0
$\begingroup$

If I want to know the azimuth (initial heading) to another point on a sphere I use the formula $$\tag{1} \tan(\theta) = \frac{\sin(\Delta\lambda)\cos(\varphi_2)}{\cos(\varphi_1)\sin(\varphi_2)-\sin(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda)} $$ where $\theta$ is the initial bearing to point 2, $\varphi_1$ and $\varphi_2$ are the latitudes of point one and two respectively and $\Delta\lambda$ is the difference in longitudes of the two points. To get the azimuth I then use the $atan2$-function and insert $$\tag{2} \sin(\Delta\lambda)\cos(\varphi_2) $$ for $X$ and $$\tag{3} \cos(\varphi_1)\sin(\varphi_2)-\sin(\varphi_1)\cos(\varphi_2)\cos(\Delta\lambda) $$ for $Y$. I know that $\tan(\alpha) = \frac{X}{Y}$ is used to calculate the heading $\alpha$ of a vector. However, I can't figure out why equation 2 is my $X$ and equation 3 my $Y$. Where do these formulae come from?

$\endgroup$
0
$\begingroup$

Let $p$, $p'$ be two unit vectors, directed from the Earth center to points $P$ and $P'$ on the sphere, and $n$ the analogous unit vector for the North Pole.

To compute the heading from $P$ to $P'$ you must set up a coordinate system in a plane perpendicular to $p$, with the $y$-axis pointing towards the North Pole. This can be simply done by constructing two unit vectors $x$ and $y$ as follows: $$ y={p\times n\over|p\times n|}\times p,\quad x=y\times p. $$ A vector $t$ in the same plane pointing towards $P'$ is: $$ t=(p\times p')\times p $$ and its coordinates are then its projections along vectors $x$ and $y$: $$ t_x=t\cdot x,\quad t_y=t\cdot y. $$ If you now express the coordinates of $p$, $p'$ as a function of their latitude and longitude (and of course $n=(0,0,1)$), you can find explicit expressions for $t_x$ and $t_y$ and should recover your formulas.

EDIT.

If $\phi$, $\phi'$ are the latitudes, and $\lambda$, $\lambda'$ the longitudes of points $P$ and $P'$, we have: $$ p=(\cos\phi\cos\lambda,\cos\phi\sin\lambda,\sin\phi),\quad p'=(\cos\phi'\cos\lambda',\cos\phi'\sin\lambda',\sin\phi'),\quad n=(0,0,1). $$ It follows that: $$ p\times n=(\cos\phi\sin\lambda,-\cos\phi\cos\lambda,0), \quad |p\times n|=\cos\phi, $$ whence $$ y=(-\cos\lambda\sin\phi,-\sin\lambda\sin\phi,\cos\phi), \quad x=(-\sin\lambda,\cos\lambda,0). $$ Moreover: $$ t=(p\times p')\times p=-p(p'\cdot p)+p'=\\ \big( \sin ^2\phi \cos \phi' \cos \lambda' +\cos \phi^2\cos \phi' \sin \lambda \sin (\lambda-\lambda') -\sin\phi\cos \phi \sin \phi' \cos \lambda,\\ \sin ^2\phi\cos \phi' \sin \lambda' -\cos^2 \phi\cos \phi' \cos \lambda \sin (\lambda-\lambda') -\sin\phi \cos \phi \sin \phi' \sin \lambda,\\ \cos \phi (\cos \phi \sin \phi'-\sin \phi \cos \phi' \cos (\lambda-\lambda'))\big), $$ and finally: $$ t_x=-\cos\phi'\sin(\lambda-\lambda'), \quad t_y=-\cos\phi'\sin\phi\cos(\lambda-\lambda')+\cos\phi\sin\phi'. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.