Central angle

A central angle is an angle which vertex is the center of a circle, and whose sides pass through a pair of points on the circle, thereby subtending an arc between those two points whose angle is (by definition) equal to the central angle itself. It is also known as the arc segment's angular distance.

Coordinates
On a sphere or ellipsoid, the central angle is delineated along a great circle. The usually provided coordinates of a point on a sphere/ellipsoid is its conjugate latitude ("Lat"), $$\phi\,\!$$, and longitude ("Long"), $$\lambda\,\!$$. The "point", $$\widehat{\sigma}\,\!$$, is actually—relative to the great circle it is being measured on—the transverse colatitude ("TvL"), and the central angle/angular distance is the difference between two TvLs, $$\Delta\widehat{\sigma}\,\!$$.

Calculation of TvL
The calculation of $$\widehat{\sigma}_s\,\!$$ and $$\widehat{\sigma}_f\,\!$$ can be found using a common subroutine:


 * $$V_s,V_f,V_w,V_c:\mathrm{\;Standpoint,\ forepoint,\ working,\ coworking\ values};\,\!$$
 * $$\widehat{\alpha}_w:\mathrm{\;Orthodromic\ azimuth\ at\ \widehat{\sigma}_w};\,\!$$


 * $${}_{\color{white}.}\!\begin{pmatrix}\operatorname{sgn}(V)=|V|\!\cdot

V^{-1};\quad\overrightarrow{\operatorname{sgn}}(V)=\operatorname{sgn}\big(\operatorname{sgn}(V)+\frac{1}{2}\big)\\{}_{(\,\operatorname{sgn}(0)=0;\qquad\overrightarrow{\operatorname{sgn}}(0)=+1\,)}\end{pmatrix}{}_{\color{white}.}\!\!\,\!$$


 * $$\Delta\lambda=\lambda_f-\lambda_s;\,\!$$


 * $${}_{\color{white}.}\!\left(\mbox{If } \phi_s=\phi_f=0\mbox{, then }\;\widehat{\sigma}_s=\frac{\pi-|\Delta\lambda|}{2},\;\widehat{\sigma}_f=\frac{\pi+|\Delta\lambda|}{2}\right){}_{\color{white}.}\!\!\,\!$$


 * $$\begin{align}\phi_w=\phi_s;\;

&\phi_c=\phi_f\!\!:\mbox{Get}\;\widehat{\sigma}_w\!\!:\\ &\widehat{\sigma}_s=\widehat{\sigma}_w\!\cdot\overrightarrow{\mbox{sgn}}(S\!B_w)+\pi\!\cdot\overrightarrow{\mbox{sgn}}(\widehat{\sigma}_w)\mbox{sgn}(1-\overrightarrow{\mbox{sgn}}(S\!B_w));\end{align}\,\!$$


 * $$\begin{align}\phi_w=\phi_f;\;

&\phi_c=\phi_s\!\!:\mbox{Get}\;\widehat{\sigma}_w\!\!:\\ &\widehat{\sigma}_f=\widehat{\sigma}_w\!\cdot\overrightarrow{\mbox{sgn}}(-S\!B_w)+\pi\!\cdot\overrightarrow{\mbox{sgn}}(\widehat{\sigma}_w)\mbox{sgn}(1-\overrightarrow{\mbox{sgn}}(-S\!B_w))\\ &\qquad\qquad\qquad\qquad\quad+2\pi\!\cdot\mbox{sgn}(1-\overrightarrow{\mbox{sgn}}(\widehat{\sigma}_w-\widehat{\sigma}_s));\end{align}\,\!$$ _____________________________________________________________________

