Orthodromes

Orthodromes on the globe between Los Angeles and London

The shortest path on the spherical surface between point A and B is an orthodrome.

The Orthodrome ( Greek orthos for "straight", dromos for "run") is the shortest connection between two points on a spherical surface .

The orthodrome is a geodesic for the special case of a spherical surface. The orthodrome is always part of a great circle . In aviation , one usually flies along these orthodromes in order to be able to cover the smallest flight distance. The colloquially often used other synonyms is a straight line .

calculation

The formulas from spherical trigonometry are the basis for the following calculations .

Variables used	meaning
${\displaystyle \,\phi }$	Geographic latitude
${\displaystyle \,\lambda }$	Geographical longitude
${\displaystyle \,A(\phi _{A},\lambda _{A})}$	Starting point
${\displaystyle \,B(\phi _{B},\lambda _{B})}$	End point
${\displaystyle \,P_{N}(\phi _{N},\lambda _{N})}$	Northernmost point of the orthodromes
${\displaystyle \,\alpha }$	Course angle at A
${\displaystyle \,\beta }$	Course angle at B
${\displaystyle \,\zeta }$	Central angle (distance AB, expressed as angle)

The direction west is negative, the direction east is positive; is positive for latitudes in the northern hemisphere and negative for the southern hemisphere. ${\displaystyle \,\lambda }$ ${\displaystyle \,\phi }$

route

The distance can be specified as the angle as follows:

${\displaystyle \,\zeta =\arccos \left(\sin(\phi _{A})\cdot \sin(\phi _{B})+\cos(\phi _{A})\cdot \cos(\phi _{B})\cdot \cos(\lambda _{B}-\lambda _{A})\right)}$

To calculate the distance between the two points, it has to be multiplied by the earth's radius (around 6,370 km) (for in radians ; if it is given in degrees, it must also be multiplied by °). ${\displaystyle \zeta }$ ${\displaystyle \zeta }$ ${\displaystyle \zeta }$ ${\displaystyle \pi /180}$

The angle can be calculated using the scalar product of the position vectors of and . The above formula results from transformations with the help of geometric addition theorems for sine and cosine. Alternatively, the formula can be derived by applying the side cosine law of spherical trigonometry to the triangle formed from the points and and the north pole. ${\displaystyle \zeta }$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle B}$

Course angles and legal courses

Heading angle

${\displaystyle \alpha =\arccos \left({\frac {\sin(\phi _{B})-\sin(\phi _{A})\cdot \cos(\zeta )}{\cos(\phi _{A})\cdot \sin(\zeta )}}\right)}$

${\displaystyle \beta =\arccos \left({\frac {\sin(\phi _{A})-\sin(\phi _{B})\cdot \cos(\zeta )}{\cos(\phi _{B})\cdot \sin(\zeta )}}\right)}$

The two parameters and can also be determined directly from the latitude and longitude or and or : ${\displaystyle \alpha }$ ${\displaystyle \beta }$ ${\displaystyle \phi _{A}}$ ${\displaystyle \phi _{B}}$ ${\displaystyle \lambda _{A}}$ ${\displaystyle \lambda _{B}}$

${\displaystyle \alpha =\arccos \left({\frac {\cos(\phi _{A})\cdot \sin(\phi _{B})-\cos(\lambda _{A}-\lambda _{B})\cdot \cos(\phi _{B})\cdot \sin(\phi _{A})}{\sqrt {1-(\cos(\lambda _{A}-\lambda _{B})\cdot \cos(\phi _{A})\cdot \cos(\phi _{B})+\sin(\phi _{A})\cdot \sin(\phi _{B}))^{2}}}}\right)}$

${\displaystyle \beta =\arccos \left({\frac {\cos(\phi _{B})\cdot \sin(\phi _{A})-\cos(\lambda _{A}-\lambda _{B})\cdot \cos(\phi _{A})\cdot \sin(\phi _{B})}{\sqrt {1-(\cos(\lambda _{A}-\lambda _{B})\cdot \cos(\phi _{A})\cdot \cos(\phi _{B})+\sin(\phi _{A})\cdot \sin(\phi _{B}))^{2}}}}\right)}$

legal courses A → B

${\displaystyle \,rwK_{A}=\alpha }$

${\displaystyle \,rwK_{B}=180^{\circ }-\beta }$

legal courses B → A

${\displaystyle \,rwK_{B}=360^{\circ }-\beta }$

${\displaystyle \,rwK_{A}=180^{\circ }+\alpha }$

Northernmost point

In a gnomonic projection , orthodromes are always shown as a straight line

Calculation of the northernmost point of an orthodrome for a starting point A and a starting course angle α:

