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 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:

Read more about this topic:  Central Angle

Famous quotes containing the word distance:

    “Midway the lake we took on board two manly-looking middle-aged men.... I talked with one of them, telling him that I had come all this distance partly to see where the white pine, the Eastern stuff of which our houses are built, grew, but that on this and a previous excursion into another part of Maine I had found it a scarce tree; and I asked him where I must look for it. With a smile, he answered that he could hardly tell me.”
    —Henry David Thoreau (1817–1862)