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 * $$\begin{matrix}S\!A_w=\cos(\phi_c)\sin(\Delta\lambda);\qquad\qquad\qquad\qquad\qquad\qquad\;\;\\S\!B_w=\sin(\phi_w+\phi_c)\sin^2(\frac{\Delta\lambda}{2})+\sin(\phi_c-\phi_w)\cos^2(\frac{\Delta\lambda}{2});\end{matrix}\,\!$$
 * $$\left(\,\sin^2(\Delta\widehat{\sigma})={S\!A_w}^2+{S\!B_w}^2;\quad|\tan(\widehat{a}_w)|=\left|\frac{S\!A_w}{S\!B_w}\right|\,\right)\,\!$$
 * $$\begin{matrix}\widehat{\sigma}_w\!\!\!&=&\!\!\!\arctan\big(|\sec(\widehat{a}_w)|\tan(\phi_w)\big)=\arctan\!\left(\left|\frac{\sin(\Delta\widehat{\sigma})}{S\!B_w}\right|\tan(\phi_w)\right),\\&=&\!\!\!\!\!\!\arctan\!\left(\frac{\sqrt{{S\!A_w}^2+{S\!B_w}^2}}{|S\!B_w|}\tan(\phi_w)\right).\qquad\qquad\qquad\qquad\qquad\end{matrix}$$
 * }
 * $$\begin{matrix}\widehat{\sigma}_w\!\!\!&=&\!\!\!\arctan\big(|\sec(\widehat{a}_w)|\tan(\phi_w)\big)=\arctan\!\left(\left|\frac{\sin(\Delta\widehat{\sigma})}{S\!B_w}\right|\tan(\phi_w)\right),\\&=&\!\!\!\!\!\!\arctan\!\left(\frac{\sqrt{{S\!A_w}^2+{S\!B_w}^2}}{|S\!B_w|}\tan(\phi_w)\right).\qquad\qquad\qquad\qquad\qquad\end{matrix}$$
 * }
 * $$\begin{matrix}\widehat{\sigma}_w\!\!\!&=&\!\!\!\arctan\big(|\sec(\widehat{a}_w)|\tan(\phi_w)\big)=\arctan\!\left(\left|\frac{\sin(\Delta\widehat{\sigma})}{S\!B_w}\right|\tan(\phi_w)\right),\\&=&\!\!\!\!\!\!\arctan\!\left(\frac{\sqrt{{S\!A_w}^2+{S\!B_w}^2}}{|S\!B_w|}\tan(\phi_w)\right).\qquad\qquad\qquad\qquad\qquad\end{matrix}$$
 * }

Each point has at least two values, both a forward and reverse value.

Occupying great circle
The arc path, $$\scriptstyle{\widehat{\Alpha}}\,\!$$, tracing the great circle that a central angle occupies, is measured as that great circle's azimuth at the equator, introducing an important property of spherical geometry, Clairaut's constant:


 * $$\sin(\widehat{\Alpha})=\Big|\cos(\phi_w)\sin(\widehat{\alpha}_w)\Big|;\,\!$$

From this and relationships to $$\widehat{\sigma}\,\!$$,


 * $$\begin{align}\widehat{\Alpha}

&=\Big|\arcsin\big(\cos(\phi_w)\sin(\widehat{\alpha}_w)\big)\Big|\!\!\!&&=\Big|\arccos\left(\frac{\sin(\phi_w)}{\sin(\widehat{\sigma}_w)}\right)\Big|,\\ &=\Big|\arctan\big(\cos(\widehat{\sigma}_w)\tan(\widehat{\alpha}_w)\big)\Big|\!\!\!&&=\Big|\arctan\big(\sin(\widehat{\alpha}_w)\sin(\widehat{\sigma}_w)\cot(\phi_w)\big)\Big|.\end{align}\,\!$$

Angular distance formulary
The angular distance can be calculated either directly as the TvL difference, or via the common coordinates (here, either SAw, SBw value set can be used):


 * $$\begin{align}{}_{\color{white}.}\\\Delta\widehat{\sigma}

&=\widehat{\sigma}_f\;-\;\widehat{\sigma}_s,\\ &=\arcsin\!\left(\sqrt{{S\!A}^2+{S\!B}^2}\,\right),\\ &\quad{}^{\mathit{(can\,only\,find\,the\,first\,quadrant,\,i.e.,\;up\,to\,90^\circ)}}\\ &=\arccos\!\Big(\sin(\phi_s)\sin(\phi_f)+\cos(\phi_s)\cos(\phi_f)\cos(\Delta\lambda)\,\Big),\\ &\quad{}^{\mathit{(not\,recommended\,for\,small\,angles,\;due\,to\,rounding\,error)}}\\ &=\arctan\!\left(\frac{\sqrt{{S\!A}^2+{S\!B}^2}}{\sin(\phi_s)\sin(\phi_f)+\cos(\phi_s)\cos(\phi_f)\cos(\Delta\lambda)}\right),\\{}^{\color{white}.}\end{align}\,\!$$

