Derivation of great-circle distance formula using of HCR's Inverse cosine formula (Minimum distance between any two arbitrary points on the globe given latitudes and longitudes)
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This document discusses the application of H.C. Rajpoot's inverse cosine formula to determine the minimum distance between two points on the surface of a sphere (great-circle distance). It provides a formula for calculating this distance based on the latitudes and longitudes of the points and explains its utility in various scenarios, including the Global Positioning System. The conclusion emphasizes the formula's accuracy and its capability to compute distances on both small and large spheres.
![Copyright© H.C. Rajpoot
𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐇𝐂𝐑′
𝐬 𝐈𝐧𝐯𝐞𝐫𝐬𝐞 𝐂𝐨𝐬𝐢𝐧𝐞 𝐅𝐨𝐫𝐦𝐮𝐥𝐚
𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝐚𝐧𝐲 𝐭𝐰𝐨 𝐩𝐨𝐢𝐧𝐭𝐬 𝐨𝐧 𝐚 𝐬𝐩𝐡𝐞𝐫𝐞
Harish Chandra Rajpoot Aug, 2016
M.M.M. University of Technology, Gorakhpur-273010 (UP), India
We know that the great circle is a circle whose plane passes through the centre of sphere. The part of a great circle
on a sphere is known as a great circle arc. The length of minor great circle arc (i.e. less than half great circle)
joining any two arbitrary points on a sphere of finite radius is the minimum distance between those points. Here
we are interested in finding out the minimum distance or great
circle distance between any two arbitrary points on a spherical
surface of finite radius (like globe) for the given values of
latitudes & longitudes.
Let there be any two arbitrary points 𝑨(𝝓𝟏 , 𝝀𝟏) & 𝑩(𝝓𝟐 , 𝝀𝟐)
lying on the surface of sphere of radius 𝑹 & centre at the point
O. The angles of latitude 𝝓𝟏 & 𝝓𝟐 are measured from the
equator plane & the angles of longitude 𝝀𝟏 & 𝝀𝟐 are
measured from a reference plane OPQ in the anticlockwise
direction. Here, we are to find out the length of great circle
arc AB joining the given points A & B. Draw the great circle
arcs AD and BC passing through the points A & B
respectively which intersect each other at the peak (pole)
point P & intersect the equatorial line orthogonally (at 90𝑜
) at
the points D & C respectively. (As shown by the dashed arcs
PD & PC in the figure 1)
Join the points A, B, C & D by the dashed straight lines
through the interior of sphere to get a plane quadrilateral
ABCD and great circle arc BD.
Now, the angle between the orthogonal great circle arcs BC
& CD, subtending the angles 𝛼 = 𝜙2 & 𝛽 = 𝜆2 − 𝜆1
respectively at the centre O of the sphere, meeting each other
at the common end point C, is 𝜃 = 𝜋
2
⁄ . Now, consider the
tetrahedron OBCD formed by joining the points B, C and D to the centre O (see the fig-1). A diahedral angle is
the angle between two intersecting planes measured in a plane perpendicular to the both the intersecting planes.
Now, the diahedral angle say 𝜃 between the lateral triangular faces BOC and COD is given by HCR’s Inverse
Cosine Formula [1] as follows
cos 𝜃 =
cos 𝛼′ − cos 𝛼 cos 𝛽
sin 𝛼 sin 𝛽
… … … … … … (1)
where, is 𝜃 is the diahedral angle between lateral triangular faces BOC and COD intersecting each other at the
line OC which is equal to the angle between great circle arcs BC and CD intersecting each other perpendicularly
i.e. 𝜃 =
𝜋
2
which is opposite to 𝛼′ = ∠𝐵𝑂𝐷 and
𝛼′
= ∠𝐵𝑂𝐷, 𝛼 = ∠𝐵𝑂𝐶, 𝛽 = ∠𝐶𝑂𝐷 are the vertex angles of triangular faces BOD, BOC and COD respectively
meeting at the apex O of tetrahedron OBCD and angle 𝛼′
is always opposite to the diahedral angle 𝜃 (as shown in
the above fig-1 and fig-2 below).