${\displaystyle \,\phi _{N}=\arccos \left(\sin(|\alpha _{A}|)\cdot \cos(\phi _{A})\right)}$

${\displaystyle \lambda _{N}=\lambda _{A}+\operatorname {sgn} (\alpha _{A})\cdot \left|\arccos \left({\frac {\tan(\phi _{A})}{\tan(\phi _{N})}}\right)\right|}$

Example calculation of the distance Berlin – Tokyo

Geographic coordinates of the start and end points:

• Berlin
  • 52 ° 31 '0 "N = 52.517 °
  • 13 ° 24 '0 "E = 13.40 °
• Tokyo
  • 35 ° 42 '0 "N = 35.70 °
  • 139 ° 46 '0 "E = 139.767 °

Angle calculation

${\displaystyle \,\phi _{A}=52{,}517^{\circ }}$

${\displaystyle \,\lambda _{A}=13{,}40^{\circ }}$

${\displaystyle \,\phi _{B}=35{,}70^{\circ }}$

${\displaystyle \,\lambda _{B}=139{,}767^{\circ }}$

${\displaystyle {\begin{aligned}\,\zeta &=\arccos {\Big (}\sin(\phi _{A})\sin(\phi _{B})+\cos(\phi _{A})\cos(\phi _{B})\cos(\lambda _{B}-\lambda _{A}){\Big )}\\&=\arccos {\Big (}\sin(52{,}517^{\circ })\sin(35{,}70^{\circ })+\cos(52{,}517^{\circ })\cos(35{,}70^{\circ })\cos(139{,}767^{\circ }-13{,}40^{\circ }){\Big )}\\&=\arccos(0{,}79353\cdot 0{,}58354+0{,}60853\cdot 0{,}81208\cdot (-0{,}59296))\\&=\arccos(0{,}1700)\\&=80{,}212^{\circ }\end{aligned}}}$

or in radians ${\displaystyle \,\zeta =1{,}400}$

Route calculation

To simplify matters, we assume a globe with a circumference of 40,000 km or a radius of 6,370 km.

${\displaystyle {\begin{aligned}L&={\frac {\zeta }{360^{\circ }}}\cdot 40\,000\ \mathrm {km} \\&={\frac {80{,}212^{\circ }}{360^{\circ }}}\cdot 40\,000\ \mathrm {km} \\&=8912\ \mathrm {km} \end{aligned}}}$

Or for in radians : ${\displaystyle \,\zeta }$

${\displaystyle {\begin{aligned}L&=\zeta \cdot 6370\ \mathrm {km} \\&=8918\ \mathrm {km} \end{aligned}}}$

Due to the idealized geospatial data, these are of course only two approximations. They only differ by 6 km because the rounded earth radius 6,370 km results in a circumference of the globe of almost 40,024 km instead of 40,000 km. The actual distance between the two specified points in Berlin and Tokyo, when using the WGS84 - reference ellipsoid are calculated at 8941.2 km more accurately, ie with a deviation of about 23 km or 0.26% compared to the second approximation.

More precise formula for calculating the distance on earth

The distance between two locations on earth can be calculated with an accuracy of 50 meters using the following formulas, see also Thaddeus Vincenty . It is not based on a sphere, but on the WGS84 ellipsoid. If coordinates of another reference ellipsoid are used, the parameters (radius) and ( flattening ) must be adjusted. ${\displaystyle a}$ ${\displaystyle f}$

Let and be the latitude and longitude of location A, and the latitude and longitude of location B in degrees. The distance between the two locations is calculated as follows: ${\displaystyle \phi _{A}}$ ${\displaystyle \lambda _{A}}$ ${\displaystyle \phi _{B}}$ ${\displaystyle \lambda _{B}}$

