
r
r A A .. r r B B == R R22 sin (sin (δδAA) ) cos cos ((λ λ AA ) sin () sin (δδBB) ) cos cos ((λ λ BB ) ) ++ R R22sin (sin (δδAA) sin () sin (λ λ AA ) sin ) sin (( δδBB) sin () sin ( λ λ BB))
+
+ R R22 cos (cos (δδAA) cos ) cos ((δδBB) ) --- --- (9)(9)
r
r A A .. r r B B == R R22 sin (sin (δδAA) ) sin sin ((δδBB) { cos () { cos ( λ λ AA ) cos () cos ( λ λ BB ) ) + sin + sin (( λ λ AA ) sin () sin (λ λ BB ) }) }
+
+ R R22 cos (cos (δδAA) cos ) cos ((δδBB) ) --- --- (9a)(9a)
r
r
r A A .. r r B B == R R22 { sin ({ sin (δδAA) ) sin sin (( δδBB) ) cos cos ((λ λ AA --λ λ BB) + cos () + cos (δδAA) ) cos cos ((δδBB) ) } } --- --- (10)(10)
However from the principle of vector
However from the principle of vector dot product;dot product; r
r A A .. r r B B = |= | r r A A | | ||rr B B | | cos cos ((αα) ) --- --- (11)(11)
Using equation (2) Using equation (2)
α
α = cos= cos-1-1( (( ( r r A A .. r r B B ) / ) / R R22) ) --- --- (12)(12)
Substituting equation (12) in equation (11) Substituting equation (12) in equation (11)
α
α= cos= cos-1-1
{
{
sin (sin (δδAA) ) sin sin ((δδBB) ) cos cos ((λ λ AA --λ λ BB) + cos () + cos ( δδAA) ) cos cos ((δδBB))}
}
--- (13)--- (13)If
If αα is in radians then the length of the arc fromis in radians then the length of the arc from AA toto BB is given byis given by
D
D == R Rαα --- (14)--- (14)
Where Where R
R ≈ 6371 km ≈ 6371 km ≈ 3959 miles ≈ 3959 miles (the Mean Radius of Earth)(the Mean Radius of Earth)
Therefore the Great-Circle Distance between two points on the
Therefore the Great-Circle Distance between two points on the earth surface is given by;earth surface is given by; D
D == R R coscos-1-1
{
{
sin (sin ( δδAA) ) sin sin ((δδBB) ) cos cos ((λ λ AA --λ λ BB) + cos () + cos (δδAA) ) cos cos ((δδBB))}
}
--- (15)--- (15)USING ACTUAL LATITUDE INSTEAD OF CO-LATITUDE USING ACTUAL LATITUDE INSTEAD OF CO-LATITUDE
It is to be noted that the above equations uses the co-latitudes (
It is to be noted that the above equations uses the co-latitudes (δδAA andandδδBB) rather than the actual) rather than the actual
latitudes of the locations. Equation (15) is valid for all possible locations on the earth. However, if we latitudes of the locations. Equation (15) is valid for all possible locations on the earth. However, if we replace the co-latitude with actual latitude values
replace the co-latitude with actual latitude values σσAA andand σσBB of the pointsof the points AA andand BB respectively, therespectively, the
following cases are need to be
Case (i):
Case (i): Both the location ABoth the location A andand BB are on the northern hemisphere (i.e. above Equator). Henceare on the northern hemisphere (i.e. above Equator). Hence from the equation (1)
from the equation (1) δδ == 90˚90˚--σσ
Hence; Hence; D
D == R R coscos-1-1
{
{
sin (sin (90˚90˚--σσAA) ) sin sin ((90˚90˚--σσBB) ) cos cos ((λ λ AA --λ λ BB) + cos () + cos (90˚90˚--σσAA) ) cos cos ((90˚90˚--σσBB))}
}
D
D == R R coscos-1-1
{
{
cos (cos (σσAA) cos () cos (σσBB) ) cos cos ((λ λ AA --λ λ BB) + sin () + sin (σσAA) ) sin sin (( σσBB))}
}
--- (16)--- (16)Case (ii):
Case (ii): BothBoth AA andand BB are on the southern hemisphere (i.e. below Equator). Hence from theare on the southern hemisphere (i.e. below Equator). Hence from the equation (1)
equation (1) δδ= 90˚= 90˚++σσ
Hence; Hence; D
D == R R -1-1
{
{
sin (sin (90˚90˚++σσAA) ) sin sin ((90˚90˚++ σσBB) ) cos cos ((λ λ AA --λ λ BB) + cos () + cos (90˚90˚++σσAA) ) cos cos ((90˚90˚++σσBB))}
}
D
D == R R -1-1
{
{
cos (cos (σσAA) ) cos cos (( σσBB) ) cos cos ((λ λ AA --λ λ BB) ) + + (- (- sin sin ((σσAA) ) (- sin () ) (- sin (σσBB))))}
}
D
D == R R coscos-1-1
{
{
cos (cos (σσAA) cos () cos (σσBB) ) cos cos ((λ λ AA --λ λ BB) + sin () + sin (σσAA) ) sin sin (( σσBB))}
}
--- (17)--- (17)Case (iii):
Case (iii): WhenWhen AA is above the Equator andis above the Equator and BB is below the Equatoris below the Equator Using equation (1);
Using equation (1); D
D == R R coscos-1-1
{
{
sin (sin (90˚90˚--σσAA) ) sin sin ((90˚90˚++σσBB) cos () cos (λ λ AA --λ λ BB) ) + + cos cos ((90˚90˚--σσAA) ) cos cos ((90˚90˚++ σσBB))}
}
D
D == R R coscos-1-1
{
{
cos (cos (σσAA) cos () cos (σσBB) ) cos cos ((λ λ AA --λ λ BB) ) - - sin sin ((σσAA) ) sin sin ((σσBB))}
}
--- (18)--- (18)Case (iv):
Case (iv): WhenWhen AA is below the Equator andis below the Equator and BB is above the Equatoris above the Equator Using equation (1);
Using equation (1); D
D == R R coscos-1-1
{
{
sin (sin (90˚90˚++σσAA) ) sin sin ((90˚90˚--σσBB) cos () cos (λ λ AA --λ λ BB) ) + + cos cos ((90˚90˚++ σσAA) ) cos cos ((90˚90˚--σσBB))}
}
D
Considering all the above cases the e
Considering all the above cases the equations (16), (17), (18) and (19) quations (16), (17), (18) and (19) can be generalized as can be generalized as follows;follows;
D
D == R R coscos-1-1
{
{
cos (cos (σσAA) cos () cos (σσBB) ) cos cos ((λ λ AA --λ λ BB)) ±± sin (sin (σσAA) ) sin sin (( σσBB))}
}
---(20)---(20)[ Use
[ Use ++ when bothwhen both AA andand BB are on the same hemisphere and are on the same hemisphere and useuse –– whenwhen AA andand BB are on differentare on different hemispheres. ]
hemispheres. ]
Note:
Note:If eitherIf either AA oror BB is on the Equator then it is on the Equator then it doesn’t matter whether you use equation (16), (17), (doesn’t matter whether you use equation (16), (17), ( 18) or18) or (19) since