Figure 1: The dashed great circle arcs PD and PC passing through
two given points A & B, intersecting each other at the peak (pole)
point P, meet the equator orthogonally at the points D & C
respectively on a spherical surface of finite radius 𝑹.](https://image.slidesharecdn.com/applicationofhcrsinversecosineformula-230122233711-c910c54c/85/Derivation-of-great-circle-distance-formula-using-of-HCR-s-Inverse-cosine-formula-Minimum-distance-between-any-two-arbitrary-points-on-the-globe-given-latitudes-and-longitudes-1-320.jpg)
![Copyright© H.C. Rajpoot
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐯𝐞 𝐄𝐱𝐚𝐦𝐩𝐥𝐞
Consider any two arbitrary points A & B having respective angles of latitude 𝝓𝟏 = 𝟒𝟎𝒐
& 𝝓𝟐 = 𝟕𝟓𝒐
& the
difference of angles of longitude 𝚫𝝀 = 𝝀𝟐 − 𝝀𝟏 = 𝟓𝟓𝒐
on a sphere of radius 25 cm. The minimum distance
between given points lying on the sphere is obtained by substituting the corresponding values in the above great-
circle distance formula as follows
dmin = 𝑅 cos−1(cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2)
= 25 cos−1(cos 40𝑜
cos 75𝑜
cos(55𝑜) + sin 40𝑜
sin 75𝑜)
= 25cos−1(0.7346063699582707) ≈ 18.64274952833712 𝑐𝑚
The above result also shows that the points A & B divide the perimeter = 2𝜋(25) ≈ 157.07963267948966𝑐𝑚
of the great circle in two great circles arcs (one is minor arc AB of length ≈ 18.64274952833712 𝑐𝑚 & other
is major arc AB of length ≈ 138.43688315115253 𝑐𝑚) into a ratio ≈
18.64274952833712
138.43688315115253
⁄ ≈ 𝟏: 𝟕. 𝟒
Conclusion: It can be concluded that the analytic formula of great-circle distance derived here directly gives the
correct values of the great-circle distance between any two arbitrary points on the sphere because there is no
approximation in the formula. This is extremely useful formula to compute the minimum distance between any
two arbitrary points lying on a sphere of finite radius which is equally applicable in global positioning system.
This formula is the most general formula to calculate the geographical distance between any two points on the
globe for the given latitudes & longitudes. This is a high precision formula which gives the correct values for all
the distances on the tiny sphere as well as the large spheres such as Earth, and other giant planets assuming them
the perfect spheres.
Note: Above articles had been derived & illustrated by Mr H.C. Rajpoot (B Tech, Mechanical Engineering)
M.M.M. University of Technology, Gorakhpur-273010 (UP) India Aug, 2016
Email: rajpootharishchandra@gmail.com
Author’s Home Page: https://notionpress.com/author/HarishChandraRajpoot
References:
[1]: H C Rajpoot. (2014). HCR’s Inverse Cosine Formula, (Solution of internal & external angles of a tetrahedron).
Academia.edu.
Link:https://www.academia.edu/9649896/HCRs_Inverse_Cosine_Formula_Solution_of_internal_and_exte
rnal_angles_of_a_tetrahedron_n](https://image.slidesharecdn.com/applicationofhcrsinversecosineformula-230122233711-c910c54c/85/Derivation-of-great-circle-distance-formula-using-of-HCR-s-Inverse-cosine-formula-Minimum-distance-between-any-two-arbitrary-points-on-the-globe-given-latitudes-and-longitudes-5-320.jpg)
