RE: Azimuth in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg95105] RE: [mg95071] Azimuth in Mathematica
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 6 Jan 2009 04:13:32 -0500 (EST)
- Reply-to: hanlonr at cox.net
Use the two argument form of ArcTan
az[ptA_, ptB_] := Pi/2 - ArcTan @@ (ptB - ptA)
ptA = {0, 0};
ptB = {1, 0};
ptC = {0, -1};
{#[[1]], #[[2]], az @@ ##} & /@
DeleteCases[Tuples[{ptA, ptB, ptC}, 2], {x_, x_}]
{{{0, 0}, {1, 0}, Pi/2},
{{0, 0}, {0, -1}, Pi},
{{1, 0}, {0, 0}, -(Pi/2)},
{{1, 0}, {0, -1}, (5*Pi)/4},
{{0, -1}, {0, 0}, 0},
{{0, -1}, {1, 0}, Pi/4}}
Bob Hanlon
On Mon, Jan 5, 2009 at 6:47 AM , TL wrote:
> Is there a function in Mathematica that would return the azimuth
> between two points with known coordinates, 0 being North?
> For example having A(0,0), B(1,0) - the azimuth should be 90 degrees
> its like ArcTan[dx/dy], but I need the quadrant taken into account as
> well and I need it to handle 90 and 270 degrees properly too
>
> /it doesn't need to convert the angles to degrees, rads is ok/