and, using half-angles,


 * $$\begin{align}{}_{\color{white}.}\\

&=2\arcsin\!\left(\sqrt{\sin^2\!\left(\frac{\phi_f-\phi_s}{2}\right)+\cos(\phi_s)\cos(\phi_f)\sin^2\!\left(\frac{\Delta\lambda}{2}\right)}\,\right),\\ &=2\arccos\!\left(\sqrt{\cos^2\!\left(\frac{\phi_f-\phi_s}{2}\right)-\cos(\phi_s)\cos(\phi_f)\sin^2\!\left(\frac{\Delta\lambda}{2}\right)}\,\right),\\ &=2\arctan\!\left(\sqrt{\frac{\sin^2\left(\frac{\phi_f-\phi_s}{2}\right)+\cos(\phi_s)\cos(\phi_f)\sin^2\Big(\frac{\Delta\lambda}{2}\Big)}{\cos^2\left(\frac{\phi_f-\phi_s}{2}\right)-\cos(\phi_s)\cos(\phi_f)\sin^2\!\Big(\frac{\Delta\lambda}{2}\Big)}}\,\right).\\{}^{\color{white}.}\end{align}\,\!$$

It can, as well, be found by means of finding the chord length via Cartesian subtraction :


 * $$\begin{align}

&\Delta{X}=\cos(\phi_f)\cos(\lambda_f) - \cos(\phi_s)\cos(\lambda_s);\\ &\Delta{Y}=\cos(\phi_f)\sin(\lambda_f) - \cos(\phi_s)\sin(\lambda_s);\\ &\Delta{Z}=\sin(\phi_f) - \sin(\phi_s);\\ &C_h=\sqrt{(\Delta{X})^2+(\Delta{Y})^2+(\Delta{Z})^2};\\ &\Delta\widehat{\sigma}=2\arcsin\left(\frac{C_h}{2}\right).\end{align}\,\!$$

Also, by using Cartesian products rather than differences, the origin of the spherical cosine for sides becomes apparent:

$$\begin{align} {\scriptstyle{\Pi}}X&=\cos(\phi_s)\cos(\phi_f)\cos(\lambda_s)\cos(\lambda_f);\\ {\scriptstyle{\Pi}}Y&=\cos(\phi_s)\cos(\phi_f)\sin(\lambda_s)\sin(\lambda_f);\\ {\scriptstyle{\Pi}}Z&=\sin(\phi_s)\sin(\phi_f);\\ \frac{{\scriptstyle{\Pi}}X\!\!+\!{\scriptstyle{\Pi}}Y}{\cos(\phi_s)\cos(\phi_f)}&=\cos(\lambda_s)\cos(\lambda_f)+\sin(\lambda_s)\sin(\lambda_f)=\cos(\Delta\lambda);\\ \Delta\widehat{\sigma}&=\arccos\Big({\scriptstyle{\Pi}}X+{\scriptstyle{\Pi}}Y+{\scriptstyle{\Pi}}Z\Big) =\arccos\Big({\scriptstyle{\Pi}}Z+\big({\scriptstyle{\Pi}}X+{\scriptstyle{\Pi}}Y\big)\Big),\\ &=\arccos\Big(\sin(\phi_s)\sin(\phi_f)+\cos(\phi_s)\cos(\phi_f)\cos(\Delta\lambda)\Big).\end{align}\,\!$$

There is also a logarithmical form:


 * $${}_{\color{white}.}\;\mathbb{N}=\exp\left(\ln\!\left(\frac{\cos\left(\frac{\phi_f-\phi_s}{2}\right)}{\sin\left(\frac{\phi_s+\phi_f}{2}\right)}\right)-\ln\left(\tan\Big(\frac{|\Delta\lambda|}{2}\Big)\right)\right);\,\!$$
 * $${}_{\color{white}.}\;\mathbb{D}=\exp\left(\ln\!\left(\frac{\sin\left(\frac{|\phi_f-\phi_s|}{2}\right)}{\cos\left(\frac{\phi_s+\phi_f}{2}\right)}\right)-\ln\left(\tan\Big(\frac{|\Delta\lambda|}{2}\Big)\right)\right);\,\!$$

$${}_{\color{white}.}\quad\!\Delta\widehat{\sigma}=2\arctan\!\left(\,\left|\exp\left(\ln\!\left(\frac{\sin(\arctan(\mathbb{N}))}{\sin(\arctan(\mathbb{D}))}\right)+\ln\left(\tan\Big(\frac{|\phi_f-\phi_s|}{2}\Big)\right)\right)\right|\,\right).\,\!$$