Derivation of great-circle distance formula using of HCR's Inverse cosine formula (Minimum distance between any two arbitrary points on the globe given latitudes and longitudes)
- 1. Copyright© H.C. Rajpoot 𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐇𝐂𝐑′ 𝐬 𝐈𝐧𝐯𝐞𝐫𝐬𝐞 𝐂𝐨𝐬𝐢𝐧𝐞 𝐅𝐨𝐫𝐦𝐮𝐥𝐚 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝐚𝐧𝐲 𝐭𝐰𝐨 𝐩𝐨𝐢𝐧𝐭𝐬 𝐨𝐧 𝐚 𝐬𝐩𝐡𝐞𝐫𝐞 Harish Chandra Rajpoot Aug, 2016 M.M.M. University of Technology, Gorakhpur-273010 (UP), India We know that the great circle is a circle whose plane passes through the centre of sphere. The part of a great circle on a sphere is known as a great circle arc. The length of minor great circle arc (i.e. less than half great circle) joining any two arbitrary points on a sphere of finite radius is the minimum distance between those points. Here we are interested in finding out the minimum distance or great circle distance between any two arbitrary points on a spherical surface of finite radius (like globe) for the given values of latitudes & longitudes. Let there be any two arbitrary points 𝑨(𝝓𝟏 , 𝝀𝟏) & 𝑩(𝝓𝟐 , 𝝀𝟐) lying on the surface of sphere of radius 𝑹 & centre at the point O. The angles of latitude 𝝓𝟏 & 𝝓𝟐 are measured from the equator plane & the angles of longitude 𝝀𝟏 & 𝝀𝟐 are measured from a reference plane OPQ in the anticlockwise direction. Here, we are to find out the length of great circle arc AB joining the given points A & B. Draw the great circle arcs AD and BC passing through the points A & B respectively which intersect each other at the peak (pole) point P & intersect the equatorial line orthogonally (at 90𝑜 ) at the points D & C respectively. (As shown by the dashed arcs PD & PC in the figure 1) Join the points A, B, C & D by the dashed straight lines through the interior of sphere to get a plane quadrilateral ABCD and great circle arc BD. Now, the angle between the orthogonal great circle arcs BC & CD, subtending the angles 𝛼 = 𝜙2 & 𝛽 = 𝜆2 − 𝜆1 respectively at the centre O of the sphere, meeting each other at the common end point C, is 𝜃 = 𝜋 2 ⁄ . Now, consider the tetrahedron OBCD formed by joining the points B, C and D to the centre O (see the fig-1). A diahedral angle is the angle between two intersecting planes measured in a plane perpendicular to the both the intersecting planes. Now, the diahedral angle say 𝜃 between the lateral triangular faces BOC and COD is given by HCR’s Inverse Cosine Formula [1] as follows cos 𝜃 = cos 𝛼′ − cos 𝛼 cos 𝛽 sin 𝛼 sin 𝛽 … … … … … … (1) where, is 𝜃 is the diahedral angle between lateral triangular faces BOC and COD intersecting each other at the line OC which is equal to the angle between great circle arcs BC and CD intersecting each other perpendicularly i.e. 𝜃 = 𝜋 2 which is opposite to 𝛼′ = ∠𝐵𝑂𝐷 and 𝛼′ = ∠𝐵𝑂𝐷, 𝛼 = ∠𝐵𝑂𝐶, 𝛽 = ∠𝐶𝑂𝐷 are the vertex angles of triangular faces BOD, BOC and COD respectively meeting at the apex O of tetrahedron OBCD and angle 𝛼′ is always opposite to the diahedral angle 𝜃 (as shown in the above fig-1 and fig-2 below). Figure 1: The dashed great circle arcs PD and PC passing through two given points A & B, intersecting each other at the peak (pole) point P, meet the equator orthogonally at the points D & C respectively on a spherical surface of finite radius 𝑹.
