Central Angle - Angular Distance Formulary

Angular Distance Formulary

The angular distance can be calculated either directly as the TvL difference, or via the common coordinates (here, either SA_w, SB_w 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:

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