Flattening of the earth: ${\displaystyle f={\frac {1}{298{,}257\,223\,563}}}$

Equatorial radius of the earth: ${\displaystyle a=6378{,}137\ \mathrm {km} }$

${\displaystyle F={\frac {\phi _{A}+\phi _{B}}{2}}}$, , ${\displaystyle G={\frac {\phi _{A}-\phi _{B}}{2}}}$ ${\displaystyle l={\frac {\lambda _{A}-\lambda _{B}}{2}}}$

First, the rough distance D is determined:

${\displaystyle S=(\sin {G})^{2}\cdot (\cos {l})^{2}+(\cos {F})^{2}\cdot (\sin {l})^{2}}$

${\displaystyle C=(\cos {G})^{2}\cdot (\cos {l})^{2}+(\sin {F})^{2}\cdot (\sin {l})^{2}}$

${\displaystyle w=\arctan {\sqrt {\frac {S}{C}}}}$

${\displaystyle D=2\cdot w\cdot a}$

It is to be used in radians . ${\displaystyle w}$

The distance is corrected by the factors and : ${\displaystyle D}$ ${\displaystyle H_{1}}$ ${\displaystyle H_{2}}$

${\displaystyle T={\frac {\sqrt {S\cdot C}}{w}}}$

${\displaystyle H_{1}={\frac {3\cdot T-1}{2\cdot C}}}$

${\displaystyle H_{2}={\frac {3\cdot T+1}{2\cdot S}}}$

The distance in kilometers is then calculated as follows: ${\displaystyle s}$

${\displaystyle s=D\cdot \left(1+f\cdot H_{1}\cdot (\sin {F})^{2}\cdot (\cos {G})^{2}-f\cdot H_{2}\cdot (\cos {F})^{2}\cdot (\sin {G})^{2}\right)}$

Calculation example Berlin - Tokyo

${\displaystyle {\begin{array}{lcl}\phi _{A}&=&52{,}516666666666667^{\circ }\\\lambda _{A}&=&13{,}400^{\circ }\\\phi _{B}&=&35{,}700^{\circ }\\\lambda _{B}&=&139{,}766666666666667^{\circ }\\\\f&=&0{,}00335281066474748\\a&=&6378{,}137\ \mathrm {km} \\F&=&44{,}108333333333333^{\circ }\\G&=&8{,}408333333333333^{\circ }\\l&=&-63{,}183333333333333^{\circ }\\S&=&0{,}41498261872684\\C&=&0{,}58501738127316\\w&=&0{,}699965690768276\\D&=&8928{,}9541420394\ \mathrm {km} \\T&=&0{,}70391883329502\\H_{1}&=&0{,}95019099899696\\H_{2}&=&3{,}74926124548527\\\\s&=&8941{,}20250458698\ \mathrm {km} \end{array}}}$

The distance has therefore been determined to be 8,941.2 km with an accuracy of about 50 m. ${\displaystyle s}$

Loxodrome

Comparison of Loxodrome (red) and Orthodrome (blue)

path	Lox.	Orth.	Diff.
NY-MO	8359 km	7511 km	10.1%
NY-DA	6207 km	6150 km	00.9%
DA-MO	6596 km	6509 km	01.3%

Loxodrome extension relative to the orthodrome along the 50th parallel in percent.

When navigating from point A to B with a compass , the Loxodrome is more suitable, as it always crosses the meridians at the same angle, so you can simply maintain the (compass) course that has been set.

For short distances, a loxodrome is only slightly longer than an orthodrome. At high latitudes and at distances below 30 degrees of longitude, the relative difference in length is less than 1%. Then it increases significantly. A journey along the 50th parallel over 180 degrees of longitude is 45% longer than the journey over a great circle, which then passes over the pole.

See also

• Geodesic line (geodesic)
• Widening

Web links

• Calculation of distance and opening price for any spheroids (English)
• Great Circle Mapper - Great Circle mapper including ETOPS ranges
• Distance calculation between two places on earth - based on an ideal sphere with a radius of 6378.388 km.

swell

Formula for more precise distance calculation:

• J. Meeus: Astronomical Algorithms. 2nd Edition. Willmann-Bell, Richmond 2000, ISBN 0-943396-61-1 , p. 85.
• Loxodrome