- 2. Copyright© H.C. Rajpoot Now, substituting the corresponding values i.e. 𝜃 = 𝜋 2 , 𝛼′ = ∠𝐵𝑂𝐷, 𝛼 = ∠𝐵𝑂𝐶 = 𝜙2 and 𝛽 = ∠𝐶𝑂𝐷 = 𝜆2 − 𝜆1 in the above Eq(1) (HCR’s Inverse Cosine Formula) as follows cos 𝜋 2 = cos ∠𝐵𝑂𝐷 − cos 𝜙2 cos(𝜆2 − 𝜆1) sin 𝜙2 sin(𝜆2 − 𝜆1) cos ∠𝐵𝑂𝐷 − cos 𝜙2 cos(𝜆2 − 𝜆1) = 0 cos ∠𝐵𝑂𝐷 = cos 𝜙2 cos(𝜆2 − 𝜆1) … … … … … . (2) Similarly, from the figure-2, the diahedral angle say 𝛾 between the lateral triangular faces BOD and COD of tetrahedron OBCD is obtained by substituting the corresponding values i.e. 𝜃 = 𝛾 which is opposite to 𝛼′ = ∠𝐵𝑂𝐶 = 𝜙2, 𝛼 = ∠𝐵𝑂𝐷 and 𝛽 = ∠𝐶𝑂𝐷 = 𝜆2 − 𝜆1 in the above Eq(1) ( HCR’s Inverse Cosine Formula) as follows cos 𝛾 = cos 𝜙2 − cos ∠𝐵𝑂𝐷 cos(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 sin(𝜆2 − 𝜆1) = cos 𝜙2 − cos 𝜙2 cos(𝜆2 − 𝜆1) cos(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 sin(𝜆2 − 𝜆1) (Setting value of cos ∠𝐵𝑂𝐷 from Eq(2)) = cos 𝜙2 (1 − cos2(𝜆2 − 𝜆1)) sin ∠𝐵𝑂𝐷 sin(𝜆2 − 𝜆1) = cos 𝜙2 sin2(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 sin(𝜆2 − 𝜆1) = cos 𝜙2 sin(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 cos 𝛾 = cos 𝜙2 sin(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 … … … … … . (3) Now, from the figure-3, consider the tetrahedron OABD formed by joining the points A, B and D to the centre O (also shown in the above fig-1). Now, the diahedral angle say 𝛿 between the lateral triangular faces AOD and BOD is obtained by substituting the corresponding values i.e. 𝜃 = 𝛿 which is opposite to 𝛼′ = ∠𝐴𝑂𝐵, 𝛼 = ∠𝐴𝑂𝐷 = 𝜙1 and 𝛽 = ∠𝐵𝑂𝐷 in the above Eq(1) ( HCR’s Inverse Cosine Formula) as follows cos 𝛿 = cos ∠𝐴𝑂𝐵 − cos 𝜙1 cos ∠𝐵𝑂𝐷 sin 𝜙1 sin ∠𝐵𝑂𝐷 cos(90𝑜 − 𝛾) = cos ∠𝐴𝑂𝐵 − cos 𝜙1 cos ∠𝐵𝑂𝐷 sin 𝜙1 sin ∠𝐵𝑂𝐷 (∵ 𝛾 + 𝛿 = 90𝑜) Figure 2: The diahedral angle between the triangular faces BOC and COD of tetrahedron OBCD is equal to the angle between great arcs BC and CD which intersect each other at right angle because the great arc BC intersects equatorial line (i.e. great arc CD) at right angle. Figure 3: The diahedral angle 𝜹 between the triangular faces AOD and BOD is equal to the angle between great arcs AD and BD which is opposite to the face angle ∠𝑨𝑶𝑩 of tetrahedron OABD.
- 3. Copyright© H.C. Rajpoot sin 𝛾 = cos ∠𝐴𝑂𝐵 − cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) sin 𝜙1 sin ∠𝐵𝑂𝐷 (Setting value of cos ∠𝐵𝑂𝐷 from Eq(2)) sin 𝜙1 sin 𝛾 sin ∠𝐵𝑂𝐷 = cos ∠𝐴𝑂𝐵 − cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) cos ∠𝐴𝑂𝐵 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝛾 sin ∠𝐵𝑂𝐷 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin ∠𝐵𝑂𝐷 √sin2 𝛾 (∵ 0 ≤ 𝛾 ≤ 𝜋) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin ∠𝐵𝑂𝐷 √1 − cos2 𝛾 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin ∠𝐵𝑂𝐷 √1 − ( cos 𝜙2 sin(𝜆2 − 𝜆1) sin ∠𝐵𝑂𝐷 ) 2 ( from Eq(3)) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin ∠𝐵𝑂𝐷 √ sin2 ∠𝐵𝑂𝐷 − cos2 𝜙2 sin2(𝜆2 − 𝜆1) sin2 ∠𝐵𝑂𝐷 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √1 − cos2 ∠𝐵𝑂𝐷 − cos2 𝜙2 sin2(𝜆2 − 𝜆1) Substituting the value of cos ∠𝐵𝑂𝐷 from the above Eq(2), cos ∠𝐴𝑂𝐵 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √1 − (cos 𝜙2 cos(𝜆2 − 𝜆1))2 − cos2 𝜙2 sin2(𝜆2 − 𝜆1) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √1 − cos2 𝜙2 cos2(𝜆2 − 𝜆1) − cos2 𝜙2 sin2(𝜆2 − 𝜆1) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √1 − cos2 𝜙2 (sin2(𝜆2 − 𝜆1) + cos2(𝜆2 − 𝜆1)) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √1 − cos2 𝜙2 (1) = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 √sin2 𝜙2 = cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2 (∵ 0 ≤ 𝜙2 ≤ 𝜋) ⇒ ∠𝑨𝑶𝑩 = cos−1(cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2) … … … … … . (4) The above ∠𝐴𝑂𝐵 is the angle subtended at the centre O by the great circle arc AB joining the given points A and B lying on the sphere (as shown in the above fig-1). Therefore the minimum distance between two points A and B on the sphere will be equal to the great circle arc AB given as follows The length of great circle arc AB = Central angle × Radius of sphere = ∠𝑨𝑶𝑩 × 𝑹 = cos−1(cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2) × 𝑅 = 𝑅 cos−1(cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2) ∴ 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝐭𝐡𝐞 𝐩𝐨𝐢𝐧𝐭𝐬 𝐀(𝛟𝟏, 𝛌𝟏) & 𝐁(𝛟𝟐, 𝛌𝟐) 𝐥𝐲𝐢𝐧𝐠 𝐨𝐧 𝐬𝐩𝐡𝐞𝐫𝐞 𝐨𝐟 𝐫𝐚𝐝𝐢𝐮𝐬 𝐑, 𝐝𝐦𝐢𝐧 = 𝑹 𝐜𝐨𝐬−𝟏(𝐜𝐨𝐬 𝝓𝟏 𝐜𝐨𝐬 𝝓𝟐 𝐜𝐨𝐬(𝝀𝟐 − 𝝀𝟏) + 𝐬𝐢𝐧 𝝓𝟏 𝐬𝐢𝐧 𝝓𝟐) ∀ 𝟎 ≤ 𝝓𝟏 , 𝝓𝟐, |𝝀𝟐 − 𝝀𝟏| ≤ 𝝅 The above formula is called Great-circle distance formula which gives the minimum distance between any two arbitrary points lying on the sphere for given latitudes and longitudes.
- 4. Copyright© H.C. Rajpoot NOTE: It’s worth noticing that the above formula of Great-circle distance has symmetry i.e. if 𝜙1 & 𝜙2 and 𝜆1 & 𝜆2 are interchanged, the formula remains unchanged. It also implies that if the locations of two points for given values of latitude & longitude are interchanged, the distance between them does not change at all. Since the equator plane divides the sphere into two equal hemispheres hence the above formula is applicable to find out the minimum distance between any two arbitrary points lying on any of two hemispheres. So for the convenience, the equator plane of the sphere should be taken in such a way that the given points lie on one of the two hemispheres resulting from division of sphere by the reference equator plane. Since the maximum value of cos−1(𝑥) is π hence the max. of min. distance between two points on a sphere is = 𝑅(π) = 𝛑𝐑 = 𝐡𝐚𝐥𝐟 𝐨𝐟 𝐭𝐡𝐞 𝐩𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫 𝐨𝐟 𝐚 𝐠𝐫𝐞𝐚𝐭 𝐜𝐢𝐫𝐜𝐥𝐞 𝐩𝐚𝐬𝐬𝐢𝐧𝐠 𝐭𝐡𝐫𝐨𝐮𝐠𝐡 𝐠𝐢𝐯𝐞𝐧 𝐩𝐨𝐢𝐧𝐭𝐬 Case 1: If both the given points lie on the equator of the sphere then substituting 𝜙1 = 𝜙2 = 0 in the great-circle distance formula, we obtain 𝐝𝐦𝐢𝐧 = 𝑅 cos−1(cos 0 cos 0 cos(𝜆2 − 𝜆1) + sin 0 sin 0) = 𝑅 cos−1(cos(𝜆2 − 𝜆1)) = 𝑅|𝜆2 − 𝜆1| = 𝑹|∆𝝀| The above result shows that the minimum distance between any two points lying on the equator of the sphere depends only on the difference of longitudes of two given points & the radius of the sphere. In this case, the minimum distance between such two points is simply the product of radius 𝑅 and the angle ∆𝜆 between them measured on the equatorial plane of sphere. This If both the given points lie diametrically opposite on the equator of the sphere then substituting |∆𝝀| = 𝝅 in above expression, the minimum distance between such points 𝑅|∆𝜆| = 𝑅(𝜋) = 𝝅𝑹 = half of the perimeter of a great circle passing through thegiven points Case 2: If both the given points lie on a great circle arc normal to the equator of the sphere then substituting 𝜆2 − 𝜆1 = ∆𝜆 = 0 in the formula of great-circle distance, we obtain 𝐝𝐦𝐢𝐧 = 𝑅 cos−1(cos 𝜙1 cos 𝜙2 cos(0) + sin 𝜙1 sin 𝜙2) = 𝑅 cos−1(cos 𝜙1 cos 𝜙2 + sin 𝜙1 sin 𝜙2) = 𝑅 cos−1(cos(𝜙1 − 𝜙2)) = 𝑹|𝝓𝟏 − 𝝓𝟐| In this case, the minimum distance between such two points is simply the product of radius 𝑅 and the angle |𝜙1 − 𝜙2| between them measured on plane normal to the equatorial plane of sphere. Case 3: If both the given points lie at the same latitude of the sphere then substituting 𝜙1 = 𝜙2 = 𝜙 in the formula of great-circle distance, we obtain 𝐝𝐦𝐢𝐧 = 𝑅 cos−1(cos 𝜙 cos 𝜙 cos(𝜆2 − 𝜆1) + sin 𝜙 sin 𝜙) = 𝑹 𝐜𝐨𝐬−𝟏(𝐜𝐨𝐬𝟐 𝝓 𝐜𝐨𝐬(𝝀𝟐 − 𝝀𝟏) + 𝐬𝐢𝐧𝟐 𝝓) In this case, the minimum distance between such two points is dependent on the radius and both the latitude, and longitudes.
- 5. Copyright© H.C. Rajpoot 𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐯𝐞 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 Consider any two arbitrary points A & B having respective angles of latitude 𝝓𝟏 = 𝟒𝟎𝒐 & 𝝓𝟐 = 𝟕𝟓𝒐 & the difference of angles of longitude 𝚫𝝀 = 𝝀𝟐 − 𝝀𝟏 = 𝟓𝟓𝒐 on a sphere of radius 25 cm. The minimum distance between given points lying on the sphere is obtained by substituting the corresponding values in the above great- circle distance formula as follows dmin = 𝑅 cos−1(cos 𝜙1 cos 𝜙2 cos(𝜆2 − 𝜆1) + sin 𝜙1 sin 𝜙2) = 25 cos−1(cos 40𝑜 cos 75𝑜 cos(55𝑜) + sin 40𝑜 sin 75𝑜) = 25cos−1(0.7346063699582707) ≈ 18.64274952833712 𝑐𝑚 The above result also shows that the points A & B divide the perimeter = 2𝜋(25) ≈ 157.07963267948966𝑐𝑚 of the great circle in two great circles arcs (one is minor arc AB of length ≈ 18.64274952833712 𝑐𝑚 & other is major arc AB of length ≈ 138.43688315115253 𝑐𝑚) into a ratio ≈ 18.64274952833712 138.43688315115253 ⁄ ≈ 𝟏: 𝟕. 𝟒 Conclusion: It can be concluded that the analytic formula of great-circle distance derived here directly gives the correct values of the great-circle distance between any two arbitrary points on the sphere because there is no approximation in the formula. This is extremely useful formula to compute the minimum distance between any two arbitrary points lying on a sphere of finite radius which is equally applicable in global positioning system. This formula is the most general formula to calculate the geographical distance between any two points on the globe for the given latitudes & longitudes. This is a high precision formula which gives the correct values for all the distances on the tiny sphere as well as the large spheres such as Earth, and other giant planets assuming them the perfect spheres. Note: Above articles had been derived & illustrated by Mr H.C. Rajpoot (B Tech, Mechanical Engineering) M.M.M. University of Technology, Gorakhpur-273010 (UP) India Aug, 2016 Email: rajpootharishchandra@gmail.com Author’s Home Page: https://notionpress.com/author/HarishChandraRajpoot References: [1]: H C Rajpoot. (2014). HCR’s Inverse Cosine Formula, (Solution of internal & external angles of a tetrahedron). Academia.edu. Link:https://www.academia.edu/9649896/HCRs_Inverse_Cosine_Formula_Solution_of_internal_and_exte rnal_angles_of_a_tetrahedron_n